[Mike Gale]

The point is that GR time dilation makes perfect sense. What's missing in your theory is a recipe for the new dilation. The only thing we have established so far is that it depends on the distance of the observer from M, but not on M itself. (Because, if it depended on M then it would be one and the same with GR dilation.)
Note that 0-gravity means M=0. There is no GR dilation in that case: dT=dt at all points in space.

[timey (OP)]

No - the 3rd time dilation is an ether type scenario and isn't observed by anyone except in the form of an accelerative force that increases at reduced h from M.
It doesn't need to be described as it already is being described within the current maths.

What does need to be re-described (for the purpose of my model), is GR time dilation, where the values remain the same, but for the reason that all m at h from M experiences this increased rate of time equally at any given elevation of h from M.

This differs from conventional relativity in that GR time dilation is no longer implying that time gets faster at h from M.  It states that time will only be getting faster for m at h from M.
The 3rd time dilation will be getting slower at h from M.
Each will be the equal and opposite of each other at any h from M.

[Mike Gale]

But it is already true that all m at h from M experience the same GR time dilation for a given value of M. You can make m as big or small as you like, as long as m>=0. I think you're talking about another dilation that depends on both m and M when the observer is colocated with m.

[timey (OP)]

But physics makes sums of m at h from M on the basis of a caesium atoms frequency always being 9,192,631,770Hz at any gravity potential...

No - I'm looking to unite the standard model with gravity.

[Mike Gale]

No. The number of oscillations per second (or the number of seconds per oscillation) depends on altitude. You can interpret that as time dilation or variable light speed. The net effect is the same.

[timey (OP)]

So why this obsession with observer dependency?
And if the atom really has a higher frequency at altitude it must have a higher energy...

[Mike Gale]

You need an observer to experience time. A clock is an observer. An observer is a clock. A clock at altitude runs faster because it has more potential energy.

[timey (OP)]

But time does not need an observer to operate.  Observer or not it's still does its thing.

[Mike Gale]

No. Time is meaningless without an observer to experience it. The observer doesn't have to be sentient though. Anything that oscillates will do. An atom for example.

[timey (OP)]

A clock at altitude runs faster because it has more potential energy.

Exactly... And all other mass at altitude has more potential energy which is why it experiences time at the same rate that the clock is ticking.  All particles will have more energy and higher frequency at altitude.

So this GR time dilation does not have to be associated with the open space surrounding mass that is the g-field being caused by M.


In reply to in between post, yes - all particles will oscillate.  The observation of this oscillation will be dependent on where the observer is observing from, but only because an observer holds his observation relative to the rate that his own clock is ticking.

[timey (OP)]

You may now mention that the pe=m times something or another equation, (it would take me time to look it up,) would cause different values of mass to have more or less potential energy at higher altitude than they would at lower altitude, and if the time dilation was potential energy related differing mass values would be subject to different degrees of time dilation at same altitude, which is not what we observe...
...and occilations must remain proportional to each other in every coordinate at h from M, to satisfy the equivalence principle.

[Mike Gale]

You don't need an observer to make the atom oscillate (at least not in a classical sense - some QM theorists would beg to differ.) The oscillation has consequences for the atom independent of any external observer. The point is that the atom is an observer of time and its mass can be arbitrarily small so that it has no significant impact on the passage of time as perceived by any other observer.

[timey (OP)]

Fair do's.  I am in agreement.
However it is the m in relation to M that causes it to have additional potential energy.  I don't see the atom gaining energy on its own steam, but as per being in a relationship with M.

But as per post above?

[Mike Gale]

The equation you're thinking of is PE=mgh. An object's experience of time at altitude does not depend on its mass. Time dilation depends on the ratio of PE to m (i.e. gh.) Note that g=GM/h². (These are all classical approximations of course.)

[timey (OP)]

Well in that case all the math can remain more or less the same.
But the interpretation of the reasons for the math have changed with the time dilation interpretation of the acceleration of gravity, where now one calculates time as stopped in the 0 gravity field, and we have a temporally deduced dilation of space where Hubble's velocity related interpretation of the red shift distance correlation is replaced by this temporal description were wavelength in the g-field is time dilation related, not distance related, and we can go look at adding energy to the particles of the black body of the black body experiment and apply the increase in occilations time dilation being experienced by the black body as being the cause of the emissions of higher energy and frequency photons...

[Mike Gale]

But time does not stop when g=0 (indeed, what would it stop relative to?) and concepts like velocity, frequency and energy are meaningless if time stands still. (We're talking about stationary reference frames of course so we can ignore SR.)

[timey (OP)]

My model would agree with you in that time does not stop for m in a 0 gravity field, for the reason that if m is there it is not a 0 gravity field, and you can't measure a 0 gravity field of m is not there.
m in a near non zero gravity field would have a fast rate, while the g-field itself would have a slow rate.

[Mike Gale]

What about light, for which m=0? You are suggesting that the passage of time depends on the value of m. That is demonstrably not true.

[timey (OP)]

Ah - well you see where lights frequency increases where it gets closer to M, where an m's frequency decreases when it gets closer to M is going in the opposing direction, therefore my model's says that light doesn't have potential energy because it has no mass, and it only experiences the acceleration or deceleration of the 3rd time dilation otherwise known as the acceleration, or deceleration of gravity.

The light gravitationally shifts and the extra/lesser length of wavelength is 3rd time dilation related,

[Mike Gale]

PE is a classical concept. It doesn't make sense for objects travelling at or near light speed or for objects that have no mass.
An observer perceives lower frequencies (i.e. red shift) when peering down a gravity well. You can interpret that as observer time speeding up due to its own mass rather than local time slowing down due to the remote object's proximity to the gravitating mass, but a third party of different mass perceives the same red shift as long as all observers are much lighter than the gravitating mass.