- #1

- 261

- 9

*g*, equivalent to the Lorentz factor γ for a clock travelling at a relative speed

*v*?

- B
- Thread starter jeremyfiennes
- Start date

- #1

- 261

- 9

- #2

Vanadium 50

Staff Emeritus

Science Advisor

Education Advisor

- 26,147

- 9,532

- #3

PAllen

Science Advisor

- 8,258

- 1,513

No correspondence, because gravitational time dilation is a function of potential difference not g. However if you assume Rindler observers, and your reference is a clock with acceleration of g, then the time rate for one ‘higher’ by h is faster by a factor of 1 + gh. This is also true to first order for the surface of planet.

[edit: in units with c=1. In common units, 1 + gh/c^{2} ]

[edit: in units with c=1. In common units, 1 + gh/c

Last edited:

- #4

- 261

- 9

- #5

- 3,598

- 1,434

There's no way to answer that question without more information.g. By what factor does B run slower than A?

For example, if you compare a clock sitting on the surface of the Earth to a clock sitting on the surface of a world with twice the radius and 4 times the mass, they will run at different rates (with the on on the larger world running slower) even though both clocks are at 1g.

- #6

- 7,306

- 6,385

It depends on the gravitational potential (usually denoted ##\phi##), not the gravitational acceleration (usually denoted ##g##). So your question has no answer as asked.g. By what factor does B run slower than A?

The rate at which a clock at Schwarzschild coordinate ##r## (assuming that it's outside the mass, therefore) ticks compared to a clock at infinity is ##\sqrt{1-2GM/c^2r}=\sqrt{1-2\phi/c^2}##. The approximations @PAllen gave derive from this under various circumstances.

Last edited:

- #7

- #8

- 7,306

- 6,385

Indeed. Start with a light pulse of frequency f at one height and send it upwards, convert it to a mass, drop the mass, and convert it back into energy. The light needs to have lost the same amount of energy on the upwards leg as the mass gained on the downwards leg, or else we have an energy-creating device here. Thus gravitational redshift, which is the same as gravitational time dilation.You actually do not need the general expression to derive the approximations. Just using the equivalence principle will work perfectly fine.

- #9

PAllen

Science Advisor

- 8,258

- 1,513

You can also just take the exact Rindler case, and note that by local Lorentz character of any GR manifold, that for a near stationary case in GR, it must be equivalent to first order to the Rindler case in SR.Indeed. Start with a light pulse of frequency f at one height and send it upwards, convert it to a mass, drop the mass, and convert it back into energy. The light needs to have lost the same amount of energy on the upwards leg as the mass gained on the downwards leg, or else we have an energy-creating device here. Thus gravitational redshift, which is the same as gravitational time dilation.

Note, the formulas I gave are exact for Rindler observers.

- #10

- 261

- 9

Ok. Thanks. Nice clear reply. I've got it now. Not as simple as I had thought.There's no way to answer that question without more information.

For example, if you compare a clock sitting on the surface of the Earth to a clock sitting on the surface of a world with twice the radius and 4 times the mass, they will run at different rates (with the on on the larger world running slower) even though both clocks are at 1g.