1. From what i understood.

Electromagnetic radiation/photons, for a gravitational field, have basically just a gravitational wave traveling with them. The true origin of that gravitational wave is the accelerated mass at the moment of creation of the EM wave. In practice, its not possible to make EM waves with out that gravitational wave. Same thing with photons. Abstractly you could have naked EM waves, but that would be a pathological solution. Its right all this?

A gas of photons has the gravitational field of the sum of all its gravitational waves. This shouldn't behave like the exterior of the Schwarzschild solution. We shouldn't get newton at the limit from this. Right?

When a photon falls in a black hole.... Depends how we model matter in the singularity. If we consider that all the stuff are still there somewhere in the singularity. Then photons add momentum to a black hole, not strictly speaking mass. I mean, that if you change reference frame, so that the black hole is immobile again, you see again the original black hole. If you throw several photons, the momentum gets out again as gravitational waves....

I tried to read wikipedia on this, but it was too complicated. What i'm doing wrong with the above?

2.

3. Its right all this?
No. Where did you read that?

4. What i found was too complicated...

explain more...

5. Originally Posted by Quantum immortal
explain more...
About what? Nothing in your post appears to be correct so it is hard to know what you are trying to ask.

6. what are properties of the gravitational field of EM radiation? It can't be the Schwarzschild solution in a different reference frame. If you change reference frame, the radiation can have arbitrarily low energy. Its gravitational field must be something like a gravitational wave or frame dragging. This way it can tend to vanish, if you try to catch up the EM wave.

How the gravitational field of a photon compares with a moving mass? What are the differences? Is there a unified expression of the metric for both massive and massless particles?

A gas of photons, behave externally like the Schwarzschild solution? How this happen? There gravitational fields add up to the Schwarzschild solution?

With black holes. A single photon drops in a black hole. Its momentum is transfered to the black hole. We change reference frame so that the black hole is immobile again. What is the mass of the black hole now?

7. This thread is teetering on the edge of the trashcan. Quantum, only you can save it. Please try to make a lick of sense.

8. Originally Posted by Quantum immortal
what are properties of the gravitational field of EM radiation? It can't be the Schwarzschild solution in a different reference frame. If you change reference frame, the radiation can have arbitrarily low energy. Its gravitational field must be something like a gravitational wave or frame dragging.
You are making up stuff under the guise of asking questions.

This way it can tend to vanish, if you try to catch up the EM wave.
Nonsense.

How the gravitational field of a photon compares with a moving mass? What are the differences? Is there a unified expression of the metric for both massive and massless particles?
No, there isn't and your questions do not make sense (again).

A gas of photons, behave externally like the Schwarzschild solution? How this happen?
What gives you this bright idea? Please don't answer, it was just a rhetorical question.

There gravitational fields add up to the Schwarzschild solution?
Not even close. I also notice that your English is devolving and regressing as well.

With black holes. A single photon drops in a black hole. Its momentum is transfered to the black hole. We change reference frame so that the black hole is immobile again. What is the mass of the black hole now?
You are outright trolling. Again.

9. BP

I want a human readable explanation about the photons gravitational field. Is this clear enough?

Usually people will just say that electromagnetic radiation has a gravitational field. From that, i assumed that it was like a moving mass. I think a lot of people understould it like that. But Its not too difficult to figure out that this is not possible. If you change reference frame, you can make the EM radiation arbitrarily week, until its undetectable.

I think i understould, that the gravitational field of EM radiation actually resembles a gravitational wave. You can change reference frame, so that the gravitational wave becomes arbitrarily week. This is consistent with above, that the EM radiation it self seams to become weaker if you change reference frame. Both should seam to vanish proportionally if you change in a certain reference frame.

Then i assume that what i say above is true. So what happens then with a ball of photons? Again i think the usual interpretation would have said that outside of the ball, its like the field of a point mass, with all the energy of the photons. But this doesn't follow trivially from above. Either its just wrong, or some involved explanation for how the outside field looks like a field from a point mass.

I'm asking this question in relation to cosmology...

Lets take the example of a volume with particles and antiparticles. Just before annihilating, we have a normal looking gravitational fied outside. When they annihilate, we have photons and gravitons inside the volume and a normal looking field outside. Is this stable or unstable? Is there some explanation how the fields change as they go outside?

I mean, the gravitational fields of a point mass and a gravitational wave are very different.

About the black hole. Its less important here. Basically, can a single photon add rest mass to a black hole? Or it just gives it momentum? If several photons fall in a black hole, with total momentum of zero, is there an excess of energy that flies off?

10. Originally Posted by Quantum immortal
So what happens then with a ball of photons? Again i think the usual interpretation would have said that outside of the ball, its like the field of a point mass, with all the energy of the photons. But this doesn't follow trivially from above. Either its just wrong, or some involved explanation for how the outside field looks like a field from a point mass.
A confined ball of photon gas will have an exterior Schwarzschild geometry.

11. ....ok

far away, gravitationaly a single photon looks like Schwarzshild?

12. Originally Posted by Quantum immortal
....ok

far away, gravitationaly a single photon looks like Schwarzshild?
Perhaps I should have clarified. I was referring to a spherical container containing a photon gas that is stationary in a timelike frame of reference.

One way you possibly could establish the metric of a lightlike object is to consider a Lorentz transformation of a Schwarzschild blackhole to a very high velocity while simultaneously decreasing the mass such that the energy remains constant. Then in the limit of one should have a metric describing a lightlike object. I believe this will result in a Weyl tensor with lightlike components.

13. your missing the electrical charge, when you put a photon through a black hole it instantly looses its electrical charge. When we are talking about moving photons from electromagnetic field, to a field that has no electrons, then the photon will act quiet differently.
We know that there is no electrical forces working in black hole, because they can not substance light. The velocity of the photon is an effect caused by the bending of space, one of many properties of a black hole that can cause irregularities in the phonon's regular behaviour.

14. what are properties of the gravitational field of EM radiation? It can't be the Schwarzschild solution in a different reference frame.
To find the gravitational field of EM radiation, you first write down its energy-momentum tensor

and then solve the Einstein equations for the metric. I don't have an explicit solution at hand right now, but I don't think it will involve gravitational waves by default unless there is a non-vanishing reduced quadrupole moment present. In other words - it depends on the exact scenario in question. I don't know what happens when you consider only a single, isolated photon, which is a completely different scenario again.