Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: jeffreyH on 22/02/2017 20:17:24
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If w is wavelength, f is frequency and p is momentum then can we say that energy = wfp?
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Read the first paragraph.
http://www.school-for-champions.com/science/waves_units.htm#.WK4XlbxFDIY (http://www.school-for-champions.com/science/waves_units.htm#.WK4XlbxFDIY)
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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.
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No.
Kinetic energy =mv2/2
v=fw
p=mv
so fwp = mv2
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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)
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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.
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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.
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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.
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No.
Kinetic energy =mv2/2
v=fw
p=mv
so fwp = mv2
Yes you are right, of course. That was obvious really.
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So then if we reformulate as E = fwp/2 do we gain any advantage? Or is this still invalid?
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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?
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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
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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.
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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).
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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.
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Yes. You can interpret variable light speed as time dilation in the absence of spatial dilation. Or the other way around.
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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.
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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.
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So mass accelerating towards earth at a rate of 9.807 metres per second squared is observer dependant?
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So mass accelerating towards earth at a rate of 9.807 metres per second squared is observer dependant?
Yes, because mass depends on velocity.
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But free fall is indicating that velocity isn't mass dependent, so why would mass be velocity dependent?
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You're confusing the classical concept of free fall with the relativistic one. The equation of motion v=at holds in both cases. It is only the momentum (or energy) associated with v that is different.
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Well since relativistic effects have been experimentally confirmed at speeds under 30 miles an hour, then I think we can ditch the classical approach
And if it is momentum that changes with v = a, then in the case of free fall surely the energy change in momentum is due to that which is the cause of the acceleration, and not that which is being accelerated.
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It's a bit of both. Mass increases with velocity and velocity increases with acceleration. That's why GR theorists prefer 4-momentum. It reminds them that relativistic mass is a consequence of spacetime dilation. It's typical insider terminology though. They want to make you feel stupid if you prefer to think in terms of relativistic mass. It's so passe.
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Space time dilation of space, or space time dilation of time?
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Both. That's what SR teaches us.
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I've taken last few posts to your thread not to clog up Jeff's because we are getting off topic, but there's something I need to ask you.
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However... which I think that is relevant to this thread, I have said:
And if it is momentum that changes with v = a, then in the case of free fall surely the energy change in momentum is due to that which is the cause of the acceleration, and not that which is being accelerated.
... and you have said:
It's a bit of both. Mass increases with velocity and velocity increases with acceleration.
So the fact that velocity increases with acceleration is due to the energy of the accelerant, surely?
How can it be anything to do with the mass value when any value of mass accelerates at same rate in free fall?
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Free fall is really a classical concept. The relativistic equivalent is uniform motion along geodesics. The difference is that Newton "makes no hypothesis" about the cause of the force whereas Einstein attributes it to spacetime dilation.
Geodesics are easy to visualize when the test mass is spiralling around the gravitating mass. Falling straight down is a bit of a mind bender. GR theorists like to think of it as space flowing towards the central mass, dragging the test mass with it. (Or is it away from? I can never remember.) The alternative view is space getting squashed, but it's harder to account for motion in that case because you have to represent the test mass as a wave (or rather two of them, one in each reference frame.) In either case, it's much easier to understand if you can think of it in mathematical terms.
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Neither of you know what I am getting at since I haven't declared intent.
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The statement that a body falls to the Earth with an acceleration of 9.8m/s/s is an approximation that is only correct if it is falling from a short distance away the further away it starts from the weaker the gravitational attraction.
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Neither of you know what I am getting at since I haven't declared intent.
Jeff this is true - but the post below gives an inkling...
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.
Free fall is really a classical concept. The relativistic equivalent is uniform motion along geodesics. The difference is that Newton "makes no hypothesis" about the cause of the force whereas Einstein attributes it to spacetime dilation.
Geodesics are easy to visualize when the test mass is spiralling around the gravitating mass. Falling straight down is a bit of a mind bender. GR theorists like to think of it as space flowing towards the central mass, dragging the test mass with it. (Or is it away from? I can never remember.) The alternative view is space getting squashed, but it's harder to account for motion in that case because you have to represent the test mass as a wave (or rather two of them, one in each reference frame.) In either case, it's much easier to understand if you can think of it in mathematical terms.
Mike - If the accelerative force is temporally derived, then one can explain the acceleration in physical terms as well as being able to calculate.
Edit: This would mean that the g-field is temporally spatial, instead of spatially spatial, and lights wavelength will get shorter, and it's frequency will escalate in the greater g-field closer to M.
Test particles with mass will be accelerated into the greater g-field by the accelerative phenomenon of seconds becoming shorter in the g-field, but being of rest mass - they will themselves be inherent with longer wavelengths and a decreasing frequency as they are accelerated by the temporally derived g-field into the lower gravity potential.
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Neither of you know what I am getting at since I haven't declared intent.
Are you going to? I am intrigued.
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The statement that a body falls to the Earth with an acceleration of 9.8m/s/s is an approximation that is only correct if it is falling from a short distance away the further away it starts from the weaker the gravitational attraction.
Of course. It's just lazy language. Acceleration is obviously not constant in free fall.
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Mike - If the accelerative force is temporally derived, then one can explain the acceleration in physical terms as well as being able to calculate.
I think you meant to say that acceleration changes over time. That is correct for the free fall case, but it doesn't make the math any easier. Quite the contrary. It's probably easier to think of acceleration as a function of position rather than time because it doesn't change if you hold your position. I think that's the point you were trying to make in your edit.
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No - I'm suggesting that acceleration is due to time dilation. That it is changes in the rate of time inherent to the g-field itself that causes the phenomenon.
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Neither of you know what I am getting at since I haven't declared intent.
Are you going to? I am intrigued.
I still need to review a few things first. Then I'll be writing it up.
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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!
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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?
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No - I'm suggesting that acceleration is due to time dilation. That it is changes in the rate of time inherent to the g-field itself that causes the phenomenon.
It amounts to the same thing. If you give me something that depends on time, I can multiply by c and turn it into a spatial dependency. The question is whether the something actually changes over time when it's sitting still. If not then it is a spatial dependency.
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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.
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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.
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Unless you allow h to change, too. But that leads to more trouble I think. Specifically, conservation of energy.
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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.
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https://www.quora.com/What-happens-to-the-Higgs-field-boson-inside-a-black-hole (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.
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Leaving you with a contravention of the conservation of energy law, no?
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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.
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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.
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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.
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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.
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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.
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This would also be instructive for anyone interested.
http://hyperphysics.phy-astr.gsu.edu/hbase/Kinetic/kintem.html (http://hyperphysics.phy-astr.gsu.edu/hbase/Kinetic/kintem.html)
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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 Etotal 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.
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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.