Question

The space and time coordinates for two events as measured in a frame S are as follows: Event 1: x₁ = x₀, t₁ = x₀/c Event 2: x₂ = 2x₀, t₂ = x₀/2c a. There exists a frame in which these events occur at the same time. Find the velocity of this frame with respect to S. b. What is the value of t at which both events occur in the new frame?

          The space and time coordinates for two events as measured in a frame S are as follows:
Event 1: x₁ = x₀, t₁ = x₀/c
Event 2: x₂ = 2x₀, t₂ = x₀/2c
a. There exists a frame in which these events occur at the same time. Find the velocity of this frame with respect to S.
b. What is the value of t at which both events occur in the new frame?
        
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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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The space and time coordinates for two events as measured in a frame S are as follows: Event 1: x₁ = x₀, t₁ = x₀/c Event 2: x₂ = 2x₀, t₂ = x₀/2c a. There exists a frame in which these events occur at the same time. Find the velocity of this frame with respect to S. b. What is the value of t at which both events occur in the new frame?
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Transcript

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00:01 All right, so let's say an event is defined by the following coordinates.
00:03 X1 is equal to something x not, and then t1 is defined as x not over c.
00:10 X2 is equal to 2x0, and then t2 is x not over 2c.
00:18 And so we want to find, we're told that there's another frame in which these events are simultaneous.
00:24 So delta t in this frame is just going to be, we'll call it basically one half, x not over c and delta x in this frame is just going to be equal to x not.
00:37 So then delta t prime in our new frame is going to be gamma times delta t minus v delta x over c squared.
00:46 And we're told that this should be equal to zero, right? and so what we have is, you know, if we solve for that, what we get is that delta t is times c is equal to v over c times delta x and so for solving for v we can see that it's going to be like c squared um times delta t over delta x and if we plug in those numbers all right so let's say delta t we can see is one half c or one half x not over c so we have c squared times one half x not over c and then this is divided by x not so these cancel.
01:34 This cancels.
01:35 We can see that this is just half the speed of light.
01:39 And then we want to know what is the value of t at which both events occur in the new frame.
01:47 All right...
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