8

In special relativity, the proper time is defined as $$d\tau^2 = c^2t^2-(x^2+y^2+z^2).$$ One usually introduces a matrix $\eta$ to represent it. I have seen two sign conventions. One has three minuses: $$\eta_3=\left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{array}\right),$$ and the other one has only one, so $\eta_1=-\eta_3$. Thus, $\tau^2=x^\mu\eta_{3,\mu\nu}x^\nu=-x^\mu\eta_{1,\mu\nu}x^\nu$, summing over repeated indices.

I have been told this is an issue in the physics community, and that relativity typically uses the $(-,+,+,+)$ East coast sign convention while particle physics/quantum field theory often uses the $(+,-,-,-)$ West coast sign convention. Why is that?

Qmechanic
  • 201,751
MickG
  • 307

3 Answers3

6

Relativists tend to use the proper time, $d\tau$, and the proper distance, $ds$, interchangably. If you're working with proper time you'd expect the equation for it to look like:

$$ d\tau^2 = dt^2 + \text{other terms} $$

while if you're working with proper distance you expect:

$$ ds^2 = dx^2 + dy^2 + dz^2 + \text{other terms} $$

The sign problem comes about because the spacetime signature requires that $ds^2 = -c^2d\tau^2 $. So you end up with:

$$ d\tau^2 = -\frac{ds^2}{c^2} = dt^2 - \tfrac{1}{c^2}\left( dx^2 + dy^2 + dz^2 \right) $$

or:

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

Both describe identical physics, so which you use is just personal preference.

John Rennie
  • 355,118
  • 5
    I find it interesting the Thomas Moore uses the (+---) convention in his Traveler's Guide to Spacetime and the (-+++) convention in A Workbook for General Relativity. Maybe we should ask him why he changed. – Bill N Mar 25 '15 at 14:03
  • Would you say that particle physicists tend to use +--- because they are dealing with proper time along a particle's trajectory (which is timelike or null) and relativists tend to use -+++ because they deal more with cosmology than particles and they would be interested in spacelike intervals (such as the radius of an AdS, for example)? –  Jul 14 '18 at 11:24
  • @DvijMankad: I think the particle physics convention stems from the fact that energy & mass are much more important quantities for them than time. For the $(+ - - -)$ convention, you have $m^2 = p^\mu m_\mu$, which makes calculations easier than having to carry around a bunch of minus signs whenever you contract a 4-vector. – Michael Seifert Jul 14 '18 at 12:11
  • @MichaelSeifert Oh, I see. That makes sense. I thought that since the Lagrangian involves the $\int d\tau$ term for a particle so that might be the reason. –  Jul 14 '18 at 12:46
2

Could you provide a simple reason for these two conventions?

The reason behind the (-,+,+,+) convention (the "mostly plus metric") is that a positive length in 3 dimensional space (e.g., the distance from my head to my toes) should still be a positive length in 4 dimensional space-time. Why should the distance from my head to my toes all of a sudden become negative just because I begin to consider special relativity?!

The reason behind the (+,-,-,-) convention (the "mostly minus metric") is that... well... There is no good reason. Clearly, you can see what side of the debate I am on.

Also, an argument from authority: Schwinger preferred the mostly-plus metric. Everybody cool uses the mostly-plus metric. For example Wald says: "We choose to use the metric signature - + + + because it is generally much more convenient than the alternative choice + - - - in that it induces a positive definite (rather than negative definite) metric on spacelike hypersurfaces."

Also, when you decide to look at thermal field theory or when you decide to wick-rotate to a euclidean field theory, if you are using the mostly-plus metric everything works out more smoothly. Whereas if you are using the mostly-minus metric everything ends up having a stupid minus sign introduced for no reason.

hft
  • 19,536
  • 2
    Perhaps the mostly minus metric is because this way $\tau^2$ doesn't have a minus in front of it, that is $\tau^2=x^\mu\eta_{\mu\nu}x^\nu$ without a minus? – MickG Mar 25 '15 at 08:34
  • Just a thought :). – MickG Mar 25 '15 at 08:35
  • 2
    It's psychological with me; I just can't bear to see $-dt^2 + \dots$. I've always been a $(1,-1, -1, -1)$ physicist. – JamalS Mar 25 '15 at 08:57
  • 8
    The (or, at least, one) reason for $(+---)$ is so that $p^\mu p_\mu = m^2$ instead of $-m^2$. Which is just as good of a reason as the argument about distance - why should particles have imaginary mass just because you begin to consider special relativity? ;-) Fix that and you get my upvote. – David Z Mar 25 '15 at 09:36
  • 1
    Nooo... unlike "m" the thing denoted by "p" has internal structure so $p^2=-m^2$ doesn't imply anything like "imaginary p"... In fact, that is nonsensical. To recap: $p^2\equiv p^\mu p_\mu=-E^2+|\vec p|^2=-m^2$. The mass $m$ is positive like usual, as is the energy $E$ and the square of the 3-momentum $|\vec p|^2$ and the magnitude of the 3-momentum. – hft Mar 25 '15 at 23:58
0

I think it depends on what you are interested in. For example, in my work considering time dependent fields in field theory coupled with gravity like inflation or quintessence, in the +--- signature you obtain mostly positive terms, intead of having to work with negative kinetic terms all the time. So do as you wish.

MickG
  • 307
D. Varela
  • 1