Lectures in physics Part 1: Mechanics Przemysław Borys 7.11.2013
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Lectures in physics Part 1: Mechanics Przemysław Borys 7.11.2013
Lectures in physics
Part 1: Mechanics
Przemysław Borys
7.11.2013
1. Scalars and vectors.
In physics we use different quantities to describe the observed phenomena. Among them the
most important are scalars and vectors. Scalars are simply the numbers, which describe for
example the temperature, mass, density, energy. Other physical phenomena cannot be described in
terms of simple numbers. An example of such quantity is the velocity. It is not sufficient to say that
a car moves at 50km/h, we need to supplement this idea by the direction he moves to. A quantity
which consists of a magnitude and the direction information is called a vector.
A vector can be of two types. One type of vector is a bound vector, and the other is a free
vector. The first type of vector has prescribed coordinate of the initial and terminal point, while the
other doesn't. An example of the second case could be a translation vector of a coordinate system.
An example of the first case can be a force which acts on a specific point of a body.
In general it is possible to summarize that a vector consists of a magnitude, direction, and
sense, which determines whether the direction pointed by a vector is taken forward, or backward.
The sense is denoted by the arrow position on the line segment in the graphical vector
representation.
1.1. Basic notation, vector addition, vector multiplication by a scalar
In the Eulcidean space we can represent vectors in two basic ways. One is to use component
notation, another is to use the unit vectors. Let us first consider the former case.
In the figure above we can see a vector (a thick arrow) which can be enclosed by a cuboid in the
coordinate system. The x component equals „2”, the y component equals „3”, the z component
equals „4”. Therefore we can denote the vector by
[]
2
3
4
, or [2,3,4]T for short, where the „T”
symbol denotes a transposition, which transforms a row into a column, and vice versa.
To understand the second way of denoting vectors, we shall first learn the rules for vector
addition and subtraction. Before we do this, I emphasise a general remmark, which should be
apparent at this point. Vectors and scalars are different mathematical objects! It is impossible to put
an equality between a scalar and a vector, as is often done by students, who write equations like
=ma , where the left hand side cannot be evaluated by the right hand side (which has no
F
arrow!) because it is impossible to set the direction of F, i.e. the information on the right hand side
is insufficient! This becomes even more evident in the component notation, where on the left hand
side the vector is represented by three numbers in an Euclidean space, while on the right hand side
we have only one number! .
The vector addition in terms of components can be easily understood. It simply refers to the
summation of the corresponding components. For example:
[][][]
1
2 3
21=3
3
4 7
Such an addition rule can also be represented by an useful geometrical construction, known
as the parallelogram rule:
In the above case we sum
=[ 6,5]T .
A
B=[1,3]T [5,2]T =C
Another „easy” operation, which can be done on vectors is the vector multiplication by a
scalar number. This is related to the enlargement of the vector length by a specific amount. To do
this, we simply multiply all the vector components by the specified scalar, for example:
[][]
1
3
3 2=6
3
9
A careful reader may ask now: how do we define the vector length, that was multiplied
above? For two dimensional vectors, the components x, and y determine sides of a right triangle,
and therefore we can use the Pythagorean theorem: ∣A∣= A2x A2y . In three dimensions, we can
first apply the Pythagorean theorem to the XY plane, and then to the resulting vector and the Z axis,
which results in ∣A∣= A2x A2y A2z like below:
Having the rule for multiplication by a scalar, the vector addition procedure can easily
become generalized to vector subtraction because
, i.e. it can be done based
A−
B=
A−B
upon the vector addition formula for a negative vector B (i..e a vector with inverse sense, or
components multiplied by (-1)). So, to calculate
A−
B in the above case, we would write
A−
B=[1,3]T −[5,2]T =[1,3]T [−5,−2]T =[−4,1] T . A geometrical construction can be the
following:
As mentioned before, the vector representation can be put in terms of such called unit vectors.
What are the unit vectors? They are vectors of length „1”, which point along the x, y, and z axes.
Their names are i, j, and k, respectively:
As can be seen from the above figure, the unit vector, when multiplied, can resemble any
possible vector along the related axis. Then, making use of the vector addition formula, we can
„compose” any possible vector from the vectors based upon the unit vectors. For example,
[][][][][] [] []
1 1
0 0
1
0
0
=
=
2
3
2 0
2 0
0
1
0 = i 2 j3 k
3 0
0 3
0
0
1
This gives the second representation for vectors, which is quite useful for writing.
1.2. Dot product
The dot product between vectors
A and
B can be written as
=∣A∣∣B∣cos
A⋅B
where α is the angle between vectors
A and
B . It can be easily seen that when the vectors are
colinear (aligned along a single line, α=0), the dot product is just the multiplication of their length,
i.e.
A⋅
B=∣A∣∣B∣ . In case when the vectors are orthogonal (the angle between them is straight,
α=90), the dot product results in 0. This can be useful for designing a very useful dot product
representation:
A x i A y j Az k ⋅ B x i B y jB z k =.
A⋅B=
k =.
A x B x i⋅j A y B y j⋅j Az B z k⋅
Ax B x A y B y Az B z
which is just because the product of orthogonal unit vectors, i.e. i⋅j , i⋅k , j⋅k , all reduce to zero.
Of course such simple relations hold only in the Euclidean space and if some of you would try to
study the physics of general relativity and curved space-time, he would encounter a more general
formula for the dot product, which can be written as:
ij
A⋅
B=∑i , j g Ai B j
which makes use of the, such called, metric tensor g, which takes a diagonal representation for the
Euclidean space, where gij=1 for i=j and gij=0 for i≠j (in fact it also differs in that it also considers
the fourth coordinate—the time, but we leave this aside for now).
The simple interpretation of the dot product is by means of the vector projection on another vector.
Consider the following figure:
Here we can see that to calculate the projection of vector A to vector B, one needs to multiply the
length of A by the cosine of the angle between the two vectors. It is immediately seen that if we
multiply the projection by the length of vector B, we end up in the dot product formula:
proj A ,
B ∣B∣=∣A∣∣B∣ cos =
A⋅
B
1.3. The cross product
The previous section gave us the information about a type of multiplication between two
vectors, which gives a scalar result. There is also a type of multiplication between two vectors,
=
which gives a vector as a result. This is called the cross product C
A×
B . It is defined as an
operation that returns
a) a vector that is perpendicular to the two operands, and its direction is determined by the right
hand rule (see figure below)
b) a vector with magnitude equal to ∣C∣=∣A∣∣B∣sin
The b) requirement means that the magnitude of the resulting vector equals to the area of the
parallelogram that is determined by these vectors (see fig. Below), where
S=∣A∣ h=∣A∣∣B∣sin .
As we did in case of the dot product, also in case of the cross product we can consider the
behaviour of the product acting on the unit vectors. The angles between unit vectors are eighter 0, π/
2, π, 3π/2. The sine of such angle can be easily evaluated and gives either 0, 1, or -1. The
magnitude of the cross product between components equals simply to the product of their
magnitudes, because they form a rectangle, not a parallelogram. The cross product between two unit
vectors (e.g. i , j ) should give a perpendicular vector, which is the remaining unit vector (in our
example k ). Taking this together, we can say:
A×
B= A x i A y j Az k × B x iB y jB z k =.
A y B z− Az B y i Az B x − Ax B z j A x B y −A y B x k
If you know what is a determinant of a 2x2 matrix, you immediately recognize that
∣
∣∣
A Az A z
A×
B= y
i
By Bz
Bz
∣ ∣
∣
Ax Ax Ay
j
k
Bx
Bx B y
and further, considering the Laplace expansion, one finally recognizes that
∣
∣
i
j k
A×
B= A x Ay A z
Bx B y Bz
which gives the most compact and easy to memorize formula for calculating the cross product
provided, that you know the way to evaluate it (the Sarrus method, see fig. Below)
As you can see, to evaluate a determinant of the 3x3 matrix, one needs to rewrite the first
two rows under the determinant, and then draw the diagonal lines from upper left corner to lower
right and from upper right corner to lower left. The symbols encountered along the lines should be
multiplied and added together with symbols multiplied along the other lines. The lines pointing
from left to right are taken with a positive sign, and those pointing from right to left are taken
negative..
Questions
1. Describe vector addition and subtraction and define the dot product, and cross product
2. Discuss the derivation of the formulas for dot and cross product from the formulas
containing sine and cosine. Give interpretation of these operations.
Reference
E. Karaśkiewicz, Zarys teorii wektorów i tensorów. PWN, 1974.
2. Introduction to the differential calculus
If you remember the school definition of the velocity, you probably recall something like
v =
s
t
where s is the displacement vector, and t is the associated time interval. However, this is
not a strict definition. Consider an accelerating car that between 0 and 10s accelerates from 0 to
50km/h. If you would evaluate velocity at time equal to 1s with t=1s , you would obtain
approximately v =2.50km /h (calculating the distance by s=
at 2
). If you would take
2
t=5s , you would obtain v =0.50km/h (because of the zero contribution from the past). For
t=0.1s you would obtain
v =4.5km /h , quite close to what would be predicted by the
formula v =at .
How it is possible to have such different velocities at single time instant? It is not possible.
What we have calculated is NOT the instantenous velocity. It is just an average velocity. And the
average can differ depending upon the period, that we take into account for averaging. To obtain the
„real” instantenous velocity, we need to take the time interval to be as narrow as possible, to not
include any „historical” contributions:
v =lim t 0
s d s
=
= s '
t dt
2.1. Differentiation
Such a limit difference quotient is called a derivative and can be denoted using the
d
dt
or ... ' symbol. It is possible to calculate such derivatives directly. For example,
d t 2
t t 2−t 2
t 22t t− t 2−t 2
=limt 0
=limt 0
=2t
dt
t
t
You will probably learn more such examples durring your math classes. For the purposes of
physics we don't need however to digest this in details. You only need to know the key idea. Based
on such considerations, it is possible to build a basic table of derivatives:
Function
Derivative
t
n
nt
n−1
cos t
− sint
sin t
cos t
e kt
ke kt
ln ∣t∣
1
t
tg t
1
cos 2 t
ctg t
−
1
2
sin t
The construction of the difference quotient gives a geometrical interpretation (it describes
the tangent of the angle of a tangent line to the curve, from basic trigonometry for the right triangle
below we have tg =
dy
):
dx
The difference quotient allows to have certain properties of the differentiation. For example:
af t ' =lim t 0
af t t −af t
f t t − f t
=a lim t 0
=a f t '
t
t
Other properties can be found as well which results in the following table:
af t'=a f t ' (derivative of a function multiplied by constant)
f tg t'= f t' g t' (derivative of a sum of two functions)
f t g t' = f t ' g t f t g t' (derivative of a product)
'
f t f t ' g t − f t g t'
=
(derivative of a fraction)
g t
g 2 t
f g t '=
df g dg t
dg
dt
(chain rule)
Examples:
2t 25sin t '=2t 2' 5sin t' =2t 2 ' 5sint ' =4t5cos t
t 2 sin t '=t 2 ' sin tt 2 sin t'=2t sint t 2 cos t
'
t 2 ' sin t−t 2 sin t' 2tsin t −t 2 cos t
t2
=
=
2
2
sin t
sin t
sin t
sint 2 ' =cos t 2t 2 ' =2t cos t 2
2.2. Integration
Integration (indefinite integration) is an inverse procedure to differentiation. We denote
the integral using the following notation:
∫ f t dt= F t
where a condition must be met that:
F t '= f t .
Looking at the table of derivatives it is possible to construct a table of integrals that fullfills the
above condition. For example t n '=nt n−1 ; substituting k =n−1 ; n=k 1 we obtain
'
t
k1
'=k 1t
k
, so
t k1
=t k , which gives the first relation in the table below:
k 1
Function
Integral
∫ t n dt
t n1
C
n1
∫ cost dt
sin t
C
∫ sin t dt
−
cos t
C
∫ e kt dt
e kt
C
k
∫ t dt
1
ln ∣t∣C
∫ cos 2 t dt
1
tg tC
1
−ctg tC
∫ sin2 t dt
Where the „+C” component originates from the idea that the derivative of a constant euqls zero.
Similarly to the differentiation, integration also obeys certain laws:
∫ af t dt=a ∫ f t dt (integral of a function multiplied by constant)
∫ f t g t dt=∫ f t dt∫ g t dt (integral of a sum of two functions)
∫ f t' g t dt = f t g t−∫ f t g t ' dt (integration by parts)
∫ f g t g t' dt=∫ f g dg (integration by substitution)
Examples:
∫ t sin t dt=
[
]
f =t
g ' =sin t =−tcost cos t dt=−tcos tsin tC
∫
f '=1 g =−cos t
∫ 2t sint 2 dt=∫ sin t 2 2t dt=.
[
]
dg
=2t dg =2tdt =.
dt
∫ sin g dg=−cos g C=−cos t 2C
g=t 2,
Another type of integral is the, such called, definite integral. This type of integral is defined
by the relation
b
∫a
where:
f t dt=F b−F a
F t=∫ f t dt
is the indefinite integral of f(t).1 In the course of the math classes, you will learn an important
identity between the definite integral and a Riemann sum, which in most of the cases is the basic
tool to construct integrals in physical problems:
b
∫a
n
f t dt=lim min t 0 ∑i =1 f i t i
i
i
where i is the point within the interval t i , and the intervals t i cover the domain from ta
to tb. Due to this relation to Riemann sum, it is possible to give a geometrical interpretation to the
definite integral: it gives the area below the integrated curve (see fig. below).
Reference:
W. Żakowski, Matematyka dla kandydatów na wyższe uczelnie
1 We can prove this by considering a difference quotient (differentiating the definite integral):
t t
∫a
t
t t
f t dt−∫a f t dt =∫t
f t dt= f t c t , where the last equality results from the mean
value theorem (the value of the integral cannot be larger than f max t and cannot be lower than f min t ,
but there must be some point t c for which the equality holds). If t 0 , the result becomes equal to f(t)dt,
which gives the integrated function. Thus the definite integral is strictly related to indefinite, because the
differentiation gives the same results, i.e. integrand is the same function.
3. Kinematics
3.1. Displacement versus path
To understand the difference between displacement and path, consider the following figure:
The path length is equal to the sum of all traversed segments of this zig-zag trajectory. It is a scalar
quantity that has no direction. On the other hand, the displacement s is defined as the distance
between the ends of trajectory and has a direction, and sense (which depends upon the choice of the
initial and terminal points—the vector should point to the terminal position).
The discrimination between path length and displacement is essential do discriminate
between two different physical objects: the average speed and average velocity. Speed is a scalar
quantity related to the path length l, while velocity is a vector quantity and relates to the
displacement s :
〈 v 〉=
〈
v 〉=
l
(average speed)
t
s
(average velocity)
t
3.2. Instantenous velocity
As we have mentioned in the mathemathical introduction, the average speed and velocity is
of limited use in practical applications. It is more convenient to consider instantenous speed and
velocity, which are defined by:
v=
dl
(speed)
dt
d s
v =
(velocity)
dt
3.3. Acceleration
In cases where the velocity is not constant, we also would like to measure its change. This can be
done by means of the acceleration, which gives the rate of change of the velocity:
dv
a=
dt
3.4. Relations for the distance in an accelerated motion on a straight line
It is important to obtain some results from the definitions of instantenous velocities and
accelerations on a straight line. When the motion takes place on a straight line it is not necessery to
use vectors, and we can use scalar equations, which we can easily differentiate and integrate. Now,
let us see, what can be learned from the definition of acceleration, proposed above:
dv
∣∣⋅dt
dt
adt=dv
∫ adt=∫ dv
a=
v=a ∫ dt
v=atv 0
The v0 term above is just an integration constant set to provide initial velocity when t=0. We can
now put the last formula for the velocity into the definition of the velocity:
ds
dt
ds
atv 0= ∣∣⋅dt
dt
atdt v 0dt =ds
∫ at dt∫ v 0 dt=s
v=
s=a ∫ t dtv 0∫ dt
at 2
s= v 0 ts0
2
where we have again an integration constant s0, which serves to have the initial displacement equal
to the s0 rather than to 0. This way we have obtained a „famous” result for the distance travelled in
an accelerated motion, which you probably know from the earlier education. However, this time
you don't have to memorize it! It is sufficient to know what is the velocity, and what is acceleration
to derive the formula without any problems. Moreover, you are now able to calculate the distance
travelled by an object that follows an acceleration which is not constant! Something you couldn't
handle earlier at school.
3.5. Angular velocity and acceleration
When an object does not move forward, but instead rotates, it is convenient to measure its angular
position φ instead of its position s.The angular position can undergo changes, i.e. a body can pass
some angular distance Δφ in a specific time Δt, giving rise to the average angular velocity ω
〈 〉=
t
As previously, it is possible to define the instantenous angular velocity and angular acceleration ε,
which measures the rate of change of the velocity. The formulas are dual to those from transitional
motion:
d
dt
d
=
dt
=
Also, exactly as we did for the translational kinematics, we can calculate the distance φ travelled in
an accelerated motion:
d
∣∣⋅dt
dt
d =dt
∫ d =∫ dt
= t0
d
=
∣∣⋅dt
dt
d =dt
d = t dt 0 dt
∫ d =∫ t dt0∫ dt
t2
=
0 t 0
2
=
3.6. Relation between the angular velocity and acceleration, and tangential velocity and
acceleration
The angular displacement covered by a body on radius r corresponds to a path length
of l= r . This gives the possibility to calculate an associated tangential velocity and
acceleration of a rotating body related to the angular quantities:
l= r (path length travelled)
dl
d
=r
dt
dt (tangential velocity)
v=r
dv
d
=r
dt
dt (tangential acceleration)
a=r
3.7. Centripetal acceleration
While performing rotations at constant angular velocity one could say that there is no
acceleration at all. However, velocity is a vectorial quantity. Since the direction of the velocity
changes, there must be some corresponding acceleration. This is called the centripetal acceleration.
Consider the following figure:
As the body rotates, the velocity vector rotates (it also moves its initial point, but this is not
relevant, and is actually not included in the definition of the velocity). Superposing the two initial
v vector which points to the center of the
points (on the right on fig. above), we can see a net
rotation circle. What is important, the angle between subsequent velocity vectors is exactly the same
as between the subsequent radii that touch to the moving body. Therefore, we can estimate the
length of v by using the formula for the arc length of an arc with radius equal to v. This would
give
v= v
To obtain the acceleration, we divide v by t which results in:
a=
v= v
t
... or 'VW', if you find it easier to memorize. Utilizing the fact that v =r , we can represent the
above formula in more common representations:
a=
v2
= 2 r
r
This can be also represented in a vectorial manner:
a =×
×
r
r is perpendicular to the plane of those two vectors, and points in the
where the term ×
×
r points to the center of the
direction of the rotation, and then the cross product ×
rotation, and since the angles between the vectors undergoing cross producta are zero, the
magnitude of the resulting vector is 2 r .
3.8. Motion in noninertial reference frames
Until now, we have considered our formulas for cases of the, such called, inertial reference
frames. An inertial reference frame is such a frame that does not accelerate, i.e., as we will learn
soon, it is a reference frame that is not subject to the action of any net force. In case of the
noninertial reference frames, the reference frame itself can move with acceleration a, which
creates certain additional effects, which are not normally seen in inertial reference frames.
To start, let us take the position vector of a moving body with respect to an inertial
coordinate system (see fig. below):
r = r0 r1
e r= e0 r 0 e1 r 1
In the above, the symbol e denotes an unit vector aligned with the vector r . To obtain the
formula for the velocity, we need to calculate the first derivative of the above formula. We already
know how to differentiate functions of time, but how to differentiate the unit vectors? The unit
vectors cannot change their magnitude, so they can only rotate. The angular velocity vector gives
the speed of rotation, and is perpendicular to the plane of rotation. The cross product ×
e is
perpendicular to
and e , so it is aligned into the direction to which e changes. The
magnitude of the change equals r e dt=dt , because the radius of rotation re=1. Taking this all
together,
d
e
=×
e
dt
Now we are ready to calculate the velocity and acceleration in a noninertial reference frame
by utilizing the above formula and the formula for a derivative of a product:
r =e0 r 0e1 r 1
dr d e
dr
d r d e0
v =
=
r 0e0 0 1 r 1 e1 1
dt
dt
dt
dt
dt
v =
0×e0 r 0e0 v 01×e1 r 1 e1 v 1
This can be rewritten in a compact notation by taking into account that r =e r :
v =0× r0 v0
1×r1v1
0 ×r0 is the circumferential velocity of the noninertial reference frame around the
Here v
cir0 =
1×r1 is the
origin of coordinates in the inertial reference frame, and respectively vcir1 =
circumferential velocity of the considered body around the origin of coordinates in the noninertial
reference frame.
Now we can differentiate the velocity to calculate the accelerations.
d
d e
d r d e
dv
0
×e0 r 0
0× 0 r 0
0 ×e0 0 0 v 0 e0 0
dt
dt
dt
dt
dt
d
1
d e1
d r 1 d e1
dv 1
×e1 r 1
1×
r
1×e1
v e
dt
dt 1
dt
dt 1 1 dt
a=
this gives:
a =0× r0
0 ×
0 ×r0
0×v 0
0 ×v0 a0
1× r1
1×
1× r1
1× v1
1× v1a1
In practical considerations we usually consider noninertial frames which either separate from the
inertial reference frame at constant acceleration without rotations, or we consider a frame that
rotates while its origin of coordinates is immobilized inside of the inertial reference frame. In such
cases we can simplify the above by rejecting the terms dependent on 0 (multiplied by things
that depend upon r 0 ):
a = a01×r1
1×
1× r12 1× v1 a1
The last formula can be written in a convenient form as
a
= a0acir1 ac acor a1
Where: a0 is the linear acceleration of the noninertial reference frame with respect to the inertial
reference frame; a cir =1×r1 is the circumferential acceleration of the body in the noninertial
2
reference frame ; a c =
1×
1 ×
r 1=− 1 r1 is the centripetal acceleration of the body in the
1×v 1 is the Coriolis acceleration; and a1 is the linear
noninertial reference frame; a cor =2
acceleration of the body with respect to the noninertial refernce frame.
3.9. Coriolis acceleration
The reader may ask now. What is the Coriolis acceleration which we have derived in
previous section? Why does it depend on the velocity? Consider the situation of a black body that
moves along the radius of the rotating plate below:
The body angular position remains fixed in the inertial refrence frame, i.e. it doesnt rotate
with the plate, it only moves along the radius (the solid line). However, when observed from the
noninertial rotating reference frame, one can see a curved trajectory (the dashed line), and not a
straight line! The body, as it moves otside of the center gains larger and larger circumferential
velocity. To enlarge velocity, one needs acceleration, and this is where the Coriolis effect occurs.
This effect is responsible for bending of the plane of oscillations of the Foucault pendulum,
as can be seen in the following figure:
It is also extremly important for the weather science since wind near the Earth surface is curved
according to the Coriolis force which—in cas of air masses moving from opposite directions, can be
the reason for a cyclone formation.
The origin of Coriolis acceleration can be thought of as being resulting from two effects (fig.
below): first of all, as the body moves along the radius, it gains circumferential velocity in a rate of
dv =dR
dv
dR
= = v
dt
dt
Second, (the figure on the right below), the velocity vector is changing direction, exactly in the
same manner as it was in the consideration of the centripetal acceleration. This gives another term
v , hence the factor „2” in front of the Coriolis acceleration.
3.10. The projectile motion
How does a cannon ball move after it is shot from the cannon? The motion can be decomposed into
two perpendicular components that can be considered separately. One component is the vertical
motion subject to the gravitational acceleration g, while the second does not receive any
acceleration (we assume the air resistance is negligible, and the cannon ball, or the projectile,
moves at constant horizontal velocity).
The components of the initial velocity can be specified as:
v 0x =v 0 cos
v 0y=v 0 sin
The maximum height can be calculated by:
{
{
gt2
H max=v 0y t−
2
v 0y
g=
t
H max=v 0y t−
H max =
t=
gt2
2
v 0y
g
v 20y v 20 sin 2
=
2g
2g
The time to reach this height follows from:
v 0y
t
v 0y
t=
g
g=
Therefore, the range R, travelled in twice this time (the time of falling down is identical to the time
of rising up), we have:
2v0x v 0y 2 v 20 cos sin
R=
=
g
g
4. Relativistic kinematics
4.1. The origin of the whole confusion: the Michelson-Morley experiment
The Michelson Morley experiment is one of the best recognized experiments that demonstrate the
independence of the speed of light value on the speed of the inertial reference frame where it is
measured. Michelson was born in Strzelno (currently Poland, previously Poland did not exist and
was these times partitioned among Russia, Prussia and Austria), while Morley was an American.
In this figure the incident light beam splits on the dichroic mirror and proceeds to two normal,
reflecting mirrors, that are located in the same distance from the beam splitting point. Therefore, in
case when the reference frame velocity equals zero, the light beams follow the same distance and
leave the interferometer in phase, and no interference can be detected.
In case when the reference frame moves with velocity v, the vertical ray should travel at a
velocity equal to c= v 2v 2up v up= c 2−v 2 (pythagorean theorem), while the horizontal ray
should move at a velocity equal to v horiz =c±v . The time of travel for the upper beam is equal to
(consider only the vertical component of the velocity and vertical distance traveled)
t up
L
= 2 2
2 c −v
2L
t up = 2 2
c −v
In case of the horizontal beam, the situation is different,
t horiz=
L
L
2Lc
= 2 2
c−v cv c −v
The formulas differ and the beams need different time to pass their distance which means that they
will meet out of phase! They should interefere!
The details of the ray trajectories: they escape the dichroic mirror in different position when they
move thowards the detector and it results in a „double slit-like” behaviour.
But nothing like that was confirmed in the experiment. The researchers aligned their interferometer
along the Earth's orbit to have a contribution of 30km/s of the Earth's orbital speed to the device. In
such case, the waves should travell the interferometer at L=1m in t horiz−t up =3.3⋅10−17 s . This
corresponds to a displacement of 10nm which should cause a detectable change in the interference
pattern for visible light (it is about 0.02λ, and becomes twice as large after rotating the
interferometer by 90o). However, the investigators couldn't detect any interference patterns.
They repeated this experiment with multiple mirrors to enlarge the interferometer arms tenfold,
to11m. Still no interference was detected. This could either mean that the aether in which the light
was supposed to spread is centered at Earth and rotates with the eard around the Sun, or that the
light speed is independent of the velocity of the intertial coordinate frame. This was very awkward,
and it was not easy to accept these facts and the relativity theory which explains them.
4.2. Developing the Lorentz transformations
If the speed of light is constant in any refrence frame, then the light wave must form a sphere in
each inertial coordinate system. Therefore, the equations for the wave front in a moving and resting
coordinate frame must have the same form. Let us look if this really occurs in a classical
formulation, where we say that
x '=x−vt and t'=t (the galilean transformation), where x and t is
the position and time in the stationary inertial refrence frame (IRF) and x' and t' is the position and
time in the moving IRF:
{
x 2 y 2 z 2 =c 2 t 2
x −vt 2 y 2z 2=c 2 t ' 2
Expanding the bracked on the left hand side of the bottom expression, and substituting t'=t we
obtain
{
x 2 y 2z 2=c 2 t 2
x 2−2xvtv 2 t 2 y 2z 2=c 2 t 2
These two expressions describe different wave fronts in terms of the unprimed coordinates! The
classical approach does not work! We must modify something. A reasoning could be made that if
the speed of light remains constant, and the path changes in the Michelson-Morley experiment, then
maybe something happens to the time so that c=
l
t
remains unchanged. Almost any function
can be expanded into a power series, and the simplest form of a power series, for small
corrections, can be represented by a linear dependency. So we assume t ' =tlx . This has also
the great consequence that the IRF still moves on a straight line with constant velocity—
something that could be disrupted by a more cumbersome time transformation. Now the equations
in primed and original coordinates read:
{
x 2 y 2z 2=c 2 t 2
x 2−2xvtv 2 t 2 y 2z 2=c 2 t 22c 2 lxtc 2 l 2 x 2
Rearranging terms, we obtain:
{
2
2
2
2 2
x y z =c t
x 2 1−c 2 l 2 −2 x t v c 2 l y 2 z 2=c 2 t 2 1−
2
v
2
c
The term 2xt(v+c2l) on the left hand side does not fit to the sphere equation. We can eliminate it by
assuming l=−
v
2 . After this modification, we end up in the equations:
c
{
x 2 y 2 z 2=c2 t 2
x 2 1−
v2
v2
2
2
2 2
y
z
=c
t
1−
c2
c2
We can immediately see that to obtain the original equation we only need to divide the time and
position transformation by the factor
1−
v2
. Then the terms
2
c
2
1−
v
2
c
in the spherical wave
equation will not occur and it will be a sphere in deed, and not an ellipsoid. In general, the resulting
transformations read:
x '=
x−vt
1−
t−
t '=
v2
c2
v
x
c2
1−
v2
c2
and they are called the Lorentz transformations of the space-time. These are the transformations
that we must use when considering motion of bodies at speeds close to the speed of light. It can be
immediatelly seen that these transformations reduce to the classical galilean transformations for
small velocities, when the ratio
v
≪1 , resulting in x'=x-vt and t'=t.
c
To transform from the primed coordinate system to the non-primed coordinate system, we repeat
the above reasoning, but now the direction of the velocity becomes negative, so:
x=
x '−−v t '
=
x 'vt '
−v 2
v2
1−
c2
c2
−v
v
t '− 2 x ' t ' 2 x '
c
c
t=
=
2
−v
v2
1− 2
1− 2
c
c
1−
i.e. this reduces to a change in the sign of the operation in the numerator of the transformation.
4.3. Relativistic summation of velocities
We already know about differentiation. Thus, we can ask about the representation of a velocity
under transformation of the coordinates. Consider two IRF-s: IRF1 that is stationary and IRF2 that
moves with velocity u. What would be the velocity w in IRF1 of a body that moves at velocity v in
IRF2?
w=
dx
dt
By means of the total derivative for the Lorentz transformation from the primed coordinate system
to the nonprimed (the velocity has then a negative sign):
dx=
dt=
dx ' vdt '
2
1−
v
c2
dt '
v
dx '
c2
v2
c2
1−
So, you probably ask now what is a total derivative? It is a way to express the increment of a
multiargument function, eg.
f x ,t =
f
f
df
df
x
t≈ x t
x
t
dx
dt
Ok, having the formula for dx and for dt, we can divide them to obtain the result:
dx dx ' vdt '
=
dt
v
dt ' 2 dx '
c
dx '
v
dt ' dt '
w=
dt '
v dx '
1 2
c dt '
uv
w=
vu
1 2
c
w=
This formula, as you can easily check, does not allow to cross the speed of light. If you move in an
IRF at speed almost equal to c, and then you try to accelerate and increase the velocity by Δc, you
end in:
w=
c c
c c
=
=c
c c c c
1 2
c
c
4.4. Space-time interval
Consider a body that moves in the x direction with velocity v. Let us calculate some quantity:
2
2
2
s = x ' −c t '
2
2
v
x
2
c2
2 x−v t
2
s =
−c
v2
v2
1− 2
1− 2
c
c
v
v2
2
t
−2
x
t
x2
2
2
2
2
4
x −2v x tv t
c
c
s 2=
−c2
2
2
v
v
1− 2
1− 2
c
c
2
v
s 2= 1− 2 x 2−c 2 t = x 2−c t 2
c
t−
The calculations shows that the value of Δs is the same no matter the coordinate system we use for
calculations. Therefore it is called an invariant of the Lorentz transformations. It is interesting to
notice that the interval can be positive or negative which corresponds to time-like or space-like
events. The negative intervals are time-like (i.e. they can have a causal relations and one must
preceed the other in any reference frame) while the positive space-like events are separated by a
distance that is too large even for the light to pass in the corresponding time interval and therefore
they are not related causually.
4.5. Time dilation
Consider now how does a clock of a body behave as it moves. Let us observe an astronaut that
travels at velocity v. We observe him at time t=0 in position x=0, and thenafter, at time t=Δt we see
him in the position x=vΔt (because he moves at speed v). Let us transform these quantities to the
moving coordinate frame to calculate the time increase of the astronaut:
t ' =t ' 2−t ' 1=
1
2
v
1− 2
c
[
t−
t ' = 1−
v
v
v t − 0− 2 0
2
c
c
]
v2
t
c2
So, the time increments in the moving coordinate frame are smaller than in the stationary
coordinate frame. This leads to certain „apparent” paradoxes due to our limited imagination, an
example being the twins paradox, but we will discuss that later.
4.6. Lorentz contraction
We have found that the time runs slower for the moving astronaut. Consider now an astronaut that
travels from Earth to the Alpha Centauri. He has 4 light years to go. But if his clock runs slower
than ours, then the velocity that he would preceive after the travel, i.e. s/t' would be larger than the
velocity that we see on Earth! This would become even more awkward when he would travel at a
speed close to the speed of light, because then, evidently, the speed at which the Alpha Centauri
approaches him, would exceed the speed of light!
Such problems however do not occur because the moving bodies become contracted. So for
example when we travel to the Alpha Centauri and our clocks run slower, then at the same time we
observe a contracted distance to the target star and we need less time for the travel.
Let us again consider some body, for example the rocket of the astronaut. In his primed coordinates
the rocket can take coordinates x1'=0, t1'=0 and x2'=L, t2'=0 (t1'=t2', because the length makes sense
as a difference in end positions only at the same moment of time). Transforming this to the
stationary reference frame, we obtain:
x 1=0
t 1=0
Lv⋅0
x 2=
=
v2
1− 2
c
v
0 2 L
c
t2 =
=
v2
1− 2
c
L
1−
1−
v2
c2
v
L
c2
v2
c2
which means that the x2 coordinate is obtained too late compared to x1. To correct this, we need to
consider where was the second end of the rocket t2-t1 before. This corresponds to a correction of the
distance by v(t2-t1), i.e.:
x 2=
L
x2 =
v2
1− 2
c
−vt 2
v2
L 1− 2
c
v2
c2
v2
x 2=L 1− 2
c
1−
Which means that the length of the rocket becomes contracted for the stationary observer with
respect to who it moves with velocity v. This holds in general to any object of length L0, whose
length contracts to L as it starts to move:
L=L0
v2
1− 2
c
This effect is known under the name of the Lorentz contraction.
4.7. The twins paradox
There is a well known paradox in the special relativity. We consider two twin astronauts. One
remains on Earth while the second makes a distant travel at speed of light and comes back to his
brother. After the travel, the brother got old, his hair is gray, but the other twin, due to the time
dilation did not become older, and is still very young.
How can this be if the same reasoning could be undertaken from the point of view of the astronaut
in the rocket, who can see his brother leaving him at speed close to speed of light, and aging slower
than the astronaut!
But the brother on Earth does not change the inertial reference frames. The astronaut does. There
is no reason to assume that something wrong happens to the clocks of the brother on Earth, but the
astronaut clock's may behave strange because they flip from one inertial reference frame to another
and we know there is a lot of unusual things at high velocities, like for example broken simultanity
of events. I.e. being back on earth and seeing the other brother young (two events) does not have to
be simultaneous after moving back to the brother's coordinate system!!!
To understand what happens, consider a simplified problem of two twins, where one of them travels
to some distant star system (say, 50ly away). The first brother looks on the second brother, who
moves at speed 0.9999c (i.e. indistinguishable from c for the observer), and reaches the star after 50
years. However, the astronaut has a different perspective. In his (moving) coordinate system, the
distant star approaches him at 0.9999c, and he remains at rest. The distance between the distant star
L=L0 1−0.9999 2 . So the astronaut is
and the Earth contracts due to the Lorentz contraction!
L L
close to the distant star after t= = 0 1−0.99992=t earth 1−0.99992 . So the escapade took
v v
him much less time than could be seen by his brother. After he got to the desired position, he stops
immediatelly, to remain in the point of interest. We Loretnz-transform the event
L ,t earth 1−0.99992 to the stationary brother's inertial reference frame to obtain the time
increment euqal to:
t earth 1−0.99992
t final−t initial =
v
L
2
c
1−0.99992
t=t earth
v
L
2
c
−
1−0.99992
0
(please note that we need to consider the difference between the initial and the terminal time
calculated with respect to the moving frame coordinates).
Questions
1. Tell the difference between instanteneous velocity, average velocity, and speed.
2. Define the acceleration and derive the expression for distance travelled in a motion with
constant acceleration.
3. Define angular velocity and angular acceleration, derive the formula for the travelled angle
if angular acceleration is constant.
4. Derive the formula for a centripetal acceleration.
5. *Derive the general formula for accelerations in noninertial reference frames (by
differentiation of the radius vector, including the differentiations of unit vectors).
6. Interpret the Coriolis acceleration.
7. Derive the reach and maximum height in a projectile motion.
8. Describe the Michelson-Morley experiment.
9. Derive* or at least explain the origin of the Lorentz transformations.
10. Derive the relativistic summation of velocities.
11. Describe the idea of a space time interval.
12. Derive the time dilation formula.
13. Derive the Lorentz contraction formula.
14. *Describe the twin's paradox.
Refrences:
1. A. A. Michelson, E. W. Morley, On the relative motion of the Earth and the lumuniferous
Ether, Am. J. Sci. 203, 22,1887.
2. R. Feynamann, The Feynamann lectures in physics, Addison-Wesley, 1977.
3. J. K. Goyal, K. P. Gupta, Theory of relativity, Krishna Prakashan Media, 1975.
4. S. Kończak, Podstawy fizyki, Wyd. Pol. Śl., 1998.
5. Dynamics
5.1. Newton's first law of dynamics
The first law of Newton states that a body remains at rest (or moves with a constant velocity along
a straight line) when there is no force acting on the body or all of the acting forces cancel each
other.
This formulation actually defines the idea of a force: it is a phenomenon that can start the motion.
So if we push a body, we act with a force; if the body slides down from a mountain, it receives a
gravitational force; if the body becomes attracted to a magnet, it receives a magnetic force, and so
on.
5.2. Newton's second law of dynamics
The most widely used law of dynamics is certainly the second law of dynamics. This law relates the
body acceleration to the forces that act on the body:
n
m
a =∑i=1 F i
Which means that the vectorial acceleration of the body, multiplied by the mass m, equals to the
vectorial sum of all forces
F i which act on the body. The first law of dynamics is a direct
consequence of the second law when the right hand side reduces to zero. In such case the body
acceleration equals zero, and it moves along a straight line with constant velocity (i.e. it cannot
change the velocity due to a zero acceleration).
It is interesting to note that the heavier is the body (the larger mass m), the lower the acceleration
subject to a given force. Therefore, mass can be considered a measure of inertia (or resistance) to
the changes in motion.
5.3. Newton's third law of dynamics
If a given body acts on another body with force
, then the second body reacts to the first
F
body with a force of the same magnitude, but opposite sign, i.e. −
F . An example of this
situation is the gravitational force that is pushing us downwards to the center of the Earth where we
actually don't move because of the resistance due to the normal force of the surface that supports
our weight.This also holds in noninertial reference frames (figure out an example!).
It is important to stress that the reaction force is applied to a different body than the original
force! It cannot be introduced into the summation of the second law of dynamics, because it is
related to some different mass m!
5.4. The momentum
An important quantity in the mechanics is the momentum,
p =m v
It is a measure of the „amount of motion”. For example a body of mass m, that moves at speed v,
has the same amount of motion as the body of mass 2m, that moves at speed v/2. Due to the
definition, one can immediatelly realize that the idea of momentum allows to express the second
law of dynamics in a compact way by:
n
d p
=∑i =1 F i
dt
This can be easily checked after substituting
p =m v to the above equation. This equation has an
important consequence. In case when the system is not subject to any external net force, the right
hand side of the equation remains equal to zero. This means that the changes in the momentum are
equal to zero. As a consequence, we can formulate the momentum conservation principle: the net
momentum of a system remains constant if there are no external forces acting on the system.
The momentum conservation principle is very useful in calculating the results of the interactions
between bodies. For example if we have a boy that weights m=70kg and jumps at velocity v=3m/s
into a resting boat of mass m=30kg, we can calculate the resulting speed of the boat with the boy
inside:
70kg⋅3m/ s30kg⋅0m/ s=70kg30kg⋅v
v=2.1m/ s
5.5. Work
Work can be defined, as the force exerted on a body throughout a specified path. In differential
terms we say:
⋅d l
dW = F
This can be expressed in terms of the force and displacement components by expanding the dot
product:
dW =F x dxF y dyF z dz
Such formulation is not suitable for the integration because it contains too many differentials.
However, having the trajectory parametried by functions
we can say that dx=
x= x t ,
y= y t , z =z t ,
dx t
dy t
dz t
dt = x ' t dt , dy=
dt= y ' t dt , dz =
dt=z ' t dt . In
dt
dt
dt
such case it is possible to integrate the work equation with respect to the parameter t (which can be
the time for example):
t1
W =∫t F x x t , y t , z t x ' tF y x t , y t , z t y ' t F z x t , y t , z t z ' t dt
0
This formulation is known in mathematics as the line integral. It takes a particularlly simple form
in two-dimensional cases, when the trajectory is prescribed by
y= f x :
x1
W =∫x F x x , f xF y x , f x f ' x dx .
0
To interpret work in simple terms, we could say that it measures the effort, that we need to put into
some task. For example if we want to push a heavy block over 1m then we need a smaller effort
than to push it over 10m. Similarly, when we need to drop coal to the basement and we have 5 tons,
we need to make more effort than to drop 2 tons... How can we relate the later to the work? Don't
you wave with the spade on a nonzero distance when performing the task?
It is also important to comment the obvious observation that when you support a heavy object on
your arms, you become tierd, however you dont move it on any distance! How it is possible to spent
work on such task??? The answer is in the microscopic behaviour of the muscles, where the muscle
fibres constantly shake moving there and forth. They actuall perform work, and burn the energetic
chemicals like ATP. We just cannot observe it on a macroscopic level.
5.6. Mechanical energy
Energy can be considered as the ability to perform work.For example, if you climb up on a
mountain with your car, then the car obtains the ability to perform work without burning fuel. It can
just roll down. This kind of energy is called the potential gravitational energy. To obtain a relation
for its magnitude, consider the work done against the gravitational force (Q=mg) when climbing up:
h1
h1
0
0
W =∫h Qdh=∫h mg dh=mg h 1−h 0
If we would take the reference height h0 equal to 0, then the relation for the work would read
W=mgh. This is the expression for the potential energy:
E p =mgh
It goes without saying that when rolling down, the potential energy decreases, and the car
accelerates. The potential energy converts into a kinetic energy. Just take a look. The gravitational
force F sets now the acceleration of the rolling car:
2
2
l
l
l
t
v
∫ m
∫ m d v ⋅dl=
∫ m d v ⋅v dt=∫ mv⋅d v = mv 2 − mv1
⋅dl=
W =∫l F
a⋅dl=
l
l
t
v
dt
dt
2
2
2
2
2
2
2
1
1
1
1
1
Again, if we set the reference (initial) velocity equal to zero, we end up in a formula for the kinetic
energy which reads:
E K=
mv 2
2
5.7. Power
It is often important to consider how much work is done in a specific amount of time. This is
measured by the power:
p=
dW
dt
The power can be easily calculated for case when an object moves against some constant force (for
example a car driving against the air resistance).:
p=
F
⋅ds
⋅ds
dW F
⋅v
=
=
=F
dt
dt
dt
This formula can be used for example to calculate the power required by a car to drive at certain
speed in presence of the air resistance, which can be described by the formula:
F =C d A
v2
2
where C d is the aerodynamic coefficient, for cars equal to 0.2-0.3 (assume 0.3); =1.225
kg
m3
is the air density, A is the cross section area of the car surface that attacks the stream of air (car
width equals 1.8m, height 1.7m, so A=1.8·1.7m2) and v=200km/h is the velocity of the object.
Subject to this data, F=1735N. This requires a power p=Fv which equals to 96kW (or 131hp,
horsepower). The requirement could be smaller if the aerodynamics of the car was better.
To put the concept of power in a right context, few examples of power consumption. You already
have an example of the car. A truck can consume up to 500hp (370kW). A train locomotive
consumes 2.5MW (3400hp). A nuclear power plant can produce up to 8GW.
5.9. The second law of dynamics in special relativity
To understand the second law of dynamics in the special relativity, one needs to consider the
formulation:
= d p
F
dt
together with the specification of the momentum itself in the special relativity. To understand the
latter, consider a relativistic collision of two pool balls:
Let the ball on the left be called „A” and the ball on the right „B”. Because the collision is
symmetrical, in the table reference system we would expect the same velocity component of the
balls in the y direction after collision.
But the momentum should be conserved in any inertial coordinate system! If the balls move at
relativistic speeds, there occurs the time dilation. So, for example, after collision the observer of
„A” can see the dashed line to move away at velocity v (e.g. 10 meters in each second) and the
same should be found by the observer „B”. However, from the point of view of „A”, observer „B”
does not move that fast. If the y component of the velocity is insignificant, one can omit the Lorentz
contraction in the y direction. But the time dilation applies. The observer „B” from point of view of
1s
„A” will travell the 10m in a dilated second, i.e. in
1−v 2 / c 2
. So the momentum, of „B” would
be equal to:
.
sB
p B =mv B=m =m
tB
sA
tA
2
1−
=mv A 1−
v BA
c
2
v BA
c
2
2
and would be different from the momentum of the „A” ball:
p A=m v A
So the momentum would not be conserved! The rescue could be to postulate that the mass
transforms with velocity as:
m=
m0
2
1−
v BA
c2
where m0 is the rest mass, and m is the relativistic mass but because such relativistic mass is an
apparent construct (for example it is not a source of gravitation—you cannot expect for example
that an accelerated electron, which mass increases and length contracts will change into a black
hole), we prefer to say that the relativistic formula for the momentum is in general not p=mv but
p=
mv
1−
v2
c2
where m stands for the normal rest mass. The second law of dynamics in such case reads:
d
F=
[ ]
mv
v2
1− 2
c
dt
Reference: M. Fowler, UVa Physics: Relativistic Dynamics, 2008.
5.10. Conservative and non-conservative forces
The work done by the forces,
W =∫ F r ⋅d r
Can depend on the trajectory (for example the work done against the frictional forces, which we
will meet later in the course),
or it can be independent of the trajectory, and then the above integral can be evaluated as a definite
integral using „potential”, where the potential at the end point subtracted by the potential at the
start point gives the work done in the movement:
In this example the work done to lift the ball is the same. However, on the right hand side you have
to push harder, but this is not consumed by gravitation! This „extra” energy is conserved into a
nonzero kinetic energy—in case B the velocity is nonzero!
6. Friction forces
The first person in history who investigated the phenomena of friction was Leonardo daVinci. He
has observerd that the friction force is proportional to weight and that it is independent of the
contact surface. He didn't distinguish however between the static and kinetic types of friction.
Currently, we make frequently use of the three laws of friction, two of which were formulated by
Amontons, and one which was formulated by Coulomb (a very talented sciencist, mostly known for
his work on electrc charges):
1. The magnitude of the friction force is proportional to the normal force by which the body
acts on the surface (Amontons). This can be formalized by defining the friction coefficient
as the ratio of the friction force F to the normal force N:
=
F
N
2. The friction force does not depend upon the size of the contact surface (Amontons),
3. The kinetic friction does not depend upon the speed of sliding (Coulomb).
6.1. Static and kinetic friction in sliding
The earliest models to explain the phenomenon of friction considered the climbing of the asperities
of one body on the asperities of the other body. This is the typical explanation that you frequently
learn in the high school. For example, Euler in 1748 considered the following picture:
In this case, the component mg sin of Q must be equal to
coefficient :
F cos which results in a friction
F cos =mg sin
F cos =Q sin
sin
F =Q
cos
F =Q
=tg
Earlier, in 1737 Belidor considered climbing of spherical asperities
A figure from the Belidor work from 1737 in Architecture Hydraulique
which resulted in a friction coefficient equal to 0,35 which he thought to be the universal friction
coefficient (i.e. equal for any considered body).
The theories based upon asperity deformation can give some insight into the friction phenomena but
do not explain all of them. For example, there exists a material—mica—which is smooth on a
molecular level and still displays friction! To overcome such difficulties, Bowden and Tabor have
proposed an adhesive theory of friction, where they assume that bodies—when connected—adhere
to each other due to molecular interactions which need to be broken in order to start the motion
which results in friction.
Within the framework of this theory it is very easy to understand the apparent independence of
friction on the contact area of the bodies. This is because the bodies do not contact on their whole
length but rather this occurs on spiky asperities:
After being loaded by the force, the asperities begin to deform by melting plastically. They do so
until the load divided by the contact area is larger than the melting pressure:
F
P melt
A
As the area grows and grows due to the plastic flow, the limit of the melting pressure is nearer and
nearer, and finally, the area growth is stopped. So, no matter the apparent contact area due to
geometrical dimensions of an object—the real contact area limits at
A=
F
.
P melt
This theory also explains the relation between the kinetic and static coefficient of friction (the
coefficient of friction K that is measured in motion is typically lower than the coefficient S
measured in static position). It is thought that the formation of adhesive contacts between bodies
requires a diffusive crossing of a layer of contaminants and lubricants, which requires time. This
time is lacking in motion and the contacts occur to a smaller degree than they occur in the static
position.
In fact, Coulomb did a series of measurements of the friction in wood, where he actually could
measure the growth of the friction coefficient with time!
6.2. Rolling friction
To understand the phenomena involved in the rolling friction, consider the figure below:
As the body rolls, it deforms on an asperity and receives a braking force (also energy is consumed
to provide the deformation). When the body passes the asperity, it starts to act on the body in
different direction, and the elastic energy is converted to the kinetic energy. Such process is never
ideal, and there occur losses of energy (for example the rubber may not expand fast enough to
release all of its elastic energy; some energy is lost for the internal rearangements of the rubber
molecules in the process of internal friction, etc.). In effect, the body is loosing energy durring the
roll and it slows down. This is the rolling friction.
To understand this phenomenon to a better degree, consider a pure deformation of the rolling body
without any obstacles:
It can be seen that due to the deformation, the support point of the body moves by from the
center and the normal force N gives now a nonzero torque = N which opposes the motion (
is the projection of R on the perpendicular direction to N)! The torque that must be applied to
the body to overcome this braking torque equals
F t R=
F t= N
R
F t =t N
t =
R
which gives a great conclusion that the rolling friction force should decrease with the radius of the
rolling body (provided the deformation size remains fixed).
Questions
1. Define and explain the three laws of dynamics.
2. Define and interpret the idea of the momentum and specify the momentum conservation
principle. When is the momentum not conserved and how does it change?
3. Derive the formula for work on a curvilinear trajectory in a 3D force field.
4. Derive the potential and kinetic mechanical energy
5. Define the power and give the limit for a force exerted by an object of specified power at
given constant velocity.
6. Derive* or specify and comment on the 2nd law of dynamics in Special Relativity.
7. What is the difference between conservative and non-conservative forces.
8. Define the three laws of friction.
9. Explain why does the friction not depend upon the contact area (Bowden-Tabor theory).
10. Explain the origin of the rolling friction.
Reference: P. Borys, Foton 106, 4-26,2009, F.P. Bowden, D. Tabor, An introduction to tribology,
Anchor press, 1973; C. M. Mate, Tribology on the small scale, Oxford Univ. Press, 2008.
7. The rigid body dynamics
7.1. Torque
Did you ever try to remove a nut from the screw?
It can become quite easy when you have a long arm of the key and is extremly difficult with a
small key.
It seems that the driving force for rotations is not just force as was in the transitional dynamics but
some other quantity, that depends also on the distance from the rotation axis. This quantity is called
the torque and is defined by
R× F
=
where R is the radius from the rotation axis where the force F is applied.
7.2. The moment of inertia and 2nd law of dynamics
Think about rotating a stick:
In the upper case it is much easier to start the rotations than in the bottom case. Why is that the
case? It seems that not only mass is what opposes rotations, but also the radius of the mass from the
rotation axis. The further the mass from the rotation axis, the larger linear velocity it must have to
move and the larger acceleration is required to establish this velocity, and consequently, a larger
force is required for establishing the acceleration.
Consider mass m on radius R from the rotation axis. To provide an angular acceleration =
a
,
R
the second law of dynamics states:
m R= F
2
m R =RF=
I =
which can be also written in a more general vectorial manner as the second law of dynamics for
the rotational motion of a rigid body:
I =
where I is the moment of inertia, that was equal to mR2 for a point mass and for a more general
rigid body it is the sum of the moments of inertia of all mass points:
N
I =∑i=1 m i R2i
where mi is the mass of the i-th point and Ri is its distance from the rotation axis. This summation
formula is often written in a continous limit of infinitely small masses mi.
2
I =∫M r dm
where the integration takes part over the whole considered mass m (typically parametrized by the
position parameter in relation to density). For example, a moment of inertia for a ring (or a hollow
cylinder) can be calculated by the following:
An infinitely small mass point can be identified by dm=Mdα/2π. Therefore:
R2 M
d
2
2
I =MR
2
I =∫0
Another example is to calculate the moment of inertia of a stick with respect to one of its ends:
L
I =∫0 M x 2
3 L
∣
dx
L
Mx
ML
I=
=
3L 0
3
2
Having the formula for a ring, we can calculate the moment of inertia for a solid disc (or a solid
cylinder):
Rmax
I =∫0
2
MR
2 R dr
R2max
2M R
3
R dr
2 ∫0
R max
4
2M R max
1
I= 2
I = M R 2max
2
Rmax 4
I=
max
Having the formula for a solid cylinder, we can derive a formula for a hollow cylinder (outer radius
r2 , inner radius r1, mass of the removed part—m, mass of the remaining part—M):
mM r 22 m1 r 21
−
2
2
r 21
m= M m 2
r2
2
2
2
Mr 1mr 2mr1 mr12 M r 21r 22
I=
−
=
2
2
2
I=
Where the middle relation follows from the proportion between mass and cross-sectional area of the
inner cylinder and of the full cylinder with radius r2.
Other important formulas for the moment of inertia include (Dietłaf, Jaworski: Reference in
Physics):
Body
Moment of inertia
Stick of length L, mass M along its end
ML2/3
Stick of length L, mass M along its center
ML2/12
Hollow cylinder, radius R, mass M, along its rotation axis
MR2
Solid cylinder, radius R, mass M, along its rotation axis
MR2/2
Ball, mass M, radius R along its center
2/5MR2
Sphere, mass M, radius R along its center
2/3MR2
7.3 Steiner's theorem
Once we have a moment of inertia given for the rotation axis that is passing the center of the mass
(ICM), we can calculate the moment of inertia of a body along any other parallel axis separated by
radius r by:
I =I CM mr 2
To prove this relation, consider a general form of the ICM:
N
I CM =∑i=0 mi x 2i y i2
Now let us shift the rotation axis to r in x direction:
N
N
N
N
I =∑i=0 mi [ x ir 2 y 2i ]=∑ i=0 mi x 2i y 2i ∑i=0 m i r 2 ∑ i=0 mi r xi =I CM mr 2
because the last smmation vanishes since it is calculated with respect to the mass center (i.e. it has
the same amount of positive mixi components as of the negative components).
Reference: Burton P., Kinematics and dynamics of planar machinery, Prentice Hall 1979.
7.4. Angular momentum
Consider the second law of dynamics in absence of any forces:
I
=0
d
I
=0
dt
This can be integrated with respect to time to give a certain conserved quantity L:
L=I =const
This quantity is called the angular momentum. It is a dual quantity to the normal momentum of
translational motion, where the mass is replaced by the moment of inertai and velocity is replaced
by the angular velocity.
The conservation of angular momentum is important for many phenomena. The most widely
recognized is the spinning ice skater, who spins faster after contracting her arms and legs towards
the rotation axis. Another example of the conservation of angular momentum is the rotation of stars.
As the interstellar matter contracts to form a star it is almost immobilized but there are always some
assymetries and a nonzero rotation. As the moment of inertia greatly decreases (the radius shrinks
after the star is formed), the star starts to rotate. It rotates extremely fast after a terminal contraction
to a neutron star which can have a diameter of only 10km instead of the original hunderths of
astronomical units of the planetary gas (1AU=150mln km). In fact, neutron stars can perform
hunderths of rotations per second. This causes their shape to become elliptic.
The angular momentum is also used for the construction of a giroscope, a device that is (for
example) being used to show the horizon in the airplanes. It consists of an extremly fast rotating
disc and it has thus a large angular momentum. The torques experienced during the flight cause
changes in the angular momentum:
d L
=
=
R× F
dt
But if the L by itself is very large, then small changes dL cannot change the direction of the artificial
horizon:
Here, ΔL corresponds for example to the change in the angular momentum after 10 evolutions of
spinning around the long axis of the airplane. Assuming that the speed of rotations does not change
(i.e. the magnitude of the vector L=Iω remains constant), one can derive an equation for the
precession:
=
d
L d I
d I e dL e dL
p× L
=
=
=
= e L p×e =0 p ×
L =
dt
dt
dt
dt
dt
(recall the derivative of the unit vector that we have considered when deriving accelerations in
noninertial reference frames).
7.5. Energy in the rotational motion
As the parts of the body rotate, they posess certain speeds around the rotation axis. Such speeds
give contributions to the kinetic energy of the movememnt:
n
E k =∑i=1
n
mi v 2i
mi r i2 2 2 n
I 2
=∑i=1
= ∑i=1 mi r 2i =
2
2
2
2
7.6. Work in the rotational motion
When the mass points are displaced under a torque around the rotation axis, a work can be
calculated by:
⋅dl=Fdl
dW = F
cos
For a force that acts in the direction of the rotation (the component of torque), we say α=0, and
dl=dφR. Therefore:
dW =FRd = d
In such case
2
W =∫ d
1
7.7. Summary of the similarities between translational and rotational motion
translational
rotational
Position x
Angle φ
Velocity v=dx/dt
Angular velocity ω=dφ/dt
Acceleration a=dv/dt
Angular acceleration ε=dω/dt
Mass m
Moment of inertia I
Second law of dynamics: F=ma
Second law of dynamics: τ=Iε
Momentum: p=mv
Angular momentum: L=Iω
Work: dW=F∙ds
Work: dW=τdφ
Power: P=Fv
Power: P=τω
2
Kinetic energy: E=mv /2
7.8. Collisions in 2D
Consider two rigid bodies that collide in two dimensions.
Kinetic energy: E=Iω2/2
n and exert a force in point p. The interaction force
The objects collide in the collision plane
obeys the 2nd law of dynamics by:
d p
=F
dt
Now taking into account the velocity changes under the influence of a force
m
a=
F
d
v F
=
dt m
we can say
d
v d p
=
dt mdt
cancelling the time infinitesimal and integrating we obtain:
p
m
p
v 2=v 1
m
v =
Similar reasoning can be performed for the rotational motion:
d
=
dt
d
I
=r × F
dt
d
r × dp
I
=
dt
dt
I d =
r × dp
r × p
=
I
I
so:
2=
1
r × p
I
Now we can consider two bodies that approach each other at velocities v 1A , v1B , 1A , 1B . The
resulting equations are:
p
v2A= vA1
m
p
v2B= vB1 −
m
AP
p
r ×
A2 =1A
IA
BP
p
r ×
B2 =1B−
IB
because this is a 2D problem, we can simplify the cross product (and neglect the z components of
the vectors—angular velocity will become a scalar in further reasoning):
∣
∣
i
j
k
rx
ry 0
p x p y 0 =k r x p y −r y p x = k r⊥⋅ p
i j k
rx ry 0
where r ⊥ stands for the r vector with interchanged x and y coordinates, and the new x coordinate
is taken negative. This makes then:
p
m
p
v 2B=v 1B−
m
AP
p
r ⋅
A2 =1A ⊥
IA
BP
p
r ⋅
B2 =1B ⊥
IB
v 2A=v 1A
This makes a system of 4 equations and 5 unknown variables. To deal with this, we introduce after
Newton the coefficient of restitution e, which describes the degree of elasticity of the collision. A
completely elastic collision sets e=1, while a plastic collision sets e=0. There is a lot of
experimental work available with coefficients of restitution of real bodies (e.g. cars, where
e=
1
). The coefficient is defined by:
v [ m/ s ]
2
v AP
v 2BP ⋅n =−e
v 1AP −v BP
n
2 −
1 ⋅
where the velocities are not the velocities of the center of the mass, but the velocities of points of
the body that touch the collision site p:
AP
v 2 =v 2
2×r
AP
However, we don't treat angular velocity as a vector. But we know that it should point up from the
considered XY plane, in the direction of the k unit vector. So:
AP
k ×r AP = r AP
r AP
x j−r y i =
⊥
AP
AP
v 2 = v2r ⊥
with similar relation for body „B” and for pre-collisional velocities (index „1”). (Notice: the angular
velocity in the above is taken without a vector, just as a magnitude! The rotation direction is
v AP
v 2BP n
perpendicular to rAP and is determined correctly by r⊥AP . Now we can subtract
2 −
to obtain:
v 2AP−v 2BP n =.
v A2⋅n2A r AP
−v B2⋅n− 2B r BP
=.
⊥ n
⊥ n
AP
AP
BP p n
p n
r ⊥ r ⊥ ⋅
r AP
p⋅n
p⋅n
A
A AP
B
B AP
⊥ r ⊥ ⋅
v 1⋅n
1 r ⊥ n
−v 1⋅n
−1 r ⊥ n
=.
mA
IA
mB
IB
AP
2
BP
2
p p r ⊥ ⋅n p pr ⊥ ⋅n
v 1AP−v 2AP n
mA
IA
mB
IB
p= p n . Evaluating the transmitted magnitude of the
where we made use of the identity
momentum, we obtain:
AP
BP
−1ev 1 −v 1 ⋅n
p=
AP
2
BP
2
1
1 r ⊥ ⋅n r ⊥ ⋅n
ma mb
IA
IB
Having p in terms of the pre-collisional velocities, it is straightforward to determine the postcollisional velocities and angular velocities.
Questions
1. What object replaces forces in rotational dynamics? Could you give some reasons?
2. Derive the second law of rotational dynamics for mass m rotating at radius r, subject to a
force F. Define the moment of inertia.
3. Derive the moment of inertia for a solid cylinder.
4. Prove the Steiner's theorem.
5. From second law of dynamics derive the principle of angular momentum conservation.
Derive the equation of precession.
6. Derive the formula for work and energy in rotational motion.
Reference: Chris Hecker, Game Developer, 11, 1997; J. Wierciński, A. Reza, Wypadki drogowe.
Vademecum biegłego sądowego. IES, 2010.
8. Analytical mechanics
8.1. Lagrange equations
Consider a special form of an energy function, that is called Lagrangian:
L=E K − E P
Which is made of the difference between the kinetic energy and the potential energy. For example if
we consider a translational motion in the x direction, the Lagrangian can have the form:
2
L=
where
m ẋ
− E p x
2
ẋ stands for the time derivative of x, i.e.
respect to x or
ẋ=v . In such case, the derivatives of L with
ẋ take a special form:
∂ L −∂ E p x
=
= F x
∂x
∂x
∂L
=m ẋ= p
∂ ẋ
Where the first equality resembles the formulation of a potential force. I.e. when you slide down by
dx to change the potential by dEp, a work has been done dW=-dEp, so the force must have been
involved by -dEp=Fdx, i.e.
F=
−dE p
.
dx
Comparing this to the second law of dynamics, we can state that:
∂L d ∂ L
=
∂ x dt ∂ ẋ
Which is called the Lagrange equation of mechanics. It can be proved by the variational calculus
that this form of equation would hold for any pair of the conjugated coordinates that occur in L. The
generalized momenta are typically denoted by letters p, while the generalized coordinates are
typically denoted with letters q. We will see this notation in the chapter devoted to the Hamilton
mechanics. For now, as an example you could consider angular motion with:
2
L=
2
I
I ̇
−E P =
−E P , so that the equation would read:
2
2
∂L d ∂ L
=
∂ dt ∂ ̇
8.1.1. The Euler-Lagrange formula for variational problem
Consider a motion along a trajectory
y x that is perturbed to Y x= y x x .
Let us find a trajectory that minimizes the following integral:
b
∫a
f x , y , y ' dx min
Substituting the perturbed trajectory, we say:
b
∫a
b
∫a
f x , Y ,Y ' dx min
f x , y ,Y ' ' dx min
To find the trajectory that minimizes the integral, we differentiate it with respect to and
compare to zero:
∫a ∂∂ fy ∂∂ yf ' ' dx=0
b
The second term can be integrated by parts taking into account that the perturbed trajectory must
start and finish in the unperturbed points (i.e. a =b=0 ):
b
∫a
∂f
d ∂f
−
dx=0
∂ y dx ∂ y '
Since the perturbation can be any possible function, the interior of the brace must vanish, which
gives rise to the Euler-Lagrange equation:
∂f d ∂f
=
∂ y dx ∂ y '
Which is the same equation that we considered for the mechanics, just represented in the domain of
x, and not t... So the Lagrange dynamics describes a motion along a trajectory of minimum kinetic
energy and maximum potential energy.
8.1.2. Example of the Lagrange formalism application
The greatest utility of the Lagrangian formulation of the mechanics is to describe a motion subject
to a system of constraints. Example application for a bead fixed on a rotating circle made from wire:
The only degree of freedom of this bead is the angle . The kinetic energy and the potential
energy can be described in terms of this angle:
mv2tangent mv2circumferential
2
2
2
2 2
˙
m R sin2
T =mR 2
2
2
E p =mgR 1−cos
T=
Using that, we can construct Lagrangian:
L=mR 2
˙2 m 2 R2 sin2
−mgR 1−cos
2
2
and then put it into the equations for:
∂L d ∂ L
=
∂ dt ∂ ̇
which reduces to:
mR 2 2 sin cos −mg R sin =mR2 ̈
̈= 2 cos −g / Rsin
that can be integrated.
8.2. Hamilton equations
It is possible to construct another function, related to the Lagrangian. It is called the Hamiltonian:
H = p q̇−L=2 E k − E k − E p =E k E p
(for example: if
p=m ẋ and q̇= ẋ ,
p q̇=m ẍ 2 )2. So Hamiltonian is the sum of the kinetic
and potential energy. It is interesting to calculate the derivatives of H with respect to p and q.
∂ H −∂ L
d ∂L
d
=
=−
=− p=− ṗ
∂q
∂q
dt ∂ q̇
dt
∂H
∂ q̇ ∂ L ∂ q̇
∂ q̇
∂ q̇
=q̇ p
−
= q̇ p
−p
=q̇
∂p
∂ p ∂ q̇ ∂ p
∂p
∂p
So, the Hamilton mechanics, while being based upon the Lagrange mechanics, offers very useful
equations for calculating the evolution of the generalized coordinates:
∂H
∂q
∂H
q̇=
∂p
ṗ=−
Questions
1. Explain the Lagrange formulation of mechanics.
2. Explain the Hamilton formulation of mechanics.
Refrence: J. R. Taylor, Classical Mechanics, Univ. Sci. Books, 2005.
9. Gravitation
Between every two mass bodies in the space there exists a force of gravitation. Newton postulated
a simple form of this force with
2 Remember: p are the generalized momentum coordinates, and q are the generalized position coordinates.
=G Mm r
F
r2
This force is always attractive and depends upon the mass M of one body, mass m of the second
body, and upon the distance between these bodies r. The symbol r denotes the unit vector along
the line connecting the two bodies, and G=6.67∙10-11Nm2/kg2.
This force is different from any other forces because it acts on each mass point causing an uniform
acceleration of bodies in space:
Mm
r
2
r
GM
a = 2 r
r
m
a =G
It works independent of the mass m of a body. If you could shut down into a freely falling container,
everything inside would fall with the same acceleration and you would not notice any inertial
effects. You could conclude that you are in the inertial reference frame. This was the conclusion of
Albert Einstein: he said that the freely falling coordinate system is an inertial refrence frame.
Because it is an inertial reference frame, even the light inside of such container will „fall” in the
direction of the attracting mass. This is the basis of a process called „gravitational lensing” of light.
However, to calculate a proper deflection angle it is not sufficient to calculate (as did Henry
Cavendish in 1784):
[
z=b tan t
GM
b
GM
b
t=atan z /b
v=∫−∞ a dt=∫−∞ 2 2
dt=∫−∞ 2 2
dz=
2
2
2
2
dz /b
bdz
b z b z
b z c b z
dt=
= 2 2
2
1 z /b b z
∞
∞
∞
]
/ 2
bGM
bc
∫− / 2 dt
GM
b b tan t bc
2
2
2
=
dt
/ 2
∫−/ 2
cos2 tsin 2 t
cos 2 t
=
GM
bc
∫−/ 2 cos t dt= 2GM
bc
/ 2
where z is the position of the photon along its straight line trajectory and b is the collision
parameter (minimum distance to mass M) and then to divide =
v
, because we would obtain
c
an angle that is twice too small (in such calculations we assume the ray moves almost along a
straight line and the conditions for the force equation depend only on z, not on b, but this is not the
source of the problem). Because the gravitational time dilation plays a role (it slows down the light
for the distant observer, exactly the same happens in a glass lens), Einstein derived identical
angular correction due to the time dilation by itself) as well as the space curvature, one needs to use
the formalism of the general relativity to obtain the final result:
=
4GM
2
c b
To give some hint against this correction, consider the following figure:
In the next chapter we derive a formula for gravitational time dilation,
= t 1−
2GM
rc 2
Which means that the time runs slower in presence of gravitation, and the light (apparently) moves
faster (in fact the distance also undergoes relativistic modification in this coordinate system, and
velocity is constant). However, the light can pass from point A to point B in shorter time than when
it moved on a straight line, just as it will be shown by us in the future in the derivation of Snell's law
by the Fermat principle. So, we could try to use the Snell's law for refraction:
n 1 sin 1=n 2 sin 2
together with a formula for the „gravitational refraction index”,
n=
v0
=
vn
c
c / 1−
2GM
rc 2
= 1−
2GM
rc 2
Now consider the point in the figure, where the angles are shown. For that point we can write the
Snell's law as:
n sin =n
dn
dz sin d
dz
Making use of the formula for the sine of a sum of the angles, we have:
n sin =nn ' dz sin cos d cos sin d
n sin ≈nn ' dz sin cos d
The derivative of n can be calculated easily after its simplification:
n= 1−
2GM
GM
GM
≈1− 2 =1− 2 2 2
2
rc
rc
c b z
−GM
n ' = 2 2 2 3/2
c z b
Now substituting such n', and assuming that n is close to 1 (the major contribution originates from
n'), we have:
−GMz
sin dzcos d
c z 2 b2 3/ 2
−GMz
bdz
zd
0= 2 2 2 3/ 2 2 2 2 2
c z b z b z b
GMb
d = 2 2 2 3/ 2 dz
c z b
0=
2
which gives identical contribution as we had in the corpuscular model! (the only difference is the
extra c factor in the denominator which results from the fact that we calculate the angle directly, not
the change of the corpuscular velocity, as previously) Assuming the equivalence principle, both
effects should act additively, thus giving an angle twice larger than predicted by the classical theory.
9.1. Gravitational time dilation
Consider a body that falls down from infinity onto a mass object M. It gains the energy:
E=
GMm
r
which converts into kinetic energy:
Ek =
mv 2
2
Equating these two energies we obtain the velocity of a freely falling body:
v=
2GM
r
Now, we can use this velocity in the time dilation formula to obtain the gravitational time dilation!
= t 1−
v2
2GM
= t 1−
2
c
rc 2
i.e. the proper time in gravitational field ( ) runs slower than for a distant observer ( t ).
9.2. Orbital speed
Consider a body of mass m that enters an orbit around a body of mass M. To do this, the body must
attain an equilibrium between the centrifugal force and the gravitational force:
mv 2 GMm
= 2
r
r
GM
v=
=v I
r
So, for example, if Earth's mass is 6⋅10 24 kg , radius equals 6400km, one obtains v=7.9km/s.
9.3. Escape velocity
A higher velocity is required to leave the orbit of a stellar body. To calculate this velocity one needs
to consider the equality of the kinetic energy and of the potential energy of the body:
mv 2 GMm
=
2
r
2GM
v=
=v II = 2 v I
r
For the escape from the Earth this makes vII=11.2km/s.
To escape from the Solar system, one needs ( M sun =2⋅1030 kg ,
R Earth=1Au=150mln km )
vII=42.2km/s, but much of this velocity is given by the orbital speed of Earth around the Sun
(vI=29.8km/s), so the actual speed to leave the Solar System from the Earth surface is lower:
mv 2 GM Em m 42.2km/ s−29.8km/ s 2
=
2
RE
2
2GM E
v 2=
12.4km/ s 2
RE
v=
2GMe
12.4km /s 2
RE
v III =16.7km / s
9.4 Gravity assist and Hohmann transfer maneuver
Questions
1. Define the Newton's law of gravitation and tell why is this force so different than the other
forces? What was the Einstein's conclusion?
2. Define the gravitational lensing formula. What is the contribution of relativistic effects?
3. Derive the gravitational time dilation.
4. Derive the expression for an orbital speed.
5. Derive the expression for an escape velocity (from gravitational field).
10. Chaotic dynamics
Chaos refers to a set of phenomena which are characterized by deterministic mechanics but their
evolution is unpredictable due to extreme sensitivity to the initial conditions. Because we can't
assure infinite precision of the initial conditions (we always have some measurement error), we
have only limited possibilities to predict the evolution of the system. An example is the trajectory of
a mass point between circular scatterers:
The solid line and the dashed line start from a single point, but the dashed line has slightly different
initial angle. After three collisions the trajectories become unrelated! This is what we mean by
unpredictability. If the angular difference would be smaller, the error would require more collisions
to grow to the magnitude that decorrelates the two movements. But this would happen, and it would
not require too many collisions, because the error grows exponentially like
= 0 exp t
where is called the Lyapunov exponent. Such an exponent can be easily visualised for some
simple biophysical systems. For example, the logistic equation, which describes the reproduction
of a species (with number of individuals equal to nk in k-th generation) in the limited avialability of
resources is written as:
n k1=ank 1−nk
For certain values of a this equation is chaotic, i.e. oversensitive to the initial conditions. Consider
a=3.95. Take n0=0.2500 and then consider the case with n'0=0.2501. The trajectories are
summarized in the following table:
k
n
n'
|n-n'|
0
0.2500
0.2501
0.0001
1
0.7406
0.7408
0.0002
2
0.7588
0.7584
0.0004
3
0.7230
0.7237
0.0008
4
0.7912
0.7898
0.0014
5
0.6527
0.6558
0.0031
6
0.8955
0.8917
0.0038
7
0.3698
0.3816
0.0118
8
0.9205
0.9321
0.0116
9
0.2889
0.2500
0.0389
10
0.8115
0.7407
0.0709
11
0.6041
0.7587
0.1546
12
0.9447
0.7232
0.2215
13
0.2064
0.7908
0.5844
It is interesting to plot the differences between n and n' for subsequent k generations to observe the
exponential departure of one trajectory from another:
=∣n k −n ' k∣
k
In fact, such chaotic systems are used in the computer science as random number generators.
Questions:
1. Explain the meaning of Lyapunov exponent.
11. Fluid dynamics
11.1. What is a fluid?
The term fluids refers to all the substances that are able to flow and do not maintain a fixed shape.
The most common examples of fluids are gases and liquids. The fluid dynamics addressed to gases
reveals its importance for example in the studies of aerodynamics, while addressed to liquids
describes the flow of liquids in pipes, rivers, or in the human vienes.
Reference: D. Giancoli, Physics.
11.2. Fluid statics: Pascal principle and the pressure
The pressure p of a fluid is defined as the force F that is exerted by a surface element s of the
fluid:
p=
F
s
The Pascal principle states that the pressure applied to the fluid changes increases the pressure
inside of the fluid everywhere by the same amount. This is very useful for the construction of a
hydraulic lift:
In this figure, a small weight exerts a small force on a piston with small surface. On the other hand,
the piston on the right has a large surface, k-times larger than the surface on the left. The force
exerted on this piston is:
p 1= p2
F1 F2
=
S 1 S2
F1 F2
=
S1 k S1
F 2=kF 1
and it may become sufficient to overcome the gravitational force of a car's weight.
11.3. Fluid statics: the Archimedes law
Before discussing the Archimedes law, let us consider the pressure distribution in a vertical tube of
fluid. The pressure results in general from the weight of the fluid that remains above the considered
level:
It is clearly seen that on height h, the weght equals to mg while m=ρV, where V=hS. As a result, the
pressure varies with depth of the fluid as:
p= g h
Knowing that the pressure varies with depth of the fluid it is easy to imagine that the pressure acting
on the bottom of any object immersed in the fluid will be larger than the pressure acting on its top.
The resulting force will push the body upwards and it is called the buoyant force. Take look at the
following figure:
The net force originating from the pressure equals to:
F =F 2−F 1= g h 2 S − g h1 S = g h S = V g
which in other words means, that the buoyant force equals to the weight of the water pushed out
by the immersed object. This is the law of Archimedes.
11.4. Fluid dynamics: the streamline and laminar flow
In case when the fluid is in motion, and does not move too fast, it can be described in terms of a
such called laminar flow. In laminar flow the flow of the fluid can be decomposed into
streamlines, i.e. if you consider a cross-section of a fluid, then each element of the cross-section
follows an unique trajectory (a streamline) surrounded by the neighbouring streamlines and does
not cross the surrounding streamlines. The word „laminar” means „in layers”.
The opposite of the laminar flow is the turbulent flow where eddies occur, which absorb a lot of
energy due to internal friction (viscosity). This can be seen on the wind streamlines approaching a
wing profile (the upper streamlines are going in a laminar flow, while the lower become turbulent):
The origin of the vorticies in the lower case is the „vacuum” region after the wing profile which
sucks down the streamlines and causes their rotation.
The condition for a change from laminar flow to a turbulent one is that the Reynolds number, which
characterizes the system, should be sufficiently large, typically larger than 4000. This number
expresses the ratio of the inertial force to the viscous force ( F = A
v
, where is the
l
viscosity coefficient, A is the contact area between a body and a fluid, v is the velocity of the fluid, l
is the thickness of the nonzero fluid velocity profile—you will learn about this in details in the
following paragraphs). Consequently, the Reynolds number may be evaluated as:
L
v
3 L
m 2
L
F
ma
t
t
t
ℜ= inertial =
=
=
=
2
F viscous
v
L
L
L
A
L2
L2
L
Lt
Lt
m
Now, expressing time by the velocity definition as t=
ℜ=
L
, we obtain:
v
Lv
Where L is the characteristic length of the system, v is the speed of flow, and ρ is the density of the
fluid. How does it work? Obviously, to sustain a laminar flow, the damping in the system should be
large to attenuate any vortices. This requires a viscous force that is sufficiently large as compared to
the inertial force and results in a small Reynolds number. Otherwise, the vortices are not attenuated
and the Reynolds number is large.
11.5. The continuity equation
Consider an ideal fluid (nonviscous, incompressible fluid) that flows in a pipe which contracts its
diameter as shown in the figure:
Now imagine that we push some amount of fluid on the wide entry of the pipe. Let the volume be
equalt to V =A1 x1 . The same amount of fluid must exit on the right hand side: V =A2 x 2 .
In this situation:
A1 x 1=A2 x 2
We can divide the above by the corresponding time increment to obtain:
A1
x1
x2
= A2
t
t
A1 v 1= A 2 v 2
which is called the continuity equation. This is one of the reasons why rivers are fast in the
mountain, where they are not so wide, while they slow down a lot when they pass the lowlands.
The continuity equation can also be written in 3D. To start, let us rewrite the above in form:
A2 v 2 −A1 v 1=0
In 3D a similar expression can be written for all perpendicular axes. Consider a cubic element of
side lengths x , y , z . Then:
1
y z v x x x−v x x x z v y y y−v y y x y v z z z −v z z=0∣⋅
x yz
results in:
v x x x− v x x v y y y− v y y v z z z −v z z
=0
x
y
z
which for an infinitely small volume element with x y z 0 reduces to:
∂v x ∂v y ∂ v z
=0
∂x ∂y ∂z
which can be also written using the nabla operator
∇= i
∂ ∂ ∂
j
k
as:
∂x
∂y
∂z
∇⋅v =0
or
div v =0
because div (the divergence) is the name for a dot product between nabla operator and a vector.
11.6. Bernoulli equation
Consider now a flow that changes the altitude, pressure and velocity:
The action of pressure p1 on the left hand side overcomes the pressure p2 on the right hand side
resulting in a net pressure difference that creates a force which drives the motion of the fluid. The
fluid also changes the energetical state: after the fluid moves by the filled part from the left, a filled
part on the right occurs. Instead of a fluid at velocity v1 and height h1 we obtain the same mass of
the fluid in the streamline at velocity v2 on height h2. In such case, the work done by the forces on
both ends of the streamline serves to allow the change of the energetical state of the streamline (the
work-energy theorem), i.e.:
mv 22 m v 21
−
m g h 2− m g h1
2
2
x 2 A2 v 22 x 1 A1 v 21
p 1 A1 x 1 − p2 A 2 x 2=
−
x 2 A2 g h 2− x 1 A1 g h 1
2
2
V v 22 V v12
p1 V − p2 V =
−
V g h2− V g h1
2
2
p1 A1 x1− p 2 A2 x 2 =
Dividing by V and rearranging the terms results in the Bernoulli equation for the
incompressible, nonviscous fluid:
p 1
v 21
v2
g h1 = p 2 2 g h 2=const
2
2
11.7. Viscosity
Consider two surfaces lubricated with a syrpup between them:
It seems reasonable that the larger the upper body, the more difficult it is to maintain its movement.
Similarly, the higher velocity, the more difficult to move. All these effects are due to the internal
friction of the syrup (the fluid in general). The internal friction relates to breaking of the cohesive
bonds of the medium, i.e. the bonds that keep the molecules of the medium together. Newton
proposed a following formula to describe the viscous drag:
F = A
v
l
where F is the force of the viscous drag, A is the surface of the sliding body, v is its velocity and l is
the thickness of the fluid layer. is a proportionality coefficient, called viscosity.
Why does the Newtonian viscous drag depend upon the surface? Well, the larger the surface, the
more cohesive bonds we need to break. Why does it depend upon the velocity? To explain this we
need to understand the velocity profile of the fluid under the moving surface:
The velocity of a fluid adjacent to the moving surface has the velocity of the moving surface. The
velocity of a fluid layer adjacent to the resting surface has zero velocity. In between we have
intermediate velocities. As the body moves, the adjacent layers of fluid slide over each other. The
higher the velocity of the upper plate, the faster the layers move with respect to each other and the
more cohesive bonds are broken per unit time, and the larger is the momentum loss of the fast layer
( F=
dp
). The force is thus proportional to the velocity.
dt
If the thickness of the fluid increases, the velocity changes decrease and the fluid layer adjacent to
the upper body receives less breaking of the cohesive bonds, and a smaller momentum loss. The
force is thus inversely proportional to the fluid thickness.
11.8 The Poiseuille law
Knowing the Newton's formula for the viscosity, we can derive a formula that describes a flow of a
viscous fluid in a pipe. Consider the following figure:
The situation is not exactly newtonian, because the subsequent layers of the fluid grow with radius.
But if we consider only a small thickness of the fluid between R and R+dR, the change in the layer
area is not significant, and we can use the differential formulation:
F = A
dv
= Av '
dl
Then we can build an equation:
dF =F R− F RdR
dF = A R v ' R− A RdR v ' RdR
dF =− d [ A Rv ' R]
Because the force is driven by the pressure (which is constant at the whole pipe cross-section), we
can write:
p 2 R dR=− d [ A R v ' R]
Integration of the both sides gives:
p R2=− A R v ' R
The integration constant is zero, because the velocity has a maximum for R=0 (i.e. v'(0)=0).
Because
A R=2 R L , we can write
p R2=− 2 R L v ' R
− p R
v ' R=
2L
− p R 2
v R=
C
4L
2
2
p l −R
v R=
4L
Where the integration constant was taken such to guarantee v(l)=0. The flow in a pipe is expressed
as the volume passing it per unit time. It can be calculated by:
l
Q=∫0 v R2 R dR
p l 2
Q=
l R−R 3 dR
∫
0
2 L
p l 4 l 4 l 4 p
Q=
− =
2 L 2 4
8L
[
]
which is the Poiseuille equation.
11.9. The Navier Stokes equation for incompressible fluid
If we consider a fluid element in motion, we can try to describe the motion in presence of a pressure
gradient. We start with the second law of dynamics:
m
dv
=F x− F x x =− F
dt
dv
V
=− F
dt
dv
F
p
dp
=−
=−
=−
dt
x A
x
dx
because we can say—in one dimensional case—that the volume subject to movement V =A x ,
where A is the cross-sectional area of the moving fluid. A three dimensional generalization of the
spatial derivative is the gradient:
dv
=−grad p
dt
The derivative on the left hand side depends upon the change of the velocity with time, but also it
varies with position, and since the fluid is in movement, the different velocity from different
position will enter the current position after moving with velocity v. This gives in 1D:
dv ∂v
∂v
= v
dt ∂t
∂x
.
and in 3D, using the divergence:
d
v ∂
v
= v ∇⋅v
dt ∂t
Expanding the time derivative in the second law of Newton, we obtain the Euler equation3:
∂ v
1
v ∇⋅
v =− grad p
∂t
In case, when we would also like to consider other forces than the pressure gradient, we must put
them on the right hand side of the second law of dynamics. A particularly interesting force is the
viscosity, defined by the Newton's formula. Let us try to calculate the F per unit volume acting
on the fluid layer in the direction x if the velocity gradient is in direction y:
F = A[v ' y y−v ' y]
v ' y y −v ' y
F
=
V
y
F
= v ' ' y
V
i.e. it is the second derivative of the velocity times the viscosity. The same can be said for the
velocity profile in the z direction. A bit more complicated is to investigate the effect of a velocity
change in the direction of flow, x. I will try to motivate it with help of some hand waving. Take a
look on the following figure:
3 Ref: Landau, Lifszyc, Hydrodynamics, p. 16
Please note that due to continuity equation (or conservation of the fluid's volume), as the fluid slows
down, it must expand. If in a time interval t the streamline propagates in the x direction to a
smaller distance, then having a fixed width of the streamline, it must expand in the y direction by
the same amount, i.e. the change in the velocity in y direction equals to the change of the velocity in
the x dierection:
dv y −dv x
. Parametrizing time t=v x dx , we can modify this relation to
=
dt
dt
dv y −dv x
which drives the viscosity on the plane perpendicular to the motion,
=
dx
dx
F = A
dv y
.
dx
The gradient is measured „to the left”, so against the direction of the x axis, and therefore we can
get rid of the minus in the relation between
dv y −dv x
.
=
dx
dx
Generalizing this to three dimensions requires the use of a laplacian
2
∇=
∂2
∂2
∂2
2
2
2 :
∂ x ∂ y ∂z
F
2
= ∇ v
V
which, after adding another term to the Euler's equation gives the Navier-Stokes equation for an
incompressible fluid4:
∂ v
1
2
v ∇⋅
v = [ grad p ∇ v ]
∂t
11.10. Stokes law
4 Ref: Landau, Lifszyc, Hydrodynamics, p. 67 (Eq. 15.7)
The Stokes law is one of the most frequently used formulas of the fluid dynamics. However, its
derivation is very difficult. In this section I will sketch the steps of the derivation without going into
the mathematical details, which are too difficult for an introductory course in physics.
In case when the movement is laminar, for low Reynolds numbers, the velocity profile along the
flow is stationary and the left hand side of the Navier Stokes equation disappears:
∇ 2 v − grad p=0
Taking a curl of this equation results in a condition for the velocity:
∇ 2 curl v =0
(the curl of a gradient disappears, and laplacian can be interchanged in order with curl due to the
linearity). This must be supplemented by a continuity equation, which in the 3D form reads:
div v =0
The above two equations were solved by Stokes to give a velocity profile around a sphere of radius
r falling in a viscous incompressible fluid (actually, he considered a dual problem of a sphere which
remains at rest in a fluid that moves at velocity v), which reads:
v r =v cos 1−
3
3R R
2r 2r 3
v =−v sin 1−
3R R 3
−
4r 4r 3
where R is the distance from the sphere's center, and is the angle with respect to the direction
of falling. The derivation is extremly difficult, but if you want to check the correctness of these
formulas, just check the form of the vector operators for spherical coordinates and see how are the
two initial equations satisfied and put R=r, and
R=∞ to check the boundary conditions of the
flow around the sphere.
Plugging this equation into the Navier-Stokes equation gives the pressure distribution, which
contributes to the drag force acting on the sphere. Another contribution comes from the Newton's
viscosity of the flow near the surface, e.g.:
∂v r
∂r
∂ v
F 2= A
∂r
F 1=A
Taking into account all the forces, and integrating them along the surface gives finally the Stokes
law:
=6 r v
F
11.11. A general expression for the drag at higher velocities
If the velocities are sufficiently large (which happens for almost any velocity in a less dense
environment, like in aerodynamics), the linear relationship between the drag force and velocity
changes to a quadratic dependency:
1
2
F = C d Av
2
Where CD is the drag coefficient (for example for cars in the air it is equal to 0.2-0.3), is the
density of the fluid, A is the projection area of the object that moves in the fluid in the direction of
the velocity, v is the velocity of the object.
To understand this force, consider a deflection of air on the surface of the car. It follows a circular
path, so a centripetal force must have been supplied ( F =
mv 2
). This gives an idea of the origin
R
of quadratic dependency. Another observation is that the Bernoulli equation also gives the pressure
difference in terms of the kinetic energy difference of the fluid layers. This is also quadratic in
character.
The drag force is one of the limitation on the speed of the cars which have a certain amount of the
horsepower. Consider a car with a power of p=100hp, density of the air equal 1.27kg m-3, width
equal to 1.7m, height equal to 1.5m, aerodynamic coefficient CD=0.3.
p=Fv
1
p= C d A v 3
2
2p
2⋅74.57kW
v =3
=
=53.5 m/s=192.7km /h
C d w h 1.27kg/ m3⋅1.7m⋅1.5m
Questions
1. Define the pressure and Pascal principle. Explain the operation of hydraulic lift.
2. Derive the Archimedes buoyant force acting on volume V.
3. Discuss the idea of laminar and turbulent flow. Derive an equation for the Reynolds number.
4. Derive the continuity equation in 1D or in 3D*.
5. Derive the Bernoulli equation from work-energy theorem.
6. Define the newtonian viscosity and comment on the contributions from all coefficients.
Write down the Stokes law and the formula for the drag at higher velocities.
7. Derive* or at least write down and comment on the Poiseuille equation.
8. Derive the Euler equation from the second law of dynamics. What term one should add to
the Euler equation to account for viscosity and obtain the Navier-Stokes equation?
9. *Comment on the origin of the viscous term in the Navier-Stokes equation.
Reference: Landau, Lifszyc, Hydrodynamics
12. Harmonic oscillations
12.1. Hooke's law
Hooke's law is used to describe the behaviour of elastic bodies, for example of a spring. It says that
the elasticity force is proportional to the deformation of the elastic body, and the direction of the
elastic force is opposite to the deformation:
F =−kx
where k is the elasticity coefficient.
12.2. Oscillations of a spring
Consider a spring put horizontally on a table, with one end attached to the wall, and with mass m
attached to its other end.
The equation of motion for the mass m is the following:
ma=−kx
m ẍ=−kx
ẍ=−20 x
k
0=
m
The last two equations define the, such called, harmonic oscillator differential equation. Why?
Just take a look on the time derivative of periodic sin and cos functions (A and B are constants):
A sin 0 t ' ' = 0 Acos 0 t ' =−20 A sin 0 t
B cos 0 t ' '=− 0 B sin 0 t' =− 20 B cos 0 t
These functions evidently satisfy the harmonic oscillator equation! They are the solution of this
equation. Thus, the motion of a body that hangs on a spring can be described by:
x= Asin 0 tB cos 0 t
12.3. Damped oscillations
Now let us consider the same spring as above but doing oscillations in a viscous medium. We know
that there will appear a viscous drag force
F =−6 r v . The minus sign implies that the
direction of the drag force is the opposite to the direction of the velocity. In general, to consider the
non-spherical objects, we can assume
F =−bv . The second law of dynamics reads then:
ma=−kx −bv
m ẍ=−k x−b ẋ
This is typically rearranged into:
ẍ2 ẋ 20 x=0
k
m
b
2 =
m
0=
The equations of type
form
ẍa ẋbx=0 are typically solved by searching for the solution of the
x= A e rx . We will do so. Substituting such a function into differential equation yields:
r 2 e rx2 r e rx 02 e rx =0
r 22 r 20 =0
The last line gives a quadratic equation for r:
=4 2−4 02
r 1,2=−± 2− 20=−±i 20−2 =−±i damped
In such case the formula for x reads:
x=e−t Ae i
damped
t
Be−i
damped
t
Because of the Euler formula:
e i =cos −i sin
we can express the inner of the bracket as a combination of trigonometric functions:
x=e−t Asin damped tB cos damped t
which limits the solution to the domain of real (non-complex, non-imaginary) functions and is the
typical expression to describe the damped harmonic motion.
12.4. Forced oscillations
If we supply the dynamics equation of the damped harmonic motion with external forcing driven at
frequency , (e.g. F =F 0 sin t we can rewrite the law of dynamics as:
m ẍ=−kx−b ẋ F 0 cos t
or:
ẍ=− 20 x−2 ẋ
F0
cos t
m
The resulting oscillations run in the rythm of so we only need to find the amplitude of the
function
x= A cos t :
−A2 cos t=− 20 A cos t2 Asin t
A 20−2 cos t−2 A sin t =
F0
cos t
m
F0
cos t
m
Now, for t=0, it turns out that
A 20−2 cos −2 A sin =
F0
m
The above equation can be related in geometrical terms to a following straight triangle:
where from te Pythagorean theorem we have:
2
2
0
2 2
2
2
2
A [ − ] 4A =
F 20
m2
F0
m
A=
2
[0 −2 ]4 2 2
12.5. The mathematical pendulum
Consider a following pendulum. It is composed of mass m, which hangs on a weightless string of
length L. The component of force that drives the rotation is mg sin . This results in a torque that
acts on the arm length L. The equation for motion reads thus:
I =−mg sin L
d2
m L2 2 =−mg sin L
dt
d2
g
=− sin
2
L
dt
2
d
g
≈−
2
L
dt
The last of these equations is the harmonic oscillator equation with frequency of oscillations equal
to 0=
g
.
L
12.6. An oscillating disc suspended on three strings
Another example of a pendulum, useful for the determination of the moment of inertia of a body is
a disc plate suspended on three strings. The second law of dynamics for the rotational motion reads:
I =R mg sin
I ≈ R mg
d 2 Rmg
=
I
dt 2
It is possible to relate the angle to because:
L = R
R
=
L
so:
d2
m g
=R 2
2
I L
dt
so:
0=R
m g
I L
12.6. The wave equation
To derive the wave equation we will consider a simple case of tranverse waves (i.e. oscillating at
right angle to the propagation direction, as opposed to longitudinal waves which oscillate in the
direction of propagation) generated on a rubber garden hose. Take look on the following figure:
In the vertical directions we can write the second law of dynamics as:
m
A dx
∂2 y
= F y x dx −F y x
∂ t2
∂2 y
= Asin [ xdx]− Asin [ x ]
∂ t2
( is the tension within the hose). If the angles are small, the sine function can be approximated
by a tangent, while tangent is simply the derivative tg =
∂y
:
∂x
∂2 y
∂ y xdx
∂ yx
dx 2 =
−
∂x
∂x
∂t
∂ y xdx ∂ y x
−
∂2 y
∂x
∂x
=
2
dx
∂t
2
∂ y ∂2 y
=
∂ t 2 ∂ x2
which is the mechanical wave equation, with speed of propagation v =
∂x
=
, so the
∂t
general form of a wave equation reads:
2
∂2 y
2∂ y
=v
2
2
∂t
∂x
Questions
1. Derive the harmonic oscillations of a spring from the Hooke's law. (formulate and show the
solution of the differential equation).
2. Solve the damped oscillator equation.
3. Derive* or write down and comment the formula for the amplitude of forced oscillations.
Do you know about applications of this phenomenon to the radio?
4. Derive the harmonic oscillator equation for a mathematical pendulum.
5. Derive the harmonic oscillator equation for a disc suspended on three strings.
6. Derive the wave equation for a vibrating string.
13. Diffusion
13.1. Fick's first law
Consider two compartments of particles, divided by a permeable wall, as in the figure below:
In these two compartments, there is a different amount of particles. Half of particles in each
compartment should move to the right, while the other half moves to the left. Taking a time step
which is sufficiently large for the particles to pass the width of the compartment, we can say, that
the flow through the barrier is proportional to:
N x N x x
N x x
N x
−
−
2
N
2
2
− x S x
S x − x 2 c x x −c x
J=
=
=
=
t S
t S
2t
x
2t
x
Taking a limit of this expression results in the First Fick's Law for one dimensional case:
J =−D
where,
D=lim t 0
dc
dx
x2
.
2t
13.2. Second Fick's law
Consider a single cell of width x which is subject to a flux entry from both sides with surface
area equal to S :
The increase in the number of particles in this compartment depends upon the net flux entering
inside:
N
= S [ J x −J x x ]
t
N
=− S [ J x x− J x ]
t
N
=−[ J x x −J x ]
t S
N
J x x− J x
=−
t S x
x
∂ c −∂ J
=
∂t
∂x
∂c
∂2 c
=D 2
∂t
∂x
[
]
The last equation is called the Second Fick's Law and it describes the propagation of the
concentration distribution subject to diffusional driving force.
Reference: Feynmann's lectures on physics.
13. Scaling of the variance in a diffusion process
statistics, expected value (on an axample series where its easy to see repeating symbols and
introduce probability) variance, scaling ov variance in diffusion
In statistics, we can characterize a random sample by its mean, and variance (or standard deviation).
The mean value above is denoted by x , and can be calculated by means of simple arithmetical
mean:
x =
1
N
N
∑i −1 x i
But the mean value does not completely characterize the series of data. It is also worthty to know
how much are the samples departing from the mean value (the average length of the blue line).
Because the blue lines can have a length that could be either positive or negative (summing plus and
minus in an average gives zero!), we rather calculate an average of squares of such distances, i.e.
the mean square distance to the mean:
2
=
N
1
2
xi −x
∑
i=1
N
Where actually after detailed analysis it turns out that we should write this as:
2 =
N
1
x i− x 2
∑
i=1
N −1
because we don't have an exact value of the mean x but only its approximation (so the
uncertainty of x increases 2 ).
To proceed, we need a probabilistic version of the equations for averages. We can derive the
suitable expressions most easily after developing some intuition on an example. Consider a data set:
{1,2,4,3,1,2,3,5,3,2,3,2,1,4,5,2,1,2,3,4}
We can notice only five symbols
si , that repeat multiple times, e.g. N(1)=4, N(2)=5, N(3)=5,
N(4)=3, N(5)=2. The total number of samples equals 19, so we can calculate an average of any
possible function of the sample value:
f x = 1
N
5
N
∑i=1 f x i
f x =∑ f s i N
i=1
si
N
5
f x ≈ ∑ f s i P si
i=1
The last formula can be generalized to a continuus distribution of sample values
si . In such case,
because a probability to obtain an exact value is extremely small, we make use of probability
density function p(s) such, that P s∈ s0, s 0ds= p s ds . Under such assumptions, we change
summation to integration:
∞
f x =∑ f s i p s i si f x=∫ f x p x dx
i
−∞
Notice, that it is not necessery to distinguish symbols with a separate letter s—the integration goes
over the whole domain of the measured data.
Let us now come back to the diffusion equation:
2
∂p
∂ p
=D
2
∂t
∂x
Let us multiply it by x2 and integrate over the whole space:
2
∞
∂ ∞
2
2 ∂ p
∂
x
p
dx
=D
x
dx
∫
∫
−∞
∂t −∞
∂ x2
Now, on the left hand side, we have a time derivative of variance, while the right hand side needs
to be integrated twice by parts to obtain meaningful results. Durring the integration by parts keep in
mind that the probability to find anything at plus/minus infinity is nonexistent, e.g. the term
∞
uv∣−∞ vanishes:
∂ 2
∂p
= Dx 2
∂t
∂x
∞
∣
∞
−∞
∞
−2Dxp∣−∞ 2D ∫−∞ p dx
∂ 2
=2D
∂t
Integrating the last equation we obtain:
2 =2Dt
Which means that during diffusion, the particles move randomly in such way, that they occuppy
more and more space, and the squared deviation from the initial position grows linearly in time. (the
standard deviation, which is a square root of variance, grows like a square root of time,
= 2Dt ).
13.4 Diffusion as a result of a random walk process
Consider a process of randomly wandering particles which jump at a probability p to the right on
the x-axis and at a probability q to the left. If we disallow trapping of particles at fixed position, the
situation looks like in the following picture:
In such case we can say that all particles that were found in position x after a time step escape to
adjacent positions (the condition that p+q=1), while for the particle probability density at x we can
write a following master equation:
p x ,t t = p p x− x ,t q p x x ,t
This can be expanded into the Taylor series. The left hand side to the accuracy of the first term of
expansion, the right hand side to the second term of expansions, since the first term vanishes for
p=q:
p x ,t
∂p
∂p
∂2 p
∂p
∂2 p
2
t= p p x , t − p
x p
x
q
p
x
,
tq
xq
x2
∂t
∂x
∂x
2 ∂ x2
2 ∂ x2
Because p+q=1, we can simplify the above to:
∂p
∂ p x ∂2 p x 2
=− p−q
∂t
∂ x t ∂ x 2 2 t
or to:
∂p
∂p
∂2 p
=−v d
D
∂t
∂x
∂ x2
x
v d =lim t 0 p−q
,
t
x2
D=lim t 0
2 t
Refernces:
E. Zauderer, Partial differential equations of applied mathematics. Wiley, 2006.
J. Klafter, R. Metzler: A random walk's guide to anomalous diffusion.Phys. Rep. 339, 2000.
13.5. Diffusion as a consequence of a stochastic integral of the Langevin equation
The diffusion results from a microscopic brownian motion of the diffusing particles. A brownian
motion describes an irregular, apparently random trajectory which results from the thermal
collisions with vibrating particles of the environment. The changes in the trajectory can be thought
of as random, and therefore they are described by a Langevin equation, which is the second law of
dynamics in presence of a gaussian random force :
m ẍ=− ẋF x t
in the limit of large damping, the inertial term m ẍ becomes small compared to the other terms of
the equation, and thus we write:
1
ẋ= F x t
This can be rewritten as:
1
dx= F dt t dt
In the stochastic calculus we introduce a process that is an integrated gaussian process,
t
t=∫0 s ds which is called the Wiener process. If t tends to 0, then d =t dt . This
allows to write:
1
dx= F dt d
The Wiener process is delta-correlated, i.e.:
d i t d j t=ij dt
which means that the correlation is nonzero only for the correlation between the process and itself.
Now, we can consider a possible function f(x). We can expand it into a Taylor series to say:
df x = f ' x dx
1
2
f ' ' x dx
2
We can substitute dx by the Langevin equation:
df x =
f ' x
[ F dt d ] 1 2 f ' ' x [ F 2 dt 22F d dt 2 d 2 ]
2
Due to the correlation, d 2=dt , so leaving aside in the above equations the terms that are
proportional to greater powers of infinitesimals (e.g. Because dt 2 ≪dt , d dt≪dt ), we can
write:
df x =
2
f 'x
[ Fdt d ] 2 f ' ' x dt
2
Taking the average of this equation it turns out that the average of Wiener increments is zero, so:
∂〈 f x 〉
f ' xF
2
=〈
〉〈
f ' ' x〉
∂t
2
Because the average of f is calculated with respect to the probability distribution p(x) as
∞
〈 f 〉=∫−∞ f x p x dx , we can state (note, on the left hand side f(x) does not depend upon t):
x ,t
F
dx=∫−∞
∫−∞ f x ∂ p∂t
∞
∞
p x ,t
2
2
∞ p x ,t ∂ f
∂f
dx∫−∞
dx
∂x
2 2
∂x2
Applying integration by parts, noticing that the probability density vanishes at plus minus infinity,
we have:
∞
∫−∞
f x
∞
∞
∂ p x ,t
f x ∂ F p x , t
f x 2 ∂2 p x , t
dx=−∫−∞
dx∫−∞
dx
∂t
∂x
2 2
∂ x2
Since this equation is valid for any possible f(x) (we have stated nothing about its identity), the
following condition must be satisfied:
∂ p x , t
1 ∂ F x p x , t 2 ∂ 2 p x ,t
dx=−
2
∂t
∂x
2
∂ x2
Which is the drift-diffusion equation (the Smoluchowski equation). An extremly useful equation in
the field of statistical physics. The diffusion coefficient relates to the damping constant and to
the noise strength by
2
D= 2 .
2
Questions
1. Derive the first Fick's law
2. Derive the second Fick's law
3. Derive the formula for the variance scaling with time in a diffusion process.
4. *Derive the diffusion equation from the random walk.
5. *Derive the diffusion equation from the Langevin equation.
Reference: Lectures in Theoretical Biophysics . K. Schulten and I. Kosztin, Department of Physics
and Beckman Institute University of Illinois
P. Borys, PhD thesis.
14. Optics
14.1. Huygens principle
The front of a light beam in terms of the Huygens principle can be viewed as a collection of
spherical waves:
The elementary waves form a flat wave front that propagates in the direction of a beam. This
principle can be used to explain the law of reflection and of refraction.
14.2. The law of reflection
The law of reflection can be derived by considering the following picture of huygens waves that fall
on a plane:
In this figure5 I have presented the adjacent beams separated by a wavelength λ.If we consider the
advances of the wavefront at intervals when it moves by the value of λ, then after reflection, a
spherical wave is formed that spreads in multiple directions as a circle which radius grows by λ
each time step, but the points that belonged to a single wavefront before reflection converge to a
new wavefront after reflection (a line that is tangent to their „Huygens circles”). This newly created
wavefront is at the same angle with respect to the surface's normal, as the original wavefront.
14.3. The law of refraction
To derive the law of refraction we shall again consider a wavefront that falls on a surface, but this
time the surface is permeable to the light. However, the light velocity inside of the material is
different (for example slower, v=nc, where n is the refraction coefficient) that outside which results
in a difference in the wavelengths of the waves:
5 Note! To make such drawing on the blackboard (in the notebook)—draw the wavefront lines first (incident and
reflected), since otherwise it is impossible to draw circles with correct, repeating radii!
It is immediatelly seen6 that to maintain a consistent wavefront (e.g. to assure that the wavefront
elements before refraction rejoin after refraction), the light beam must bend. Consider the triangle
with one edge denoted by δ. We can write:
1
=sin 1
2
=sin 2
1 sin 1
=
2 sin 2
v 1 t sin 1
=
v 2 t sin 2
v 1 sin 1
=
v 2 sin 2
c /n1 sin 1
=
c /n2 sin 2
n 2 sin 1
=
n1 sin 2
The last equality defines the refraction law, first established experimentally by Willebrod Snell.
14.4. The laws of relection and refraction by the Fermat principle
6 Note! To make such drawing on the blackboard (in the notebook)—draw the wavefront lines first (incident and
reflected), since otherwise it is impossible to draw circles with correct, repeating radii!
The laws of refraction and reflection can also be derived by the Fermat principle which states that
the light follows a trajectory between points A and B which consumes an extremal (e.g. Minimum)
amount of time. Consider the following images of reflection and of refraction:
To find a path that minimizes the time in case of reflection, we must find a point x, that minimizes:
t=v l 1l 2=
c
x 2 a 2 d −x 2 b2
n
Differentiating this with respect to x and comparing to zero, we obtain a condition:
0=
2 d − x
2x
−
2
2
2 x a 2 d − x2b2
x
d−x
=
2
2
x a d − x2b2
sin 1=sin 2
Doing the same for the refractive case, keeping in mind that the velocity differs, we minimize a
condition:
t=
l1 l2 1
= n x 2 a 2n2 d −x 2b 2
v1 v2 c 1
which gives similar results but the sines remain multiplied by the refraction index (which does not
cancel on both sides):
n1 x
x 2a
=
2
n2 d −x
d −x 2b 2
n 1 sin 1=n 2 sin 2
14.5. Diffraction on a diffraction grating
When a beam of light enters a pinhole surrounded by an impermeable material, then the light
exhibits behaviour that is known as diffraction, or bending of light on a scattering object:
Applying the Huygens principle to the pinholes we can observe some patterns of wave
amplification. There are directions where maxima of the wave from one slit meet the maxima of the
wave fron the other slit. On the other hand it is also possible that maxima of the wave produced on
one slit meet the minima of the wave produced on the other slit, and the signal disappears. This is
the diffraction effect, caused by interference of waves.
We will derive it for a double slit system, but these results are applicabile to any periodic diffraction
grating, since if you'd have more slits spaced by the same distance from eachother, then the phase
difference between subsequent rays emiting from these slits (e.g. between slit 3 and 2, or slit 4 and
3) would be the same as between slit 1 and 2 in the considered case.
To obtain a destructive interference (where minimum meets maximum), we must have:
d sin = 2n1
2
To have a constructive interference (a maximum meets a maximum), we must have:
d sin =n
Which gives the condition for the subsequent diffraction fringles. It can be easily seen that the
angles become small when the ratio
is small. So, when λ <<d, then we essentially do not
d
observe interference as a result of diffraction. In such circumstances we can handle the light beam
as a geometrical object—a straight line that propagates in space—and this is the regime of the such
called geometrical optics.
14.6. Derivation of the diffraction pattern on a single slit
It is possible to consider a superposition of all elementary waves that form within the slit, not only
those from the edges. To do this, we must consider a formula for the phase shift for any point
located at position y from the upper edge. If y=d, we know that the wave delayed by
L=d sin
Which can be converted to an angular shift by:
d =L
2 2d
=
sin
in such case, if the elementary waves can be described by:
E=
E0
cos [ kr −t y ]
d
(we take the amplitude per unit length, which is suitable for integration). Integrating this over the
whole slit, we obtain
E0
2 sin
cos kr− t
y dy
d
d /2
E0
2 sin
E=
sin kr −t
y
2 d sin
−d / 2
d /2
E =∫−d / 2
[
[
]
]
To evaluate the definite integration, we recall a formula for difference of sines:
sin −sin =2 sin
[ ] [ ]
−
cos
2
2
which resuluts in Fraunhofer diffraction pattern:
E=
E0
d sin
sin
cos [ kr −t ]
d sin
[
]
Which (ignoring the constant prefactor) takes a following dependency on
=0.1 :
d
This pattern is crossing zero when the sine contains a multiplicity of , so
the maxima, we substitute =d sin :
E=
E 0 cos [ kr− t ]
sin
[ ]
And calculate the derivative with respect to :
dE E 0 cos [ kr −t ]
=
d
cos
[ ] −sin [ ]
2
The numerator equals zero when:
[ ]
=tan
Since tangent is a periodic function, and grows very fast to infinity
d sin
=n To find
The solutions
x=tan x , except for x=0, occur approximately at the end of period of the function,
i.e. for
x=2n1
, n=1,2 , ...
2
so:
=2n1 n
2
=2n1
2
d sin =2n1
2
and d sin =0 .
14.7. Diffraction on an obstacle
On the laboratory practice we try to calculate the hair width using diffraction methods. The situation
is quite similar to the diffraction on a single slit, it is an inverse of that. In general, the wave front
carries an intensity
E 0 that spreads over the beam width w. Then, after the diffraction takes place,
the Huygens waves that would normally pass the center, where an obstacle (e.g. the hair) is located
is blocked. Therefore, the diffraction pattern, calculated in 14.6. (for a radiation amplitude per unit
length
E 0 /w ) should be subtracted from the intensity occuring from the whole beam of width w
(the pattern in 14.6. with d=w):
E=
{ [
E 0 cos [ kr− t ]
w sin
d sin
sin
−sin
w sin
] [
]}
The resulting pattern looks like the following:
You should notice that the maxima and minima coincide with maxima and minima for the
Fraunhofer pattern for diffraction on a single slit, but they are extremely small in amplitude, much
darker than the main fringle for =0 . Therefore a measurement of such diffraction should be
carried out in rather dark conditions.
14.6. Thin lens equation.
Consider a spherical surface that accepts a light beam7:
7 Be very careful with this drawing on the lecture/on the exam! If it is inaccurate, you may identify the angles
incorrectly.
Here we can write Snell's law for the angled beam as:
n 1 sin 1=n 2 sin 2
To express the angles we can draw a supporting picture:
since 1= and 2=− , for small angles, where sin , we have:
n1 =n2 −
n 1 n2 = n2−n1
expressing the angles as ratios of arc length s to the corresponding radii, we have:
s
s
s
n 1 n2 =n 2−n 1
p
o
r
n 1 n 2 n2 −n1
=
p o
r
Now take a look at a thick lens:
We can write two equations:
1 n n−1
=
p o
r1
n
1 1−n
=
p' o'
r2
We can rewrite this:
1 n n−1
=
p o
r1
n
1 1−n
=
L−o o '
r2
going with L to zero and adding these eqations we obtain a thin lens equation:
1 1
1 1
=n−1 −
p o'
r1 r2
Which can be also written in more compact form as
1 1 1
=
p o f
where p is the position of the source object with respect to the lens, o is the position of the image
formed by the lens, and f is the focal length of a lens. This equation allows to determine some rules
for constructions in geometrical optics. Namely
●
The ray that enters the lens in parallel to the observation axis, (i.e. it crosses the axis at
infinity, so
●
p=∞ ) exits in such way that o=f, which follows from the lens equation.
The ray that exits the lens in parallel to the observation axis enters the lens at p=f (by the
same reasoning)
●
The ray that is crossing the center of a convex lens exits the lens unchanged due to
symmetrical refraction on entrance and on exit. This is summarized in the figure below:
14.7. Optical instruments
A simple magnifying glass is constructed of only a single lens and forms a diverging set of rays
which resemble a magnified virtual object at position which resembles the distance of a good sight
(larger than 25cm).
The magnification of the magnifying glass can be referred from the lens equation for case when the
eye is put close to the lens so, that the whole „good sight length” is on the virtual side of the lens.
Then:
∣
1
1 1
= ⋅25cm
−25cm p f
25cm 25cm
−1
=
p
f
25cm
−1m=
f
25cm
m=1
f
A microscope is formed from two lenses (an objective and an ocular (or eyepiece)), which work in
the following way:
A real object sents light through the objective lens and forms a real image. Then, an occular lens
takes this image on the principles of a magnifying glass to form a virtual image that is observed by
the eye. The magnification of the microscope follows from the magnification of the objective lens,
which results from the triangles in bold, i.e. mob=
magnifying glass (the occular), which is moc =1
m=mob moc=
s
and from the magnification of the
f1
25cm
, so in gneral we have
f2
s
25cm
1
f1
f2
A telescope is constructed in similar way to a microscope but it is designed to magnify objects that
are located infinitely far (or simply: very far) from the observer. The magnification relates to a
change in angles of the incoming rays:
(for sake of explanation of the „magical” ray bending in the focus between lenses i have introduced
the path of real rays in red and magenta, as seen by the geometrical optics standard constructions—
i.e. such bending is formed of two different rays! Red line changes to magenta). Taking a look at the
blue triangles, we can conclude that
h
f1
h
2≈tan 2=
f2
f
m= 2 ≈ 1
1 f 2
1≈tan 1=
It should be also noticed that the image becomes inverted: the rays that originally were directed
upwards, are now directed downwards.
How large can be the magnification of a telescope? It could be constructed as large as possible, but
in astronomy a limit is set by the diameter of the eyepupil (7mm). Astronomical objects are often
dark (e.g. the nebulae, distant galaxies) so it is favourable to increase the amount of light entering
the eye, and certainly not to decrease it. This can be accompanied when the radius of the cone
entering the eyepupil is not larger than 7mm:
h2 h1
=
f2 f1
f 1 h 2 7mm
m= = =
f 2 h1
D
where D=2h1 is the diameter of the telescope's objective.
14.8. Polarization
The electromagnetic wave is a transverse wave with perpendicular directions for the oscillations of
the magnetic field vector and electric field vector (we will learn about it when digesting the
Maxwell equations). The electromagnetic wave may look like the following:
In this example the E field oscillates in a single plane. We say that such wave has a flat polarization.
Typically a light beam is composed of many different modes of oscillations, so it is unpolarized. It
may become flat polarized after passing it through a device that is called a polarizer. Such device
allows to pass only waves with fixed polarization and damps out other components.
Except for the flat polariation it is possible to have circular polarization. Imagine two electrical
fields that are shifted by 90o with respect to each other:
It can be easily seen that the resulting vector (vectorial sum of these waves) circulates around the
axis. This is circular polarization.
14.10. Birefringence
The polarization matters a lot for materials which are birefringent.In such materials there is
typically an uniaxial symmetry, i.e. there is one axis which is symmetrical to rotations to certain
degree. The directions perpendicular to the optical axis are isotropic. If there is an increased order in
particular direction, it is possible to have a different (e.g. smaller) electron density, and
consequently the light wave does not interact with matter to the same degree (acceleration of
charges) and a different (e.g. larger) value of the speed of light is possible (maximum value for the
vacuum) in the direction perpendicular to the symmetry axis (perpendicular, because light is a
transverse wave with oscillations perpendicular to the direction of motion).
The ordinary ray oscillates in the direction perpendicular to the optiacl axis and has a velocity that
does not depend upon direction. The extraordinary ray vibrates in the direction that contains
components paralell to the optical axis and these components move at larger velocity than the rest:
As a result, the extraordinary ray is bent inside of the crystal. The birefringent materials are of great
user for the construction of advanced microscopes, the differential image contrast microscopes,
which use birefringent crystals to split the beam into two narrowly spaced beams that pass through
the sample and are then recombined to a single beam to cause a destructive interference if the
beams are identical, and constructive interference when there is a difference in beams (e.g. when an
edge of two material phases occurs in the sample).
Questions
1. Derive the refraction and reflection law using Huygens principle.
2. Derive the refraction and reflection law using Fermat principle.
3. Derive the conditions for maxima and minima in case of a driffraction on a diffraction
grating.
4. Derive the equation for the intensity distribution in Fraunhofer (single slit) diffraction.
5. Derive the thin lens equation.
6. Explain the operation of the magnifying glass, of the telescope and of the microscope.
7. What is the circular polarization?
8. Explain the origin of the extraordinary ray in birefringent materials.