Say I use the metric signature $(-+++)$. Then $\partial_a=(\partial_0,\partial_i)=(-\partial^0,\partial^i)$, but $\partial^a=(\partial^0,\partial^i)=(-\partial_0,\partial_i)$.

The same goes for $p^a$ and $p_a$, I suppose. I know that we need to contract, say, $x^a$ with $p_a$ to give $-x^0p_0+x^ip_i$. So my question is: Does this convention of whether to put ``$-$'' on quantities with superscripts or subscripts apply to all quantities? $x^a$ and $x_a$? $A^a$ and $A_a$?

  • $\begingroup$ Thanks for the edit. I think you cleared my confusion already. $\endgroup$
    – user46348
    May 15 '14 at 18:09

In general, when replacing a free index with a specific one, no signs ever get introduced: \begin{align} \partial_a & \to (\partial_0, \partial_i) \\ \partial^a & \to (\partial^0, \partial^i). \end{align} This holds for all signatures. (On a side note, I'm being pedantic about not using "$=$" signs for a reason - a tensor is not equal to a single, albeit unspecified, component of itself.)

The only question, then, is the relation between $\partial_0$ and $\partial^0$, and between $\partial_i$ and $\partial^i$. In general, again without regard to signature, we have \begin{align} \partial_a & = g_{ab} \partial^b \\ \partial^a & = g^{ab} \partial_b. \end{align} In special relativity, $g_{ab} = \eta_{ab}$ and $g^{ab} = \eta^{ab}$, where now we need to agree on a signature to conclude \begin{align} \partial_0 & = -\partial^0 & \partial_i & = \partial^i \\ \partial^0 & = -\partial_0 & \partial^i & = \partial_i. \end{align}

Actually, your worry about where to put the signs means you've put in an extra. In fact, we have $$ x^a p_a = x^0 p_0 + x^1 p_1 + x^2 p_2 + x^3 p_3. $$ This is exactly the same as $$ x_a p^a = x_0 p^0 + x_1 p^1 + x_2 p^2 + x_3 p^3. $$ The above holds in GR in any signature. If you are in SR and have the $(-,+,+,+)$ signature, then you can write $$ x^a p_a = -x^0 p^0 + x^1 p^1 + x^2 p^2 + x^3 p^3 $$ or $$ x^a p_a = -x_0 p_0 + x_1 p_1 + x_2 p_2 + x_3 p_3. $$ Note the negative term only enters when you insist on using the same indices (either upper or lower) for both $x$ and $p$. Also note that these two formulas only hold in SR, where swapping indices at worst introduced a negative. In general, $$ x^a p_a = g_{ab} x^a p^b, $$ which can have $16$ terms in the sum.

  • $\begingroup$ Clear, concise and complete. +1 $\endgroup$ May 15 '14 at 18:51

Using your metric signature, the metric is: $$\eta_{\mu\nu}=\begin{pmatrix}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}$$

The generic position vector is defined as (in Cartesian) $$x^{\mu}=\begin{pmatrix}t\\x\\y\\z\end{pmatrix}$$ And the quantity $x_{\mu}=\eta_{\mu\nu}x^{\nu}$. So you can see that $x_\mu$ becomes the one with a minus sign on $x_0$.

Then, the derivative is defined as: $$\partial_\mu=\frac{\partial}{\partial x^\mu}$$ So you can see that $\partial_0=\frac{\partial}{\partial t}$ is positive. When you apply the metric, $$\partial^\mu=\eta^{\mu\nu}\partial_\nu=\eta^{\mu\nu}\frac{\partial}{\partial x^\nu}=\frac{\partial}{\partial x_\mu}$$ and thus $\partial^0$ becomes negative.

So to answer your question, yes the convention applies. Changing a superscript to a subscript is done through applying the metric. $A_a=\eta_{ab}A^b$ and $A^a=\eta^{ab}A_b$ and $\eta^{ab}\eta_{ab}=4$.


Yes, indeed it does.


Have you taken a maths course on Metric Spaces perhaps?

A Metric Space, roughly, is something where we have a notion of how to measure things.

So, for example, Euclidean Space is a very fimiliar space, you measure the square of the magnitude of a vector to be

$$ \vec{x} \cdot \vec{x} = (x^1)^2 + (x^2)^2 + (x^3)^2 = (x_1)^2 + (x_2)^2 + (x_3)^2 $$

This could also be written using a metric as

$$ \vec{x} \cdot \vec{x} = g_{ij} x^i x^j $$

where $i = 1,2,3$ and $j = 1,2,3$. Here I'm using Einstein Summation Convention, such that $g_{ij} x^i x^j$ with the $i$ and $j$ repeated like that, one upper and one lower, is shorthand for

$$ g_{ij} x^i x^j = \sum_{i=1}^3 \sum_{j=1}^3 g_{ij} x^i x^j$$


So our metric, $g_{ij}$ is just

$$ g_{ij} = \mbox{diag}(+1,+1,+1) $$

that is, when $i=j$ we get $g_{ii} = + 1$, while if $i \neq j$ we get zero, $g_{ij}\vert_{i\neq j} =0$.


You'll then see that when raising and lowering indices using the metric like

$$ x^i = g^{ij} x_j \quad \mbox{ or } \quad x_i = g_{ij} x^j $$

That the raised one $x^i$ and the lowered one $x_j$ are the same, since the $g_{ij}$ are each just $+1$.

So that was nice and easy.


Now when we move to Minkowski Space for Special Relativity, it gets a little more complicated. The metric components (the equivalent of the $g_{ij}$) are no longer all $+1$. We usually use $\eta^{\mu \nu}$ for the Minkowski Space metric, where

$$ \eta^{\mu \nu} = \mbox{diag}(-1,+1,+1,+1) $$

in your convention. (It should be noted that not everyone uses the same set of +1 and -1, so you should always check this and make sure. It can be very fustrating when something doesn't work out because the other guy was using a different sign!).

Anyway, a vector, like $A^{\mu}$ as you ask, can be lowered with the metric tensor like

$$ A_{\mu} = g_{\mu \nu} A^{\nu} $$

So since by definition

$$ A_{\mu} = (A^0, A^i) $$

when we lower the index we get

$$ A_{\mu} = g_{\mu \nu} A^{\nu} = (g_{0 \nu} A^{\nu}, g_{i \nu} A^{\nu}) = \left( \sum_{\nu = 0}^3 g_{0 \nu} A^{\nu}, \sum_{\nu = 0}^3 g_{i \nu} A^{\nu} \right) $$

But since all the $g_{\mu \nu}$ are zero when $\mu \neq \nu$, the only ones in the sum that are not zero are

$$ A_{\mu} = g_{\mu \nu} A^{\nu} = (g_{0 0} A^{0}, g_{i i} A^{i}) $$

and we know this from the definition of our metric tensor above, namely

$$ A_{\mu} = g_{\mu \nu} A^{\nu} = ((-1) A^{0}, (+1) A^{i}) = (-A^{0}, A^{i}) $$


You can use the metric tensor to raise and lower any index, such as, for example,

$$ T_{\mu \nu \rho}^{\; \; \; \; \sigma} = g_{\mu \alpha} g_{\nu \beta} g^{\sigma \gamma} T^{\alpha \beta}_{\; \; \; \rho \gamma} $$

  • $\begingroup$ The metric of GR is not a metric in the sense of metric spaces, it is a metric in the sense of pseudo-Riemannian manifolds (en.wikipedia.org/wiki/…). $\endgroup$ May 16 '14 at 6:24
  • $\begingroup$ @joshphysics : I am not exactly sure what you are trying to say, but if it is that a Riemannian manifold is not a metric space as it has a metric tensor as opposed to a metric function, then that would be incorrect. See the fifth sentence of second paragraph of the wikipedia article on Metric Tensor. Moreover, the tangent space to our GR Lorentzian manifold is indeed a metric space, namely it is Minkowski space locally, which is a pseudo-Euclidean space. See the introductory paragraph inMetric Space. $\endgroup$
    – Flint72
    May 16 '14 at 13:32
  • $\begingroup$ On a connected, Riemannian manifold $M$, one can use the metric tensor to define a distance function $d:M\times M\to \mathbb R$ such that the pair $(M,d)$ is a metric space. In fact $d(p,q)$ if defined as the greatest lower bound of lengths of piecewise smooth curve segments from $p$ to $q$. On a pseudo-Riemannian manifold however, I'm relatively certain that the same cannot be done because the metric tensor has indefinite signature. In any case, it's important to note that the metric tensor is not a metric on the manifold in the sense of metric spaces. $\endgroup$ May 18 '14 at 2:50
  • $\begingroup$ See, for example, the bullet point at the end of this nLab article: ncatlab.org/nlab/show/pseudo-Riemannian+metric $\endgroup$ May 18 '14 at 2:59

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