Energy & Mass of a Photon

Question

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

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

Answer 1

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 $E = mc^2$.

Answer 2

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.