Does a photon curve space-time, i.e., produce a gravitational field?


by redtree
Tags: curve, field, gravitational, photon, produce, spacetime
redtree
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#1
May3-08, 02:12 PM
P: 64
Does a photon curve space-time, i.e., produce a gravitational field? Is the degree of curvature a function of its energy?
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jnorman
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#2
May3-08, 06:04 PM
P: 308
no. a photon exhibits no locality.
dst
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#3
May3-08, 06:06 PM
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Quote Quote by jnorman View Post
no. a photon exhibits no locality.
How are you reading this?

pmb_phy
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#4
May4-08, 11:38 AM
P: 2,955

Does a photon curve space-time, i.e., produce a gravitational field?


Quote Quote by jnorman View Post
no. a photon exhibits no locality.
Yo jnorman! How's it going?

That a photon's position cannot be determined with infinite precision to without leaving its energy totally unknown does not dictate that a photon can't generate a gravitational field. I don't see why they'd be mutually exclusive. One can always give a region of space in which the particle is contained within and as such the energy will be between certain but finite limits. Since a photon has energy it also has active gravitational mass and therefore can generate a gravitational field.

Pete
jnorman
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#5
May4-08, 03:51 PM
P: 308
hi pmb - i think we have had this discussion before, eh? once a single photon is emitted, it is essentially everywhere in the universe - ie, it displays no locality (ie, its position cannot be defined with ANY precision). since it has no defined position, it, ergo, cannot exhibit any local effect on gravitational field. as i recall, a light beam can, however, generate observable field effects. as always, feel free to correct me...
yuiop
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#6
May4-08, 05:42 PM
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Quote Quote by jnorman View Post
hi pmb - i think we have had this discussion before, eh? once a single photon is emitted, it is essentially everywhere in the universe - ie, it displays no locality (ie, its position cannot be defined with ANY precision). since it has no defined position, it, ergo, cannot exhibit any local effect on gravitational field. as i recall, a light beam can, however, generate observable field effects. as always, feel free to correct me...
Please explain why a photon generates no observable gravitational field effect while a light beam can. Isn't a light beam just a group of more than one photon? Are you specifically referring to a laser beam?

If we place a laser at one end of a sealed tube and a target at the opposite end of the tube and all the photons from the laser arrive at the target within the expected time, then does that not imply a high probablity of the photons being within the tube during the journey from the laser to the target and imply some sort of locality? I find the quick tour of the universe by the photons between the emmiter and the target hard to visualise, especially if we limit the photons to the speed of light :P As I understand it the locality of a photon has a probabilty distribution that puts a high probability on the position ofa photon being in a certain area and a very small probabilty of it being almost anywhere else. I think it is a bit like the position of an electron that belongs to an atom. There is a small possibility of the electron being almost anywhere in the universe, but a much higher probability of the electron being somewhere near the atom it "belongs" to. With this analogy we could say that an electron has no locality and therefore no influence on the gravitational field which I think you will agree is nonsense.
LURCH
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#7
May5-08, 11:30 AM
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I think we can all agree that it is not possible (at present) for us to measure the gravitational field of a photon. So, all of this discussion must be somewhat speculative. The correct answer to this question is not known with any high degree of certainty, is it?

I am pretty sure the photon does not have a gravitational field. I say this because a photon has no mass other than relativistic mass. If relativistic mass is a source of gravity, apparent paradoxes arise. If these paradoxes cannot be resolved, they serve as proof that relativistic mass cannot be a source of gravity, and gravity only proceeds from rest mass. Since a photon has no rest mass, it has no gravity. (That's my speculation.)
peter0302
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#8
May5-08, 02:31 PM
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Since photons respond to gravity. as we can see from gravitational red/blue shifting as well as from gravitational lensing, there is no reason not to believe the photons themselves have a gravitational field.

And, Lurch, a photon may not have an invariant mass, but it does have an invariant energy, i.e., hf, so I don't see what paraxoes would arise from considering hf as equivalent to mass-energy.
mitesh9
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#9
May5-08, 02:42 PM
P: 58
As Einstein Said "The mass tells the space how to curve, and the space(-time curvature) tells the mass how to move", which may translate to "every thing that has mass create a curvature in spacetime and everything that follows the curved spacetime (i.e. responds to gravity) has mass".

Since, the photons respond to gravity, they have mass (let it be reletivistic) and hence should curve spacetime.
atom888
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#10
May5-08, 09:08 PM
P: 92
Quote Quote by redtree View Post
Does a photon curve space-time, i.e., produce a gravitational field? Is the degree of curvature a function of its energy?
Well, i'm an ether fan so I have to say no on this. lol don't quote me
Phrak
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#11
May5-08, 11:31 PM
P: 4,513
Spatial curvature is described by 10 numbers; roughtly how time curves into space, space into space, and finally time into time (10 = four spacetime compontents taken two at a time, disregarding order taken). An electromagnetic wave is unique in that, unlike massive matter, all of it's energy is it's momentum. It will curve spacetime differently than massive objects. But you asked about photons. Photons propagate as waves (but so do the massive objects). So now we have to mix quantum mechanics with general relativity. Good luck with that, as no one has measured the effects of a single photon, or even a flock of them, on spacetime curvature. But in theory one can treat a photon as a propagating wave with a bandwidth and spatial extent having a well defined energy at each point in spacetme. If you want to know how such an electromagnetic field would bend spacetime you need one of the smart guys around here, because I have very little clue, sorry.
Phrak
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#12
May5-08, 11:59 PM
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Quote Quote by LURCH View Post
I am pretty sure the photon does not have a gravitational field. I say this because a photon has no mass other than relativistic mass.
Place equal amounts of matter and anitmatter in a box on a scale. It's a very good box; it's very reflective, and light doesn't get in or out. Allow all the stuff to annihilate to photons. Does the box change weight?
xantox
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#13
May6-08, 07:35 AM
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Quote Quote by LURCH View Post
I am pretty sure the photon does not have a gravitational field. I say this because a photon has no mass other than relativistic mass. If relativistic mass is a source of gravity, apparent paradoxes arise. If these paradoxes cannot be resolved, they serve as proof that relativistic mass cannot be a source of gravity, and gravity only proceeds from rest mass. Since a photon has no rest mass, it has no gravity. (That's my speculation.)
http://www.physicsforums.com/showthread.php?t=154391
In general relativity, gravity is coupled to energy density and momentum flow, not only mass like in newtonian gravity. On this basis, an electromagnetic wave (general relativity does not consider light in terms of photons) will exert its own gravity, though extremely weak and not currently measurable. Gravity exerted by massive bodies is much higher because of their huge energy content (see the c squared term in the Einstein formula).

Quote Quote by mitesh9 View Post
As Einstein Said "The mass tells the space how to curve, and the space(-time curvature) tells the mass how to move", which may translate to "every thing that has mass create a curvature in spacetime and everything that follows the curved spacetime (i.e. responds to gravity) has mass".
This quote was not by Einstein but by John Wheeler, and he said "matter", as a generic term, not "mass" (see C. W. Misner, K. Thorne, J. Wheeler, "Gravitation", W. H. Freeman (1973), page 5). Also it is just a pedagogical way to summarize the meaning of general relativity, one must always refer to the actual formulas to determine what the theory precisely says.
mitesh9
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#14
May6-08, 07:37 AM
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Quote Quote by Phrak View Post
Place equal amounts of matter and anitmatter in a box on a scale. It's a very good box; it's very reflective, and light doesn't get in or out. Allow all the stuff to annihilate to photons. Does the box change weight?
Good one!!!
pmb_phy
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#15
May6-08, 07:58 AM
P: 2,955
Quote Quote by jnorman View Post
hi pmb - i think we have had this discussion before, eh? once a single photon is emitted, it is essentially everywhere in the universe - ie, it displays no locality (ie, its position cannot be defined with ANY precision).
To be precise there is no quantum theory of gravity (or relativistic quantum mechanics) so we're really all taking an educated guess. However I see no reason to assert that a photon is everywhere in the universe. Quantum mechanics certainly makes no such assertion. All that can be said is that for each quantum state of any particle there is an associated wave function. The physical interpretation of that wave function is that the magnitude squared of the function represents the probability density of finding the particle in a particular region. Only when the exact value of the momentum is determined will the probability density be uniform and thus the chances of finding it anywhere in the universe will be zero. However that comes from non-relativistic quantum mechanics. Relativity restricts the speed of a particle to less than the speed of light and therefore the probability density can never be uniform. And even this interpretation of pronability refers only to essembles of identical experiments, not to individual experiments. There is no restriction on the limits of accuracy placed on each single measurement.

Best wishes

Pete
pmb_phy
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#16
May6-08, 08:00 AM
P: 2,955
Quote Quote by xantox View Post
This quote was not by Einstein but by John Wheeler, and he said "matter", as a generic term, not "mass" (see C. W. Misner, K. Thorne, J. Wheeler, "Gravitation", W. H. Freeman (1973), page 5). Also it is just a pedagogical way to summarize the meaning of general relativity, one must always refer to the actual formulas to determine what the theory precisely says.
Wheeler made such statements in various places and using different terms each time. In Exploring Black Holes he phrased it using the term mass rather than matter. Due to the way the authors definined mass in that book I protested but Wheeler was adament about it.

Pete
Wizardsblade
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#17
May6-08, 08:13 AM
P: 145
Quote Quote by Phrak View Post
Place equal amounts of matter and anitmatter in a box on a scale. It's a very good box; it's very reflective, and light doesn't get in or out. Allow all the stuff to annihilate to photons. Does the box change weight?
We know that light pulled in by gravity so we know that at least some weight will remain if not all.

Or as one of my physics professors said, weigh a flashlight, turn it on till the batteries die then weigh it again now that all the light is out of it =)
pmb_phy
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#18
May6-08, 08:19 AM
P: 2,955
Quote Quote by LURCH View Post
I think we can all agree that it is not possible (at present) for us to measure the gravitational field of a photon. So, all of this discussion must be somewhat speculative. The correct answer to this question is not known with any high degree of certainty, is it?
You're assuming the the mass-energy of the photon in question is so small as to be neglegible. I see no reason to make that assertion. I see no reason that a photon with large enough mass-energy can't generate a very strong and measureable gravitational field. Especially since the magnitude of the gravitational field is frame dependant and thus one can always transform to a new frame where the mass-energy is as large as one would like.
I am pretty sure the photon does not have a gravitational field. I say this because a photon has no mass other than relativistic mass. If relativistic mass is a source of gravity, apparent paradoxes arise.
That's the problem with using the term relativistic mass since it seems to imply that it is different than mass. Its well known that mass is the source of gravity. simply turn to page 404 in Gravitation by Misner, Thorne and Wheeler (MTW) to see that. I.e. the authors state on that page in the second paragraph that
Mass is the source of gravity.
The context of that statement, given Eq. (17.1), the authors are really refering to what you call relativistic mass (and which I and MTW simply call mass). MTW also use the term mass-energy to refer to the same thing. In actuallity MTW are rwefering to the mass density in that expression and are speaking about the T00 component of the stress-energy-momentum (SEM) tensor. And there are no paradoxes exist which can't be resolved under close scrutiny.
If these paradoxes cannot be resolved, they serve as proof that relativistic mass cannot be a source of gravity, and gravity only proceeds from rest mass. Since a photon has no rest mass, it has no gravity. (That's my speculation.)
The source of gravity is not rest mass. Why would you even think that??? Its well known that the mathematical quantity which acts as the source of gravity is the SEM tensor which does not vanish for a quantum of light. Even Einstein said that mass is completely determined by the SEM tensor. Since the SEM tensor has energy, momentum and stress terms it follows that each is a source of gravity. A photon has both energy and momentum.

Best wishes

Pete


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