I don't know if the following is correct, i want to compute the following derivative

$$\frac{\partial }{\partial (\partial_{\mu}A_{\nu})}\left(\partial^{\alpha}A^{\beta}\partial_{\alpha}A_{\beta} \right)$$

So I first introduce the metric in order to lower the first two indices

$$\partial^{\alpha}A^{\beta}\partial_{\alpha}A_{\beta}=\eta^{\alpha\gamma}\eta^{\beta\delta}\partial_{\gamma}A_{\delta}\partial_{\alpha}A_{\beta} $$

and then

$$\begin{align}\frac{\partial }{\partial (\partial_{\mu}A_{\nu})}\left(\eta^{\alpha\gamma}\eta^{\beta\delta}\partial_{\gamma}A_{\delta}\partial_{\alpha}A_{\beta}\right) &= \eta^{\alpha\gamma}\eta^{\beta\delta}\frac{\partial }{\partial (\partial_{\mu}A_{\nu})}\left(\partial_{\gamma}A_{\delta}\partial_{\alpha}A_{\beta}\right) \\ &=\eta^{\alpha\gamma}\eta^{\beta\delta}\left(\partial_{\alpha}A_{\beta}\delta_{\mu \gamma}\delta_{\delta \nu}+\partial_{\gamma}A_{\delta}\delta_{\alpha\mu}\delta_{\beta\nu}\right) \\ &=\eta^{\alpha\mu}\eta^{\beta\nu}\partial_{\alpha}A_{\beta}+\eta^{\mu\gamma}\eta^{\nu\delta}\partial_{\gamma}A_{\delta} \\ &=2\partial^{\mu}A^{\nu}\end{align}$$

I was troubled because in the metric sometimes the first index is the summed one and other times the second. Is it correct that $A^{\alpha}=\eta^{\alpha \beta}A_{\beta}=\eta^{\beta \alpha}A_{\beta}$?

  • 7
    $\begingroup$ Yup you got it! $\endgroup$ Mar 8, 2013 at 20:49
  • 1
    $\begingroup$ Thanks, I was troubled because in the metric sometimes the first index is the summed one and other times the second. Is this right? $A^{\alpha}=\eta^{\alpha \beta}A_{\beta}=\eta^{\beta \alpha}A_{\beta}$ $\endgroup$
    – J L
    Mar 8, 2013 at 21:12
  • 1
    $\begingroup$ Yes; the Minkowski metric is symmetric, so you can switch the two indices. $\endgroup$ Mar 8, 2013 at 21:22
  • 3
    $\begingroup$ Hi Nivalth - questions that just ask if your work is correct or not aren't really the kind of thing we do here, but luckily you mentioned what was confusing you in a comment, so I edited it into the question. @joshphysics you might want to turn your comments into an answer. $\endgroup$
    – David Z
    Mar 10, 2013 at 5:33

1 Answer 1


Well, as the people of this answer had said your calculations are quite right. But I would like to write a quick answer about metric tensor.

I) An Geometric Intuition

The metric tensor is (roughly speaking) a bilinear map which produces a particular scalar called a line element, which is simply the value of the norm of differential line element vectors, i.e.

enter image description here

$$ ds^{2}\equiv g\Bigg(dx^{\mu}\frac{\partial \vec{r}}{\partial x^{\mu}},dx^{\nu}\frac{\partial \vec{r}}{\partial x^{\nu}}\Bigg) := \|d\vec{r}\|^{2} =: \langle d\vec{r},d\vec{r}\rangle = \sum^{3}_{\mu=0}\sum^{3}_{\nu=0} g_{\mu\nu}dx^{\mu}dx^{\nu} \tag{3}$$

Now, in general metric tensors aren't easy matrices like $\delta_{ij}$ and $\eta_{\mu\nu}$. In fact the metric tensor can become a tensor field which varies through space (and then the geometry varies pointwise).

To describe this general behaviour of "a tensor field which varies through space (and then the geometry varies pointwise)" we need the manifold mathematical framework (which is beyond the scope of this answer).

But, now, suppose you have then the metric tensor:

$$ ds^{2}= \sum^{3}_{\nu=0}\sum^{3}_{\mu=0} g_{\nu\mu}dx^{\nu}dx^{\mu} \tag{4}$$

Then you can say without any doubts that:

$$ds^{2} = \langle d\vec{r},d\vec{r}\rangle =\sum^{3}_{\nu=0}\sum^{3}_{\mu=0} g_{\nu\mu}dx^{\nu}dx^{\mu} \tag{5}$$

But you already know that,

$$\langle d\vec{r},d\vec{r}\rangle = \sum^{3}_{\mu=0}\sum^{3}_{\nu=0} g_{\mu\nu}dx^{\mu}dx^{\nu} \tag{6}$$

Then, you can conclude:

$$ ds^{2} = \langle d\vec{r},d\vec{r}\rangle = \sum^{3}_{\mu=0}\sum^{3}_{\nu=0} g_{\mu\nu}dx^{\mu}dx^{\nu} = \sum^{3}_{\nu=0}\sum^{3}_{\mu=0} g_{\nu\mu}dx^{\nu}dx^{\mu} = \langle d\vec{r},d\vec{r}\rangle = ds^{2} \tag{7} $$

And then:

$$ g_{\mu\nu} = g_{\nu \mu} \tag{8}$$

Otherwise the geometry defined by the metric tensor will be "different", i.e., different line elements for a permute of metric tensor indexes $ g_{\mu\nu} \neq g_{\nu\mu} $ (which if not true indeed).

In particular we have for the Minkowski components of metric tensor: $\eta_{\mu\nu} = \eta_{\nu\mu}$

II) A non-trivial example

Now, consider then the line element of Kerr Geometry:

$$ds^2 = \displaystyle -\Bigg(1-\frac{2Mr}{\rho^2}\Bigg)dt^2 - \frac{2Marsin^2(\theta)}{\rho^2}[dt d\phi+d\phi dt] +$$

$$\displaystyle +\frac{\rho^2}{\Delta}dr^2+\rho^2d\theta^2+ \frac{sin^2(\theta)}{\rho^2}\Big[(r^2+a^2)^2 - a^2\Delta sin^2(\theta) \Big]d\phi^2 \tag{9}$$

This geometry describes the spacetime "outside" of a rotating object like a star,planet or (eventually) a black hole. Now, you notice then an interesting term of this line element:

$$ \Theta_{t\phi} = - \frac{2Marsin^2(\theta)}{\rho^2}[dt d\phi+d\phi dt] \tag{10} $$

This term is responsable for a physical effect called frame dragging which basically says that a particle will gain angular momentum just for staying close to this black hole region.

So, if the metric tensor wouldn't exibited the property $(8)$ you won't never be able to write the sum: $[dt d\phi + d\phi dt]$, because the terms of metric tensor would be:

$$\Theta_{t\phi} - \Theta_{\phi t} \neq 0 \tag{11}$$

So, then the geometry wouldn't be "the same".

III) The precise definition

The metric tensor actually is a covariant $(0,2)-$tensor, defined on a (differentiable) manifold $M$ to inject the basic notions of geometry like length and angles. This map can be visualized as:

$$\begin{array}{rl} g :T_{p}M \times T_{p}M &\to \mathbb{R} \\ (X,Y) &\mapsto g(X,Y) \end{array}$$

This tensor is defined like:

$i) All-the-properties-of-a-bilinear-functional $

$ii) g(X,Y) = g(Y,X)$

$iii) g(X,Y) = 0$ $ \forall X \in T_{p}M$ $\implies Y=0$ (positive definite)

Acctually the metric tensor isn't distinguible of an inner product. And because of that we can say that metric tensor gives the norm of vectors (which is an element of $\mathbb{R}$), and then the notion of arc lenght, then the line element and finally the geometry of space i.e.

$$ g(X,X) = ds^{2} \tag{12}$$

IV) The precise definition (For Spacetime)

Now, if you are in spacetime the properties of the metric tensor changes drastically because you do not have a positive definite metric (i.e. the property $iii)$ above), due to the fundamental character of Minkowski metric:

$$ ds^{2} = -c^{2}dt^{2} + d \textbf{x}^{2} \tag{13} $$

You can say then that the "inner product isn't well defined" and this is true from the point of view of linear algebra. And note that Minkowski metric is correct by....Special Relativity! Which is a physical framework and is totattly correct. So, because of these facts the metric tensor have what is called a Lorentzian Signature and all of Riemannian Geometry have to be extended to Pseudo-Riemannian Geometry, which is the framework of General Relativity.

$i) All-the-properties-of-a-bilinear-functional $

$ii) g(X,Y) = g(Y,X)$

$iii) g(X,X) > 0 \to X $ (spacelike vector)

$iv) g(X,X) = 0 \to X $ (lightlike vector)

$v) g(X,X) < 0 \to X $ (timelike vector)

Lorentzian signature "Cracks" the vectors on that three types above, and this is awesome because we can differentiate physically directions in space and time.


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