Our discussion in the previous blog post raised the following question:
Paradox 1. Static observers in a homogeneous uniform gravitational field will see some of an incoming wave packet to reflect back, but freely falling observers see it to propagate freely in a flat Minkowski space.
Solution. There is no such thing as a uniform gravitational field in the sense that we would need for the paradox!
If we try to define such a field, let us put there two static observers connected with a tight rope. Both observers will feel a constant acceleration g.
But it is a well-known fact that in a Minkowski space, a tight rope set between rockets that both have an acceleration g will break. QED.
Optical gravity claims that to a global observer in the Schwarzschild solution, a gravitational potential well will appear as an optically dense zone.
Suppose that we have a string or a rope under tension. Its weight per meter is analogous to the optical density of a zone in space.
Suppose the string is in the left-right direction and that the weight of the string increases as we go right.
. • ● ■ weights
|---------------------------------------------------|
string under tension
The setup is equivalent to a string of uniform weight to which we attach individual weights whose mass increases as we go right.
Suppose that we have a simple "bump" wave packet in the string proceeding right. The bump is a gaussian wave packet which extends infinitely to the left and the right. The diagram below does not show this infinite extension.
___ ---->
______/ \______
Some of the energy in the bump will be spent on moving the attached weights. Each weight will act as a new source of waves:
● weight
___ original wave ---->
______ / \_____
<--- \___/ left-moving new wave
\___/ --> right-moving new wave
In the diagram above, we have greatly exaggerated the new waves. The inertia of the weight resists the up and down movement of the string. Therefore, the effect of a weight is roughly equivalent to:
1. at the position of the weight, first pull the string down with a finger so that you exert an impulse p down;
2. then pull it up and exert a impulse 2p up;
3. pull down again and exert an impulse p down.
Our pulling procedure will make new waves to proceed to both the left and the right.
The new waves will deform the original wave. How do they deform it?
___ original wave
_______________/ \_______
____ ______ ________
\___/ \___/ new wave 1
________ ___ _________
\___/ \___/ new wave 2
In the diagram above, we have drawn two examples of new waves. We have drawn both the left-moving and the right-moving new wave. When we sum these waves, we see that there is kind of a "sonic boom" progressing along with the original wave.
If the weight of the string increases to the right, then the sonic boom wave will be very steep on its right side, but much less steep on the left side.
The sum of the original gaussian wave plus the sonic boom wave will appear to be lower amplitude on the left side than the original wave.
The Fresnel formula can be used to calculate the reflection if there is a sharp change in the optical density. If we have a smooth change, maybe we can use the Fresnel formula if we divide the path into parts of one wavelength?
If we have two steps in the optical density within one wavelength, maybe the reflection should be calculated by adding up the steps and using the Fresnel formula?
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