# Minkowski Metric Sign Convention

1. Jun 9, 2012

### PLuz

Hello,

I believe this is a really stupid question but I can't seem to figure it out. So given a Minkowski spacetime one can choose either the convention (-+++) or (+---).

Supposedly it's the same. But given the example of the four momentum:

Choosing (+---) in a momentarily comoving frame the only non null component is $p^0$. Taking the inner product $<p,p>$ one gets $p^2=m^2c^2$ which is the expected result.

Choosing the other convention, (-+++), doing the same calculation one gets $p^2=-m^2c^2$.

So one has to add a minus sign...do we have to always add this minus sign by hand in every inner product with whatever vectors or something is missing me?

2. Jun 9, 2012

### Muphrid

Yes, this is the same way that a positive magnitude of a spacetime interval will mean timelike in one convention and spacelike in the other. There is no physical difference between the two, because the systems are internally consistent.

3. Jun 9, 2012

### Staff: Mentor

As Muphrid mentioned, either convention is internally consistent and both correctly represent the physics. You just have to be consistent with the convention you are using in the signs of all of your dot products. Timelike intervals squared are positive under one convention and negative under the other.

Personally, I use the following approach to keep the convention straight.

If I am using the (-+++) convention then I will write the metric as:
$ds^2= -c^2 dt^2 + dx^2 + dy^2 + dz^2$

If I am using the (+---) convention then I will write the metric as:
$c^2 d\tau^2= c^2 dt^2 - dx^2 - dy^2 - dz^2$

4. Jun 11, 2012

### PLuz

I understood what both told me, thank you once again for the fast answer, but something is still not clear to me.

Given two vector fields, say $V$ and $U$. If i'm working in the convention (+---) the dot product is given by: $<V,U>=\eta_{\mu \nu} V^{\mu}U^{\nu}=V^0 U^0 -V^i U^j$
, where the roman letters mean only the spacial components and $\eta_{\mu \nu}$ is the Minkowski metric.

Choosing the other convention (-+++) the dot product is given by $<V,U>=\eta_{\mu \nu} V^{\mu}U^{\nu}=-V^0 U^0 +V^i U^j$

So you said that both are correct and I just have to stick with one in the calculations, yes I understand that, but if $V=U=P$ where $P$ is the 4 momentum. They state diferent things. In one convention we find $P^2=m^2c^2$ and in the other $P^2=-m^2c^2$, in a momentarily comoving frame.

So if I want them to state the same, expected, result, in the (-+++) convention I have to change the dot product to $<V,U>=-\eta_{\mu \nu} V^{\mu}U^{\nu}=V^0 U^0 -V^i U^j$. Can I do this?Is this incorrect?

Last edited: Jun 11, 2012
5. Jun 11, 2012

### yenchin

In the (+,-,-,-) convention $<V,U>=\eta_{\mu \nu} V^{\mu}U^{\nu}=V^0 U^0 -V^i U^j$, note that there is no minus sign in front of the metric. Since you *already* use (+,-,-,-) convention your metric $\eta_{\mu \nu}$ already is the diagonal matrix [1,-1,-1,-1].

6. Jun 11, 2012

### Staff: Mentor

That is correct. In one convention timelike intervals squared are positive, and under the other convention timelike intervals squared are negative. So in both cases P² is a timelike interval squared.

This is incorrect to the best of my knowledge. The dot product is not changed, just the intrepretation of positive or negative as timelike.

7. Jun 11, 2012

### PLuz

OK, now I understand perfectly well what all of you told me. Thank you very much.

Can anyone then explain me what is writen in wikipedia:

"en.wikipedia.org/wiki/Four-momentum"

in the second section "Minkowski Norm" because they put the minus sign...

8. Jun 11, 2012

### Staff: Mentor

No, they didn't. Look carefully, the dot product is defined as normal. They used the convention that negative intervals squared are timelike. Therefore m²c² is not equal to the interval squared but the negative of the interval squared. If they had redefined the dot product then they would not have needed the minus sign.