# Advantages of using different metric signatures in relativity and QFT

I am studying General Relativity and some basic QFTs. It bothers me a lot that different books use different metric signatures, i.e. $(-+++)$ and $(+---).$ Can anyone tell me the advantages of using both of them, or why some people prefer the former while some prefer the latter? I would like to know which one should I while tackling different problems.

• From my own limited experience, I think if you just keep your notes in whatever form you like to use say - + + + irrespective of the books you use. I think it's just a convention, but someone with more knowledge than I might correct me on that – user108787 Oct 5 '16 at 16:42
• I think that it is really just a "convention"; in the sense that it has no physical significance. Expressions look different when using one against the other and it is pure preference I guess which one is used. This blog post math.columbia.edu/~woit/wordpress/?p=7773 lists and discusses some "pros" and "cons". In some fields one is popular (-+++ in GR) and in some the other (+--- QFT). You might save a minus here and there depending on what you do and what metric you use but I think one should use the sign convention that is used most in the particular fields to avoid confusion. – N0va Oct 5 '16 at 17:01
• High energy physics community mostly uses the mostly minus convention. Whereas GR community uses mostly plus convention. But physics is independent on what the metric signature. – AMS Oct 5 '16 at 17:17

$(+,-,-,-)$

The particle physicist's convention is useful if you like your relevant quantities to be positive. Timelike 4-vectors have positive magnitude in this formalism, so you don't have to include extra minus signs in your formulas. Spacelike vectors come out as negative, which is satisfying because events separated by a spacelike distance cannot be causally related.

$(-,+,+,+)$

The relativist's convention is satisfying if you like to keep your 3-vector formalism free from the minus signs. Also, raising and lowering the indices in flat spacetime simply amounts to flipping the sign of the time component. Furthermore, this convention is preferable when working in spaces of general dimensionality, for example in string theory or when using AdS/CFT correspondence. You always have just one minus sign in the metric, no matter how many spatial dimensions you're working with. So the determinant of the metric tensor, for example, would always be $-1$. In the other formalism, it would be $(-1)^{d+1}$, where $d$ stands for number of spacetime dimensions.

P.S. For your own notes, it's generally a good idea to pick a certain system of units and conventions and stick to them whenever they're not obscuring what you're doing.

This is a summary of the reference East versus West Coast Sign Conventions provided in the comments above by M.J. Steil, which I wrote to learn the advantages of the sign convention used, with the E convention seemingly superior.

You can use either + - - - (W) or - + + + (E).

With E, spatial coordinates are treated the same in both relativistic quantum field theory and non-relativistic case QFT. If you use W as far as space goes, you end up using a negative definite metric, when a positive definite metric would be preferable.

Using E, if you try to create a QFT, well-defined, analytically extendable to imaginary time, your result will be in the standard 4D Euclidean metric.

If you use W , again a negative definite metric appears, whereas you really would prefer a positive definite version again.

With E, the Clifford algebra Cliff (3,1) is the algebra of real four by four matrices. You can choose your gamma-matrices to be real matrices and work with a real spinor representation (AKA the Majorana representation).

Instead, if you choose to use W, Cliff(1,3) is the algebra of two by two quaternionic matrices, which introduces unnecessary complications.

In physics, imaginary numbers are as commonly used as real ones, and we can work away with them until the end of the calculation, when they have to be converted back to real numbers. This leads to different definitions and notations, for the same concepts, being used by mathematicians and physicists. This may result in confusion if you are reading up on the same subject with 2 books, one written by a mathematician and the other by a physicist, but this is an old problem, and not limited to sign conventions, unfortunately.

When discussing advantages and disadvantages of East-Coast (-,+,+,+) and West-Coast (+,-,-,-) metric signatures, it should be kept in mind that the West-Coast metric signature (+,-,-,-) is the natural signature of space-time.

Combining Albert Einstein's mass-energy equivalence $$E=mc^2$$ and the relativistic energy invariant $$E^2/c^2 - p⃗^2 = m_0^2 c^2$$, and measuning length in light-seconds (c=1) yields: $$E^2 = m^2 = m_0^2 + p⃗^2 = m_0^2 + p_1^2 + p_2^2 + p_3^2$$

Hence the rest mass $$m_0$$ is the fourth component of the momentum vector, and the energy $$E$$ or mass $$m$$ is the total length of the momentum vector.

This sum of four squares can be written, according to Leonhard Euler's four-squares-identity, as the product of two sums of each four squares:

$$(m_0^2 + p_1^2 + p_2^2 + p_3^2) = (r_0^2 + r_1^2 + r_2^2 + r_3^2)(s_0^2 + s_1^2 + s_2^2 + s_3^2 )$$

Therein the components are given as follows:

$$m_0 = (r_0 s_0 - r_1 s_1 - r_2 s_2 - r_3 s_3)$$

$$p_1 = (r_0 s_1 + r_1 s_0 + r_2 s_3 - r_3 s_2)$$

$$p_2 = (r_0 s_2 - r_1 s_3 + r_2 s_0 + r_3 s_1)$$

$$p_3 = (r_0 s_3 + r_1 s_2 - r_2 s_1 + r_3 s_0)$$

The proof is by simple algebraic evaluation.

This is the quaternion multiplication rule.

Please note that the first row of quaternion multiplication

i.e. the metric product $$m_0 = r⃗∙s⃗$$, shows the

West-Coast metric signature (+,-,-,-)

## protected by Qmechanic♦Oct 5 '16 at 18:59

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).