## Thursday, July 28, 2011

### Finding the delay of a clock without using the Lorentz transformations, part 2

In my previous post, I discussed how Jill would measure the time delay of a clock she was approaching at .6c without using the Lorentz transformations. I don't know if this has been done, but experiments like this could certainly serve as another validation of SR. However, in particular I'm trying to point out that Jill, regardless of whether she is moving, does not measure Jack's clock to be going faster.

I just edited the previous post to add some more information, including about how Jack can use the same reasoning to show the time delay in Jill clock is by a factor of .8, without using the Lorentz transform. This post will be about how Jill can make the same observation for clock1 (although the calculation is different), and an observer at clock1 would be able to make the reciprocal observation about Jill. Details are below the fold.

Since Jill and clock1 are moving away from each other, rather than toward each other, the diagram is different (and actually simpler). Jill still sees her clock move from 0 to 8 on her journey from clock1 to Jack, however, at the end of the trip cloc1 is read 4. That means Jill sees two of her own seconds to pass for every second that passes on clock1, or that the images of consecutive seconds on clock1 are two light-seconds apart for Jill.

So, if clock1 waits for t seconds between sending the image of 0 and sending the image of 1, the separation distance between the image of 0 and the image of 1 will be 1.6ct. Since 1.6ct = 2 ls, we get t=1.25 seconds. Thus, the fraction of seconds as measured by clock1 to seconds Jill measures for clock is 1/1.25, which is again .8. This means Jill measures clock1 to tick off 6.4 seconds on her trip between clock1 and Jack, the same as she measured for Jack.

In the reciprocal, when an observer (Jerry) at clock1 sees Jill pass Jack, Jerry has seen 8 seconds pass on Jill's clock, but 16 seconds pass on clock1 (the ten for the trip itself plus another 6 to see the image). So Jerry sees Jill's clock move at half the rate his clock does. Jerry can make the same calculations Jill does to measure Jill's clock ticking off .8 seconds for each of his.

Read more!

I just edited the previous post to add some more information, including about how Jack can use the same reasoning to show the time delay in Jill clock is by a factor of .8, without using the Lorentz transform. This post will be about how Jill can make the same observation for clock1 (although the calculation is different), and an observer at clock1 would be able to make the reciprocal observation about Jill. Details are below the fold.

Since Jill and clock1 are moving away from each other, rather than toward each other, the diagram is different (and actually simpler). Jill still sees her clock move from 0 to 8 on her journey from clock1 to Jack, however, at the end of the trip cloc1 is read 4. That means Jill sees two of her own seconds to pass for every second that passes on clock1, or that the images of consecutive seconds on clock1 are two light-seconds apart for Jill.

So, if clock1 waits for t seconds between sending the image of 0 and sending the image of 1, the separation distance between the image of 0 and the image of 1 will be 1.6ct. Since 1.6ct = 2 ls, we get t=1.25 seconds. Thus, the fraction of seconds as measured by clock1 to seconds Jill measures for clock is 1/1.25, which is again .8. This means Jill measures clock1 to tick off 6.4 seconds on her trip between clock1 and Jack, the same as she measured for Jack.

In the reciprocal, when an observer (Jerry) at clock1 sees Jill pass Jack, Jerry has seen 8 seconds pass on Jill's clock, but 16 seconds pass on clock1 (the ten for the trip itself plus another 6 to see the image). So Jerry sees Jill's clock move at half the rate his clock does. Jerry can make the same calculations Jill does to measure Jill's clock ticking off .8 seconds for each of his.

Read more!

## Tuesday, July 26, 2011

### Finding the delay of a clock without using the Lorentz transformations

For my readers who are not following the almost 2000-comment thread, aintnuthin and I are discussing Special Relativity. In particular, there is a question about whether a specific number is a prediction of what Jill measures, or a prediction of what Jill predicts Jack will measure.

The basic scenario: clock1 and Jack are at rest, sitting six light-seconds apart, and Jack has a clock (clock2) synchronized to clock1. Jill, holding a clock, passes by clock1 traveling inertially at .6c and synchronizes her clock to clock1 (so they now both read 0), and them passes by Jack. When Jill passes Jack, her clock reads 8 seconds and Jack's clock reads 10 seconds. If you use the Lorentz Transformations (LT) from the view that Jill's inertial state is the rest frame, Jill gets 6.4 seconds for clock2. The disagreement is over whether the 6.4 seconds is supposed to be what jack sees on his clock, as far as I can tell. My answer is below the fold.

My response is that the 6.4 seconds is the time Jill measures for clock2, not the time Jack measures for clock2. You can show it is the former with basic algebra. First, because of light-speed delay, Jill sees jack's clock to read -6 when Jill passes clock1. As Jill passes Jack, her clock has gained 8 seconds while Jack’s has gained 16 seconds. Jill can use that and her relative velocity of .6c to tell how much time passes on Jack's clock for her, without using the LT. I will load a diagram to help illustrate this.

This diagram is based on clock2 sending out an image reading -6 and then an image reading -4, and Jill receiving those images 1 second apart in time. Jill can measure how far apart the images were when they were sent, and therefore how much time passed in Jill's frame between when the first image was sent and when the second image was sent.

In between the times when clock2 reads -6 and clock2 reads -4, the image of clock2 reading -6 travels at c (as measured by Jill), while the clock2 itself travels at .6c.

To forestall an objection, none of this is an explanation for the time delay. It is only a measure of the time delay that Jill can make without using the LT.

Read more!

The basic scenario: clock1 and Jack are at rest, sitting six light-seconds apart, and Jack has a clock (clock2) synchronized to clock1. Jill, holding a clock, passes by clock1 traveling inertially at .6c and synchronizes her clock to clock1 (so they now both read 0), and them passes by Jack. When Jill passes Jack, her clock reads 8 seconds and Jack's clock reads 10 seconds. If you use the Lorentz Transformations (LT) from the view that Jill's inertial state is the rest frame, Jill gets 6.4 seconds for clock2. The disagreement is over whether the 6.4 seconds is supposed to be what jack sees on his clock, as far as I can tell. My answer is below the fold.

My response is that the 6.4 seconds is the time Jill measures for clock2, not the time Jack measures for clock2. You can show it is the former with basic algebra. First, because of light-speed delay, Jill sees jack's clock to read -6 when Jill passes clock1. As Jill passes Jack, her clock has gained 8 seconds while Jack’s has gained 16 seconds. Jill can use that and her relative velocity of .6c to tell how much time passes on Jack's clock for her, without using the LT. I will load a diagram to help illustrate this.

This diagram is based on clock2 sending out an image reading -6 and then an image reading -4, and Jill receiving those images 1 second apart in time. Jill can measure how far apart the images were when they were sent, and therefore how much time passed in Jill's frame between when the first image was sent and when the second image was sent.

In between the times when clock2 reads -6 and clock2 reads -4, the image of clock2 reading -6 travels at c (as measured by Jill), while the clock2 itself travels at .6c.

**(Edit: adding sentnces) That means the rate of separation between the image of clock1 readin -6 and the actual clock 1 is c-.6c, or .4c. Thus, the ratio of the distance traveled by the image of clock1 reading -6 to the distance traveled by clock1 is c/.4c, or 1/.4 (End of edit).**Since the image of clock2 reading -6 and the image of clock2 reading -4 are 1 light-second (ls) apart, the total distance from where clock2 generates the image of -6 and where that image is when clock2 generates the image of -4 is 1/.4 which is 2.5 ls. Since the image of clock2 travels at c, the image takes 2.5 seconds to travel 2.5 ls. So, Jill measures 2.5 seconds to pass while Jack’s clock moves from -6 to -4, or two seconds. 2/2.5 = .8, so for every second Jill observes on her clock, she observes .8 seconds to pass on Jack's clock, the amount predicted by the LT. Over the course of all 8 seconds Jill observes on her clock, this becomes 8 * .8 = 6.4 seconds. Again, this is what she observes to pass on the clocks, not a prediction of what Jack observes.To forestall an objection, none of this is an explanation for the time delay. It is only a measure of the time delay that Jill can make without using the LT.

**(Edit: adding sentnces) I also want to note that Jack can use the exact same process to measure the time delay for Jill, without using the Lorentz equations. Jack does not see the image of Jill passing clock1 until 6 seconds after it happens, when his clock read 6. So, Jack sees the entire trip in 4 seconds. In that time, Jill's clock moves from 0 to 8. Jack can use the same logic as above to measure Jill's clock to tick off .8 seconds for every seconds of his. (End of edit).**Read more!

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