Special Relativistic time dilation and length contraction derived

This thread is for those who want to learn how a couple of the well-known effects from Einstein's Special Theory of Relativity can be derived from first principles. The derivation of time dilation and length contraction presented here is based on a simple thought experiment.

We start with the postulates of Special Relativity:

1. The laws of physics take the same form in all inertial reference frames.
2. The speed of light is constant for all observers.

These are the only assumptions we will make in order to derive the time dilation and length contraction effects predicted by the theory. We will also need some algebra.

The situation that the derivation will be based on is as follows:

There are two cars, which we will call BLUE and GREEN, on a straight road. The green car travels at speed v relative to the blue car, in a straight line along the road. We will assume that v is less than the speed of light, c.

We are interested in two different reference frames, so we need two different coordinate systems. The blue car denotes distances measured along the road as x, and times as t. The green car denotes distances by a different coordinate, x', and times by t'. We make no assumptions initially about the relationship between t and t' or x and x'. That is, the time intervals or distances between any given pair of events may be different when measured by the blue car, as compared to when they are measured by the green car.

The blue car chooses its coordinate system such that all distances to other objects on the road are measured from the blue car itself. This means that, according to BLUE, its own x coordinate is always x=0. Similarly, the green car measures distances from its own location, so that the position of GREEN at all times is x'=0. We assume that the x and x' axes both point along the road in the same direction. We know nothing about the relative scales of the axes yet.

In the equations which follow, keep in mind that any quantity labelled with a prime (') is measured in the reference frame of the GREEN car, while any quantity without a prime is measured in the frame of the BLUE car.

A diagram of the situation is given below. The diagram shows both reference frames - the points of view of both the BLUE and the GREEN car.

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When the GREEN car reaches the point x=A, where A is some constant distance away from the blue car, the BLUE car emits a pulse of light in the direction of the green car. This pulse of light is represented as a pink blob in the diagram. We will call the emission time of the pulse t=t'=0. It is assumed that the clocks used by both cars to measure time are mechanically identical, tick at the same rate when at rest, and keep time accurately.

We will look at where the green car is when the light pulse hits it, and also consider the time it takes the light to travel between the cars, from the point of view of both the blue and the green cars.

First, we write down expressions for the positions of each of the cars and the light pulse in each reference frame. Here we use the basic constant-speed equation \(x = x_0 + vt\) repeatedly, where \(x_0\) is the position of an object at t=0, and v is its velocity in the x direction. This equation is presumed to hold in both reference frames, due to the first postulate of Special Relativity.

BLUE frame (top diagram)

BLUE car: x(t) = 0 ........(1)
GREEN car: x(t) = A + vt ..(2)
light pulse: x(t) = ct ....(3)

In this frame, the blue car is at rest, the green car moves in the positive x direction at speed v, and the light pulse moves in the positive x direction at speed c. Note that, at t=0, the blue car and the light pulse are both at x=0, and the green car is at x=A.

GREEN frame (bottom diagram)

BLUE car: x'(t') = -B' - vt' .....(4)
GREEN car: x'(t') = 0 ............(5)
light pulse: x'(t') = -B' + ct' ..(6)

In this frame, the GREEN car is at rest. The BLUE car moves in the negative x' direction. We assume that the speed of the blue car as seen by the green car is the same as the speed of the green car as seen by the blue car, but in the opposite direction. Hence, the quantity v has the same magnitude in the GREEN frame as in the BLUE frame, and it needs no prime (') to distinguish it.

On the other hand, since we don't know anything about the distance scales yet, we cannot assume that the distance between GREEN and BLUE is the same in both frames. Therefore, in the GREEN frame, we designate the position at time t'=0 as x'=-B'. We don't know how the constant B' relates to A yet.

Finally, we use the second postulate of special relativity to say that in both frames, the speed of the light pulse must be the same, so c, like v, is the same in both sets of equations above, and c needs no prime. (Another way of saying this is that c=c' and v=v').

Also note that in equations (4), (5) and (6), both the distances and times are measured in the GREEN frame, which is why x' is expressed as a function of t' (and not t). Remember, we don't yet know if t' is the same as t or not.
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We now consider the time of arrival of the light pulse at the green car.

BLUE frame

The pulse arrives when the x coordinates of the pulse and the green car are the same, which means, from equations (2) and (3):

A + vt = ct

Solving for t, we get:

Time of arrival of pulse in BLUE frame: \(t = \frac{A}{c - v} .......(7)\)

Where is the green car when the pulse arrives? Plug the value of t from equation (7) into quation (2) we get:

Position of GREEN car when pulse arrives: \(x = A + v\left[\frac{A}{c - v}\right] = \frac{A}{1 - v/c} ....(8)\)

GREEN frame

In this frame, the pulse arrives when the x' coordinates of the pulse and the green car are the ame. From equations (5) and (6) we get:

0 = -B' + ct'

Solving for t':

Time of arrival of pulse in GREEN frame: t' = B'/c .......(9)

Using equation (4):

Position of BLUE car when pulse arrives at the GREEN car:
\(x' = -B' - vt' = -B' -\frac{vB'}{c} = -B'\left(1 + \frac{v}{c}\right)....(10)\)
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Now, let's look at the DISTANCES between the two cars, both when the pulse is emitted and when it is received.

When the pulse is emitted, we have:

BLUE car: x = 0, x' = -B'
GREEN car: x = A, x' = 0

The distance between the cars is as follows:

In BLUE frame: d(0) = A - 0 = A ........(11)
In GREEN frame d'(0) = 0 - (-B') = B' ..(12)

When the pulse is received, we have:

BLUE car: x = 0, x' = -B'(1 + v/c), from equation (10), above.
GREEN car: x = A/(1 - v/c), x' = 0, from equation (8), above.

The distance between the cars is as follows:

In BLUE frame:
\(d(t) = 0 - \left[-B'\left(1 + \frac{v}{c}\right)\right] = B'\left(1 + \frac{v}{c}\right) .......(13)\)
In GREEN frame:
\(d'(t') = \frac{A}{1 - v/c} - 0 = \frac{A}{1 - v/c} ...........(14)\)
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We are now in a position to relate the length scales of the x and x' axes. We write:

\(d'(0) = \gamma d(0), .....(15)\)
\(d'(t') = \gamma d(t) ...(16)\)

where \(\gamma\) is some constant to be determined. In words, these equations say "The distance d' between the two cars, measured in the GREEN frame, are equal to \(\gamma\) times the distance d between the cars measured in the BLUE frame." They are taken at two specific times, corresponding to two specific events (when the light pulse is emitted and received). We assume that the scale relationship is the same at time t as it was at time 0.

Using (15), (11) and (12) we get:

\(B' = \gamma A ....(17)\)

Also, from equations (13), (14) and (16) we get:

\(\frac{A}{1 - v/c} = \gamma B'\left(1 + \frac{v}{c}\right) ....(18)\)

Replacing B' in (18) with the expression given in (17) we get:

\(\frac{A}{1 - v/c} = \gamma^2 A \left(1 + \frac{v}{c}\right).\)

The constant A cancels out on both sides. Rearranging we get:

\(\gamma^2 = \frac{1}{(1 - v/c)(1 + v/c)}\)

\(\gamma^2 = \frac{1}{1 - (v/c)^2}\)

\(\gamma = \frac{1}{\sqrt{1 - (v/c)^2}} ....(19)\)

Since v is less than c, \(\gamma\) is a number greater than 1. \(\gamma\) is the well-known relativistic or "Lorentz" factor. Note that it depends only on the relative speed v between the objects in question (in this case the two cars), since c is a constant according to the second relativistic postulate.

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Length contraction

Re-writing equation (16), with the value of \(\gamma\) from equation (19):

\(d'(t') = \frac{d(t)}{\sqrt{1 - (v/c)^2}} .... (20)\)

In other words, the distance between the cars, as measured by the GREEN car, is GREATER than the distance between the cars as measured by the blue car. Since we are talking about the same physical distance here, this must mean that the GREEN car's rulers, which the GREEN observer uses to measure the distance, are contracted compared to the BLUE car's rulers. Shorter rulers mean that GREEN measures a longer distance.

This is relativistic length contraction.
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Time dilation

How far did the light pulse travel? It covered the distance between the cars. We need to take into account that the cars were already a certain distance apart when the pulse was emitted. According to each observer:

BLUE observer:

\(d(0) = A ..............(11)\)
\(d(t) = B'\left(1 + \frac{v}{c}\right) ....(13)\)
So distance travelled by pulse = \(B'\left(1 + \frac{v}{c}\right) - A .... (21)\)

Since the speed of the pulse was c, the time taken to cover the distance, as measured on A's clock is the result in (21) divided by c.

GREEN observer:

\(d'(0) = B' ............(12)\)
\(d'(t') = \frac{A}{1 - v/c} ..(14)\)
So distance travelled by pulse = \(\frac{A}{1 - v/c} - B' .... (22)\)

The speed in this frame is also c, so the the time taken in this frame is the result in (22) divided by c.

Using \(B' = \gamma A\), equation (21) reduces to:

distance(BLUE) = \(B'\left(1 + \frac{v}{c}\right) - A = A \left[\gamma \left(1 + \frac{v}{v}\right) - 1\right]\)

And equation (22) reduces to:

distance(GREEN) = \(\frac{A}{1 - v/c} - B' = B' \left[\frac{1}{\gamma (1 - v/c)} - 1\right]\)

Taking the ratio of equations (22) and (21), we get:

distance(GREEN) / distance(BLUE) = time(GREEN)/time(BLUE)

\(= B'\left[\frac{1}{\gamma(1 - v/c)} - 1\right] / A \left[\gamma (1 + v/c) - 1\right] ....(23)\)

Recall from (19) that

\(\frac{1}{\gamma^2} = 1 - (v/c)^2 = (1 + v/c)(1 - v/c)\)

Rearranging, we get:

\(\frac{1}{\gamma (1 - v/c)} = \gamma (1 + v/c) .... (24)\)

This allows us to cancel the terms in square brackets in equation (23), leaving:

time(GREEN) / time(BLUE) = \(B'/A = \gamma\)

or

time(GREEN) = \(\gamma\) time(BLUE)

This equation says that the travel time for the pulse, as indicated by the GREEN observer's clock, was longer than the travel time as measured on the BLUE observer's clock. In other words, the GREEN clock runs slow compared to the BLUE clock. This is relativistic time dilation.

Copyright: James R (2004)

 

It's not different, draqon. I thought I'd post it because:

(a) some people might want to learn something.
(b) some people have been arguing, for example, that length contraction or time dilation does not exist.

This is a proof that if the postulates of special relativity are correct, then length contraction and time dilation automatically follow. Thus, if anybody wants to attack special relativity, he will need to attack one or both of the postulates. Attempting to dispute something like length contraction further down the line is a complete waste of time, as I have posted a completely unassailable proof that it follows logically from the postulates.