Proof of forbidden free electron-photon absortion using spacetime diagram I'm trying to solve this exercise from the book "Modern Classical Physics" by Kip Thorne:

Show, using spacetime diagrams and also using frame-independent calculations, that the law of conservation of 4-momentum forbids a photon to be absorbed by an electron

I managed to prove this using coordinates: choose the Lorent's frame in which the electron is at rest, in this frame its 4-momentum is simply $\vec{p_e}=(m,0,0,0)$ and let the photon be moving from left to right, so its 4-momentum is $\vec{p_\gamma}=(\epsilon,\epsilon,0,0)$. Using conservation of 4-momentum, after the absortion the 4-momentum of the electron should be $\vec{p_e} = (m+\epsilon,\epsilon,0,0)$, but this generates a contradiction, because for any massive particle we have that $\vec{p}^2= -m^2$, but if $\vec{p_e}^2=-m^2-2m\epsilon-\epsilon^2 +\epsilon^2=-m^2$, we conclude that $2m\epsilon=0$, so either the mass of the electron is zero or the energy of the photon is zero, contradiction!
My problem is, how can I prove that this reaction is forbidden using a spacetime diagram? I tried to draw one, but I can't see where the contradiction will arise from it. 
 A: If you draw some electron worldline inside the past light cone suddenly transforming into a photon line (on the light cone) and another electron worldline inside the future light cone, you should notice that it doesn't satisfy both energy and momentum conservation (bold symbols are four-vectors):
$$\tag{1}
\boldsymbol{p}_A = \boldsymbol{p}_B + \boldsymbol{k}.
$$
This equation could be written as this:
$$\tag{2}
\boldsymbol{p}_A - \boldsymbol{p}_B = \boldsymbol{k}.
$$
The four-vector $-\, \boldsymbol{p}_B$ could be interpreted as an electron moving to the past, in the past light cone.  Now draw the worldlines in the total momentum frame.  You should notice that there's a problem!
Also, take the invariant square of equation (2) ($\circ$ is the four-vector scalar product):
$$\tag{3}
p_A^2 + p_B^2 - 2 \,\boldsymbol{p}_A \circ \boldsymbol{p}_B = k^2 \equiv 0.$$
Then $p_A^2 = p_B^2 = m^2$ (the electron stays an electron!) so (3) reduces to
$$\tag{4}
m^2 = \boldsymbol{p}_A \circ \boldsymbol{p}_B.
$$
This equation cannot be true, unless $\boldsymbol{p}_A \equiv \boldsymbol{p}_B$ (which implies $\boldsymbol{k} = 0$, i.e no photon!)
