# Does $p=mc$ hold for photons?

Known that $E=hf$, $p=hf/c=h/\lambda$, then if $p=mc$, where $m$ is the (relativistic) mass, then $E=mc^2$ follows directly as an algebraic fact. Is this the case?

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As you may know, photons do not have mass.

Relating relativistic momentum and relativistic energy, we get:

$E^2 = p^2c^2+(mc^2)^2$.

where $E$ is energy, $p$ is momentum, $m$ is mass and $c$ is the speed of light.

As mass is zero, $E=pc$.

Now, we know that $E=hf$. Then we get the momentum for photon.

Note that there is a term called effective inertial mass. Photon does have it.

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Can you provide a reference for this "effective inertial mass" term. It's not clear to me how it fits into your answer, and I'm having trouble finding information about it through Google searches. – Geoffrey Jan 2 '15 at 19:52

Here's another way to think about it (personally, I think this addresses the question most directly):

$E = hf$ and $p = \frac{hf}{c}$ both apply to photons. What those get you is simply that $E = pc$, so you can conclude that $E = pc$ should be valid for photons. And it is.

Now, your question is worded to ask whether you can start with $p = mc$ and plug in $E = pc$ to get $E = mc^2$. But I think what you really want to know is, can you start with $E = mc^2$ and use it with $E = pc$ to derive $p = mc$?

The answer is, of course, no. $E = mc^2$ doesn't apply to photons. In fact, there is no case in which $E = mc^2$ and $E = pc$ both apply to the same object. So you can never validly combine them. The former is for objects at rest, for which $p = 0$, and the latter is for massless objects, for which $m = 0$, and which always move at the speed of light. As others have shown, they're both special cases of $E^2 = p^2 c^2 + m^2 c^4$.

Incidentally, I can't think of a single physical system for which $p = mc$ is satisfied.

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$p=m v=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}} v$, so $p=m c$ for a photon. If you want $p=m_0 c$, you need $v = \frac{1}{\sqrt{2}} c$. Sorry, I do not find it necessary to say that $E$ is not always equal to $m c^2$. – BartekChom Sep 16 '15 at 19:01

If we start with $E^2=(cp)^2+(c^2m)^2$ as a relativistically correct equation, in units of energy (squared) then we can try to extract $v$ from it, rather than trying to put $v$ into other equations and hope it all works out.

So let's divide everything by $E^2$, we get $1=(cp/E)^2+(c^2m/E)^2$ which is also correct. And since we usually aren't worried about $E=0$, the division by $E$ usually isn't much to worry about. But now everything is dimensionless, if we want to get a nice result about speeds we can multiply by $c^2$ to get:

$c^2=(c^2p/E)^2+(c^3m/E)^2$, or

$(c^2p/E)^2=c^2-(c^3m/E)^2$.

Now the thing on the left has units of velocity (squared), and it's something less than $c^2$ in magnitude. That's not a coincidence because it actually is the velocity[1]. $v=c^2p/E$ (in 2d or 3d space put vector arrows over $v$ and $p$ everywhere, so $\vec{v}=c^2\vec{p}/E$) is a totally correct equation for velocity in terms of $p$ and $E$, just use:

$\vec{v}=c^2\vec{p}/E$.

If instead you want to use $E$ and $m$ (instead of $E$ and $\vec{p}$), then we have:

$v^2=c^2-(c^3m/E)^2$. (Note, you only get $v$ not $\vec{v}$ because $E$ and $m$ are scalars, you can't get a direction).

Now, maybe from Newtonian Physics you are used to expressing $v$ in terms of $\vec{p}$ and $m$, this can also be done. Recall our starting point $E^2=(c\vec{p})^2+(c^2m)^2$, so $E=\sqrt{(c\vec{p})^2+(c^2m)^2}$. Then take our correct equation for $\vec{v}$ in terms of $\vec{p}$ and $E$

$\vec{v}=c^2\vec{p}/E$

and put in that expression for $E$ to get:

$\vec{v}=\frac{c^2\vec{p}}{\sqrt{(c\vec{p})^2+(c^2m)^2}}$.

So in summary, we express $v$ in terms of (any two of) $E$, $m$, and $\vec{p}$.

For $E$ and $m$ we get $v^2=c^2-(c^3m/E)^2$.

For $E$ and $\vec{p}$ we get $\vec{v}=c^2\vec{p}/E$.

For $m$ and $\vec{p}$ we get $\vec{v}=\frac{c^2\vec{p}}{\sqrt{(c\vec{p})^2+(c^2m)^2}}$.

No limits, nothing wrong, no infinities, no division by zero, and no exceptions.

Some people prefer to work with $\vec{\beta}=\vec{v}/c$, instead of $\vec{v}$ and indeed the equations look nicer that way.

For $E$ and $m$ we get $\beta^2=1-(c^2m/E)^2$.

For $E$ and $\vec{p}$ we get $\vec{\beta}=c\vec{p}/E$.

For $m$ and $\vec{p}$ we get $\vec{\beta}=\frac{c\vec{p}}{\sqrt{(c\vec{p})^2+(c^2m)^2}}$.

[1] If this all seemed unbelievable, notice that the worldline of a particle has a unit (in the minkowski geometry) tangent. For a massless particle they move at c, and all the equations work. For massive particles, if you scale that unit tangent by the rest mass, that spacetime vector is the actual energy-momentum spacetime vector $(E,c\vec{p})$, so the tangent line actually goes $E$ units in time for every $pc$ units in space, we divide by $E$ to see what happens in unit time, so in unit time it goes $pc/E$ units in space. The factor of $c$ is literally just to get this ratio in units of speed.

edit The middle equation $\vec{v}=c^2\vec{p}/E$ is the simplest, and indeed if you solve for $\vec{p}$, you get $\vec{p}=(\frac{E}{c^2})\vec{v}$. And yes old equations can sometimes work by replacing $m$ with $\frac{E}{c^2}$, but since $\frac{E}{c^2}$ is not constant ($E$ changes when $p$ changes) not all the old equations are equivalent. For example in Newtonian physics you might right $F=mdv/dt$ or $F=d(mv)/dt$, however if you replaced $m$ with $\frac{E}{c^2}$ in those two equations you'd get two different equations. So there is no simple cookbook for going from an arbitrary newtonian equation to a correct relativistic equation. You just have to learn the correct relativstic equations.

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According to Special Relativity the relativistic energy for a particle is: $E^2= m^2c^4+p^2c^2$

The invariant quantity under relativistic transformations is the rest mass $m$ of the particle.

For a photon $m=0$

Using some simple algebra it is found $E=pc$ for a photon.

You will see this preserves the frequency and energy relationship.

The error in the question is that momentum $p$ is always related to mass and velocity ($p=mv$ where $c$ is placed in as $v$ for the photon), whereas for a massless particle this does not apply.

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As a further elaboration, let's look at this from the angle of relativistic momentum.

Recall that momentum, in relativistic mechanics, is not a linear function of velocity as it is in Newtonian mechanics where $p = mv$. In relativistic mechanics:

$p = \gamma m v$

$m$ is the invariant mass

$\gamma = \dfrac{1}{\sqrt{1 - \dfrac{v^2}{c^2}}}$

Clearly, for a non-zero $m$, $p \rightarrow \infty$ as $v \rightarrow c$

Now, please keep in mind that $p$ in the relativistic energy relationship is not just $mv$ but is the relativistic momentum $\gamma m v$:

$E^2 = (\gamma m v c)^2 + (mc^2)^2$

From this, it's clear that the relativistic energy is:

$E = \gamma m c^2$

So, if we fix $E$ and let $m \rightarrow 0$, we find that $v \rightarrow c$ in the limit.

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