Simultaneity and Spatial Separation

[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.

 

[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.

 

[fdsa1234]

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

The speed of light?

 

[Matterwave]

The speed of light?

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?

 

[fdsa1234]

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?

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

 

[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.

 

[Matterwave]

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.

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.

 

[fdsa1234]

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.

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

 

[Matterwave]

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

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?

 

[fdsa1234]

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?

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?

 

[Matterwave]

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?

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'##?

 

[fdsa1234]

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##?.

Should it not also equal 3m?

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

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'##?

##\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?

 

[Matterwave]

Should it not also equal 3m?

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

Yes, this is correct.

##\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?

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

 

[fdsa1234]

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

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

 

[Matterwave]

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

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?

 

[fdsa1234]

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?

So,

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

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

 

[Matterwave]

So,

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

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

Looks right to me. :)

 

[fdsa1234]

Looks right to me. :)

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?

 

[Matterwave]

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?

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##.