Can anyone explain what metric signature is?

I have a basic knowledge regarding tensors, btw.

Also, how is it related to fundamental understanding of general relativity?

  • $\begingroup$ Do you know what the metric tensor is, or should we explain that before explain what it's signature means? $\endgroup$ Apr 27, 2012 at 15:00
  • $\begingroup$ @JohnRennie I sort of know what the metric tensor is, but if metric tensor is explained at least briefly, I will appreciate :) $\endgroup$
    – user27515
    Apr 27, 2012 at 15:03

2 Answers 2


In relativity there is an invariant called the proper time, $\tau$. It's an invariant in the sense that all observers will agree on it's value. In special relativity the proper time is defined as:

$$\mathrm d\tau^2 = \mathrm ds^2 = c^2~\mathrm dt^2 - \mathrm dx^2 - \mathrm dy^2 - \mathrm dz^2$$ or $$\mathrm d\tau^2 = -\mathrm ds^2 = -c^2~\mathrm dt^2 + \mathrm dx^2 + \mathrm dy^2 + \mathrm dz^2$$

You see both sign conventions and I've never been sure which is more generally accepted. Anyhow, you can write the equation for $\mathrm ds^2$ as a matrix equation using:

$$\mathrm ds^2 = g_{\alpha\beta}x^\alpha x^\beta$$

where $x$ is the vector $(t, x, y, z)$ and $g$ is the matrix:

$$\left( \begin{matrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}\right)$$

The matrix is called the metric tensor (or a representation of it) and the signature is the number of positive and negative values on the leading diagonal. In this case it's (1,3) or you often just add together the negative and positive numbers to give, in this case, just 2.

Exactly the same equation is used in general relativity, but the matrix representing the metric tensor is more complicated and generally not diagonal, so you have to diagonalise it to calculate the signature. The Wikipedia article goes into this in more detail.

The reason we're interested in the signature is that we expect spacetime to have one timelike co-ordinate and three spacelike co-ordinates, so we expect the signature to be always (1,3).

  • 2
    $\begingroup$ You missed a $d$ in $c^2dt^2$. But more importantly: I think it's pretty universal that the sign of proper time matches the sign of coordinate time. $c^2d\tau^2 = c^2dt^2-dx^2-dy^2-dz^2$. The issue with the metric signature is just about whether $c^2d\tau^2 = \pm ds^2$. In one signature, the Minkowski metric tensor has diagonal elements $-1,1,1,1$ and in the other signature it has diagonal elements $1,-1,-1,-1$. The latter is more common in particle physics, the former in GR and fundamental QFT, I think. $\endgroup$
    – David Z
    Apr 27, 2012 at 23:12
  • 1
    $\begingroup$ You say often just add the numbers to signature of 2 but I have never before heard the signature of Minkowski spacetime referred to as 2. I have always seen the (1,3) notation (or -,+,+,+ or +,-,-,-). It doesn't make sense to add the diagonal elements and say the signature is 2. A flat 2 dimensional plane space would also have signature 2 by that method which certainly is not the same as (1,3). $\endgroup$
    – FrankH
    Mar 19, 2013 at 19:17

Signature is defined as the sign of eigenvalues, which means you must first diagonalize the metric. Afterwards flipping the signature then just means multiplying the metric in eigenbasis by $-1$. If $+g_{\mu\nu}$ has the set of eigenvalues as $(\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4})$ then $-g_{\mu\nu}$ will have $(-\lambda_{1},-\lambda_{2},-\lambda_{3},-\lambda_{4})$ as eigenvalues. It may seem like this could have some unprecedented effect on the underlying physics but it doesn't. This happens because of the way Einstein Field Equation is defined. $$R_{\mu\nu} -\frac{1}{2}g_{\mu\nu}R = 8\pi G T_{\mu\nu}$$ Both $\pm g_{\mu\nu}$ satisfy the einstein field equation ($G_{\mu\nu} = 8\pi G T_{\mu\nu}$). So, whichever choice you make has no effect on the underlying physics and the covariant equations derived from them. This happens because with the choice of positive sign in front of metric i.e. $+g_{\mu\nu}$ we have $+R_{\mu\nu}$ and $+T_{\mu\nu}$, whereas $R$ stays the invarient. Under the transformation $+g_{\mu\nu} \rightarrow -g_{\mu\nu}$, we have $+R_{\mu\nu} \rightarrow -R_{\mu\nu}$ and $+T_{\mu\nu} \rightarrow -T_{\mu\nu}$:

$$-R_{\mu\nu} +\frac{1}{2}g_{\mu\nu}R = -8\pi G T_{\mu\nu}$$


$$R_{\mu\nu} -\frac{1}{2}g_{\mu\nu}R = 8\pi G T_{\mu\nu}$$

I hope this clears up your doubt.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.