General Relativity - Circular Orbit around Earth

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Homework Help Overview

The discussion revolves around calculating proper time in different frames for a particle in a circular orbit around Earth, as described by general relativity. The problem involves understanding the metric for spacetime and the implications of light traveling in this context.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the metric for circular orbits and question the implications of differentials in polar coordinates. There are discussions about the relationship between proper time and coordinate time, particularly for light pulses.

Discussion Status

Several participants have provided insights into the mathematical relationships involved, particularly regarding the metric and the implications of light traveling at speed c. There is ongoing exploration of how proper time relates to the different frames of reference, with some participants questioning assumptions and clarifying definitions.

Contextual Notes

Participants note the constraints of the problem, including the need to consider the effects of general relativity and the specific conditions of circular orbits. There is an emphasis on the conceptual understanding of time experienced by different observers, particularly in the context of light and proper time.

  • #31
Recall that the correct expression for the light pulse is ds^2 = -\left( 1 - \frac{2GM}{c^2r} \right) c^2 dt^2 + \left( 1 + \frac{2GM}{c^2r} \right) dr^2

##ds = 0##, but ##dt \neq 0## and ##dr \neq 0##.

But you just need to find ##\Delta \tau = \int d\tau## for the total round trip of the light. So, you don't need to worry about ##dt## or ##dr##.
 
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  • #32
TSny said:
Recall that the correct expression for the light pulse is ds^2 = -\left( 1 - \frac{2GM}{c^2r} \right) c^2 dt^2 + \left( 1 + \frac{2GM}{c^2r} \right) dr^2

##ds = 0##, but ##dt \neq 0## and ##dr \neq 0##.

But you just need to find ##\Delta \tau = \int d\tau## for the total round trip of the light. So, you don't need to worry about ##dt## or ##dr##.

Yeah, but ##d\tau = 0## so ##\int d\tau = 0##. So from the light's perspective, no time has passed at all. Does this even make sense?
 
  • #33
unscientific said:
Yeah, but ##d\tau = 0## so ##\int d\tau = 0##. So from the light's perspective, no time has passed at all.
That's right. Of course, no physical observer can travel at the speed of light. But, in a thought experiment, you can imagine the observer moving at 0.99999...c (add as many 9's as you wish). The time elapsed for this observer will approach zero. It's the ultimate "twin paradox" scenario where the traveling twin would not age at all!
Does this even make sense?
It's strange, but it's a natural consequence of SR and GR.
 
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