[Pmb]

A photon can interact with a gravitational field because it has 'passive' mass.  Can you please explain the term 'passive' as it implies it has no effect?

Active gravitational mass refers to the mass that generates a gravitational field. Passive gravitational mass is the mass that gravity acts on.

A photon can not interact with the gravitational field.

Sure it does. What would make you believe otherwise? There is an example of a pulse of light generating a gravitational field. Why would you think that light can generate a grvitational field but photons wouldn't?

  It only appears to because it follows the geodesics of curved space time as caused by gravity.

That is incorrect. Consider a uniform gravitational field. The field will deflect matter, including light. However a uniform gravitational field has zero spacetime curvature, yet there can be a non-gravitational field present with a suitable change in spacetime coordinates from an inertial frame.

[lightarrow]

We are discussing about photons. Can a photon generate a gravitational field?

yes.

And which is the quantum theory of gravity which says this?
Hint: there still isn't any accepted quantum theory of gravity...

[lightarrow]

I think it is debatable whether a photon has momentum.

No, it's not debatable, it's ordinary physics.

[Pmb]

And which is the quantum theory of gravity which says this?
Hint: there still isn't any accepted quantum theory of gravity...

First off all qu7antum theories of gravity must reduce to GR in any acceptable theory. What I've done above is to use the approximation that a photon is a point particle with zero proper mass. Clifford Will uses this same approximation in his book Was Einstein RightI recommend that you take the following advice from Exploring Black Holes, by Taylor and Wheeler if you read the acknowledgement you'll see this

Phillip Morrison made several suggestions and convinced us not to invoke that weird quantum particle, the photon, in a treatment of classical theory of relativity (except in some exercises).

You're asking me about a quantum notion in a classial theory. I do know that people I've talked to about this, and friends/acquantances who are experts in their field, think of photons just like I do. It can be show, with classical GR, that a g-field which has been created by a source consisting of beam of EM radiation (a "pencil of light" as they call it) deflects particles. A pulse of light can be thought of as a collection of photons and the result is an interaction of the photon and g-field.  How could you accept that a beam of light is deflected but a single photon can't? This is all explained in such text books such as Tolman's book.

[JP]

Consider a uniform gravitational field. The field will deflect matter, including light. However a uniform gravitational field has zero spacetime curvature, yet there can be a non-gravitational field present with a suitable change in spacetime coordinates from an inertial frame.

Hmm... In Newtonian gravity, a uniform field won't deflect matter, since it has zero gradient.  From what I know of GR, the influence of gravity can be computed from gradients of the field as well.  A uniform field in GR should still have zero gradient and no influence, shouldn't it?  Or am I missing something?

By the way, I agree with you that a photon should be a source of gravity.  In addition to arguing that light/EM fields are energy and a source of gravity, you can also put light in a box, as I mentioned above.  The box has an effective mass and should be a source of gravity.  In all these cases light generates gravity because it has energy.  Since light's energy comes in packets called photons...

[Pmb]

Hmm... In Newtonian gravity, a uniform field won't deflect matter, since it has zero gradient.

A uniform field has a non-zero gradient in Newtonian gravity. If the gravitational potential has phi = gz then F = -m*grad phi = -mg.

[JP]

Ah, got it.  I was thinking gravitational potential. 

[MikeS]

A photon can interact with a gravitational field because it has 'passive' mass.  Can you please explain the term 'passive' as it implies it has no effect?

Active gravitational mass refers to the mass that generates a gravitational field. Passive gravitational mass is the mass that gravity acts on.

A photon can not interact with the gravitational field.

Sure it does. What would make you believe otherwise? There is an example of a pulse of light generating a gravitational field. Why would you think that light can generate a grvitational field but photons wouldn't?

  It only appears to because it follows the geodesics of curved space time as caused by gravity.

That is incorrect. Consider a uniform gravitational field. The field will deflect matter, including light. However a uniform gravitational field has zero spacetime curvature, yet there can be a non-gravitational field present with a suitable change in spacetime coordinates from an inertial frame.

This is a circular argument. 

If you are referring to to the experiment you mention in a later post where you say that a pencil of light gravitationally deflects particles.  The only reference I have found to such an experiment is one dated 1931. I don't wish to knock the experiment but I would not have thought they had the technology back then of carrying out such an experiment and obtaining unambiguous results.  I don't think that light can generate a gravitational field.  If you have references to any modern papers I would appreciate links to them.

I thought this was part of General Relativity and generally accepted.

There is no such thing as a uniform gravitational field from any reference frame other than the free falling reference frame.  If it were uniform there would be no gravity.  I agree it would have zero spacetime curvature.  The free falling reference frame only appears to have zero curvature (acceleration) exactly because it is free falling.  From any other reference frame light is following a geodesic in curved space-time.
Gravity causes space-time to be non linear (curved).  From the viewpoint of an observer free falling in curved space-time an object falling with them will appear to be stationary, likewise a beam of light will have a constant frequency.

[lightarrow]

First off all quantum theories of gravity must reduce to GR in any acceptable theory. What I've done above is to use the approximation that a photon is a point particle with zero proper mass.

So you also have found a model for the photon! Fantastic!
You didn't say that all this is your New Theory, however...

You're asking me about a quantum notion in a classial theory. I do know that people I've talked to about this, and friends/acquantances who are experts in their field, think of photons just like I do.

In some cases you can treat a photon in that way, but not in all.

It can be show, with classical GR, that a g-field which has been created by a source consisting of beam of EM radiation (a "pencil of light" as they call it) deflects particles.

Yes, and there isn't even need of talking of photons, since in that case you have a region of space with non zero energy density and so Einstein's equation tells us that there must be curvature, but unfortunately this has nothing to do with a single photon creating curvature.

A pulse of light can be thought of as a collection of photons and the result is an interaction of the photon and g-field.  How could you accept that a beam of light is deflected but a single photon can't?

Because the total effect is not the simple sum of the single effects. For example, even if a single photon has zero mass, a system of 2 photons can have non zero mass.

[yor_on]

"What is the mechanism that allows gravity to latch onto a photon?"

Think of a photon as being 'something', whatever that is. It is defined by it's momentum which is a effect from its 'speed' and 'energy containment'. If it is 'timeless' which anything existing at that speed, consisting of matter, would be to us then nothing inside this SpaceTime would exist for it. This is assuming Einsteins definition of its speed as being 'c'. But it is a photon, nothing really 'material' so maybe that definition has nothing to do with it.

But any which way, we have yet to observe 'aging' photons from our frames of reference. And if it, as we observe it, won't age then it can't expend 'energy'. That means that, assuming a propagation, the photon is to us 'frozen in time', its 'metabolism' the same from source to sink.

Gravity is weird, we can play with it as thinking of it as a 'density', or 'gradients'. Gradients is probably a better description from the photons 'point of view'. What gravity does is that it lends you 'energy' or take it away, depending on your motion relative that gravity's potential. So when a photon moves against the gravity potential it redshifts, to the observer losing energy, or if moving with the potential blue shifts, gaining energy. But those effects are observer dependent, defined by your relation relative the 'photon' we assume you to observe. You might be able to see it the same way as in classical physics? Two cars colliding, or a car colliding with a wall, will give you a different energy, here letting the 'gravity' at each point of that photons propagation represent some 'energy' relative you. Not a good example but I'll let it stay for now

The point is that this timeless photon can't lose 'energy', but if it were to work against gravity's potential it would have to, intrinsically. And the only way it can avoid losing that energy is to move in geodesics. Gravity could be seen as when you blend colors, you see this mix of swirling colors as you blend, that's a little like gravity's potential on a 2-d plane as it in a way is a mix defined by the matter/mass, 'energy', and 'motion' locally. But you could also see gravity as something static, existing throughout SpaceTime, as some logical 'number space' defining its 'strength', using 'times arrow' to constantly adjust the numbers describing each point. Or as 'holes' and 'dips', whatever gravity is it gives only one 'direction' from where you won't expend energy, and the photon always follow that 'geodesic straight line'.

It's a weird subject.

==

There is one more thing. If we were to assume that 'gravity' could be represented by a collection of numbers, defining each point in SpaceTime, times arrow constantly redefining them. Would they then be observer dependent? I think they would.

[Pmb]

There is no such thing as a uniform gravitational field from any reference frame other than the free falling reference frame.

Sure there is. Here's a Newtonian example whose size is finite in extent - http://home.comcast.net/~peter.m.brown/gr/grav_cavity.htm

  If it were uniform there would be no gravity.

That depends on how you look at it. Einstein viewed a gravitational field which resides as being defined by the metric tensor and Christoffel symbols.

Here are some references

[1] Principle of Equivalence, F. Rohrlich, Ann. Phys. 22, 169-191, (1963), page 173/
[2] Radiation from a Uniformly Accelerated Charge, David G. Boulware, Ann. Phys., 124, (1980), page174.
[3] Relativistic solutions to the falling body in a uniform gravitational field, Carl G. Adler, Robert W. Brehme, Am. J. Phys. 59 (3), March 1991.
[4] Gravitation, Charles. W. Misner, Kip S. Thorne, John Archibald Wheeler, (1973), sect 6.6.
[5] The uniformly accelerated reference frame, J. Dwayne Hamilton, Am. J. Phys., 46(1), Jan. 1978.

[Pmb]

So you also have found a model for the photon! Fantastic!
You didn't say that all this is your New Theory, however...

I don't know what you mean. I havent' found anything "new". Here's an interesting article - "The mass of a gas of massless photons," H. Kolbenstvedt, Am. J. Phys. 63 (1), January 1995

Yes, and there isn't even need of talking of photons, since in that case you have a region of space with non zero energy density and so Einstein's equation tells us that there must be curvature, but unfortunately this has nothing to do with a single photon creating curvature.

What makes you believe that?

[Pmb]

From the viewpoint of an observer free falling in curved space-time an object falling with them will appear to be stationary, likewise a beam of light will have a constant frequency.[/color]

Relativity states that that cannot happen. Also, measurements from different frames will yield measure different  values

[MikeS]

There is no such thing as a uniform gravitational field from any reference frame other than the free falling reference frame.

Sure there is. Here's a Newtonian example whose size is finite in extent - http://home.comcast.net/~peter.m.brown/gr/grav_cavity.htm

 
Ok, point taken but it could be argued this is still a free falling reference frame, it just appears to be static but it is still moving (falling) through space-time regardless of apparent motion.  Just the same as any gravitating body may appear to be stationary but is still moving through space-time.

[MikeS]

From the viewpoint of an observer free falling in curved space-time an object falling with them will appear to be stationary,[/color] likewise a beam of light will have a constant frequency.[/color]

Relativity states that that cannot happen. Also, measurements from different frames will yield measure different  values

"From the viewpoint of an observer" meaning from 'their' reference frame.  I agree that different reference frames will yield different results.
Why can't that happen?  Can you explain please.

[lightarrow]

I don't know what you mean. I havent' found anything "new". Here's an interesting article - "The mass of a gas of massless photons," H. Kolbenstvedt, Am. J. Phys. 63 (1), January 1995

I haven't read the article. What does it says, essentially?

Yes, and there isn't even need of talking of photons, since in that case you have a region of space with non zero energy density and so Einstein's equation tells us that there must be curvature, but unfortunately this has nothing to do with a single photon creating curvature.

What makes you believe that?

Curvature cannot be frame-dependent. A photon's energy instead is frame-dependent.

[Pmb]

[quotre]Curvature cannot be frame-dependent. A photon's energy instead s frame-dependent.
[/quote]
It sure can, at least that is the way that Einstein always used it. You're using a different definition of gravity than Einstein did. Einstein defined the gravitational force by the Christoffel symboles and the metric tensor. You and many others associate the presence of a gravitational field with a non=vanishing Reimann tensor.

It follows from Einstein's definition of "gravitational field." The curvature is a tensor so althought its components change with the frame of reference so do observer's observations. E.g. consider a photon. In one frame it has a certain energy but in another frame moving relative to the first. This is why Einstein said, in his 1916 review paper on GR, that if you change frames of reference you can go from a frame with no g-field to one that does have a g-field.

Forget that paper. I was thinking of something else, i.e. a paper by Rindler. I'll get that reference later.

Pete

[lightarrow]

This is why Einstein said, in his 1916 review paper on GR, that if you change frames of reference you can go from a frame with no g-field to one that does have a g-field.

Do you have a link (not in german, possibly  [])?

[yor_on]

By Tilman Sauer; "The first comprehensive overview of the final version of the general theory of relativity was published by Einstein in 1916 after several expositions of preliminary versions and latest revisions of the theory in November 1915. A historical account of this review paper is given, of its prehistory, including a discussion of Einstein's collaboration with Marcel Grossmann, and of its immediate reception."

And S. N. Bose: The Principle of Relativity.

Author: Einstein, Albert, 1879-1955; Minkowski, Hermann, 1864-1909 Subject: Relativity (Physics) Publisher: Calcutta University of Calcutta

[Pmb]

Do you have a link (not in german, possibly  [])?

I found on on the internet. See http://www.alberteinstein.info/gallery/pdf/CP6Doc30_English_pp146-200.pdf

Read page 156, especially where he says We thus find the ocurance of a gravitational field connected with the space-time variability of the g_s.

Recall the article I referenced above, http://xxx.lanl.gov/ftp/physics/papers/0204/0204044.pdf. It has all the relevant statements made by Einstein on this topic, at least all I could find.