Momentum of photon close the sun or black hole

[exponent137]

We know, that photon change direction for 1.75'', when it flies close to the sun. This angle is proportional to radius of the sun. If we calculate clasically the angle is 0.82'' (arc second).

Which is momentum, which is given to sun (or black hole) by the photon change of direction. Is the same as change of momentum of the photon. Must we calculate change of momenum of the photon clasicaly or relativistically?

 

[pervect]

We know, that photon change direction for 1.75'', when it flies close to the sun. This angle is proportional to radius of the sun. If we calculate clasically the angle is 0.82'' (arc second).

Which is momentum, which is given to sun (or black hole) by the photon change of direction. Is the same as change of momentum of the photon. Must we calculate change of momenum of the photon clasicaly or relativistically?

The good news is that the system of the sun + photon does have a conserved total momentum, as long as the sun and photon are essentially alone in the universe (the technical name for the requirement is that they are in an asymptotically flat space-time).

The bad news is that it's not particularly easy to calculate this momeuntm. To be really technically correct, I think you'd have to use the ADM momentum.

While it is mostly concerned with mass, http://en.wikipedia.org/wiki/Mass_in_General_Relativity talks a little bit about momentum in General relativity. Note that the article talks about how time translation symmetries generate conserved energies. The parallel argument is that space translation symetries generate conserved momenta.

The following argument is a little suspect, but I think it might work. The Komar energy of the photon is a little easier to deal with than the ADM energy - essentially, it just gets multiplied by a "red-shift" factor, which is equal to the square root of g00, the metric coefficient for time. There isn't any Komar momentum defined that I'm aware of - Komar energy is defined for static systems, and the photon isn't static - but one might guess that the momentum of the photon similarly gets multiplied by the same red-shift factor that the energy was multiplied by, i.e. the square root of g00.

If you're not familiar with momentum in special relativity, you might want to read a little bit about it here, first, before reading the next remark:
http://en.wikipedia.org/wiki/4-momentum

Basically, we expect that the length of the ADM energy-momentum 4-vector will be zero, because the invariant mass of a photon is zero, and the ADM energy-momentum transforms a lot like the standard 4-vectors in special relativity (see above). We also expect that the ADM energy should be equal to the Komar energy. Given this, we can guess that the ADM momentum gets multiplied by the same "red-shift" factor as the Komar energy.

 

[MeJennifer]

We know, that photon change direction for 1.75'', when it flies close to the sun. This angle is proportional to radius of the sun. If we calculate clasically the angle is 0.82'' (arc second).

Which is momentum, which is given to sun (or black hole) by the photon change of direction. Is the same as change of momentum of the photon. Must we calculate change of momenum of the photon clasicaly or relativistically?

The photon does not change direction at all, it just appears so from our frame of reference. In reality the path of the photon is a geodesic and thus it simply follows the curvature of space.

 

[pervect]

The photon does not change direction at all, it just appears so from our frame of reference. In reality the path of the photon is a geodesic and thus it simply follows the curvature of space.

While a photon is too small to cause an effect that can actually be observed, note that suns do wobble when large planets orbit them. In fact, this has acatually been observed. This is true in both GR and in Newtonian gravity.

Photons also have momentum, just like planets do. However, the amount is usually small enough that it can be ignored in practice. In theory, though, there is no difference between the wobble caused by an orbiting planet and the wobble caused by orbiting "photons" - it's just that the amount of momentum in the photons is very small.

 

[exponent137]

If we look from distance in flat space, light-ray change direction for 1.75''.
1. I suppose that frequency of this photon is not changed in -+ infinity.
2. We looked sun from the distance (from flat space). I suppose that this distant momentum change equals momentum change of photon.
3. I suppose that momentum change of photon is the same as 1.75'' momentum change?
4. It is very unrealistically that this photon does not cause momentum change.
5. Can mass of sun be defined from large distance so that we ignore general relativity: If we observe sun from distance we see some velocity (because of photon impact). We know momentum change of photon (mcp). Then
mcp=m(sun) * v(sun)?

 

[Meir Achuz]

I agree with that.

 

[pervect]

If we look from distance in flat space, light-ray change direction for 1.75''.
1. I suppose that frequency of this photon is not changed in -+ infinity.
2. We looked sun from the distance (from flat space). I suppose that this distant momentum change equals momentum change of photon.
3. I suppose that momentum change of photon is the same as 1.75'' momentum change?
4. It is very unrealistically that this photon does not cause momentum change.
5. Can mass of sun be defined from large distance so that we ignore general relativity: If we observe sun from distance we see some velocity (because of photon impact). We know momentum change of photon (mcp). Then
mcp=m(sun) * v(sun)?

I think I mostly agree too, with the provision that the frequency of the photon does vary as the photon falls towards the sun. This is where the redshift factor I mentioned comes into play. If one only considers the photon "at infinity", one can neglect this. If one wants to compute the momentum of the system when the photon is at some position other than infinity, one needs to take into account the redshift factor. Essentially I was arguing earlier that one needs to use the energy of the photon "at infinity" to compute its momentum when the photon is not at infinty, i.e. close to the sun, rather than the locally measured energy of the photon . The locally measured energy of the photon will be higher than the energy-at-infinty, due to gravitational blueshift. It's probably simpler and safer to avoid considering this issue, though, and only treat the case where the photon is at infinity.

In order to sucessfully assign a momentum to the sun, one still needs asymptotically flat space-time, or a close enough approximation of the same. One will automatically have this if one idealizes the problem so that the photon and the sun are alone in an infinite vacuum space-time.

 

[MeJennifer]

I think I mostly agree too, with the provision that the frequency of the photon does vary as the photon falls towards the sun.

Yes from any other frame of reference than the photon's one.

However I consider it incorrect to assert that gravity influences the frequency of the photon itself. And neither does gravity influences it's momentum.
Space-time curvature and gravitational "pull" does not influence frequencies in the local frame, and neither does it influence the object's momentum. The observed effects from other frames of references are relativistic.

 

[exponent137]

I thougth that frequency of photon is equal from -infinity to +infinity. In mid time path of photon is curved by sun.

 

[pervect]

I thougth that frequency of photon is equal from -infinity to +infinity. In mid time path of photon is curved by sun.

It depends on how you measure it, exactly. If you use local clocks, the frequency of the photon changes as the photon gets closer to the sun. This is usually called "Gravitational redshift", and was demonstrated in the Harvard tower experiment. Note that in this case, the locally measured frequency of the photon would increase as it neared the sun, so it would be blueshift rather than redshift.

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/gratim.html

If you use a coordinate based approach, rather than the simpler local approach, in the Schwarzschld coordinate system the energy-at-infinty E₀ is constant for the photon during its orbit. Note, however, that E⁰ varies during the orbit (E⁰ = g⁰⁰ E₀) and that E = sqrt(|E₀ E⁰|) also varies. In particular, we know that g⁰⁰ = 1/(1-2GM/rc^2) for the Schwarzschild coordinate system.

[add] I've been talking about energies here, but E = h v, so the two are proportional. The fact that clocks run at different rates causes a large amount of the complexity here - one of the questions here is "whose clock do you use to measure the frequency of the photon". A more subtle, but equally important, question is how does one measure the frequency of a photon when that photon is not at the same location as the observer. This gets into issues of parallel transport, and of the relativity of simultaneity. The easiest solution is to avoid the difficulty by always measuring the frequency of the photon with the clock of a local observer. When this approach is adopted, the frequency of the photon does vary as it gets closer to the sun.

 

[WhyIsItSo]

While a photon is too small to cause an effect that can actually be observed, note that suns do wobble when large planets orbit them. In fact, this has acatually been observed. This is true in both GR and in Newtonian gravity.

Photons also have momentum, just like planets do. However, the amount is usually small enough that it can be ignored in practice. In theory, though, there is no difference between the wobble caused by an orbiting planet and the wobble caused by orbiting "photons" - it's just that the amount of momentum in the photons is very small.

I read the Wiki you pointed to earlier. All the formulas for momentum appear (to my limited understanding) to contain a unit for mass.

Since a photon is massless, I for one am none the wiser as to where a photon's momentum is coming from. In fact, I would have thought MeJenifer was correct is stating the photon doesn't change course at all, rather is merely following the "straight" path of local space-time.

What am I missing?

 

[selfAdjoint]

I read the Wiki you pointed to earlier. All the formulas for momentum appear (to my limited understanding) to contain a unit for mass.

Since a photon is massless, I for one am none the wiser as to where a photon's momentum is coming from. In fact, I would have thought MeJenifer was correct is stating the photon doesn't change course at all, rather is merely following the "straight" path of local space-time.

What am I missing?

A photon's momentum is given by Einstein's other 1905 equation: \mathbf{p} = h\nu\mathbf{n}. where \nu is the photon's frequency, h is Planck's constant, and \mathbf{n} is a unit vector in the direction of the photon's motion. As you see it doesn't require a mass.

 

[NateTG]

I read the Wiki you pointed to earlier. All the formulas for momentum appear (to my limited understanding) to contain a unit for mass.

Since a photon is massless, I for one am none the wiser as to where a photon's momentum is coming from. In fact, I would have thought MeJenifer was correct is stating the photon doesn't change course at all, rather is merely following the "straight" path of local space-time.

What am I missing?

In classical physics, momentum is indeed tied to rest mass. However in relativity (both special and general) momentum, mass, and such are a bit different.

 

[WhyIsItSo]

Hmm, does that mean momentum does not imply inertia?

selfAdjoint, does that equation mean light would be defracted by passing near the sun? Just like passing through some medium?

 

[jtbell]

In classical physics, momentum is indeed tied to rest mass.

In classical electromagnetism, electromagnetic fields carry momentum as well as energy.

 

[pervect]

The idea that it is necesary to have mass in order to have momentum is an (incorrect) carry-over from Newtonian mechanics.

Momeuntum in relativity is not equal to mass*velocity. It's easy enough to learn the new defintions if one does not "cling" to outdated notions inherited from Newtonian mechanics.

 

[WhyIsItSo]

The idea that it is necesary to have mass in order to have momentum is an (incorrect) carry-over from Newtonian mechanics.

Momeuntum in relativity is not equal to mass*velocity. It's easy enough to learn the new defintions if one does not "cling" to outdated notions inherited from Newtonian mechanics.

This is an issue about which I am suffering persistent confusion.

In terms of relativity, can you provide a verbal definition of momentum.

I'm stumbling over my mental "tie" between momentum and mass. I also thought inertia and momentum were intimately related (if not actually the same thing). Very, very confused here.

 

[pervect]

Probably the most fundamental defintion of momentum is that due to Noether's theorem.

http://en.wikipedia.org/wiki/Noether's_theorem

The important thing about momentum is that it is a conserved quantity. Noether's theorem guarantees that any physical system (one based on the sort of physics we use, the sort of physics that can be derived from an action principle) that has space translation symmetry will have a conserved quantity.

Note that Newtonian physics, special relativity, AND general relativity can all be expressed in terms of an action principle, if desired, so therefore Noether's theorem applies to all of them. (It is more difficult to find the necessary symmetries in GR, however, even though it does satisfy the "action principle" requirement).

We can also see via Noether's theorem why energy and momentum are so closely related, something that we can't see easily in any other manner.

Momentum is just a consequence of space translation invariance, and energy is just a consequence of time translation invariance. Thus momentum and energy "mix" together in relativity the same way that time and space do.

I'm not sure if you are familiar with how time and space "mix", the classic illustration of this can be found in Taylor & Wheeler's space-time physics. Basically, space and time "mix" together to from an invariant, the Lorentz interval, in very much the way that north-south and east-west distances "mix" together to from an invariant which we simply call "distance". The major (and important) difference is one of sign. The Euclidean distance formula is

ds^2 = dx^2 + dy^2

The Lorentz formula is

ds^2 = dx^2 - dt^2

Anyway, Noether's theorem is the abstract defintion that most clearly defines momentum. It's possible to approach the material in a somewhat less abstract manner, which is more understandable, perhaps, but less fundamental. The approach one takes here is that "momentum must be a conserved quantity".

One can then show that in Newtonian mechanics mv is such a conserved quantity, and call it momentum.

In SR, one can then show that mv does not give rise to a conserved quantity, but that mv / sqrt(1-(v/c)^2) does give rise to a conserved quantity. Without Noether's theorem this seems almost "accidental", with Noether's theorem one can give a "reason" why momentum is conserved.

Now the fact that mv / sqrt(1-(v/c)^2) doesn't , unfortunately, directly address the momentum of light. One can see that m is zero, and so is sqrt(1-(v/c)^2), so that we have 0/0. The correct formula for the momentum of light turns out to be p = E/c.

Of course you've probably already seen these equations. The basic difficulty is that you just seem to refuse to accept them. Basically you seem to have learned Newtonian mechanics "too well", and can't do the unlearning of Newtonian mechanics needed to learn relativity. It's really not hard - you just have to decide to not believe that p = mv anymore, and replace it with the relativistic versions: p = E/c for light, and p=mv/sqrt(1-(v/c)^2) for particles with a rest mass. Note that both of these equations satisfy the more fundanmental relationship

E^2 - (pc)^ = (m c^2)^2

This fundamental relationship says that (mc^2) is a Lorentz invariant of the energy-momentum 4-vector. It can bea viewed as a direct consequence of Noether's theorem and the invariance of the Lorentz interval, ds^2 = dx^2 + dy^2 + dz^2 - dt^2

 

[exponent137]

Gravitational force of photon on the sun

I continue with this problem. (I believe that photon act on the sun with force)

A photon, which flies far distant from the sun, act on the sun with force F=k/r^2. k is proportionality factor and r is distance from the sun. Change of photon's (sun's) momentum is proportional to (sin(theta))^2, and theta is proportional to 1/r. But, if we has current of photons extended to infinity, uniformly dense and it flies in one direction, it act on the sun with infinite force. (Calculation show that this force is logarithmically divergent with r. F=integral of (k dr/r))

But, this is doubtful. Probabably something special which make calculation converent should be respected in calculation? I used some corrections to this approximation, but all are divergent.

 

[pervect]

The problems with this Newtonian analysis should become apparent when one remembers that light deflects differently (twice as much) in GR than it does with the Newtonian analysis.

You might get a correct order of magitude estimate, but don't be surprised if the detailed answers are off by a factor of as much as 2:1 with this sort of approach.

Basically, if you want to do GR, you have to learn it first.

 

[cesiumfrog]

Even classically, electromagnetic waves carry momentum. On the back of an envelope you can calculate semi-classically, using conservation of energy and momentum, how much a photon's energy (and hence wavelength) changes when it is reflected by a mirror. The same applies for photons that are deflected around gravitational masses or, at the other end of the scale, scattered by individual gas atoms. You can also estimate how much momentum is transferred to the mirror.. obviously in the case of the sun, the result is experimentally completely negligible (consider the mass-energy ratio of a photon compared to a star).

Basically, if you want to do GR, you have to learn it first.

So true, more should heed that.

 

[exponent137]

The problems with this Newtonian analysis should become apparent when one remembers that light deflects differently (twice as much) in GR than it does with the Newtonian analysis.

You might get a correct order of magitude estimate, but don't be surprised if the detailed answers are off by a factor of as much as 2:1 with this sort of approach.

Basically, if you want to do GR, you have to learn it first.

Angle of deflection theta is proportional to 1/r in Newtonian mechanics (NM) or in general relativity (GR). This is valid at big r's. I this light ray changes direction by angle theta, it gives momentum by a factor –2(sin(theta/2))^2 of input momentum of photon. At small theta's this factor is theta^2.
I suppose that this calculation is the same in NM and in GR. Uniform current of photons is proportional to r dr. So integral r/r^2 dr is ln (r) and this is divergent at big r's. I look this current from infinity, where space is asimptotically flat.
Or, verification of the first step: Is change of photon's momentum, in primary direction of photon, really proportional precisely to -2(sin(theta/2))^2? Is this true in NM and GR?
Where I am wrong?

 

[pervect]

Angle of deflection theta is proportional to 1/r in Newtonian mechanics (NM) or in general relativity (GR). This is valid at big r's. I this light ray changes direction by angle theta, it gives momentum by a factor –2(sin(theta/2))^2 of input momentum of photon. At small theta's this factor is theta^2.
I suppose that this calculation is the same in NM and in GR. Uniform current of photons is proportional to r dr. So integral r/r^2 dr is ln (r) and this is divergent at big r's. I look this current from infinity, where space is asimptotically flat.
Or, verification of the first step: Is change of photon's momentum, in primary direction of photon, really proportional precisely to -2(sin(theta/2))^2? Is this true in NM and GR?
Where I am wrong?

The angle of deflection is inversely proportional to the impact parameter (I asssume that's what you mean by radius), but the proportionality constant is different in GR than it is in Newtonian mechanics.

The detailed explanation is somewhat technical, not knowing how much you know I'm not going to get into it unless you twist my arm some, except to mention that it has a lot to do with the non-flatness of space (and not just space-time!) near a gravitating body.

The existence of a difference is quite well known, that is why it is possible to test general relativity by looking at light deflection of starlight by the sun. GR predicts twice the deflection angle.

The importance of this depends on the accuracy of your calculations. If you are looking for no better than a 2:1 accuracy, it's probably OK to use the Newtonian approximation, but if you are looking for better than this, you need a much more sophisticated approach than what you've been doing.

Your analsyis that the effect diverges for an infinite beam of light is probably OK, as a mere 2:1 error won't affect the answer. This observation suggests that the problem cannot be formulated without including the gravitational contributions of the light to the metric - that the simple approximation of ignoring the light won't work for any finite intensity of the light.

 

[pmb_phy]

The photon does not change direction at all, it just appears so from our frame of reference. In reality the path of the photon is a geodesic and thus it simply follows the curvature of space.

He's referring to the change in the spatial trajectory which can obviously occur when the worldline is a geodesic. So the photon does change direction, it changes spatial direction.

Pete

 

[MeJennifer]

He's referring to the change in the spatial trajectory which can obviously occur when the worldline is a geodesic. So the photon does change direction, it changes spatial direction.

Pete

That would imply the assertion that there is absolute space which in contrary to the theory of GR.
In my view something that travels on a geodesic does not change direction at all.

Obviously the directional change can be observed but this is simply a relativistic effect. Claiming that a photon changes direction in space due to a gravitational field is simply Newtonian thinking, similar to a claim that a free falling object under the influence of the gravitational field of the Earth accelerates and changes direction in space. The term space is clearly misused here. Remember that space is not flat near a gravitational field, so don't treat it as flat and claim a change in direction!

In GR a photon never changes direction, it simply cannot. It can only be emitted and absorbed and it always travels on a geodesic.

 

[pervect]

That would imply the assertion that there is absolute space which in contrary to the theory of GR.
In my view something that travels on a geodesic does not change direction at all.

Obviously the directional change can be observed but this is simply a relativistic effect. Claiming that a photon changes direction in space due to a gravitational field is simply Newtonian thinking, similar to a claim that a free falling object under the influence of the gravitational field of the Earth accelerates and changes direction in space. The term space is clearly misused here. Remember that space is not flat near a gravitational field, so don't treat it as flat and claim a change in direction!

In GR a photon never changes direction, it simply cannot. It can only be emitted and absorbed and it always travels on a geodesic.

Geodesics are not quite "straight lines". They are simply the closest that one can get to a straight line in a curved geometry.

As far as light deflection goes, this is one of the classical tests of GR. See for instance http://en.wikipedia.org/wiki/Tests_of_general_relativity

Deflection of light by the Sun

The first observation of light deflection was performed by noting the change in position of stars as they passed near the Sun on the celestial sphere. The observations were performed by Sir Arthur Eddington and his collaborators during a total solar eclipse,[2] so that the stars near the sun could be observed. Observations were made simultaneously in the city of Sobral, Ceará, Brazil and in the west coast of Africa [citation needed]. The result was considered spectacular news and made the front page of most major newspapers. It made Einstein and his theory of general relativity world famous.
number.

For a non-wiki reference, try http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/grel.html

Or do a google search for "gravitational light deflection" or "gravitational lensing".

Since the deflection of light is well documented and experimentally confirmed, I think you need to adjust your understanding to match the evidence, rather than the evidence to match your understanding.

 

[MeJennifer]

Since the deflection of light is well documented and experimentally confirmed, I think you need to adjust your understanding to match the evidence, rather than the evidence to match your understanding.

What you are doing is to pretend that curved space is flat and then conclude that light is deflected.
Instead light simply follows a geodesic in curved space, it does not change direction at all. And if we understand that space is curved we would realize immediatelty that it is not the light that is deflected!

Really it is you who seem to measure everything from the perspective of flat space.

The first observation of light deflection was performed by noting the change in position of stars as they passed near the Sun on the celestial sphere.

It is because space is curved that we measure the apparent change in position, not because light is deflected.

Light is "deflected", free falling objects "accelerate", the frequency of light "changes". Such statements simply describe Newtonian thinking, they add 0 to an understanding of GR, on the contrary they simply add to the confusion.
How can you expect people to understand GR if you stick to flat space viewpoints and ignore relativistic effects?

Lightdoes not deflect, free falling objects do not accelerate, and frequencies do not change under the influence of gravity. These effects can be explained by space curvature and relativistic effects alone. So why not do that?

 

[exponent137]

The angle of deflection is inversely proportional to the impact parameter (I asssume that's what you mean by radius), but the proportionality constant is different in GR than it is in Newtonian mechanics.

Your analsyis that the effect diverges for an infinite beam of light is probably OK, as a mere 2:1 error won't affect the answer. This observation suggests that the problem cannot be formulated without including the gravitational contributions of the light to the metric - that the simple approximation of ignoring the light won't work for any finite intensity of the light.

I assume that intensity of light (j) is finite, but it occupies all the volume in space. I suppose that you thought so.
But, this is as cylindricaly hole inside of some photon mass. (I know that it the case of spherical hole -for instance hole inside of center of Earth-we do not feel gravity of mass of Earth, which is around of us.) I suposse that this is true also cylindircal hole. So in this case light around do not affect to light in hole.
However, this infinite force, which I calculated is anti-intuitive. I think that there is something which make this calcualtion finite instead of logaritmically divergent.
I please for hints.

 

[pervect]

I don't think you are going to get very far in General Relativity until you are able to deal with gravity as a curvature of space-time rather than as a force.

When you can do this, articles like

http://www.springerlink.com/content/x346w7x336224520/

"On the gravitational field of a massless particle", Journal of General Relativity and Gravitation

may be helpful to you. Note that when talk about the "Gravitational field" in this article, they mean the Riemann tensor, which describes the curvature of space-time and can be regarded as the "tidal force" experienced by a particular class of observers.

If you try to work the problem using gravity as a "force", you won't be doing General relativity. You'll be doing some sort of Newtonian approximation at best, or possibly some non-standard 'personal theory' :-(.

 

[exponent137]

I will ask new question about photon, which give momentum to sun. More precisely, I will ask how it gives momentum to a black hole.

When photon's direction is a little curved it gives its lost momentum to the black hole. but its momentum must travel to black hole as gravitational waving (GW). GW's speed is speed of light. but distances to horizon are more and more extended (because of dx/(1-2GM/r c^2)). Then the horizon must be traversed.
So, how fast really GW reaches the blach hole that it gives it momentum?

 

[exponent137]

In SR, one can then show that mv does not give rise to a conserved quantity, but that mv / sqrt(1-(v/c)^2) does give rise to a conserved quantity. Without Noether's theorem this seems almost "accidental", with Noether's theorem one can give a "reason" why momentum is conserved.

Now the fact that mv / sqrt(1-(v/c)^2) doesn't , unfortunately, directly address the momentum of light. One can see that m is zero, and so is sqrt(1-(v/c)^2), so that we have 0/0. The correct formula for the momentum of light turns out to be p = E/c.

Wait a minute. So E is not a vector, so if E is not changed, p is not changed. So photon which flies close to the sun (or Black hole) does not excange momentum, because E in -oo and in +oo is almost the same. So if E is the same, photon would not excange momentum, despite that its angle is changed?? Is this formula E/c precise?

I don't think you are going to get very far in General Relativity until you are able to deal with gravity as a curvature of space-time rather than as a force.

I ask about exchanged momentum. If its time derivative is force for me, what is the problem? Infinite force is infinitite derivative of exchanged momentum.

 

[gravity guru]

this question is for pervect, wouldn"t the frequency of the photon be determined by the time it takes light to travel from the photons location in the sun to the sun"s center, the closer to the center the higher the frequency? gravity guru.

 

[Yor_on]

Yes from any other frame of reference than the photon's one.

However I consider it incorrect to assert that gravity influences the frequency of the photon itself. And neither does gravity influences it's momentum.
Space-time curvature and gravitational "pull" does not influence frequencies in the local frame, and neither does it influence the object's momentum. The observed effects from other frames of references are relativistic.

To me it seems the opposite Jennifer :)
It's like everything is relative, I'm guessing you are thinking of the 'gold_standard' of photons when you state your opinion. But whenever we observe something we are interfering and drawing a result from that interference. A gravitational well will change the properties of a photon when observed by us resting outside that well.

But it is a very irritating point that photon, as it doesn't even have a physical existence in spacetime although we observe them at all times :)
That is, it doesn't occupy any place at all.
So how many photons can we superimpose in the same location then? An infinite amount?
And to me it all comes back to if there is something 'material' called a photon or if it is just us, not having the slightest idea about what it is.

 

[exponent137]

I ask the same question once again, because it was not clear answer.

I think if a photon flies over the horizon of a black hole, it gives it its momentum. If everything is only a curved space, than black hole can be moved only with a help of matter which moves slower than photons. It seem unlogically.

The second argument is photon which flies from starr close to sun and then to us. We can imagine a sun like a black box and we see, that momentum of a photon is changed. So momentum of a sun is changed.

But probably there are equations which can answer on my question?

 

[Jonathan Scott]

In simple cases like this, momentum of the overall system is conserved. In order to describe gravitational interactions accurately, we have to allow for the curvature of space, so we have to choose a coordinate system to map what is happening. The most practical coordinate system for the space around a single dominant central mass is an isotropic one, where the scale factor between each direction in local space and the corresponding coordinate in the coordinate space is the same, and the coordinate speed of light differs from the local speed of light only by a scalar factor (rather than a tensor).

According to Special Relativity, a photon of energy E and velocity \mathbf{v} has momentum \mathbf{p} = E \mathbf{v}/c^2 with magnitude E/c. If it is deflected overall by something through a small angle \alpha, expressed in radians, then the change in momentum has magnitude p \alpha perpendicular to the original path, so the change in momentum of the central mass is equal and in the opposite direction.

The same logic applies for gravity around a dominant central mass, described in isotropic coordinates. In that case, the total energy E remains a constant (as in Newtonian gravity, where the change in potential energy matches the change in kinetic energy), and the only slight difference is that c is now the coordinate speed of light, which varies slightly with potential, approximately as (1-2Gm/rc^2) times the standard value.

More generally, for weak fields described using isotropic coordinates, the following relativistic equation holds for all test objects, from bricks to photons, traveling in any direction, including radially inward or outward and tangentially, or anywhere in between:

\frac{d\mathbf{p}}{dt} = \frac{E}{c^2} \, \mathbf{g} \left ( 1 + \frac{v^2}{c^2} \right )

where \mathbf{g} is the effective Newtonian gravitational field. Since E is constant, we can also divide by it to get an equation of motion which does not depend on the energy of the test object:

\frac{d}{dt} \left ( \frac{\mathbf{v}}{c^2} \right ) = \frac{1}{c^2} \, \mathbf{g} \left ( 1 + \frac{v^2}{c^2} \right )

Note that even for a photon this still says that the downwards momentum is increasing at a rate which only depends on the field, regardless of its direction of travel. For example, if the photon travels downwards, then its momentum is E/c in the downwards direction, which increases with time because c decreases with potential.

The specific calculations change near a black hole, and there is a "photon" sphere radius (at Schwarzschild radial coordinate 3Gm/c^2) outside the black hole at which the curvature of the photon's path puts it into a circular (unstable) orbit. However, the general principle remains the same that (from the point of view of a suitable coordinate system) momentum is still conserved for the total system.

 

[exponent137]

Thanks.
Now the next question:
How tiny fractions of momentum the photon is giving to the sun? Here the principle of uncertainty comes in play.
I calculated a formula

If delta(L) is recalculated in delta(phi) and we respect tiny changes of momentum

delta(phi) ^3 >= Rs * lambda/(2pi Rb)

Rb is the radius of sun as black hole = 3E3 m (only for a short record.)
Rs is the radius of the sun = 7E8 m.
lamba is a wave lenght, for instance 3E-7 m.
delta(phi) is change of angle around the sun.
delta(L) is change of photons path, where principle of uncertainty is still valid.
This is not small angle (1.5e-2 rad)
But if we include radio waves, this angle become close to 2 pi.

So, how it is with the principle of uncertainty in this example.

 

[Jonathan Scott]

Thanks.
Now the next question:
How tiny fractions of momentum the photon is giving to the sun? Here the principle of uncertainty comes in play.
I calculated a formula

If delta(L) is recalculated in delta(phi) and we respect tiny changes of momentum

delta(phi) ^3 >= Rs * lambda/(2pi Rb)

Rb is the radius of sun as black hole = 3E3 m (only for a short record.)
Rs is the radius of the sun = 7E8 m.
lamba is a wave lenght, for instance 3E-7 m.
delta(phi) is change of angle around the sun.
delta(L) is change of photons path, where principle of uncertainty is still valid.
This is not small angle (1.5e-2 rad)
But if we include radio waves, this angle become close to 2 pi.

So, how it is with the principle of uncertainty in this example.

Sorry, I have no idea what you're talking about. Mixing QM and gravity rarely makes sense anyway, but I don't even know what you are trying to say.

 

[exponent137]

Sorry, I have no idea what you're talking about.

According to general relativity, a photon gives its momentum to sun continuously. So, on a very short path it gives very small momentum to sun. But this is in contradiction with the principle of uncertainty. So I calculated the smallest rates of momentum, which are given by the photon and these rates are much larger than in the area of Planck's distances.
If this is not enough, I can explain derivation more precisely...