Is the Massless Photon Explanation for E=mc2 Valid?

[Herman Trivilino]

Definitions don't have to be demonstrated. They just have to be adopted. Clearly $E^2 - p^2$ is well-defined and positive if $E > p$, so it can be used to define a positive constant (its square root).

To adopt this as a definition of $m$ for a massive particle, don't we have to start with expressions for $E$ and $p$ that don't involve $m$? It seems circular to define $E$ and $p$ in terms of $m$ and then use $\sqrt{E^2-p^2}$ to define $m$.

 

[SiennaTheGr8]

To adopt this as a definition of $m$ for a massive particle, don't we have to start with expressions for $E$ and $p$ that don't involve $m$? It seems circular to define $E$ and $p$ in terms of $m$ and then use $\sqrt{E^2-p^2}$ to define $m$.

That was my point a few posts up: this only works if you first define $E$ and $\vec p$ without recourse to $m$ (or proper time, by the way), and then show that they constitute the components of a four-vector. Then you can define $m$ as the norm of the four-momentum.

Would you do this using Noether's theorem / least-action principle / Lagrangian mechanics?

 

[PeterDonis]

don't we have to start with expressions for $E$ and $p$ that don't involve $m$?

Yes, and that's what the method robphy described does. $E$ and $p$ are defined by projecting the object's 4-momentum onto the basis vectors of an inertial frame.

Or, alternatively, you can just define $m$ as the norm of the 4-momentum, which doesn't involve $E$ and $p$ (the norm of a 4-vector exists independently of any choice of inertial frame).

this only works if you first define $E$ and $\vec p$ without recourse to $m$ (or proper time, by the way), and then show that they constitute the components of a four-vector.

You have this backwards. You define the 4-vector first--the 4-momentum of the object. Then you define $E$ and $p$ as its components in a particular inertial frame, and $m$ as its norm.

 

[john 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, and that the general equation is

E^2 =m^2c^4 +(pc)^2. I had cited a derivation of E = m gamma c^2 from thought experiments. I was reminded that one of those thought experiments must have involved mass (It did.) and so gave a specific, rather than general, result. The general equation can, however, be reconciled from E = gamma mc^2 and p = gamma mv. I believe another earlier reply indicated that this is not kosher, because 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?

A side question. Is it possible to copy and paste from a word document having Equation Editor objects inserted. It would surely help keep my notes less awkward.

[Moderator's note: post moved from a separate thread to this one since this is what it's referring to.]

 

[PeterDonis]

@john t , at this point it would be very helpful if you would do two things:

(1) Quote the specific parts of the specific posts that you are referring to. The Quote and Reply buttons at the lower right of each post will help you do that.

(2) Use the PF LaTeX feature to post the math of the derivation you are talking about, instead of using an attachment. A previous post by @jtbell gave you the pointer for help with that.

Until you do those things, we can't really respond to your post #34 (which, as the note shows, I moved here from the new thread you started, since it belongs here).

Is it possible to copy and paste from a word document having Equation Editor objects inserted.

Unfortunately I don't think it is. AFAIK Equation Editor doesn't use LaTeX under the hood. Learning the PF LaTeX feature is well worth the time and effort if you are going to post here.

 

[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.

 

[Herman Trivilino]

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}=0If 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.