Spacetime & Rotating Planet: Velocities Impact

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What affect does spacetime have on a rotating planet? (Velocities in the center differ from surface velocities)
 
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I don't know the answer but here's what i think the answer might be...

Spacetime does not, and can not, affect different parts of the same object differently.

Here’s why I think it can’t.

Time is relative based on a frame of reference. Which means, a planet (the planet) has only 1 frame of spacetime reference – regardless if it is spinning or not. It experiences time as whole – as a single frame.

The only way time differs is when spacetime is observed from, at least, two reference frames. It’s only the perception of time that differs – not time itself. Time doesn’t really change the way you’re thinking it does, like energy. The only thing that changes is how time is perceived – relatively speaking.

In other words, a person standing on the surface of the Earth will perceive time at a slower rate than a person at the center – but it’s only perception. This does not imply that the Earth itself is experiencing different times. It only implies that, from the reference frame of someone on the earth’s surface comparing itself against the frame of person in the core, that time slows down.

Bottom line, time is experienced by Earth at a constant rate, surface to core, - not a varible rate. This is regardless of the Earth "standing still", rotating, orbiting, or arching through the cosmos.
 
LawrenceM said:
I don't know the answer but here's what i think the answer might be...

Spacetime does not, and can not, affect different parts of the same object differently.

Here’s why I think it can’t.

Time is relative based on a frame of reference. Which means, a planet (the planet) has only 1 frame of spacetime reference – regardless if it is spinning or not. It experiences time as whole – as a single frame.

The only way time differs is when spacetime is observed from, at least, two reference frames. It’s only the perception of time that differs – not time itself. Time doesn’t really change the way you’re thinking it does, like energy. The only thing that changes is how time is perceived – relatively speaking.

In other words, a person standing on the surface of the Earth will perceive time at a slower rate than a person at the center – but it’s only perception. This does not imply that the Earth itself is experiencing different times. It only implies that, from the reference frame of someone on the earth’s surface comparing itself against the frame of person in the core, that time slows down.

Bottom line, time is experienced by Earth at a constant rate, surface to core, - not a varible rate. This is regardless of the Earth "standing still", rotating, orbiting, or arching through the cosmos.

This is wrong.

For a quantitative analysis, see

https://www.physicsforums.com/showthread.php?p=1543402#post1543402.
 
Is it possible that this would be cumulative and could possibly wind the Earth matter like a spring? To put it another way, if you had a stick the length of the diameter of the Earth and spun it at the same rate for the same amount of time as the earth, would the stick appear as a spiral? (I am wondering if this is why our galaxy looks the way it does.)
 
I know these examples are different because of gravitational time effects but would either be true? I attempted to understand the mathematical analysis but came up waaaayyyy short.(2x9=4 right?) Any help appreciated.
 
I asked a question here, probably over 15 years ago on entanglement and I appreciated the thoughtful answers I received back then. The intervening years haven't made me any more knowledgeable in physics, so forgive my naïveté ! If a have a piece of paper in an area of high gravity, lets say near a black hole, and I draw a triangle on this paper and 'measure' the angles of the triangle, will they add to 180 degrees? How about if I'm looking at this paper outside of the (reasonable)...
From $$0 = \delta(g^{\alpha\mu}g_{\mu\nu}) = g^{\alpha\mu} \delta g_{\mu\nu} + g_{\mu\nu} \delta g^{\alpha\mu}$$ we have $$g^{\alpha\mu} \delta g_{\mu\nu} = -g_{\mu\nu} \delta g^{\alpha\mu} \,\, . $$ Multiply both sides by ##g_{\alpha\beta}## to get $$\delta g_{\beta\nu} = -g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \qquad(*)$$ (This is Dirac's eq. (26.9) in "GTR".) On the other hand, the variation ##\delta g^{\alpha\mu} = \bar{g}^{\alpha\mu} - g^{\alpha\mu}## should be a tensor...

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