Special relativity

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The Postulates of Relativity

The principle of relativity: The laws of physics are the same in any inertial reference frame, i.e. there is no preferred frame in which to do physics.
The invariance of the speed of light: The speed of light in vaccuum (denoted by 'c') has the same value in any inertial reference frame.

The Lorentz Transformation

Derivation

Special relativity is concerned with how coordinates transform between reference frames that are in relative motion at a constant velocity. Suppose that two frames S and S' (with coordinates (x,y,z,t) and $LaTeX: (\tilde{x},\tilde{y},\tilde{z},\tilde{t})$ , respectively) have their axes aligned at time t=0 and S' is traveling with speed v (relative to S) along the positive x-axis. A quick and ``intuitive" calculation for finding the coordinates of an event in the frame S' in terms of the coordinates in S yields the Galilean transformation:

$LaTeX: </dd></dl> <p>\begin{align} \tilde{x} &= x - vt \\ </p> <pre>\tilde{y} &= y \\ \tilde{z} &= z \\ \tilde{t} &= t. </pre> <p>\end{align}$

However, this is not consistent with the invariance of the speed of light. For suppose we consider the path of a beam of light which was shot from the origin at time t = 0. Then in S we have that $x=ct$. Plugging this into the above transformation yields $LaTeX: \tilde{x} = (c-v)\tilde{t}$ , so that the beam has velocity c - v in S'.

The correct transformation rules should still have $LaTeX: \tilde{x}$ depend linearly on x - vt. Thus we look for a $LaTeX: \gamma$ such that

$LaTeX: \tilde{x} = \gamma (x-vt)$

Note that, because of isotropy of space and time, $LaTeX: \gamma$ should not depend on x or $t$ but may depend on $v$. Because of the first postulate of relativity, we can equally consider S' to be at rest and S to be moving along the x-axis with velocity -v. Therefore if $LaTeX: \tilde{x} = \gamma (x-vt)$ then we should have that

$LaTeX: x = \gamma (\tilde{x} + v\tilde{t})$

Now we use the invariance of the speed of light: if we shoot a light beam along the $x-$axis from the origin at time 0, then we have that x = ct and $LaTeX: \tilde{x} = c\tilde{t}$ . Plugging these into the above equations gives $LaTeX: c\tilde{t} = \gamma (ct - vt)$ and $LaTeX: ct = \gamma(c\tilde{t} + v\tilde{t})$ . This implies that

$LaTeX: \begin{align} ct &= \gamma(\gamma (ct - vt) + \frac{v}{c}\gamma(ct-vt)) \\ c &= \gamma^2(c - \frac{v^2}{c}) \\ \gamma &= \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \end{align}$

We can now solve for $LaTeX: \tilde{t}$ in terms of x and t:

$LaTeX: \begin{align} t' &= \frac{1}{v}(\frac{1}{\gamma}x - \tilde{x}) \\ &= \frac{1}{v}(\frac{1}{\gamma}x - \gamma(x-vt)) \\ &= \gamma(t + (\frac{1}{v\gamma^2} - 1)x) \\ &= \gamma(t - \frac{v}{c^2}x) \end{align}$

Therefore, the proper transformation, called the Lorentz transformation, is given by:

$LaTeX: \begin{align} \tilde{x} &= \gamma(x-vt) \\ \tilde{y} &= y \\ \tilde{z} &= z \\ \tilde{t} &= \gamma(t - \frac{v}{c^2}x) \end{align}$

or in matrix form by:

$LaTeX: \left(\begin{array}{c} c\tilde{t} \\ \tilde{x} \\ \tilde{y} \\ \tilde{z}\end{array}\right) = \left(\begin{array}{cccc}\gamma & -\beta\gamma & 0 & 0 \\-\beta\gamma & \gamma & 0 & 0 \\0 & 0 & 1 & 0 \\0 & 0 & 0 & 1\end{array}\right)\left(\begin{array}{c} ct \\ x \\ y \\ z\end{array}\right)$

where $LaTeX: \beta = \frac{v}{c}$ . We define the rapidity $LaTeX: \phi$ by the relation $LaTeX: e^\phi = \gamma(1 + \beta)$ . Then $LaTeX: e^{-\phi} = \frac{1}{\gamma}\frac{1}{1+\beta} = \gamma(\frac{1}{\gamma^2(1+\beta)}) = \gamma(\frac{1- \beta^2}{1+\beta}) = \gamma(1-\beta)$ . Thus $LaTeX: \cosh\phi = \frac{e^\phi + e^{-\phi}}{2} = \gamma$ and $LaTeX: \sinh\phi = \frac{e^\phi - e^{-\phi}}{2} = \gamma\beta$ . Substituting these in for the matrix form of the Lorentz transformation yields

$LaTeX: \left(\begin{array}{c} c\tilde{t} \\ \tilde{x} \\ \tilde{y} \\ \tilde{z}\end{array}\right) = \left(\begin{array}{cccc}\cosh\phi & -\sinh\phi & 0 & 0 \\-\sinh\phi & \cosh\phi & 0 & 0 \\0 & 0 & 1 & 0 \\0 & 0 & 0 & 1\end{array}\right)\left(\begin{array}{c} ct \\ x \\ y \\ z\end{array}\right).$

Notice the resemblance between the above matrix and a matrix corresponding to a rotation.

Length Contraction and Time Dilation

From the Lorentz transformation we can derive the Lorentz contraction and time dilation. Consider a rod of length $LaTeX: L_0$ (relative to itself) at rest in frame S' such that one end is at $LaTeX: \tilde{x} = 0$ and the other end is at $LaTeX: \tilde{x} = L$ . When t = 0 one end will be at the origin and the other end will be at L, where $LaTeX: L_0 = \gamma L$ , giving the rod a length of $LaTeX: \frac{1}{\gamma}L_0 < L_0$ in S. Now consider a clock in S' which is fixed at x' = 0. Then its position in S is x = vt so that $LaTeX: \tilde{t} = \gamma(t - \frac{v^2}{c^2}t) = \frac{1}{\gamma}$ . Thus $LaTeX: t = \gamma \tilde{t} > \tilde{t}$ so that the clock is running slowly with respect to S.

Four-vectors

It is straightforward to verify that the Lorentz transformation preserves the quantity $LaTeX: (ct)^2 - x^2 - y^2 - z^2$ . Thus we can define the invariant bilinear form $LaTeX: g((t, x, y, z), (\tilde{t},\tilde{x},\tilde{y},\tilde{z})) = c^2t\tilde{t} - x\tilde{x} - y\tilde{y} - z\tilde{z}$ . The vector space $LaTeX: R^4$ with this bilinear form is called Minkowski space. Elements of Minkowski space are called four-vectors. By convention, we write four-vectors as $LaTeX: x^\mu = (x^0, x^1, x^2, x^3, x^4)$ where $LaTeX: x^0 = ct$ and $LaTeX: x^1, x^2, x^3$ are the spatial coordinates. Let $LaTeX: x_\mu$ denote the covector defined by $LaTeX: x_\mu(y^\nu) = g_{\mu \nu} x^\mu y^\nu$ . Thus in the standard dual-basis, we have that $LaTeX: x_\mu = (c^2 x^0, -x^1, -x^2, -x^3)$ .

Given a four-vector $LaTeX: x^\mu$ , we define the four-velocity to be $LaTeX: \frac{dx^\mu}{d\tau}$ . The proper time, $LaTeX: \tau$ , can be thought of as the time measured by a clock which is attached to the object in motion. Because of time dilation, we have that $LaTeX: x^0 = ct = c\gamma\tau$ where the speed on which $LaTeX: \gamma$ depends is $LaTeX: v = \sqrt{(\frac{dx^1}{dt})^2 + (\frac{dx^2}{dt})^2 + (\frac{dx^3}{dt})^2}$ . We have that $LaTeX: \frac{dx^\mu}{d\tau} = (c\gamma, \frac{dx^1}{dt}\frac{dt}{d\tau}, \frac{dx^2}{dt}\frac{dt}{d\tau}, \frac{dx^3}{dt}\frac{dt}{d\tau}) = \gamma(c, \frac{dx^1}{dt}, \frac{dx^2}{dt}, \frac{dx^3}{dt})$ . The norm squared of this four-vector is then $LaTeX: \gamma^2(c^2 - v^2) = c^2$ .