Special Relativity/Relativistic dynamics

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Special Relativity

1 Relativistic Dynamics

[edit] Relativistic Dynamics

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

[edit] 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, $δ S = 0$ ) 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 $d s$ , 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 $d s$ :

$ds = \sqrt{c^2 dt^2 - dx^2 - dy^2 - dz^2}$

After pulling out $c d t$ from the square root, and noting that $\frac{dx^2+dy^2+dz^2}{dt^2} = v^2$ , 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:

$S = \int L dt$

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 $κ$ . 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, $L = \frac{1}{2}mv^2$ . To compare the two Lagrangian, we perform a Taylor expansion on the square root:

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

The first term, $c κ$ , 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 $κ = - m c$ .

Therefore, the relativistic Lagrangian is:

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

[edit] 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:

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

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

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

$H = \Sigma p_i v_i - L = \gamma mv^2 + mc^2\sqrt{1-v^2/c^2}$

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

After some algebra, we obtain:

$H = γ m c 2$

For an object at rest, $γ = 1$ , 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:

$E 2 = p 2 c 2 + m 2 c 4$

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

[edit] Mass

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

$m = γ m 0$

However, for more contemporary interpretations of Special Relativity, mass, $m 0$ is considered an invariant quantity for all reference frames and $γ m 0$ 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.

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