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Topic: Einstein's factor of 2 in starlight deflection
Replies: 12   Last Post: Jul 14, 2014 3:17 PM

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Posts: 67
Registered: 3/17/12
Re: Einstein's factor of 2 in starlight deflection
Posted: Jul 14, 2014 3:17 PM
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What becomes obvious when you look at the circle above and view the one radius tangent you quickly will make out that the gap bc is about half a radius. If we reduce the tangent and adjacent arc to half a radius then the gap bc is around a quarter. From there on the gap bc tails off at the rate of the square of the difference between tangent x and the radius. If the circle were to be infinite
there would be no gap bc and the circumference would assume it was a straight line.

In the following part we, again, assume the Sun is in its Schwarzschild state.

On the subject of the gravitational nudges amounting to half a Planck radius we can assume that a free seeking lightbeam will undergo a similar experience.
If you take a circle to represent the gravitational field of the Sun and draw an equatorial line, ac, from nine o'clock to
three o'clock with 0 at the circle's centre we can then do the following:
at a point one quarter of a radius up the circumference from c we mark as point b. Draw a line from b back to a; line ab becomes the intended straight line
flight path of our lightbeam had it been free from gravitational influence. For later convenience extend the the line ab to outside of the circle to a point x.

So, in the first half of the lightbeam's trip gravity pulls it down to the grazing incidence of the Sun represented by point 0, amounting to a shift from the straight line scenario of one eighth a radius length of our circle.

If we then take another point measuring one radius of the circumference down the circle from point b to point d we find, eventually, that this point, d, is the exit point from the Sun's gravitational field of the lightbeam. If you draw a chord between point b & d you will find that is the direction of exit of our light beam's final escape from the gravitational field.

Now measure the angle between chord bd and bx, dbx, and you will find it measures two radians. Our lightbeam has shifted one radius length from its
original straight line, ab, but has been deflected by two radians.

The nature of the second half of the lightbeam's journey goes something like this:

Having got to the grazing incidence at point 0 on our circle, the lightbeam would have done the following: It starts off from point a, at very nearly zero gravitational deflection. By the time it has got to the grazing incidence at 0
it will be receiving 90 degrees gravitational deflection to its side. In fact for a moment it is in orbit. As it pulls away from the orbital it begins to experience greater and greater deflection to its side. The more it escapes the
greater is the deflection. In the first half of its journey, a0, the deflecting influence of gravity goes from zero to 90 degrees, averaging 45 degrees.
On the outward bound journey, 0d, it will experience deflection influence to its side from between 90 and 180 degrees, theoretically, averaging 135 degrees of deflection.

And yet all gravity has had to do is nudge the lightbeam over by a total of one Planck radius for all this to happen. As with the orbital the direction of travel is constantly corrected by tangential change, but in the orbital the progress is uniform and constant. With our lightbeam the curvature becomes greater and greater in the second phase as explained above.

The flightpath of the lightbeam is nudged over every Planck unit of time by half the Compton wavelength of the Sun's mass but when it proceeds to the next Planck time unit it does so in a corrected direction just as it did on the orbital. It appears to to take on the same mechanical principle as the situation in free fall. But that's all it does;
appearances are deceptive.

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