[Mike Gale]

Now that I have a little more time here is why this is not valid for photons. If E = wfp then since for the photon p = h/w then the equation becomes E = wfh/w = fh. So that we end up with fh = h/w. This then gives fw = h/h meaning that fw = 1. This cannot be true.
Scrub that it should be fh = hc/w. Which gives fw/c = h/h which is valid. Not the direction I was going in but interesting. I think I need a holiday!

You dropped a c: fh=hc/w. Oh. Never mind. I see you caught yourself.

[Mike Gale]

Ok so here is a conundrum. We can say that w = E/pf. If we want to define dw = ? for a remote frame in a far lower potential then what changes on the right hand side in order for us to calculate the change in wavelength? Is it everything in proportion? Just frequency? Energy or momentum? Will wf still equal c? Is this due to time dilation?

SC metric says that the perceived value of c at a remote location is different than the local one, but the product wf is always equal to whatever you perceive c to be. However, E=hf so perceived energy is smaller than local energy if you're looking down a gravity well and larger if you're lookup up.

[Mike Gale]

Unless you allow h to change, too. But that leads to more trouble I think. Specifically, conservation of energy.

[jeffreyH (OP)]

Ok so here is a conundrum. We can say that w = E/pf. If we want to define dw = ? for a remote frame in a far lower potential then what changes on the right hand side in order for us to calculate the change in wavelength? Is it everything in proportion? Just frequency? Energy or momentum? Will wf still equal c? Is this due to time dilation?

SC metric says that the perceived value of c at a remote location is different than the local one, but the product wf is always equal to whatever you perceive c to be. However, E=hf so perceived energy is smaller than local energy if you're looking down a gravity well and larger if you're lookup up.

So what if we state E = m(wf)²? At the event horizon we then have no rest energy. How can we have rest mass without rest energy. The perception from a remote frame is that all particles lose rest  energy at the horizon.
In which case gamma is of no consequence at that point so the infinity disappears.

[jeffreyH (OP)]

https://www.quora.com/What-happens-to-the-Higgs-field-boson-inside-a-black-hole
Since inside a black hole the Higg's field may break down then it is exactly at the event horizon that rest mass may disappear.

[timey]

Leaving you with a contravention of the conservation of energy law, no?

[jeffreyH (OP)]

That was the point of the question on kinetic energy. Only I made an untenable point about negative kinetic energy. This is different and involves invariant mass.

[timey]

Even worse!  If invariant mass disappears that is a more defined contravention.  With kinetic energy, there is the possibility that one can say that kinetic energy is not inherent to the mass itself, but with invariant mass, ie: rest mass, disappearing at the event horizon - this is a clear cut and well defined contravention of the conservation of energy law.

[Mike Gale]

So what if we state E = m(wf)²? At the event horizon we then have no rest energy. How can we have rest mass without rest energy. The perception from a remote frame is that all particles lose rest  energy at the horizon.
In which case gamma is of no consequence at that point so the infinity disappears.

Mass loses its meaning at the horizon because escape velocity is light speed. Same problem with the Higgs field. I suppose that's one reason GR theorists prefer 4-momentum. In practical terms, it means that the particle nature of mass is lost at the horizon and all that remains is a wave function, for which all points on the horizon are equally probable. In other words, kinetic energy is transformed into entropy.

[timey]

Mike - I can appreciate that you yourself are no doubt just writing in shorthand, but for anyone reading who might be misled:

Entropy is a state of function, therefore mass does not become entropy.
To calculate entropy one must divide by T = temperature.

The conundrum concerning black holes is that they do not have significant heat signature.  Mass that falls into a BH will reduce the temperature by the inverse square law proportional to the additional mass, which is the exact opposite of that which occurs in non-black hole physics.

This is what leads BH's to a contravention of the conservation of energy law, and the second law of thermodynamics.

Bekenstein suggested that a BH's entropy should be proportional to it's event horizon, and the conundrum was solved by Hawking's in the form of Bekenstein Hawking's radiation.

Nobody understands what happens to mass, or what it becomes when it enters the event horizon.

[Mike Gale]

Entropy is a state of function, therefore mass does not become entropy.
To calculate entropy one must divide by T = temperature.

Entropy is a statistical concept. It makes no sense when there is only one test mass at a known location. However, its location becomes more and more uncertain with increasing velocity. That uncertainty can be expressed as entropy. Note that temperature is a measure of kinetic energy. Google Hans Beth if you want to learn more about this.

[jeffreyH (OP)]

This would also be instructive for anyone interested.
http://hyperphysics.phy-astr.gsu.edu/hbase/Kinetic/kintem.html

[Mike Gale]

So what if we state E = m(wf)²? At the event horizon we then have no rest energy. How can we have rest mass without rest energy. The perception from a remote frame is that all particles lose rest  energy at the horizon.
In which case gamma is of no consequence at that point so the infinity disappears.

Don't forget about relativistic mass. Gamma diverges to infinity at the horizon so m=inf when c=0. Assuming free fall of course. If the mass is suspended somehow, then E_total can certainly be zero from the distant perspective. In that case, I think you're right. There will be rest mass with no rest energy. The rest mass still has energy in the local reference frame though.

[Mike Gale]

I think c approaches zero faster than m approaches infinity so it may well be true that E=0 at the horizon (in the free fall case.) Another reason black holes are black I suppose. I wonder what the gravitational potential is in that case. It must be huge, but maybe it doesn't matter if there is no rest mass energy. Perhaps the force of gravity needs to be expressed as a function of rest energy rather than rest mass. That is, G/c² is invariant rather than just G.
Here's another interesting effect. Radial distance in the local reference frame grows on approach to the horizon (as coordinate velocity increases.) From that perspective, the gravitating mass is accelerating away and reaches light speed when the test mass arrives at the horizon (at the end of time.) There has to be a point of inflection, where the acceleration of the test mass is insufficient to overcome the expansion of space. Gravity would be repulsive beyond that limit.