# Simultaneity and Spatial Separation

1. Sep 21, 2014

### fdsa1234

1.

Two events occur simultaneously in an inertial reference frame, separated by a distance of 3 metres. Within a different inertial frame that is moving with respect to the first, one event occurs 10^-8 seconds later than the other.

(a) In the moving frame, what is the spatial distance between the two events?

(b) What is the speed of the second frame relative to the first?

2./3.
I'm lost here. I've tried solving for delta(x) in:

delta(t)' = gamma * ( delta(t) - v/c^2 * delta(x) )

but I'm stumped.

We've just begun doing relativity work in my class, but my professor doesn't do examples during lecture, making the work quite challenging to do on my own. We've been shown a few of the entry-level relativity equations (time dilation, invariant intervals, and the like), but I have no idea how to do this problem. Any hints in the right direction would be greatly appreciated.

2. Sep 21, 2014

### Matterwave

Hint for part a: are there quantities in special relativity that are independent of reference frame?

Hint for part b: Once you have part a, you will know both the x,t of event 1 and x,t of event 2 in both frames of reference. You can then look at which Lorentz transformation (what v) will give you this transformation between the 2.

3. Sep 21, 2014

### fdsa1234

The speed of light?

4. Sep 21, 2014

### Matterwave

Yes, that's one thing. I was thinking of something else though. Something more similar to a distance. What's an invariant distance in special relativity?

5. Sep 21, 2014

### fdsa1234

Hmm... ct? I thought that the answer might be 3 +/- 1 metre (one or the other, not both), but I'm not sure...

6. Sep 21, 2014

### fdsa1234

3 +/- 3, that is. I think it's +3? So, the new separation is 6 metres? Or am I completely off here...?

Apologies for the double post; I exceeded the post edit time limit.

7. Sep 21, 2014

### Matterwave

What would be your reasoning to think that one frame will differ by 3 whole meters when the time is only different by 10^-8 seconds?

When you make a Lorentz transformation between two frames, what stays constant? In a Euclidean 3-space, when you make a rotation, the distance $d^2=x^2+y^2+z^2$ is a constant. In Special relativity there's an object that looks very much like that that is constant when you make a Lorentz transformation.

I would be very surprised if you haven't learned this concept, for you to be asked this question.

8. Sep 21, 2014

### fdsa1234

The invariant interval $I = -c^2 * delta(t)^2 + d^2$?

9. Sep 21, 2014

### Matterwave

Ah, yes, but easier to just write $\Delta s^2=-c^2 \Delta t^2+\Delta x^2$ since we only care about 1 spatial dimension in this problem. Can you see how to use this to solve part a?

10. Sep 21, 2014

### fdsa1234

I see. I hadn't thought to use that; the context it was used in the lecture was for four-vectors with multiple spatial dimensions.

So, $\Delta s^2=-c^2(10^-8)^2 + (3)^2$ (that should be 10^-8, but it won't show up properly)? From that I get $\Delta s^2 = (-1) * (3 * 10^8)^2 * (10^-8)^2 + (3)^2 = 0$...? Is this wrong, or does that mean that the two events occur in the same place?

11. Sep 21, 2014

### Matterwave

You've mixed up the two frames (also, for latex, you need to use {} wrappers if you want a superscript to include more than 1 character).

In the first frame, the two events are simultaneous, meaning $\Delta t=0$ and the two events are separated by a distance of 3 meters $\Delta x=3 m$ what is $\Delta s$?. In the second frame the two events are not simultaneous $\Delta t'=10^{-8}s$. What is $\Delta s'$? And then what is $\Delta x'$?

12. Sep 21, 2014

### fdsa1234

Should it not also equal 3m?

$\Delta s^2 = -c^2(0)^2 + (3)^2$?

$\Delta s'^2=-c^2(10^-8)^2 + \Delta x'^2$

But how can I solve that if both $\Delta s'$ and $\Delta x'$ are unknown?

Last edited: Sep 21, 2014
13. Sep 21, 2014

### Matterwave

Yes, this is correct.

Ah, but $\Delta s'$ is known! Recall, why did we care about it in the first place?

14. Sep 21, 2014

### fdsa1234

That's what I thought. Is $\Delta s'$ also equal to 3 because it is invariant?

15. Sep 21, 2014

### Matterwave

Yes, it is also equal to 3 meters (always remember your units). So $\Delta s = \Delta s' = 3m$, then can I now finish part a?

16. Sep 21, 2014

### fdsa1234

So,

$(3)^2 = -c^2(10^-8)^2 + \Delta x'^2$

I'm getting $x' = 4.2425.$ Please tell me that's correct.

17. Sep 21, 2014

### Matterwave

Looks right to me. :)

18. Sep 21, 2014

### fdsa1234

Thank you so much. For part b), can I just plug numbers into a Lorentz equation, say, $t' = γ(t - vx/c^2)$? And I'm assuming the resulting speed should be a significant fraction of c?

19. Sep 22, 2014

### Matterwave

Yes, this would work, but you should understand why. Physically, what are we doing when we use this equation?
Also, it might be easier, now that you have $\Delta x'$ to use the $x'=\gamma(x-vt)$ equation to figure out $v$.