- #1
stallm
- 7
- 0
Hi
In the Schwarzschild metric, the proper time is given by
[itex]c^{2}dτ^{2} = (1- \frac{2\Phi}{c^2})c^2 dt^2 - r^2 dθ^2[/itex]
with where [itex]\Phi[/itex] is the gravitational potential. I have left out the [itex]d\phi[/itex] and [itex]dr[/itex] terms.
If there is a particle moving in a circle of radius [itex] R [/itex] at constant angular velocity ω, the proper time experienced by the particle completing one whole orbit will be
[itex] τ = (1- \frac{2\Phi(R)}{c^2} - R^2 \frac{ω^2}{c^2}) t [/itex]
where t is the period of the orbit. If the particle were to remain stationary for time t instead it would experience proper time
[itex] τ = (1- \frac{2\Phi(R)}{c^2}) t [/itex]
Which is more than the value for the circular orbit. Since the circular motion is inertial and the stationary motion isn't, doesn't this violate the principle of most proper time?
Thanks
In the Schwarzschild metric, the proper time is given by
[itex]c^{2}dτ^{2} = (1- \frac{2\Phi}{c^2})c^2 dt^2 - r^2 dθ^2[/itex]
with where [itex]\Phi[/itex] is the gravitational potential. I have left out the [itex]d\phi[/itex] and [itex]dr[/itex] terms.
If there is a particle moving in a circle of radius [itex] R [/itex] at constant angular velocity ω, the proper time experienced by the particle completing one whole orbit will be
[itex] τ = (1- \frac{2\Phi(R)}{c^2} - R^2 \frac{ω^2}{c^2}) t [/itex]
where t is the period of the orbit. If the particle were to remain stationary for time t instead it would experience proper time
[itex] τ = (1- \frac{2\Phi(R)}{c^2}) t [/itex]
Which is more than the value for the circular orbit. Since the circular motion is inertial and the stationary motion isn't, doesn't this violate the principle of most proper time?
Thanks