Is the Massless Photon Explanation for E=mc2 Valid?

[sweet springs]

That still leaves me looking for a derivation of p = gamma v and a derivation of the equation for E^2.

Inner product of force and velocity equals to increase rate of energy, so
$$\frac{dE}{dt}=\mathbf{v}\cdot\frac{d\mathbf{p}}{dt}$$
If you know E you can get p from this relation and vice versa.

 

[Mister T]

This question relates to answers I got on a much earlier posting. I had given my opinion for the reason that photons have energy, though massless, is that E = m gamma c^2 is irrelevant for photons, because the denominator of gamma is zero and the equation is invalid. What I was told was that this is not the reason,

The relation $E=\gamma mc^2$ is valid only when $v<c$ because that condition is necessary for its derivation. So attempting to use it for particles with $v=c$ is not valid because it can't be derived under that condition, not because of some feature of its content.

 

[sweet springs]

From the formula #36

$$\frac{dE^2}{dt}-\frac{E}{c^2}(\frac{v_x}{p_x}\frac{d\ p^2_xc^2}{dt}+\frac{v_y}{p_y}\frac{d\ p^2_yc^2}{dt}+\frac{v_z}{p_z}\frac{d\ p^2_zc^2}{dt})=0$$

We may say from homogeneity of direction
$\frac{v_x}{p_x}=\frac{v_y}{p_y}=\frac{v_z}{p_z}:=\frac{1}{f(v)}$ or $\mathbf{p}=f(v)\mathbf{v}$
$$\frac{dE^2}{dt}-\frac{E}{f(v)c^2}\frac{d\ p^2c^2}{dt}=0$$If we can assume $E=f(v)c^2$　(!) or $\mathbf{p}=\frac{E}{c^2}\mathbf{v}$, $E^2-p^2c^2 = const.=f(0)^2c^4$ and $f(v)^2 (1-\frac{v^2}{c^2})=const. =f(0)^2$
$$f(v)=\frac{f(0)}{\sqrt{1-\frac{v^2}{c^2}}}$$

This is just a arbitrary assumption but seems probable.

 

[sweet springs]

Similarly to p, we may say $E=E(v)$ so from the top second equation
$$\frac{dE^2}{dt}-\frac{E(v)}{f(v)c^2}\frac{d\ p^2c^2}{dt}=0$$
This equation should not depend on value of v (?) so $$\frac{E(v)}{f(v)c^2}=const.=\frac{E(0)}{f(0)c^2}$$ and we may postulate const.=1.

 

[robphy]

Inner product of force and velocity equals to increase rate of energy, so
$$\frac{dE}{dt}=\mathbf{v}\cdot\frac{d\mathbf{p}}{dt}$$
If you know E you can get p from this relation and vice versa.

You'll need some more information [which should be made explicit] to pin down the constant of integration.
This expression arises in the work-energy theorem
and gives the definition of the kinetic-energy [not the total energy, i.e. the time-component of the 4-momentum].

 

[robphy]

In my opinion, when talking about trying to the get to the [massless] photon as some limit of a massive particle,
the limiting sequence should be specified.

Consider this energy-momentum diagram of a particle with rest-mass 4:

The tip of the 4-momentum (at P) of a massive particle lies on a particular hyperbola [called the mass-shell].
The tip of the 4-momentum of a massless particle, like the photon, lies on the null-cone [along the cyan-dotted line].

To get from this massive particle to a massless particle,

• one could try to "make the particle move faster" by following the "fixedm" sequence along its mass-shell.
  But this won't achieve a massless particle because you are always on this m=4 mass-shell,
  which asymptotically approaches the null-cone (in any frame) but never reaches the null-cone [where m=0].
• one could try to follow "fixedE" to the null-cone,
  which is sequence of faster particles with decreasing mass but increasing relativistic-momentum, but all with the same relativistic-energy in this frame.
• one could try to follow "fixedp" to the null-cone,
  which is sequence of faster particles with decreasing mass and decreasing relativistic energy, but all with the same relativistic-momentum in this frame.
• or one could try some other sequence with varying v and varying E,p, and m.

So, I think that one should really spell out the limiting procedure,
accompanied with a physical model of how to take that limit.

 

[Dale]

I am going from a specific to a general equation. But I am doing so, by stipulating additional conditions. Why is this not justified? Is the math logic bad?

Yes, stipulating additional conditions makes your solution even more specific. To generalize an equation you have to make fewer stipulations.