# Is this valid?

1. Oct 12, 2011

### Gunthi

$$\nabla_a[(-g)^{\frac{1}{2}}T^a] = T^a\nabla_a[(-g)^{\frac{1}{2}}]+(-g)^{\frac{1}{2}}\nabla_aT^a$$

I just realized that I don't quite understand how a tensor density behaves when multiplied by a vector. I'm trying to find some clues in D'Inverno's book but I'm getting more confused.

2. Oct 12, 2011

### WannabeNewton

$\triangledown _{\mu }(-g)^{1/2}= 0$ so you can pull it out of the covariant derivative. The main difference between tensor densities and tensors with regards to the covariant derivative is that with the former you have an extra term involving the weight of the density.

3. Oct 12, 2011

### Gunthi

So the covariant derivative is distributive like the Lie derivative? How could I prove that?

Thanks

4. Oct 12, 2011

### WannabeNewton

Yes $\triangledown _{\alpha }(S\otimes T) = T\otimes (\triangledown_{\alpha }S ) + S\otimes (\triangledown _{\alpha }T)$. One rather easy way (but a tad bit mechanical) to prove it, in component form, would be to take the general definition of the covariant derivative in terms of the christoffel symbol and use the fact that the product of an (m,n) tensor with an (k, l) tensor gives some (m + k, n + l) tensor.

5. Oct 12, 2011

### Sam Gralla

You're just asking about the Leibnitz rule. I'm pretty sure every derivative ever invented satisfies the Liebnitz rule. (If it didn't, I wouldn't want to call it a derivative.)

For example see Wald where he defines derivatives as maps on tensors that satisfy a number of properties, among them the Leibnitz rule. (The covariant derivative is one such derivative. And keep in mind that from Wald's point of view tensor densities are just tensors chosen with respect to particular coordinate systems.)