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

In summary, the conversation discusses whether a photon can produce a gravitational field and if its degree of curvature is dependent on its energy. There is disagreement on whether a photon has locality and can generate observable gravitational field effects. Some mention the difficulty in measuring the gravitational field of a photon and the concept of relativistic mass. It is suggested that photons may have a gravitational field since they respond to gravity and have energy. The conversation also touches on the idea of treating a photon as an electromagnetic wave to understand its effect on spacetime curvature.
  • #1
redtree
285
13
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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  • #2
no. a photon exhibits no locality.
 
  • #3
jnorman said:
no. a photon exhibits no locality.

How are you reading this?
 
  • #4
jnorman said:
no. a photon exhibits no locality.
Yo jnorman! How's it going? :smile:

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
 
  • #5
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...
 
  • #6
jnorman said:
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.
 
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  • #7
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.)
 
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  • #8
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.
 
  • #9
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.
 
  • #10
redtree said:
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
 
  • #11
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.
 
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  • #12
LURCH said:
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?
 
  • #13
LURCH said:
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.)

https://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).

mitesh9 said:
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.
 
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  • #14
Phrak said:
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!
 
  • #15
jnorman said:
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
 
  • #16
xantox said:
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
 
  • #17
Phrak said:
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 =)
 
  • #18
LURCH said:
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 referring 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
 
  • #19
Phrak said:
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?
If you consider the box of matter to be a closed system then photons cannot escape from the box. The walls of the box will then either reflect the photons and thus transfering momentum (and thus weight) to the walls or the walls absorb the energy of the photon with a corresponding increase in the weight of the box. Even a box of photons will generate a gravitational field.

The reason a beam of light, which consists of photons, generates a gravitational field is because each photon generates its own field. The contributions from each photon go into increasing the gravitational field. The exact relation is not linearly dependant though. However if the field is weak then the increase in sum of the contributions of each photon add up linearly.

Pete
 
  • #20
pmb_phy said:
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.

Also keep in mind that all this is asssumes that a measurement has been made and the interaction of a particle with the photon's gravitational field might be considered as such an measurement.

Best wishes

Pete

Also keep in mind that all this is asssumes that a measurement has been made and the interaction of a particle with the photon's gravitational field might be considered as such an measurement.

Pete
 
  • #21
pmb_phy said:
However I see no reason to assert that a photon is everywhere in the universe.

i begin with the notion that a photon moves at C. at C, distance has no meaning - there is no distance between things. ergo, a photon is essentially everywhere at once. and of course, our old general rule - you cannot say anything about a photon in between the time it is emitted and the time it is absorbed...

again, please feel free to knock that down.
 
  • #22
jnorman said:
i begin with the notion that a photon moves at C. at C, distance has no meaning - there is no distance between things. ergo, a photon is essentially everywhere at once. and of course, our old general rule - you cannot say anything about a photon in between the time it is emitted and the time it is absorbed...

again, please feel free to knock that down.

If it will not be considered an attempt to "knock that down", let me say I have a mirror and a photon emitting device (say Laser Gun), separated by a distance of 10c km. I also have a fluorescent screen, that detects the photon impact. Now if I shoot a single photon from the Laser Gun, can I not come to know that where would that photon be after 5 seconds? Of course I can know, that after 5 seconds, it will be halfway between the source and the mirror (especially, when we know that the light travels in a straight line)! I can confirm that by putting the screen at halfway distance. Mind you that, I knew where the photon would be even before putting the screen, an I just put the screen for the sole purpose of confirmation.
 
  • #23
mitesh9 said:
a distance of 10c km
"10c km"? What does that mean?

Edit: I think I know what you mean...but please use normal conventions and notation.
 
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  • #24
pmb_phy said:
In Exploring Black Holes he phrased it using the term mass rather than matter.
OK, let's forget that version, as it is of course confusing.
 
  • #25
pmb_phy said:
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

Wheeler invented the "geon" that is a self gravitating photon that orbits itself. Clearly that demonstrates that at least Wheeler believes that light has active gravitational mass.
 
  • #26
This is all every nice, but to which parts of the stress-energy tensor does light contribute?
 
  • #27
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 =)
It would have to weigh slightly less. 2 D batteries contain about 150,000 joules of energy. m = E/c^2, so the flashlight will weigh 1.68x10^-12 kg less when it's drained.
 
  • #29
Anything with four-momentum can curve spacetime. The momentum of light is a nonzero null vector (it has zero magnitude, but is not itself zero), but the stress-energy tensor is proportional to the outer product of the momentum with itself, not the inner product. If T doesn't vanish, the Einstein tensor doesn't vanish by Einstein's equation, which means that spacetime can't be flat.
 
  • #30
jnorman said:
a photon is essentially everywhere at once.

It is possible to "trap" photons in high-Q cavities for long times, long enough that the "same" photon (imagine a Fock state with one photon in the cavity) can interact with several different systems (atoms, qubits) before it escapes. Now, it is of course meaningless to ask where the photon is in the cavity, but we DO know that it is IN the cavity meaning it is definately not "everywhere".
See e.g. the work by Haroche et al.

The same thing is true for e.g. single photon sources/detectors where the photons are quite well "localized".
 
  • #31
I have read on a number of occasions that two parallel light beams do not gravitationally attract each other while two anti-parallel light beams do gravitationally attract each other. Is that true? If it is true, why does it work one way but not the other?
 
  • #32
It is true. One may understand it more easily in linearized general relativity, where the first case arise by cancellation of the gravitomagnetic and gravitoelectric accelerations, while in the second they do add.
 
  • #33
kev said:
I have read on a number of occasions that two parallel light beams do not gravitationally attract each other while two anti-parallel light beams do gravitationally attract each other. Is that true? If it is true, why does it work one way but not the other?

If two parallel light beams pass through a gravitational field, do they remain parallel?

[Clarification: What do you mean by "anti-parallel"? I've seen reference to opposite vs. same direction, but not the way you put it.]

Regards,

Bill
 
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  • #34
Antenna Guy said:
If two parallel light beams pass through a gravitational field, do they remain parallel?

[Clarification: What do you mean by "anti-parallel"? I've seen reference to opposite vs. same direction, but not the way you put it.]

Regards,

Bill
I always assumed the references to "anti-parallel" in this context to mean parallel beams with the photons in each beam moving in opposite directions as opposed to to two parallel beams that have photons moving in the same direction. That is just my interpretation. I assume we mean the same thing?

It might be added that it can be predicted that particles with rest mass behave in a similar way, with beams of massive particles gravitationally attracting each other more weakly at higher velocites when they are comoving and more strongly when the beams of massive particles are moving in opposite directions.
 
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  • #35
kev said:
I always assumed the references to "anti-parallel" in this context to mean parallel beams with the photons in each beam moving in opposite directions as opposed to to two parallel beams that have photons moving in the same direction. That is just my interpretation. I assume we mean the same thing?

I read "anti-parallel" as not in same direction, and was wondering if you were including velocity components in opposite directions (rather than just velocities - as I had seen before).

Regards,

Bill
 

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