Show that \(E^{2}-p^{2} c^{2}=E^{2}-p^{2} c^{2},\) that is, that \(E^{2}-p^{2} c^{2}\) is a Lorentz invariant. Hint: Look at derivation showing that the space-time interval is a Lorentz invariant.

Lorentz transformations are mathematical transformations that relate the space and time coordinates of an event in one inertial frame of reference to those in another inertial frame of reference moving with a constant relative velocity. The Lorentz transformations for space and time coordinates are given by: \begin{align} x' &= \gamma(x - vt) \\ t' &= \gamma(t - \frac{vx}{c^2}) \end{align} where \(x\) and \(t\) are the space and time coordinates in the original frame, \(x'\) and \(t'\) are the space and time coordinates in the moving frame, \(v\) is the relative velocity between the frames, \(c\) is the speed of light, and \(\gamma\) is the Lorentz factor, defined as \(\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\).

The energy and momentum of a particle can be combined into a four-momentum vector \(P = (E/c, \vec{p})\), where \(E\) is the energy of the particle, \(c\) is the speed of light, and \(\vec{p}\) is the momentum vector of the particle. The squared magnitude of this four-vector is given by: \begin{equation} |P|^2 = \frac{E^2}{c^2} - |\vec{p}|^2 \end{equation}

Under a Lorentz transformation along the \(x\) axis, the energy and momentum of the particle in the moving frame, \(E'\) and \(\vec{p}\,'\), are related to those in the original frame via: \begin{align} E' &= \gamma\left(E - \frac{v}{c^2}pc_x\right) \\ p'_x &= \gamma(p_x - v\frac{E}{c^2}) \\ p'_y &= p_y \\ p'_z &= p_z \end{align} where \(pc_x\) is the \(x\) component of the momentum vector \(\vec{p}\), and \(p_x\), \(p_y\), \(p_z\) are the components of the momentum vector in the original frame.

Now we want to show that \(E^2 - p^2c^2 = E'^2 - p'^2c^2\). Using the expressions for \(E'\) and \(\vec{p}\,'\) from step 3 and the squared magnitude of the four-vector from step 2, we compute: \begin{align} E'^2 - p'^2c^2 &= \left(\gamma\left(E - \frac{v}{c^2}pc_x\right)\right)^2 - c^2\left(\gamma^2(p_x - v\frac{E}{c^2})^2 + p_y^2 + p_z^2\right) \\ &= \gamma^2\left(E^2 - 2E\frac{v}{c^2}pc_x + (\frac{v}{c^2}pc_x)^2 - (p_x^2 - 2p_xv\frac{E}{c^2} + v^2\frac{E^2}{c^4})c^2 - p_y^2c^2 - p_z^2c^2\right) \\ &= \gamma^2\left(E^2 - p_x^2c^2 - p_y^2c^2 - p_z^2c^2\right) \\ &= \gamma^2(E^2 - p^2c^2) \end{align} Since \(\gamma^2(E^2 - p^2c^2) = E^2 - p^2c^2\), this expression is invariant under Lorentz transformations, and we have shown that \(E^2 - p^2c^2\) is a Lorentz invariant.