# 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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Thank you for the answer. The equation E^2=(mc^2)^2 is one I recognize but it is not in my physics book (fund. of modern physics by Resnick et al.) I'll have to see how it is deduced, if at all deduced. – Nick Rosencrantz Jun 26 '12 at 7:15
@Nick Rosencrantz For a motivating analogy to the relativistic energy equation it is useful to look at it as m^2 = E^2-p^2 (i have suppressed the constants) in this case the invariant quantity is placed on the left. Notice the similarity between this and invariant length in say 2-dim euclidean geometry s^2 = x^2 + y^2. The difference being that the "time-like" component in the SR equation is assumed to be imaginary- as squaring produces a negative quantity (which i moved w/ algebra) So the point is it can be intuitive to look at this like length of a 4-dim vector with an imaginary component. – jaskey13 Jun 26 '12 at 18:32

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&velocity ($p=mv$ where $c$ is placed in as $v$ for the photon)- which for a massless particle this does not apply.

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I see i was late in my post as my answer is the same as @Zee Malasia. Oh well... – jaskey13 Jun 26 '12 at 2:31
I attempted to delete as my answer is redundant but it seems i can't? – jaskey13 Jun 26 '12 at 2:33
No need, there's nothing wrong with having the same thing explained in different ways. – David Zaslavsky Jun 26 '12 at 3:14
It's a good answer and good to have two elaborations. I didn't know the E^2 equation was needed to explain this. – Nick Rosencrantz Jun 26 '12 at 7:16

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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 That's interesting and somewhat more familiar way of presenting the relativistic energy to an amateur. Thanks. – Nick Rosencrantz Jun 26 '12 at 17:55

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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 Addessing your last point about $p=mc$ there is the trivial case where $\gamma v = c$, which happens to hold for $v=c/\sqrt{2}$. So for any massive object there is one particular speed (not $c$!) such that $p=mc$. Of course this is hardly interesting. – Michael Brown Jan 17 at 13:43 Part of the problem is notation. Einstein never wrote $E=mc^2$. He was very careful to use a different symbol, $E_o$ for rest energy. The $E$ in $E=pc$ and the $E$ in the ubiquitous, and incorrect, $E=mc^2$ are not, and cannot be, the same $E$. One refers specifically to rest energy and the other one does not. Different concepts need different symbols as Einstein fully knew. Too bad most textbook/popular authors didn't listen. – user11266 Jan 17 at 15:28 @JoeH I'd say they are the same energy, just for different systems. $E = mc^2$ is fine if you specify that it only applies to an object at rest; I don't believe it's practical or necessary to use a different symbol to designate the energy of an object at rest. Analogously, I don't believe it's necessary to use different symbols for the $x$ in $x = vt$ and the one in $x = \frac{1}{2}at^2$, even though one could make your argument that they are not the same $x$. – David Zaslavsky♦ Jan 17 at 19:47 My argument doesn't apply to your kinematic example since in both cases, $x$ (which really should be $\Delta x$ since there is no explicit assumption that $x_o$ is zero) refers to the same concept, spatial displacement. Rest energy and total particle energy are significantly different concepts, at least Einstein thought so. – user11266 Jan 18 at 4:11 @JoeH Of course you're technically right about $\Delta x$. Anyway, I would apply the same reasoning to your statement: the argument doesn't apply to the energy formulas because in both cases, $E$ refers to the same concept, total energy. Einstein may have considered energy at rest to be something fundamentally different from energy while in motion, but I do not. (At some level this is a matter of opinion, of course) – David Zaslavsky♦ Jan 18 at 4:22