# Relativity and Momentum of photons

I read somewhere that as the speed of a particle approaches c, its mass tends to infinity. Light is composed of particles (photons). If that is the case, then why don't we feel an enormous weight whenever light falls on us?

• I would like to simply comment that the statement "a particle's mass tends to infinity as its speed approaches c" is a relic of one of the worst conventions ever created by modern physics, the "relativistic mass." This mass is defined as the Lorentz factor times the rest mass. It is far, far better to think of mass as an invariant quantity and simply write γm everywhere, for the invariant mass m. You will then not have the problem you asked about. – flevinBombastus Feb 24 at 15:32
• Strongly related or perhaps a duplicate: physics.stackexchange.com/questions/2229/… – dmckee Feb 24 at 23:03

as the speed of a particle approaches c, its mass tends to infinity.

While statements like this appear frequently in the literature (pop-sci and textbooks),
it needs more modern terms and a more precise description of the limiting process.

Consider this energy-momentum diagram of a particle with rest-mass $$M$$.

The tip of the 4-momentum (at P) of a massive particle lies on a particular hyperbola [called the mass-shell], where the rest-mass is constant.

The tip of the 4-momentum of a massless particle, like the photon, lies on the null-cone [along the cyan-dotted line].

To get from this massive particle with rest-mass $$m=M$$ to a massless particle $$m=0$$,

• one could try to "make the particle move faster" by following the "fixedm" sequence along its mass-shell, while keeping its rest-mass constant.
This is actually your first sentence.
As others have mentioned, "mass tends to infinity"
implicitly uses an archaic and deprecated term "relativistic mass"
(which is perhaps-mistakenly attributed to Einstein,
although Einstein distanced himself from it ---- [see ref]).
A modern interpretation of "[relativistic] mass tends of infinity"
is that the "relativistic-energy tends to infinity",
which amounts to increasing the upward component in the plot above
(say by following the fixedm process).

But this won't achieve a massless particle because you are always on this mass-shell (keeping rest-mass $$m$$ constant with value $$M$$), which asymptotically approaches the null-cone (in any frame)
but never reaches the null-cone [where $$m=0$$].

• one could try to follow "fixedE" to the null-cone, which is sequence of faster particles with decreasing mass but increasing relativistic-momentum, but all with the same relativistic-energy in this frame.
In this limiting process, one reaches the null-cone and results in a massless particle with the same relativistic energy as the original massive particle.

• one could try to follow "fixedp" to the null-cone, which is sequence of faster particles with decreasing mass and decreasing relativistic energy, but all with the same relativistic-momentum in this frame.
In this limiting process, one reaches the null-cone and results in a massless particle with the same relativistic momentum as the original massive particle.

• or one could try some other sequence with varying velocity $$v$$ and varying $$E$$, $$p$$, and $$m$$ to reach the null-cone.

So, I think that one should really spell out the limiting procedure, accompanied with a physical model of how to take that limit.

ref:
Einstein Never Approved of Relativistic Mass
The Physics Teacher 47, 336 (2009); https://doi.org/10.1119/1.3204111
Eugene Hecht

The Concept of Mass in the Einstein Year
https://arxiv.org/abs/hep-ph/0602037
L.B. Okun

On the Abuse and Use of Relativistic Mass
https://arxiv.org/abs/physics/0504110
Gary Oas

as the speed of a particle approaches c, its mass tends to infinity.

This statement is true only for particles with a non-zero rest mass $$m_0$$. If such a particle moves with velocity $$v$$ it will have the total mass $$m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}} \tag{1}$$ From this formula you can indeed see, that for $$v$$ approaching the speed of light $$c$$, the mass $$m$$ goes to $$\infty$$.

But for photons the situation is different, because they have rest-mass $$m_0 = 0$$. Then equation (1) yields $$m = \frac{0}{0}$$ which is mathematical nonsense.

But you can at least rearrange equation (1):

$$m \sqrt{1 - \frac{v^2}{c^2}} = m_0$$

For photons you can set $$m_0 = 0$$ and $$v = c$$, and you get

$$m \cdot 0 = 0$$

This simply means that a photon can have any mass $$m$$. But it does not mean $$m = \infty$$

• This answer is largely misleading. Photons don't have any mass, i.e., their mass is simply zero. It is totally incorrect to say that they can have any mass $m$. Also, the kind of "re-arrangement" you are proposing is not allowed. First and foremost, even $m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$ is not a legal formula. The concept of relativistic mass is banned in physics for good. [...] – Feynmans Out for Grumpy Cat Feb 24 at 15:41
• [...] Nonetheless, for massive particles, you can write $E=\frac{m_0c^2}{\sqrt{1-\frac{v^2}{c^2}}}$ (where I shouldn't be writing $m_0$ to denote the invariant (rest) mass, I should only be writing $m$ as that is the only legal mass in physics). But, this is strictly a formula only for massive particles. For massless particles, you simply cannot write such a formula. The "analogous" formula for massless particles is simply $E=pc$ with no reference to mass whatsoever. – Feynmans Out for Grumpy Cat Feb 24 at 15:41
• @DvijMankad I agree that the old nomenclature of rest mass and relativistic mass has been abandoned in physics by now, and for a good reason. But nevertheless it still persists in popular science literature. The statement quoted by the OP is obviously using this old wording. That's why I continued to use this wording, to solve his question. Using the new meaning of mass here, in my opinion would have added more confusion than enlightening. – Thomas Fritsch Feb 24 at 15:49