Do photons obey the 1/r^2 gravity law?

[swle]

Do photons obey the 1/r^2 gravity law?

This was a question that came up in our recent PhySoc meeting and no-one present seemed to know the answer (lecturers included!)

Does anyone have a unequivocal answer (and preferably citing a source)?

I have done a quick search through old posts, but I couldn't see anything.

Thanks!

 

[K^2]

Do you mean gravity produced by photons or gravity experienced by photons?

 

[Barry_G]

Do you mean gravity produced by photons or gravity experienced by photons?

Gravity produced by photons? Aren't they supposed to be without mass?

Besides word "obey" I think strongly suggests it's about gravity experienced by photons.

 

[Barry_G]

Do photons obey the 1/r^2 gravity law?

This was a question that came up in our recent PhySoc meeting and no-one present seemed to know the answer (lecturers included!)

Does anyone have a unequivocal answer (and preferably citing a source)?

I have done a quick search through old posts, but I couldn't see anything.

Thanks!

Since photons obeying gravity is about them following geodesics and curvature of space-time rather than obeying Newton's law and force of gravity, if you are looking for some explicit statement about it I think you should better ask whether curvatures of space-time curve according to inverse square law, and if they do then it would follow from that photon trajectories are influenced accordingly.

 

[HallsofIvy]

In order to talk about properties of photons, we are going to have to use general relativity. And in general relativity there is no "1/r^2" law.

 

[Dickfore]

The "1/r² gravity law" is called Newton's Law of Universal Gravitation. Strictly speaking, it is invalid within GR. Instead, it is supplanted by Einstein's field equations:
<br /> R_{\mu \nu} - \frac{1}{2} R \, g_{\mu \nu} = \frac{8 \pi G}{c^4} T_{\mu \nu}<br />
(neglecting the cosmological constant term), where R_{\mu \nu} is the Ricci curvature tensor, R = R^{\mu}_{\mu} = g^{\mu \nu} R_{\mu \nu} is the scalar curvature, T_{\mu \nu} is the stress-energy tensor, c is the speed of light in vacuum, and G is the Universal Gravitational Constant that also enters in Newton's Law.
In the limit of "weak gravitational fields" (when the metric tensor may be approximated as g_{00} = 1 + 2 \phi/c^2, g_{0 i} = 0, g_{i k} = -\delta_{i k}, where \phi is the scalar gravitational potential, and this sets the condition what is meant by a weak gravitational field), the 00 component of Einstein's equations reduces to the Poisson equation for the gravitational potential:
<br /> \nabla^2 \phi = 4\pi G \, \rho<br />
where \rho is the mass density.

Because electromagnetic radiation ("photons") contribute to the stress-energy tensor, I would say they contribute to the curvature of spacetime. However, if you take the trace of Einestein's field equations:
<br /> g^{\mu \nu} R_{\mu \nu} - \frac{1}{2} R g^{\mu \nu} g_{\mu \nu} = \frac{8 \pi G}{c^4} g^{\mu \nu} T_{\mu \nu}<br />
<br /> R = -\frac{8 \pi G}{c^4} \, T^{\mu}_{\mu}<br />
and you remember that the electromagnetic stress-energy tensor is traceless, you see that electromagnetic fields do not contribute to the scalar curvature of spacetime!

Additionally, as claimed by previous posters, light follows geodesics (according to the rules of geometric optics) in curved spacetime. Or, if you write down Maxwell's equations, you need to use the covariant derivative:
<br /> D_{\nu} F^{\mu \nu} = -\mu_0 \, J^{\mu}<br />
<br /> F_{\mu \nu} = D_{\mu} A_{\nu} - D_{\nu} A_{\mu} = \partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}<br />
that is defined through the Christoffel symbols, which know about the spacetime curvature.

In any case, propagation of light is affected by the curvature of spacetime. This was experimentally proven by observing the deflection of starlight during a total eclipse of the Sun.

 

[bobc2]

Do photons obey the 1/r^2 gravity law?

This was a question that came up in our recent PhySoc meeting and no-one present seemed to know the answer (lecturers included!)

Does anyone have a unequivocal answer (and preferably citing a source)?

I have done a quick search through old posts, but I couldn't see anything.

Thanks!

Yes. For example a photon in the gravitational force field of the Earth would result in a force of:

f = G(m1)(m2)/r^2

where G is Newton's gravitational constant, m1 is the mass of the earth, and m2 is the mass of the photon

f = G(m1)(0)/r^2 = 0

 

[pervect]

There's a famous result that shows that light deflects twice as much due to gravity than "Newtonian" predictions. This was one of the first tests of General Relativity.

It's not too clear how to interpret the original question so that it's testable by experiment, but if you take it as an expression of "is light deflection calculated to have the samve value in GR as it is in Newtonian theory" the answer is no.

 

[Bill_K]

There's a famous result that shows that light deflects twice as much due to gravity than "Newtonian" predictions. This was one of the first tests of General Relativity.

The famous "Newtonian calculation" that gets half the correct value has no legitimate basis, it's simply wrong. In Newtonian theory there is no light deflection at all, because light obeys Maxwell's Equations and travels in straight lines.

 

[pervect]

The famous "Newtonian calculation" that gets half the correct value has no legitimate basis, it's simply wrong. In Newtonian theory there is no light deflection at all, because light obeys Maxwell's Equations and travels in straight lines.

The only good derivation I can think of for the "Newtonian" light deflection is in the context of the PPN approximation.

Here you can recover the "Newtonian" results by setting the PPN paramater gamma=0. I believe this corresponds to the predictions of Newton-Cartan theory, a geometrical reformulation of Newtonian gravity. I think you can still characterize Newton-Cartan theory with a 1/r^2 force law.

I'm not sure of the history of all this - but at least at the time, people seem to be convinced of the validity of the test.

[add]
I suppose I should mention http://en.wikipedia.org/w/index.php?title=Parameterized_post-Newtonian_formalism&oldid=523184874 for the other readers of this thread who might not all be familiar with the acronym PPN. Anyway, because gamma is the only PPN pamarmeter important to light deflection, I view this as convincing demonstration that "space curvature" is the explanation for the extra deflection of light in GR, as gamma can be interpreted as the "amount of space curvature per unit mass". Occasionally this line of argument of mine causes some argument, I think it's reasonable though.

 

[bcrowell]

The 1/r^2 law is a Newtonian approximation to general relativity. As others have noted, there is no 1/r^2 law in GR.

A photon gas inside a spherical mirror produces the same external gravitational field as an equivalent amount of mass-energy in any other form: https://www.physicsforums.com/showpost.php?p=2956775&postcount=15 If the fields are weak, then the Newtonian approximation is good, and the field is approximately 1/r^2.

Gravity produced by photons? Aren't they supposed to be without mass?

FAQ: If light is massless, why is it affected by gravity?

General relativity describes gravity as curvature of spacetime. Any sufficiently small particle (massive or massless) traveling through a curved spacetime moves along a geodesic, which means a "line" that is as straight as possible.

Another thing to realize is that "mass" has a specialized technical meaning in relativity; it means m=\sqrt{E^2-p^2} (in units where c=1). When we say that a photon is massless, that's what we mean. But mass in GR doesn't have all the properties you might think. For example, mass isn't additive, and a box full of photons has a nonzero contribution to its mass coming from the photons, even though the photons individually have zero mass.

FAQ: Does light produce gravitational fields?

The short answer is yes. General relativity predicts this, and experiments confirm it, albeit in a somewhat more indirect manner than one could have hoped for.

Theory first. GR says that gravitational fields are described by curvature of spacetime, and that this curvature is caused by the stress-energy tensor. The stress-energy tensor is a 4x4 matrix whose 16 entries measure the density of mass-energy, the pressure, the flux of mass-energy, and the shear stress. In any frame of reference, an electromagnetic field has a nonvanishing mass-energy density and pressure, so it is predicted to act as a source of gravitational fields.

There are some common sources of confusion. (1) Light has a vanishing rest mass, so it might seem that it would not create gravitational fields. But the stress-energy tensor has a component that measures mass-energy density, not mass density. (2) One can come up with all kinds of goofy results by taking E=mc^2 and saying that a light wave with energy E should make the same gravitational field as a lump of mass E/c^2. Although this kind of approach sometimes suffices to produce order-of-magnitude estimates, it will not give correct results in general, because the source of gravitational fields in GR is not a scalar mass-energy density, it's the whole stress-energy tensor. However, there is one case of interest where this does happen to work. If a photon gas of total mass E is contained inside a spherical mirror, then the external spacetime is exactly the Schwarzschild solution for a mass E/c^2. The external field has a contribution from the photons that is double this amount, but half of that is canceled by the pressure at the mirror.

Experimentally, there are a couple of different ways that I know of in which light has been tested as a gravitational source. An order of magnitude estimate based on E=mc^2 tells us that the gravitational field made by an electromagnetic field is going to be extremely weak unless the EM field is extremely intense.

One place to look for extremely intense EM fields is inside atomic nuclei. Nuclei get a small but nonnegligible fraction of their rest mass from the static electric fields of the protons. According to GR, the pressure and energy density of these E fields should act as a source of gravitational fields. If it didn't, then nuclei with different atomic numbers and atomic masses would not all create gravitational fields in proportion to their rest masses, and this would cause violations of Newton's third law by gravitational forces. Experiments involving Cavendish balances[Kreuzer 1968] and lunar laser ranging[Bartlett 1986] find no such violations, establishing that static electric fields do act as sources of gravitational fields, and that the strength of these fields is as predicted by GR, to extremely high precision. The interpretation of these experiments as a test of GR is discussed in [Will 1976] and in section 3.7.3 of [Will 2006]; in terms of the PPN formalism, if E fields did not act as gravitational sources as predicted by GR, we would have nonzero values of the PPN zeta parameters, which measure nonconservation of momentum.

Another place to look for extremely intense EM fields is in the early universe. Simple scaling arguments show that as the universe expands, nonrelativistic matter becomes a more and more important source of gravitational fields compared to highly relativistic sources such as the cosmic microwave background. Early enough in time, light should therefore have been the dominant source of gravity. Calculations of nuclear reactions in the early, radiation-dominated universe predict certain abundances of hydrogen, helium, and deuterium. In particular, the relative abundance of helium and deuterium is a sensitive test of the relationships among a, a', and a'', where a is the scale-factor of the universe. The observed abundances confirm these relationships to a precision of about 5 percent.[Steigman 2007]

Kreuzer, Phys. Rev. 169 (1968) 1007

Bartlett and van Buren, Phys. Rev. Lett. 57 (1986) 21

Will, "Active mass in relativistic gravity - Theoretical interpretation of the Kreuzer experiment," Ap. J. 204 (1976) 234, available online at http://articles.adsabs.harvard.edu//full/1976ApJ...204..224W/0000224.000.html

Will, "The Confrontation between General Relativity and Experiment," http://relativity.livingreviews.org/Articles/lrr-2006-3/ , 2006

Steigman, Ann. Rev. Nucl. Part. Sci. 57 (2007) 463

 

[K^2]

Gravity produced by photons? Aren't they supposed to be without mass?

Gravity isn't generated by mass. It's generated by stress-energy tensor, which is certainly non-zero for photons, which have both energy and momentum. Photons do generate gravity in GR.

The "1/r2 gravity law" is called Newton's Law of Universal Gravitation. Strictly speaking, it is invalid within GR.

Strictly speaking? Sure. But it does hold loosely within reasonable limits, and will be followed by light as well as matter. If you pretend that photons are particles traveling at speed c having a mass p/c, and you are looking at "acceleration" due to gravity in perpendicular direction, you'll only be a factor of 2 off. So 1/r² dependence can still be said to hold. It's an entirely valid question to ask.

Of course, the answer comes in with many caveats. But you are not going to be able to explain them all to a person who asks such a question.

 

[Barry_G]

The 1/r^2 law is a Newtonian approximation to general relativity. As others have noted, there is no 1/r^2 law in GR.

Two beams of light are passing next to some massive planet or a star, where one beam is at double the distance away than the other. Will trajectory of the closer beam be four times as influenced compared to further away beam?

If the fields are weak, then the Newtonian approximation is good, and the field is approximately 1/r^2.

What do you mean by "weak", anything less than black hole? But isn't that strength proportional to square of the distance? How do you measure the strength of gravity field?

FAQ: If light is massless, why is it affected by gravity?

Perhaps because photons actually do have mass?

For example, mass isn't additive, and a box full of photons has a nonzero contribution to its mass coming from the photons, even though the photons individually have zero mass.

Mass is supposed to be additive, and you just said yourself photons actually do contribute to the mass of the box.

One place to look for extremely intense EM fields is inside atomic nuclei. Nuclei get a small but nonnegligible fraction of their rest mass from the static electric fields of the protons. According to GR, the pressure and energy density of these E fields should act as a source of gravitational fields. If it didn't, then nuclei with different atomic numbers and atomic masses would not all create gravitational fields in proportion to their rest masses, and this would cause violations of Newton's third law by gravitational forces. Experiments involving Cavendish balances[Kreuzer 1968] and lunar laser ranging[Bartlett 1986] find no such violations, establishing that static electric fields do act as sources of gravitational fields, and that the strength of these fields is as predicted by GR, to extremely high precision.

I'm not sure what you just said. Are atomic masses exact sum of their protons, neutrons and electrons masses or not?

 

[Barry_G]

Gravity isn't generated by mass. It's generated by stress-energy tensor, which is certainly non-zero for photons, which have both energy and momentum.

I'd say gravity is a property proportional to what we call mass, rather than "generated" by mass. Anyway, your statement sounds as if GR invalidates all the rest of the physics. You surely meant to say that specifically and only in theory of General Relativity gravity is not 'generated' by mass?

How do you know this is not just semantics? How do you know what you call stress-energy tensor is not just another name for mass? How do you measure stress-energy tensor, and how does it relate to measurements of mass?

Photons do generate gravity in GR.

Isn't gravity just another name for mass? I mean when we measure mass are we not actually measuring the force of gravity? How do you separate one from another? Is there any reason to conclude mass and gravity are not one and the same, one property rather than two separate ones?

Strictly speaking? Sure. But it does hold loosely within reasonable limits, and will be followed by light as well as matter. If you pretend that photons are particles traveling at speed c having a mass p/c, and you are looking at "acceleration" due to gravity in perpendicular direction, you'll only be a factor of 2 off. So 1/r² dependence can still be said to hold. It's an entirely valid question to ask.

So this perpendicular acceleration of photons toward some mass they pass next to, is two times less or two times more compared to would be for, say, electrons?

 

[K^2]

I'd say gravity is a property proportional to what we call mass, rather than "generated" by mass. Anyway, your statement sounds as if GR invalidates all the rest of the physics. You surely meant to say that specifically and only in theory of General Relativity gravity is not 'generated' by mass?

How do you know this is not just semantics? How do you know what you call stress-energy tensor is not just another name for mass? How do you measure stress-energy tensor, and how does it relate to measurements of mass?

Short answer is that mass doesn't transform correctly under change of frame of reference. The only quantity that does transform correctly, and happens to be the conserved quantity under the transformations, is the stress-energy tensor.

 

[Barry_G]

Short answer is that mass doesn't transform correctly under change of frame of reference. The only quantity that does transform correctly, and happens to be the conserved quantity under the transformations, is the stress-energy tensor.

But that doesn't mean stress-energy tensor does not refer to the same property as what we call mass in the rest of the physics. It seems to me they are the same, where GR equations just embed some "corrections" to the classical concept of mass, just like SR equations do, but they all refer to the same thing, the same property. Wouldn't you agree? In any case, how do we measure this stress-energy tensor? How do we even get to compare the two?

 

[WannabeNewton]

But that doesn't mean stress-energy tensor does not refer to the same property as what we call mass in the rest of the physics. It seems to me they are the same, where GR equations just embed some "corrections" to the classical concept of mass, just like SR equations do, but they all refer to the same thing, the same property. Wouldn't you agree? In any case, how do we measure this stress-energy tensor? How do we even get to compare the two?

How can they be the same? The SET is a 2 - tensor field and mass density is a scalar field. The point is that there are factors other than just mass that contribute to curvature in general. If you consider the Newtonian limit then the 00 component will get the usual, familiar result involving the scalar potential and mass density.

 

[K^2]

where GR equations just embed some "corrections" to the classical concept of mass

GR equations aren't corrections. They are a total overhaul. While classical approximation exists for many phenomena, such as planetary orbits and to an extent even frame dragging, via gravitomagnetic interactions, there are also situations where classical gravity fails absolutely. Gravitational field of a photon is one such example. Photon distorts space-time despite having zero mass.

 

[Barry_G]

How can they be the same? The SET is a 2 - tensor field and mass density is a scalar field.

Their description is not the same, but both stress–energy tensor and mass refer to the same physical phenomena.

http://en.wikipedia.org/wiki/Stress-energy_tensor : The stress-energy tensor is the source of the gravitational field in the Einstein field equations of general relativity, just as mass density is the source of such a field in Newtonian gravity.

The point is that there are factors other than just mass that contribute to curvature in general.

Like what for example? You can not say "other than mass" if you are referring to GR because there is no concept of mass in that theory to start with, and instead there is stress-energy tensor.

 

[Barry_G]

GR equations aren't corrections. They are a total overhaul. While classical approximation exists for many phenomena, such as planetary orbits and to an extent even frame dragging, via gravitomagnetic interactions, there are also situations where classical gravity fails absolutely. Gravitational field of a photon is one such example. Photon distorts space-time despite having zero mass.

You can not say "despite having zero mass" because in GR there is no concept of mass to start with. It's like saying in USA there are no lifts because there are elevators. And when you say photons have a property that you quantify with the description called "stress–energy tensor", translated to classical physics you just said photons have mass.

http://en.wikipedia.org/wiki/Stress-energy_tensor : The stress-energy tensor is the source of the gravitational field in the Einstein field equations of general relativity, just as mass density is the source of such a field in Newtonian gravity.


Can we measure this "stress-energy_tensor", how do we measure it?

Short answer is that mass doesn't transform correctly under change of frame of reference.

Can you give me some practical example relating to some actual measurements?

If you pretend that photons are particles traveling at speed c having a mass p/c, and you are looking at "acceleration" due to gravity in perpendicular direction, you'll only be a factor of 2 off.

Why pretend? A photon is an elementary particle, is it not? Is it any less real or any less of a particle than electron for example? We can emit individual single photons in double-slit experiments, and whenever we actually get to measure them we measure discrete energy quanta, just like we do when we measure electrons, right? Where by "measure" I mean when they impact a sensor at specific location with specific energy.

Now, as you said, when a photon passes next to some mass it will experience acceleration towards that mass, but this acceleration will be off by the factor of two. What experiment are you referring to, can you point me to some paper or internet article where I can see some actual numbers regarding this perpendicular acceleration of photons?

 

[K^2]

You can not say "despite having zero mass" because in GR there is no concept of mass to start with.

Of course there is. m=\frac{\sqrt{p_{\mu}p^{\mu}}}{c}. You still have forces and inertia in GR, and there is inertial mass, which is equal to \gamma m, and you can show that this gives rise to fictitious force we call gravity.

You have a great number of misconceptions on the subject. You should probably read a good textbook on Special Relativity.

translated to classical physics you just said photons have mass

p^{\mu}=(\frac{\hbar \omega}{c}, \hbar k_x, \hbar k_y, \hbar k_z)
p_{\mu}p^{\mu} = \frac{\hbar^2 \omega^2}{c^2} - \hbar^2 k^2 = 0

Photon's mass is zero.

Why pretend? A photon is an elementary particle, is it not? Is it any less real or any less of a particle than electron for example?

Not "pretend it's a particle", but "pretend it has mass p/c". Like I said earlier, photon is a massless particle, and should not be affected by gravity at all under Newtonian gravity.

Can we measure this "stress-energy_tensor", how do we measure it?

Indirectly.

Can you give me some practical example relating to some actual measurements?

Basically, every experiment that tests General Relativity.

 

[Barry_G]

Of course there is. m=\frac{\sqrt{p_{\mu}p^{\mu}}}{c}. You still have forces and inertia in GR, and there is inertial mass, which is equal to \gamma m, and you can show that this gives rise to fictitious force we call gravity.

Can you point Wikipedia article or some other reference where I can see that is indeed General Relativity equation and in what context was given?

What forces are there in General Relativity, can you point some reference so I can see you are not just imagining things?

You have a great number of misconceptions on the subject. You should probably read a good textbook on Special Relativity.

Special Relativity? Are we not talking about stress–energy tensor and General Relativity?

And while we at it, do photons have mass in Special Relativity?

p^{\mu}=(\frac{\hbar \omega}{c}, \hbar k_x, \hbar k_y, \hbar k_z)
p_{\mu}p^{\mu} = \frac{\hbar^2 \omega^2}{c^2} - \hbar^2 k^2 = 0

Photon's mass is zero.

Is that GR or SR? Reference?

Indirectly.

For example? Can you be more specific what are you actually referring to or provide some reference?

Basically, every experiment that tests General Relativity.

You said "mass doesn't transform correctly under change of frame of reference". Can you be more specific what are you actually referring to or provide some reference?

 

[haushofer]

Your reference would be any book on gr. The inner product in the expression for m is with respect to the general metric g_mu\nu.

 

[andrien]

you can see some basics of GR here,
http://en.wikipedia.org/wiki/Introduction_to_general_relativity
photon's mass is zero,there has been various tests to verify inverse square law.one can see some here
https://docs.google.com/viewer?a=v&q=cache:GhpnEZBV3R8J:www.princeton.edu/~romalis/PHYS312/Coulomb%2520Ref/TuCoulomb.pdf+upper+limit+on+photon+mass+pdf&hl=en&gl=in&pid=bl&srcid=ADGEESiuc8NGdoELeIbdt20LFBwqQygVZlW_G84ItdaN44ouYUXkXB3fIgBV6VZbAjdW5jd-O3n9dKoW9ionvb_uc0M-7qE_uDEIv-6OSgLvRoLVZ5BSG448XKi2oB6yF_E32mzjRQIK&sig=AHIEtbQfjcdCpAwstrj_0-0qNki_LePXWQ

 

[Barry_G]

Your reference would be any book on gr. The inner product in the expression for m is with respect to the general metric g_mu\nu.

What are you replaying to? What are you trying to say?

I'm talking about this:

http://en.wikipedia.org/wiki/Stress-energy_tensor : The stress-energy tensor is the source of the gravitational field in the Einstein field equations of general relativity, just as mass density is the source of such a field in Newtonian gravity.


It's one or the other, depending on what theory you are talking about, but not both, as they both describe the same physical phenomena. Ok? Now, what is your point? And there is whole internet one click away, so please point some link to confirm whatever is you are trying to say, especially if you are going to be vague as your last reply.

 

[Barry_G]

you can see some basics of GR here,

What are your replying to? What question is that supposed to answer?

photon's mass is zero,there has been various tests to verify inverse square law.one can see some here

...a nonzero photon mass could be so small that present-day experiments cannot probe it.

The experimental results just serve to set an upper bound to the photon mass...

 

[K^2]

If photon mass was non-zero, light could travel at different velocities. That would cause many problems. For starters, your GPS would not work.

We also have field theory to support the massless photon. That whole deal with Higgs boson? It has to do with massless photon.

So both relativistic quantum field theory and General Relativity require a massless photon, and these are two theories that have the best experimental support, some predictions tested to better than 10^-11.

Can you point Wikipedia article or some other reference where I can see that is indeed General Relativity equation and in what context was given?

http://en.wikipedia.org/wiki/Four-momentum

What forces are there in General Relativity, can you point some reference so I can see you are not just imagining things?

You are being rude despite being wrong in just about every way with several people pointing to references proving so.

But here is the reference on forces.
http://en.wikipedia.org/wiki/Four-force

You said "mass doesn't transform correctly under change of frame of reference". Can you be more specific what are you actually referring to or provide some reference?

Mass is an invariant quantity. In a moving reference frame, inertial mass increases. If gravitational mass doesn't increase to match, you are going to have inconsistencies even at the level of classical gravity. So the source of gravity is energy, not mass. Fact that almost all, if not all, of mass is dynamically generated is also a proof of that. Energy is frame dependent, but what's energy in one frame is momentum in another. So gravity has to depend on four-momentum. But four-momentum density doesn't transform correctly, because densities aren't invariant under coordinate transformations. So you have to go to a stress-energy tensor, which gives you four-momentum density given a reference frame.

This is the very foundation of General Relativity. And you really should just sit down and read a book. However, before you do that, you need to learn Special Relativity, because you definitely don't know any of that either.

More importantly, when you don't understand subject, don't try to behave like you do. You can ask questions about why things the way they are, but you need to understand in advance that there are good reasons. You are not going to invent anything new until you properly understand the existing theory developed over centuries by smarter people than anyone here. Every question you ask has been asked, answered, and found to be answered satisfactory. Most, over a century ago.

This isn't some fringe theory we are talking about. This is well established science, and it is well-established for a reason.

 

[Barry_G]

If photon mass was non-zero, light could travel at different velocities. That would cause many problems. For starters, your GPS would not work.

We also have field theory to support the massless photon. That whole deal with Higgs boson? It has to do with massless photon.

So both relativistic quantum field theory and General Relativity require a massless photon, and these are two theories that have the best experimental support, some predictions tested to better than 10^-11.

http://en.wikipedia.org/wiki/Photon#Experimental_checks_on_photon_mass : The photon is currently understood to be strictly massless, but this is an experimental question. If the photon is not a strictly massless particle, it would not move at the exact speed of light... Relativity would be unaffected by this


http://www.princeton.edu/~romalis/PHYS312/Coulomb Ref/TuCoulomb.pdf : a nonzero photon mass could be so small that present-day experiments cannot probe it. The experimental results just serve to set an upper bound to the photon mass...

http://en.wikipedia.org/wiki/Four-momentum

That article is about Special Relativity, and I don't see your equation there.

You are being rude despite being wrong in just about every way with several people pointing to references proving so.

Your condescending attitude is rude. What do you imagine I am wrong about when I only made few statements all of which relating to this Wikipedia quote:

http://en.wikipedia.org/wiki/Stress-energy_tensor : The stress-energy tensor is the source of the gravitational field in the Einstein field equations of general relativity, just as mass density is the source of such a field in Newtonian gravity.

It is not about whether I am wrong, but whether you can back up your claims.

So the source of gravity is energy, not mass.

According to?

Fact that almost all, if not all, of mass is dynamically generated is also a proof of that.

What are you talking about, what experiment proves mass is dynamically generated?

And you really should just sit down and read a book. However, before you do that, you need to learn Special Relativity, because you definitely don't know any of that either.

What we started to talk about is stress-energy tensor and your statement that photons generate gravity. So if you would get back on the topic and point out reference to some experiment which measured this gravity or stress-energy tensor of a photon. Can you?

More importantly, when you don't understand subject, don't try to behave like you do.

When you can't answer a question, don't try to change the subject.

 

[K^2]

You keep asking questions about standard results given in every text-book. When I give you derivations of these, you ask more questions. What sort of level of knowledge am I supposed to expect? I can't teach you the entirety of Special Relativity and a big chunk of General Relativity in a hand-full of posts. You are expected to have basic understanding of these subjects if you want to have a discussion in this part of the forum. Not only do you lack these, but you continue to behave like it's my fault.

If the photon is not a strictly massless particle, it would not move at the exact speed of light... Relativity would be unaffected by this

Relativity, sure. How about Electrodynamics, Quantum Electrodynamics, and Quantum Chromodynamics? Fact that they rely on massless photon and produce good results doesn't matter?

That article is about Special Relativity, and I don't see your equation there.

It's the second equation on the page. Out of 9 total. I shouldn't need to point something like that out to you.

According to?

What do you want me to even say here? I've given you a standard result and an argument. You chose to ignore it. Assumption that invariant mass is source of gravity would violate equivalence. That's all that needs to be said. Can you look up equivalence principle yourself?

What are you talking about, what experiment proves mass is dynamically generated?

Every single experiment in nuclear physics since discovery of a quark. Masses of constituents of a nucleon are not even 10% of the mass of the nucleon. If gravitational mass of a proton was sum of gravitational masses of quarks, matter would be 1-2 orders of magnitude lighter. The mass of matter is mostly dynamic. That's well established.

And don't even bother asking for reference before you at least try to look up masses of nucleons and quarks.

Your condescending attitude is rude. What do you imagine I am wrong about when I only made few statements all of which relating to this Wikipedia quote:

You do not understand what a stress-energy tensor is or where it comes from. You do not understand what is mass, four-momentum, how they are related, and why it has to be the later and not the former that is the source of gravity. Yet you want me to explain to you the quote that hinges on all of these terms. I've given you the simple explanation. You are not happy with it. You want to understand it. That's a good thing, certainly. But you can't expect me to explain entire SR to you in here. You'll have to learn it on your own.

 

[Barry_G]

You do not understand... But you can't expect me to explain entire SR to you in here. You'll have to learn it on your own.

You don't understand that I do not want you to explain anything, and if possible stop talking about me. You already made more than enough interesting claims, now I would like you to back them up with some references, if you can. First of all this one, from post #12:

"Photons do generate gravity in GR."

No more explanations please, just a link to some experiment where we can see what measurements you base that claim on. Is there such experiment? Has anyone ever measured this gravity of photons you speak of? Just show me, that's all.

 

[K^2]

http://en.wikipedia.org/wiki/Electromagnetic_stress–energy_tensor

 

[Barry_G]

http://en.wikipedia.org/wiki/Electromagnetic_stress–energy_tensor

No mention of any experiment or measurements there, they don't even mention photons. Plus, photons have zero electric charge, and they are magnetically neutral too, so I am sorry to inform you but that is nowhere near to even begin to support your claim how photons generate gravity. Never mind, just one more thing, what is supposed to be the strength of photon gravity field?

 

[PeterDonis]

a nonzero photon mass could be so small that present-day experiments cannot probe it. The experimental results just serve to set an upper bound to the photon mass...

Yes, that's true. The current upper bound (at least as of 2003, this is the latest experiment I'm aware of) is 10^-51 grams:

http://www.aip.org/pnu/2003/split/625-2.html

But note that that is an upper bound: the results, as with all previous experiments done to test this, are consistent with the photon being massless.

Also, you appear to be confused about the usage of the term "mass". When we say a photon has zero mass, we mean invariant mass:

http://en.wikipedia.org/wiki/Invariant_mass

When you say that "mass" is what produces gravity, you really mean (whether you realize it or not) energy:

http://en.wikipedia.org/wiki/Mass–energy_equivalence

The stress-energy tensor is just the correct relativistic generalization of "energy produces gravity": it includes contributions from energy, momentum, pressure, and stress, because you have to include all those things for the source of gravity to transform correctly when you change coordinates.

So there is nothing contradictory about saying that photons have zero mass (meaning invariant mass) but still produce gravity (because they have energy, or more precisely a nonzero stress-energy tensor).

As far as experiments being done to directly test that photons produce gravity, their gravity is far too weak to directly detect. But that's also true of plenty of objects that have nonzero invariant mass, like atoms. I believe modern Cavendish-type experiments can detect the gravity from masses on the order of a kilogram, but I haven't been able to find a reference online.

 

[HomogenousCow]

Keep it civil people.
In relativity we have to write all equations in a coordinate independent manner, called general covariance. Some scalars are not coordinate independent, some are, for example the inner product between two four vectors is, mass density measured in an arbitrary frame is not.
Thus we require an entity that is coordinate independent to describe the effect of matter on space time, the Stress-Energy tensor is chosen for Einstein's Relativity. The Stress-Energy tensor is a second order tensor that describes the flux of four-momentum across a slice of constant four-space. For example the 00 component is the flux of the 0th component of the momentum four vector, energy across a slice of constant 0, time, hence it is just the energy density.
Also, be aware that in GR, when we speak of the photon we are being very informal about it, GR is not a quantum theory and only a quantum theory can quantize the EM field, thus putting a good definition on what we mean by a "Photon".
Now here's where your question gets answered, I will put this in two parts, the effect of "photons" on spacetime, and the effect of spacetime on the "photons"
In GR we usually speak of photons as particles, however GR is a classical theory, hence we still have to formally treat photons in their usual field sense. An electromagnetic field has a stress energy tensor known as the electromagnetic-stress-energy tensor. Simply feed this into the Einstein field equation, (actually, not so simple, I don't think we actually have any closed form solutions of this) and "in principle", you can get the metric for that system.
2.If you already have a prescribed metric for the system, you can use this to solve Maxwell's equations in curved space time. (Again, "In principal", whenever curved space time is involved things do indeed get very nonlinear, and even worse, coupled)

 

[Barry_G]

http://en.wikipedia.org/wiki/Mass–energy_equivalence

Article says: "Through all such conversions, however mass remains conserved..."

It seems to me they say mass can be singled out and distinguished from the rest of the energy. And while speaking of mass conservation in that sense, what about positron-electron annihilation? They both have mass and yet they produce nothing else but photon which supposedly has no mass. That doesn't sound as if mass is conserved in the way that part of the article seems to suggests.

So there is nothing contradictory about saying that photons have zero mass (meaning invariant mass) but still produce gravity (because they have energy, or more precisely a nonzero stress-energy tensor).

Yeah, it's ambiguous enough not to be contradictory.

As far as experiments being done to directly test that photons produce gravity, their gravity is far too weak to directly detect.

And so my point is that your statement "photons produce gravity" is just about as valid as if I say "photons have mass".

 

[Barry_G]

An electromagnetic field has a stress energy tensor known as the electromagnetic-stress-energy tensor. Simply feed this into the Einstein field equation...

If photons are electrically and magnetically neutral, what numbers do we feed in that equation?

 

[K^2]

No mention of any experiment or measurements there, they don't even mention photons. Plus, photons have zero electric charge, and they are magnetically neutral too, so I am sorry to inform you but that is nowhere near to even begin to support your claim how photons generate gravity. Never mind, just one more thing, what is supposed to be the strength of photon gravity field?

You do realize that a photon IS electromagnetic field, right? That isn't actually news to you, I hope. You see what happens when you ask for references without explanation? You don't understand it. Then I have to explain it. Then you don't understand that either. Where is an end to this? At what point do you accept that you need to step back and learn some fundamental theory?

It seems to me they say mass can be singled out and distinguished from the rest of the energy.

Yes. It's the energy that remains after you take away all of the kinetic energy, if you'd like to think of it that way. Yes, it's certainly special. But it's importance lies in particle propagation. Not in how it generates gravity. Gravity is generated by all of the energy.

Again, all of this is something you should know before you enter a discussion about gravity. SR is a prerequisite to GR, and we can't get past your confusion on that subject.

And so my point is that your statement "photons produce gravity" is just about as valid as if I say "photons have mass".

No. The difference is that if we say, "Photons have mass, but it's so absolutely tiny as not to be detectable by any experiment we have conducted, nor to show up as a side-effect in absolutely any theory," we can also say, "Who cares?" Our theory is more precise than experiment can refine with a massless photon. If photon mass has not manifested itself in all of this, the only sane assumption is to assume it is zero and keep going until it becomes a problem.

Photon producing no gravity would say that general relativity is absolutely wrong. It would fly in the face of all that we know about gravity. Possible? Technically. So are leprechauns. Technically.

And while speaking of mass conservation in that sense, what about positron-electron annihilation?

Is e-p annihilation described by Lorentz transformation? "Mass conservation in that sense" has nothing to do with annihilation processes. Mass is not generally a conserved quantity. It's merely a frame-invariant one.

 

[HomogenousCow]

Oh, Now I see where the confusion has come in.
Please re-read my post, a Photon is a quantum entity and thus cannot be described by a classical theory such as GR. Photons are the quantized "bits" of the electromagnetic field in quantum mechanics, in classical theory we still regard the EM field as a field. The Photon being changeless is irrelevant in general relativity.

 

[PeterDonis]

Article says: "Through all such conversions, however mass remains conserved..."

By "mass" here, the article means (at least, assuming it was actually trying to be correct) either total energy, or the total invariant mass of the *system*, as opposed to its components. Invariant mass is not additive: objects with zero invariant mass can form a system that has nonzero invariant mass. See below.

It seems to me they say mass can be singled out and distinguished from the rest of the energy.

No, they're not (at least, not if they're trying to be correct). See above.

And while speaking of mass conservation in that sense, what about positron-electron annihilation? They both have mass and yet they produce nothing else but photon which supposedly has no mass.

This is an example of how invariant mass is not additive. The total invariant mass of the system is conserved; it's the same before and after the annihilation. This is the relativistic version of saying that total energy is conserved; if we do our analysis in the center of mass frame of the total system, the invariant mass *is* the total energy (with a factor of c^2 in there if you want to measure mass and energy in different units). But stating it as conservation of the system's invariant mass let's us transform to other reference frames while the conservation law continues to hold.

Now look at the details in the center of mass frame. Before the annihilation, we have an electron and a positron, which are moving towards each other with equal and opposite velocity. The total energy of the system is:

* electron invariant mass + positron invariant mass + electron kinetic energy + positron kinetic energy

and the total momentum of the system is zero, because the momenta of the electron and positron are equal and opposite.

After the annihilation, we have two photons, which are moving in opposite directions. The total momentum is again zero (the photon momenta are equal and opposite and so cancel), and the total energy is

* photon 1 energy + photon 2 energy

By conservation of energy, we must have

* electron invariant mass + positron invariant mass + electron kinetic energy + positron kinetic energy = photon 1 energy + photon 2 energy = total energy of system = total invariant mass of system (because we're in the center of mass frame)

So the invariant mass of the system is conserved, even though the invariant masses of the components change.

Yeah, it's ambiguous enough not to be contradictory.

No, it's using terminology precisely enough to make it clear how it's not contradictory, because the term "mass" can mean different things. The apparent "contradiction" only arises if one is sloppy about terminology.

And so my point is that your statement "photons produce gravity" is just about as valid as if I say "photons have mass".

But the difference is that "photons produce gravity" is unambiguous, while "photons have mass" is ambiguous; it depends on what you mean by "mass". Photons have zero invariant mass, but nonzero energy. So "photons have mass" can be true or false depending on how you interpret it, while "photons produce gravity" is unambiguous and true. It's you who are insisting on ambiguous terminology, not me.

 

[HomogenousCow]

Why do all threads end this way...

 

[K^2]

HomogenousCow, if we have a single photon, and all we do not consider how it interacts with anything else, it is fully described by Maxwell's Equations. Quantization, and indeed, linearity, are not necessary until we start considering a second particle in the same space. There should not be a problem with describing a photon in space-time it itself curves in a manner that is 100% consistent with both GR and QM.

As soon as you throw in a second photon, or any other particle, yes, we start making assumptions.

 

[HomogenousCow]

HomogenousCow, if we have a single photon, and all we do not consider how it interacts with anything else, it is fully described by Maxwell's Equations. Quantization, and indeed, linearity, are not necessary until we start considering a second particle in the same space. There should not be a problem with describing a photon in space-time it itself curves in a manner that is 100% consistent with both GR and QM.

As soon as you throw in a second photon, or any other particle, yes, we start making assumptions.

I apologize, my knowledge of photons is not all that thorough.
I am a bit confused here myself, in GR we can calculate classical trajectories for photons (null paths), however isn't this a direct violation of quantum mechanics?

 

[K^2]

I am a bit confused here myself, in GR we can calculate classical trajectories for photons (null paths), however isn't this a direct violation of quantum mechanics?

While we can't do full-on quantum gravity, we can do field theory in curved space-time. So if you have a macroscopic massive body that produces curvature which we can assume to be otherwise irrelevant on quantum level, we can talk about trajectories of quantized particles in the resulting space-time.

And the reason this works is actually quite interesting. Are you familiar with path-integral formulation of field theory? Let's look at classical optics. Light is said to take the shortest path. But that's if you think of light as beams. Furthermore, how would it know which path is shortest? Enter wave optics. You can look at light propagation and note that it's equivalent to point sources throughout space emitting spherical waves. Each source excites the neighbors, and so it propagates.

But what path does light take in such description? All of them. From every point it goes in all possible directions, and from each following point it goes in all directions available there. However, what happens when we add all of these paths together? The phase of the EM wave will depend on distance traveled up to that point. With infinitely many ways to travel from one point to another, aren't all of the phases going to be random? Almost. The paths that are local minima will have zero derivative with respect to perturbation, and so will have infinitely many paths of infinitesimal variation with identical length. All of the random phases cancel out. The phase corresponding to shortest path persists. And we get optics.

Turns out, same thing works for particles in field theory. Things are a little more complicated because the relationship between k and ω is a bit more complex, but you still have some frequency with respect to proper time. That means that distance along the world-line of a path is the distance that will affect final phase. And that means the shortest space-time distance is the path the particle will take. Ergo, particles of quantum field theory follow geodesics of GR.

Problems happen when we decide that we want these particles to influence the curvature of space-time. The whole thing becomes non-linear, axioms of QFT fail, and we have no theory.

There might be a way to formulate a self-consistent non-linear field theory, but what are you going to do with it, when all of your QFT framework depends on quantization, which, in turn, depends on linearity?

 

[HomogenousCow]

I am familiar with the path integral formulation (Although I dread it), but as I understand we have to do a functional integral over all possible paths with the phase being the action of that path. I do understand your point, it is a good approximation to just assume the stationary path since it is there that the phases add constructively.
On a side note, what is the current status of quantum gravity? How is string theory coming along?

 

[pervect]

If photons are electrically and magnetically neutral, what numbers do we feed in that equation?

The components of the stress energy tensor most relevant to light are momentum density, energy density, and pressure.


Wiki's treatment of the stress-energy tensor http://en.wikipedia.org/w/index.php?title=Stress–energy_tensor&oldid=517465899 is not bad.

The following diagram is from the wiki and has all the components of the tensor.

http://upload.wikimedia.org/wikipedia/commons/3/37/StressEnergyTensor.svg

 

[andrien]

just realized barry is not the opener of thread

 

[Barry_G]

You do realize that a photon IS electromagnetic field, right?

Field? I'd say eight fields, two electric and six of them magnetic, but no, that's not what mainstream theory would tell you. A photon is quanta of electromagnetic radiation, and despite the name, despite there are, I mean could very well be, magnetic and electric fields constituting a photon, it is still electrically and magnetically neutral, which means its electric and magnetic field is measured to be zero. Ok?

Electron, for example, is not neutrally charged, it has both electric and magnetic fields measured to be greater than zero, so are you trying to say electron gravity field is due to not only its intrinsic mass but also due to its electric and magnetic fields?

That isn't actually news to you, I hope. You see what happens when you ask for references without explanation? You don't understand it. Then I have to explain it. Then you don't understand that either. Where is an end to this? At what point do you accept that you need to step back and learn some fundamental theory?

I asked about experiments and actual measurements, never mind that. Now, let me explain what is happening here. You claimed photons generate gravity and then you referred to electromagnetic stress–energy tensor as the source for this photon gravity field. So what you need to address is how do you imagine photon electric and magnetic zero charge have anything to do with electromagnetic stress–energy tensor, not to mention any gravity even.

Don't explain, please, just take your claims and put them into practice. Use electromagnetic stress–energy tensor and calculate the strength of a single photon gravity field, if you can. Just show me, that's all.

Yes. It's the energy that remains after you take away all of the kinetic energy, if you'd like to think of it that way. Yes, it's certainly special. But it's importance lies in particle propagation. Not in how it generates gravity. Gravity is generated by all of the energy.

Mhm, so what's the number, what is the strength of a single photon gravity field?

Again, all of this is something you should know before you enter a discussion about gravity. SR is a prerequisite to GR, and we can't get past your confusion on that subject.

This is forum where people talk about stuff and ask questions, it's not a competition or some vanity contest, wake up! Besides, no one is forcing you to talk to me or back up your claims, suit yourself.

No. The difference is that if we say, "Photons have mass, but it's so absolutely tiny as not to be detectable by any experiment we have conducted, nor to show up as a side-effect in absolutely any theory," we can also say, "Who cares?" Our theory is more precise than experiment can refine with a massless photon. If photon mass has not manifested itself in all of this, the only sane assumption is to assume it is zero and keep going until it becomes a problem.

I care, because I am nice, friendly and caring person. Plus, photon mass has manifested itself as much as photon gravity field, they are concepts describing one and the same physical phenomena.

http://en.wikipedia.org/wiki/Stress-energy_tensor : The stress-energy tensor is the source of the gravitational field in the Einstein field equations of general relativity, just as mass density is the source of such a field in Newtonian gravity.

Photon producing no gravity would say that general relativity is absolutely wrong. It would fly in the face of all that we know about gravity. Possible? Technically. So are leprechauns. Technically.

I did not say photons do not have gravity field, they obviously do. What I said is how your statement "photons produce gravity" is just about as valid as if I say "photons have mass".

http://en.wikipedia.org/wiki/Photon#Experimental_checks_on_photon_mass : The photon is currently understood to be strictly massless, but this is an experimental question. If the photon is not a strictly massless particle, it would not move at the exact speed of light... Relativity would be unaffected by this

 

[andrien]

photons are result of quantizing the electromagnetic field using say creation and annihilation operator.They don't have charge,so they are not affected by EM field.if you are using stress energy tensor then it need to expressed for a single photon.It is however possible to interpret but it is certainly useful to go with the EM field rather than a single photon.

 

[Barry_G]

No, it's using terminology precisely enough to make it clear how it's not contradictory, because the term "mass" can mean different things. The apparent "contradiction" only arises if one is sloppy about terminology.

The beginning of wisdom is to call things by their right names.
- Chinese Proverb

But the difference is that "photons produce gravity" is unambiguous, while "photons have mass" is ambiguous; it depends on what you mean by "mass". Photons have zero invariant mass, but nonzero energy. So "photons have mass" can be true or false depending on how you interpret it, while "photons produce gravity" is unambiguous and true. It's you who are insisting on ambiguous terminology, not me.

By "mass" I of course mean intrinsic mass. The one that is given for elementary particles where, for example, mass of an electron is 9.109x10^-31 kg, and for photon is estimated to be less than 1x10^-18 eV/c^2, according to Wikipedia. Where 1x10^-18 eV/c^2 = 1.783x10^-54 kg, I think.

I don't agree I am being ambiguous, at least I am trying to be very specific, and what I am saying here is that energy is ambiguous concept, especially if you can not pinpoint what part of that energy belongs to frequency, velocity or intrinsic mass. Now, if you'll forgive me I'd prefer to ignore the rest of your post so we can concentrate on what we started to talk about. So let me ask you, if not measured, has anyone calculated what is supposed to be the strength of a single photon gravity field? What's the number?

 

[WannabeNewton]

Don't explain, please, just take your claims and put them into practice. Use electromagnetic stress–energy tensor and calculate the strength of a single photon gravity field, if you can. Just show me, that's all.

This is obviously easier said than done to be shown in full in a thread. Why don't you try it yourself? L_{EM} = -\sqrt{-g}g^{ac}g^{bd}\triangledown_{[a}A_{b]}\triangledown_{[c}A_{d]} so take the total lagrangian density that includes this matter field lagrangian density and the einstein lagrangian density, vary the respective action to obtain the field equations and solve it if you want (=D). I don't see why you have a problem with the idea that the maxwell field can contribute to curvature. It isn't just mass density that contributes to the curvature. Look up the lens thirring effect as a correction to Newtonian mechanics where the OTHER parts of the energy momentum tensor contribute to the non vanishing of the gravito - magnetic field.