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## Karl's model

Hello Sarge Originally Posted by sarge johannes--you say;
They don't (not global, only local and the Sun-Earth is not local:-). Photons in a gravitational field do NOT travel with C. So there is no paradox. Your paradox is based upon the assumption that photons travel at C. We have already had the discussion of the bending of light, when photons travel at C there can be no bending. It is simple, photon in a gravitational field do not travel at constant speed, therefore there is bending and no paradox as you proposed.

Best - Johannes

johannes ----- The speed of photons IS C. The velocity is not. So to get around the bending let the photon travel along a line connecting the center of the sun and the center of earth. Then the velocity will also be C.

Secondly, will you please tell me , if as you contend light travels at speeds different from C then what happens to the other variables in C(or v)=freq x wavelength?

best, sarge

I have been busy the last days, so I could not respond earlier.

As for your claim "The speed of photons IS C." - we are in disagreement here.

I give you my argument here: So IF the speed of the photon is constant, then the only paths that a photon can follow around a large mass is either a circular orbit (with the large mass at the center) or a straight line. Neither a circular path nor a linear path are associated with bending of light, i.e. a constant speed of light does not allow bending of light around a large mass...

Best - Johannes Reply With Quote

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## Johaqnnes --- our basic difference appears to be in the first of the three sections. I am assumming each diagram is a seperate proof. You will have to bear with me as I figure out how to physically reply using the computer. I may end up hand writing this and mailing it to Joe. No I do not agree with you at all that your conclusion that Vp . Ap =zero. I won't bother at all with the cylinder cordinates but will address you first and third proof. No response is needed until I'm through. ----- best sarge Reply With Quote

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## johannes --- If I follow you right in the first of your three sections Your statement that Vp.Ap =0 applies only at closest point of approach. Then Ap is a sine function everywere else.and at the closest point of appr5oach is at a maximum. At great distance the sine = 0. However we should ignore the first two diagrams because the angle in the third is the correct angle to work with along with the geodesic of the first diagram.

The vector addition you want is the velocity vector along the geodesic (=C. a scalar) and the the velocity along the radial (Vescape . sine theta, a varable as a function of R) using the third diagram angle.. Would you be will ing to redraw and discuss that diagram?

In other words, with that diagram, at distance dVe/dr = minimum, at theta = zero, dVe/dr = maximum.
In standard radial cordinates that is what happens.

Ve is given by my escape velocity using the standard form for the gravitational field of the earth when writing an energy equation but when writing a force equation there is a factor of 2 that must be added to compensate for the gravitational flux of the photon having that of a clyndrical source. It is really a two body problem.

Also the rest mass of the photon is = to hf (a scalar) , not zero and must be considered in any force equation..

The problem, in polar cordinates, is more complicated than any of your examples. I can only do it with velocity vectors and do not see how to write it in terms of forces but if you care to look at maxwell then it is analogous to the calculating the force at a point on the radial between a point charge and a charged conducting line .

---sarge
Last edited by sarge; 07-14-2012 at 06:25 AM. Reply With Quote

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## Also johannes, referencing post 3 above, our only real basic disagreements are;
1)I say that a photon has a rest mass.
2) that ALONG the geodesic it travels a set of closely spaced observers would all measure it's speed as C.
3) That a factor of 2 would occur due to the dissapation if the gravitational flux from the photon being distributed from a cylindrical surface area and be proportional to a weakening at distance according to 1 /2Pi R while the gravitational flux from a spherical volume dissipates at a rate of 1/4 Pi R.

Can't you see that if the photon is traveling at C then it has a speed exceeding the escape speed and will not go into a circular orbit. No math necessary to understand that.

Sorry that I'm sort of ignoring your arguments because you are not dealing with a wave model like this. We disagree at a more basic level than mechanics and that is all the bending shows.

best, sarge
Last edited by sarge; 07-16-2012 at 08:18 AM. Reply With Quote

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