Dirac equation more natural in $O(1,3)$ than $O(3,1)$?

Question

In physics, the Lorentz group $O(1,3)$ is of central importance, being the setting for electromagnetism and special relativity.

According to Wikipedia:

Some texts use $O(3,1)$ for the Lorentz group; however, $O(1,3)$ is prevalent in quantum field theory because the geometric properties of the Dirac equation are more natural in $O(1,3)$.

Why is Wikipedia claiming that the geometric properties of the Dirac equation are more natural in $O(1,3)$? What does it mean for the Dirac equation to be more natural in $O(1,3)$ than $O(3,1)$? What geometry property is it? What is Wikipedia trying to say?

Here we define $O(m,n)$ via the metric $$ g = \mathrm{diag}(\underbrace{1,\ldots,1}_{m},\underbrace{-1,\ldots,-1}_{n}) $$ such that the inner product of vectors defined through the metric $g$ in the vector space $\mathbb{R}^{p,q}$ is an invariant. So $O(m,n)$ consists of general linear $GL[m+n,\mathbb{R}]$ matrices $A$ satisfying $g^{-1}A^{\mathrm{Tr}}g=A^{-1}$, where $A^{\mathrm{Tr}}$ is the transpose of $A$.

Answer 1

You are missing the physical information carried by the corresponding 4-vectors. The two versions just differ by "what remains positive". In QFT we often deal with collisions described in momentum space, where the zeroth component carries the interpretation of energy if you compute the norm of a four-momentum vector of a particle at rest, $$p=(E,\vec{0})$$ you get $$p^2 = E^2$$ which should be equal to the rest mass $m^2c^4$ which we also want to be positive. This is general for a massive particle, whether it is described by the Klein-Gordon Eq. or Dirac. So as you can see the time axis carries a lot of relevant information for collisions.

On the other case when you are studying general relativity (GR) the other signature is probably more common. Again the only reason being that the space part probably displays more relevant phenomena in this branch. Here time will carry a minus sign while the spatial part will be positive so metrics will usually have the form $$ds^2 = -dt^2 + \sum_{ij}dx_i dx_j $$ and as you can see making the infinitesimal time interval $dt=0$ gives immediately a positive "proper" infinitesimal distance.

At the end of the day these are conventions, as people have already mentioned in the comments $O(1,3)$ is exactly the same as $O(3,1)$.

Answer 2

The reason for this is the clifford algebra, which the gamma matrices are required to obey:

$$\{\gamma^{\mu},\gamma^{\nu}\} = 2\eta^{\mu\nu}I_{4}$$

If you choose the $(+, -,-,-)$ signature, then $\gamma^{0}$ can be diagonal and real, and you can choose the standard "the blocks of the matrices are pauli spin matrices" representation for the $\gamma^{i}$. Otherwise, you have to compromise on all of this. (the easy answer is to just multiply all of your gammas by $i$, I guess, but "not carrying factors of i" seems more natural to me)

Answer 3

From the question it looks as though you are asking about the difference between adopting metric signature $(-1,+1,+1,+1)$ verses $(+1,-1,-1,-1)$. I found it more natural to do general relativity with $(-1,1,1,1)$. But there is now so much particle physics and quantum field theory that adopts $(+1,-1,-1,-1)$ that it is a pain to avoid that choice, because if you did then you would be forever worrying about signs when comparing your work with textbooks and research literature.

A reason to prefer $(-,+,+,+)$ is that then the dot product between 4-momenta has a positive dot product between 3-momenta, which makes sense. Also, one minus sign is better than three. It makes the consideration of spatial geometry more natural in GR, I would say. On the other hand a reason to prefer $(+,-,-,-)$ is that then the invariant $p_\mu p^\mu$ comes out positive for 4-momentum. More generally, one more often deals with timelike than spacelike vectors when considering evolution of particles, collisions, fields and the like, so a positive invariant for the timelike case can mean you have fewer signs to worry about in the end.

After writing the above I found that I was in agreement with another answer: that of ohneVal. So there you have it.