Gravitational field of photons

[espen180]

According to L. Susskind in his GR lecture series, a photon has a gravitational field around it. This is intuitive to me as well. The photon has energy, therefore it curves space-time. I have searched, but I cannot find any sources which discuss the nature of the field.

If we call the x-axis in rectangular coordinates the trajectory of the photon, what is its gravitational field?

 

[Dale]

Since we lack a quantum theory of gravity it is kind of hard to answer this question. However, the general class of solutions modeling the spacetime curvature of massless radiation is called pp-waves.

 

[edpell]

I would love to see a diagram of equal gravitational potential around the photon. The effect of gravity can never precede the photon in the direction of travel. And the wake it leaves behind dies-off quickly due to the photon moving on. And the photon is not a point particle (quantum mechanics) it is more a distribution that falls off with distance from the peak. No point no singularity problems.

 

[Dale]

I would love to see a diagram of equal gravitational potential around the photon.

I am not expert in this, but my understanding is that gravitational potential energy is generally only defined for static spacetimes, and pp-waves are not static.

 

[jnorman]

as a photon has no location between the time it is emitted and the time it is absorbed, it seems counterintuitive that there would be any gravitational field associated with it. i believe susskind is incorrect.

 

[Dmitry67]

Imagine ideally reflective sphere containing matter and antimatter. Matter is source of gravity. Then matter and antimatter annihilates and it is converted into photon gas inside the sphere. GR insists that gravitational field cannot magically ‘disappear’, hence, photon gas *MUST* create a gravitational field.

In fact, situation is more complicated:
Before the annihilation gravity is created by:
A. Matter and anti-matter
B. Mass of the sphere

After the annihilation:
C. Pressure of the photon gas inside the sphere.
D. Mass of the sphere
E. Tension of the sphere, created by the pressure of the photon gas.


I believe A+B=C+D+E, so the far observer should not notice anything (is it correct?)

 

[Dmitry67]

Also, I am not sure what happens if there is no sphere.
And what is % of gravity created by C, D and E?

 

[edpell]

as a photon has no location between the time it is emitted and the time it is absorbed, it seems counterintuitive that there would be any gravitational field associated with it. i believe susskind is incorrect.

huh? If I have a bench top setup to emit photons on the left and have them travel over the table and be detected on the right I know at all times approximately where the photon is. Let say there is a vacuum over the table to make this easier. The photon moves about 1 foot per nanosecond. We can plot the photons position as a function of time. We know where it is.

 

[edpell]

Imagine ideally reflective sphere containing matter and antimatter. Matter is source of gravity. Then matter and antimatter annihilates and it is converted into photon gas inside the sphere. GR insists that gravitational field cannot magically ‘disappear’, hence, photon gas *MUST* create a gravitational field.

In fact, situation is more complicated:
Before the annihilation gravity is created by:
A. Matter and anti-matter
B. Mass of the sphere

After the annihilation:
C. Pressure of the photon gas inside the sphere.
D. Mass of the sphere
E. Tension of the sphere, created by the pressure of the photon gas.


I believe A+B=C+D+E, so the far observer should not notice anything (is it correct?)

Doesn't the energy of the photons count for something?

 

[Dmitry67]

No, energy (=relativistic mass) is not a source of gravity in GR.
So objects moving close to c do not generate MORE gravity because they are 'heavier'. This is a common misconception (that relativistic mass=energy creates gravity). I was also a victim of that misconception before I came here.

 

[espen180]

No, energy (=relativistic mass) is not a source of gravity in GR.
So objects moving close to c do not generate MORE gravity because they are 'heavier'. This is a common misconception (that relativistic mass=energy creates gravity). I was also a victim of that misconception before I came here.

Energy density is a component of the stress-energy tensor and therefore it contributes to spacetime curvature. This curvature is the source of the gravitational field. This is how I understand it, and Susskind also makes this argument in his lectures.

 

[edpell]

Dmitry67 and espen180, am I correct in says that the two of you are holding exact opposite positions on this?

 

[espen180]

First off, energy is not relativistiv mass. Relativistic mass is m=γm₀, while energy relates to mass through E=mc². Dmitry seems to be saying that energy is not a source of gravity, but the way I percieve it is that GR does not differentiate between mass and energy density, since they are equivalent through E=mc².

Second, Dmitry seems to be saying that the momentum density and momoentum flux is independent of mass, though this is not the case, as p=γmv

If I have interpreted his posts correctly, then yes, we are at a disagreement.

 

[Truth_Seeker]

Check this out :
http://www.springerlink.com/content/b87v628331668g2r/fulltext.pdf

 

[espen180]

Check this out :
http://www.springerlink.com/content/b87v628331668g2r/fulltext.pdf

My personal a priori assumptions about the field were similar to what I read in the abstract.

I also wonder if a gravitational magnetic field analogous to the magnetic field around a moving charge would be possible.

 

[Hepth]

I also wonder if a gravitational magnetic field analogous to the magnetic field around a moving charge would be possible.

see "frame-dragging", "gravitomagnetism".

 

[espen180]

Yes, I am aware of both concepts and their applications. I was speculating about the photon's properties, not the general existence of such a field.

 

[Dmitry67]

First off, energy is not relativistiv mass. Relativistic mass is m=γm₀, while energy relates to mass through E=mc². Dmitry seems to be saying that energy is not a source of gravity, but the way I percieve it is that GR does not differentiate between mass and energy density, since they are equivalent through E=mc².

Second, Dmitry seems to be saying that the momentum density and momoentum flux is independent of mass, though this is not the case, as p=γmv

If I have interpreted his posts correctly, then yes, we are at a disagreement.

For photon gas, there is momentum flux, but for photons momentum is NOT p=ymv !
Forget about y and check the particular case of photon gas with m0=0 and v=c

 

[espen180]

You have to know the energy of the photon, that is, its frequency. The momentum is given by p=hf/c.

 

[jnorman]

edpell - inre location of a photon - no, you do not know where a single photon is. here is a clip from david snoke's paper on photons:

"Note that the uncertainty principle also is essentially a wave property, and has nothing to do with “wave-particle duality.” This is a fact known to all students who learn the mathematics of Fourier analysis, usually in the sophomore or junior year. It is also stressed in introductory graduate quantum mechanics. The momentum of a wave corresponds to its wavelength. A wave with a single wavelength is by definition an extended wave which fills all space, and therefore can be assigned no single position in space. "
ref - http://philsci-archive.pitt.edu/archive/00001504/01/photon.doc

 

[yuiop]

I can not find a reference, but I have read on a number of occasions that

(a)photons traveling in the same direction have no mutual gravitational effect on each other and
(b)photons traveling in opposite directions do attract each other.

It is also worth considering the behaviour of two massive objects to see what can be learnt. Two small masses (m1 and m2) moving in the same direction with a velocity v relative to an observer and separated by a distance d, will have a transverse force of G*m1*m2/d^2*sqrt(1-v^2/c^2). In the limit as v goes to c the transverse force goes towards zero which agrees with claim (a). The transverse acceleration of m2 towards m1 is proportional to G*m1/d^2*(1-v^2) which goes towards zero even more rapidly as v goes to c.

A single photon or a system of two photons traveling in the same direction has no rest mass and all the energy is kinetic and this is a hint that gravitational attraction is a function of rest mass energy, rather than kinetic energy or relativistic mass. It can be shown that a system of photons traveling in opposite directions have a centre of mass and a definable rest mass and this is consistent with claim (b).

A photon gas with many photons traveling in random directions will also have a definable rest mass and a gravitational field as mentioned by previous posters.

 

[Dmitry67]

A single photon or a system of two photons traveling in the same direction has no rest mass and all the energy is kinetic and this is a hint that gravitational attraction is a function of rest mass energy, rather than kinitic energy or relativistic mass. It can be shown that a system of photons traveling in opposite directions have a centre of mass and a definable rest mass and this is consistent with claim (b).

A photon gas with many photons traveling in random directions will also have a definable rest mass and a gravitational field as mentioned by previous posters.

It makes sense, thank you
However, I am confused.

Say, there are 2 intensive (but quite short) light beams going from different galaxies towards each other.

Both beams do not gravitate, based on what you say.

But for the short period of time, beams intersect in some area, creating sort of 'photon gas', which, as you say, does gravitate.

So, does gravity 'appear from nothing' during the time when they ipass thru each other and then fades?
Also, the area where they intersect is frame-dependent.

 

[yuiop]

It makes sense, thank you
However, I am confused.

Say, there are 2 intensive (but quite short) light beams going from different galaxies towards each other.

Both beams do not gravitate, based on what you say.

Photons going in the same direction in a single beam do not mutually gravitate, but it is possible that the bunch of photons as a whole will gravitationaly interact with other particles that they pass on the way. I gave the example of two masses m1 and m2 that seem to have a weaker mutual gravitational interaction according to an observer that they are moving with respect to. If a third larger mass m3 is introduced, that is stationary with respect to the observer, the gravitational interaction of the combined mass (m1+m2) with m3 is actually stronger than when m1 and m2 have zero horizontal motion relative to m3. This conclusion comes from a study of the "relativistic submarine" by Matsas. As you can see it is not a simple topic! The strength of gravity of moving systems is highly directional and dynamic.

But for the short period of time, beams intersect in some area, creating sort of 'photon gas', which, as you say, does gravitate.

So, does gravity 'appear from nothing' during the time when they ipass thru each other and then fades?
Also, the area where they intersect is frame-dependent.

My guess is that even when the two beams are far apart, they can still be treated as one "system", just as two normal masses far apart can have a mutual zero momentum frame. There is the problem that by definition the two bunches of photons can not possibly be "aware" of each other until they intersect and I am not sure how to resolve that. Sorry I can not be more helpful here and I can only try and give a few hopefully helpful hints. I am as interested as you are to know if there is a resolution to this.

 

[jtbell]

Keep in mind, GR is a classical theory, not a quantum one. We don't yet have a generally accepted theory of quantum gravity. As far as I know, GR calculations of the gravitational interaction of light consider the light as classical electromagnetic radiation, using the classical notions of energy density and momentum density of the electromagnetic field.

 

[Dmitry67]

But in the questions we discussed (light in the box, light splashes in opposite directions etc) above we did not use any quantum stuff.

 

[jtbell]

Some people in this thread seemed to me to be asking about photons in the sense of discrete QM particles, as opposed to classical "spashes of light."

 

[Rasalhague]

First off, energy is not relativistiv mass. Relativistic mass is m=γm₀, while energy relates to mass through E=mc².

Some authors, e.g. Taylor and Wheeler in Spacetime Physics, do use the word "energy" for p⁰, the time component of the energy-momentum 4-vector, which is gamma times rest mass, with the appropriate scaling factor: $E = p^0 = \gamma m_0 \, c^2$. They deprecate the term "relativistic mass", and give the name "mass" (="rest mass") to this vector's magnitude. For Taylor and Wheeler, $E = m \,c^2$, or equivalently for them $E = m_0 \,c^2$, means: rest energy equals mass; in a frame where the system is at rest (where its 3-momentum vanishes), the time component of its energy-momentum 4-vector equals the magnitude of its energy-momentum 4-vector.

Others do as you do and give the name "relativistic mass" to $\gamma m_0 \, c^2$, which Taylor and Wheeler call energy.

Someone posted link to a good article on this subject here a while ago. I thought I'd bookmarked it, but I can't find it now. I think the gist of it was that Einstein himself, except in some early papers, always made it clear that for him the famous equation meant $E = m_0 \,c^2$, rest energy equals mass (times the appropriate scaling factor).

 

[yuiop]

Some authors, e.g. Taylor and Wheeler in Spacetime Physics, do use the word "energy" for p⁰, the time component of the energy-momentum 4-vector, which is gamma times rest mass, with the appropriate scaling factor: $E = p^0 = \gamma m_0 \, c^2$. They deprecate the term "relativistic mass", and give the name "mass" (="rest mass") to this vector's magnitude. For Taylor and Wheeler, $E = m \,c^2$, or equivalently for them $E = m_0 \,c^2$, means: rest energy equals mass; in a frame where the system is at rest (where its 3-momentum vanishes), the time component of its energy-momentum 4-vector equals the magnitude of its energy-momentum 4-vector.

Others do as you do and give the name "relativistic mass" to $\gamma m_0 \, c^2$, which Taylor and Wheeler call energy.

Someone posted link to a good article on this subject here a while ago. I thought I'd bookmarked it, but I can't find it now. I think the gist of it was that Einstein himself, except in some early papers, always made it clear that for him the famous equation meant $E = m_0 \,c^2$, rest energy equals mass (times the appropriate scaling factor).

There are many types of energy.

$$Rest\ Energy = m(c^2)$$

$$Total\ Energy = m(\gamma c^2)$$

$$Momentum\ Energy = m(\gamma v c})$$

$$Kinetic\ Energy = m(\gamma c^2-c^2)$$

Note that none of the above expressions imply that the mass m has changed in any way, just as the Newtonian expression for kinetic energy (KE):

$$KE = m(v^2/2)$$

does not imply any change in the mass with motion. Note the modern usage does not use the symbols $m_0$ for rest mass and m for relativistic mass. The unqualified symbol "m" and the unqualified term "mass" simply mean rest mass and generally in modern GR, there is not considerd to be any other kind of mass.

P.S. These comments are not aimed at anyone in particular. Just general info .

 

[Rasalhague]

The only one of these terms that I hadn't heard of is "momentum energy". Taylor and Wheeler just call this momentum (i.e. 3-momentum). I see Blandford & Thorne call the vector 4-momentum, $\vec{p}$. They call its time component "energy $\varepsilon$". This energy minus its length (i.e. mass) they call "energy $E$" (well, at least they were thoughtful enough to give them different symbols, if not different names...), and its spatial components they call "momentum $\textbf{p}$". They call the first row of components of a matrix representing the stress-energy tensor "energy flux". Presumbly the other rows are "momentum flux".

 

[yuiop]

The only one of these terms that I hadn't heard of is "momentum energy". Taylor and Wheeler just call this momentum (i.e. 3-momentum).

The term $m(\gamma v c)}$ found in the expression for total energy [itex]E = \sqrt{[(mc^2)^2+( m \gamma v c)^2}][/tex] has units of energy, so it seems reasonable to call it momentum energy.[/itex]