Special relativity effects in general relativity

  • #26
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This is the wrong way to look at GR. While it can be made rigorous in stationary space-times, it is certainly not what the conceptual core of GR says. The same goes for everything you say following the quoted statement.

I explained the pitfalls of your line of thinking in a different post (see the 2nd paragraph): https://www.physicsforums.com/showpost.php?p=4770244&postcount=13
Ok, I guess I'm wrong, but kinematical time dilation and gravitational time dilation do add up? When I said 'upward acceleration' I really meant upward force, like in this video: .

So when a rocket travels, with its engines firing so it could fight off what we used to call gravity, it travels to the Sun in some kind of path. It undergoes time dilation because of velocity and gravitational time dilation due to its position. If my line of reasoning is correct? Or you may explain to me, what is the cause, or condition, that determines why gravitational time dilation arises. An object that is stationary on the surface of Earth is said to accelerate in GR, right?
 
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  • #27
WannabeNewton
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Or you may explain to me, what is the cause, or condition, that determines why gravitational time dilation arises.
It arises from acceleration. This is why gravitational time dilation vanishes for freely falling frames but exists in the frame of an observer stationary in a gravitational field, as per the equivalence principle. You don't need space-time curvature in order to have gravitational time dilation also for this reason but if you do have space-time curvature then it will indirectly affect the closed form expression for gravitational time dilation in those coordinate systems wherein it does show up.

An object that is stationary on the surface of Earth is said to accelerate in GR, right?
Yes.
 
  • #28
A.T.
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Ok, I guess I'm wrong, but kinematical time dilation and gravitational time dilation do add up?
They rather multiply.

When I said 'upward acceleration' I really meant upward force, like in this video: .
You talked about "canceling out". In that video there is no "canceling of forces" in Einstein's model, just in Newton's.


what is the cause, or condition, that determines why gravitational time dilation arises.
Gravitational time dilation arises between different positions in a gravitational field (along the potential gradient), or in an accelerating reference frame (along the frame's acceleration).

An object that is stationary on the surface of Earth is said to accelerate in GR, right?
It has proper acceleration, like the apple on a branch. Put a smart phone on a table, and it shows 1g upwards proper acceleration.
 
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  • #29
WannabeNewton
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You still may do so, I think it will help me. Especially the comparision between SR effects in flat space time vs curved space time
Here are some examples from past threads of mine:

https://www.physicsforums.com/showthread.php?t=717211
https://www.physicsforums.com/showthread.php?t=718299
https://www.physicsforums.com/showthread.php?t=730009

Let me know if you want more examples. I spent a considerable amount of time last summer solving a lot of the exercises in "General Relativity: An Introduction for Physicists"-Hobson so I have a slew of examples I can provide you with if need be.

Here's another fun one: say we have an observer moving to the right across a linear platform at constant velocity ##v## relative to some inertial frame, with the platform accelerating upwards at the rate ##g##. Then it is well known that a gyroscope comoving with the observer will be observed to precess in the observer's rest frame defined by the axes obtained by boosting the axes of a local Rindler frame fixed to the platform to the right at velocity ##v##. There are numerous ways to explain why this happens using only SR, the most physically appealing of which is a standard relativity of simultaneity argument.

But say we wanted to use the equivalence principle instead. Then we can basically treat this as a GR problem (although it isn't a GR problem per say since there is no space-time curvature and this is also a computationally trivial problem). Imagine we are in Rindler coordinates with a field of Rindler observers ##e_t = \frac{1}{z}\partial_t## stationary in this gravitational field and a metric ##ds^2 = -g^2 z^2 dt^2 + dx^2 + dy^2 + dz^2## where ##z = \frac{1}{g}## represents in Rindler coordinates the stationary position of the observer standing on the platform. Now we introduce a field of observers obtained by boosting along ##x## with velocity ##v## (##\nabla v = 0##) from the global Rindler frame so that ##e_{t'} = \Lambda_{t'}{}{}^{\mu}e_{\mu} = \gamma(e_{t} + ve_{x})## and ##e_{x'} =\Lambda_{x'}{}{}^{\mu}e_{\mu}= \gamma(e_x + ve_t)##; then ##\xi^{\flat} = gze^t + ve^x## where ##\xi = \partial_t + v\partial_x##.

Note first that ##[e_{t'},e_{i'}] = 0## hence if we consider a gyroscope comoving with ##e_{t'}## at ##z = \frac{1}{g}## then the gyroscope precesses relative to the axes ##\{e_{i'}\}## with angular velocity ##\omega^{\alpha}|_{z = g^{-1}}\partial_{\alpha} = -\frac{1}{2}(\xi_{\mu}\xi^{\mu})^{-1}\epsilon^{\alpha\beta\gamma\delta}\xi_{\beta}\partial_{ \gamma}\xi_{\delta}|_{z = g^{-1}}\partial_{\alpha} = \frac{1}{2}\gamma^2 vg \partial_y##. It's very easy to interpret this precession through the equivalence principle if we consider the metric ##g_{\mu'\nu'}## in the coordinates centered on the observer moving across the platform because ##g_{t'x'} = \Lambda_{t'}{}{}^{\mu} \Lambda_{x'}{}{}^{\nu}g_{\mu\nu} = \gamma^2 v (1 - g^2 z^2)## so there's a gravitomagnetic field ##\vec{\nabla}\times\vec{g}## proportional to ##\gamma^2 v g^2 z## hence the gyroscope at ##z = \frac{1}{g}## precesses at a rate proportional to ##\gamma^2 vg## due to this gravitomagnetic field.
 
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  • #30
PeterDonis
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kinematical time dilation and gravitational time dilation do add up?
Under certain limited kinds of scenarios, yes, you can think of them this way. But rockets moving between the Earth and the Sun are not really suited to this kind of thinking.

The key point about gravitational time dilation is that it is between two objects which are both *stationary*, as WBN pointed out. For example, if you are standing on the ground and I am standing at the top of a tall tower, there is gravitational time dilation between us--your clocks "run slower" than mine in an invariant sense, as we can both verify by, for example, exchanging light signals and measuring their round-trip travel times.

What you are calling "kinematical time dilation" is due to relative motion, and obviously two objects which are in relative motion can't both be stationary. For example, if you're standing on the ground and I jump out of an airplane high above you, I'm not stationary (though you are, because you are at rest relative to the Earth, which is the source of the gravitational field). So there is no invariant way to specify our "time dilation" relative to each other (gravitational or otherwise).

But now consider this scenario: you are standing on the ground, and I am orbiting the Earth in such a way that I pass directly over you on each orbit. Now there *is* a way for us to compare our "rates of time flow" in an invariant sense: we each measure how much time elapses, by our own clocks, between my successive passages overhead. In other words, since my motion is periodic, it can be treated as "stationary" for this purpose. And in this scenario, my "rate of time flow" relative to you will be the sum of two effects: my altitude (which makes my clock run faster than yours), and my orbital velocity (which makes my clock run slower than yours). These effects work in opposite directions, so the net effect will depend on their relative magnitudes; if you work the numbers, my clock will run slower than yours for low Earth orbits, but for higher orbits, my clock will run faster (the break point is, IIRC, an orbit at 1.5 Earth radii).

When I said 'upward acceleration' I really meant upward force
Yes, but you said "curved spacetime upward acceleration". The upward acceleration isn't due to curved spacetime; it's due to the rocket engine, or whatever else is exerting the force (if you're standing on the surface of the Earth, it's the Earth exerting force on you and pushing you upward).
 
  • #31
WannabeNewton
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Ok, I guess I'm wrong, but kinematical time dilation and gravitational time dilation do add up?
Sort of, at least in simple circumstances. For example if we have an observer in circular free fall orbit in Schwarzschild space-time then the observer's "gamma factor" in Schwarzschild coordinates will be ##\gamma = (1 - 2M/R - R^2 \omega^2)^{-1/2}## where the ##2M/R## accounts for the gravitational part and the ##R^2 \omega^2## accounts for the kinematical part. Then ##\gamma^{-1}## represents the rate at which the circularly orbiting observer's clock ticks relative to a clock at spatial infinity.
 

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