You are gazing up at the sky one day when you see something crazy. Two identical spaceships are traveling towards each other with a velocity close to the speed of light, and they crash! One of the spaceships is made solely from anti-matter, therefore all matter is converted to photons (all with the same wavelength) in the collision.
Let's set this up the way we usually do:
- Planet frame (t,x): At rest on the planet
- Spaceship frame (t',x'): An onserver just behind and in the same frame as what we will call spaceship A.
We have one reasercher in each frame, trying to measure the wavelength of the photons.
Let's begin by finding the expressions for the momenergy four vectors for the two spaceships.
\(P_\mu(A) = (E_A,\vec{p}_A)=(\gamma_Am_A,\gamma_Am_A\vec{v}_A)\)
\(P_\mu(B) = (E_B,\vec{p}_B)= (E_A,-\vec{p}_A)= (\gamma_Am_A,-\gamma_Am_A\vec{v}_A)\)
where \(\gamma\) is the gamma factor, \(\vec{v}\) is the velocity and \(m\) is the mass of the spaceship, each corresponding to spaceship A or B, depending on their notation. The marked system would yield similar expressions, only for marked values (values in the spaceship frame). We know that we can double check that this is correct by transforming between the frames of reference, using the matrix-transformation \(P'_mu = c_{\mu\nu}P_\nu\).
If we want to write down the momenergy for photons we'll have to rething these expressions. Previously, we wrote expressions for the momenergy four-vectors using \(E=\gamma m\). But photons do not have mass, and so we would only get \(E=0\), which is unfortunate, as we know photons have energy. Instead, we can write
\(E_f = \frac{h}{\lambda_f}\)
where \(E_f\) is the energy of a photon, \(\lambda_f\) is its wavelength and \(h\) is Plancks constant, equal to \(2\pi\) when using natural units.
So that was energy. But we hit the same issue when adressing the momentum, \(\vec{p}=\gamma m\vec{v}\). We have a look at the length of the 4-vector momenergy \(P_\mu = (E,\vec{p})\):
\(P=\sqrt{P_\mu P^\mu}=\sqrt{E^2-p^2}\)
(this is the normal way of finding the length of a vector, only for a four-vector we get the negative sign under the root). By inserting our regular expressions, \(E = \gamma m\) and \(\vec{p} = \gamma m\vec{v}\), we get
\(P=\sqrt{(\gamma m)^2-(\gamma mv)^2}=\gamma m \sqrt{1-v^2}=m\) (Recall that \(\gamma = \frac{1}{1-v^2/c^2}\), where c=1)
By rewriting this we get an expression for the momentum of the photon, with m=0:
\(P=\sqrt{E_f^2-p_f^2}=m_f=0\)
\(p_f=\pm\sqrt{E_f^2}=\pm E_f\)
So the momentum of a photon is equal to its energy, and the direction of the photon decides weather this is positive or negative. Therefore we can write:
\(P^\gamma_\mu = (E_f,p_f,0,0) = (E_f,E_f,0,0)\)
for a photon only moving in the positive x-direction, where E is the energy of the photon.
If we now imagine that in the collision of the spaceships, only two photons were emmited in the opposite direction of each other, we could write
\(P_\mu(f_x)+P_\mu(f_{-x}) = P_\mu(A)+P_\mu(B)\)
as we have conservation of momenergy in the planet frame. We use the expressions for momenergy of the spaceships and of the photons, and get that
\(E_{f_x} + E_{f_{-x}} = 2E_A\) and \(E_{f_x} - E_{f_{-x}} = 0\)
which shows is that \(E_f = E_{f_x} = E_{f_{-x}} \). In other words the energy of both the photons is \(E_A\), which means they have the same energy as seen from the planet frame. This makes sense as all the mass in a collision between matter and antimatter is turned into energy.
Now we assume that these two photons are emmited with an angle \(\theta\) off the x-axis. We can use basic trigonometry and see that our new expressions for momenergy of a photon would become \(P^\gamma_\mu = (E_f,E\cos{\theta},E\sin{\theta},0)\). Through the same argument as before, the two photons must have the same energy, but opposite directions seen from the planet frame.
We can generalize this, and imagine that a lot of photons are emitted in all possible directions, but through conservation of momenergy, we have that for any photon emitted there is always another photon emitted in the opposite direction with the same energy- if not the energy would not be conserved.
We can use the fact that all photons are emitted with the same wavelength in order to find the energy of one single photon, which will be the same as the energy in the planet frame:
\(E_f = \frac{E_{tot}}{N}=\frac{2E_A}{N}\)
Where N is the total number of photons emmited in the explosion, and \(E_A=\gamma_Am_A\) (recall that this refers to spaceship A). As we also have an expression for the wavelength, \(\lambda = h/E\), we find that the wavelegth of a single photon (and thus the whole explosion) is 164 meters? Something has gone wrong here, perhaps with the conversion of units? Unfortunately we cannot seem to figure out exactly what at the moment, but it's safe to say that realistic wavelengths should be between 380 and 700 nm, so we are pretty far off.
We'd spend more time solving this, but unfortunately time is not something we have a lot of, and we must press onwards!