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1.a. Classical Mechanics
1.b. The Michelson-Morley experiment
1.c. Einstein's resolution of the contradiction
1.d. Consequences of special relativity
1.e. Doppler shift for light
1.f. Lorentz transformation
1.g. Velocity transformation
1.h. Relativistic momentum
1.i. Relativistic energy
The central equation of classical dynamics is the third law of Newton:
F=dp/dt
where
p=mv,
is the momentum.
This form of Newton's equation is advantageous compared to the one
expressed by velocity, because one can generalize it to include the
physics of velocities close to that of light, relativity.
Newton's equation has a very important property, Galilei invariance
x -> = x' = x - v t ,
ux ->
u'x =
ux - v
t -> t' = t.
Newton's equation is unchanged if we perform these substitutions, i.e. we observe the same system from a moving coordinate system.
How does the distance between two points x1 and x2 transform?
x1 - x2 -> x'1 - x'2 = x1 - x2,
because time is absolute, t' = t. In other words the distance between two points is an absolute quantity, as well.
If one measures the speed of life, according to Newton, in a coordinate system moving parallel to the light ray:
c -> c' = c + v
Then we face an apparent contradiction with Maxwell's theory. The Maxwell equations imply that speed of electromagnetic radiation (light) in empty space is always c. At the end of the nineteenth century, the only way pysicists were able to resolve this contradiction was to assume that space is not empty, but filled with a hypothetical fluid, the ether. It is easy to laugh at this idea today, but at that time both Newton's mechanics and Maxwell's electromagnetic theory were assumed to be absolute laws of nature. We know now that neither of them are. Einstein discovered the limitations of classical mechanics in 1905, further limitations were uncovered by the founders of quantum mechanics about which we learn later. Dirac discovered the limitations of classical electrodynamics in 1929.
How to prove the existence of ether? Perform measurements of the speed of light in moving coordinate systems. Problem how to do this. Speed of light is c=3¥ 108 m/s. The fastest one could move in Michelson's time (who performed the pioneering experiment of measuring the speed of light in moving coordinate systems) was around 100 km/h µ 30m/s. (train). This is seven orders of magnitude smaller than the speed of light. I.e. one needed to detect a deviation of about 1/107. In fact, the situation is even worse. Most relativistic effects are proportional to v2/c2, thus they would be of O(10-14). This was way beyond the experimental capabilities of anyone at that time. So why don't we use the earth as a moving lab? But even the earth's speed around the sun, 3¥ 104 m/s is paltry compared to the speed of light. To detect a deviation of 1/104 Michelson needed to use the power of interferometry.
Michelson interferometer has two perpendicular arms. Light is
split
into two beams by a semitransparent mirror. The two light rays, one
travelling
along the direction of motion of the earth, one perpendicularly are
coherent.
So after they are reflected back to the mirror they interfere
constructively
or destructively, depending on their relative phase. The interference
pattern
of the combined rays is detected though an optical device.
Let us calculate the time it takes for these light rays to come
back
to the semitransparent mirror where they were originally separated,
assuming all the time that an ether exists and light
travels with a velocity of c only in ether. The time a light
ray takes traveling parallel to the direction of motion of the earth
and
going back and forth along an arm of length L1 is
t1 = L1 / (v+c) + L1 / (c-v) = 2 L1 / (c2-v2)
This is so beause the speed of light is c in the system tied to
the
ether, so it changes to c+v and c-v when the light ray is observed in
a coordinate system
moving with a velocity, v (which is the velocity of earth on its
orbit).
Now in the perpendicular arm, of length L2, of the
interferometer the light ray again moves with velocity c in the
ether's
system, but if the movement is perpendicular in the earth's system
then
it's velocity vector is at an angle, slightly eviating from the
perpendicular, in the ether's system. So the speed in the coordinate
system moving with earth (travelling in both directions) is
c'= (c2-v2)1/2.
As a result, the time it takes to travel back and forth between the two mirrors is
t2 = 2 L2 / (c2 - v2)1/2
The time difference between the return of the two coherent light rays is
t1 - t2 = 2 L1 / (c2-v2)- 2 L2 / (c2-v2)1/2
Unlike it is indicated in the book it is impossible to make two
arms
to have equal length to a precision of up to a fraction of the
wavelength
of the light, so we must assume that L1 and
L2 are not equal.
Fortunately, the result of the experiment is not affected by this.
Michelson and Morley slowly rotated the arms of the interferometer
by
90o. Then the role of the two arms got exchanged. The time
difference
between the return of the two light rays became
t'1 - t'2 = 2 L2 / (c2-v2) + 2 L1 / (c2-v2)1/2
Michelson and Morley watched the change of fringes in their
observing
microscope during the rotation. By this they checked whether
t'1- t'2 was different
from t1 - t2 or not. That difference is
t1 - t2 - t'1 + t'2= 2L1/(c2-v2) - 2L2/(c2-v2)1/2+ 2L2/(c2-v2) -
2L1/(c2 -v2)1/2 = 2(L1 + L2) [1/(c2-v2) - 1/(c2-v2)1/2 ]
Using v << c we obtain an approximate form of the square bracket if we expand in v/c.
1/(c2-v2) = 1/c2 + v2/c4+ ...
1/( c2-v2)1/2= 1/c2 + v2/2c4 + ...
In other words
1/( c2-v2) - 1/(c2-v2)1/2= v2/2c4 + ...
Now if we substitute this into the equation for the time difference we obtain
t1 - t2 - t'1 + t'2= 2(L1 + L2)v2/2c4 + ...
The neglected terms are of O(v4/c6 ) and as such are negligible compared to the leading term. The time difference leads to a path difference,
l-l'= c (t1 - t2 - t'1 + t'2)= (L1 + L2)v2/c3 + ...
where we asskume in leading order that the speed of light in all
systems is c. This approximation brings in a relative error of O(
v2/c2).
The path difference divided by the wave length provides the number of
wavelength
of deviation between the two positions of the interferometer. Using
actual
numbers, as described in the book, Michelson and Morley expected 0.4
wavelength
difference and there was none.
Lorentz proposed in the late nineteenth century, to save the ether concept, that objects, due to a hitherto unknown molecular force, physically contracted in the direction of the motion by a factor of
1/g =(1-v2/c2)1/2
Notice that such an extra factor would make the time difference
t1- t2 - t'1 +
t'2
cancel exactly and explain the result of the experiment.
Einstein came up with a different, revolutionary idea. He
discarded
the ether and set up two postulates which, are in complete agreement
with
Maxwell's equations, but contradict to Newton's law and Galilean
relativity.
The deviation from Newton's law becomes perceptible only at speeds
that
are not negligible compared to the speed of light. In other words,
Einstein's
theory of relativity reduces to the Galilei-Newton theory at low
speeds.
This is is the first appearance of an all important rule, called the
correspondence
principle, which is a constraint on theories trying to explain
new
phenomena. The correspondence principle sounds almost trivial, but it
is
a strong constraint on new theories. It states that a new theory
must
yield the laws of the old theory in the classical limit. The
classical
limit is the limit when the laws of the old theory are valid. In our
particular
case it is the domain of low velocities. Using a more precise
formulation
we can say that the new dynamics, built on Einstein's principles must
yield
Newton's dynamics provided powers of v/c are neglected. The
correspondence
principle is applicable to general relativity, quantum mechanics,
quantum
field theory, string theory, etc. as well.
The two postulates of Einstein are:
1. All laws of physics have the same form in every inertial reference frame.
2. The speed of light is c in every inertial reference frame.
The first postulate is not different than the postulate of Galileo. The second postulate contradicts tto Newtonians dynamics as we will see below.
Synchronization of clocks: This is needed to be able to measure
time
consistently. At every point in a given inertial frame we can set up
clocks
that run synchronized. Send a light signal from point A to point B
and
start the clock at the same time at A. Then start the clock at point
B
at the time when the light signal arrives, but start it at time
l/c,
where l is the distance between the two points. It is easy to
see
that this definition is self consistant. I.e go around a loop in
space from point to point and get a new definition of time at point
A. The new definition will coincide with the original one.
Simultaneity: An event is defined by a location and a time. Two events that are simulaneous in an inertial reference frame may be not simultaneous in another inertial frame. Note that in the Newtonian world the two events would be simultaneous in every inertial system. Using Einstein's postulates we can see however, that they are no longer simultaneous in moving systems.
Say two events are two lightning bolts at the ends of a moving freight car, which are observed by a person (observer A) sitting on the ground (rest system of earth) smack in the middle of the car at the moment when the lightning bolts hit. Assume that the two lighting bolts are simultaneous according to observer A. This person would assertain the simultaneity of the two lightning bolts by measuring the times when the light signals reach him/her. If the two light rays starting from the two bolts reach the observer at the same time. then he knows that the two ends of the car were at the same distance from him/her (he measures that exactly half the car passed by him). Thus, he/she establishes that the two bolts were simultaneous. The same event observed by a person (observer B) sitting on the car, in the middle, look different. By the time the signal arrives from the front end of the car to observer A, observer B traveled past this point, so he encounters the lightning bolt sooner. On the other hand the light ray from the bolt at the end of the car has just arrived to observer A, so it could not reach observer B who has moved ahead already. Consequently, the two events are not simultaneous for observer B. Clearly, we assumed here that the speed of light is the same in both systems. Without this assumption, using the Galilei transformation, observer B would have concluded that the two ebents were simultaneous.
Time dilation: define proper time as the time
between
two events in a coordinate system where the two events happen at
the same
point in space. Say the proper period of a pendulum is the period
in the
coordinate system (the proper frame), where the pendulum is at rest.
We
will find the relation, using Einstein's second postulate, between
proper
time and the time measured between the same two events in a reference
frame
moving with velocity v compared to the proper frame. Again,
using Galilean invarance the time difference between the two events
would be the same in every coordinate system.
Again, imagine a train moving with velocity v. A light
signal is sent from the
floor of the train straight up towards a mirror, which is at a
distance
L. Since the velocity of light is c in every coordinate
system anobserver
on the train (obsever A) finds that the light signal returns after
time
t=2L/c. In other words we have L= c t /2. The time,
t, is the proper time difference
between two events: 1) the start of the light signal and 2) its
later
detection at the same point.
Ler us examine now the same two events from the point of view of
an obsrver (obsever B) at rest in the earth based coordiante system,
i.e., for a person wo is standing besides the tracks on which the
train runs. As we saw earlier that in general, time difference
between two events depends on the inertial frame in which the
observation is made. So we must assume that the time difference
between the two events will be different, t', for observer B.
What will observer B see? Cerainly, observer
B will see the lightray going up, bouncing off from the mirror and
bouncing back to the train. Of course, during that time the train has
moveda distance of x=t'v. The path of the lightray is composed
of two sections
of equal lenght, t'c/2 each. This is so because it takes the
light ray
of speed c to go up to the mirror t'/2 seconds, so the
distance is L'=t'c/2.
Then the three distances, t'c/2, t'v/2, and L=tc/2d
form a right triangle.
The Pythagorean theorem is valid:
(t'c/2)2 = (t'v/2)2 + (tc/2)2
This is simply solved for t'. It gives
t' = tg = t/(1 - v2/c2)1/2.
The multiplier of t,
g = 1/(1 - v2/c2)1/2.
is called sometimes the Lorentz factor. It will keep popping up in
discussions
of problems in relativity. The usual notation is the Greek letter
g (gamma).
The conclusion from our thought experiement is that the time interval between the two events (start and arrival of the light ray) is different in the proper system (train), where the two events happen at the same point then in th esystem bound to earth, in which the two events do not happen at the same point (not a proper frame). The time interval measured in the proper frame is always the shortest. In moving frames it is dilated by the Lorentz factor, g, which depends on the relative velocity of the two systems.
An obvious consequence of the relation of proper time and of time
as
seen from a moving coordinate system is that according go special
realtivity
one cannot reach speeds larger than c. The Lorentz factor would
become
complex, physically meaningless at v>c, strictly speaking
even at v=c.
If special relativity is valid then these velocities are
inadmissible.
Since the Lorentz factor, g, is always
larger than 1, t'>t.
As we seen above, time intervals , as seen form a moving
coordinate system are longer.
This effect is called time dilation. A moving pendulum seems
to slow down
because its period, T is longer in a moving coordinate system.
Of course
at any speeed that can be reached by a macroscopic object,
inclulding
a rocket, is so much smaller than the speed of light that the
deviation
of g from 1 is hardly obsrvable. Take for
example a space shuttle that circles
the earth. It takes 90 minutes for a satellite in a low orbit to
circle the
globe. The girth of the earth is 40,000 km. So the speed is v
= 40,000x1000/60¥90,
about 7500m/s. the speed of light is c=3¥108m/s. Then the Lorentz factor
is
g = 1/(1 - v2/c2)1/2= 1 + v2/2c2 + ... = 1 +(7500/3 ¥10-8)2/2 = 1 + 3.1¥10 -10,
only very slightly larger than 1. Thus the change of the period of
the
pendulum on the shuttle is practically not measurable.
Microscopic objects, such as an elementary particle, can easily
reach
speeds that are close (but never equal) to c. Take e.g.,
muons, elementary
particles that are not quite stable, but have an average lifetime of
2.2
ms. This does not mean that every single
muon lives exactly 2.2ms, but
it means that if one has N muons at rest. then in 2.2ms the mumber of muons
decreases to N/e. In another 2.2ms.
the number of muons further decreases
to N/e2, etc. The lifetime is of course defined in
the rest frame of the
muons. Having such a lifetime, they would travel c ¥2.2 ms=650m. as an
average,
before they decayed to electons and other particles. Cosmic radiation
is constituted from various extremely high energy particles, mostly
protons that constantly bombard the upper athmosphere of earth. These
particles collide with atomic nuclei of the upper atmosphere and a
large number of particles, among other muons, are produced in these
collisions. These muons are produced at an altitude of 10,000m, or
more. Still, a large percentage of these muons arrives to the surface
of earth before decaying. How can that happen? The muons travel
10,000/650µ15 lifetimes before they
arrive at the detectors. Their numbers should decrease by a factor of
e15 µ3,000,000 by the time
they arrive. But, a large fraction of them still arrive. The
explanation is simple: Relativity tells us that the lifetime of the
muon in an earth-bound coordinate system is larger by a factor of
g. If the speed of the muons is 0.999c, or
more, as frequently happens, then the Lorentz factor is
1/(1-0.9992)1/2. Write 0.999 as 1-0.001 and
remember that (1-x)2=1-2x+... to get that g = 1/0.0022µ22.4. In other words, the lifetime of
muons as seen by the observer on earth is T'=22.4¥2.2msµ50ms.
during which time the muon traveling at close to the velocity of
light, will travel c¥50 ms=14,500m. No wonder that a large percentage of
the muons reaches the surface of earth. Of course, using just muons,
produced by cosmic radiation, it would be difficult to establish the
exact form of the time dilation formula, because we know neither the
exact altitude of their production, nor their exact velocity.
Experiments performed on muons produced in accelerators, where the
location and velocity of muons is very well known confirm the formula
for time dilation precisely.
Length Contraction: After discussing how a time interval
changes when one observes an event from different inertial systems,
let us discuss how the proper distance between points or the
proper length of an object change when obseved from a moving
coordinate frame. The proper distance is defined by measuring the
distance between two points in a coordinate system in which they are
at rest.
The simplest way to understand length contraction is to follow the book and consider the trip of a spaceship to a star, which is at a distance L from earth. We will assume that the star is not moving compared to the earth. Then L is the proper distance between the endpoints of the trip.
Observer A, sitting on earth measures the time needed for the trip as T=L/v, where v is the velocity of the spaceship. Of course, Observer B, traveling on the spaceship, measures a different duration, T'=L'/v, for the trip. Notice that observer B sees the star appoaching at exactly velocity v. We do not assume that the distance measured by observer B is also L. Dividing the two equations we obtain
T/T'=L/L'
Now the two events, departure and arrival, happen at the same
point (at the ship) for observer B. Therefore T' is the proper
time difference between the two events. Consequently,
T=T'g. Then we have
L'=L T'/T=L/g=
L(1-v2/c2)1/2 < L
It is obvious that the distance between the stars seems shorter for
observer B, who is moving with velocity v compared to the
proper system (where the distance is L). This phenomenon is called
length contraction.
If v << c then the contraction is negligible, in agreement with the correspondence principle.
Length contraction explains why muons arrive to the surface of the
earth from the point of view of an observer traveling with the muons.
For such and observer the lifetime of the muon is unchanged, 2.2ms.
On the other hand this observer does not have to travel the distance
10,000m from the upper atmosphere where the muon was produced, but
only the contracted length, L/ g, which is
possible with the speed close to that of light.
Problem: Suppose a star is 5 light years away from earth. A light year is a distance, how far light travels in one year. 1 ly=3¥ 108m s-1 ¥365 days ¥ (3600¥24 s/days) = 9.46 ¥ 1015m
Suppose now, that a spaceship travels at a speed 0.8 c . How far away does the star seem to be for the observer traveling on the spaceship?
According to the length contraction formula, the distance in the moving system is
L'= 5 ly / g =5 ly ¥(1-0.82)1/2 = 0.6 ¥ 5 ly= 3 ly.
There is a substantial difference between Doppler shifts for
light
and sound. Sound propagates in a medium, while light, as we learned
before,
does not. There is a prefered coordinate system for sound, the rest
system
of tne medium conducting sound waves. There is no such preferred
system
for light. All inertial systems are equivalent.
Let us assume that a light source, at rest in inertial frame S,
emits
light of frequency f. The wave length of light is l= c/f.
Now, consider observer B moving toward observer A with velocity
v.
Observer B, whose rest sytem we will label as S', observes the same
light ray
and finds that its frequency is f' and its wavelength is l'=c/f'.
This is so because according to the postulate of Einstein the
speed
of light is c in every inertial system.
Observer B rushes against the light wavefronts emitted in system
S.
The speed of light for her/him is of course c, so during a
complete
period the light wave, T', the wave front progresses a
distance of
l1=cT'. During the same time observer B moves
towards the light source by a distance of
l2=vT'.
So the distance between two successive wave fronts, as seen by
observer
B, is the difference of these two distances l '=l1-l2=(c-v)T'.
. Then the frequency , as discussed earlier, is
f'= c / l' = c / T' (c-v) = c / (c-v) g T = f c / (c-v) g.
where we used f=1/T, in frame S. We also used the formula
for time dilation
T'= Tg, where g, as before is the Lorentz factor, g=1/(1 -
v2/c2)1/2. Substituting this
form of g
into the above expression for f', we obtain the final formula
for the relativistic
Doppler shift:
f' = f [ (1+ v/c) / (1 - v/c) ]1/2
It is reassuring that this formula is symmetric. In other words
the
inverse relation
f= f' [ (1 - v/c) / (1+ v/c) ]1/2
is exactly the same as the previous relation except the sign of
the
speed of the coordinate system has been reversed. This shows that
there
is nothing that distinguishes any of these systems, the formulas for
Doppler
shift are identical in them. In other words it does not make any
difference
whether the light source approaches an observer or the observer
approaches
the light source. The observer would not be able to distinguish them.
This
is consistent with Einstain's postulate that all inertial frames are
equivalent.
This is not so for the propagation of sound waves, for which the
formula
for Doppler shift in the two coordinate systems would be different
because the two coordinate systems are distinguishable: in one the
medium is at rest, while in the other one it is not.
The relativistic Doppler shift plays an important role in
astrophysics.
According to the Doppler shift formula objects receding from the
observer
seem to have lower frequencies than they have in their rest system.
Now
light emitted by stars contains well defined, and well known spectral
lines. We will learn about spectral lines later in the quarter.
Analyzing spectral lines astronomers discovered that the lines
do
not have the same frequencies as similar lines on earth. The ratios
of
various frequencies of light coming from a single star are the same
as
those on earth. The inescapable conclusion of astronomers was that
these
stars are moving with a relativistic velocity compared to earth.
Curiously
most frquencies were shifted downwards, i.e stars look more red as
seen
from earth as they are in their rest system. This is called a red
shift. Stars having a red shifted spectrum must move away from
us. In other words, we observe that most stars recede from us.
In the 1920's and 30's Edwin Hubbble analyzed light from a large
numer
of stars. He had a way of estimating the distance of stars from
earth.
When he correlated the distance of stars with their red shift he
found
to his amazement that stars that are farther away have larger red
shifts,
consequently recede from earth faster. He also found that the
correlation
between velocitiy and distance is linear. The inescapable conclusion
is
that the universe is expanding uniformly. At the largest distances
the
speed of expansion is close to that of light. It is best to imagine
the
universe as a balloon in the process of being blown up. Imagine that
the
volume of the balloon expands proportionally to the surface, so that
pairs
of specks inside recede from each other everywhere, depending on
their distance
from each other. This scenario of expanding universe was the first
indication
for the, now universally accepted, big bang theory of the universe.
We saw earlier how the Galilei transformation connects
coordinates
and time. We would like to obtain similar relativistic formulas that
supercede
Galilei transformation. The mathematical form of the new
transformation
was invented by Lorentz, therefore it is called Lorentz
transformation.
Obviously, Lorentz transfromations are more complicated than Galilei
transformations,
because time transforms nontrivially.
One feature of these transformations
is sure, they should also be linear relations, because otherwise
there
would be preferred coordinates and times (the derivative of x'
would have
zeros at some values of x and t, where these are
coordinates measured in
systems S' and S, respectively). Thus we can start from relations
x' = a x + b t,
and
t'=c x + d t.
Here we assumed that at t=t'=0 the origins of the two
coordinate systems
coincide. In other words, when t=0 then x=0 also implies x'=0. This
just defines when we start the clocks in the two coordinate
systems.
We assume that coordinate system S' moves with velocity v to
the
right. Then a point moving to the right with velocity v in
system S (i.e
it satisfies x=vt) must be at rest in coordinate system S',
i.e., if it
was at the origin of that system at t'= 0, then it stays
there. In other
words, we must have
0 = a v t + b t.
Solving this equation for b, we obtain b=-v a. Then we
have the transformation
formula
x'= a ( x - v t ).
Note that this formula is the same as the galilei transformation
formula,
except for the multiplier a. Since at low speeds, according to the
correspondence
principle Lorentz transformations must reduce to the galilei
transformations,
a must approach 1 when v approaches zero. Since coordinate
systems S and
S' cannot be distinguished, and according to Einstein, the laws of
physics
look the same in both systems, the inverse Lorentz transformation
must
have the form
x = a ( x' + v t'),
because in coordinate system S' the coordinate system S moves to the
left
with velocity v. Here, we tacitly assumed that the constant a
does not
depend on the sign of v. The validity of this can be checked
after we will
complete the construction of Lorentz transformation formulas. In
other
words, we should check the self-consistency of the formulas.
Substituting
x' into the expression for x we obtain
x = a [ a (x - v t ) + v t' ].
Now we can express t' as
t' = a [ t - (a 2-1) x / ( a v ) ].
The time difference between two events that take place at the same
coordinate
in coordinate system S (S is the proper system for these two events,
consequently t is the proper time) is
t'1 - t'2= a ( t1 - t2 ),
since x1=x2. But we know how time
differences transform. The time differenc
is system S' must be g (the Lorentz
factor)
times the proper time difference, t1-t2. It is
dilated by a factor g in a system moving
with velocity v compared to the proper system. Consequently, a=g
and the final formulas for a Lorentz transformation relating the
coordinates of two coordinate systems moving with respect to each
other with a velocity v along : the x axis are
x' = g ( x - v t ),
y' = y,
z' = z,
t' = g( t
- v x / c2).
In the last equation we used that (g2-1)/g
=gv2/c2.
Now we can see that since a= g the
assumption concerning the independence of the
constant a on the sign of v is valid.
Inspecting the Lorentz transformation formulas we can observe
that indeed time transfors highly nontrivially, as predicted.
Notice that
when velocity v is small the transformation formula for time reduces
to
t'=t, in agreement with the correspondence principle.
We can easily invert the Lorentz transformation formulas. They
are just linear eqiations. we need to solve these equations for the
variables
x, y, z, and t. We obtain
x = g ( x' +v t' ),
y = y',
z = z',
t= g ( t'
+v x' / c2
),
which formulas coincide with the direct transformation formulas,
provided
the primed and unprimed coordinates are exchanged and v is
cahnged to -v.
This shows that the formulas satisfy Einstein's principle of
equivalence
of inertial systems. The transformation laws have the same form in
every
coordinate system
The transformation rules for the y and z coordinates follow from the following simple considerations: Suppose, in general, that the transformation rule is
y'=a y + b x + c t
Now, b=c=0, because the system should be symmetric for the exchange of x to -x and t to -t., because there is nothing that could distinguish left and right along the x axis. But then we should also have a=1, because the system S' and S should have symmetric relations. If a would not be equal to 1, then the expression of y by y' would be different and the two coordinate systems would be distinguishable.
To see how velocities transform one needs to see first how infinitesimal distances transform when the time change is also infinitesimal. In other words one needs a differential form of the Lorentz transformation formula
dx' = g ( dx - v dt ),
dy' = dy,
dz' = dz,
dt' = g (
dt - v dx / c2).
Dividing the first three equations by the fourth we obtain
ux' = (ux - v) / (1-u
x v /
c2).
uy' =
uy /
g (1-u
x v /
c2).
uz' =
uz /
g ( 1-u
x v /
c2).
Where both the denominator and the numerator on the right hand side were divided by dt. We also itroduced the notation uI and u' I for the components of the velocities in the unprimed and primed coordinate systems, respectively. Obviously, when the velocities are much smaller than the velocity of light this formula reduces to the Galilei transformation, because the denominators are equal to 1. At the other limit, when ux=c, then we obtain that u'x=c,, as well.
Let us find the inverse transformation for the velocities. Multiplying by the denominator we obtain
ux' (1-uxv / c2) = u x - v.
Expressing ux from the above equation we obtain
ux = (u'x+ v)/(1+u'xv / c2).
This is exactly the same equation as the expression of u'x by ux, except that the velocities are exchanged and the sign of v is changed. This is consistent with the requirements that the two coordinate systems are completely equivalent.
It is also easy to show that once a velocity is smaller than c, as it should be, it will also smaller than c in every coordinate system. See what is the condition for u'x < c:
(ux- v)/(1-uxv / c2) < c
Expressing ux we obtain the following
constraint for it:
ux < 2c/(1 + v/c).
Since the right hans side is larger than c, ux certainly satisfies this constraint, consequently u'x is certainly smaller than c.
In other words, if one adds two velocities, the velocity of the object and the valocity of the coordinate system, where both may be close to the velocity of light, still the resulting velocity will be smaller than the velocity of light. This would not happen if we used the the Galilei transformation formula for velocities, because then velocities would just be added.
Take an example. Suppose ux=0.8 c and v = 0.8 c.. Then we have
u'x = (0.8 c+ 0.8 c)/( 1 + 0.8x0.8)= 1.6 c/1.64 = 0.976c < c.
We wish to generalize the notion of momentum to satisfy the
requirements of relativity. We must define momentum so that it is
consistently conserved. In other words we seek a quantity that is
conserved in every inertial frame once it is conserved in one frame.
It is easy to show that momentum defined along traditional lines,
p=mv, does not satisfy this constraint. Take two objects of
equal mass, m. Let them have velocities equal in magnitude
(v), but opposite in direction. Then the Newtonian momenta,
along the direction of the motion are p1=mv and
p2=-mv. The total momentum,
P=p1+p2=0. This is the center of mass
system (cms). Supose that the collision is completely inelastic, e.g.
the two objects stick together after the collision and stay at rest
in the center of mass system. Momentum is of course conserved in the
cms.
Then, if momentum is defined correctly then the total momentum should
be conserved in every coordinate system. Let us now try to calculate
the momenta in the rest system of one of the objects.
A Lorentz transformation of velocity v along the direction of
motion takes one of the particles into its rest system system S',
because using the formula for velocity transformation at u=v
we obtain
u'= (u - v) / (1- u v / c2) = 0,
so the velocity of this object is indeed zero in the moving
coordinate system.
For the particle of velocity -v we obtain in system S'
u'=(- v - v) / (1+ v2 / c2) = - 2 v / (1 + v 2/c2).
Then the total momentum of the two objects before collision would be
P = -2mv/(1+v 2/c2).
To obtain the total momentum after the collision let us transform the velocity of the combined system from the center of mass to system S'. This can be simply done because the velocity of center of mass is zero so in the moving system it will be
v'=(0-v)/(1-0v/c2)=-v.
Since the total mass is 2m, we would conclude that after the collison in system S' the total momentum is P'=-2mv. This, however, does not agree with the total momentum in system S' before the collison. We can conclude that the Newtonian momentum is not conserved in S' in spite of its conservation in the cms. Thus, the Newtonian momentum needs to be modified. Of course, we can allow only modifications that will reduce to the formula p=mv at low velocities.
The way to find the correct form of mmomentum is to try defining momentum as p=mvf(v2), with an unkown function, f, and then go through the above derivation again and find an equation for f based on the requirement that momentum is also conservedin system S'. We are not going through the algebra, though it is not very hard, just quote the result
f(v2)= g.
The correct relativistic definition of the momentum vector is p = m g v.
Now we can write down the correct definition of Newton's law as
F = dp/dt,
where one should use the relativistic momentum.
As an example let us solve the problem of constant force. We know that the Newtonian solution of constant acceleration is not allowed. A constant acceleration would eventually lead to velocities larger than c. Assuming one dimensional motion, let us calculate the derivative of the momentum using the chain rule (the momentum depends on time only through the dependence of the velocity on time):
F = dp / dt = (dp / dv) (dv / dt) = m [ g+ (v2 / c2) g3 ] a = m a g3.
Expressing a we obtain
a = (F / m) ( 1 - v2 / c2 )3/2.
showing that only at low velocities is he accleration nearly
constant. It decreases when the velocity approaches c.
The above equation is a differential equation for the velocity. This
equation can easily be solved if we notice that a simple
rearrangement of the relativistic Newtons equation gives
F dt = dp = ( dp / dv) dv.
Using the initial condition that the velocity is zero at t=0 we obtain
p / m = v g = F t / m.
Note that here F and m are constants. We can solve this equation for v to get
v = c F t / (F2 t2+ m2 c2)1/2.
It is easy to see that while at small time v= a t = F t/m, a it should be, at large t v indeed approaches c.
If we wish a useful definition for relativistic energy, then it should be conserved. Energy is conserved only if the work done by forces equals to the increase of energy. In other words the energy is equal to the integral of the force along the path
Ú F dx = Ú dp/dt dx.
We can use the relation
F dx = dp / dt dx = m (d(vg) / dt ) dx = m d(vg) dx / dt = m v d(vg) / dv dv.
Now it is easy to establish by differentiation that
v d(vg) / dv = c2 dg.
Then finally we obtain
F dx = m c2 dg.
Integrating both sides with the condition that the velocity is zero at the beginning of the movement and v is at the end we obtain the expression for the kinetic energy
K = m c2 g - m c2.
Now this expression looks very different from the Newtonian expression for kinetic energy, but it is not really that different. Let us assume that the velocity attained by the object is much smaller than the velocity of light. in that case ve can expand g in a binomial series with respect to the small quantity v2/c2. The power is -1/2, so the result is
g = 1 + 1/2 v2/c2 + 3/8 v4/c4 + ...
Substituting this expression into the relativistic form of the kinetic energy we obtain
K = m/2 v2 + 3m/8 v4/c2 + ...
Clearly, the first term is identical to the nonrelativistic kinetic energy while the next term is much smaller provided v2/c2 << 1.
Let us define the total energy as the kinetic energy plus m c2 that we will call the rest energy. Then the total energy is
E = m c2 g .
The term mc2 is only a constant, its addition shifts the energy scale only. This is the famous expression of energy by Einstein. In fact as we will see below the conservation law concerning energy is valid only for the total energy only.
In the nuclear age we learned that elements are not indestructable, they can be transformed into each other in nuclear reaction. A simple example is the proton (Hydrogen nucleus), usually denoted by p, and the neutron is neutral counterpart, which is the constituent of all atomic nuclei but Hydrogen. These two particles constitute the nucleus of deuterium, or heavy hydrogen, called deuteron. The proton and the neutron attract each other (there is an attractive force between them), so upon approaching each other (very slowly, with practically zero velocity, as it is usually said adiabatically) then they produce energy. Since energy is conserved the total energy of the proton-neutron system is decreasing. Finally, by the time they get close enough to form a deuteron they give off a large amount of energy. Assuming the deuteron is at rest energy conservation requires that the total energy of the deuteron is negative. Einstein explained such phenomena in a much simpler way. The total energy of the proton and the neutron at rest is larger then the total energy of the deuteron at rest
mp c2 + mn c2 > md c2.
The energy difference between the initial and final states is exactly the energy freed in the reaction when the proton and the neutron join into a deuteron. This energy difference is called the binding energy of the deuteron. It amounts to about a tenth of a percent of the total rest energy of the proton plus neutron system. The final conclusion of Einstein is that energy and matter are not in general conserved indpendently. Under certain circumstances matter can be converted to energy and energy to matter. Of course at the time when Einstein discovered the expression E = mc2g for energy, atomic nuclei were just bein discovered, Nothing was known about their properies.
Another, more practical example is the fusion of two deuterons
into a helium nucleus that consists from two protons and two
neutrons, exactly the same stuff the two deuterons are made of. Let
us try to calculate how much energy is gained from the fusion of a
kilogram of deuterium into helium. The easiest way to answer this
question is to figure out first what percentage of matter is
transformed into energy and then calculate how much rest energy that
percentage of 1 kg has.
The binding energy of deuteron, which is defined as
bD = mp c2 + mn c2 - mD c2 = 2.23MeV.
Most of the time energy in physics is measured in electronvolts. One electronvolt is the energy a unit charge (e) gains when it is accelerated through 1V potential difference. 1eV=1.6¥10-19Joule. 1MeV is 1 million eV. The binding energy of the helium nucleus is
bHe = 2 mp c2 + 2 mn c2 - mHe c2 = 28.28MeV.
Then the energy freed in each reaction d + d Æ He4 is
bHe - 2 bd = 28.28 - 2¥2.23 = 23.82 MeV.
The force is attractive between the two deuterons. An attractive potential is negative. Then imagine that one of the deuterons creates a potential well, a kind of a pit, for the other deuteron (and vica versa). Then on approach, the second deuteron falls into the potential well created by the first and some kinetic energy is created, while the potential energy is negative. This negative potential energy, the depth of the well is the binding energy.
The total mass energy of two deuterons is approximately four atomic mass energy units (they are made from four nucleons, i.e. protons and neutrons, which have very similar masses ), 4¥935 MeV. Then the fraction of mass transformed into energy is
f = 23.82/4¥935 = 6.37¥10-3.
This fraction, less than a percent, seems to be small but when the corresponding mass energy is calculated for macroscopic quantities it represents a tremendous amount of energy. Suppose now that this fraction of 1kg is transformed into energy. How much energy will be produced? Can we light a bulb with it? To get energy from mass we need to take
1kg¥0.00637 c2 = 0.00637¥ 9¥1016J = 5.8¥1014J.
This is the same as the production of 1000MW power plant that can supply a large city for 5.8¥108s = 20 years.
There is an important relation between the relativistic momentum and relativistic energy:
E2 = (pc)2 + (mc2)2. (1)
This relation can be easily proven if we use the expressions E=mc2g and p=mvg.
Substituting into eq. (1) above we obtain
m2c4g2 = m2c2 v2g2 + m2c4.
Carrying the momentum term to the left side and dividing by m2c2 we obtain
(c2 - v2)g2 = c2,
which is a trivial relation, if we use the explicit form of g.
In the particular case, when the object has zero mass, as some particles, like the photon, the quantum of electromagnetic radiation does,then the relation between energy and momentum simplifies into E = pc.
Relativistic energy conservation means that the sum of relativistic energies is conserved. In an elastic collision e.g. we have
E1 + E2 = E'1 + E'2,
where Ei = mi c2 (1 - vi2 )-1/2.
Take a simple example of the decay of a pi meson or pion into a muon and neutrino. This is the way charged pions decay. We would like to find out the momentum of the emitted muon and neutrino. Neutrinos are, according to current observations, massless, jsut like photons. In the rest system of the pion their momanta are equal and opposite, pm=p and pn=-p Then using eq. (1) (energy conservation) tells us that
mp c2 = (p2c2+ mm c2)1/2 + pc.
We can calculate the momentum from this relation by rearranging the equation and squaring to get
(mp c2- pc)2 = p2c2+ mm c2.
Calculating the square on the left hand side allows us to eliminate the quadratic term in p and we are left with a linear equation for it which gives
p = c (mp2 - mm2) / 2mp.
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