# Relativity calculation, Lorentz transformations

1. Jan 27, 2010

### Dawei

1. The problem statement, all variables and given/known data

Two flashes of light strike at the same time, at the two orange circles on the diagram. The green train is traveling at a constant 150 kmh relative to the grey platform. The train is 1 km long.

As measured by someone at point F on the grey platform, how much time passes between the arrival of the waves at F’ on the train? (basically, the 2nd and 4th picture):

http://farm5.static.flickr.com/4008/4310911490_7ccb6e5ebf_o.jpg

2. Relevant equations

x = γ(x’ + vt’)

t’ = γ(t – vx/c2)

γ = (1 – v2/c2)-1/2

3. The attempt at a solution

Since they will arrive at exactly the same time for an observer on the train at F’, t = 0. t' will be the difference.

t’ = γ [0 – vγ(vt’)/c^2]

where v = 41.66666, and γ = 1.00000000000001

I’d like to hear any feedback on if I’m on the right track before trying to solve for t’.

Last edited: Jan 27, 2010
2. Jan 27, 2010

### vela

Staff Emeritus
When you say "at the same time," which reference frame are you talking about? From your diagram, it appears you're saying the events were simultaneous in the rest frame, but in your analysis, you seem to be assuming they were simultaneous in the moving frame.

3. Jan 27, 2010

### Dawei

Sorry, yes, the diagram is correct, the two waves will strike F on the platform at the same time.

4. Jan 27, 2010

### vela

Staff Emeritus
In this case, you don't need to use relativity at all because you're only interested from the point of view of the rest frame observer.

5. Jan 28, 2010

### Dawei

But the observer is in a different reference frame than F', so wouldn't there be a difference?

The observer is on the rest frame trying to observe the time difference of the two wave fronts striking F' on the moving frame.

I mean I feel like I could use classical physics if I were only interested in the observer at F', but since the train is moving there will be a time difference that must be accounted for no?

6. Jan 28, 2010

### Altabeh

This is not correct! Since the train is moving along, say, positive x-axis, and that F' is travelling towards then the signal emanating from G' reaches F' sooner than the signal emanating from E' does (if we know that F' is exactly located at the center of the distance E'G'.)

Actually all you need is to find

$$\Delta t' = \gamma(\Delta t - v\Delta x/c^2)$$.

where $$\Delta t = |t_{G'}-t_{E'}|$$, and $$\Delta t' = |t'_{G'}-t'_{E'}|$$ which is not known to us. Because both orange circles flash at the same time and F observes that two flashes occur simultaneously at the time of emanating, so for F the signals reach F' at the same time and thus $$\Delta t' =0$$.

AB

Last edited: Jan 28, 2010
7. Jan 28, 2010

### vela

Staff Emeritus
You wrote, "As measured by someone at point F on the grey platform, how much time passes between the arrival of the waves at F’ on the train?" To me, what that means is the observer at F looks at his watch and records what it reads at the instants of picture 2 and 4, and then he subtracts the two numbers to find the time elapsed between the two events. Is this how you're interpreting the problem?

8. Jan 28, 2010

### Altabeh

I think the question asks about the delay that signals emanating from G and G' have in hitting F' from F's point of view. If this is the case, then the difference is obviously zero. But otherwise you must be right!

AB

9. Jan 28, 2010

### Dawei

Yes, that is how I am reading the problem.

So I'm thinking "T1" = time it took for light to reach F' from G' = 500 m / (c + v)
"T2" = time it took for light to reach F' from E' = 500 m / (c - v)

Time difference = ΔT = T1 - T2 = 4.6 x 10^-13 seconds.

That would be the time difference observed by an observer on the train at F'.

But what about the observer at F? I'm calling this ΔT', and this is where I'm guessing I use Lorentz. But what are x and x' in this example?

10. Jan 28, 2010

### vela

Staff Emeritus
This time difference you found is what the observer at rest measures. For observer F, he will calculate xF', the position of the person on the train, and xG, the position of the wavefront emitted from G as:

$$x_{F'} = vt$$
$$x_{G} = 500-ct$$

Setting the two equal and solving for t gives the time observer F measures, which works out to what you called T1. Similarly, your T2 is what observer F measures for the second wavefront catching up to observer F'.