If light has energy, does it have mass?

Question

Light has energy, so does that mean it has mass due to what $E=mc^2$ tells us? And if light does have mass, then how does it still travel at the speed of causality? I feel like I'm getting confused between relativistic mass and rest mass or I'm misinterpreting $E=mc^2$.

Answer 1

In special relativity mass can be interpreted as the energy of a system in the (inertial) frame that minimizes the energy (the "rest frame") or equivalently as the absolute value of the square root of the norm of the total energy-momentum four vector of the system.

For a single photon/beam of light, there is no minimum energy frame, but the greatest lower bound of the energy is zero, so a photon/beam of light has zero mass.

However in a system of, for example, two photons taking antiparallel paths there is a frame that minimizes the energy of the system and that minimum energy is non-zero and so the system is not massless. This means that a system consisting entirely of photons can have a non-zero mass, even though individual photons are necessarily massless.

Answer 2

$E = mc^2 = \gamma m_0 c^2$, where $m_0$ is the rest mass and $m = \gamma m_0$ is the 'relativistic mass' (Note that this is not a 'real' mass - it's a mathematical trick to make the relativistic momentum formula look the same as it does in classical mechanics - $p = mv$ is still true in special rel if m is relativistic mass). The term is misleading, however, because $KE \neq mv^2$, $KE = (1-\gamma)m_0c^2$.

Let's look in detail at what happens to $E = \gamma m_0c^2$ as $v \rightarrow c$ and $m_0 \rightarrow 0$.

$E = \frac{m_0}{\sqrt{1-v^2/c^2}}c^2 \rightarrow \frac{0}{0}c^2$

Now that 0/0 term is nasty - the mathematicians call it 'indeterminate form' because it doesn't actually have any meaning. In other words, the mathematics says "I have no heckin' idea what the energy is," we need to use something else to figure it out.

This mathematics is also a strong motivation that any particle traveling at c must have zero mass - if the numerator were anything other than zero, $m_0/0$ would cease to be indeterminate and become undefined - indeterminate means that the equation is unhelpful, but undefined means that the equation is violated. Hence any luminal particle must have zero rest mass.

Light has no rest mass, but it is possible to think of it as having a 'relativistic mass' by applying $m = E/c^2$ with no regard to the gamma factor. However, this does not mean that light has mass in any meaningful way - it is simply a measure of how much mass a stationary particle would give up to be entirely converted into the light.

TL;DR: Light has no rest mass, but we can assign it a 'mass' based on $E=mc^2$. However, it is not 'real mass' in the sense that it cannot be used to accurately calculate the mechanical properties of photons in the way you'd expect.

Side note: Most physicists prefer the more complete formula $E^2 = p^2c^2 + m_0^2c^4$, where $E$ is the particle's energy, $p$ is the particle's momentum, and $m_0$ is the particle's rest mass. For a stationary, massive particle, $p=0 \Rightarrow E^2 = m_0^2c^4 \Rightarrow E = m_0c^2$, and for a massless particle, $E^2 = p^2c^2 \Rightarrow E = pc$, which is the fundamental relationship for light.