Four-momentum and Relativistic Lagrangian

Conservation of Energy & Momentum

In an inertial frame {$S$}, let $E$ be the total energy and $(p_x,p_y,p_z)$ be the relativistic momentum of a massive object. {$S'$} is another inertial frame moving at speed $v$ relative to {$S$} on the x-direction. Let $$\gamma \equiv \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$ $\gamma$ is a constant as {$S$} and {$S'$} are inertial frame. On the x-direction in {$S$}, $$\begin{align} \frac{dx'}{d\tau} &= \frac{d}{d\tau}(\gamma(x - vt)) \\ &= \gamma \frac{dx}{d\tau} - \gamma v \frac{dt}{d\tau} \\ mc\left(\frac{dx'}{d\tau} \right) &= \gamma c\left( m\frac{dx}{d\tau} \right) - \gamma \left(\frac{v}{c} \right) \left(mc^2\frac{dt}{d\tau}\right) \end{align}$$ As time in observing frame is related to proper time by $$dt = \frac{d \tau}{\sqrt{1-\frac{v^2}{c^2}}}$$ where $u$ is the speed of the object. Thus, $$\frac{dt}{d\tau} = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \equiv \gamma_o$$ in which $\gamma_o$ may varies with time if the object is accelerating. So, $$\frac{dt}{d\tau}(mc^2) = \gamma_o mc^2 = E$$ Therefore, $$\bbox[5px,border:2px solid #666] { p_x'c = \gamma p_xc - \beta\gamma E }$$ For y-direction, $$\begin{align} \frac{dy'}{d\tau} &= \frac{dy}{d\tau} \\ m \left(\frac{dy'}{d\tau} \right) c &= m\left(\frac{dy}{d\tau}\right)c \\ p_y' c &= p_y c \end{align}$$ Similarly, for z-direction $$p_z' c = p_z c$$ For energy, $$\begin{align} (E')^2 &= (mc2)^2 + (p'c)^2 \\ &= (mc^2)^2 + (\gamma p_x - \gamma \beta E)^2 + (p_y c)^2 + (p_z c)^2 \\ &= E^2 - (p_x c)^2 + ((\gamma p_x c)^2 + (\gamma \beta E)^2 - 2 \gamma^2\beta p_x c) \end{align}$$ As $\beta = \frac{v}{c}$, $$(\gamma \beta E)^2 = \frac{\frac{v^2}{c^2}E}{1-\frac{v^2}{c^2}} = -E^2 + \gamma^2 E^2$$ Substitute back, $$\begin{align} E'^2 &= \gamma^2(p_x^2c^2 + E^2 -2vp_x) - p_x^2c^2 \\ &= \frac{1}{1-\frac{v^2}{c^2}}(p_x^2c^2 + E^2 -2vp_x - p_x^2c^2 + v^2p_x^2) \\ &= \frac{(E - \frac{v}{c}p_xc)^2}{1-\frac{v^2}{c^2}} \end{align}$$ Taking square root, we have $$\bbox[5px,border:2px solid #666] { E' = \gamma (E - \beta p_xc) }$$ When there is no net force in {$S$}, $$\begin{cases} \sum_i \Delta p_{i\, x} = 0 \\[2ex] \sum_i \Delta p_{i\, y} = 0 \\[2ex] \sum_i \Delta p_{i\, z} = 0 \\[2ex] \sum_i \Delta E_i = 0 \\[2ex] \end{cases}$$ Then in {$S$}, $$\begin{cases} \sum_i \Delta p_{i\, x}' = \gamma c\sum_i \Delta p_{i\, x} - \beta\gamma \sum_i \Delta E_i = 0 \\[2ex] \sum_i \Delta p_{i\, y}' = \sum_i \Delta p_{i\, y} = 0 \\[2ex] \sum_i \Delta p_{i\, z}' = \sum_i \Delta p_{i\, z} = 0 \\[2ex] \sum_i \Delta E_{i}' = \gamma (\sum_i \Delta E_i - \beta c\sum_i \Delta p_{i\, x}) = 0 \\[2ex] \end{cases}$$ Hence, momentum and energy are also conserved in other inertial frames.

Four Momentum

The momentum and energy under Lorentz transformation are given by $$\begin{cases} E' = \gamma E - \beta\gamma p_xc \\[2ex] p_x'c = \gamma p_xc - \beta\gamma E \\[2ex] p_y' c = p_y c \\[2ex] p_z' c = p_z c \\[2ex] \end{cases}$$ One may express in Lorentz transformation matrix $${ \begin{bmatrix} E'/c \\ p_x' \\ p_y' \\ p_z' \end{bmatrix} = \begin{bmatrix} \gamma &-\beta\gamma & 0 & 0 \\ -\beta\gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} E/c \\ p_x \\ p_y \\ p_z \end{bmatrix} }$$ As the vector $$\begin{bmatrix} E/c \\ p_x \\ p_y \\ p_z \end{bmatrix}$$ satisfies the Lorentz transformation matrix, it is defined as the four-momentum. $$p = \begin{bmatrix} p^0 &p^1 & p^2 & p^3 \\ \end{bmatrix} = \begin{bmatrix} E/c &p_x & p_y & p_z \\ \end{bmatrix}$$ The inner product of the four-momentum $$\begin{bmatrix} E/c & p_x & p_y & p_z \end{bmatrix} \begin{bmatrix} E/c \\ p_x \\ p_y \\ p_z \end{bmatrix} = - \frac{E^2}{c^2} + (p)^2 = - m^2c^2$$ which is a constant. Thus, four-momentum inner product is an invariant in Lorentz transformation.

Derivation of Relativistic Lagrangian

An action is defined as $$S \equiv \int^{t_2}_{t_1} Ldt$$ in which $L$ is the Lagrangian satisfying the relation $$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q_j}}\right) = \frac{\partial L}{\partial q_j}$$ By Noether's theorem, if $S$ has a symmetry under transformation of $s$, then the quantity $$\sum_i\frac{\partial L}{\partial \dot{q_i}}\epsilon(q)$$ is conserved.
Let the transformation be translation. As the transformation $\epsilon(q) = q' - q$ is arbitrary and does not depend on time, $\frac{\partial L}{\partial q}$ must be constant over time under the transformation. As $p^\mu$ is conserved under translational transformation in an inertial frame, where $\mu = 0, 1, 2, 3$. The Lagrangian can be defined using $p^\mu$ $$p^\mu = \frac{\partial L}{\partial \dot{x^\mu}}$$ As $$\delta S = \sum_\mu\frac{\partial L}{\partial \dot{x^\mu}}\delta x^\mu =\sum_\mu p^\mu \delta x^\mu$$ and $$\delta S =\sum_\mu \frac{\partial S}{\partial x^\mu}\delta x^\mu$$ we have $$p^\mu = \frac{\partial S}{\partial x^\mu}$$ Then, we can find $S$ $$\begin{align} S &= \sum_\mu \int p^\mu \,dx^\mu \\ &= \sum_\mu \int p^\mu \frac{dx^\mu}{d\tau}\left(\frac{d\tau}{dt}\right) dt \\ &= \int p^2 \left(\frac{1}{m\gamma}\right) dt \\ &= - \int mc^2 \sqrt{1-\frac{v^2}{c^2}}dt \end{align}$$ Thus, the relativistic Lagrangian is $$\bbox[5px,border:2px solid #666] { L = -mc^2 \sqrt{1-\frac{v^2}{c^2}} }$$

$\gamma$ can be expressed in terms of differentiating proper time $\tau$ with respect to local time $t$ $$\frac{d\tau}{dt} = \sqrt{1-\frac{v^2}{c^2}}.$$ By $ds^2 = -c^2d\tau^2$, $$d\tau = \frac{1}{c}\sqrt{\eta_{\alpha\beta}dx^\alpha dx^\beta} = \frac{1}{c}\sqrt{\eta_{\alpha\beta}\frac{dx^\alpha}{dt} \frac{dx^\beta}{dt}}dt.$$ Thus, the Lagrangian can be expressed as $$L = -mc\sqrt{\eta_{\alpha\beta}\frac{dx^\alpha}{dt} \frac{dx^\beta}{dt}}.$$ Upon integration, it is the distance (square root of line element) of the Minkowski space. Distance is an extrema, the shortest path between two points. We see that we may use definition of distance to define the Lagrangian.