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On the subject of SR compatibility, the SC metric (old or new) for free fall (radial or orbital) can be expressed as:ds^2 = ( c dt / gamma )^2 - ( dr gamma )^2 - ( r d(angle) )^2Where (Lorentz's) gamma is defined in terms of escape velocity (classical or relativistic, depending on which metric you subscribe to) and proper light speed (c.) Note that escape velocity is equal to coordinate velocity (dr/dt) in the radial free fall case, but not the orbital one. Set ds=0 to find the coordinate velocity of light.

I'm pretty certain that my diagram, that places my model's variable 'times' for the speed of light into a Newtonian geometric, also results in Pythagoras.

Quote from: Mike Gale on 10/02/2017 02:15:32On the subject of SR compatibility, the SC metric (old or new) for free fall (radial or orbital) can be expressed as:ds^2 = ( c dt / gamma )^2 - ( dr gamma )^2 - ( r d(angle) )^2Where (Lorentz's) gamma is defined in terms of escape velocity (classical or relativistic, depending on which metric you subscribe to) and proper light speed (c.) Note that escape velocity is equal to coordinate velocity (dr/dt) in the radial free fall case, but not the orbital one. Set ds=0 to find the coordinate velocity of light.So with ds set to zero we end up with ( c dt / gamma )^2 = ( dr gamma )^2 + ( r d(angle) )^2. We are back to Pythagoras.

Quote from: timey on 11/02/2017 13:26:57I'm pretty certain that my diagram, that places my model's variable 'times' for the speed of light into a Newtonian geometric, also results in Pythagoras.Well with ( c dt / gamma )^2 = ( dr gamma )^2 + ( r d(angle) )^2 we end up with a distance divided by gamma on the left side. On the right is a distance multiplied by gamma. As one expands the other contracts. It's bigger on the inside.

What do you think that means for the holographic principle and Beckenstein's entropy?

If we look at kinetic energy we can say that1/2 gamma m dv^2 - (GM/dr) (gamma m) = 0is of interest. If we set a value for the right hand side in a strong field that would be interesting.

No that is not what I meant. If you progress through the equation you end up with 1/2 gamma m a^2 - v^2/2 (gamma m). So that the particle has a force moving it but the gravitational field appears to have no such force since it reduces to a kinetic energy term. Which is why people convince themselves that the gravitational force is fictitious.