Calculating spacetime intervals between events


by maker2807
Tags: interval, space, time
maker2807
maker2807 is offline
#1
Feb2-11, 02:26 PM
P: 3
Hi, I am new to this forum so hello to everybody. I have this problem to solve:

1. The problem statement, all variables and given/known data
Q is moving away from P at speed 4/5 c. After 3 years (in Q's frame of reference) he turns around (assuming that turning takes no time) and is moving back to P at speed 4/5 c.
What are spacetime intervals between:
(A) Start and turn
(B) Turn and arriving back
(C) Start and arriving back

c = speed of light

2. Relevant equations



3. The attempt at a solution
I suppose that problem isn't very difficult, but I'm not sure about my solution. First of all, I'm not sure if I'm supposed to calculate s or s2. The second problem is, that I'm not sure if I should use t = 3 years (time in Q's frame of reference) or t = 5 years (time in P's frame of reference calculated from time dilatation formula).
For (A), I think that r = 4 ly (distance travelled at 4/5c for 5 years in P's frame of reference) and t is either 5 or 3 years. Then result should be easily obtained from the equation. Am I right?
For (B), I think it's the same as (A), because speed is the same as well as time. Or should I change the speed sign to minus because of different direction?
For (C) should the distance be zero because of event's occuring at the same place? And what about time? Should it be 6 years or 10 years?

Thanks in advance for your responses.
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collinsmark
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#2
Feb3-11, 09:15 PM
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Hello maker2807,

Welcome to Physics Forums!
Quote Quote by maker2807 View Post
I suppose that problem isn't very difficult, but I'm not sure about my solution. First of all, I'm not sure if I'm supposed to calculate s or s2.
I'm guessing you are asked to give Δs. Because all events in this problem (the start, the turn and the end) are time-like separated, you can expect that (Δs)2 is negative, thus Δs is an imaginary number.

Or it may be possible that you are supposed to give Δτ (the Greek letter tau, not to be confused with regular time Δt) as your answer, where (Δs)2 = -c2τ)2. So you can expect Δτ to be real and positive for time-like separated events. τ is called "proper-time" and is the time elapsed in the moving frame (or whatever particular frame of reference is for the world-line under consideration).

But to know for sure, you should check you coursework or ask your instructor.
The second problem is, that I'm not sure if I should use t = 3 years (time in Q's frame of reference) or t = 5 years (time in P's frame of reference calculated from time dilatation formula).
Just pick one frame of reference and stick with it.

Or better yet, solve for the space-time interval in both frames of reference individually. Space-time intervals are independent of the frames of reference, so if you worked the problem right, you'll get the same answer either way. You can use it to double check your work.
For (A), I think that r = 4 ly (distance travelled at 4/5c for 5 years in P's frame of reference) and t is either 5 or 3 years. Then result should be easily obtained from the equation. Am I right?
Okay, let's start by keeping things in terms of the "stationary" frame P. You've already calculated that Δr = 4 ly (this is spacial distance that Q moved, relative to frame P). So how much time elapsed in frame P? That's the time that you want to use.

Now double check you work, by solving the problem in terms of the "moving" frame Q. From the problem statement, you know that the time elapsed (Δt') in the Q frame is 3 y. So what is Δr', the spacial distance that Q traveled relative to the Q frame of reference?

A word of advice: Don't just start plugging numbers into formulae all willy-nilly, when solving for Δr'. You're looking for the spacial distance that Q traveled relative the moving frame of reference Q. Think about it for a moment if you have to.
For (B), I think it's the same as (A), because speed is the same as well as time.
Sounds reasonable to me.
Or should I change the speed sign to minus because of different direction?
You can do that too if you want! The answer comes out the same. [If b = a2, what is (-a)2? ]
For (C) should the distance be zero because of event's occuring at the same place?
The space-time difference is not zero, because even though the spacial location is the same, each object moved through time (and Q moved through both space and time).
And what about time? Should it be 6 years or 10 years?
And keep in mind that P and Q traveled through different world-lines, so the total space time intervals from start to arriving back will be different for Q than it is for P.

But the problem is asking for the space time interval directly between two events. One event is the start, and the other event is arriving back. So imagine a straight space-time line drawn between the two events. Does that follow Q's or P's world-line?

By the way, this entire problem might be a perfect problem to solve by using a space-time diagram. If you are not familiar with space-time diagrams, I would suggest looking in to them and how to use them.
maker2807
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#3
Feb4-11, 04:09 AM
P: 3
Thank you very much for your explanation, I think I am getting into it. I found out answers to the problems A and B and I hope they are correct. Would you mind checking them? However, I'm still quite stuck with part C.

For part (A), let's start with P's frame of reference. I'll be using ly units for simplification (I hope it's not incorrect). Time elapsed between start and turn for P is 5 years. Spatial distance between these two events is 5yrs*(4/5)c = 4 ly. So s^2 = (4 ly)^2 - (5yrs*c)^2 = -9 ly^2 and s = i*3ly. Checking it in Q's frame of reference brought me to following idea - Q in his frame of reference didn't move at all. For Q it seems, that everything around moved. So spatial distance for Q is 0 and time distance is 3 years. Then s^2 = 0 - (3yrs*c)^2 = -9 ly^2 which is the same as for P's frame of reference.

For part (B),(in P's frame) spatial distance is 4ly and time difference is 5 years we get the same s^2 = -9 ly^2 and in Q's frame, spatial distance is 0 and time difference is 3 years and again we get s^2 = - 9ly^2

For part (C), on the beginning of your post, you wrote that space-time intervals are independent of the frames, but later you wrote that total space time intervals will be different for Q than it is for P. Why?
One idea came to my mind - to take absolute spatial distance into account rather than relative - this way, the results stay same in both frames - For P's frame, spatial distance is 10yrs*(4/5)c = 8ly and time difference is 10 years, so we get s^2 = 64 ly^2 - 100 ly^2 = -36 ly^2 and for Q's frame, spatial distace is still 0 and time difference is 6 years, so s^2 = -36 ly^2 and s = i*6ly. This even makes sense to me in a way that spacetime interval between start and return should be (logically) the sum of spacetime interval between start and turn and interval between turn and coming back.

But if we stay with relative spatial distance, then we get different spacetime intervals - for both P and Q relative distance between start and coming back is 0. However time elapsed is 10 years for P and 6 years for Q so we get s^2 = - 100ly^2 (for P) or s^2= -36 ly^2 for Q.

Is either of this proceedings correct?
Thank you again.

collinsmark
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#4
Feb4-11, 04:53 PM
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Calculating spacetime intervals between events


Quote Quote by maker2807 View Post
For part (A), let's start with P's frame of reference. I'll be using ly units for simplification (I hope it's not incorrect). Time elapsed between start and turn for P is 5 years. Spatial distance between these two events is 5yrs*(4/5)c = 4 ly. So s^2 = (4 ly)^2 - (5yrs*c)^2 = -9 ly^2 and s = i*3ly. Checking it in Q's frame of reference brought me to following idea - Q in his frame of reference didn't move at all. For Q it seems, that everything around moved. So spatial distance for Q is 0 and time distance is 3 years. Then s^2 = 0 - (3yrs*c)^2 = -9 ly^2 which is the same as for P's frame of reference.
Very nice*

*see below regarding s vs. τ.
For part (B),(in P's frame) spatial distance is 4ly and time difference is 5 years we get the same s^2 = -9 ly^2 and in Q's frame, spatial distance is 0 and time difference is 3 years and again we get s^2 = - 9ly^2
Also very nice.

So now you you'll need to figure out if you are supposed to give your answers in terms of s or τ. (Once again, τ here is the Greek letter tau, not to be confused with t.) Both s and τ are measures of spacetime intervals. And both are just different ways of expressing the same thing. They only really differ by the fact that s has space-like units and τ has time-like units; and s is real when dealing with space-like separated events and τ is real when dealing with time-like separated events. Since all the events in this problem are time-like separated, it sort of makes more sense to express the space-time intervals in terms of τ. But I'm not sure if that's what you're supposed to do, so you'll have to check your notes, textbook, or ask your instructor to know for sure.
For part (C), on the beginning of your post, you wrote that space-time intervals are independent of the frames, but later you wrote that total space time intervals will be different for Q than it is for P. Why?
P's path through spacetime is different than Q's path through spacetime. They travel along different world-lines.

Let's step away from special relativity just for the moment, and consider a triangle. Suppose the triangle has three vertices (corners), labeled A, B, and C. Is the distance AB + BC equal to distance AC? No, of course not. The distances are different because the path taken is different.

Okay, now let's bring special relativity into the mix. A, B, and C are spacetime events. A is the point in spacetime that Q began its journey. B is the point in spacetime that Q turned around. C is the point in spacetime that Q returned.

In this example, A and C are not separated in spacial dimensions but they are separated in the time dimension. Event B is separated from A and C in both space and time.

And the interesting thing about special relativity, which you should probably get used to, is that the interval AB + BC is actually shorter than the direct path AC. Interval AC is actually a longer total path than the combination of AB and BC. That fact doesn't doesn't really matter that much for this problem, but it is something to keep in mind.
One idea came to my mind - to take absolute spatial distance into account rather than relative - this way, the results stay same in both frames - For P's frame, spatial distance is 10yrs*(4/5)c = 8ly and time difference is 10 years, so we get s^2 = 64 ly^2 - 100 ly^2 = -36 ly^2 and for Q's frame, spatial distace is still 0 and time difference is 6 years, so s^2 = -36 ly^2 and s = i*6ly. This even makes sense to me in a way that spacetime interval between start and return should be (logically) the sum of spacetime interval between start and turn and interval between turn and coming back.

But if we stay with relative spatial distance, then we get different spacetime intervals - for both P and Q relative distance between start and coming back is 0. However time elapsed is 10 years for P and 6 years for Q so we get s^2 = - 100ly^2 (for P) or s^2= -36 ly^2 for Q.
The correct answer is in there somewhere. But it's up to you to determine which of your possible answers is the correct one.

Is part (C) of the original problem asking for the spacetime interval AB+BC or is it asking for AC?

Once again, I think the correct answer might be much more obvious to you if you drew out the events and intervals on a spacetime diagram. Spacetime diagrams were invented for problems like this.
maker2807
maker2807 is offline
#5
Feb11-11, 09:21 AM
P: 3
Hello, I figured it out and submitted to my teacher. It was correct. So thank you very much for help.


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