I decided to teach myself relativity over the Christmas holiday, and I've gotten a bit stuck.

Coordinates in space time can be defined by a collection of coordinates,

$$ x^0 = ct \\ x^1 = x \\ x^2 = y \\ x^3 = z $$

This collection is denoted $x^{\mu}$. My linear algebra kicks in here, and I see that it is a column vector:

$$x^{\mu} =\begin{pmatrix}ct\\x\\y\\z\end{pmatrix}$$

Next they define the metric tensor by a matrix,

$$\eta_{\mu\nu} =\begin{pmatrix}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}$$

And then say that the definition of the dot product is:

$$A\cdot B = \eta_{\mu\nu}A^{\mu}B^{\nu}$$

This leads be to believe that $\nu$ denotes another column vector, but of what ?

It is then used in the definition of the space time interval,

$$ds^2 = \eta_{\mu\nu}dx^{\mu}dx^{\nu}$$

What is this $x^{\nu}$ ?

  • 1
    $\begingroup$ $\nu$ is the same thing as $\mu$. $\endgroup$
    – Kyle Kanos
    Dec 18, 2013 at 20:26
  • $\begingroup$ Then why do they introduce the new notation? Does it define a different set of coordinates? $\endgroup$ Dec 18, 2013 at 20:29
  • $\begingroup$ $\nu$ could have the value $0$, $1$, $2$ or $3$. Unless there is both a sub- and superscript in the same term, because then there is a sum over $0$, $1$, $2$ and $3$. It is conventional to have this definition for greek symbols, whereas latin letters such as $i$ and $j$ run from $1$ to $3$. $\endgroup$
    – Hunter
    Dec 18, 2013 at 20:31
  • $\begingroup$ $\mu$ and $\nu$ are independent indices. $\eta_{\mu\nu}A^\mu B^\nu$ is a shorthand notation for the double sum $\sum_{\mu=0}^3\sum_{\nu=0}^3\eta_{\mu\nu}A^\mu B^\nu$. It's called the Einstein summation convention. $\endgroup$
    – Pulsar
    Dec 18, 2013 at 20:42

1 Answer 1


As Kyle says, $\nu$ is just a (free) index. You can use any letter.

More precisely, $x^\mu$ is the $\mu$-component of of the vector $\mathbf{x}=(x_1,x_2,\dots,x_n)$. And $x^\nu$ is the $\nu$-component of of the vector $\mathbf{x}=(x_1,x_2,\dots,x_n)$. So you can see that the vector is the same, $\mathbf{x}$, you just name the components with a different index.

  • $\begingroup$ So, $x^\nu$ is the $\nu$-term of the same vector? $\endgroup$ Dec 18, 2013 at 20:34
  • 1
    $\begingroup$ @NictraSavios Yes. $\endgroup$
    – jinawee
    Dec 18, 2013 at 20:36
  • 1
    $\begingroup$ I see, one last question, was I right about the terms being columns? (Since vectors are just columns, or more accurately, rank-1 tensors) $\endgroup$ Dec 18, 2013 at 20:38
  • 2
    $\begingroup$ @NictraSavios In most cases, yes ($x_\mu$ would be a row vector). But in general, I think that they are defined as tangent vectors on a manifold (but I don't know much differential geometry). $\endgroup$
    – jinawee
    Dec 18, 2013 at 20:45
  • 2
    $\begingroup$ @NictraSavios: Only as components. If you think of a change of basis in terms of the matrix of the components of the new basis vectors written as columns, then components of column vectors change with the inverse of this matrix, hence "contravariant" (upper-index), while components of row vectors don't invert, hence "covary" with the change of basis (lower-index). But vectors and covectors are not just their components, although they have different equivalent formulations (e.g., vectors in tangent space as derivations for functions on a manifold). $\endgroup$
    – Stan Liou
    Dec 18, 2013 at 20:53

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