Speed of the light and dilation of time

jcsd

He said 'on average'. To be honest you don't even need to write a single equation to see why he must be correct - if it wasn't the case the two observers wouldn't measure a difference in the time interval.

GAsahi

This is the case being debated, please stop adding your own case. Would you please stay on topic?

PAllen

I disagree with you. As normally used (Doppler measured in vacuo for light actually received by direct path), the pattern of doppler for circular and stationary (NOT CENTRAL) observer are different. Further, the very phrase 'at the same time' is seriously problematic. For who? The stationary observer? The circular moving observer? They obviously differ as to 'same time', and for the circular traveler, there is no unique way to even define global simultaneity.

I agree with what Gwellsjr said about doppler and about temporal relationships. #113 summarizes what is true for OP scenario.

GAsahi

"Average" has nothing to do with it, when A measures redshift, B measures redshift. When A measures blueshift, B measures blueshift. gjwellsjr has claimed (at least twice) that A measures redshift while B measures blueshift. I can't believe that it is already past 100 posts and people don't get it.

GAsahi

Yes, you have tried this hair splitting earlier. The point is that they both measure the same type (red OR blue) shift, NOT opposite shifts as gjwellsjr incorrectly claimed.

jcsd

Yes "average" does have everything to do with it as that is what he said, I'm not sure redshift or blueshift had even been mentioned at this point.

The two observers have clocks that run at different speed and no objective "when".

PAllen

People don't get it because it is wrong (except for your scenario of carefully inserted sequence of mirrors, and measuring only on that path). In any normal discussion, one refers to doppler measured through empty space by direct light path. In this case the doppler pattern is not symmetric between the two observers. As to 'same time', I noted in another response that this is not even meaningful.

GAsahi

It is a red herring.


It has been mentioned, please go back and read the thread. Start reading at post 20.

GAsahi

Irrelevant, do you really think that the observers measure opposite shifts (one red, the other one blue)? Why is it so difficult for you to admit that you are defending an error?


Hair splitting: throughout the revolution, both observers only see one type of shift if they maintain the direction of emitting signals.

jcsd

It's not a red herring, because that's what was said in the post that you objected to!




So I should start reading the thread at post 20, to see if redshift or blueshift was mentioned before post 20?

GAsahi

I objected to his incorrect claim that the observers measure opposite type of shifts.





Just start reading at post 20, ok?

PAllen

Because I am not in error. If we place on, each world line I describe for OP scenario, a monochromatic light source and a spectrograph recording light from the other, the recordings on the spectrographs would not have the same shape. What could be clearer than that?

GAsahi

You mean one would be blueshifted and the other one would be redshifted? Yes or No?

PAllen

Both choices are wrong. Each graph would be a smooth, periodic function, but the shape would be different. If you try to compare the graphs at a meeting point, besides the difference in shape, the length of the graphs would be different (assuming identically constructed devices), because of the difference in accumulated proper time. Trying to say which point on one graph corresponds to which point on the other would get into exactly the issues I raised in #113, which is a summary of what I think the important features are.

GAsahi

You are not answering the question, you are just attempting to evade it. Obviously there is a redshift from the emitted frequency since the observers are separating. Is the shift the same (towards red) or not? Yes or No?

PAllen

I am not evading it in any way. Your choice itself is wrong. Both graphs show redshift and blue shift portions, as stated many times. But the shape is different (as is the length). The circular observer's graph will have a wider period of blue shift compared to the stationary observer's graph.

GAsahi

Sure you are, you have been playing this game for quite a few posts now.


No, they do not. Let me tighten the condition such that you can't try to slip through: the observers emit beams of light in ONE direction only, so there is NO switching from redshift to blueshift, it is redshift throughout. Now, please answer the question.