# Partial derivative of $x^2$

Tags:
1. Aug 31, 2015

### shooride

How can I figure out $\partial_\mu x^2$ on the manifold $(M,g)$? I thought that it should be $2x_\mu$, but I think I'm wrong and the answer is $2x_\mu+x^\nu x^\lambda \partial_\mu g_{\nu\lambda}$, right?! In particular, it seems to me, we can't write $\partial_\mu=g_{\mu\nu}\partial^{\nu}$. However, we can raise or lower the indices of the covariant derivative with metric, I mean $\nabla^\mu=g^{\mu\nu}\nabla_\nu$. Is this true because partial derivative is not a tensor but covariant derivative is? Could you please explain it to me?

2. Aug 31, 2015

### Orodruin

Staff Emeritus
What do you mean by $x^2$? If $x^\mu$ are coordinates, then $x^2$ is a coordinate dependent statement and you must specify the coordinate system in which this is true in order to make any sense of it. Generally, there is no such thing as a position vector on a manifold.

3. Aug 31, 2015

### shooride

Can one consider $x^2$ as the distance between two points $x^\mu$ and $x^\mu_0=(0,0,\dots)$?

4. Aug 31, 2015

### Orodruin

Staff Emeritus
No.

5. Aug 31, 2015

### shooride

Sorry, I meant square of distance! $(x^\mu x_\mu)^{1/2}$ is just the Euclidean distance?!

6. Aug 31, 2015

### Orodruin

Staff Emeritus
There is no such concept in a general manifold.