Find Velocity of Frame for Simultaneous Occurrence of Events A & B

AI Thread Summary
To find the velocity of a reference frame where events A and B occur simultaneously, it is essential to recognize that they are separated by a lightlike interval, implying a causal relationship. The discussion highlights the confusion around achieving simultaneous occurrence in different frames, especially considering that a frame cannot move at the speed of light. It is clarified that while events can be perceived as simultaneous in a different frame, they must still adhere to the invariant nature of spacetime intervals. The conversation also touches on the distinction between lightlike, spacelike, and timelike intervals, emphasizing that spacelike separation prevents events from occurring at the same location. Ultimately, the problem requires careful consideration of the physical implications of the intervals involved.
khfrekek92
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Homework Statement


events A(ct,x)=(1m, 2m) and B(ct,x)=(3m,4m) are separated in a lightlike interval. I need to find the velocity of a separate reference frame in which they BOTH occur at the same time.

Next I need to find the velocity of the frame where they will occur at the same LOCATION.

Homework Equations


t=(gamme)t'


The Attempt at a Solution


To solve for this, I need to find where the t' for both events is equal. But when i set tb'=ta', everything cancels out and I end up getting 1=3. Is there any other way of doing this that I am missing? I am getting the same problem with the location part as well.

Thanks in advance!
 
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Or am I just doing this completely wrong?
 
This problem appears to have been designed to force you to think about the physical situation it describes before you jump into the equations and formulas. Think about why we use the word "lightlike" to describe some intervals.
 
Hmmm... Well Lightlike means they are only in a causal relationship if you use light. So would the frame of reference have to moving at speed c? I thought you can't have a frame moving at speed c?
 
khfrekek92 said:
Hmmm... Well Lightlike means they are only in a causal relationship if you use light. So would the frame of reference have to moving at speed c? I thought you can't have a frame moving at speed c?

That's a bingo!
 
Oh! Duh, why didn't I think of that? So if they were in a spacelike interval, you still wouldn't be able to have them happen at the same time, would you?
 
khfrekek92 said:
Hmmm... Well Lightlike means they are only in a causal relationship if you use light. So would the frame of reference have to moving at speed c? I thought you can't have a frame moving at speed c?

A better description of lightlike separation might be that a flash of light could have been emitted at one event and absorbed at the other... But this definition will bring you to the same result.
 
Remember that ds^2 is an invariant. This means that if ds^2=0 in one frame, then it is 0 in all frames. Further, this means that ds^2 can never change signs. 2 events which occur at the same time must have ds^2<0 (assuming metric (+ - - -)), and 2 events which occur at the same point in space must have ds^2>0.
 
khfrekek92 said:
Oh! Duh, why didn't I think of that? So if they were in a spacelike interval, you still wouldn't be able to have them happen at the same time, would you?

Either you have it backwards or you're typing too quickly: spacelike separation means that they'll be always be at different locations, but you can find a reference frame in which they happen at the same time or either one before the other.
 
  • #10
I just did a similar problem, but with runners in a time-like interval. The problem was that two runners were racing, one runner was given a time handicap, find an inertial frame where the handicap doesn't exist. You don't have to going at the speed of light for something to APPEAR to happen at the same time from another frame. The question was pretty open-ended (Jackson...), but I basically said/found that if these are regular human runners you'll have to be going much faster for relativistic effects so just ignore velocity addition (or travel perpendicular to running motion), and there was a definite speed at which an observer in K' claims the runners started at the same time. Just wanted to throw this bit of info at you to add to confusion :p
 
  • #11
Haha! Thanks so much guys, I finally got all of them done (there was one problem for finding equal times AND distances for each of the three intervals.) Thanks so much for all of your help!

-Zack
 
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