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

In summary: A(ct,x) and B(ct,x), separated by a lightlike interval.-Zack needs to find the velocity of a separate reference frame in which they both occur at the same time.-He also needs to find the velocity of the frame where they will occur at the same LOCATION.-The Attempt at a Solution-To solve for this, Zack needs to find where the t' for both events is equal.-But when i set tb'=ta', everything cancels out and Zack ends up getting 1=3.-Is there any other way of doing this that Zack is missing?-Zack is getting the
  • #1
khfrekek92
88
0

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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  • #2
Or am I just doing this completely wrong?
 
  • #3
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.
 
  • #4
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?
 
  • #5
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!
 
  • #6
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?
 
  • #7
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.
 
  • #8
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.
 
  • #9
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
 

What is the definition of "Find Velocity of Frame for Simultaneous Occurrence of Events A & B"?

The velocity of frame for simultaneous occurrence of events A & B refers to the speed at which an observer must travel in order to see events A and B happening at the same time.

Why is it important to find the velocity of frame for simultaneous occurrence of events A & B?

Finding the velocity of frame for simultaneous occurrence of events A & B can help us understand how different observers perceive the same events. It also plays a crucial role in theories of relativity and can help us make accurate measurements and predictions about the timing of events.

How is the velocity of frame for simultaneous occurrence of events A & B calculated?

The velocity of frame for simultaneous occurrence of events A & B can be calculated using the Lorentz transformation equations, which take into account the relative motion and distance between the observer and the events.

What factors can affect the velocity of frame for simultaneous occurrence of events A & B?

The velocity of frame for simultaneous occurrence of events A & B can be affected by the relative speed and distance between the observer and the events, as well as the direction of motion and the observer's frame of reference.

How does the velocity of frame for simultaneous occurrence of events A & B relate to the speed of light?

According to the theory of relativity, the velocity of frame for simultaneous occurrence of events A & B cannot exceed the speed of light. As an object approaches the speed of light, the frame of reference for simultaneous events changes and time dilation occurs, making it impossible for events A and B to be seen as happening at the same time by an outside observer.

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