Special Relativity/Relativistic dynamics

Relativistic Dynamics

To derive quantities in relativistic dynamics, it is perhaps most straightforward to use Lagrangian mechanics and Principle of Least Action.

Introduction

Recall the Principle of Least Action, which states that a mechanical system should have a quantity called the action $S$. Such quantity is minimised (in other words, $\delta S=0$) for the actual motion of the system.

The action of a relativistic system should be:

• a scalar: that means Lorentz transformations will not affect this quantity
• an integral of which the integrand is a first-order differential

The only quantity that satisfies the two criteria above is the space-time interval $ds$, or a scalar multiple thereof. In short, we can conclude that the action must have the following form:

$S=\kappa \int ds$

Recall the definition of the space-time interval $ds$:

$ds=\sqrt{c^{2}d{t}^2-d{x}^2-d{y}^2-d{z}^2}$

After pulling out $cdt$ from the square root, and noting that $\frac{d{x}^2+d{y}^2+d{z}^2}{d{t}^2}=v^2$, we have:

$ds=cdt\sqrt{1-v^2/c^2}$

Hence:

$S=c\kappa \int \sqrt{1-v^2/c^2}dt$

Now, the action integral can be expressed as a time integral of the Lagrangian between two fixed time:

$S=\int Ldt$

Then we can just read off the Lagrangian:

$L=c\kappa \sqrt{1-v^2/c^2}$

What is remaining now is determining the expression for $\kappa$. At this point we should note that for low velocity $v$, this relativistic expression for the Lagrangian should resemble that of the classical free Lagrangian, $L=\frac{1}{2}m{v}^2$. To compare the two Lagrangian, we perform a Taylor expansion on the square root:

$L=c\kappa (1-\frac{v^2}{2c^2}+o(v^2))$

The first term, $c\kappa$, is a constant. That will not affect the equations of motion (see Euler-Lagrange Equation, for example). The second term, after expanding out, is $-\kappa \frac{v^2}{2c}$. To reduce to the classical limit, we can put $\kappa =-mc$.

Therefore, the relativistic Lagrangian is:

$L=-m{c}^2\sqrt{1-v^2/c^2}$

Momentum and Energy

Recall that the canonical momentum is given by $p_i=\frac{\partial L}{\partial v_i}$, and rewriting $v^2=v_i{v}^i$ (Einstein summation notation employed), we have:

$p_i=\frac{1}{2}\frac{2m{v}_i}{\sqrt{1-v^2/c^2}}=\gamma m{v}_i$

where $\gamma =\frac{1}{\sqrt{1-v^2/c^2}}$.

With the canonical momenta defined, we can now construct the Hamiltonian:

$H=\mathrm{\Sigma }{p}_i{v}_i-L=\gamma m{v}^2+m{c}^2\sqrt{1-v^2/c^2}$

Since the Hamiltonian is invariant with time, it represents the energy.

After some algebra, we obtain:

$H=\gamma m{c}^2$

For an object at rest, $\gamma =1$, and the equation above reduces to the famous mass-energy equivalence relation:

$E=m{c}^2$

Two additional expressions relating momentum and energy can be derived from the expressions above:

$E^2=p^2{c}^2+m^2{c}^4$

$\overrightarrow{p}=E\frac{\overrightarrow{v}}{c^2}$

Mass

At the original version of Special Theory of Relativity as proposed by Einstein, the relativistic mass is given as follows:

$m=\gamma m_0$

However, for more contemporary interpretations of Special Relativity, mass, $m_0$ is considered an invariant quantity for all reference frames and $\gamma m_0$ is used instead of $m$. This convention allows the physicist to keep track of whether it is inertial mass or gravitational mass that is being considered.