While I was solving a problem from my homework I arrived at the conclusion that a photon has zero momentum. I know the most probable thing is that I'm wrong and I did a bad calculation, yet in what I'm interested in is what would it mean for a photon to have zero momentum. My intuition says that in most have an infinite wavelength, therefore zero frequency, so there wouldn't be any photon. So (in my mind) in order for a photon to exist, it must have momentum. I wonder what are your thoughts on this.
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Ok, so the problem was the following:
Suppose we have a collision between a positron (A) of mass $m$ and an electron (B) where the positron moves along the x-axis and the electron is at rest. When these collide they annihilate generating two photons C and D, where $\angle(\vec{p_A},\vec{p_C}) = \varphi$. Write $E_C$ in terms of $E_A$, $m$ and $\varphi$. (I'm translating because my mother language isn't english, so sorry if my translation is clunky.)
I'm going to write my attempt at solving this problem, showing where I made my mistake and why I wrongly arrived at the conclusion that a photon can have zero momentum.
I proved that $p_D^2 = p_A^2 + p_C^2 - 2p_A p_C \text{cos} \varphi$. First we have two energies, those being the initial one and the final one. $E_i = E_A + E_B$ while $E_f = E_C + E_D$, energy is conserved $\Rightarrow E_i = E_f$ $\Rightarrow E_A + E_B = E_C + E_D \quad \therefore E_C = E_A + E_B - E_D$. Momentum is also conserved $\Rightarrow \vec{p_A} = \vec{p_C} + \vec{p_D} \Rightarrow \vec{p_C} = \vec{p_A} - \vec{p_D}$ $$p_D^2 = p_A^2 + ||\vec{p_A} - \vec{p_D}||^2 - 2p_A ||\vec{p_A} - \vec{p_D}|| \text{cos}\varphi$$ Here's where I made my mistake, I assumed that a) $a \cdot b = ||a|| ||b||$ b) $||a-b|| = ||a|| - ||b||$ and c) $||a|| = ||b|| \Rightarrow a = b$ (both $a$ and $b$ are vectors). I'll proceed with my attempt at the solution. $$\Rightarrow p_D^2 = p_A^2 + p_A^2 - 2p_A p_D + p_D^2 - 2p_A(p_A - p_D) \text{cos}\varphi$$ $$\Rightarrow 0 = p_A - p_D - p_A \text{cos}\varphi + p_D \text{cos}\varphi$$ $$\therefore p_A = p_D$$ $$\therefore p_C = 0$$ As you can see, with mistakes a), b) and c) I arrived at the conclusion that a photon can have zero momentum which of course is wrong, because in order to have zero momentum the photon's wavelength must be infinite (because of $p_\gamma = \frac{h}{\lambda}$), or the photon's frequency must be zero, which is another way to say that it isn't a photon, that (in this case) C is nothing.
At first I couldn't see a problem with my math, and I wondered if it was posible that a photon had zero momentum; I'm a student (I'm at my fifth semestre of studying physics), so I have no idea if my intuition was correct. Thus I decided to post my question to this site to see what others had to say about my reasoning and theirs, but as you can see, I saw that my math was wrong and arrived at the wrong conclusion.