# Quick one-line working on Geodesic Deviation Equation

1. Feb 28, 2015

### unscientific

Taken from my lecturer's notes on GR:

I'm trying to understand what goes on from 2nd to 3rd line:

$$N^\beta \nabla_\beta (T^\mu \nabla_\mu T^\alpha) - N^\beta \nabla_\beta T^\mu \nabla_\mu T^\alpha = -T^\beta \nabla_\beta N^\mu \nabla_\mu T^\alpha$$

Using commutator relation $T^v \nabla_v N^\alpha = N^v\nabla_v T^\alpha$ we swap the 'N' for the 'T' in the last term of the second line:

$$N^\beta \nabla_\beta (T^\mu \nabla_\mu T^\alpha) - T^\beta \nabla_\beta N^\mu \nabla_\mu T^\alpha = -T^\beta \nabla_\beta N^\mu \nabla_\mu T^\alpha$$

$$N^\beta \nabla_\beta \left(\frac{D T^\alpha}{D\lambda}\right) - T^\beta \nabla_\beta N^\mu \nabla_\mu T^\alpha = -T^\beta \nabla_\beta N^\mu \nabla_\mu T^\alpha$$

This means that $\nabla_\beta \left(\frac{D T^\alpha}{D\lambda}\right) = 0$. Why is this so?

2. Feb 28, 2015

### ChrisVer

May I ask what is T and N?
Nevertheless, it looks like the thing you have in the parenthesis $(T^a \nabla_a T^b)$ should be the geodesic equation?

3. Feb 28, 2015

### George Jones

Staff Emeritus
Use the product rule on the second term of the second line.

4. Feb 28, 2015

### unscientific

That would just take me back to the first line.

5. Feb 28, 2015

### unscientific

$T$ and $N$ are tangent vectors and normal vectors correspondingly, with $T$ describing movement along a geodesic and $N$ describing movement from one geodesic to another.

I thought the geodesic equation is $\frac{d^2x^\beta}{d\tau^2} + \Gamma^\beta _{\mu v} \frac{d x^\mu}{d\tau} \frac{d x^v}{d\tau}$?

I reduced the term in the brackets to $\nabla_\beta \left(\frac{D T^\alpha}{D\lambda}\right)$. I'm trying to understand why it is zero.

6. Feb 28, 2015

### ChrisVer

The geodesic equation by writting $U^a = \frac{dx^a}{d \tau}$ (the velocity vector/ is it tangent to the trajectory??) can be written in the form : $U^a D_a U^b=0$ (that's why that expression even though I didn't know what it stands for reminded me of the geodesics equation)

7. Feb 28, 2015

### unscientific

For a curve on space-time $x^\alpha = x^\alpha (\lambda)$, we define the absolute derivative of a vector $V^\mu$ along that path to be expression $\frac{D V^\mu}{D\lambda} = T^\alpha \nabla_\alpha V^\mu$. For $V^\mu = T^\mu$, we get $\frac{D T^\mu}{D\lambda} = T^\alpha \nabla_\alpha T^\mu$

If the vector $T^\mu$ is parallely transported along the path, $\frac{D T^\mu}{D\lambda} =0$. This is obviously true for the tangent vector $T^\alpha$ ,so $\frac{D T^\alpha}{D\lambda} =0$

Using $T^\alpha = \frac{dx^\alpha}{d\lambda}$ and expanding, it simply gives the geodesic equation:

$$\frac{d^2 x^\alpha}{d \lambda^2} + \Gamma^{\alpha}_{\mu v} \frac{dx^\mu}{d\lambda} \frac{dx^v}{d\lambda} = 0$$

Last edited: Feb 28, 2015
8. Feb 28, 2015

### George Jones

Staff Emeritus