# General relativity question - geodesics.

1. Apr 27, 2010

### 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?

Last edited: Apr 27, 2010
2. Apr 27, 2010

### 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?

Last edited: Apr 27, 2010
3. Apr 27, 2010

### 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

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

Last edited: Apr 27, 2010
4. Apr 27, 2010

### 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)

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

I got (1) direct from the EL equation

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

Last edited: Apr 27, 2010
5. Apr 27, 2010

### 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.

Last edited: Apr 27, 2010
6. Apr 27, 2010

### 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

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

7. Apr 27, 2010

### Cyosis

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

Last edited: Apr 27, 2010