- #36
exponent137
- 565
- 34
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.
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.
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