I was just arguing why I treat the redshift from A to O as purely gravitational time dilation, but not the redshift B to O. The difference is that A and O are static, so there's no contribution from a varying distance. B's redshift on the other hand does have such a contribution. I mentioned the...
Wikipedia, here and here. Zero dot product between image and projection direction, not between image and the vector you project onto. I think that's the usual definition.
Synge 1960, I think. I came across this result a few times.
That's not my definition - at least I'm not aware of it. A has...
To the contrary. It is most important to know clearly what your coordinates are and what they are not. I did not say that coordinates are something very physical, I think you misrepresented my stance there grossly. But "those frames you're using" are not some numbers, but local inertial frames...
I think an orthogonal projection onto U gives a vector parallel to U.
True, but I don't try to calculate redshifts. The time dilation factor I'm talking about is without the Doppler factor. An observer at infintiy would not attribute the redshift to time dilation alone. He would do this for A's...
Right. So if you insist that time dilation is just some coordinate issue, I'm not talking about time dilation. Call it "transverse Doppler effect" or something, it doesn't matter.
Ok, but: if time dilation were just a coordinate artifact (you did't claim that, I know), we would not bother...
That's good.
No Problem. For me, there is an operational definition and a measurement that indicates that u's clock is "ticking slower" as judged by the reference observer. And that is time dilation. If you insist on defining it in terms of a global coordinate chart - well, I think I would...
Pardon me, but you lost me here. I showed exactly two contributions, one gravitational and one kinematical, the product of which happens to be the total time dilation, the reciprocal of ##u^t## namely. As an answer, you say I didn't show that [the reciprocal of] ##u^t## factors into two...
Sorry, my entry point in this thread was Dale Spam's #22 , I never read the OP and did not intend to start a parallel discussion. Concerning the OP, the answer is easy: proper accelerations don't matter courtesy of the clock hypothesis. Gravitational fields do matter, however, but there are none...
No, not really. I knew beforehand that the dilation must be the product of a gravitational and a kinematical component by the way it works: A static observer sees something at his position and relays it to the reference observer. Whatever he sees will look even slower for the reference observer...
Maybe I do. I always try to see things from the perspective of their operational implementation and, correspondingly, try to find a covariant, geometric expression for what is happening. Coordinates come into the game sometimes abstractly as a calculation tool, but more often by their physical...
Exactly. That is true for all static spacetimes in static coordinates.
But it wasn't a general statement. It was made for static spacetimes with a certain reference, as stated.
You're right, of course. The product of two four velocites is their respective gamme factor, not its inverse. Here's...
Ok, but see my answer to your third reply. In my opinion, the coordinate time is relevant only if is connected with something physically interesting.
There seem to be misunderstandings. I used exactly the one(s) you used also. I just replaced the coordinate velocity with the relative velocity...
Sorry, I can't follow. I don't see why a decomposition should use, of all things, a coordinate velocity.
The thing is: ##\gamma(U,v)= f(U) \, g(v)##. f is a function of the potential only, and g is a function of the relative velocity only. That is a geometric, frame-independent, unique...
Ah, I see. But they use the coordinate velocity dx/dt. The decomposition is in terms of relative velocity to the local static observer. If I'm not mistaken, that is ##v=v_{rel} =v_{co}/ \sqrt{1-2U}##. So you have
##\sqrt{1-2U-v_{co}^2}=\sqrt{1-2U-(1-2U)v^2}=\sqrt{1-2U}\sqrt{1-v^2}##
It's just...