A train is driving at 0.8c relative to the rails.

On the train, a car is driving with 0.8c relative to the train, in the same direction as the train.

How fast is the car driving relative to the rails?

I know the formula for velocity addition in Special Relativity, but I'm trying to derive the formula with basic steps.

So I reasoned: if 1 second passes on the rails, 0.6 seconds pass on the train. So relative to the train, the car drives 0.6 * 0.8 = 0.48 light seconds. So that's... 0.48 * 0.6 = 0.288 light seconds relative to the rails?

The train has driven 0.8 light seconds relative to the rails. That makes a total of 0.8 + 0.288 = 1.088 light seconds in 1 second? That can't be right. Yet I can't find my error. Can someone help me?

• You can simply add the rapidities. Feb 10 at 14:29

The most common mistake is to overlook the relativity of simultaneity, and that's what happened here. At any given ground-frame time, train clocks at the front of the train show an earlier time than clocks at the back. Therefore, the train-frame time the car spends moving, given that it moves for 1 second of ground-frame time, is less than what you calculated.

Getting the right train-frame time is not trivial since you don't know the ground-frame speed. You could introduce that speed as an unknown, or try it the other way around: assume the car drives for 1 second in the train frame (and remember that ground-frame clocks near the front show a later time).

• The clocks inside the train would only tick differently if the train was accelerating. This is not the case. Every inertial reference frame is equally valid and, therefore, train, rails and car can be considered stationary in their own reference frames, on their reference frames, the other reference frames have the speeds described in the problem. Feb 9 at 19:54
• @ViniciusAraujoRitzmann I don't know what you mean by "tick differently". I only said that the clocks of the train frame are not synchronized with the clocks of the ground frame, which is true as you surely know. Feb 9 at 20:21
• Yes, but what I mean is their time do not pass at diffrent rates. They might not be synchronized at first, for instance one can start at the 3 seconds mark and the other at the 5 seconds mark, but "how long" a second lasts is the same. One second at the front and back of the train lasts the same amount of time, this difference of two seconds will never change unless the train accelates. The way you said the front and back clocks show different times made it look like they are not ticking at the same rates. I am sorry if you did not mean that. Feb 9 at 20:48
• @ViniciusAraujoRitzmann I was trying to preserve the spirit of OP's approach to the problem, which involved reasoning from the fact that wrt ground clocks, train clocks run slow. Of course that's not true in an absolute sense but I think OP knows that. Feb 9 at 22:02
• Very late reaction, but thanks for your answer! It was very helpful :) Apr 4 at 8:09

I believe your mistake was not considering that the car moves relative to the train as well, so the time measured inside the car is different from that measured inside the train.

There are three reference frames in this problem. The rails, the train and the car. To know how the rail frame sees the car, you need to switch frames twice (rail to train, train to car). Remember: $${\gamma} = \frac{1}{\sqrt{1-\beta^2}}$$ and $$\beta = \frac{v}{c}$$ $$t_{train} = \gamma_{train}\left(t_{rails} - \beta_{train} x_{rails}\right)$$ $$t_{car} = \gamma_{car}\left(t_{train} - \beta_{car} x_{train}\right)$$ Since both are moving evenly and the time coordinate is measured in meters: $$x=\beta t$$. Then: $$t_{train} = \gamma_{train}\left(t_{rails} - \beta_{train}^2t_{rails}\right)=\gamma_{train}\left(1 - \beta_{train}^2\right)t_{rails}$$ $$t_{car} = \gamma_{car}\left(t_{train} - \beta_{car}^2t_{train}\right)=\gamma_{car}\left(1 - \beta_{car}^2\right)t_{train}$$ Now substitute one into the other: $$t_{car} =\gamma_{car}\gamma_{train}\left(1 - \beta_{car}^2\right)\left(1 - \beta_{train}^2\right)t_{rails}$$ Replacing with numbers:

$$t_{car} =\frac{1}{1-0.8^2}\left(1 - 0.8^2\right)^{2}t_{rails}$$ Because $$\gamma_{car} = \gamma_{train}$$ since they have the same relative speed $$v_{train} = v_{car}$$. Therefore: $$t_{car} = 0.36t_{rails}$$

This means that for every second that passes on the rails, 0.36 seconds pass inside the car, as seen by someone on the rails. That is, someone on the rails believes that inside the car, time passes more slowly.

However, the reverse is not true (the car does not see the time on the rails pass faster). Someone inside the car believes to be at rest, so for the car are the train and the track that are moving. Someone inside the car would also measure that time on the rail passes more slowly.

Hope this helps. In addition, there is a very good channel on youtube about special and general relativity called "eigenchris", I believe that there you can find classes that will help you too.

EDIT: I edited the answer because I noticed that I forgot to pass the beta coefficient forward in the equations and that, in addition, time here is measured in meters, so it is ordinary time multiplied by the speed of light , what I had not remembered before. I noticed that there was a problem in the previous equations because if the movement were in the -x direction, there would be no time dilation, which would not be correct.

• Very late reaction, but thanks for taking the time to answer :) Apr 4 at 8:11

Edited February 24 to change notation from u, v′ and v to Einstein’s 1905 v, w and V. Retain Tipler’s numbers and results. Tipler’s notation: V, u′ and u.

Einstein 1905 EDoMB On the Electrodynamics of Moving Bodies - Section §5

https://www.fourmilab.ch/etexts/einstein/specrel/www/

Paul A Tipler, Physics, Vol 3 Chapter 39-5, numbers shown: 0.8c and 0.8c .

Einstein: Frame K ( upper case ) with variables x and t; frame k ( lower case ) with variables x′ and t′.

We use axis X with variables x and t and axis X′ ( prime ) with variables x′ and t′.

Axis X is the track on which the train runs with speed-v and axis X′ is the bed of the train, on which bed the bullet moves with speed-w.

speed-v :: train speed on the track based on x and t

speed-w :: bullet speed on the train, based on x′ and t′

speed-V :: compound bullet speed on the track based on x and t

Your train moves on axis X with speed v = 0.8c

v = x/t = 0.8c → x = 0.8 × t

Thus axis X′ with origin x′ = 0 ( the train ) moves on the track with speed v.

Axis X′ ( the moving train car ) fires off a bullet from x′ = 0 with speed w = 0.8c, as judged on the train.

w = x′ / t′ = 0.8c → x′ = 0.8 × t′

Equations of “transformation” as appear in EDoMB-Section §3:

t′ = gamma × ( t – vx/c² )

x′ = gamma × ( x – vt )

Let’s not ask how this transformation system was derived. Let’s just substitute into w = x′ / t′ :

w = x′ / t′ = gamma × ( x – vt ) ÷ gamma × ( t – vx/c² )

Speeds w and V are the same thing, just from different points of view. If we transform w as a function of x′ and t′ to a function of x and t, we have speed-V, bullet speed as judged on the track.

The right hand side is a function of x and t. Rewrite it.

Gamma divides out.

w = ( x – vt ) ÷ ( t – vx/c² )

w × ( t – vx/c² ) = ( x – vt )

wt – wvx/c² = x – vt

wt + vt = x + wvx/c²

( w + v ) × t = ( 1 + wv/c² ) × x

x / t = ( w + v ) / ( 1 + wv/c² )

Shown Section §5. It is the standard form in use today.

V = x / t

V = ( w + v ) / ( 1 + wv/c² )

Note: this is quotient, V = x / t is not train speed v = x / t, nor is it bullet speed on the train w = x′ / t′ . It is bullet speed on the track using track values x and t rather than car values x′ and ′t.

With v = 0.8c and w = 0.8c :

V = x / t = ( 0.8c + 0.8c ) / ( 1 + 0.8c × 0.8c /c² )

V = x / t = 1.6c / ( 1 + 0.64 )

V = x / t = 1.6c / 1.64

V = x / t = 0.98c

What’s going on here? There are two things in motion: a train, and a bullet on the train. There are three judgments: speed of the train on the track, speed-v; speed of the bullet on the train, speed-w ; speed of the bullet on the track, speed-V. Classically, V = w + v.

It is not classical because the clock on the train runs slowly. Since speed involves a quotient, x / t, the transforms simplify:

t′ = gamma × ( t – vx/c² )

x′ = gamma × ( x – vt )

Simplify

t′ = t – vx/c²

x′ = x – vt

This algebra says that meter sticks on track and train are identical, but the clock on the train is slower than the clock on the train. Then bullet speed per unit time judged on the train is higher speed than would be judged by a ‘proper’ clock on the train.

Does the train’s slow clock yield w = 0.8c ?

V = v + w

Unacceptable: V = 0.8c + 0.8c = 1.6c

Question: If the second of these speeds, w = bullet-on-train, is judged by a track clock, would we have w = 0.18c ? The sum is then:

Acceptable: V = 0.8c + 0.18c = 0.98c

The above is not shown. A perfectly obvious question has not been asked or answered. Not in Einstein, Tipler or Wiki.