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This is chapter 37 problem number 58.
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We have a s frame, and then we have an s prime frame.
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This s prime frame is moving with respect to this s frame with the speed velocity b.
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So we have two events happening in this problem.
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We're going to look at it from the point of view of the s frame and then a s frame.
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Frame.
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So let's say event one in the s frame and then in the s prime frame and event two, in the s frame and an s prime frame.
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Let's make a table here.
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Now for the first event, let's call the position of the first event based on the s frame x1, t1, the time in the s frame.
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And then the s prime frame, let's call it x1 prime and t1 prime.
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For event two, let's call x2 and t2, x2 prime and t2 prime.
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Now, there are a couple of information that is given to us in the problem.
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For example, the time from the s frame, the time difference between the two events is 1 .8 seconds.
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So basically t2 minus t1 is 1 .8 seconds in s frame.
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And in the s prime frame, the difference in time is given to us as 2 .15 seconds again in this prime frame.
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And also one of those things that is given to us, well, by the way, we're really.
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Doing the assumption of, you know, t1 equals t1 prime, let's say the initial occurrence is like at zero, and then x prime equals x is going to be equal zero, which means the origins kind of coincide, coin side when t equals zero.
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So actually x1 and x2 in the problem are given to us as the same because the events are happening at the same spot as far as the s frame is concerned.
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So we also have this information that is useful that we're going to make use of it in the future.
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Now, we could go and write down the transformation from one frame to the other.
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Let's see what's asked first.
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The distance between the positions of two events in the s prime frame is what's asked in the problem.
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So we're basically looking for x2 prime minus x1 prime.
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This is what we're looking for.
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In order to get an answer to this, as you know, we're going to have to know what v is first.
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We need to know what b is, and then we can go and look at the difference between the positions of the two events in the s prime frame.
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So in order to get to v, let's do.
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What t1 prime is going to be equal to gamma t1 minus v x1 over c squared, right? so what t2 prime is going to be gamma t2 minus v x2 over c squared.
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So now if, well, remember t2 prime minus t1 prime was given to us as 2 .15.
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Let's take advantage of that, which means the right -hand side of this minus the right -hand side of the first equation.
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So gamma t2 minus vx2 over c -squared minus gamma t -1 minus vx -1 over c -squared.
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Now i'm going to group them, gamma t2 minus vx2 over c -squared minus t -1 plus vx -1 over c squared.
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Look at these two terms.
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There is a t2 minus t1 embedded here.
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So gamma t2 minus t1 since we know what this is.
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Minus, so we have v over c squared for both, v over c squared.
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Let's group these x2 minus x1.
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Remember, this term, we know that it is zero because the two vents are happening at the same point for the s frame.
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So x2 equals x1.
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From here, this term, the entire term is going to be zero...