Would the photon lose all its energy at infinity?

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Offline Bill S

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Re: Would the photon lose all its energy at infinity?
« Reply #100 on: 26/02/2015 18:46:51 »
Quote from: John
you send a 511keV photon down into a black hole, and the black hole mass increases by 511keV/c˛. No energy is acquired by the descending photon. In similar vein no energy is lost by the ascending photon.

There is something about this oft repeated assertion that puzzles me (so what’s new?)

If the 511keV photon [goes] down into a black hole, and the black hole mass increases by 511keV/c˛. That’s fine, and I wouldn’t argue with that; but does that mean that conservation of energy would prevent the photon from gaining, or losing, energy, as long as that energy came from, or passed to, the energy of the black hole?  In either case the total energy would remain constant.  Wouldn’t it?

NB, this thought came to me while dog walking, and by the time I came home PhysBang had posted #99, the two seemed to be linked.
There never was nothing.

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Offline yor_on

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Re: Would the photon lose all its energy at infinity?
« Reply #101 on: 26/02/2015 19:51:49 »
Try this one Bill.

E. Noether's Discovery of the Deep Connection Between Symmetries and Conservation Laws: By Nina Byers It discusses "`Proper' and `Improper' Conservation Laws", and refers also to gravity.

"In special relativity these theories have a `proper energy theorem' in the sense of Hilbert and we will show how `proper energy theorems' give a principle of local energy conservation. In general relativity, on the other hand, the proper energy theorem becomes improper in that the energy-momentum tensor for which the theorem holds is gauge dependent. As will be shown below, there is transfer of energy to and from the gravitational field and it has not meaning to speak of a definite localization of the energy of the gravitational field in space. Consequently we do not have a principle of local energy conservation in spacetime regions in which there exist gravitational fields."

Maybe this one will help too?
As that is what it hinges on.

What is a gauge?
« Last Edit: 26/02/2015 20:05:12 by yor_on »
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Offline David Cooper

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Re: Would the photon lose all its energy at infinity?
« Reply #102 on: 26/02/2015 22:12:54 »
c = √(1/ε0μ0)

That is where you were wrong until you corrected it in a later post.
It isn't wrong. What's the square root of a sixteenth? A quarter. And what's one divided by the square root of sixteen? A quarter.

I too thought c = √(1/ε0μ0) was different from c = 1/√(ε0μ0), but it isn't - they are equivalent, so John made no error. When PMB was misled into saying John had got this wrong, his dyslexia made him read c = √(1/ε0μ0) as c = √(ε0μ0), so his "correction" was an honest mistake. As usual this is just one great big misunderstanding.

John's appears to have been right all the way through this thread (and certainly on the main issue) - there is no energy loss to the photon as it climbs out of a gravity well and no gain when it enters one.
« Last Edit: 26/02/2015 22:18:17 by David Cooper »

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Offline PmbPhy

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Re: Would the photon lose all its energy at infinity?
« Reply #103 on: 26/02/2015 22:21:06 »
Quote from: David Cooper
I too thought c = √(1/ε0μ0) was different from c = 1/√(ε0μ0), but it isn't - they are equivalent, so John made no error. When PMB was misled into saying John had got this wrong, his dyslexia made him read c = √(1/ε0μ0) as c = √(ε0μ0), so his "correction" was an honest mistake. As usual this is just one great big misunderstanding.
Thanks, David. Much appreciated. It wasn't so much my dyslexia as it was John expressing the speed of light in a manner which it's never expressed and not pointing this fact out when Jeff also objected to it.

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Offline yor_on

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Re: Would the photon lose all its energy at infinity?
« Reply #104 on: 27/02/2015 00:06:30 »
The interesting thing, to me then, is still how one should define it losing energy in a expansion. Because as I've argued earlier, that's not 'observer dependent', at least not in the terms I'm used to think about it. I can use lights duality for it, but if anyone has another idea how to define it? Most probably the definition I use is the correct one though, and if it would be it should also be a proof of a real duality existing, or 'coexisting' if one like :)
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Offline yor_on

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Re: Would the photon lose all its energy at infinity?
« Reply #105 on: 27/02/2015 00:56:14 »
Light slows in a gravitational field. Unless gravitons are physically real, there is nothing in a gravitational field to take the place of atoms in other media. What slows light in a gravitational field?
The altered properties of space. In mechanics a shear wave travels at a speed v = √(G/ρ) where G is the shear modulus of elasticity and ρ is density. In electrodynamics the an electromagnetic wave travels at a speed c = √(1/ε0μ0) where ε0 is the permittivity of space and μ0 is the permeability.

John, How do you think there?
Are you arguing different permittivity to a perfect vacuum? Using a electromagnetic field to define it by I can accept, but the definition of permittivity is one where there is no friction and resistance, namely a vacuum. So either you mean that a electromagnetic field is a vacuum? Or that the vacuum we set as the standard to measure all other permittivity against varies? (or maybe both? Reading you again? although that would be highly contradictorily as it assumes Maxwell to be wrong, while still using his definitions)

"the parameter ε0 is a measurement-system constant. Its presence in the equations now used to define electromagnetic quantities is the result of the so-called "rationalization" process described below.

But the method of allocating a value to it is a consequence of the result that Maxwell's equations predict that, in free space, electromagnetic waves move with the speed of light. Understanding why ε0 has the value it does requires a brief understanding of the history."

And if you use Plank scale.

" Planck normalized to 1 the Coulomb force constant 1/(4πε0) (as does the cgs system of units). This sets the Planck impedance, ZP equal to Z0/4π, where Z0 is the characteristic impedance of free space.

Normalizing the permittivity of free space ε0 to 1: Sets the permeability of free space µ0 = 1, (because c = 1).

Sets the unit impedance or unit resistance to the characteristic impedance of free space, ZP = Z0 (or sets the characteristic impedance of free space Z0 to 1).
radius r). "


Finally "Both ε0 and μ0 appear in the Maxwell curl equations, so there are two free parameters, their product and their ratio. The two curl equations determine the propagation velocity of an electromagnetic wave in vacuum (c = 1/sqrt(μ0ε0)), and the ratio is related to the magnitude of E over H (sqrt(μ0/ε0) = 377 ohms). " By Bob S.

https://en.wikipedia.org/wiki/Vacuum_permittivity
https://en.wikipedia.org/wiki/Planck_units
https://www.physicsforums.com/threads/permittivity-and-permeability-of-free-space.122121/

==

the point is that a vacuum classically is 'empty'. A free charge can move through a vacuum without a EM field needed for it to propagate in. Quantum electrodynamics using the concept of a vacuum consisting of EM still have to follow that definition, although rationalizing it away as an effect of a 'vacuum ground state' equivalent to the classical counterpart without really explaining how it comes to be. And gravity is not electromagnetic. If you want to think of it as gravitons then this is interesting. http://www.physlink.com/Education/AskExperts/ae658.cfm

And QED is not really about waves, it's about quanta. so when it uses this concept it's conceptually different from defining it as waves, (although if you think as me the duality still should be there) But I'm having trouble following your reasoning here.
« Last Edit: 27/02/2015 02:49:40 by yor_on »
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Offline jeffreyH

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Re: Would the photon lose all its energy at infinity?
« Reply #106 on: 27/02/2015 02:19:19 »
Quote from: David Cooper
I too thought c = √(1/ε0μ0) was different from c = 1/√(ε0μ0), but it isn't - they are equivalent, so John made no error. When PMB was misled into saying John had got this wrong, his dyslexia made him read c = √(1/ε0μ0) as c = √(ε0μ0), so his "correction" was an honest mistake. As usual this is just one great big misunderstanding.
Thanks, David. Much appreciated. It wasn't so much my dyslexia as it was John expressing the speed of light in a manner which it's never expressed and not pointing this fact out when Jeff also objected to it.

My objection is to the fact that this is OK with unity as a numerator but 15/SQRT(16) for instance is not the same as SQRT(15/16). So there is a distinction. It can matter. So to pose it in the correct form is not trivial. Especially if the numerator could be variable. In this case it is not so doesn't matter.
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Offline jeffreyH

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Re: Would the photon lose all its energy at infinity?
« Reply #107 on: 27/02/2015 02:31:04 »
See for example hyperphysics where you can read this: "The conservation of energy principle is one of the foundation principles of all science disciplines. In varied areas of science there will be primary equations which can be seen to be just an appropriate reformulation of the principle of conservation of energy". You gave a Lagrangian for a massive particle in a gravitational field, wherein the first portion is the kinetic energy, and the mgz is the potential energy. This is not appropriate for a photon, because the photon is massless, and it's all kinetic energy. If you throw a massive particle upwards, kinetic energy is converted into potential energy. When all the kinetic energy is converted into potential energy, the particle has reached its highest point, and then it start falling back down, converting potential energy into kinetic energy. When you send a photon upwards, it doesn't slow down and stop. Instead, it speeds up. If you don't believe me, contact Don Koks, the editor of the Baez/PhysFAQ website, who said this:

"Now use the Equivalence Principle to infer that in the room you are sitting in right now on Earth, where real gravity is present and you aren't really accelerating (we'll neglect Earth's rotation!), light and time must behave in the same way to a high approximation: light speeds up as it ascends from floor to ceiling (it doesn't slow down, as apparently quoted on your discussion site), and it slows down as it descends from ceiling to floor; it's not like a ball that slows on the way up and goes faster on the way down..."   
In General Relativity, we are free to use systems of coordinates in which the coordinate speed of light over finite distances can change. This is one way to represent the change in the energy of light from the effect of gravity on that light. In other systems of coordinates, we use the change in frequency of the light to represent the change in energy of the light due to gravity.

Because the light can only be represented as kinetic energy, the only way to represent the change in energy is in the kinetic energy, either through speed or frequency. When it comes to an absorption event, in the system of coordinates in which the absorber is at rest, the light is absorbed at a higher or lower frequency depending on the way that gravity has changed the photon.

In a standard application of this, in a photon traveling away from or towards the Earth, for example, there is no concern about the conservation of energy as the energy of the entire system is conserved.

Yes its the whole system's energy. As light is all kinetic energy you have a different system to that of other lower velocity particles. I am actually impressed by John for once I must admit. He actually did know what the Lagrangian was.
« Last Edit: 27/02/2015 09:43:39 by evan_au »
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Offline jeffreyH

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Re: Would the photon lose all its energy at infinity?
« Reply #108 on: 27/02/2015 02:53:56 »
Try this one Bill.

E. Noether's Discovery of the Deep Connection Between Symmetries and Conservation Laws: By Nina Byers It discusses "`Proper' and `Improper' Conservation Laws", and refers also to gravity.

"In special relativity these theories have a `proper energy theorem' in the sense of Hilbert and we will show how `proper energy theorems' give a principle of local energy conservation. In general relativity, on the other hand, the proper energy theorem becomes improper in that the energy-momentum tensor for which the theorem holds is gauge dependent. As will be shown below, there is transfer of energy to and from the gravitational field and it has not meaning to speak of a definite localization of the energy of the gravitational field in space. Consequently we do not have a principle of local energy conservation in spacetime regions in which there exist gravitational fields."

Maybe this one will help too?
As that is what it hinges on.

What is a gauge?

Great post. I am just working through conservation laws and symmetries. The important point to note is "As will be shown below, there is transfer of energy to and from the gravitational field and it has not meaning to speak of a definite localization of the energy of the gravitational field in space. Consequently we do not have a principle of local energy conservation in spacetime regions in which there exist gravitational fields." When viewed in respect of the kinetic nature of the photon with zero rest mass the transfer is from the photon to the gravitational field. John declares that the energy is radiated away into space. How does this conserve energy in the system?
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Offline jeffreyH

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Re: Would the photon lose all its energy at infinity?
« Reply #109 on: 27/02/2015 02:59:36 »
One more thing. John talks of a mass deficit when a brick falls freely in a gravitational field. How does this square with the fact that a mass weighs more at lower altitudes. Surely it should get lighter in that case? Energy is undefined to all intents and purposes and is simply a vehicle in equations of force and therefore motion. Energy changes in a system and therefore the total mass. How can we even say that mass can be defined when energy is a critical component.
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Offline yor_on

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Re: Would the photon lose all its energy at infinity?
« Reply #110 on: 27/02/2015 04:01:35 »
Are you thinking of what should be defined as rest mass there Jeffrey? And I thought John defined a photon as intrinsically being the same, no matter observer dependencies? It makes sense to me too defining it that way, although in a measurement also depending on the circumstances under which it is measured. You write that "energy is radiated away into space.". Think I have to reread the whole thread here :)

All photons leaving a 'gravity well' as a sun should redshift from the view of a thought up observer in that sun. But we're still 'visited' by photons traveling since the beginning of the Big Bang, Billions of light years, and they seem the same as any other 'photons', as far as I know?  That is, if we adapt it for age-(distance) redshifts ('expansion' being one cause) taking place. If we want to take it to its extreme we'll have to assume some final eruption from a star, soon to become a 'black hole'. Would that mean that some photons then lose their energy and 'die out' propagating? As the gravity well (sun) balance on this edge of becoming a black hole? I think that is called a 'tired light theory' myself? And as gravity's reach is defined as 'infinite' you then can argue the same for its whole 'propagation'.

"A slightly different kind of supernova explosion occurs when even larger, hotter stars (blue giants and blue supergiants) reach the end of their short, dramatic lives. These stars are hot enough to burn not just hydrogen and helium as fuel, but also carbon, oxygen and silicon. Eventually, the fusion in these stars forms the element iron (which is the most stable of all nuclei, and will not easily fuse into heavier elements), which effectively ends the nuclear fusion process within the star. Lacking fuel for fusion, the temperature of the star decreases and the rate of collapse due to gravity increases, ..... until it collapse completely on itself, blowing out material in a massive supernova explosion..... "
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Offline jeffreyH

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Re: Would the photon lose all its energy at infinity?
« Reply #111 on: 27/02/2015 05:00:45 »
So take the lagrangian [tex]L = \frac {1}{2}m(\dot{x}^2 + \dot{y}^2 + \dot{z}^2) - mgz[/tex]. This is not applicable to the photon. If we remove the term [tex] - mgz[/tex] then we are working with a directional velocity only [tex]\frac {1}{2}(\dot{x}^2 + \dot{y}^2 + \dot{z}^2)[/tex]. In the absence of any gravitational field we can orient the path of the photon along the z-axis so that x and y terms vanish. The only way we can maintain this path is by somehow defining a flat spacetime that does not disturb the path. In this way we can then view the effect of gravity on a wave function without the added complexity. The potential and kinetic energy terms then need a relativistic mass equation for the photon. The wave function is not usually a one particle affair. However it is a very useful exercise.

For an inertial frame the directional velocity is always c in a vacuum so this can be considered a constant. This only changes due to the gravitational field. With the inclusion of relativistic mass we can restore momentum to the photon.
« Last Edit: 27/02/2015 05:11:24 by jeffreyH »
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Offline jeffreyH

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Re: Would the photon lose all its energy at infinity?
« Reply #112 on: 27/02/2015 05:26:11 »
Thanks to a page on Pete's site we have the relativistic mass defined as m = p/c. Where p is the momentum and c is the speed of light. The momentum is defined as p = hf/c where h is Planck's constant, f is the frequency and again c is the speed of light. This is where the wave enters via its frequency used to define its relativistic mass.

In basic form we have the relation E = mc^2. Since frequency is part of the relativistic mass of the photon we have an energy frequency relationship when viewed in this way. However this is used to illustrate the point and not to indicate that E = mc^2 can be directly applied to photons. Just in case anyone complains.
« Last Edit: 27/02/2015 05:28:44 by jeffreyH »
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Offline jeffreyH

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Re: Would the photon lose all its energy at infinity?
« Reply #113 on: 27/02/2015 05:38:37 »
The confusion comes about when considering the frames of an observer. In a remote frame the energy of the photon moving into a gravitational field will appear to change. In all frames where the observer is coincident with the photon the energy will be constant. This is a given. This does show the direct relationship between time dilation and the expression of wavelength, frequency and energy as viewed by remote observers. An increase in frequency is therefore linked to the increase in the speed of change due to effects on time.
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Offline jeffreyH

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Re: Would the photon lose all its energy at infinity?
« Reply #114 on: 27/02/2015 05:40:04 »
One consequence of this is that the a percentage of the acceleration due to gravity is simply due to the increase in rate of change.
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Offline PmbPhy

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Re: Would the photon lose all its energy at infinity?
« Reply #115 on: 27/02/2015 07:40:29 »
Quote from: jeffreyH
Thanks to a page on Pete's site we have the relativistic mass defined as m = p/c.
I'm glad I was able to be of service in that way. :)

Which page are you referring to if I may I ask?

Quote from: jeffreyH
However this is used to illustrate the point and not to indicate that E = mc^2 can be directly applied to photons. Just in case anyone complains.
No. That is correct. You really can use it in that way. See:
http://home.comcast.net/~peter.m.brown/ref/relativistic_mass.htm
and notice that what you just described appears in these well known special relativity textbooks:

Relativity: Special, General and Cosmological by Rindler, Oxford Univ., Press, (2001), page 120
From Introducing Einstein's Relativity by Ray D'Inverno, Oxford Univ. Press, (1992), page 50
Special Relativity by A. P. French, MIT Press, page 20

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Offline jeffreyH

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Offline jeffreyH

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Re: Would the photon lose all its energy at infinity?
« Reply #117 on: 27/02/2015 08:54:05 »
Quote from: jeffreyH
Thanks to a page on Pete's site we have the relativistic mass defined as m = p/c.
I'm glad I was able to be of service in that way. :)

Which page are you referring to if I may I ask?

Quote from: jeffreyH
However this is used to illustrate the point and not to indicate that E = mc^2 can be directly applied to photons. Just in case anyone complains.
No. That is correct. You really can use it in that way. See:
http://home.comcast.net/~peter.m.brown/ref/relativistic_mass.htm
and notice that what you just described appears in these well known special relativity textbooks:

Relativity: Special, General and Cosmological by Rindler, Oxford Univ., Press, (2001), page 120
From Introducing Einstein's Relativity by Ray D'Inverno, Oxford Univ. Press, (1992), page 50
Special Relativity by A. P. French, MIT Press, page 20

Thanks Pete. I tell you two things I don't like. 1) Being told I only talk Pop Science nonsense and 2) I am peddling some personal theory. I have worked hard and put in the time to get to the point I am at now. This stuff isn't easy so you have to be committed to learning it. This may mean you have to read some texts through multiple times or read various texts to pick things up. To be dismissed in such an offhand way is insulting.
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Offline JohnDuffield

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Re: Would the photon lose all its energy at infinity?
« Reply #118 on: 27/02/2015 09:49:44 »
If the 511keV photon [goes] down into a black hole, and the black hole mass increases by 511keV/c˛. That’s fine, and I wouldn’t argue with that;
Good stuff. IMHO conservation of energy is just about the most important rule in physics.

but does that mean that conservation of energy would prevent the photon from gaining, or losing, energy, as long as that energy came from, or passed to, the energy of the black hole?
No. If one thing gains energy, another thing loses it. But think about it. That photon is descending at the speed of light. If it's gaining energy, how is it getting it? If it starts off 93 million miles from the black hole, it takes 8 minutes for anything to get from the black hole to the photon. And things can't get out of a black hole. There is no magical mysterious mechanism by which energy can get from the black hole to the photon instantaneously.

In either case the total energy would remain constant. Wouldn’t it?
Yes. This is the thing about gravity. There are websites and books out there that say gravity is negative energy, when it isn't. When two things fall together, energy is conserved. The books always balance. The total energy of the universe is not zero.

David: thanks re the c = √(1/ε0μ0) expression.

Quote from: yor_on
The interesting thing, to me then, is still how one should define it losing energy in a expansion...
Me too. People say the CMBR photons have redshifted a thousandfold and lost most of their energy. But nobody seems to be able to say where it's gone. By the way, I think the Nina Byers paper is the sort of thing that causes confusion. 

Quote from: yor_on
Are you arguing different permittivity to a perfect vacuum?
No. It's a perfect vacuum, but like Einstein said, a concentration of energy in the guise of a massive star "conditions" the surrounding space, altering its metrical properties, such that "the speed of light is spatially variable". Then c = √(1/ε0μ0) at one location is not the same as c = √(1/ε0μ0) at another. The permittivity and/or permeability of space at one elevation is not the same as at another. But wherever you go you measure things like c and ε0 to be the same because it's an "immersive scale change". It's a bit like a flatlander trying to measure whether his world has been stretched using a ruler that's also been stretched.   

Quote from: yor_on
the point is that a vacuum classically is 'empty'.
That's not what Maxwell or Einstein thought. They thought of space as a gin-clear ghostly elastic thing, something that has properties, something that can be stressed, something that can wave. Check out this where you can read Einstein saying a field is "a state of space". Also see the  arXiv and this too. 

Quote from: yor_on
And gravity is not electromagnetic. If you want to think of it as gravitons then this is interesting.
Einstein struggled for years trying to combine gravity and electromagnetism. As for gravitons, they're "field quanta", they're virtual particles, not real particles. It's like you divide a field up into chunks and say each one is a virtual particle. And according to Einstein, a field is a state of space. Now, how many states of space are there where an electron is? One.


Quote from: Jeffrey
He actually did know what the Lagrangian was.
My physics knowledge is extensive. That sometimes causes problems when it conflicts with some popscience book or science article, or even a textbook or paper.

Quote from: Jeffrey
Great post. I am just working through conservation laws and symmetries. The important point to note is "As will be shown below, there is transfer of energy to and from the gravitational field and it has not meaning to speak of a definite localization of the energy of the gravitational field in space. Consequently we do not have a principle of local energy conservation in spacetime regions in which there exist gravitational fields."
I think this paper causes confusion, and I would recommend that you set it aside.

Quote from: Jeffrey
When viewed in respect of the kinetic nature of the photon with zero rest mass the transfer is from the photon to the gravitational field.
There is no transfer of energy from the photon to the gravitational field.

Quote from: Jeffrey
John declares that the energy is radiated away into space. How does this conserve energy in the system?
This applies to two massive bodies, not to the photon descending into the black hole.   

Quote from: Jeffrey
One more thing. John talks of a mass deficit when a brick falls freely in a gravitational field. How does this square with the fact that a mass weighs more at lower altitudes.
Because matter is affected half as much as light. If you replace the falling brick with a 511keV/c˛ electron and compare with a descending 511keV photon, the potential energy converted into kinetic energy is half the apparent energy gain of the photon. So imagine your brick has a mass x at some elevation. You weigh it using light in some guise. Then at the lower elevation, the brick has a mass which is less than x. But when you weigh it using light, you use light at what you think is the same frequency as before, when actually it's at a lower frequency. So the brick weighs heavier. Remember that at the lower elevation, you and your clocks are all going slower. So an descending photon appears to have a higher frequency, when it doesn't. It didn't gain energy, you lost it. You have a mass deficit too.   
« Last Edit: 27/02/2015 09:53:38 by JohnDuffield »

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Offline evan_au

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Re: Would the photon lose all its energy at infinity?
« Reply #119 on: 27/02/2015 10:25:24 »
Quote from: yor_on
the vacuum we set as the standard to measure all other permittivity against
As I understand it, ε0 and μ0 are constants of the universe (like c), when you measure them locally.

In selecting the best material for a particular application, an electrical engineer is interested in the relative permittivity & permeability, εr and μr, which are both >1*.

The speed of light in this non-vacuum environment is now v= (1/√(ε0μ0))(1/√(εrμr)) = c/√(εrμr) =c/η

where η is the index of refraction (of great interest to opticians), and η=√(εrμr)

*except in certain metamaterials, under quite restrictive conditions.

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Offline JohnDuffield

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Re: Would the photon lose all its energy at infinity?
« Reply #120 on: 27/02/2015 13:19:39 »
Given that we talk about gravitational lensing, and Einstein described a gravitational field as space that is "neither homogeneous not isotropic", I think this is worth a read:

Inhomogeneous Vacuum: An Alternative Interpretation of Curved Spacetime
"The strong similarities between the light propagation in a curved spacetime and that in a medium with graded refractive index are found. It is pointed out that a curved spacetime is equivalent to an inhomogeneous vacuum for light propagation. The corresponding graded refractive index of the vacuum in a static spherically symmetrical gravitational field is derived. This result provides a simple and convenient way to analyse the gravitational lensing in astrophysics."
« Last Edit: 27/02/2015 14:04:29 by JohnDuffield »

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Offline PhysBang

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Re: Would the photon lose all its energy at infinity?
« Reply #121 on: 27/02/2015 13:32:42 »
but does that mean that conservation of energy would prevent the photon from gaining, or losing, energy, as long as that energy came from, or passed to, the energy of the black hole?
No. If one thing gains energy, another thing loses it. But think about it. That photon is descending at the speed of light. If it's gaining energy, how is it getting it? If it starts off 93 million miles from the black hole, it takes 8 minutes for anything to get from the black hole to the photon. And things can't get out of a black hole. There is no magical mysterious mechanism by which energy can get from the black hole to the photon instantaneously.
No, there is just the operation of gravity. Do you believe in Newton's third law?
Quote
In either case the total energy would remain constant. Wouldn’t it?
Yes. This is the thing about gravity. There are websites and books out there that say gravity is negative energy, when it isn't. When two things fall together, energy is conserved. The books always balance. The total energy of the universe is not zero.
Gravitational potential energy is a relative term that depends on the arbitrary choice of a zero point. Thus that can be negative. That's just a mathematical fact.
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Quote from: yor_on
Are you arguing different permittivity to a perfect vacuum?
No. It's a perfect vacuum, but like Einstein said, a concentration of energy in the guise of a massive star "conditions" the surrounding space, altering its metrical properties, such that "the speed of light is spatially variable". Then c = √(1/ε0μ0) at one location is not the same as c = √(1/ε0μ0) at another. The permittivity and/or permeability of space at one elevation is not the same as at another. But wherever you go you measure things like c and ε0 to be the same because it's an "immersive scale change". It's a bit like a flatlander trying to measure whether his world has been stretched using a ruler that's also been stretched.   
The real question is whether this can be used to do any physics. So far, I have not seen any evidence, including in the one paper that JohnDuffield cited, that one can use this equation to a description of motion that matches the observations we have of gravitational phenomena.

Quote
Quote from: yor_on
the point is that a vacuum classically is 'empty'.
That's not what Maxwell or Einstein thought. They thought of space as a gin-clear ghostly elastic thing, something that has properties, something that can be stressed, something that can wave. Check out this where you can read Einstein saying a field is "a state of space". Also see the  arXiv and this too. 
I think it better to ignore Einstein's lecture, where he said that space was filled with stress-energy and read a textbook on GR instead. As Einstein knew when he describe the aether as the stress-energy tensor, this means that space, or rather spacetime (since the tensor is described only over spacetime), is filled not with stuff but with mathematical relationships.
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Quote from: Jeffrey
He actually did know what the Lagrangian was.
My physics knowledge is extensive. That sometimes causes problems when it conflicts with some popscience book or science article, or even a textbook or paper.
Then could you please show us an example of how to calculate, say, the fall of a pencil using your vacuum permeability equation? I have yet to see an example of how your ideas actually work. Given that you identify yourself as an authority here, I would love to see the examples.

Quote from: Jeffrey
Great post. I am just working through conservation laws and symmetries. The important point to note is "As will be shown below, there is transfer of energy to and from the gravitational field and it has not meaning to speak of a definite localization of the energy of the gravitational field in space. Consequently we do not have a principle of local energy conservation in spacetime regions in which there exist gravitational fields."
I think this paper causes confusion, and I would recommend that you set it aside. [/quote]
I agree that if one reads scientific articles, then one might be tempted to come to the conclusion that JohnDuffield's claims are false. If one wants merely to believe them, then please avaoid reading scientific articles or texts.

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Offline jeffreyH

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Re: Would the photon lose all its energy at infinity?
« Reply #122 on: 27/02/2015 14:00:56 »
John, when I said you impressed me with the Lagrangian I was being flippant. I have never seen any evidence that you understand how the physics work which is apparent by the Pop Science things you say. You have enough of these now to put forward your pet theory.

The shame is you would probably make a very good science historian.
Fixation on the Einstein papers is a good definition of OCD.

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Re: Would the photon lose all its energy at infinity?
« Reply #123 on: 27/02/2015 14:09:46 »
Quote from: yor_on
the vacuum we set as the standard to measure all other permittivity against
As I understand it, ε0 and μ0 are constants of the universe (like c), when you measure them locally.

In selecting the best material for a particular application, an electrical engineer is interested in the relative permittivity & permeability, εr and μr, which are both >1*.

The speed of light in this non-vacuum environment is now v= (1/√(ε0μ0))(1/√(εrμr)) = c/√(εrμr) =c/η

where η is the index of refraction (of great interest to opticians), and η=√(εrμr)

*except in certain metamaterials, under quite restrictive conditions.

Above is an example of where the square root should be as in c/√(εrμr). It does matter and John should know this.
Fixation on the Einstein papers is a good definition of OCD.

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Re: Would the photon lose all its energy at infinity?
« Reply #124 on: 27/02/2015 14:12:55 »
Well, either you assume a different permittivity John, or you are proposing some to me unknown, not observer dependent mechanism? Because you are indeed telling me that light 'slows down'. If you're using another definition that whole post I reacted on was unnecessary. I saw Pete using time dilations to define it but in your case I still don't know what you use? I don't mind people having pets, I have them too :) Writing "The permittivity and/or permeability of space at one elevation is not the same as at another." either seem to assume that there is one observer dependent redshift and another that solely belong to the 'photon' propagating, or that you are thinking that all gravitational redshifts are outside observer dependencies?. The last one makes a joke of any idea defining a photon as intrinsically being the same in a gravitational redshift, the first one is new to me, two mechanisms for a gravitational redshift, and I think it has to be proved.
=

The point is that you don't like different 'paths' as I gather?
« Last Edit: 28/02/2015 01:31:23 by evan_au »
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Re: Would the photon lose all its energy at infinity?
« Reply #125 on: 27/02/2015 14:36:12 »
Well, either you assume a different permittivity John, or you are proposing some to me unknown, not observer dependent mechanism?
The permittivity is different at a lower location. Only when you go there, you're different too, so the permittivity appears to be unchanged.

Because you are indeed telling me that light 'slows down'.
Yes, that's what Einstein said, repeatedly. Check out the Einstein digital archives, and search on speed of light or velocity of light.   

If you're using another definition that whole post I reacted on was unnecessary. I saw Peter using time dilations to define it but in your case I still don't know what you use? And you're starting to come on as arrogant, which I find slightly surprising.
I meant to just give a straight answer that was factually correct and in line with Einstein and the evidence.

I don't mind people having pets, I have them too :) Writing "The permittivity and/or permeability of space at one elevation is not the same as at another." either seem to assume that there is one observer dependent redshift and another that solely belong to the 'photon' propagating, or that you still are thinking that all gravitational redshifts are outside observer dependencies?
I'm sorry, I'm not clear what you mean here. Please restate.

The last one makes a joke of any idea defining a photon as intrinsically being the same in a gravitational redshift, the first one is new to me, two mechanisms for a redshift, and I think it has to be proved.
We observe gravitational redshift, there's no issue in proving that. And IMHO conservation of energy says there shouldn't be any issue in proving that we observe it because we change. When you send a 511keV photon into a black hole, the black hole mass doesn't increase by a zillion tonnes. People say the descending photon is subject to an infinite blueshift, but its E=hf energy did not increase. But when you fall down, some of your mass-energy is converted into kinetic energy which is radiated away, leaving you with a mass deficit and less mass-energy. So the unchanged photon energy looks like it's increased.   

The point is that you don't like different 'paths' as I gather?
Again, I'm not clear what you mean. If you're referring to the photon taking many paths, I'm totally happy with that. I've used the analogy of a seismic wave to for that.

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Offline yor_on

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Re: Would the photon lose all its energy at infinity?
« Reply #126 on: 27/02/2015 14:46:45 »
Actually the whole premise of propagation is just a pain in the ** to me John. Using a field as I think of it, one can avoid those questions. They belong to the container society.
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Re: Would the photon lose all its energy at infinity?
« Reply #127 on: 27/02/2015 14:48:19 »
Try to read it as it stands. "Writing "The permittivity and/or permeability of space at one elevation is not the same as at another." either seem to assume that there is one observer dependent redshift and another that solely belong to the 'photon' propagating, or that you are thinking that all gravitational redshifts are outside observer dependencies?"

It's clear to me, what do you find unclear there?

Ignore 'still' btw, that was my fingers, not me :)
took it away as I saw it.
=

Let's use the full text btw, because it's the conclusions that make a difference there.

"Writing "The permittivity and/or permeability of space at one elevation is not the same as at another." either seem to assume that there is one observer dependent redshift and another that solely belong to the 'photon' propagating, or that you are thinking that all gravitational redshifts are outside observer dependencies?.

The last one makes a joke of any idea defining a photon as intrinsically being the same in a gravitational redshift, the first one is new to me, two mechanisms for a gravitational redshift, and I think it has to be proved."

« Last Edit: 27/02/2015 14:59:33 by yor_on »
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Re: Would the photon lose all its energy at infinity?
« Reply #128 on: 27/02/2015 15:15:40 »
The observer dependent redshift belongs to the equivalence principle in my thoughts. With some translations it will define our earth as equivalent to a moving (constantly uniformly accelerating) spaceship at one constant gravity (ignoring spin). That's why I think of it as observer dependent although you measure this redshift being 'at rest' with Earth. I'm open for other interpretations though, but they need to fit what I understand to be relativity, or prove my toughts incorrect. It's actually easier to measure the blueshift thinking of it :) As you don't have to 'levitate', just stay on the ground, but the principle is complementary to measuring light leaving your spaceship, going in the 'opposite direction' (red shift, as defined from some inertial observer) as well as incoming light in the direction of your motion (giving you a blueshift) That's one of the things making Einstein the most awesome thinker I've read, he's so clear in his thinking. It's easy if one just give oneself some time to think it through (that's directed to all & Pete, we seem to share a similar taste there:)
=

So to summarize, if I'm correct gravitational redshifts is observer dependent, or the equivalence principle has to be changed in some way. If it is correct, then there is no way, that I know of, giving this principle (gravitational redshifts) two different mechanisms, one observer dependent the other belonging to some 'intrinsic principle' embedded in the photon solely.

thats one

Two

No way this is 'outside' observer dependencies. If it was then we can throw away GR. As far as I understands it.
=

and yes, it makes a 'photon' intrinsically the same, which is how I first read you John.
« Last Edit: 27/02/2015 16:00:34 by yor_on »
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Re: Would the photon lose all its energy at infinity?
« Reply #129 on: 27/02/2015 16:31:55 »
Ok I think we've eliminated redshifts and their complementary time dilations from belonging to some 'intrinsic principle', but we have not eliminated a twin experiment. And it is not about a 'vacuum thickening its density' (permittivety), as far as I get it John? 

A twin experiment is between two 'clocks', of a same origin, one accelerating to uniformly move away from that origin, to then return, finding one twins clock having been slower. Now, in my definitions this is a result of frames of reference interacting, but it is not a result of one clock ticking slower (or faster). Why I define it that way has to do with a lot of things, our definitions of repeatable experiments for one. observers in different uniform motions measuring each others clocks another, Also it has to do with you measuring, you won't ever find your clocks 'pace' to become 'slower' or 'faster', and your life span is the same, no matter what you do or how 'fast' you are. That one need a lot of reading up on to see how I think. But thinking that way, locality never change its 'pace', and that fit all experiments you can do, no matter NIST.

And it's not 'light paths', if I now remember you correct?
=

the easiest way to see how I mean is to accept the definition of 'c', then define it as valid for all circumstances instead of arguing that 'c' has different values due to mass and accelerations. And make it equivalent your local 'clock'. If I would want it otherwise, still setting 'c' to my local clock (arrow), I would be forced to define 'c' to different values, not only in accelerations (and mass), but also in uniform motions. And 'c' and your local arrow is the same, ideally and experimentally, prove me wrong.

The only way you can go around the last would be arguing that a acceleration is the sole instigator of a time dilation, making any ideas of 'muon experiments' wrong, as we already have defined what a gravitational red (as well as blue) shift is. They are observer dependent, and so depending on your frame of reference (Earth, or 'free falling' uniformly moving Muon in this case) A gravitational acceleration is a 'free fall', no 'forces' acting upon you locally defined, and I define it locally. As well as the logic of defining it soley to accelerations won't work.

This one is really easy to understand. Just jump out a plane without a parachute, and see the Earth rush to meet you. Because that is what you will feel, ignoring wind and atmosphere tugging on you. It's frames of reference and in Einsteins world the Earth is constantly uniformly accelerating, at one gravity.

that doesn't mean that accelerations and mass doesn't count, everything count, mass energy, accelerations, uniform motion. To me it all seems to have to do with frames of reference interacting.
« Last Edit: 27/02/2015 17:47:50 by yor_on »
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Re: Would the photon lose all its energy at infinity?
« Reply #130 on: 28/02/2015 10:37:52 »
And that brings us to how to define a frame of reference, ideally, or from scaling. From scaling you could use some positional system, locally defined, Then try to see how that (it's still 'ideally' defined though in some sense) frame interacts with others, like NIST do in their experiments with different clocks at different locations. There is no way I see to define that local arrow I have to some (exact) position in a space and time though, as in 'there it is' :) And you will meet the same problem scaling, especially if the 'common clock' is created through interactions. That means that a 'ground beat' should exist, that's what makes repeatable experiments, and make us agree on a time for something, we being 'at rest' with, and in, some common frame of reference. But it also talks about the 'direction of time', locally measured, as something defined by those local interactions, interacting with each other on a larger scale, through 'c'. You could call it a bifurcation of Mach's principle, related to the common universe's experience of a global time, best described through our definitions of a Big Bang, applicable wherever you go in a 'infinite universe'. If we have a 'bit' then that should be the final arbiter of what a local position is, but to give it a 'time' you always will use your local clock.
=

what that state to me is that we always will be looking through tinted glasses defining a position, scaling. We have our ideal clock and ruler, locally invariant, and we all agree on it being equivalent. If we didn't repeatable experiments cease to exist. The real point is that they are equivalent, that is what have made physics. Now consider that through scaling, would you expect this to become untrue, scaling something down? The terms of how it interact may change, but I would expect this to hold true all the way down under.

What I, very tentatively, see such a reasoning doing, is giving us a local time (clock) equivalent to 'c', created through interactions in a 'commonly agreeable on universe'. That's one of the things making scaling so incredibly fascinating. You can't free yourself from your local clock and ruler though, so even if I would define a 'bit' as some smallest common nominator of 'time', as I use Planck scale for, you still would find 'time' to tick for you in that experiment, if you now could measure at that scale, which you can't. You can't prove it, because you can't free yourself from the universe you exist in, and making a experiment not using time is a very weird proposition. I would call it another reality myself, coexistent with the one we observe.
=

I would say that we have a lot of indirect proofs for it existing though. From the ideas of probabilities, to Feynman's sum over histories, to entanglements, to the ideas behind a quantum computer. That has very little to do with the idea of 'c'. But a lot to do with what may happen as you scale it down. The point to take home from such a reasoning is that the clock and ruler is you, and that everything you measure use it. Scaling doesn't change that fact, which is why it will be very hard, probably impossible, to design a experiment proving it. And you live 'in time', if time is consisting of 'bits' then what's between them won't exist for you, because that's where you are not.

Looking at it this way it is a geometry, a weird one, shaped through scaling, up into a macroscopic world where we live and 'observe'. Although I have a distinct feeling that algebra will describe it better than a geometric formulation, and that was the way I understand Einstein to have thought of it too. I think he was right.

( actually this last is about whether you want to define it locally or globally. Globally ( a 'common universe') Algebra should be your best choice. Locally I expect geometry to do just as good, possibly even clearer?. That's my opinion at least, well, for now :)
« Last Edit: 28/02/2015 16:19:53 by yor_on »
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Re: Would the photon lose all its energy at infinity?
« Reply #131 on: 28/02/2015 14:01:05 »
If you define the way free fall act as a result of accelerations, then mass does not belong to it. Mass expresses itself not as 'accelerating', locally described. A real acceleration will give you a added feeling of weight added to you, whereas a gravitational acceleration gives you the opposite,  'weightlessness'. That 'weightlessness' is the exact same phenomena that describe a uniform motion. Einstein described it through different coordinate system, from a 'global point of view' as me standing on Earth, defining the 'free falling guy' to accelerate. From the point of view of that guy falling though, instead becoming the exact same as if he was uniformly moving through a space, weightless, in a geodesic. All of this locally defined.
=

We can turn it on its head by using one frame of reference, Earths. Then all gravitational accelerations belong to local accelerations as defined from some inertial point of view. now, is this right? Naaah, Einstein used locality, but also described it from the idea of a container. That's why we need different coordinate systems.
« Last Edit: 28/02/2015 15:13:42 by yor_on »
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Re: Would the photon lose all its energy at infinity?
« Reply #132 on: 28/02/2015 14:05:59 »
The container demands you to use different coordinate systems. No way to describe this universe, not using frames of reference. But what then, is this idea of 'locality'? Well, in Einsteins reasoning you find a complementary description, as defined from the idea of something bigger, the common universe we live in. myself I stay locally anchored all the time. Because that is what defines us. I use frames of reference as he showed us, but I describe locality as what defines us. It has taken me a awful long time to express it, and that was what impressed me with Sachs (In Bills thread I think it was), that he too saw the same, a 'preferred frame', not in the mean of one frame of reference being more important, but from the aspect of the one considering yourself being the absolute best frame of reference, describing your experiences. And that's locality.

Once you realize that one, the only thing you need to do, is to take it to its extremes and see where it leads you.
=

Locally anchored, QM and relativity has a lot in common. Scaling is indeed about where a 'locality' ends, hoping to define it through quanta, or 'bits'. That is a good idea, as long as it in cooperates the other side of it, what comes before that quanta of 'motion', or 'time', or 'distance', or 'mass'. The other side don't use our arrow. Does that make you 'time less'? :) Nope, you're defined through a arrow and its outcomes, but what is 'you', is another question, that one belong to the exact same place as where we ask ourselves what 'thoughts' are? That's physics too (well to me it is:), but not one we can measure on. No way to measure that thought. The only thing we can measure on are those physical things that accompany it, but that's not the thought.

Yep, I see physics everywhere :)

And to my eyes, what stopped Einstein in his later years, and what made him argue against Entanglements and probability 'God does not play dice', was the way he thought of it, as a container. His work on a fifth dimension, unifying the universe, came from that exact position. He had the answer, but he wanted another. And that is what we all have John :) You and me, and all together. We have our pets.
« Last Edit: 28/02/2015 15:26:42 by yor_on »
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Offline jeffreyH

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Re: Would the photon lose all its energy at infinity?
« Reply #133 on: 28/02/2015 18:27:26 »
yor_on I agree with Pete that your posts do cause problems in reading a thread. Yet when I take the time to read through them, which I usually don't, you make some very interesting points. It is a shame that these get missed. I know you think an awful lot about this stuff. I just wish you posts were easier to read.
Fixation on the Einstein papers is a good definition of OCD.

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Re: Would the photon lose all its energy at infinity?
« Reply #134 on: 28/02/2015 19:26:00 »
yor_on

It's streams of posts like this that make your writing useless. Almost nobody reads your posts because its like reading a book about nothing. It's a waste of our time so why do you do this? It makes everything else in the thread much harder to follow because you push other people's posts off the page into previous pages. Nobody wants to scroll through all this garbage to look for a small gem. Now please get with it.

Note: I know how selfish you are so I know that you're more interested in your own desires and less interested in everyone else's at our expense so I know that you don't care how we feel about it. Therefore when you try to justify your poor attitude you'll fail and just be wasting our time again.

+1 (with regard to the first paragraph)

Pete, you're picking fights all over. Believing that a expertise in one subject gives one reason to act rudely, or advertise ones own site as 'better', is a mistake on TNS. At least it used to be?  Actually I don't know any site that finds the last to be acceptable? I'm getting pretty tired reading your attacks now.

I think Pete's been pushed beyond a point where all the irritation which he's held back for many years is suddenly being unleashed in a great blast as he tells everyone straight what he thinks. (I'm mindreading, but I hope in a good way.) I'm sure it'll all come good in the end, but we all need to think more carefully about how we treat each other and how we come across.
« Last Edit: 28/02/2015 19:28:29 by David Cooper »

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Offline yor_on

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Re: Would the photon lose all its energy at infinity?
« Reply #135 on: 28/02/2015 19:31:29 »
You read the rules and then you either accept them, or you leave. Is that complicated David?
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Re: Would the photon lose all its energy at infinity?
« Reply #136 on: 28/02/2015 20:06:50 »
It saddens me to see these discussions turn into personal issues. I respect each and every member here that has something of interest to offer and especially those that I've become friends with.

One faithful member stated not long ago: "We should all be better than this."

And I applaud those who ascribe to that notion. Let's all work on this goal.
"The more things change, the more they remain the same."

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Re: Would the photon lose all its energy at infinity?
« Reply #137 on: 28/02/2015 20:47:25 »
Quote from: jeffreyH
Yet when I take the time to read through them, which I usually don't, you make some very interesting points. It is a shame that these get missed.
And that's one of the worst problems that he's causing. There's simply too much "filler" to want to wade through.
« Last Edit: 08/03/2015 05:38:19 by evan_au »

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Offline David Cooper

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Re: Would the photon lose all its energy at infinity?
« Reply #138 on: 28/02/2015 20:53:53 »
You read the rules and then you either accept them, or you leave. Is that complicated David?

Which rules are being broken? You are allowed to post long, rambling essays if you wish, and Pete's allowed to express annoyance with them, and to do so in a way that's rather stronger than might be socially acceptable. As for Pete's forum, it's trying to do something different which doesn't involve competing against this one, but to work as an addition to it so that disrupted conversations can be restarted in a more controlled space and better progress can be made with them. (I am involved in one such discussion there which has proved to be almost impossible to have on any other forum due to people who can't reason insisting on trying to derail it with a bombardment of illegal objections.) It is not designed to drag anyone away from here.

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Re: Would the photon lose all its energy at infinity?
« Reply #139 on: 28/02/2015 21:55:30 »
One useful trick would be for everyone to try to start from scratch in every thread, acting as if they have no negative history with anyone else who's taking part in them. That way, whenever war breaks out, it'll be easier to look back and find out what sparked it off.

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Offline jeffreyH

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Re: Would the photon lose all its energy at infinity?
« Reply #140 on: 28/02/2015 22:50:30 »
So is it a yes the photon does lose all its energy at infinity or a no?
Fixation on the Einstein papers is a good definition of OCD.

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Re: Would the photon lose all its energy at infinity?
« Reply #141 on: 28/02/2015 23:01:26 »
So is it a yes the photon does lose all its energy at infinity or a no?
No. Of course not. Where on earth did you get that idea anyway? The energy of a photon remains constant as it moves through a gravitational field.

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Offline PhysBang

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Re: Would the photon lose all its energy at infinity?
« Reply #142 on: 28/02/2015 23:13:04 »
Look what I have to keep dealing with.
I pointed out that I thought you were ignoring the very definition you were posting. It wasn't meant as an attack, just a point.

« Last Edit: 03/03/2015 10:55:16 by evan_au »

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Offline David Cooper

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Re: Would the photon lose all its energy at infinity?
« Reply #143 on: 28/02/2015 23:36:27 »
So is it a yes the photon does lose all its energy at infinity or a no?

Think about what happens with sound. There's someone plucking a string on a double bass at the bottom of a towerblock. You are able to watch and listen from the top of the towerblock, and you can hear that the note is flat. You go down in the lift and listen again, and it doesn't sound flat any more, so you go back up to the top and find that it's flat again. When you look down at the string from up there through a powerful telescope you can see it vibrating, and it appears to be vibrating exactly in sync with the sound you are hearing. You conclude from this that the sound has not reduced in frequency as it has climbed out of the gravity well, but that it is being generated at a lower frequency in the first place and it maintains that frequency all the way up as the sound climbs towards you. If it travels to infinity, the frequency will remain constant, but if you travel out to listen to it you will hear it as continuing to get more flat the further you get out of the gravity well. This reduction in the frequency you hear will become closer and closer to zero though (meaning the note doesn't appear to get much flatter at all) and the change will soon become impossible to measure - the strongest effect is deep down in the well, but once out in deep space you are in practical terms no longer in that well, even though you technically still are.

However, the frequency could still be reduced to zero over infinite distance because of dark energy and the expansion of space. It is this expansion that could decrease the energy of a photon infinitely, but not its journey out of a gravity well. [This then leads to other questions. Where then does that energy go to? Is the loss driven by the expansion of space, or do photons help to drive the expansion of space by throwing off some of their energy?]
« Last Edit: 28/02/2015 23:47:58 by David Cooper »

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Offline Russell Crawford

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Re: Would the photon lose all its energy at infinity?
« Reply #144 on: 01/03/2015 12:33:04 »
It would depend on the property of the photon that is in review. If one is speaking of the particle property or the wave property the answer would be different. This, of course, is all speculation. It there is in fact a particle property, then the photon could lose all its energy before "infinity", whereas if one is looking at it from a wave perspective, infinity can never be reached. There is no way to evaluate an infinite wavelength because its properties cannot be known.
That said if the wavelength could be infinite, the energy could, due to gravitational redshift, become increasingly smaller in an infinite sense.
I tend to believe that the particle property may be the best interpretation of the energy of a photon and that at some point it will have a "quanta" of energy that will be confirmed. But that is just an unfounded "belief".
With regard to "where the energy goes" it would seem to me (again a guess) that it would be lost over the distance from the source of the energy and the point at which it is measured. Perhaps trapped by gravity as planets are trapped and sorted?  Perhaps it could be called energy over distance or redshift potential energy. Again, that is speculation. I hope this answer will help people think about this important subject.
« Last Edit: 01/03/2015 15:46:55 by Russell Crawford »

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Offline yor_on

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Re: Would the photon lose all its energy at infinity?
« Reply #145 on: 01/03/2015 13:23:48 »
I'm sure it'll all come good in the end, but we all need to think more carefully about how we treat each other and how we come across.
You read the rules and then you either accept them, or you leave. Is that complicated?

Which rules are being broken? You are allowed to post long, rambling essays if you wish, and Pete's allowed to express annoyance with them, and to do so in a way that's rather stronger than might be socially acceptable. As for Pete's forum, it's trying to do something different which doesn't involve competing against this one, but to work as an addition to it so that disrupted conversations can be restarted in a more controlled space and better progress can be made with them. (I am involved in one such discussion there which has proved to be almost impossible to have on any other forum due to people who can't reason insisting on trying to derail it with a bombardment of illegal objections.) It is not designed to drag anyone away from here.

Well David. I'm sorry that it has come to this, and I actually Pm:ed Pete on the rules.
There are some rules concerning our behavior on this forum that we should try to follow. Keeping it friendly is one, another is about advertising ones own site. You can look it up or ask Pete. When it comes to writing long chunks :) Yep, got carried away there, it is as Jeffrey said, I think about it, then I see something I want to make better etc etc, ad infinitum. A dangerous thing, and one that shoot me in my ** here.

That doesn't mean that it excuses breaking other rules. It's in the end a Moderator decision whether to warn or not, so let us see what happens. If there is a lack of interest from the moderators point of view, I would suggest the rules to be rewritten though, so that they better fit the way this forum is going to act in the future.
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(Well. What'da'ya'now :) pm's Internet slang for 'personal message' so it's ok any which way. Weird stuff)
« Last Edit: 03/03/2015 10:46:06 by evan_au »
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Offline yor_on

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Re: Would the photon lose all its energy at infinity?
« Reply #146 on: 01/03/2015 15:45:10 »
So is it a yes the photon does lose all its energy at infinity or a no?

Depends, and it goes back to how you define it. How does a accelerating expansion act on light? As a wave it's possible to understand, as a 'point-like photon' it's not. The only way I see it working is accepting a existing duality, in all circumstances. And that takes us once again, to a place of 'no propagation', exchanging it to excitations in a field. That is also a question of whether this duality becomes a macroscopic phenomena, quanta microscopically, or not?
=

that one is treating it 'practically' though. Theoretically I would guess no, it shouldn't lose its energy, unless we introduce this expansion acting on it. But all of this reasoning builds on you and me presuming a propagation. Exchanging that to excitations in a field it's no longer about 'unique photons propagating in a space' per se. To me it then seems to become more of a question of conservation laws. A propagation, or no propagation, that is the question :)
« Last Edit: 01/03/2015 15:59:11 by yor_on »
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Offline jccc

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Re: Would the photon lose all its energy at infinity?
« Reply #147 on: 01/03/2015 16:17:20 »
hope you guys don't mind, just my thought. i been learning from this thread.

is it possible that when atoms exited, they give off gravitational waves? like a bell knocked gives off vibrating energy?

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Offline yor_on

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Re: Would the photon lose all its energy at infinity?
« Reply #148 on: 01/03/2015 16:41:38 »
If I remember rightly Einstein had some thoughts on 'micro gravity' in the paper linked at http://physics.aps.org/story/v15/st11
Think it's called a 'Einstein-Rosen bridge' or nowadays 'wormhole'.
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Offline jeffreyH

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Re: Would the photon lose all its energy at infinity?
« Reply #149 on: 01/03/2015 16:49:54 »
So is it a yes the photon does lose all its energy at infinity or a no?

Think about what happens with sound. There's someone plucking a string on a double bass at the bottom of a towerblock. You are able to watch and listen from the top of the towerblock, and you can hear that the note is flat. You go down in the lift and listen again, and it doesn't sound flat any more, so you go back up to the top and find that it's flat again. When you look down at the string from up there through a powerful telescope you can see it vibrating, and it appears to be vibrating exactly in sync with the sound you are hearing. You conclude from this that the sound has not reduced in frequency as it has climbed out of the gravity well, but that it is being generated at a lower frequency in the first place and it maintains that frequency all the way up as the sound climbs towards you. If it travels to infinity, the frequency will remain constant, but if you travel out to listen to it you will hear it as continuing to get more flat the further you get out of the gravity well. This reduction in the frequency you hear will become closer and closer to zero though (meaning the note doesn't appear to get much flatter at all) and the change will soon become impossible to measure - the strongest effect is deep down in the well, but once out in deep space you are in practical terms no longer in that well, even though you technically still are.

However, the frequency could still be reduced to zero over infinite distance because of dark energy and the expansion of space. It is this expansion that could decrease the energy of a photon infinitely, but not its journey out of a gravity well. [This then leads to other questions. Where then does that energy go to? Is the loss driven by the expansion of space, or do photons help to drive the expansion of space by throwing off some of their energy?]

That's the sort of thing I was looking for. That sums up the situation exactly. It is unclear to many exactly what the true situation is. Thanks for this excellent post.
Fixation on the Einstein papers is a good definition of OCD.