# Why does photon emit a frequency when its rest mass is zero? [duplicate]

This one really boggles my mind.
In the equation $E=mc^2$ and $E=hv$, we can equate $hv=mc^2$. Then, $v/m=c^2/h=k$ (constant)
Therefore, $v=k.m$ So, $v$ is directly proportional to m.

Lets see the case of photons.

Photons have a rest mass =0, then its frequency should also be zero (from the above result). But this can't happen as photons emit frequency as they travel through space-time.

## marked as duplicate by ACuriousMind♦ quantum-mechanics StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jul 21 '17 at 9:16

You see, the equation $E=mc^2$ is only a special case of the original relativistic energy-momentum relation equation when the object has zero momentum, that is - its velocity is zero: $$E^2=(mc^2)^2+(pc)^2$$

Where $p$ is momentum so when a non zero rest mass object is at rest, $p=0$ and it simplifies to $E=mc^2$.

Now, for photons/light they have zero rest mass($m=0$) but they do have a value for momentum. So thus then the energy of a photon from the equation becomes: $$E=pc$$ So then the correct relation comes like: $$E=hf=pc$$ $$p=\frac{hf}{c}$$ Thus, $$p=\frac{h}{\lambda}$$ $$\lambda=\frac{h}{p}$$ Which is the De-Broglie relationship. So finally we can say that the photon can have energy even though it does not have mass, because it has momentum.

$E = mc^2$ is the rest energy, that is, the energy associated with its intrinsic mass. Photons have no rest energy as they have no mass, but that doesn't mean they don't have energy. All their energy is kinetic energy. So you cannot use just $E = mc^2$ here. The full equation that includes all energy, both the rest energy and kinetic energy, is

$$E^2 = m^2 c^4 + p^2 c^2$$

Note the quadratic: this is not a simple sum of energies. (What it really is is better seen if one takes it in Planck units: then it is $E^2 = m^2 + p^2$ which can be rearranged to $m^2 = E^2 - p^2$ which says the mass is the length of the space-time momentum of the particle. Photons have no space-time momentum. In this case their temporal momentum, which is their energy, must equal their spatial momentum. There is no excess energy, i.e. no rest mass. It's a really beautiful aspect of the geometry of Einstein's relativity, itself a wonderful thing to behold. It is just soo cool!) The first part is essentially the rest energy, the second part is essentially the kinetic energy (not entirely literally, due to the quadratic). If you take a photon, $m = 0$ but we still have $p^2 c^2$ left, and you take the square root of that puppydog and you get

$$E = pc$$

Now you can set that to $h\nu$

$$h\nu = pc$$

so

$$\nu = \frac{pc}{h}$$

If there's any momentum at all, and there must be for the photon to exist, there will be a nonzero frequency.

FWIW you can write down an equation that has a straight sum of rest energy ($E = mc^2$) and kinetic energy by a relativistic kinetic energy equation too that is analogous to Newton ($K =\frac{1}{2}mv^2$), but the latter becomes undefined for mass 0, since it is written in terms of velocity, and all photons have the same velocity ($c$). (Mathematically it gives the nonsense expression $0 \cdot \infty$.) So you need to use this as it works in all situations. The trick with photons is their kinetic energy is thus seen to be not dependent on velocity or mass, but momentum, and passing to the quantum theory, on frequency, or wavelength.