Metric Sign Convention: Effects on Klein Gordon & Dirac Equations

iangttymn
Messages
12
Reaction score
0
The two standard conventions for the Minkowski metric are diag(1,-1,-1,-1) and diag(-1,1,1,1). The physics comes out the same either way, but I'm trying to make a list of the things that change depending on the convention you use.
The Klein Gordon equation is one - with the "mostly plus" (MP) metric it is
(\partial^2 - m^2)\phi = 0
and with the "mostly minus" (MM) metric it is
(\partial^2 + m^2)\phi = 0
Another is the sign regarding which is the "positive frequency" solution to Klein Gordon/Dirac. Another is the sign on the Clifford algebra. For MP the more natural choice is
\{\gamma^{\mu}, \gamma^{\nu}\} = -2\eta^{\mu\nu}
and for MM the more natural choice is
\{\gamma^{\mu}, \gamma^{\nu}\} = 2\eta^{\mu\nu}
(you can actually make either choice for either metric but the Dirac equation only has the nice "square root" of the Klein Gordan equation form with these choices.

Can anyone point out some other things that are affected by the convention?
 
Physics news on Phys.org
Physically, the reversal of the sign means that we reverse the axes and the direction of time. If anything changes at all with the conventions, it must be in an anisotrophic space.
 
I don't mean any changes in the fundamental physics. I just mean arbitrary conventional changes, like the ones I mentioned. The Klein Gordon equation is saying exactly the same thing in both cases - you just have to write it differently. I'm just trying to get a feel for the different things that are written differently based on the decision.
 
Real Clifford algebras Cl(1,3) and Cl(3,1) are not isomorphic (complexified are isomorphic). Whether it may have any physical significance is, to my knowledge, not known. I suspect it may, but not in any of today's theories that I know.
 
Eynstone said:
Physically, the reversal of the sign means that we reverse the axes and the direction of time.
Isn't that wrong? Signature convention is different to swapping the future and past null cones.

I don't understand why anyone chose the mostly negative convention? Mostly positive is maximally consistent with ordinary Riemannian (spatial) geometry, and is perhaps even motivated by incorporating time as the imaginary component..
 
Mostly negative is being liked in QFT where time translations generator is being interpreted as "energy operator", and "energy" should be "positive". With mostly negative you do not have to flip the sign when you go from covariant P_0 to contravariant P^0.
 
OK, so this has bugged me for a while about the equivalence principle and the black hole information paradox. If black holes "evaporate" via Hawking radiation, then they cannot exist forever. So, from my external perspective, watching the person fall in, they slow down, freeze, and redshift to "nothing," but never cross the event horizon. Does the equivalence principle say my perspective is valid? If it does, is it possible that that person really never crossed the event horizon? The...
ASSUMPTIONS 1. Two identical clocks A and B in the same inertial frame are stationary relative to each other a fixed distance L apart. Time passes at the same rate for both. 2. Both clocks are able to send/receive light signals and to write/read the send/receive times into signals. 3. The speed of light is anisotropic. METHOD 1. At time t[A1] and time t[B1], clock A sends a light signal to clock B. The clock B time is unknown to A. 2. Clock B receives the signal from A at time t[B2] and...
From $$0 = \delta(g^{\alpha\mu}g_{\mu\nu}) = g^{\alpha\mu} \delta g_{\mu\nu} + g_{\mu\nu} \delta g^{\alpha\mu}$$ we have $$g^{\alpha\mu} \delta g_{\mu\nu} = -g_{\mu\nu} \delta g^{\alpha\mu} \,\, . $$ Multiply both sides by ##g_{\alpha\beta}## to get $$\delta g_{\beta\nu} = -g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \qquad(*)$$ (This is Dirac's eq. (26.9) in "GTR".) On the other hand, the variation ##\delta g^{\alpha\mu} = \bar{g}^{\alpha\mu} - g^{\alpha\mu}## should be a tensor...
Back
Top