Sunday, 22 September 2019

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


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


m2c2=gμνPμPν


(though he only was considering the flat space case gμν=ημν. Rewriting the above as:


0=gμνPμPνm2c2



Proceeding to factor this Dirac obtained an expression like:


0=(γμPμmc)(γνPν+mc)


Where {γμ,γν}=γμγν+γνγμ=2gμν. This is entirely general, though in quantum theory, the Pμ would be replaced with their quantum operators acting on a wavefunction Ψ.


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


0=(γμPμ±mc)


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:


ds2=gμνdxμdxν


The mathematical parallel to the first equation seems pretty straight forward: 0=gμνdxμdxνds2


0=(γμdxμds)(γνdxν+ds) 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



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


gμνdxμdxν=(γμdxμ)(γνdxν)


as you did. Note that the γμ 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 ds2. Unlike m2c2, 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


m2c2=gμνPμPν



Is an equation (with physical content),


ds2=gμνdxμdxν


is the expansion of ds2, which could just as well (or better) have been written g (for example), in a local basis derived from a coordinate basis.


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