Momentum of photon close the sun or black hole

[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...