If photon had mass, the emitted radiation wave energy from fusion or fission would be like this. (M-M_photon) C²=hv

Edited July 18, 201114 yr by alpha2cen

If photon had mass, the emitted radiation wave energy from fusion or fission would be like this. (M-M_photon) C²=hv

A photon at rest accordingly has no mass. But a photon can never be at rest and it does "sink" (bend) under the influence of a gravitational field and has mass-like influences/ properties such as energy of contact like a solar sail. Your equation hv = E/m does not seem to make sense in that Plank's constant h (a very small number), times a velocity (v) the speed of light or less, together being hv, could never equal the speed of light squared, right?

^.

Edited July 18, 201114 yr by pantheory

v means frequency(1/sec). Fusion or fission energy is E= M C² . If photons are released as the fusion energy, some modification will be required. I mean the mass deficiency is (M - M_photon).

What's wrong with the current equations? Some people assign a "relativistic mass" to a photon. (I personally think it causes more trouble than it's worth, since the multiple definitions of mass tend to not be used consistently) How is this different and of what use is it?

Current calculation

 

E = (M_initial - M_final) C²

M_initial ; initial mass of fusion or fission reactant.

M_final ; final mass of fusion or fission product.

 

New calculation

E=(M_initial system - M_final system) C²= ( M_initial - M_final - M_photon + M_photon released) C²

M_initial system ; initial mass of the system which contains fusion or fission reactant.

M_final system ; final mass of the system which contains fusion or fission product.

M_photon ; final photon mass of the system.

M_photon released; photon mass which is emitted to the surround.

 

Form new equation we can calculate system energy loss.

 

At a star problem we can calculate the released energy by using this equation.

It is the difference between the initial mass and the final mass of the star.

It is not the mass difference between initial fusion reactant and final fusion product.

You are measuring different things with the two equations. All you've done is include the photons as part of the final system. You can do that anyway, if it's useful to you for some reason, by defining what your system is. Generally, though, the photons are considered part of the released energy.