General relativity question - geodesics.

[ghetom]

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:

\stackrel{d}{ds} \stackrel{\partial L}{\partial \dot{x^{\mu}}} - \stackrel{\partial L}{\partial x^{\mu}} =0

I get

\ddot{z} = 2g(1+gz)\dot{t}^2

And

\dot{t} = E/(1+gz)^2 ( where E is a constant )

So

\ddot{z} = g{E^2}/(1+gz)^3

which isn't right. Alternatively,

\ddot{z}= \dot{t}^2 z'' + \ddot{t}z' ( where ' is (d/dt) )

it's stationary, so z' = 0 thus z''=\ddot{z}/\dot{t}^2=2g(1+gz)

which still isn't right. Any takers?

 

[Cyosis]

I think the two in your geodesic equation shouldn't be there. You probably missed the 1/2 in the Christoffel symbol. Moreover you made a mistake in trying to find an expression for (dt/ d\tau)^2. Can you show us how you obtained your expression?

 

[ghetom]

Yes the 2 is wrong. My apologies.

For the mu = 0 equation I have:

(d/ds)(2\dot{t}(1+gz)^2) =0

thus

\dot{t} (1+gz)^2 = constant

also

\ddot{z} = \dot{t}d\dot{z}/dt = \dot{t}(d/dt)(\dot{t}z') = \dot{t}(z'(d\dot{t})/dt +\dot{t}z'')=\dot{t}^2z''+\ddot{t}z'

 

[ghetom]

Update: I've now solved it using the Christoffel symbols, but am still confused as to why the Euler-Lagrange method hasn't worked.This is wrong:
\ddot{z} = -g{\dot{t}^2}(1+gz) (1)

It should read:

\ddot{z} =-g{\dot{t}^2}/(1+gz) (2)

I got (1) direct from the EL equation

-2\ddot{z} - {\dot{t}^2}d_{z}(1+gz)^2 = 0 (3)

 

[Cyosis]

You're probably losing terms in places where you should be summing over the indices. Using the EL-equations I end up with \ddot{z}-g(1+gz)\dot{t}^2 =0.

Your second equation is wrong as well. Did you derive that one using the geodesic equation? Either way I just calculated it both ways and I get the same result, as it should be. I can't be of further assistance unless you show me your full derivation. Also I don't think you have gotten your \dot{t} correct yet.

 

[ghetom]

Thanks. Given that's the case then z''=\ddot{z}/\dot{t}^2=g(1+gz) can't be true (since this is the wrong answer).

I can't see any thing wrong for \dot{t}:

whats wrong with:
L = (1+gz)^2dt^2-dx^2-dy^2-dz^2)

therefore by EL:

(d/ds)(2\dot{t}(1+gz)^2) =0

so

\dot{t} (1+gz)^2 = constant

 

[Cyosis]

Your Lagrangian is wrong, recall L=g_{\alpha \beta}\dot{x}^\alpha\dot{x}^\beta=(1+gz)^2 \dot{t}^2-\dot{x}^2-\dot{y}^2-\dot{z}^2.