Gravitational field of a classical "photon" - Orbiter-Forum

Linguofreak · 09-06-2012, 09:13 AM

I've been wondering about the gravitational field associated with a classical "photon" in general relativity. I use the qualifier "classical" and the quotes around "photon" here because I expect that the quantum case with a real photon would be much different.

In other words, consider a typical textbook example given for a black hole:

m = 1 solar mass
v = 0
E = sqrt((mc^2)^2 + (pc)^2) = sqrt((mc^2) + 0) = mc^2 = 1 solar mass * c^2 ~= 1.8 * 10^47 Joules
r_s = 2Gm/c^2 ~= 3.0 km

The spacetime associated with this black hole is the typical Schwarzschild metric.

Now, if we hold E constant at 1.8 * 10^47 Joules and give v some non-zero value, we get the Schwarzschild metric for a somewhat lighter black hole as experienced by an observer with velocity v relative to the hole (so that the total energy of the hole as viewed by that observer is still 1 solar mass * c^2). So what happens in the limit of v=c (and thus m=0)?

**Urwumpe** · 09-06-2012, 10:27 AM

You would get a photon with a very short wavelength:

$E=\hbar\omega=h\nu=\frac{hc}{\lambda}$

Fizyk · 09-06-2012, 11:48 AM

That is a complicated problem.
First thing, it's not that easy to define what does it mean to hold E constant ->

Mass in general relativity - Wikipedia, the free encyclopedia

It also might be hard to come up with a nice set of coordinates to write the resulting metric. The coordinates used to write the Schwarzschild metric are strongly connected to its symmetries, which will be lost.
In conclusion, I don't see any simple way to see what it would be like, but maybe there is one.

**Urwumpe** · 09-06-2012, 01:12 PM

Quote:

Originally Posted by Fizyk

That is a complicated problem.

Yes, but if you use the simplified problem, it is really just down to conservation of four-momentum.

If you are interested in the field equation of a single photon, it isn't that simple, that is correct. It also isn't describeable by the Schwarzschild metric, which assumes non-rotating, uncharged masses.

The Maxwell equations could be the better tool to describe the problem then.

Fizyk · 09-06-2012, 03:09 PM

Quote:

Originally Posted by Urwumpe

Yes, but if you use the simplified problem, it is really just down to conservation of four-momentum.

The simplified problem is also a complicated problem :p Also, I don't exactly see how the conservation of four-momentum would help here, would you elaborate?

Quote:

Originally Posted by Urwumpe

It also isn't describeable by the Schwarzschild metric, which assumes non-rotating, uncharged masses.

Sure, but still it might give some nice results for spin-0, massless particles. It would be quite interesting to find a solution for that, it's also possible that somebody has done this already.

EDIT: Indeed someone has - http://www.springerlink.com/content/x346w7x336224520/

Quote:

Originally Posted by Urwumpe

The Maxwell equations could be the better tool to describe the problem then.

For actual photons - yes, you would probably have to use Maxwell equations, but we want to simplify the problem, right?

I actually tried doing what Linguofreak suggested. I tried writing the Schwarzschild metric in cylindrical coordinates $(t,\rho,\phi,z)$ (it gets ugly) and substituting $z' = \gamma(z-vt)$ , $t' = \gamma(t-vz)$ (it gets even uglier, also it's actually a mistake as I think about it now). So far, no results which would make any sense.

It looks like I might have more luck with actually trying to assume some form of the metric and solving Einstein equations, but I'm afraid I made an assumption which will turn out to be false, and without it the whole thing gets really hard. I'll see what I'll get anyway, I'm quite close to getting the Ricci tensor for the assumed metric.

**Urwumpe** · 09-06-2012, 03:56 PM

Quote:

Originally Posted by Fizyk

The simplified problem is also a complicated problem :p Also, I don't exactly see how the conservation of four-momentum would help here, would you elaborate?

The four momentum of the stationary black hole shall remain constant, while its velocity approaches the speed of light. Thus you can have a photon with the same momentum as the black hole and if I am not completely wrong there, also the same relativistic gravity field as the black hole surrounding it - but still, the photon and the black hole would be different, since the photon has the absolute reference of its speed of light, while the black hole is only stationary in its frame of reference.

Fizyk · 09-06-2012, 04:13 PM

Quote:

Originally Posted by Urwumpe

The four momentum of the stationary black hole shall remain constant, while its velocity approaches the speed of light.

I guess you mean the "length" of the four-momentum, which is just mass (because constant four-momentum would mean the black hole still doesn't move). This would make sense, but the problem here is that I don't think there is a way to define a four-momentum of a black hole. Black holes are regions of spacetime rather than particle-like bodies, and as such can't really have a four-momentum. Also if one assumed that the whole mass of the black hole in the singularity is like a particle, it doesn't help - this "particle" is in such a strange region of spacetime that it doesn't really have anywhere to move.

Anyway, even ignoring that - constant "length" of the four-momentum is constant mass, and to get a photon we need to have a zero mass. That's why Linguofreak suggested decreasing mass with increasing velocity, so that we could get v=c, m=0. This is actually what the authors of the paper I linked in the previous post did.

Quote:

Originally Posted by Urwumpe

Thus you can have a photon with the same momentum as the black hole and if I am not completely wrong there, also the same relativistic gravity field as the black hole surrounding it

Well, from the paper I linked in the previous post it looks like the gravitational field of a photon is totally different from that of a black hole, and different in a very strange way. I didn't expect anything like that, there are some singularities in the curvature tensor.

**Urwumpe** · 09-06-2012, 04:37 PM

(Disclaimer: I just claim stuff in the field of relativity to be proved wrong about my concepts of it and learn something new. I am programmer, not a physicist)

well, what speaks against Black Holes having a momentum? They are just collapsing stars, and if I understand relativity correctly, all black holes are still collapsing, we are just unable to see it, because it happens infinitesimal slow. As the star, that existed before it, already had a momentum, the black hole should also have one. Even more, every single mass of the star, may it be one cubic meter of plasma, one atom, or even just one quark, should also have a momentum, that does not just cease to exist at one point. If the black hole has to be seen as a single black hole particle that is the sum of all particles of the star and conservation of momentum still applies... I still see no difference. It just maybe depends on where you stand when you measure the momentum of each component of the star.

Linguofreak · 09-06-2012, 05:20 PM

Quote:

Originally Posted by Urwumpe

Yes, but if you use the simplified problem, it is really just down to conservation of four-momentum.

If you are interested in the field equation of a single photon, it isn't that simple, that is correct. It also isn't describeable by the Schwarzschild metric, which assumes non-rotating, uncharged masses.

The Maxwell equations could be the better tool to describe the problem then.

That's why I put "photon" in quotes, because I'm not looking for the gravitational field associated with a wave described by the Maxwell equations, or with a particle that has spin. I'm looking for the gravitational field of a classical (that is, non-quantum) point particle with m=0 and nonzero momentum.

Quote:

Originally Posted by Fizyk

EDIT: Indeed someone has - http://www.springerlink.com/content/x346w7x336224520/

$40 for access, unfortunately.

Fizyk · 09-06-2012, 05:29 PM

Quote:

Originally Posted by Urwumpe

well, what speaks against Black Holes having a momentum? They are just collapsing stars, and if I understand relativity correctly, all black holes are still collapsing, we are just unable to see it, because it happens infinitesimal slow. As the star, that existed before it, already had a momentum, the black hole should also have one.

There are black holes as collapsing stars and there are black holes as regions of spacetime. Usually when talking about the mathematical side of the problem, the second meaning is the default one, so this is what can't really have a momentum.

As for black holes as collapsing stars - they are something more material, but that doesn't help much. The curvature of spacetime caused by their mass causes all kinds of problems.

We are used to being able to tell relative velocities of bodies that are far away from each other, for example I could in principle measure the velocity of a plane flying a few km above my head, or even velocities of some stars relative to the Sun. All of that is possible, because the spacetime in which we are conducting these measurements is approximately flat.

Basically, measuring relative velocity in general relativity is equivalent to splitting a vector into components. If we have a flat piece of paper, we can draw two vectors in one point of it, name one "x" and the second "y", and now everywhere on the paper we can split a vector into "x" and "y" components. Our two vectors define "x" and "y" directions on the whole paper in a natural way. Even if we are far from the point where the vectors were defined, we can still use parallel transport to get them to us and immediately know where is x, and where is y.

This isn't so simple, when the space(time) is curved. If instead of a flat piece of paper we had a paper sphere and drew two vectors in one point, we have a problem if someone tells us to split a vector on the other side of the sphere into "x" and "y". The directions are defined quite well in the neighborhood of our starting point, but far from it they aren't. The parallel transport trick won't work either, since the result depends on the path along which the vectors are transported.

And this is basically the problem with relative velocities and energy/momentum in general relativity. The four-velocities/four-momenta of the particles in the collapsing star may be well-defined, but there is no natural way to split them into time and space components (so, in case of four-momentum, into energy and momentum) in our frame of reference. Our frame only works in our neighborhood, to do such a splitting there we have to be there.

Quote:

Originally Posted by Linguofreak

$40 for access, unfortunately.

Yeah, I hate that. I still have an account at my univeristy, so I could access it through there. If you are on a university, try accessing the site through their computers, universities usually have free access to papers. I would send it to you somehow, but that would probably be copyright violation.

09-06-2012, 09:13 AM	#1
Linguofreak Orbinaut	Gravitational field of a classical "photon" I've been wondering about the gravitational field associated with a classical "photon" in general relativity. I use the qualifier "classical" and the quotes around "photon" here because I expect that the quantum case with a real photon would be much different. In other words, consider a typical textbook example given for a black hole: m = 1 solar mass v = 0 E = sqrt((mc^2)^2 + (pc)^2) = sqrt((mc^2) + 0) = mc^2 = 1 solar mass * c^2 ~= 1.8 * 10^47 Joules r_s = 2Gm/c^2 ~= 3.0 km The spacetime associated with this black hole is the typical Schwarzschild metric. Now, if we hold E constant at 1.8 * 10^47 Joules and give v some non-zero value, we get the Schwarzschild metric for a somewhat lighter black hole as experienced by an observer with velocity v relative to the hole (so that the total energy of the hole as viewed by that observer is still 1 solar mass * c^2). So what happens in the limit of v=c (and thus m=0)?

09-06-2012, 11:48 AM	#3
Fizyk Orbinaut	That is a complicated problem. First thing, it's not that easy to define what does it mean to hold E constant -> Mass in general relativity - Wikipedia, the free encyclopedia Mass in general relativity - Wikipedia, the free encyclopedia It also might be hard to come up with a nice set of coordinates to write the resulting metric. The coordinates used to write the Schwarzschild metric are strongly connected to its symmetries, which will be lost. In conclusion, I don't see any simple way to see what it would be like, but maybe there is one.

09-06-2012, 03:09 PM	#5
Fizyk Orbinaut	Quote: Originally Posted by Urwumpe Yes, but if you use the simplified problem, it is really just down to conservation of four-momentum. The simplified problem is also a complicated problem :p Also, I don't exactly see how the conservation of four-momentum would help here, would you elaborate? Quote: Originally Posted by Urwumpe It also isn't describeable by the Schwarzschild metric, which assumes non-rotating, uncharged masses. Sure, but still it might give some nice results for spin-0, massless particles. It would be quite interesting to find a solution for that, it's also possible that somebody has done this already. EDIT: Indeed someone has - http://www.springerlink.com/content/x346w7x336224520/ Quote: Originally Posted by Urwumpe The Maxwell equations could be the better tool to describe the problem then. For actual photons - yes, you would probably have to use Maxwell equations, but we want to simplify the problem, right? I actually tried doing what Linguofreak suggested. I tried writing the Schwarzschild metric in cylindrical coordinates $(t,\rho,\phi,z)$ (it gets ugly) and substituting $z' = \gamma(z-vt)$ , $t' = \gamma(t-vz)$ (it gets even uglier, also it's actually a mistake as I think about it now). So far, no results which would make any sense. It looks like I might have more luck with actually trying to assume some form of the metric and solving Einstein equations, but I'm afraid I made an assumption which will turn out to be false, and without it the whole thing gets really hard. I'll see what I'll get anyway, I'm quite close to getting the Ricci tensor for the assumed metric. Last edited by Fizyk; 09-06-2012 at 03:47 PM.

09-06-2012, 05:29 PM	#10
Fizyk Orbinaut	Quote: Originally Posted by Urwumpe well, what speaks against Black Holes having a momentum? They are just collapsing stars, and if I understand relativity correctly, all black holes are still collapsing, we are just unable to see it, because it happens infinitesimal slow. As the star, that existed before it, already had a momentum, the black hole should also have one. There are black holes as collapsing stars and there are black holes as regions of spacetime. Usually when talking about the mathematical side of the problem, the second meaning is the default one, so this is what can't really have a momentum. As for black holes as collapsing stars - they are something more material, but that doesn't help much. The curvature of spacetime caused by their mass causes all kinds of problems. We are used to being able to tell relative velocities of bodies that are far away from each other, for example I could in principle measure the velocity of a plane flying a few km above my head, or even velocities of some stars relative to the Sun. All of that is possible, because the spacetime in which we are conducting these measurements is approximately flat. Basically, measuring relative velocity in general relativity is equivalent to splitting a vector into components. If we have a flat piece of paper, we can draw two vectors in one point of it, name one "x" and the second "y", and now everywhere on the paper we can split a vector into "x" and "y" components. Our two vectors define "x" and "y" directions on the whole paper in a natural way. Even if we are far from the point where the vectors were defined, we can still use parallel transport to get them to us and immediately know where is x, and where is y. This isn't so simple, when the space(time) is curved. If instead of a flat piece of paper we had a paper sphere and drew two vectors in one point, we have a problem if someone tells us to split a vector on the other side of the sphere into "x" and "y". The directions are defined quite well in the neighborhood of our starting point, but far from it they aren't. The parallel transport trick won't work either, since the result depends on the path along which the vectors are transported. And this is basically the problem with relative velocities and energy/momentum in general relativity. The four-velocities/four-momenta of the particles in the collapsing star may be well-defined, but there is no natural way to split them into time and space components (so, in case of four-momentum, into energy and momentum) in our frame of reference. Our frame only works in our neighborhood, to do such a splitting there we have to be there. Quote: Originally Posted by Linguofreak $40 for access, unfortunately. Yeah, I hate that. I still have an account at my univeristy, so I could access it through there. If you are on a university, try accessing the site through their computers, universities usually have free access to papers. I would send it to you somehow, but that would probably be copyright violation. Last edited by Fizyk; 09-06-2012 at 05:32 PM.

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09-06-2012, 10:27 AM	#2
Urwumpe Donator	You would get a photon with a very short wavelength: $E=\hbar\omega=h\nu=\frac{hc}{\lambda}$

09-06-2012, 01:12 PM	#4
Urwumpe Donator	Quote: Originally Posted by Fizyk That is a complicated problem. Yes, but if you use the simplified problem, it is really just down to conservation of four-momentum. If you are interested in the field equation of a single photon, it isn't that simple, that is correct. It also isn't describeable by the Schwarzschild metric, which assumes non-rotating, uncharged masses. The Maxwell equations could be the better tool to describe the problem then.

09-06-2012, 03:56 PM	#6
Urwumpe Donator	Quote: Originally Posted by Fizyk The simplified problem is also a complicated problem :p Also, I don't exactly see how the conservation of four-momentum would help here, would you elaborate? The four momentum of the stationary black hole shall remain constant, while its velocity approaches the speed of light. Thus you can have a photon with the same momentum as the black hole and if I am not completely wrong there, also the same relativistic gravity field as the black hole surrounding it - but still, the photon and the black hole would be different, since the photon has the absolute reference of its speed of light, while the black hole is only stationary in its frame of reference.

09-06-2012, 04:13 PM	#7
Fizyk Orbinaut	Quote: Originally Posted by Urwumpe The four momentum of the stationary black hole shall remain constant, while its velocity approaches the speed of light. I guess you mean the "length" of the four-momentum, which is just mass (because constant four-momentum would mean the black hole still doesn't move). This would make sense, but the problem here is that I don't think there is a way to define a four-momentum of a black hole. Black holes are regions of spacetime rather than particle-like bodies, and as such can't really have a four-momentum. Also if one assumed that the whole mass of the black hole in the singularity is like a particle, it doesn't help - this "particle" is in such a strange region of spacetime that it doesn't really have anywhere to move. Anyway, even ignoring that - constant "length" of the four-momentum is constant mass, and to get a photon we need to have a zero mass. That's why Linguofreak suggested decreasing mass with increasing velocity, so that we could get v=c, m=0. This is actually what the authors of the paper I linked in the previous post did. Quote: Originally Posted by Urwumpe Thus you can have a photon with the same momentum as the black hole and if I am not completely wrong there, also the same relativistic gravity field as the black hole surrounding it Well, from the paper I linked in the previous post it looks like the gravitational field of a photon is totally different from that of a black hole, and different in a very strange way. I didn't expect anything like that, there are some singularities in the curvature tensor.

09-06-2012, 04:37 PM	#8
Urwumpe Donator	(Disclaimer: I just claim stuff in the field of relativity to be proved wrong about my concepts of it and learn something new. I am programmer, not a physicist) well, what speaks against Black Holes having a momentum? They are just collapsing stars, and if I understand relativity correctly, all black holes are still collapsing, we are just unable to see it, because it happens infinitesimal slow. As the star, that existed before it, already had a momentum, the black hole should also have one. Even more, every single mass of the star, may it be one cubic meter of plasma, one atom, or even just one quark, should also have a momentum, that does not just cease to exist at one point. If the black hole has to be seen as a single black hole particle that is the sum of all particles of the star and conservation of momentum still applies... I still see no difference. It just maybe depends on where you stand when you measure the momentum of each component of the star.

09-06-2012, 05:20 PM	#9
Linguofreak Orbinaut	Quote: Originally Posted by Urwumpe Yes, but if you use the simplified problem, it is really just down to conservation of four-momentum. If you are interested in the field equation of a single photon, it isn't that simple, that is correct. It also isn't describeable by the Schwarzschild metric, which assumes non-rotating, uncharged masses. The Maxwell equations could be the better tool to describe the problem then. That's why I put "photon" in quotes, because I'm not looking for the gravitational field associated with a wave described by the Maxwell equations, or with a particle that has spin. I'm looking for the gravitational field of a classical (that is, non-quantum) point particle with m=0 and nonzero momentum. Quote: Originally Posted by Fizyk EDIT: Indeed someone has - http://www.springerlink.com/content/x346w7x336224520/ $40 for access, unfortunately.