Do Photons have Mass?


Do Photons have Mass?

As an admin of the Quantum Physics facebook group, I can assure you with great certainty that this is the most popular question we get. In this article, we shall investigate this question to provide a reasonable answer.

One of the ways some people attack this question is by arguing that since gravitational attraction is caused by mass, as Newton's equation shows: $$\mathbf{F}=G\dfrac{m_1\cdot m_2}{r^2}$$ and that since light is bent due to gravity, then light must have mass. But that's actually false, Newton's equation is only an approximation and so: $$\mathbf{F}\approx G\dfrac{m_1\cdot m_2}{r^2}.$$ Furthermore, the cause of gravity is not mass but energy and momentum. And so this argument is false.

From special relativity: $$E^2=(mc^2)^2+(pc)^2$$  which can be written as: $$E^2-(pc)^2=(mc^2)^2\tag{$\star$}$$ where $E$ denotes energy, $m$ rest mass, $p$ is the momentum and $c$ is the speed of light in vacuo which is approximately $3\cdot10^8\rm m/s$.
And we know that the momentum of a photon is given by $\frac{hf}{c}$ and its energy by $E=hf$ where $h$ is Planck's constant (approx $6.62 \times 10^{-34}\rm m^2 kg / s$) and $f$ is its frequency. Therefore, the following holds: $$E=hf\quad p=\frac{hf}c\implies E=pc.$$
Thus, $E^2=(pc)^2\Rightarrow E^2-(pc)^2=0$. We put this result in $\text{($\star$)}$ to get: $$0=E^2-(pc)^2=(mc^2)^2.$$ Solving for $m$ gives $m=0$. So photons are massless. $\square$

to be continued...

3 comments:

  1. But what about the momentum... The momentum (p) is often defined by:
    p=mv
    where m is mass and v is velocity.
    So how can the photon have momentum but no mass?
    (how to write that fancy? for example:
    E^2=(mc^2)^2+(pc)^2
    which you write in a much fancier way than that)

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    Replies
    1. Hello,

      The momentum $\bf p$ is defined by $\mathbf{p}=m\mathbf{v}$ only for massive particles in non-relativistic speeds - i.e. speeds where we don't consider relativistic effects, where the gamma factor $\gamma=\sqrt{1-\frac{v^2}{c^2}}\approx 1$. You would be asking yourself, where did we get this definition? It's simply comes from the mass-momentum-energy relation given by Special Relativity: $$E^2=(mc^2)^2+(\mathbf{p}c)^2$$ We solve for $\mathbf{p}$ and we get: $$\boxed{\mathbf{p}=\dfrac{\displaystyle\sqrt{E^2-(mc^2)^2}}{c}.}$$
      For a particle who is massless, like photons, we have that $E^2-(mc^2)^2=E^2$ and thus $\mathbf{p}$ is $E/c=hf/c$.

      ~~Jeff

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