# Covariant Derivative - where does the minus sign come from?

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1. Apr 20, 2015

### unscientific

I was reading through hobson and my notes where the covariant acts on contravariant and covariant tensors as
$$\nabla_\alpha V^\mu = \partial_\alpha V^\mu + \Gamma^\mu_{\alpha \gamma} V^\gamma$$
$$\nabla_\alpha V_\mu = \partial_\alpha V_\mu - \Gamma^\gamma_{\alpha \mu} V_\gamma$$

I mean, if you simply multiply $g_{\mu \gamma}$ on both side of the equation:
$$\nabla_\alpha V^\mu g_{\mu \gamma} = \partial_\alpha V^\mu g_{\mu \gamma} + \Gamma^\mu_{\alpha \gamma} V^\gamma g_{\mu \gamma}$$
$$\nabla_\alpha V_{\gamma} = \partial_\alpha V_\gamma + \Gamma^\mu_{\alpha \gamma} V_\mu$$
Swapping $\mu$ and $\gamma$ we have
$$\nabla_\alpha V_{\mu} = \partial_\alpha V_\mu + \Gamma^\gamma_{\alpha \mu} V_\gamma$$

2. Apr 20, 2015

### Staff: Mentor

Because of the different transformation laws that are obeyed by vectors and covectors, i.e., by upper and lower index quantities.

Last edited: Apr 20, 2015
3. Apr 20, 2015

### Staff: Mentor

I don't know the answer to your question, but this part is wrong. You cannot have one upstairs and two downstairs repeated indexes.

4. Apr 20, 2015

### unscientific

I tried to look up a derivation, but couldn't find it anywhere, so I'm asking if anyone has seen a proof or working that starts from the first equation and shows how the minus sign appears.

5. Apr 20, 2015

### JorisL

Carroll's notes seem to treat this in quite some detail.
http://preposterousuniverse.com/grnotes/grnotes-three.pdf [Broken]

I haven't thoroughly checked them since I used the book that came from those.

However I suspect these show even more steps if memory serves me.

P.S. try checking those and come back if you have problems

Last edited by a moderator: May 7, 2017
6. Apr 20, 2015

### Mentz114

Good answer. Carrolls explanation is very well put together.

Last edited by a moderator: May 7, 2017
7. Apr 20, 2015

### Matterwave

Notice that there is a summation for $\gamma$ on the vector and Christoffel symbols in the original formula (as Dalespam said, this is invalid). Multiplying by $g_{\mu\gamma}$ will not work. What you can do is introduce another index and multiply by $g_{\mu\tau}$ For the second term though, you actually lower the Christoffel symbol index instead of the vector index so you will get: $$\nabla_\alpha V_\mu = \partial_\alpha V_\mu +\Gamma_{\mu\alpha\gamma}V^\gamma$$

As you can see, now you are dealing with a Christoffel symbol of the first kind ($\Gamma_{\mu\nu\gamma}\equiv g_{\mu\tau}\Gamma^{\tau}_{~~\nu\gamma}$) instead of a Christoffel symbol of the second kind that we are usually more familiar with.

EDIT: WARNING: There's an error in my first term.

Last edited: Apr 26, 2015
8. Apr 21, 2015

### Orodruin

Staff Emeritus
For some intuition, I usually first try to tell students how it works in curvilinear coordinates on the flat $\mathbb R^n$. Given a set of curvilinear coordinates $x^\mu$ we can define two sets of basis vectors $\vec E_\mu = \partial \vec r /\partial x^\mu$ and $\vec E^\mu = \nabla x^\mu$ (where $\nabla$ is the gradient and $x^\mu$ a coordinate function). These bases are neither orthogonal nor normalised, but they do obey $\vec E_\nu \cdot \vec E^\mu = \delta^\mu_\nu$ and will generalise into the coordinate bases for vectors and covectors. The Christoffel symbols are given by how the basis changes with respect to the coordinates (since this is $\mathbb R^n$, we can compare vectors at different points without problem) $\vec E^\sigma\cdot \partial_\mu \vec E_\nu = \Gamma^\sigma_{\mu\nu}$, i.e., $\partial_\mu \vec E_\nu = \Gamma^\sigma_{\mu\nu} \vec E_\sigma$. But at the same time we have
$$0 = \partial_\mu \vec E^\sigma \cdot \vec E_\nu = \vec E^\sigma \cdot \partial_\mu \vec E_\nu + \vec E_\nu \cdot \partial_\mu \vec E^\sigma = \Gamma^\sigma_{\mu\nu} + \vec E_\nu \cdot \partial_\mu \vec E^\sigma.$$
It follows that $\vec E_\nu \cdot \partial_\mu \vec E^\sigma = - \Gamma^\sigma_{\mu\nu}$, or in other terms $\partial_\mu \vec E^\sigma = -\Gamma^\sigma_{\mu\nu} \vec E^\nu$ and there you have your minus sign.

This generalises readily to curved spaces and the absolutely easiest way to see that it must be like that is to apply the covariant derivative to the scalar field $V_\mu W^\mu$ for which it reduces to a partial derivative. At the same time, the covariant derivative should obey the product rule and we would have
$$\nabla_\sigma(V_\mu W^\mu) = (\nabla_\sigma V_\mu) W^\mu + V_\mu \nabla_\sigma W^\mu = W^\mu \nabla_\sigma V_\mu + V_\mu \partial_\sigma W^\mu + \Gamma^\mu_{\rho\nu} V_\mu W^\rho = V_\mu \partial_\sigma W^\mu + W^\mu \partial_\sigma V_\mu.$$
The only way of satisfying this for arbitrary vector fields is if
$$\nabla_\sigma V_\mu = \partial_\sigma V_\mu - \Gamma^\rho_{\mu\nu} V_\rho.$$

9. Apr 24, 2015

### samalkhaiat

1) You need to be careful with the indices.

2) It is true that $$\left( \nabla_{\alpha} V^{\mu}\right) g_{\mu \tau} = \nabla_{\alpha} \left( V^{\mu} g_{\mu \tau} \right) = \nabla_{\alpha} V_{\tau} ,$$ because $\nabla_{\alpha} g_{\mu \nu} = 0$.

3) But $$\partial_{\alpha} ( V^{\mu} g_{\mu \tau} ) \neq ( \partial_{\alpha} V^{\mu} ) g_{\mu \tau} = \partial_{\alpha} ( V^{\mu} g_{\mu \tau} ) - V^{\mu} \partial_{\alpha} g_{\mu \tau} .$$ So, when you contract with $g_{\mu \tau}$, you get $$\nabla_{\alpha} V_{\tau} =\partial_{\alpha} V_{\tau} + \left( g_{\mu \tau} \Gamma^{\mu}_{\alpha \gamma} - \partial_{\alpha} g_{\tau \gamma} \right) g^{\gamma \beta} V_{\beta} .$$ Now the minus sign show up, because $$g^{\gamma \beta} \left( g_{\mu \tau} \Gamma^{\mu}_{\alpha \gamma} - \partial_{\alpha} g_{\tau \gamma} \right) = - \Gamma^{\beta}_{\alpha \tau} .$$

10. Apr 26, 2015

### Matterwave

Ah whoops, Sam's answer is correct. I was too careless in mine. Please disregard. Thanks. :)