It is much easier to demonstrate this using the representation of the Laplace operator in spherical polar coordinate:
\Delta = \frac{1}{r^2} \frac{\partial}{\partial r} r^2 \frac{\partial}{\partial r}
The very successful modern physics theories requires you to NOT distinguish identical particles (the photon that goes in and the photon that goes out) like this. Meaning, there are experiment whose result cannot be explained if we cling to the classical notion of distinguishing all the photons...
Applying a finite rotation is the same as adding up many many many small rotations. (Infinitesimally) Small rotation is related to the partial derivation:
\frac{\partial L}{\partial \phi_j}
If the Lagrangian is invariant under rotation, then the above partial derivative is zero. Using...
Spacial rotation matrices is one representation of the rotation group. Spin operators is another. You need to understand some basic representation theory and how the angular momentum operators are generators of rotation.
Essentially a photon just came into being (and traveling at c), while the equivalent amount of energy and momentum is taken away from the electron (perhaps dropping down to a lower atomic energy level).
As far as we know, this is instantaneous and probabilistic at the single electron-photon level.
Carry your atomic clock and meter stick to a spaceship that's chasing the light at any speed (subluminal, of course). You will still measure it to that value.
You seems to think that the "derivative" of the Lagrangian:
\frac{\partial L}{\partial x_j} - \frac{d}{dt} \frac{\partial L}{\partial x}
which, if x_j has dimension length has dimension of force, is the force of constrain.
NO! You are trying to minimize the Lagrangian again some constrain...
There is no such thing as a critical distance. The size of the wavefunction packets is not an intrinsic properties of eletrons/positrons, but what state they are in. And in any states, there is a finite probability that they would annihilate and turn into some other pairs (photon is not the only...
The numerical value of c is irrelevant. It is just a factor for unit conversion. The universe does not distinguish energy and mass like we do with our language. So historically we invented two units for them, but in fact there is a natural conversion factor between them that we did not discover...
Both of these statements are not correct. The latter seems to have some experimental support of late, but keep in mind that 1. the experiment has not be independently repeated and verified and 2. we have been observing supernova neutrino for many years now and they are found to be...
In my reply #2, I gave two definitions of the same quantity I. By equating them, you can compute u.
u is energy / volume. You also know how much energy is in each photon. So you can get photon / volume.
You forgot the right-hand-side of the Euler-Lagrange equation:
\frac{\partial L}{\partial x_j} - \frac{d}{dt} \frac{\partial L}{\partial \dot{x}_j} = 0
You first two equations are also wrong (check the dimension of each term, they are not the same). For example:
\frac{d}{dt}\frac{\partial...
Didn't check your math, but the argument is correct.
Not sure how did you get the "size of the radio". But dividing the power by the area of the sphere of radius d, you get power per area (Intensity):
I = P/A
Intensity is related to energy density u (energy per volume) and speed v of photon by...