Tensor algebra with derivative of the metric

[jmlaniel]

I am trying to proove that the following relation:

$$A_{\nu}$$ $$\partial_{\mu}$$ $$\partial^{\nu}$$ $$A^{\mu}$$ = $$A_{\nu}$$ $$\partial^{\mu}$$ $$\partial^{\nu}$$ $$A_{\mu}$$

The only way I found is by setting:

$$A_{\nu}$$ $$\partial_{\mu}$$ $$\partial^{\nu}$$ $$A^{\mu}$$ = $$A_{\nu}$$ $$\partial_{\mu}$$ $$\partial^{\nu}$$ $$g^{\mu \sigma}$$ $$A_{\sigma}$$

Then by imposing that the metric is the Minkowski metric (all constant), their derivatives are all equal to zero so I can get the metric out of the derivative and get :

$$A_{\nu}$$ $$\partial_{\mu}$$ $$\partial^{\nu}$$ $$A^{\mu}$$ = $$A_{\nu}$$ $$g^{\mu \sigma}$$ $$\partial_{\mu}$$ $$\partial^{\nu}$$ $$A_{\sigma}$$

Then by contracting the metric with the first derivative and renaming the indices, I can achieve my goal.

My question is the following one : can I achieve this without imposing the Minkowsky metric?

Thank!

 

[Bill_K]

These are not tensor quantities. To have an expression which does not depend on g_uv being constant (Minkowski or otherwise) you need to use covariant derivatives instead of partial derivatives.

 

[LAHLH]

Are you sure your expression holds in curved spacetime?

It is certainly true that $$A_{\nu}\nabla_{\mu}\nabla^{\nu}A^{\mu}=A_{\nu}\nabla^{\mu}\nabla^{\nu}A_{\mu}$$ which follows from metric compatability: $$\nabla_{\alpha}g^{\mu\nu}=0$$

and I thought by some freak cancellations your expression may hold for a general metric too, but I can't see it after a brief look.

You have contracted out all indices on both sides of your equations, meaning that you have a valid tensor equation( scalar on the LHS and RHS), and since your equation is valid in Minkowski, the comma goes to semi colon rule suggests the generalisation is $$A_{\nu}\nabla_{\mu}\nabla^{\nu}A^{\mu}=A_{\nu}\nabla^{\mu}\nabla^{\nu}A_{\mu}$$

 

[Bill_K]

Just because all the indices are contracted doesn't mean it's a covariant expression. All he's doing is raising and lowering μ, and to do this he needs to say the derivative of g_μν is zero. Which of course it is if covariant derivatives are used instead of partials.

 

[jmlaniel]

I agree that my title might be misleading on the fact that my particular term is actually not a tensor by definition. It is actually a part of a larger derivation based on tensor algebra. It is actually related to the lagrangian of the electromagnetic field.

I can also confirm that the derivatives in my expression are not covariant derivative (this would simplify my problem).

From what I gather from your comments, I see no other way to impose the Minkowsky metric in order to obtain the desired result.

On the other hand, since this problem is related to electromagnetism, is it directly implied that the metric is a Minkowsky one? Or is it possible to use a different metric in an electromagnetism situation?

 

[LAHLH]

Just because all the indices are contracted doesn't mean it's a covariant expression. All he's doing is raising and lowering μ, and to do this he needs to say the derivative of g_μν is zero. Which of course it is if covariant derivatives are used instead of partials.

I didn't mean to suggest it was a covariant expression, only that it is a tensor equation in Minkowski space where it holds, hence the comma goes to semicolon rule suggests $$A_{\nu}\nabla_{\mu}\nabla^{\nu}A^{\mu}=A_{\nu}\nabla^{\mu}\nabla^{\nu}A_{\mu}$$ will hold in a general curved space (which it does via met compat). Similar to how $$\partial_{\mu}T^{\mu\nu}=0$$ in flat implies $$\nabla_{\mu}T^{\mu\nu}=0$$ will hold in curved.

I was then curious to see if since $$A_{\nu}\nabla_{\mu}\nabla^{\nu}A^{\mu}=A_{\nu}\nabla^{\mu}\nabla^{\nu}A_{\mu}$$ will hold in a general spacetime, maybe when one expanded the covariant derivatives and what not some cancellations happened to occur meaning things just happened to reduce to the OP form, but I couldn't see this.

 

[atyy]

On the other hand, since this problem is related to electromagnetism, is it directly implied that the metric is a Minkowsky one? Or is it possible to use a different metric in an electromagnetism situation?

It's possible to use metrics with Maxwell's Equations other than the Minkowski's flat spacetime. However, even in flat spacetime, if partial rather than covariant derivatives are used, a given expression will be valid in only one coordinate frame (if the expression is Lorentz covariant, it will be valid in any Lorentz inertial frame, but not in non-inertial frames).

 

[WannabeNewton]

Why not start with $$A_{\nu }A^{\mu },^{\nu }_{\mu } = A_{\nu }g_{\sigma \mu }A^{\mu },^{\nu \sigma }$$

 

[jmlaniel]

I have the felling that we got lost somewhere along the way... I might have been misleading in my in two posts.

I just want to know if my algebra is correct and that I have to assume that the derivative (normal one and not covariant) of the metric must be zero in order to get the metric factor out of the two derivatives. My question is more an algebraic one more than a discussion on covariant derivative or what is a tensor.

Can anyone comment on this?

 

[LAHLH]

I have the felling that we got lost somewhere along the way... I might have been misleading in my in two posts.

I just want to know if my algebra is correct and that I have to assume that the derivative (normal one and not covariant) of the metric must be zero in order to get the metric factor out of the two derivatives. My question is more an algebraic one more than a discussion on covariant derivative or what is a tensor.

Can anyone comment on this?

You can only pass the metric through a partial derivative if the derivative of the metric is zero (i.e. flat space). So a priori the relation you posted only holds in flat space.

If one was in a curved space, one can go into a locally inertial frame where your relation holds too, and because in flat space the christoffel symbols vanish and $$\partial_{\mu}=\nabla_{\mu}$$ one can write the relation as $$A_{\nu}\nabla_{\mu}\nabla^{\nu}A^{\mu}=A_{\nu}\nabla^{\mu}\nabla^{\nu}A_{\mu}$$, but then this is a tensor relation and holds in any coordinates, so holds for a general spacetime (this is the comma goes to semi colon rule).

So if one is in a curved spacetime the relation to start with is $$A_{\nu}\nabla_{\mu}\nabla^{\nu}A^{\mu}=A_{\nu}\nabla^{\mu}\nabla^{\nu}A_{\mu}$$...this only reduces to what you have in the case of a flat space. (My first thought was maybe christoffel and things from this could cancel each other out freakishly and just so happen to the relation with partials hold in general, but I think not)

So in summary, do you have to assume flat space for your relation to hold? yes. Do you have to assume the derivative of metric is zero to pull things out and pass through things in the way your relation assumes? yes.