I Cosmic Inflation Explained: Constant Velocity of Electromagnetic Radiation

JonathanMFreedman
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C = sqrt(E/M)...this would suppose the ratio of the amount of energy vs. the amount of mass in the universe. If not, why not. If there is no mass, just energy, or much less mass at the moment of the hypothetical Big Bang, then, there C would be significantly higher, thus explaining cosmic inflation.
 
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In the context of ##E=mc^2##, the ##c^2## is nothing more than a unit conversion factor between units of energy and units of mass. Don't try to read more into it than that.
 
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You might want to take a look at the PF Rules on personal theories. You can't just toss one out and expect us to explain "why not".
 
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The initial post does indeed violate the forum rule about personal theories, so we are closing the thread here.

@Ibix’s point about ##c^2## being just a unit conversion factor is well taken.
 
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I asked a question here, probably over 15 years ago on entanglement and I appreciated the thoughtful answers I received back then. The intervening years haven't made me any more knowledgeable in physics, so forgive my naïveté ! If a have a piece of paper in an area of high gravity, lets say near a black hole, and I draw a triangle on this paper and 'measure' the angles of the triangle, will they add to 180 degrees? How about if I'm looking at this paper outside of the (reasonable)...
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...
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