Raising and lowering indices of partial derivative

1. Apr 10, 2010

Derivator

Hi,

why can I raise and lower indices of a partial derivative with the help of the metric tensor?

E.g., wh is the following possible?
($$\phi$$ is a scalar function)

$$\partial^\mu \phi = g^{\mu\nu}\partial_\nu \phi$$

--
derivator

2. Apr 10, 2010

bcrowell

Staff Emeritus
You can't. You can only raise and lower indices on covariant derivatives, not partial derivatives.

3. Apr 10, 2010

clem

In special relativity, raising the index just changes the sign of the spacelike part.
This is necessary for $$\partial_\mu\partial^\mu$$ to be the D'Alembertian.

4. Apr 10, 2010

bcrowell

Staff Emeritus
One way to resolve the apparent contradiction between my #2 and your #3 is to note that in SR there is no distinction between partial derivatives and covariant derivatives. So we could say that the partial derivatives in your d'Alembertian example are "really" covariant derivatives.

At this point I think we could use some more clarification as to what Derivator had in mind with the question. Derivator, why does it seem surprising to you that the indices can be lowered and raised in this way?

5. Apr 10, 2010

clem

Yes, what I said works only for the Minkowski metric g.

6. Apr 10, 2010

George Jones

Staff Emeritus
In, for example, spherical coordinates there is a difference.

7. Apr 11, 2010

Derivator

well, its from general relativity context and its constantly used in our practice lessons...

8. Apr 11, 2010

bcrowell

Staff Emeritus
What I don't think you've made clear is why you consider it surprising or strange that it works this way.

9. Apr 17, 2010

Derivator

Isn't the covariant and partial derivative of a scalar function the same?

-> $$\partial^\mu \phi = g^{\mu\nu}\partial_\nu \phi$$
is correct if Phi is scalar.

10. Apr 17, 2010

lotm

My understanding was that

$$\partial^\mu \phi = g^{\mu\nu}\partial_\nu \phi$$

is used to define the contravariant derivative $$\nabla^{\mu}$$ ($$\nabla$$ rather than $$\partial$$ since we're in GR, though as you've said it reduces to $$\partial$$ in the case of a scalar function).

Given a differentiable manifold, we can consider one-form fields induced upon the manifold by scalar fields: given a scalar field $$\phi$$, we get a one-form field with components $$\partial_\mu \phi$$. We can also consider vector fields induced by curves through the manifold: given a curve $$\gamma: \lambda \in \Re \rightarrow p\in M$$, we get a vector field with components $$\frac{\partial x^\mu}{\partial \lambda}$$.

From this starting point, tensor products can be used to build tensor fields of higher valences (so in addition to the (0,1) fields - the one-forms - and the (1,0) fields - the vectors - we can get tensor fields of valence (m,n)). We then select some particular (0,2) tensor field, and decide that this shall be our metric g. g will take two vectors as arguments, and deliver a scalar. That is,

$$g(X,Y) = \chi$$

or if you prefer,

$$g_{\mu\nu} X^\mu Y^\nu = \chi$$

But that means that $$g_{\mu\nu}X^\mu$$ has, in effect, an empty argument place which could be filled by a vector; i.e. it is something which will map vectors to scalars - in other words, a one-form. So the notation $$X_\nu$$ is introduced as shorthand for $$g_{\mu\nu}X^\mu$$.

The same trick, using the inverse of the metric (i.e. the (2,0) tensor field such that $$g_{\mu\nu}g^{\nu\rho} = \delta^{\rho}_{\mu}$$) will allow you to link any one-form (components $$p_{\mu}$$) with a particular vector (components $$p^\mu$$). In particular, the one-form field with components $$\partial_\mu \phi$$ has an associated vector field $$\partial^\mu \phi$$, defined by

$$\partial^\mu \phi = g^{\mu\nu}\partial_\nu \phi$$

11. Apr 17, 2010

Derivator

hm, ok.

but if $$\partial$$ is used for the partial derivative, this is only true for scalar fields Phi. Right?

(and it's always true, if $$\partial$$ denotes the covariant derivative, in which case we should better use the symbol $$\nabla$$ )