Momentum of photon close the sun or black hole

  • #26
pervect
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MeJennifer said:
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.
 
  • #27
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pervect said:
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. :smile:
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.

wikipedia said:
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? :confused:
 
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  • #28
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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 infinte 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.
 
  • #29
pervect
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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' :-(.
 
  • #30
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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?
 
  • #31
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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.
 
  • #32
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.
 
  • #33
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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.
 
  • #34
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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?
 
  • #35
Jonathan Scott
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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 [itex]E[/itex] and velocity [itex]\mathbf{v}[/itex] has momentum [itex]\mathbf{p} = E \mathbf{v}/c^2[/itex] with magnitude [itex]E/c[/itex]. If it is deflected overall by something through a small angle [itex]\alpha[/itex], expressed in radians, then the change in momentum has magnitude [itex]p \alpha[/itex] 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 [itex]E[/itex] 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 [itex]c[/itex] is now the coordinate speed of light, which varies slightly with potential, approximately as [itex](1-2Gm/rc^2)[/itex] 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, travelling in any direction, including radially inward or outward and tangentially, or anywhere in between:

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

where [itex]\mathbf{g}[/itex] is the effective Newtonian gravitational field. Since [itex]E[/itex] 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:

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

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 [itex]E/c[/itex] in the downwards direction, which increases with time because [itex]c[/itex] decreases with potential.

The specific calculations change near a black hole, and there is a "photon" sphere radius (at Schwarzschild radial coordinate [itex]3Gm/c^2[/itex]) 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.
 
  • #36
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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.
 
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  • #37
Jonathan Scott
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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.
 
  • #38
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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...
 
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