- #1
ghetom
- 17
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I'm doing some revision for a General relativity exam, and came across this question:
A Flat Earth space-time has co-ordinates (t, x, y, z), where z > 0, and a metric
ds2 = ((1 + gz)2)dt2 − dx2 − dy2 − dz2
where g is a positive constant.
Write down the geodesic equations in this space-time.
Hence, or otherwise, show that A Flat Earth physicist, stationary at a point with z = h, will measure a ‘gravitational acceleration’ of magnitude g(1 + gh)−1
---//---
I've solved the geodesics, but can't get the 'show that' at the end:
Using the Euler-Lagrange equation:
[tex] \stackrel{d}{ds} \stackrel{\partial L}{\partial \dot{x^{\mu}}} - \stackrel{\partial L}{\partial x^{\mu}} =0[/tex]
I get
[tex]\ddot{z} = 2g(1+gz)\dot{t}^2[/tex]
And
[tex]\dot{t} = E/(1+gz)^2[/tex] ( where E is a constant )
So
[tex]\ddot{z} = g{E^2}/(1+gz)^3[/tex]
which isn't right. Alternatively,
[tex]\ddot{z}= \dot{t}^2 z'' + \ddot{t}z'[/tex] ( where ' is (d/dt) )
it's stationary, so z' = 0 thus [tex]z''=\ddot{z}/\dot{t}^2=2g(1+gz)[/tex]
which still isn't right. Any takers?
A Flat Earth space-time has co-ordinates (t, x, y, z), where z > 0, and a metric
ds2 = ((1 + gz)2)dt2 − dx2 − dy2 − dz2
where g is a positive constant.
Write down the geodesic equations in this space-time.
Hence, or otherwise, show that A Flat Earth physicist, stationary at a point with z = h, will measure a ‘gravitational acceleration’ of magnitude g(1 + gh)−1
---//---
I've solved the geodesics, but can't get the 'show that' at the end:
Using the Euler-Lagrange equation:
[tex] \stackrel{d}{ds} \stackrel{\partial L}{\partial \dot{x^{\mu}}} - \stackrel{\partial L}{\partial x^{\mu}} =0[/tex]
I get
[tex]\ddot{z} = 2g(1+gz)\dot{t}^2[/tex]
And
[tex]\dot{t} = E/(1+gz)^2[/tex] ( where E is a constant )
So
[tex]\ddot{z} = g{E^2}/(1+gz)^3[/tex]
which isn't right. Alternatively,
[tex]\ddot{z}= \dot{t}^2 z'' + \ddot{t}z'[/tex] ( where ' is (d/dt) )
it's stationary, so z' = 0 thus [tex]z''=\ddot{z}/\dot{t}^2=2g(1+gz)[/tex]
which still isn't right. Any takers?
Last edited: