[jeffreyH (OP)]

If w is wavelength, f is frequency and p is momentum then can we say that energy = wfp?

[jeffreyH (OP)]

Read the first paragraph.
http://www.school-for-champions.com/science/waves_units.htm#.WK4XlbxFDIY

[jeffreyH (OP)]

Now if the velocity isn't constant we have a way of linking time dilation to particles with rest mass via the change in the wave characteristics.

[alancalverd]

No.

Kinetic energy =mv²/2

v=fw

p=mv

so fwp = mv²

[timey]

E = hv
where:
v = f*wavelength
and
wavelength = h/p
where:
p = h*vbar
and:
vbar = v/a

vbar is the bit relevant to wave frequency via v, and to time via a

In vector form:

p = hbar*k
where:
p = momentum
hbar = h/2pi
and
k = angular wave vector

Perhaps you need to consider a calculation involving components within the structure of
p = h*vbar
and
p = hbar*k
...where a and f and v, via v/a can be variable in the face of k.

(If that doesn't make sense, please accept my apologies in advance)

[Mike Gale]

If w is wavelength, f is frequency and p is momentum then can we say that energy = wfp?

That's true for a photon because wf=c in that case. In general, E^2=m^2c^4+p^2c^2.

[Mike Gale]

No.

Kinetic energy =mv²/2

v=fw

p=mv

so fwp = mv²

KE is only approximately equal to mv²/2. It is a classical concept, which must be abandoned in relativistic contexts.

[Mike Gale]

Now if the velocity isn't constant we have a way of linking time dilation to particles with rest mass via the change in the wave characteristics.

Exactly. The SC metric can be interpreted in terms of variable light speed. But it ultimately boils down to dilation of some sort, either spatial or temporal, just not both as in SR.

[jeffreyH (OP)]

No.

Kinetic energy =mv²/2

v=fw

p=mv

so fwp = mv²

Yes you are right, of course.  That was obvious really.

[jeffreyH (OP)]

So then if we reformulate as E = fwp/2 do we gain any advantage? Or is this still invalid?

[timey]

Invalid for what?
If you want to calculate a change in velocity due to time dilation then why are you looking at E without taking into account acceleration?

p = mv
and
v = f*wavelength

How would dividing fwp by 2 make an account of acceleration?

Edit:  p is already inclusive of v, which is itself inclusive of f*wavelength so perhaps this is why you are dividing by 2, but if so then you are also dividing the momentum by 2.
Why would dividing momentum in 2 have relevance?

[Mike Gale]

The factor of 1/2 only applies if m>0. It comes from:
E = sqrt(m²c⁴+ p²c²) = mc² * sqrt(1 + v²/c²) ~ mc² + mv²/2
There is no equivalent approximation for a photon. In that case m=0 and:
E = pc

[timey]

But Mike - Jeff is looking to describe a change in velocity as time dilation related for particles with rest mass and without.

Acceleration is the rate of change of velocity. Momentum is the mass times the velocity. So if you multiply the mass times the acceleration, you get the rate of change of momentum.

This rate of change is:
F = ma
and
F = momentum changes

This amounts to saying the same thing. Both are equal to mass times change in velocity divided by time. And therefore both are equal to F.

Do you think that Jeff might benefit from looking at using the Planck Einstein relation with regards to DeBroglie and somehow include vbar which is inclusive of v/a.

[Mike Gale]

I don't know what vbar is, but it is certainly true that F=ma. I think Jeff's point is that variable light speed imposes a force on a photon and that's why light rays curve in a gravitational field. He's right of course, but he's barking up the wrong tree trying to relate the momentum of a photon to that of a mass. The force on a photon is (p*dc/dt + c*dp/dt).

[timey]

I can't speak for what Jeff is doing, but I don't think it's trying to relate momentum of a photon to mass.

vbar is v/a
where:
p = h* vbar


https://en.m.wikipedia.org/wiki/Planck–Einstein_relation

But from my perspective I am interested because if variable light speed imposes a force on a photon, I am looking at that force being time dilation related.

Edit: And indeed the variable speeds of light being actually caused by that time dilation force.

[Mike Gale]

Yes. You can interpret variable light speed as time dilation in the absence of spatial dilation. Or the other way around.

[timey]

In the case of free fall it doesn't matter what m is, all m accelerates at same rate, so p = mv cannot describe momentum in the case of free fall.

[Mike Gale]

In the case of free fall it doesn't matter what m is, all m accelerates at same rate, so p = mv cannot describe momentum in the case of free fall.

Mass is observer-dependent because it depends on velocity. But photons have no mass so it's a moot point in that case. They do have momentum though and that is also observer-dependent if light speed is variable.

[timey]

So mass accelerating towards earth at a rate of 9.807 metres per second squared is observer dependant?

[Mike Gale]

So mass accelerating towards earth at a rate of 9.807 metres per second squared is observer dependant?

Yes, because mass depends on velocity.