The Naked Scientists

The Naked Scientists Forum

Author Topic: Would the photon lose all its energy at infinity?  (Read 69594 times)

Offline Bill S

  • Neilep Level Member
  • ******
  • Posts: 1828
  • Thanked: 12 times
    • View Profile
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.
 

Offline yor_on

  • Naked Science Forum GOD!
  • *******
  • Posts: 11999
  • Thanked: 4 times
  • (Ah, yes:) *a table is always good to hide under*
    • View Profile
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 »
 

Offline David Cooper

  • Neilep Level Member
  • ******
  • Posts: 1505
    • View Profile
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 »
 

Offline PmbPhy

  • Neilep Level Member
  • ******
  • Posts: 2762
  • Thanked: 38 times
    • View Profile
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.
 

Offline yor_on

  • Naked Science Forum GOD!
  • *******
  • Posts: 11999
  • Thanked: 4 times
  • (Ah, yes:) *a table is always good to hide under*
    • View Profile
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 :)
 

Offline yor_on

  • Naked Science Forum GOD!
  • *******
  • Posts: 11999
  • Thanked: 4 times
  • (Ah, yes:) *a table is always good to hide under*
    • View Profile
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 »
 

Offline jeffreyH

  • Global Moderator
  • Neilep Level Member
  • *****
  • Posts: 3926
  • Thanked: 55 times
  • The graviton sucks
    • View Profile
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.
 

Offline jeffreyH

  • Global Moderator
  • Neilep Level Member
  • *****
  • Posts: 3926
  • Thanked: 55 times
  • The graviton sucks
    • View Profile
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 »
 

Offline jeffreyH

  • Global Moderator
  • Neilep Level Member
  • *****
  • Posts: 3926
  • Thanked: 55 times
  • The graviton sucks
    • View Profile
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?
 

Offline jeffreyH

  • Global Moderator
  • Neilep Level Member
  • *****
  • Posts: 3926
  • Thanked: 55 times
  • The graviton sucks
    • View Profile
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.
 

Offline yor_on

  • Naked Science Forum GOD!
  • *******
  • Posts: 11999
  • Thanked: 4 times
  • (Ah, yes:) *a table is always good to hide under*
    • View Profile
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..... "
 

Offline jeffreyH

  • Global Moderator
  • Neilep Level Member
  • *****
  • Posts: 3926
  • Thanked: 55 times
  • The graviton sucks
    • View Profile
Re: Would the photon lose all its energy at infinity?
« Reply #111 on: 27/02/2015 05:00:45 »
So take the lagrangian bd5aabbac9bc3e5a0bc6cd4eae2fe473.gif. This is not applicable to the photon. If we remove the term d0c139858db37b5480e051c5f4a918d1.gif then we are working with a directional velocity only f7abdd2b85df907f6d6e726f7286a405.gif. 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 »
 

Offline jeffreyH

  • Global Moderator
  • Neilep Level Member
  • *****
  • Posts: 3926
  • Thanked: 55 times
  • The graviton sucks
    • View Profile
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 »
 

Offline jeffreyH

  • Global Moderator
  • Neilep Level Member
  • *****
  • Posts: 3926
  • Thanked: 55 times
  • The graviton sucks
    • View Profile
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.
 

Offline jeffreyH

  • Global Moderator
  • Neilep Level Member
  • *****
  • Posts: 3926
  • Thanked: 55 times
  • The graviton sucks
    • View Profile
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.
 

Offline PmbPhy

  • Neilep Level Member
  • ******
  • Posts: 2762
  • Thanked: 38 times
    • View Profile
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
 

Offline jeffreyH

  • Global Moderator
  • Neilep Level Member
  • *****
  • Posts: 3926
  • Thanked: 55 times
  • The graviton sucks
    • View Profile
 

Offline jeffreyH

  • Global Moderator
  • Neilep Level Member
  • *****
  • Posts: 3926
  • Thanked: 55 times
  • The graviton sucks
    • View Profile
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.
 

Offline JohnDuffield

  • Sr. Member
  • ****
  • Posts: 488
  • Thanked: 1 times
    • View Profile
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 »
 

Offline evan_au

  • Neilep Level Member
  • ******
  • Posts: 4126
  • Thanked: 247 times
    • View Profile
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.
 

Offline JohnDuffield

  • Sr. Member
  • ****
  • Posts: 488
  • Thanked: 1 times
    • View Profile
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 »
 

Offline PhysBang

  • Hero Member
  • *****
  • Posts: 588
  • Thanked: 14 times
    • View Profile
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.
Quote
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.
Quote
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.
 

Offline jeffreyH

  • Global Moderator
  • Neilep Level Member
  • *****
  • Posts: 3926
  • Thanked: 55 times
  • The graviton sucks
    • View Profile
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.
 

Offline jeffreyH

  • Global Moderator
  • Neilep Level Member
  • *****
  • Posts: 3926
  • Thanked: 55 times
  • The graviton sucks
    • View Profile
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.
 

Offline yor_on

  • Naked Science Forum GOD!
  • *******
  • Posts: 11999
  • Thanked: 4 times
  • (Ah, yes:) *a table is always good to hide under*
    • View Profile
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 »
 

The Naked Scientists Forum

Re: Would the photon lose all its energy at infinity?
« Reply #124 on: 27/02/2015 14:12:55 »

 

SMF 2.0.10 | SMF © 2015, Simple Machines
SMFAds for Free Forums