Read Section 37.1
All disciplines employ jargon: shorthand words that express particular meanings.
They save time for the professional, but can get in the way of learning by an amateur.
I'll try to alert you to some of the less obvious uses of jargon.
"Classical physics" refers to Newtonian or Maxwellian physics, the physics of our everyday world.
In other words, non-relativistic, non-quantum physics.
The presence or absence of the word "classical" can completely change the meaning of what's going on!
The symbol b (lower-case beta), in a
relativistic context always means b = v/c,
the ratio of a speed to the speed of light.
We begin with the famous Michelson-Morley experiment.
To get the basic idea behind the experiment, imagine a race between two perfectly matched swimmers in
a broad river. One swims 100 m straight upstream, turns, and swims back to the straight point; the other
swims 100 m straight across the current, and returns. Classically, the first swimmer will lose the race. (Can you prove it?)
This illustrates the idea of a Classical or Galilean (named after Galileo Galilei) transformation:
If a point in space is given by (x, y, z) measured in frame-1 and
(x', y', z') measured in frame-2 moving at a speed u along their common positive x-axis, then
x = x' + ut
y = y'
z
= z'
If this point is moving with a velocity v along the x-axis in frame-1 then frame-2 measures
vx = dx/dt = vx' + u
In the Michelson-Morley experiment, the swimmers are replaced by light beams, and optical interference methods are used to measure extremely small time differences for light traveling in perpendicular directions. The experimental fact is that there is no time difference, and therefore no "current". There is no "special" direction in which the speed of light is exactly c; it is exactly c in all directions – light does not behave according to classical transformations. By implication, there is no special medium through which light travels.
Mathematical Tools:
Most of what we need is algerbra but an important additional tool will be the binomial expansion:
(1 + Z) n = 1 + nZ + ½n(n-1)Z2 +
...
If Z << 1, then the higher order terms ( Z2 , etc. ) are very VERY small, and can be ignored.
For example, if Z = b2 and n = -½, then
(1 - b2)-½ = 1 + ½b2 + (3/8)b4 + ... = 1 + ½b2
This expansion will come up again, since b = v/c is frequently small, and expansions in powers of b are useful.
Read Section 37.2
Postulates are fundamental hypotheses, on which a theory is based. If the predictions
of the theory disagree with experiment, then one or more of the postulates must be
incorrect. Know the two postulates of relativity.
We speak of observers in relativity.
You can think of an observer as a system synchronized, recording clocks and meter sticks in a particular inertial frame. Since different observers (or different inertial frames) are in relative motion, you must imagine that one system of meter sticks and clocks can pass through another system. In general, one observer will assert that another observer's clocks are not synchronized.
We are concerned with events.
By definition, an event is a "happening" that is located at some point (x, y, z) at an instant of time, t, as observed by some observer Einstein showed that a different observer will not only see it as occurring at a different place (x', y', z') but also at a different time, t'. The successful relativistic analysis of problems demands that we carefully identify events.
The first postulate: the laws and equations of physics are the same in
all inertial frame of reference.
The second postulate: The speed of light c
is the same in all inertial frame of reference and independent of the speed of
the source.
Two observers, using coordinate systems with different origins, and with axes rotated with respect to each other, may measure completely different values for the coordinates (x, y, z) of a point in space. If they each calculate the distance Ds between two such points, where
{distance}2 = (Ds)2
= (Dx)2 + (Dy)2
+ (Dz)2 = (Ds')2
= (Dx')2 + (Dy')2
+ (Dz')2
they will get the same answer. The distance is an invariant quantity in nonrelativistic space.
In the world of high speeds, two observers may measure completely different values for the coordinates and the time (x, y, z, t) of an event. If they each calculate the spacetime-interval Dt between two events, where
{interval}2
= (Dt)2 = (cDt)2 - (Ds)2 = (Dt')2 = (cDt')2
– (Ds')2
they will get the same answer. The interval is an invariant quantity everywhere (and everywhen!).
Notice that the speed of light c acts like a conversion factor to assure that each term in the interval has the same dimensions: meters.
Example 1:
A starship leaves Earth (event 1) and travels at 95% light speed, later arriving at Proxima Centauri (event 2), which lies 4.3 light-years from Earth.
a) What are space and time separations between events 1 and 2 as measured in the Earth frame?
b) What are space and time separations between the events as measured in the starship frame?
Solution 1:
The distance between Earth and Proxima Centauri, as measured from Earth, is 4.3 light years. The starship travels at 0.95 c, so it takes a time 4.3/.95 = 4.53 years, as measured in the Earth frame. (How might that time be measured?)
The interval is the same in both frames:
(interval)2 = (4.53 light-years)2 - (4.3 light-years)2
= (20.52 - 18.49) (light-years)2
= 2.03 (light-years)2
In the starship frame, both events occur at the same place, so the space separation in that frame is zero. So the interval is entirely the time separation. The time between the events in the starship frame is
(time separation)2 = 2.03 (years)2, or
(time separation) = 1.42 years
As measured in the Earth frame, event 2 occurs 4.53 years after event 1. As measured in the starship frame, event 2 occurs only 1.42 years after event 1. We, on Earth, might conclude that the starship's clocks run slowly.
Read Section 37.3
This example illustrates time-dilation, the fact the different observers measure different time separations between two events.
We can generalize this result.
Let Dt and Ds be the time and space separation of the two events in the Earth frame, and Dt' be the time separation of the two events in the frame of an observer who is present at both events. (Ds' is zero, because in that frame both events occur at the same place.) According to the observer in the unprimed frame, the "primed" observer is traveling at speed given by
Ds = v Dt
We can substitute that expression in the equation for the invariant interval:
(Dt')2 = (cDt')2 - 0 = (Dt)2 = (cDt)2 - (Ds)2
= (cDt)2
- (vDt)2
or
Dt'
= (1 - b2)½
Dt
Notice that Dt and Dt' are the time separation between two events, as measured by two different observers. The statement that "moving clocks run slowly" is deceptively simple, and often not helpful. Here's a way to avoid mixing up the different time intervals: if the time difference between two events can be measured on a single clock that is present at both events, that clock measures the shortest time difference. Any other observer must use two different clocks, one present at each event. The shortest time difference is often called the proper time interval between the events.
A clock could be an alarm clock, beating heart, or the lifetime of an unstable particle. The difference in time intervals measured by difference observers has nothing to do with defective clocks! It's related to the structure of space-time itself.
Because the factor 1/(1 - b2)½ appears many times, this is abbreviated as g (lower case gamma):
g = 1/(1 - b2)½
Time-dilation has been tested trillions of times; for example, every time an unstable particle went down a beam pipe in an accelerator lab, time-dilation was proven. No physicist doubts the validity of this effect. The Hafele-Keating experiment in 1971 was cute in that atomic clocks had become so precise that time dilation could be measured on big, slow apparatus. Their flights were made on commercial airlines; they purchased tickets for a couple of seats for themselves, and seats for the clocks. The experiment was also one of the first experimental tests of the famous twin paradox.
Example 2:
At exactly 12:00 noon a light is
flashed from the
At noon plus 0.02 s, a firecracker
goes off in
a) How fast would a person (or clock) have to travel to be present at both events?
b) What is the value of the invariant interval between the two events?
Solution 2:
a) To be present at both events, a person would have to travel at a rate such that Ds = vDt, so
v = Ds/Dt = 2.5´108 ms-1
b = v/c = 5/6
(b) The invariant interval is given by
(Dt)2 = (cDt)2 - (Ds)2
(Dt)2 = [3´108(0.02)]2 - (5´106) = 11´1012 m2
Notice that the moving observer would measure as the time separation between the two events
Dt' = (1 – b2)½ Dt = 0.011 s
Example 3:
Same problem as above, but now Dt = 1.33´10-2 s.
Solution 3:
a) Now
v = Ds/Dt = 3.75´108 ms-1
b = v/c = (3.75)/3 = 5/4
The speed would have to be greater than light speed. But nothing can travel faster than light! So it would be impossible for an observer to be present at both events.
b) The invariant interval is given by
(Dt)2 = (cDt)2 - (Ds)2
(Dt)2 = [3´108(0.0133)]2 - (5´106)2=
-9´1012
m2
The square of the invariant interval is negative!
The result from Examples 2 and 3 can be generalized: If it is impossible for an observer to be present at two events, because it would entail traveling faster than light speed, the square of the invariant interval is negative.
(What speed is necessary if the invariant interval is zero?)
If the time part of the interval dominates {(interval)2
is positive} the interval is called time-like.
If the space part of the interval dominates {(interval)2 is negative} the interval is called space-like.
If the {(interval)2
is zero}, the interval is light-like.
Read Section 37.4
Think of what it means to measure the length of something. If it's sitting there, you have all day to put a ruler beside it, make multiple measurements, or use a variety of techniques. If the thing is flying by you, it's trickier. You can rig up multiple cameras along the line of flight, and flash them all at the same instant. Or, you might measure the speed with a radar gun, and measure the length of time it takes for the object to pass.
No matter, the rest frame (the inertial frame of reference in which the object is at rest) is special; it measures the greatest length. If you are moving with respect to the object, you measure a reduced length. As in the case of time-dilation, it is easiest to see how the length varies using the idea of the space-time interval (which does not change in any inertial frame). Let l = Ds in a frame where the length is at rest and l' = Ds' be the length measured in the moving frame where v = Ds/Dt is the velocity of the primed frame measured by the rest frame:
(Dt')2
= 0 - (l')2 =
(Dt)2 = (vl/c)2 – (l)2
or
l' = l(1 - b2)½
= l / g
Note that since g is always bigger than 1, the length l' is always less than l. You can use this to check if you’ve remembered the relationship properly.
Special Note: Sometimes, the subscript ‘0’ is used to denote proper-time and proper-length. Be very careful, it is not always clear which frame (the primed or unprimed) contains the proper (or rest) quantity.
Last Modified: 9 January 2006, GSI.