General Relativity - Circular Orbit around Earth

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Homework Statement



(a) Find the proper time in the rest frame of particle
(b) Find the proper time in the laboratory frame
(c) Find the proper time in a photon that travels from A to B in time P


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Homework Equations




The Attempt at a Solution



Part(a)
[/B]
The metric is given by:

[tex] ds^2 = -\left( 1 - \frac{2GM}{c^2r} \right) c^2 dt^2 + \left( 1 + \frac{2GM}{c^2r} \right) dr^2 [/tex]

Circular orbit implies that ##dr^2 = 0##, so
[tex]ds^2 = c^2 d\tau^2 = -\left( 1 - \frac{2GM}{c^2R} \right) c^2 dt^2 [/tex]
[tex] \left( \frac{d\tau}{dt} \right)^2 = \left( 1 - \frac{2GM}{c^2R} \right) [/tex]
[tex] \frac{d\tau}{dt} = \sqrt { \left( 1 - \frac{2GM}{c^2R} \right) } [/tex]

Since the time between event A and B is ##dt = P##, the time experienced in the rest frame must be ## d\tau = \sqrt { \left( 1 - \frac{2GM}{c^2R} \right) } P ##?

I'm not sure how to approach parts (b) and (c)..
 

Answers and Replies

  • #2
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The metric is given by:

[tex] ds^2 = -\left( 1 - \frac{2GM}{c^2r} \right) c^2 dt^2 + \left( 1 + \frac{2GM}{c^2r} \right) dr^2 [/tex]

Circular orbit implies that ##dr^2 = 0##
Does ##dx^2 + dy^2 +dz^2 = 0## for two neighboring points on the orbit?
 
  • #3
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Does ##dx^2 + dy^2 +dz^2 = 0## for two neighboring points on the orbit?
They are at the same radius, so yes.
 
  • #4
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They are at the same radius, so yes.
For general motion in the xy plane (z = 0), how would you express ##dx^2+dy^2## in polar coordinates ##(r, \theta)##?
 
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For general motion in the xy plane (z = 0), how would you express ##dx^2+dy^2## in polar coordinates ##(r, \theta)##?
[tex] r^2sin^2 \theta d\phi [/tex]
 
  • #8
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On the circular orbit of radius R, what is the value of little r?
 
  • #9
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On the circular orbit of radius R, what is the value of little r?
r = R
 
  • #10
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Right.
 
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[tex]ds^2 = -\left( 1 - \frac{2GM}{c^2R} \right) c^2 dt^2 + \left( 1 + \frac{2GM}{c^2r} \right) R^2 d\phi^2 [/tex]

Using ## d\phi = \omega dt = \frac{2\pi}{P} dt ##

[tex] ds^2 = -\left( 1 - \frac{2GM}{c^2R} \right) c^2 dt^2 + \left( 1 + \frac{2GM}{c^2 R} \right) R^2 (\frac{2\pi}{P})^2 dt^2[/tex]
 
  • #12
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Not quite. Does ##R^2P^2dt^2## have the correct dimensions of distance squared?
 
  • #13
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Not quite. Does ##R^2P^2dt^2## have the correct dimensions of distance squared?
Using ## d\phi = \omega dt = \frac{2\pi}{P} dt ##,

[tex] ds^2 = -\left( 1 - \frac{2GM}{c^2R} \right) c^2 dt^2 + \left( 1 + \frac{2GM}{c^2 R} \right) R^2 (\frac{2\pi}{P})^2 dt^2 [/tex]
[tex] -c^2 d\tau^2 = -\left( 1 - \frac{2GM}{c^2R} \right) c^2 dt^2 + \left( 1 + \frac{2GM}{c^2 R} \right) R^2 (\frac{2\pi}{P})^2 dt^2 [/tex]
[tex] \left( \frac{d\tau}{dt} \right)^2 = \left( 1 - \frac{2GM}{c^2R} \right) - \left( 1 + \frac{2GM}{c^2 R} \right) R^2 \left(\frac{2\pi}{Pc}\right)^2 [/tex]
 
  • #14
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Looks good.
 
  • #16
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What is ##dx^2 + dy^2 + dz^2## for a fixed point on the orbit?
 
  • #17
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What is ##dx^2 + dy^2 + dz^2## for a fixed point on the orbit?
For a fixed point, it is ##0##, so ##\frac{d\tau}{dt} = 1 - \frac{2GM}{c^2R}##.
 
  • #18
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...and ##ds^2 = ##?
 
  • #19
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...and ##ds^2 = ##?
Solving, we get ## \frac{d\tau}{dt} = 1 - \frac{2GM}{c^2R} ##.
 
  • #20
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For part (c), since beam of light travels at speed ##c## in all frames, is ##dr^2 = c^2 dt^2 = c^2 P^2##?
 
  • #21
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For part (c), is ##dr^2 = c^2 dt^2##?
No. ##dr/dt## for the radially traveling light pulse is just a "coordinate speed of light" and need not equal c.

But no matter what the coordinate speed of light, what is ##ds## for any two neighboring space-time events on the worldline of a light pulse?
 
  • #22
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No. ##dr/dt## for a light pulse is just a "coordinate speed of light" and need not equal c.

But no matter what the coordinate speed of light, what is ##ds## for any two neighboring space-time events on the worldline of a light pulse?
Time-like, no rest frame. so ##ds = 0##.
 
  • #23
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It's neither spacelike or timelike. It's "lightlike". Yes, ds = 0.
 
Last edited:
  • #24
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Yes.
So does this mean the equation is simply:

[tex] 0 = -\left( 1 - \frac{2GM}{c^2R} \right) c^2 dt^2 + \left( 1 + \frac{2GM}{c^2r} \right) R^2 d\phi^2 [/tex]
[tex] 0 = -\left( 1 - \frac{2GM}{c^2R} \right) c^2 dt^2 [/tex]
 
  • #25
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The light is traveling radially. So, what is ##dx^2 + dy^2 + dz^2## in polar coordinates?

But, you are interested in the proper time. How is ##ds## related to ##d\tau##?
 

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