Gravitational Time Dilation: Math Formula for Clock-Slowing Factor

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SUMMARY

The mathematical formula for gravitational time dilation is derived from the Schwarzschild metric, specifically the equation ##\sqrt{1-2GM/c^2r}##, which compares the ticking rate of a clock at a gravitational potential ##\phi## to a clock at infinity. The discussion emphasizes that gravitational time dilation depends on the gravitational potential rather than acceleration, with the approximation of clocks in different gravitational fields being valid under the equivalence principle. The conversation also highlights the relationship between gravitational redshift and time dilation, confirming that both phenomena are interconnected.

PREREQUISITES
  • Understanding of the Schwarzschild metric in general relativity
  • Familiarity with gravitational potential and its notation (##\phi##)
  • Knowledge of the equivalence principle in physics
  • Basic concepts of gravitational redshift
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  • Study the derivation of the Schwarzschild metric in general relativity
  • Explore the implications of the equivalence principle in different gravitational scenarios
  • Learn about gravitational redshift and its mathematical formulation
  • Investigate the differences between Rindler observers and stationary observers in gravitational fields
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Physicists, students of general relativity, and anyone interested in the effects of gravity on time measurement and the underlying mathematical principles.

jeremyfiennes
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What is the mathematical formula for the time dilation (clock-slowing factor) for a clock in a gravitational field g, equivalent to the Lorentz factor γ for a clock traveling at a relative speed v?
 
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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/c2 ]
 
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Not quite with you. I have a clock A in outer space where there is no gravity. And one, B, stationary in a gravitational field g. By what factor does B run slower than A?
 
jeremyfiennes said:
Not quite with you. I have a clock A in outer space where there is no gravity. And one, B, stationary in a gravitational field g. By what factor does B run slower than A?
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.
 
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jeremyfiennes said:
Not quite with you. I have a clock A in outer space where there is no gravity. And one, B, stationary in a gravitational field g. By what factor does B run slower than A?
It depends on the gravitational potential (usually denoted ##\phi##), not the gravitational acceleration (usually denoted ##g##). So your question has no answer as asked.

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.
 
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Ibix said:
The approximations @PAllen gave derive from this under various circumstances.
You actually do not need the general expression to derive the approximations. Just using the equivalence principle will work perfectly fine.
 
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Orodruin said:
You actually do not need the general expression to derive the approximations. Just using the equivalence principle will work perfectly fine.
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.
 
Ibix said:
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 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.

Note, the formulas I gave are exact for Rindler observers.
 
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Janus said:
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.
Ok. Thanks. Nice clear reply. I've got it now. Not as simple as I had thought.
 

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