Quote:
Originally Posted by rtomes
When considering photons in a gravitational field (e.g. here on Earth in the laboratory) we can divide the directions into the 6 axes and determine the results for each case. The 4 horizontal axes are all the equivalent and Einstein has already shown that the change in direction is 2x the Newtonian one. So for these 4 photons we have that the acceleration vector and the momentum are changed by 2 times Newtonian gravity (g):
b = 2g
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rtomes, you are kidding here, right? Einstein never said anything of the kind. His theorz predicted that in the case of starlight grazing the sun, the light would be bent more than would be expected classically. This does not mean that
in general your b will be 2 g.
And what 4 axis are you defining in the horizontal plane? IIRC there are only 2 orthogonal directions in one plane. Do you separate here between +x and -x?
Quote:
Originally Posted by rtomes
For the vertical photons, there is also a change in momentum predicted by Einstein. This is called "gravitational redshift" and Einstein's formula may be manipulated by using E=mc^2 and E=hf to find that b = g in both the vertically up and down cases.
df/dt = fg/c (as given by Einstein)
so b = 1/m.dp/dt = 1/m/c.dE/dt = h/m/c.df/dt = h/m/c.fg/c = Eg/E = g
That means there is no 2x factor for vertical photons, just the normal Newtonian rate of change of momentum per unit mass.
So taken over the whole sphere of directions for random photons we have 2x in 4 directions and 1x in two directions. That makes an average of 5/3x for random photon directions.
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That would maybe apply if the photon is moving past the sun.
Furthermore, you cannot just take an average like that, if you have these kinds of spatial variations, you will have to integrate over the sphere you are working with.
edited as an afterthought
There is something wrong with your definition of b, which troubled me, but I did not notice immediately.
You would like to do relativistic physics here, or at least your variation of it. You define your
pull in a totally classical way. You write:
b = (1/m) dp/dt
So classically this would be the force per unit mass, with units m s[sup]-2[sup] kg
-1, where p is the momentum of the particle mv. Naturally, you need to expand this to the relativistic framework, where still F = dp/dt holds, however, the relativistic momentum of a particle of rest mass m
0 is given by p = gamma m
0 v, with gamma the Lorentz factor (1 -v
2/c
2)
-1/2. Now, we see that the force F on a particle of rest mass m
0 depends on velocity v, i.e. you need more force to accelerate the particle as it gains more mass (the problem that you cannot reach light speed, unless you don't have mass).
Thus now the question is with b, what do we divide by, by the rest mass m
0, which would be okay because then you find that
b = d/dt ( gamma v )
or do you want to divide by the relativistic mass m (which is actually in your equation), but that leads to the problem that you have gamma before and behind the d/dt, and how this should be interpreted, at least you will end up with a b(t).
I think that b does not help us very much, there is no need to divide by the mass of the particles or of the conglomerate, as the equations of motion are totally fine with particles of various mass. This is just complicating stuff for no reason.
Quote:
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Originally Posted by rtomes
When we have a conglomerate of matter and radiation (such as the Sun's core or the Sun's outer layers) this variable b is the correct one to use as a measure of the acceleration of any part of that conglomerate. When we add together a bunch of matter and radiation, if we weight each part by its mass, then we will be summing correctly the momentum of that conglomerate. Then dividing again by the total mass we will get acceleration as a result.
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So you have mass and radiation (which you convert through E=mc
2 to mass). The problem is, however, that in words it sounds nice, but .... you do not start with dp/dt, you start with the other side of the equation, with F, because you want to calculate the acceleration part of dp/dt, and F will still be given by e.g. Newton's law of gravitation. So, you are complicating stuff here unnecessarily. And you can "sum the momentum of the conglomerate" but that does not give you acceleration if you divide it by mass, only if you divide (classically or relativistically) the time derivative of the momentum by the (classical or relativistic) mass, then you get an acceleration.