Question regarding the general form of four-vectors in particle physics

Why is there a $m_X c$ in the first element of a four-vector which (I thought) should have dimensions of energy, not momentum?

The momentum 4-vector of a particle of mass $m$ and three-velocity $mathbf{u}$ can be defined as:

$$P = (gamma m c, gamma m mathbf{u}) = left(frac{E}{c}, mathbf{p}right) tag{A}$$ or as

$$ P = (gamma m c^2, gamma m cmathbf{u}) = (E, mathbf{p}c) tag{B}$$

Note that convention B is just A multiplied by $c$; all components in A have dimensions of momentum, whereas all components in B have dimensions of energy.

Also, I've entered 3-vectors into these expressions as a shorthand to represent the 3 associated components, e.g. A should really be written:

$$P = (gamma m c, gamma m mathbf{u}) = (frac{E}{c}, p_x, p_y, p_z)$$

In the solution, they've used convention A, but you would get the same answer if you used convention B instead - it's just that all your equations would be multiplied by $c^2$ now. I suggest that you check this directly!

Instead, let's do a different check to see if we recover the $E^2 = |mathbf{p}|^2 c^2 + m^2 c^4$ relation using the invariance of the momentum four-vector norm for both cases. We're going to work out the norm squared explicitly in both cases in the rest frame of a particle of mass $m$ (so $gamma = 1$ and $mathbf{u} = 0$) and then equate them to a general expression in terms of $E$ and $mathbf{p}$.

Case A: $$m^2c^2 = frac{E^2}{c^2} - |mathbf{p}|^2 tag{1}$$ Case B: $$m^2 c^4 = E^2 - |mathbf{p}|^2 c^2 tag{2}$$

We see that equation 2 is just equation 1 multiplied by $c^2$ and that both are correct. This makes sense as convention B has just scaled the four-vector in convention A by a factor of $c$ - any equation involving the same relations between these four-vectors should then appropriately be scaled overall by some power of $c$ (which has no effect overall). The solution produces an equation involving two-fold products of these four-vectors, so using convention B will scale the equations by $c^2$.

To answer the final part of this question, you have an $m_X c$ because $E_X = m_X c^2$ in the rest frame of X.

It was my understanding that general four-vectors are written as $(E, p_x c, p_y c, p_z c)$. I thought that elements of four vectors must all have the same dimensions and that those dimensions are energy (as above).

I believe the first part of this question has now been addressed. However, I would like to add that the momentum four-vector (whichever convention you choose) is not the only four-vector that exists.

Actually, $X = (ct, x, y,z)$ is probably the most basic four-vector; this transforms as $X' = Lambda X$ under Lorentz transformations (where X' labels the transformed components of the same four-vector in the new inertial frame and $Lambda$ is the Lorentz transformation matrix).

Now, any 4-component object $A$ that transforms like $A' = Lambda A$ when $X$ transforms like $X' = Lambda X$ is a 4-vector. The elements of $A$ should all have the same dimensions, but they don't have to be energy.

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$$P_X=left(E_X,,boldsymbol{p_X}cright)=(m_Xc^2,0,0,0)$$ $$P_c=left(E_c,,boldsymbol{p_c}cright)=(E_c,{p_c}^xc,0,0)$$ So $$P_Xcdot P_c=E_cm_Xc^2-0=E_cm_Xc^2 tag{C}$$ substituting this result in (1): $$m_d^2c^2=m_X^2c^2+m_c^2c^2-2E_cm_Xc^2 tag{D}$$ $$implies E_cstackrel{color{red}{{?}}}{=}frac{m_X^2+m_c^2-m_d^2}{2m_X}$$ Well, this is definitely not the same answer as (A). So what am I missing?

Step C is correct, but step D is incorrect: under convention B, the norm squared of the momentum four-vector is also scaled by $c^2$. This is because in the rest frame of a particle of mass $m$, $P = (mc^2, mathbf{0})$ so the norm-squared is now $m^2 c^4$ (unlike $m^2 c^2$ under convention A).

If you use this norm, you will recover the correct answer - let's check:

$$P_d^2=P_X^2+P_c^2-2P_X cdot P_c tag{E}$$ is a relation between four-vectors and holds under both conventions.

Using the corrected norms and your expression for $P_X cdot P_c$ , we find:

$$m_d^2 c^4 = m_X^2 c^4 + m_c^2 c^4 - 2 E_c m_X c^2,$$ which is just

$$m_d^2c^2=m_X^2c^2+m_c^2c^2-2 E_c m_X$$ scaled by $c^2$ as expected.