# What does $\nu$ mean in relativity?

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 $x^{\nu}$ ?

• $\nu$ is the same thing as $\mu$. Dec 18, 2013 at 20:26
• Then why do they introduce the new notation? Does it define a different set of coordinates? Dec 18, 2013 at 20:29
• $\nu$ 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 $3$. Dec 18, 2013 at 20:31
• $\mu$ 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. Dec 18, 2013 at 20:42

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.
• So, $x^\nu$ is the $\nu$-term of the same vector? Dec 18, 2013 at 20:34
• @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). Dec 18, 2013 at 20:45