Equivalence between $t$ and $u$ channels Reading about QFT diagrams, I've seen examples like Bhabha scattering where the channel $u$ wasn't necessary due to the final states are distinguisable for being made of the different particles and having $t$ was enough (see answer to: Why does not Bhabha scattering contain u-channel diagram?). 
But my question is: when I have to decide between $t$ or $u$ diagram, which one should I pick? Would I obtain the same thing? 
 A: If you have:
$$ k + p \rightarrow k' + p' $$
where I'm just labeling their particles by the momenta, then the Mandelstam variables are:
$$ s = (k_{\mu} + p_{\mu})^2 $$
$$ t = (k_{\mu} - k'_{\mu})^2 = (p_{\mu} - p'_{\mu})^2$$
$$ u = (k_{\mu} - p'_{\mu})^2 = (p_{\mu} - k'_{\mu})^2$$
If $k$ and $p$ annihilate, as in:
$$ e^++e^-\rightarrow \mu^++\mu^-$$
then $s$ is invariant mass of the intermediate state.
This is also true if the process is:
$$ e^++e^-\rightarrow e^++e^-$$
However, you don't really know if the particles annihilated. They may have just scattered, in which case $t$ is the invariant mass of the exchanged virtual quanta.
Note the symmetry between $k$ and $p$ here: you can't really say the positron emitted a virtual photon and the electron absorbed it, because the electron may have emitted it and the positron absorbed it. The Feynman diagram covers both cases (note: there are separate diagrams for $\gamma$ exchange and $Z^0$ exchange, but $t$ doesn't know the difference and you have to add the amplitudes of all the diagrams).
But I digress: because all positrons are indistinguishable, you don't know if you detected the same one from your beam, or a "new" one that was created from the decay of the $s$-channel intermediate state. Hence: you must add the s and t channel amplitudes.
There is also scattering without the possibility of annihilation, e.g. inelastic electron-proton scattering:
$$ e^- + p \rightarrow e^- + X $$
Here the only channel is the t-channel (ignoring the case where $X$ contains an electron--which I've never heard of). Note that now it is really tempting to say the electron emitted a virtual photon that was absorbed by the proton causing to to break up, but that is not correct. The virtual quanta was exchanged and that's it. (That $t < 0$ means it's space-like, so time-ordering its existence doesn't really make sense).
Finally, consider scattering identical particles, say a beam of ultra-relativistic electrons with momentum $k_{\mu} = (E/c, 0,0, E/c) $ off stationary electrons in a target piece of foil, with $p_{\mu} = (mc, 0, 0, 0)$: 
$$ e^- + e^- \rightarrow e^- + e^- $$
Let's say you put a spectrometer at 90 degrees and tune it to $k'$. When you detect a scattered electron you do not now where it came from. Was it from the beam:
$$ t = (E/c-k',-k', 0, E/c)^2 $$
or from the target:
$$ u = (mc-k',-k', 0, 0)^2 $$
The correct answer is both. Any electron you detect is part from the beam and part from the target: this is the double-slit experiment, where the particle goes through both slits.
Hence you have to add the amplitudes for the t and u channel diagrams.
