viewtopic.php?p=321467#p321467
Halc wrote:Yes, except 450m/sec, not km per sec. I said it wrong the first time. Bob at the equator is moving at that speed, and is also going in a circle, but it takes a day to do a lap, so much lower acceleration. We should run the experiment for at least a day if we were to do this for real, so Bob's net velocity at the end is zero just like that of Alice.

OK. So Bob is the guy on the equator and Alice is in the centrifuge at the pole. (Poor Alice. I'd rather be Bob!) Bob's acceleration (towards the centre of the Earth. v

^{2}/r = 0.032 ms

^{-2}) can be taken to be near enough zero compared to Alice's acceleration towards the centre of the centrifuge = 10000 ms

^{-2}. (Assuming 1000g's.)

In SR terms, the Lorentz factor for both of them, relative to someone who is not moving at 450ms

^{-1} with the rotation of the Earth but not in the centrifuge (perhaps Colin, standing still in the middle of the centrifuge on the North Pole) as calculated by: 1 / (1 - v

^{2}/c

^{2})

^{0.5} is:

1 / (1 - 202500/9X10

^{16})

^{0.5} = 1 / 0.999999999998874999..

So over the course of 1 day the SR time dilation between Alice and Clive, and between Bob and Clive, due to their differences in linear velocity, would be about 9.72X10

^{-8} seconds. About 1 second every 28000 years. Perfectly measurable. I've read that caesium clocks can measure 1 second in 1.4 million years. So we could look for any time dilation between Alice and Bob and if a caesium clock can't find it, I presume we can safely say that it isn't there and you are right.

So your experiment is doable. Let's do it!

(Footnote: I always write square root here as "to the power of 0.5" because I can't figure out how to do a square-root symbol in here and if you try to write "1/2" as a superscript, using the "sup" tag, it doesn't seem to work, annoyingly.)

Bob is not in the box here, and the environment is very different where Alice is. Each knows which is which. I'm putting at least 1000 g's on poor Alice.

Centripetal acceleration = v

^{2}/r. If it was actually 1000g's on Alice, with her tangential speed of 450ms

^{-1}, that's about 10000ms

^{-2}. So the centrifuge radius would be 20.25 metres. It would do about 3.5 rotations per second.

Right, but we're not doing that. Only Alice, at the exact point where she has the same linear speed as Bob, and the same gravitational potential as well. <-- Pun mildly intended. There is no third clock somewhere else on the centrifuge.

Yes, understood.

I think it would be the same at exactly one point in the box: the one that moves at 450m/sec.

If we could do the experiment, that proposition is what we would be empirically testing.

Clearly it runs faster at the center (not moving at all), and slower further out (more V and more acceleration), so somewhere in between they must balance. If acceleration is not a factor, it balances at whatever RPM gives us 450m/sec. If acceleration matters, it balances at a smaller radius than that.

If we want 1000g, we need to pick an r that is about 22m I think, 290000x smaller than the r of Bob who is accelerating at about 1/290th of a g.

Yes, those are roughly the numbers that I got.

Your GR equation is useless since r is identical in both instances. It is computing gravitational dilation of something stationary, not of something moving. The SR equation you use in your program only computes a relative dilation for a clocks separated by h, and we have no h here. The appropriate SR equation is the simple Lorentz factor for 450m/sec in both cases. Acceleration (g) does not play into that equation.

The equation for the Schwartzschild solution of Einstein's GR equations is indeed useless because it's not applicable. If the rotation is equivalent to a gravitational force, then it's not a force that is inversely proportional to the square of 'r'. It is directly proportional to r. So to calculate the proposed effect of the acceleration I'd have to integrate from the centre of the centrifuge out to 22m, a series of small h's within which acceleration can be taken to be constant. (Or follow Schwartzschild and create a solution for Einstein's GR field equations, which is mathematically beyond me because I never studied Tensor Calculus.)

I strongly suspect that you still misunderstand what this integration process is doing and what the various values in the GUI of the application represent. But I could be wrong. Perhaps I'll post an annotated version of the GUI.

I still stand by my assessment of that. Every locally measurable way, yes, but not every measurable way. My Uranus/Earth thing bears that out. Both locally detect a similar g, but the one with the slightly higher g has considerably less actual dilation.

I explained the reason for this, when calculated using the numerical integration method, in an earlier post. And I ran the simulation to show that both methods (SR and GR) yield the same result for Uranus.

There is no local test for actual dilation or for r (the r is what differs between Earth and Uranus). David would be ecstatic if there was such a test, since it would prove that his is the only correct view.

In terms of the relationship between r and g, the difference between Earth and Uranus is that on Uranus the value of g reduces more slowly, with respect to r, than it does on Earth.