Is the Massless Photon Explanation for E=mc2 Valid?

[john t]

I have wondered how a photon could be massless, given E=mc2. I have seen explanations involving treatment of the situation through consideration of momentum. It seems to me my own explanation is valid, and I ask for comments.
Actually E= m’c2 =gmc2, where g=1/sq rt{(1/[1-(v2/c2)]}. Photons travel at v = c, so the denominator is zero, making the equation invalid at the speed of light, thus eliminating the apparent conflict.
John

 

[ZapperZ]

I have wondered how a photon could be massless, given E=mc2. I have seen explanations involving treatment of the situation through consideration of momentum. It seems to me my own explanation is valid, and I ask for comments.
Actually E= m’c2 =gmc2, where g=1/sq rt{(1/[1-(v2/c2)]}. Photons travel at v = c, so the denominator is zero, making the equation invalid at the speed of light, thus eliminating the apparent conflict.
John

Please read this:

https://www.physicsforums.com/insights/do-photons-have-mass/

... and watch this:

https://www.physicsforums.com/media/e-mc-is-incomplete.127/

Zz.

 

[john t]

This did not answer my question. Do you think my assertion is correct?

 

[stoomart]

This did not answer my question. Do you think my assertion is correct?

The video should have answered your question, because the equation used in your assertion (E=mc²) is not complete.

 

[Ibix]

Mass is the square root of the length of the energy-momentum four-vector. That is, $$\begin{eqnarray}m_0^2c^4&=&E^2-p^2c^2\end{eqnarray}$$In an object's rest frame, $p=0$, so $E_0=m_0c^2$. You can do a Lorentz transform into any other frame, and get $E=\gamma E_0$ and $p=\gamma vE_0/c^2=Ev/c^2$. You can substitute that expression for $p$ into (1) and rearrange to get your $E=\gamma m_0c^2$.

The problem is that light has no rest frame, so you can't reason your way to the $p=Ev/c^2$ expression that you need to get to your $E=\gamma m_0c^2$. So you are correct that the equation is inapplicable, but that's because the derivation relies on the assumption that there exists a rest frame for the object. This is not valid for light. The cancelling zeroes is nonsense following from an initial invalid assumption, not a solution to it.

Edit: or you could just note that $E=\gamma E_0=\gamma m_0c^2$ - but the problem is still that you can't start in a rest frame with light.

 

[Mister T]

I have wondered how a photon could be massless, given E=mc2.

$E$ is also zero.

But note that you are using $E$ as the rest energy, not the total energy. It's the total energy that's equal to $hf$, but it's the rest energy that's equal to $mc^2$.

So your confusion arises from having seen others get sloppy with their notation.

 

[john t]

Mass is the square root of the length of the energy-momentum four-vector. That is, $$\begin{eqnarray}m_0^2c^4&=&E^2-p^2c^2\end{eqnarray}$$In an object's rest frame, $p=0$, so $E_0=m_0c^2$. You can do a Lorentz transform into any other frame, and get $E=\gamma E_0$ and $p=\gamma vE_0/c^2=Ev/c^2$. You can substitute that expression for $p$ into (1) and rearrange to get your $E=\gamma m_0c^2$.

The problem is that light has no rest frame, so you can't reason your way to the $p=Ev/c^2$ expression that you need to get to your $E=\gamma m_0c^2$. So you are correct that the equation is inapplicable, but that's because the derivation relies on the assumption that there exists a rest frame for the object. This is not valid for light. The cancelling zeroes is nonsense following from an initial invalid assumption, not a solution to it.

Edit: or you could just note that $E=\gamma E_0=\gamma m_0c^2$ - but the problem is still that you can't start in a rest frame with light.

This is meant as a reply to lbix, whose comments I found the clearest. I think you are saying that the formula relating rest mass to relativistic mass is, itself, invalid because it assumes a rest mass and that it cannot even be used in getting to E=gamma mc^2 .

 

[Ibix]

The expression $m^2c^4=E^2-p^2c^2$ applies to anything. The expression $E=\gamma mc^2$ is a specialisation that you cannot derive without assuming that whatever it is has a rest frame. The rest frame of light is a contradiction in terms, so the second expression does not apply to light.

 

[john t]

Thanks. I would like to see a derivation of that equation which commenters are so familiar with. I have been working from a different derivation that seems not to bring up that equation. I first followed a published thought experiment involving a moving bouncing mass in a cylinder. The math leads to the m'=gamma mc^2 equation. Then I looked at derivation of the mass/energy relation. This goes from the mass/velocity relation to m'=(dm'/dv)[(c^2-v^2)/v. Expressing force as dp/dt or d(m'v)/dt and considering energy as force times distance leads to E=gamma mc^2. Nowhere in the chain of the derivation did the relation m2c4-E2-p2c2 appear, although consideration of momentum (its conservation) was entailed in the mass/velocity derivation. I seem to have gotten to a specialized version without recognizing the fuller expression by following the thought experiment I mentioned, which does not represent the situation of a photon.

 

[Nugatory]

I seem to have gotten to a specialized version without recognizing the fuller expression by following the thought experiment I mentioned, which does not represent the situation of a photon.

We'd have to see the exact derivation that you're talking about to know for sure... But I would bet fairly long odds that it works by considering the situation from two different frames in which the object has two different speeds, or by considering one frame in which the object changes speed. Either way, that's a special case that does not include photons because they travel at one speed, $c$, in all frames.

 

[Mister T]

Thanks. I would like to see a derivation of that equation which commenters are so familiar with.

Well, every derivation has to start with some definitions. In this case I think (although others may show me I'm wrong) that the expression $\sqrt{E^2-(pc)^2}$ defines the mass. So there are two cases. Massive particles for which $E$ is larger than $pc$ and massless particles for which $E$ is equal to $pc$.

I have been working from a different derivation that seems not to bring up that equation. I first followed a published thought experiment involving a moving bouncing mass in a cylinder. The math leads to the m'=gamma mc^2 equation.

That's a moving bouncing particle. And the particle is massive, as opposed to massless.

Then I looked at derivation of the mass/energy relation. This goes from the mass/velocity relation to m'=(dm'/dv)[(c^2-v^2)/v.

I'm not sure what the mass/velocity relation is. If it's $\gamma m=\frac{m}{\sqrt{1-(\frac{v}{c})^2}}$, that's a relation that applies to massive, as opposed to massless, particles.

Expressing force as dp/dt or d(m'v)/dt [...]

But $p$ is not in general equal to $\gamma mv$. The equality holds only for massive, as opposed to massless, particles.

 

[john t]

Well, every derivation has to start with some definitions. In this case I think (although others may show me I'm wrong) that the expression $\sqrt{E^2-(pc)^2}$ defines the mass. So there are two cases. Massive particles for which $E$ is larger than $pc$ and massless particles for which $E$ is equal to $pc$.
That's a moving bouncing particle. And the particle is massive, as opposed to massless.
I'm not sure what the mass/velocity relation is. If it's $\gamma m=\frac{m}{\sqrt{1-(\frac{v}{c})^2}}$, that's a relation that applies to massive, as opposed to massless, particles.
But $p$ is not in general equal to $\gamma mv$. The equality holds only for massive, as opposed to massless, particles.

 

[john t]

Sorry, I'm getting used to format of this site. John Nugatory replied with a comment that one would need to see the derivation I was following. Attached hopefully is an excerpt from a document I once assembled. Forgive the inclusion of all the trivial math steps

 

[SiennaTheGr8]

Is this what you want to derive?

$E^2 = (mc^2) ^2 + (pc) ^2$

If so, start with the relativistic equations for energy and momentum:

$E = \dfrac{mc^2}{\sqrt{1 - \left( \frac{v}{c} \right) ^2}}$

and

$p = \dfrac{mv}{\sqrt{1 - \left( \frac{v}{c} \right) ^2}}$.

This is actually a good algebra exercise, so I'll leave it to you. Let us know if you need help.

 

[SiennaTheGr8]

Note: deriving the energy-momentum relation that way doesn't tell you that it will be valid for the massless case. But it was already known from Maxwell that $E = pc$ for light. So it's kind of a happy accident that $E^2 = (mc^2)^2 + (pc)^2$ works for massless particles!

See David Griffiths's comments here: https://books.google.com/books?id=Wb9DYrjcoKAC&pg=PA90

 

[SiennaTheGr8]

(Also note that that Griffiths edition is a bit outdated, as neutrinos are no longer thought to be massless.)

 

[SiennaTheGr8]

Okay, here's a hint for the algebra problem I gave you:

$p = \dfrac{mv}{\sqrt{1 - \left( \frac{v}{c} \right) ^2}}$

$pc = \dfrac{mc^2 \left( \frac{v}{c} \right)}{\sqrt{1 - \left( \frac{v}{c} \right) ^2}}$

 

[Mister T]

Just as I stated, that's a derivation for a massive (as opposed to massless) particle.

Note that the very first sentence is erroneous:

An outcome of special relativity is that mass increases with velocity.

It is in fact a re-definition of mass that causes that outcome. For Einstein, it was a matter of only a few years before he realized it was not a necessary consequence. It took the literature decades to follow. By the mid-1990's we started to see it disappear from college-level introductory physics textbooks. It is now virtually gone.

But back to your original question. If one re-defines mass to make that claim true, there are several ways to do that by the way but using the most popular way, the photon can be said to have a mass of $hf/c^2$. That is not, however, what is meant when it's stated that the photon is massless.

So, in summary, according to the way mass is defined in that statement, the photon has mass.

According to the way mass is usually defined, the photon is massless.

 

[john t]

Replying to SiennatheGr8. Thanks for the algebra problem you layed before me. I was not able to eliminate v. Oh well, I am able to work it the other way around and go from the general equation and p=gamma mv to E=mc^2. Is it not more rigorous to go from the general to the specific cadse, anyway?

That still leaves me looking for a derivation of p = gamma v and a derivation of the equation for E^2. I did not find your referenced Griffiths comments very helpful.

Also can anyone tell me how to write in math notation in this forum.

 

[jtbell]

Also can anyone tell me how to write in math notation in this forum.

If you're using a desktop or notebook, starting at the top right of the page:

INFO --> Help/How-To --> LaTeX Primer

 

[SiennaTheGr8]

@john t,

The Griffiths link was merely a trusted source for my claim that it's only a happy accident that $E^2 = (mc^2)^2 + (pc)^2$ works for massless things (see where he says, "Personally, I would regard this 'argument' as a joke...").

As for the algebra problem, we start with these two equations:

1) $\, E = \dfrac{mc^2}{\sqrt{1 - \left( \frac{v}{c} \right)^2}}$

2) $\, pc = \dfrac{mc^2 \left( \frac{v}{c} \right)}{\sqrt{1 - \left( \frac{v}{c} \right)^2}}$

First square Equation 1:

3) $\, E^2 = \dfrac{(mc^2)^2}{1 - \left( \frac{v}{c} \right)^2}$

Then square Equation 2 and add $(mc^2)^2$ to both sides:

4) $\, (pc)^2 + (mc^2)^2 = (mc^2)^2 \left( \dfrac{\left( \frac{v}{c} \right)^2}{1 - \left( \frac{v}{c} \right)^2} + \dfrac{1 - \left( \frac{v}{c} \right)^2}{1 - \left( \frac{v}{c} \right)^2} \right) = \dfrac{(mc^2)^2}{1 - \left( \frac{v}{c} \right)^2} = E^2$

QED.If you want to see how Equation 1 is derived (relativistic energy), study Einstein's original $E=mc^2$ paper (it makes use of the relativistic Doppler effect, which he'd derived a few months earlier in his first paper on special relativity): https://www.fourmilab.ch/etexts/einstein/E_mc2/www/

In modern notation, the equation he ends up with is more or less this:

5) $\, E_k = E_0 \left( \dfrac{1}{\sqrt{1 - \left( \frac{v}{c} \right)^2}} - 1 \right)$,

where $E_k$ is kinetic energy and $E_0$ is rest energy (he uses the symbol $E_0$ for something else, so try not to get confused there). Then he does a binomial expansion to show that when $v \ll c$, Equation 5 reduces to:

6) $\, E_k = \dfrac{E_0 v^2}{2 c^2}$.

Comparing Equation 6 with the classical $E_k = mv^2 / 2$, he obtains:

7) $\, E_0 = mc^2$.

Since total energy $E = E_0 + E_k$, we can easily get Equation 1 from Equations 5 and 7.Finally, if you want to see how Equation 2 is derived (relativistic momentum), read about the Tolman/Lewis thought experiment here: https://books.google.com/books?id=FrgVDAAAQBAJ&pg=PA76

 

[PeterDonis]

it's kind of a happy accident that $E^2 = (mc^2)^2 + (pc)^2$ works for massless particles!

I'm not sure I agree with this statement, at least not without qualification. I see why Griffiths says it, since he starts from the assumptions of classical physics and then investigates what needs to be modified in relativity. But if you start from the assumptions of relativity to begin with, which seems more logical to me since we know the universe is in fact relativistic, not classical (to put it another way, it's just a historical accident that classical physics was formulated first, so we shouldn't treat its assumptions as logically prior), then the relation $E^2 = p^2 + m^2$ is just a definition of $m$. This is easier to see if you write it in the form $E^2 - p^2 = m^2$, which is how a lot of relativity textbooks do it. In other words, starting from the assumptions of relativity, "mass" is something we derive--it's the invariant length of the 4-momentum vector, which can be either positive (timelike, "massive") or zero (null, "massless"). The "happy accident" from this point of view is that $E = p$ (or $E = pc$ in conventional units) is the same relation that can be derived from Maxwell's Equations for light. But even that is just a consequence of the fact that Maxwell's Equations are Lorentz invariant.

 

[SiennaTheGr8]

I'm not sure I agree with this statement, at least not without qualification. I see why Griffiths says it, since he starts from the assumptions of classical physics and then investigates what needs to be modified in relativity. But if you start from the assumptions of relativity to begin with, which seems more logical to me since we know the universe is in fact relativistic, not classical (to put it another way, it's just a historical accident that classical physics was formulated first, so we shouldn't treat its assumptions as logically prior), then the relation $E^2 = p^2 + m^2$ is just a definition of $m$. This is easier to see if you write it in the form $E^2 - p^2 = m^2$, which is how a lot of relativity textbooks do it. In other words, starting from the assumptions of relativity, "mass" is something we derive--it's the invariant length of the 4-momentum vector, which can be either positive (timelike, "massive") or zero (null, "massless"). The "happy accident" from this point of view is that $E = p$ (or $E = pc$ in conventional units) is the same relation that can be derived from Maxwell's Equations for light. But even that is just a consequence of the fact that Maxwell's Equations are Lorentz invariant.

This would require a Lagrangian approach, wouldn't it? So that you could derive the four-momentum without smuggling in $m$?

 

[SiennaTheGr8]

Just to clarify what I mean, and to expand a bit:

Most primers get to the four-momentum by taking the proper-time derivative of the four-displacement and scaling by $m$. With this approach, it's a "happy accident" that the four-momentum itself works for massless systems (and it's a double-whammy, since proper time isn't defined for null intervals).

 

[Mister T]

the relation $E^2 = p^2 + m^2$ is just a definition of $m$. This is easier to see if you write it in the form $E^2 - p^2 = m^2$, which is how a lot of relativity textbooks do it. In other words, starting from the assumptions of relativity, "mass" is something we derive--it's the invariant length of the 4-momentum vector, which can be either positive (timelike, "massive") or zero (null, "massless"). The "happy accident" from this point of view is that $E = p$ (or $E = pc$ in conventional units) is the same relation that can be derived from Maxwell's Equations for light. But even that is just a consequence of the fact that Maxwell's Equations are Lorentz invariant.

How can one demonstrate that this relation can be used to define the mass of particles for which $E \neq p$?

I refer of course to massive, as opposed to massless, particles. Typically we say that $E=\gamma m$ and $p=\gamma mv$, and then demonstrate that when we substitute them into that relation we get $m$. But when we do it that way it seems we're introducing $m$ before we define it. Are we simply saying that for a massive particle $m$ is introduced, not defined? And then we introduce that relation as a definition of $m$ and demonstrate that it's valid for massive particles? (Moreover, we can use it to define the mass of composite objects that consist of a collection of particles, both massive and massless.)

 

[robphy]

I' But if you start from the assumptions of relativity to begin with,
[snip]
then the relation $E^2 = p^2 + m^2$ is just a definition of $m$. This is easier to see if you write it in the form $E^2 - p^2 = m^2$, which is how a lot of relativity textbooks do it. In other words, starting from the assumptions of relativity, "mass" is something we derive--it's the invariant length of the 4-momentum vector...

I would argue that $E^2 = p^2 + m^2$ is not a definition of $m$.
Rather, it expresses the relationship (a constraint) between $E$ and $p$ (an observer's components of the 4-momentum of that particle).

Logically speaking:

Given $P^a$ and the metric $g_{ab}$, we can form $P^aP^b g_{ab}$ and call it $m^2$ (in the +--- signature)--this defines $m$.

Next, if we are then given an observer's timelike 4-velocity $t^a$, we can then define the components of $P^a$ (E and p) for that observer... and see that they are constrained by the square-norm of $P^a$.
(For another observer and her 4-velocity, we get another set of components (E' and p') that are constrained by an analogous relation.)

 

[PeterDonis]

How can one demonstrate that this relation can be used to define the mass of particles for which $E \neq p$?

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). So the only possible wrinkle would be if there existed some particle for which $E < p$.

Or, alternatively, we can define $E$ and $p$ using the 4-momentum. See my response to robphy (coming shortly).

 

[PeterDonis]

Given $P^a$ and the metric $g_{ab}$, we can form $P^aP^b g_{ab}$ and call it $m^2$ (in the +--- signature)--this defines $m$.

Yes, and if you write down $P^a P^b g_{ab}$ in an inertial frame, it is $E^2 - p^2$. I was assuming that had been done, but you're right that this step should be made explicit.

This method also makes the commonality with the null/lightlike case easy to see. For that case, $P^a P^b g_{ab} = 0$.

This also takes care of the possibility of $E < p$, which I mentioned in my response to MisterT; that would require the vector $P^a$ to be spacelike. We can either rule that out a priori, or consider the possibility of tachyons and work out the consequences.

Next, if we are then given an observer's timelike 4-velocity $t^a$, we can then define the components of $P^a$ (E and p) for that observer.

It's not enough to choose $t^a$; you need to choose an entire tetrad, and it needs to be a tetrad field over spacetime. That is equivalent to choosing an inertial frame (at least in flat spacetime--I don't think we need to go into the complications raised by curved spacetime here).

 

[robphy]

My point is that the 4-momentum and its (invariant) mass m (using the metric) is primary, independent of observer. The role of the observer (which leads to his decomposition of $P^a$ into E and p according to him) is secondary.

Geometrically, the objects (vector and its magnitude (using the metric)) are primary; components of objects are secondary.

 

[PeterDonis]

Geometrically, the objects (vector and its magnitude (using the metric)) are primary; components of objects are secondary.

Yes, agreed.

 

[Mister T]

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.