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I know very little special relativity. I never leaned it properly, but every time I read someone saying: "if you boost in the x-direction, you get such and such" my mind goes blank! I tried understanding it but always get stuck with articles that assume that the reader knows everything.

So, what is a Lorentz boost, and how to calculate it? And why does the direction matters?

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2 Answers 2

up vote 17 down vote accepted

Lorentz boost is simply a Lorentz transformation which doesn't involve rotation. For example, Lorentz boost in the x direction looks like this:

\begin{equation} \left[ \begin{array}{cccc} \gamma & -\beta \gamma & 0 & 0 \newline -\beta \gamma & \gamma & 0 & 0 \newline 0 & 0 & 1 & 0 \newline 0 & 0 & 0 & 1 \end{array} \right] \end{equation}

where coordinates are written as (t, x, y, z) and

\begin{equation} \beta = \frac{v}{c} \end{equation} \begin{equation} \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \end{equation}

This is a linear transformation which given coordinates of an event in one reference frame allows one to determine the coordinates in a frame of reference moving with respect to the first reference frame at velocity v in the x direction.

The ones on the diagonal mean that the transformation does not change the y and z coordinates (i.e. it only affects time t and distance along the x direction). For comparison, Lorentz boost in the y direction looks like this:

\begin{equation} \left[ \begin{array}{cccc} \gamma & 0 & -\beta \gamma & 0 \newline 0 & 1 & 0 & 0 \newline -\beta \gamma & 0 & \gamma & 0 \newline 0 & 0 & 0 & 1 \end{array} \right] \end{equation}

which means that the transformation does not affect the x and z directions (i.e. it only affects time and the y direction).

In order to calculate Lorentz boost for any direction one starts by determining the following values:

\begin{equation} \gamma = \frac{1}{\sqrt{1 - \frac{v_x^2+v_y^2+v_z^2}{c^2}}} \end{equation} \begin{equation} \beta_x = \frac{v_x}{c}, \beta_y = \frac{v_y}{c}, \beta_z = \frac{v_z}{c} \end{equation}

Then the matrix form of the Lorentz boost for velocity v=(vx, vy, vz) is this:

\begin{equation} \left[ \begin{array}{cccc} L_{tt} & L_{tx} & L_{ty} & L_{tz} \newline L_{xt} & L_{xx} & L_{xy} & L_{xz} \newline L_{yt} & L_{yx} & L_{yy} & L_{yz} \newline L_{zt} & L_{zx} & L_{zy} & L_{zz} \newline \end{array} \right] \end{equation}


\begin{equation} L_{tt} = \gamma \end{equation} \begin{equation} L_{ta} = L_{at} = -\beta_a \gamma \end{equation} \begin{equation} L_{ab} = L_{ba} = (\gamma - 1) \frac{\beta_a \beta_b}{\beta_x^2 + \beta_y^2 + \beta_z^2} + \delta_{ab} = (\gamma - 1) \frac{v_a v_b}{v^2} + \delta_{ab} \end{equation}

where a and b are x, y or z and δab is the Kronecker delta.

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Did you check out the wikipedia article

Basically it is just a change of co-ordinates when you change your frame of reference from one that is at rest, to another frame which is moving w.r.t to it at a constant velocity $v$.If the changes inertial frame is moving along the x-axis of the old frame, with the y and z axis parallel to each other, it is called a lorentz boost in the x-direction. The change of co-ordinates can be found out using the lorentz transformation matrix give by Adam, or the co-ordinate transformation formula. These can be derived using the fact that the interval between two points $(ct)^2-x^2-y^2-z^2$ is lorentz invariant. Refer to chapter1 of classical theory of fields by Landau and Lifschitz.

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