Energy & Mass of a Photon [duplicate]

$$\text{Please read the whole question before answering}$$

Before I ask my question, I would like to say that "Yes, I do know a photon has no mass."

I was helping someone here on P.SE with the subject of how photons are produced. For some reason I was also thinking of Einstein at the time I helped said person, and as it often does for me, and I started thinking about $E=MC^2$ and the energy of a photon.

I looked it up later (here, here, and here) and the respective energies of the wavelengths Blue(0.445 microns), Green(0.525 microns), and Red(0.644 microns) are $4.46*10^{-19} J$, $3.78*10^{-19} J$, and $3.08*10^{-19} J$.

Now I don't know calculus or higher level physics, but it occurred to me that if something has energy, by Einstein's $E=MC^2$, it must have mass (Because the speed of light is constant). I did a quick thought experiment, and sure enough, something with no mass would have no energy $$E=MC^2$$ $$E=0⋅C^2$$ $$E = 0$$

Using SI units, since the units used are arbitrary, I did another thought experiment:

Using $E=MC^2$, what is the mass of a photon at the wavelength 0.445μm (Blue)? $$E=MC^2$$ $$4.46⋅10^{-19} \text{J} = M⋅C^2$$ $$4.46⋅10^{-19} \text{J} = M⋅(299,792,458 \text{ m/s})^2$$ $$4.46⋅10^{-19} \text{J} = M⋅(8.98755⋅10^{16} \text{ m}^2/\text{s}^2)$$ $$\text{And since we can rearrange to get } E/C^2 = M,$$ $$\frac{4.46⋅10^{-19} \text{J}}{8.98755⋅10^{16} \text{ m}^2/\text{s}^2} = M$$ $$\frac{4.46⋅10^{-19} \text{Kg⋅m}^2/\text{s}^2}{8.98755⋅10^{16} \text{ m}^2/\text{s}^2} = M$$ $$\frac{4.46⋅10^{-19} \text{Kg}}{8.98755⋅10^{16}} = M$$ $$4.96242⋅10^{-36} \text{Kg} = M$$

I don't know about you, but that does not look like a mass of 0.

Lets do another: What is the mass of a photon at frequency 0.525μm (Green)? $$E=MC^2$$ $$3.78⋅10^{-19} \text{J} = M⋅C^2$$ $$3.78⋅10^{-19} \text{J} = M⋅(299,792,458 \text{ m/s})^2$$ $$3.78⋅10^{-19} \text{J} = M⋅(8.98755⋅10^{16} \text{ m}^2/\text{s}^2)$$ $$\frac{3.78⋅10^{-19} \text{J}}{8.98755⋅10^{16} \text{ m}^2/\text{s}^2} = M$$ $$\frac{3.78⋅10^{-19} \text{Kg⋅m}^2/\text{s}^2}{8.98755⋅10^{16} \text{ m}^2/\text{s}^2} = M$$ $$\frac{3.78⋅10^{-19} \text{Kg}}{8.98755⋅10^{16}} = M$$ $$4.20581⋅10^{-36} \text{Kg} = M$$

Again, something doesn't smell right.

Honestly, I'm not sure what to make of this. Obviously, if photons had any mass at all, Earth would probably not be here (What with the sun sending hundreds of trillions of photons a year, for 4.5 billion years. We'd have been sandblasted away long ago)

So clearly something is very wrong. In order of most, to least likely:

• I messed up with something in my calculations.
• I assumed something, that shouldn't have been assumed.
• I made some, other, unknown mistake somewhere.
• Photons have special rules, when it comes to this equation.
• The equation I used is an oversimplification.
• There's something in calculus/higher level physics that explains this, and I am just oblivious to it.
• There is a debate going on over this somewhere.
• Its a conspiracy.
• Whales can fly.
• Einstein was wrong.

I got my information/double checked my information at these websites:

The equation itself.

The speed of light (1)(2)

Energy of photon (1)(2)(3)

Definition of a Joule.

Someone please shed light on this, it's been driving me crazy since 8/26/14.

TIA P.SE

marked as duplicate by Kyle Kanos, Brandon Enright, alemi, Ali, John RennieAug 29 '14 at 5:55

• You've rediscovered the mass-equivalence. You may be interested to read this Wikipedia article on orders of magnitude for mass. – Kyle Kanos Aug 28 '14 at 20:31
• $E=mc^2$ is only a portion of the equation, suitable for objects with zero momentum. Better is $E^2 = (mc^2)^2 + (pc)^2$. The photons have no rest mass, but they do have momentum. – BowlOfRed Aug 28 '14 at 20:33
• $E = mc^2$ does not apply for photons. The proper equation is $E = hv$. – Dave Coffman Aug 28 '14 at 20:33
• @KyleKanos The difference between my question and that one would be, I did a lot of thought experiments? Lol, yeah, I'll look. – CoilKid Aug 28 '14 at 21:12

Briefly, the formula $E=mc^2$ applies only particles at rest in an inertial frame of reference.

Since there is no rest frame for a photon, no inertial reference frame in which a photon is at rest, one cannot apply the formula $E=mc^2$ to a photon.

In more detail, the four-momentum of a particle has components $(\frac{E}{c}, \vec p)$

The four-momentum for a photon is, of course, light-like which means that the 'length' of the photon four-momentum is zero

$$E^2_\gamma - (p_\gamma c)^2 = 0$$

where the $\gamma$ subscript indicates the energy and momentum of a photon.

Thus, we immediately get the result

$$E_\gamma = p_\gamma c$$

The photon energy and momentum are proportional.

Now, for a time-like four-momentum, the 'length' of the four-momentum is non-zero and is proportional to the invariant mass of the particle

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

This is the origin of the relativistic energy-momentum relation

$$E = \sqrt{(pc)^2 + (mc^2)^2}$$

Note that setting the invariant mass to zero yields the photon energy-momentum relation given earlier. Thus, we say that the invariant mass of a photon is zero.

Also, note that, for non-zero invariant mass, setting the momentum equal to zero (particle is at rest) yields the famous $E = mc^2$.

• I believe that that explains almost everything for me. I'm still not sure how something with no mass can have momentum though. Perhaps I missed something glaringly obvious? – CoilKid Aug 28 '14 at 21:29
• Wait, so the magnitude of a photon of wavelength 0.445 $mu_0$ is $4.96242⋅10−36$? $4.96242⋅10−36$Whats? What exactly is magnitude here? – CoilKid Aug 28 '14 at 21:46
• @CoilKid, it's hard to shake Newtonian intuition where it would be true that zero mass implies zero momentum except for the case that the velocity is 'infinite'. In a certain sense, the speed c is to relativistic mechanics as 'infinite' speed is to Newtonian mechanics. – Alfred Centauri Aug 28 '14 at 21:46

The correct general equation is $E^2=m^2c^4+p^2c^2$, since $m=0$, $E=pc$. A photon's momentum is $p=\hbar k$, where k is the wavenumber ($k=\frac{2\pi}{\lambda}$). This would be consistent with $E=h\nu$ where $\nu$ is the frequency.

• What's with all the squares? I feel like you could greatly simplify by just taking the square root of both sides. $$E=MC^2 + PC$$ – CoilKid Aug 28 '14 at 21:16
• $\sqrt{a^2 + b^2}$ does not in the general case equal $a + b$ – BowlOfRed Aug 28 '14 at 21:17
• No? I have never heard in any class I have ever taken, that $\sqrt{a^2+b^2}$ is not equal to $a+b$. Could you please cite something explaining how that works? – CoilKid Aug 28 '14 at 21:26
• If $a$ and $b$ are both equal to $1$, then $a+b = 2$, but $\sqrt{a^2 +b^2} = 1.414...$ – BowlOfRed Aug 28 '14 at 21:29
• You should feel silly :) – Fattie Apr 26 '16 at 13:02