Factorization of the energy/mass relation (as used to obtain the Dirac equation) applied to the invariant interval

Question

In considering the Klein-Gordon equation Dirac was examining the energy equation:

$$m^{2}c^{2}=g_{\mu\nu}P^{\mu}P^{\nu}$$

(though he only was considering the flat space case $g_{\mu\nu}=\eta_{\mu\nu}$. Rewriting the above as:

$$0=g_{\mu\nu}P^{\mu}P^{\nu}-m^{2}c^{2}$$

Proceeding to factor this Dirac obtained an expression like:

$$0=\left(\gamma_{\mu}P^{\mu}-mc\right)\left(\gamma_{\nu}P^{\nu}+mc\right)$$

Where $\left\{ \gamma_{\mu},\gamma_{\nu}\right\} =\gamma_{\mu}\gamma_{\nu}+\gamma_{\nu}\gamma_{\mu}=2g_{\mu\nu}$. This is entirely general, though in quantum theory, the $P^{\mu}$ would be replaced with their quantum operators acting on a wavefunction $\Psi$.

Continuing with the general case, choosing one factor or another leads one to the Dirac equation for either positive or negative mass solutions:

$$0=\left(\gamma_{\mu}P^{\mu}\pm mc\right)$$

In choosing either + or - I would think one would have to make the same choice (for consistency) in all other areas of spacetime geometry.

For example, the infinitesimal invariant interval is generally expressed as:

$$ds^{2}=g_{\mu\nu}dx^{\mu}dx^{\nu}$$

The mathematical parallel to the first equation seems pretty straight forward: $$0=g_{\mu\nu}dx^{\mu}dx^{\nu}-ds^{2}$$

$$0=\left(\gamma_{\mu}dx^{\mu}-ds\right)\left(\gamma_{\nu}dx^{\nu}+ds\right)$$ however, in attempting to do this in another question of mine: (Is the Invariant interval S between two points independent of the path taken?) I was literally browbeaten by users. (This may have been due to my poor choice of wording)

My question is:

If the factorization of the first equation is soo completely accepted, then what is wrong with factoring the latter equation (i.e the interval)?

The answers I got didn't address any proper issues that I'm aware of, especially when compared to Dirac's factorization as above (If I took them at face value then the Dirac equation would be invalid ha!).

Answer 1

Your conclusion is wrong. $\gamma^{\mu} p_{\mu} \pm m=0$ is valid only as an operator equation. It is valid only when it acts on the on-shell field operator as in QFT, or on the particle state as in the Dirac equation.

Answer 2

You are mixing up notation and actual algebraic operations. The metric is a section of the symmetric square of the cotangent bundle, which in turn is a factor of the symmetric algebra on this bundle. If you extend the scalars from the real numbers to the Clifford algebra of the scalars product in each point, you indeed can factorise the expression

$$g_{\mu\nu}dx^{\mu}dx^{\nu} =\left(\gamma_{\mu}dx^{\mu}\right)\left(\gamma_{\nu}dx^{\nu}\right)$$

as you did. Note that the $\gamma^\mu$ take values in a different Clifford algebra at every point. I guess one can find a parameterized matrix representation so that they become smooth matrix valued functions.

The problem is with your factorisation of $ds^2$. Unlike $m^2c^2$, which actually is a square, this is just notation. It seems to suggest that it is the square of some quantity $ds$, and moreover that this $ds$ actually is the differential of some function, but neither is true in general.

Where

$$m^{2}c^{2}=g_{\mu\nu}P^{\mu}P^{\nu}$$

Is an equation (with physical content),

$$ds^{2}=g_{\mu\nu}dx^{\mu}dx^{\nu}$$

is the expansion of $ds^2$, which could just as well (or better) have been written $g$ (for example), in a local basis derived from a coordinate basis.