Question

Question #c64a4

Answer 1

Einstein's famous equation will show that photons have momentum.

Explanation:

Not many people know that the full simplified equation is `E = sqrt((mc^2)^2 + (pc)^2)`

E = energy
m = mass
c = speed of light (`3` x `10^8` m/s
p = momentum

Now because photons have no mass the `mc^2` falls away because:

`mc^2` = `0` x (`3` x `10^8)^2` = 0

So you just left with:

E = `sqrt((pc)^2)`
E = `pc`

And because the photon has a certain amount of energy depending on the specific wavelength of the light, one can say:

p = `E/c`

Thus photons have momentum.

Hope I helped :)

Answer 2

The formula is only correct in classical physics, therefore it cannot be used to attain a momentum for a particle traveling at the speed of light. Using relativistic arguments, we find `p=(hf)/c`.

Explanation:

In special relativity, the path of a massive particle (that therefore does not travel at the speed of light) in spacetime can be parameterized by the proper time of the particle. This is analogous to the pathlength parameterization of a path in space. This gives a path as a function of the proper time `tau`, so `r^mu(tau)`. (The `mu` is a tensor index and is used to indicate that `r` is a vector in Minkowski space.)

We can now define a four-vector momentum for these massive particles (particles with mass m).
`p^mu=m(dr^mu)/(d tau)`

Of course `tau` is not the only possible parameterization of the path, we can also parameterize this by the time `t` measured by an observer. For this we note that
`tau=t/gamma=tsqrt(1-((v(t))/c)^2)`
where `v` is the relative velocity (as scalar) of the particle measured by the observer and `c` the speed of light.

Now we can use this to write down a more useful equation for the momentum:
`p^mu=mgamma(dr^mu)/(dt)=(mv^mu)/sqrt(1-(v/c)^2`
Here `v^mu` is the relative four-vector velocity measured by the observer.

Now if we think about massless particles, this definition makes little sense. Massless particles travel at the speed of light. Particles that travel at the speed of light cannot be parameterized using proper time, since the notion of passing time doesn't really hold at the speed of light. You could try to use the second equation, but this results in having a `0/0` term, which doesn't help either.

Now there are two ways you could go, you could say that the momentum is undefined for massless particles, or you could try and generalize the definition used before to a definition that works for both massive and massless particles, but gives you the same result for massive particles.

The latter is of course more satisfying, so that's what we're going to try. To do this, we're going to link four-momentum to energy.

Suppose we have a massive particle traveling at a constant velocity `vec(v)=(v_x,v_y,v_z)` measured from a certain inertial frame. This gives us:
`r^mu=(ct,v_xt,v_yt,v_zt)`
Therefore:
`p^mu=m/sqrt(1-(v_x^2+v_y^2+v_z^2)/c^2)(c,v_x,v_y,v_z)`.

Now it is interesting to calculate `||p^mu||^2`. For this we note that `p^mu` is a four-vector in Minkowski space where very much like `Deltas^2=-c^2Deltat^2+Deltax^2+Deltay^2+Deltaz^2`, we have:
`||p^mu||^2=-(m/sqrt(1-(v_x^2+v_y^2+v_z^2)/c^2))^2(c^2-v_x^2-v_y^2-v_z^2)=-(mc)^2/(1-(v_x^2+v_y^2+v_z^2)/c^2)(1-(v_x^2+v_y^2+v_z^2)/c^2)=-m^2c^2`

Now we can compare this to the energy of a particle, first the rest energy `E_0=mc^2`, so `||p^mu||^2=-E_0^2/c^2`. It is also worth noting that when one only looks at the temporal component, the first component of `p^mu`, often written down as `p^0`, we see:
`p^0=mgammac=E/c`, where `E` is the total energy.

Note that `||p^mu||^2` takes into account both the space and time part of the momentum, if you take only the spatial part, you end up with something more like the classical momentum:
`p=sqrt(p_x^2+p_y^2+p_z^2)=mgammav`.
This is the quantity that we talk about when we say momentum, just like in classical physics. However this still doesn't give us a momentum for massless particles.

Since `||p^mu||^2=-(p^0)^2+p^2`, we find
`p^2=||p^mu||^2+(p^0)^2=1/c^2(E^2-E_0^2)`.

This last equation is something we can talk about with massless particles, since `E_0=mc^2=0`, we have `p^2=E^2/c^2`, so `p=E/c`. The energy of photon is known, and is given by `E=hf`, where `h` is Planck's constant and `f` the frequency of light. Therefore `p=(hf)/c`.