# Special relativity: flashes of light

1. Nov 14, 2015

### whatisreality

1. The problem statement, all variables and given/known data
In a frame of reference A lights are on the x axis at x = D and x = -D, where D = 0.6 x109. They flash simultaneously at t = 0.

There's also a frame of reference A' moving at v = 0.8c.

i) Where and when do the flashes happen in A'?

ii) Therefore when would observers at the origins of A and A' see the light?

2. Relevant equations

3. The attempt at a solution
i) Well, I have to add D to the location in x to find the location in x'. Then I have to add something else. But I'm pretty confused about what to add. As far as an observer in A' is concerned, I think the light has to travel x' = D + vt'? But I can't use that because I don't know t' or x', so that's probably wrong...
In A, both flashes happen at t = 0.6 x109 / 3x108 = 2 seconds. So I have t' as well:
t' = γt where γ is the Lorentz factor, so maybe I do have t' actually, I think it's 10/3.

Essentially, my question for the first part is whether x' = D + vt' is right.

And then the problem is that haven't I already worked out when the observer at A sees the light? It's at t =2. And for the observer at the origin of A', at t = 10/3?

2. Nov 16, 2015

### Staff: Mentor

What does the Lorentz Transformation tell you about where and when the flashes occur in A'?

Chet

3. Nov 16, 2015

### whatisreality

Using x' = γ(x - vt) , t=0, x' = -1 x 109m for the bulb at -D, and 1 x 109 for the bulb at D.

And as to when they happen - that should be at two different times. Using t' = γ( t - $\frac{vx}{c^2}$), and subbing in t = 0,
t' = -2.67 s for the bulb at +D, and 2.67 s for the bulb at -D. I think.

So if I'm in frame F', it looks like F is moving, doesn't it? And that means anything measured in F will appear shorter to me, and seem like it happened either later/ earlier, depending on the direction of movement?

4. Nov 16, 2015

### whatisreality

And about the second part: I'm pretty sure the time at the origin of A is easy to calculate. It's just t = D/c and is the same for both bulbs.

As for at the origin of A', for the bulb at D: That person thinks the flash happened at t = 2.67, and that the light travelled for t = $\frac{1 \times 10^9}{c}$- 2.67 s, but I'm really not 100% sure about subtracting 2.67 seconds! Should I be adding instead? Or just doing nothing? Does that take account of the fact that F' is moving towards the bulb?

And for the bulb at -D: it travels for $\frac{1 \times 10^9}{c}$ + $\frac{x'}{v}$ + 2.67? Maybe? Struggling to get my head round this.

5. Nov 16, 2015

### Staff: Mentor

Forget about the F frame of reference for now. In the F' frame of reference, the two flashes occur at -1 x 109m, + 2.67 s and 109, -2.67 s. Since they are traveling at the speed of light in F', at what value of t' do each of these flashes reach x' = 0?

6. Nov 16, 2015

### whatisreality

t' = $\frac{1 \times 10^9}{c}- 2.67$ for x' = D?

$\frac{1 \times 10^9}{c} + 2.67 + \frac{x'}{v}$ for x = -D. Possibly? Not sure I've got the adding/ subtracting of 2.67 the right way round. Or that I'm accounting for the movement of the frame correctly!

7. Nov 16, 2015

### Staff: Mentor

Your answer for -2.67 is correct. Since frame F' doesn't know that it is moving, your answer for +2.67 is not correct. Lost the x'/v.

8. Nov 16, 2015

### whatisreality

F' thinks the light bulb at -D is moving away from it though, doesn't it? At the v = 0.8c assigned to F' relative to F? Although by that logic I should also have subtracted x'/v for x' = D as well. That's seriously confusing.

9. Nov 16, 2015

### whatisreality

Or is the -2.67 term accounting for the movement?

10. Nov 16, 2015

### Staff: Mentor

Once the flash occurs at the indicated location and time, it doesn't matter what the light bulb does next. It is the light flash that is traveling toward x' =0 at the speed of light; the light bulb has nothing to do with this.

11. Nov 16, 2015

### whatisreality

Oh, I get it!

Thanks for wading through my messy first post and replying. I really appreciate it! :)