Why does four-momentum have the same transformation matrix as spacetime coordinates?

Question

I will outline my question in 1+1D for brevity. We can passively transform our coordinate system using a Lorentz boost; $\Lambda^{\bar{\nu}}_{\mu}x^{\mu}=x^{\bar{\nu}}$. I've seen that, by stipulating that the speed of light is the same to all observers, rather than time, we can describe this transformation using the matrix:

$$\begin{bmatrix} \bar{ct} \\ \bar{x} \\ \end{bmatrix}=\begin{bmatrix} \gamma & -\beta\gamma \\ -\beta\gamma & \gamma\\ \end{bmatrix}\begin{bmatrix} ct \\ x \\ \end{bmatrix} $$

However, what's not obvious to me is why transformations $\Lambda^{\bar{\nu}}_{\mu}p^{\mu}=p^{\bar{\nu}}$ have the same form.

$$\begin{bmatrix} \bar{E} \\ \bar{p_{x}} \\ \end{bmatrix}=\begin{bmatrix} \gamma & -\beta\gamma \\ -\beta\gamma & \gamma\\ \end{bmatrix}\begin{bmatrix} E \\ p_{x} \\ \end{bmatrix} $$ I understand that all four-vectors transform under Lorentz transformations. The part I don't understand is why the Lorentz matrix has the same form, even though we're transforming a different basis. Intuitively it doesn't make sense to me why energy and momentum are transforming as though they were spacetime coordinates.

Answer 1

four-momentum is literally the first (proper) time derivative of position, multiplied by mass, so it is a vector with the same transformation rule.

Answer 2

Matematically, it is only the derivative at both sides with respect to the invariant parameter $\tau$, and multiplication by another scalar invariant $m$. As the Lorentz matrix is a constant, that is, it doesn't depend on $\tau$, the first equation leads to the second one.

Answer 3

$p^\mu$ has to transform as $p^\mu \to \Lambda^\mu_{~~\nu}p^\nu$ if you want quantities like $$x^\mu p_\mu$$ to be lorentz invariant. For example if you want to measure the mass of a particle you might choose $$x^\mu = (1,\mathbf{0})$$ and $$p^\mu = (m,\mathbf{0}),$$ so$$x^\mu p_\mu = m$$ If this wasn't boost invariant then you'd measure a different mass for the particle in different frames.

Answer 4

As a supplement to the earlier answers, one can form a scalar $S=(tE-\vec x\cdot \vec p)$ which has the units of action. Regarding this as the [Lorentz-metric] dot-product of two 4-vectors, this implies $(E,\vec p)$ also transforms a 4-vector.