# I Massless photon

1. Jan 20, 2017

### 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

Last edited by a moderator: Jan 20, 2017
2. Jan 20, 2017

### ZapperZ

Staff Emeritus

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

... and watch this:

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

Zz.

3. Jan 21, 2017

### john t

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

4. Jan 21, 2017

### stoomart

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

5. Jan 21, 2017

### 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.

Last edited: Jan 21, 2017
6. Jan 21, 2017

### Mister T

$E$ 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.

7. Jan 24, 2017

### john t

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 .

8. Jan 24, 2017

### 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.

9. Jan 24, 2017

### 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.

10. Jan 24, 2017

### Staff: Mentor

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.

11. Jan 24, 2017

### Mister 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.

12. Jan 24, 2017

### john t

#### Attached Files:

• ###### relativistic mass and energy.docx
File size:
793.5 KB
Views:
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13. Jan 24, 2017

### 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

#### Attached Files:

• ###### relativistic mass and energy.docx
File size:
793.5 KB
Views:
36
14. Jan 24, 2017

### SiennaTheGr8

Is this what you want to derive?

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

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

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.

15. Jan 24, 2017

### 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!

16. Jan 24, 2017

### SiennaTheGr8

(Also note that that Griffiths edition is a bit outdated, as neutrinos are no longer thought to be massless.)

17. Jan 24, 2017

### SiennaTheGr8

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

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

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

18. Jan 24, 2017

### 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:

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.

19. Jan 26, 2017

### 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.

20. Jan 26, 2017

### Staff: Mentor

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

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

21. Jan 26, 2017

### 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...").

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

22. Jan 26, 2017

### Staff: Mentor

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.

23. Jan 27, 2017

### SiennaTheGr8

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

24. Jan 27, 2017

### 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).

25. Jan 27, 2017

### Mister T

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.)