Question

Inertial system S moves at constant velocity $\mathbf{v}\mathbf{=}\mathbf{\beta c}\left(\mathrm{cos\varphi }\hat{x}+\mathrm{sin\varphi }\hat{y}\right)$with respect to S. Their axes are parallel to one other, and their origins coincide at data-custom-editor="chemistry" $\mathbf{t}\mathbf{=}\mathbf{t}\mathbf{=}\mathbf{0}$, as usual. Find the Lorentz transformation matrix A.

Short answer

The Lorentz transformation matrix is:

$\left[\begin{array}{cccc}\mathrm{\gamma }& -\mathrm{\gamma \beta }\mathrm{cos\varphi }& -\mathrm{\gamma \beta }\mathrm{sin\varphi }& 0\\ -\mathrm{\gamma \beta }\mathrm{cos\varphi }& \left(\mathrm{\gamma }{\cos }^2\mathrm{\varphi }+{\sin }^2\mathrm{\varphi }\right)& \left(\mathrm{\gamma }-1\right)\mathrm{sin\varphi cos\varphi }& 0\\ -\mathrm{\gamma \beta }\mathrm{sin\varphi }& \left(\mathrm{\gamma }-1\right)\mathrm{sin\varphi cos\varphi }& \left(\mathrm{\gamma }{\sin }^2\mathrm{\varphi }+{\cos }^2\mathrm{\varphi }\right)& 0\\ 0& 0& 0& 1\end{array}\right]$

Step-by-step solution

Principle of Lorentz transformation

Lorentz Information is an important part of physisc sthat deals with the linear tranformations from a specific co-ordinate frame in space-time to a non-static frame, having a constant velocity with a respect to the former.

Find XY in terms of xy by using equation (1.29)

Consider the matrix is:

$\left(\begin{array}{c}\overline{{\mathrm{A}}_{\mathrm{y}}}\\ \overline{{\mathrm{A}}_{\mathrm{z}}}\end{array}\right)=\left(\begin{array}{cc}\cos \mathrm{\varphi }& \sin \mathrm{\varphi }\\ -\sin \mathrm{\varphi }& \cos \mathrm{\varphi }\end{array}\right)\left(\begin{array}{c}{\mathrm{A}}_{\mathrm{y}}\\ {\mathrm{A}}_{\mathrm{z}}\end{array}\right)$

If we take $\overline{{\mathrm{A}}_{\mathrm{x}}}$as X and $\overline{{\mathrm{A}}_{\mathrm{y}}}$as Y ${\mathrm{A}}_{\mathrm{x}}$and as x ${\mathrm{A}}_{\mathrm{y}}$and as y:

localid="1658826300610" $$\begin{gathered} \mathrm{X}=\cos \mathrm{\varphi x}+\sin \mathrm{\varphi y}.....\left(\mathrm{i}\right) \\ \mathrm{Y}=-\sin \mathrm{\varphi x}+\cos \mathrm{\varphi y}...\left(\mathrm{ii}\right) \end{gathered}$$

Using Lorentz-transform from equation (12.18) to get X Y in terms of xy

$$\begin{gathered} \overline{\mathrm{X}}=\mathrm{\gamma }\left(\mathrm{X}-\mathrm{vt}\right)...\left(\mathrm{iii}\right) \\ \overline{\mathrm{Y}}=\mathrm{Y}...\left(\mathrm{iv}\right) \\ \overline{\mathrm{Z}}=\mathrm{Z}...\left(\mathrm{v}\right) \\ \overline{\mathrm{t}}=\mathrm{\gamma }\left(\mathrm{t}-\frac{\mathrm{v}}{{\mathrm{c}}^2}\mathrm{X}\right)...\left(\mathrm{vi}\right) \end{gathered}$$

Now, from putting the value of equation (i) in equation (iii):

$$\begin{gathered} \overline{\mathrm{X}}=\mathrm{\gamma }\left(\mathrm{X}-\mathrm{vt}\right) \\ =\mathrm{\gamma }\left(\cos \mathrm{\varphi x}+\sin \mathrm{\varphi y}-\mathrm{\beta ct}\right)....\left(\mathrm{vii}\right) \end{gathered}$$

Equating equation (iv) with (ii) we get:

$\overline{\mathrm{Y}}=\mathrm{Y}=-\sin \mathrm{\varphi x}+\cos \mathrm{\varphi y}$

By further calculation of equation (vi):

$\overline{\mathrm{t}}=\mathrm{\gamma }\left(\mathrm{t}-\frac{\mathrm{v}}{{\mathrm{c}}^2}\mathrm{X}\right)$

Multiplying both sides with, c:

role="math" localid="1658829796222" $\begin{array}{rcl}\mathrm{c}\overline{\mathrm{t}}& =& \mathrm{c\gamma }\left(\mathrm{t}-\frac{\mathrm{v}}{{\mathrm{c}}^2}\mathrm{X}\right)\\ \mathrm{or},\mathrm{c}\overline{\mathrm{t}}& =& \mathrm{c\gamma t}-\frac{\mathrm{v}}{\mathrm{c}}\mathrm{\gamma X}\\ \mathrm{or},\mathrm{c}\overline{\mathrm{t}}& =& \mathrm{c\gamma t}-\mathrm{\beta \gamma X}\\ \mathrm{or},\mathrm{c}\overline{\mathrm{t}}& =& \mathrm{\gamma }(\mathrm{ct}-\mathrm{\beta X})....\left(\mathrm{viii}\right)\end{array}$

Putting the value from equation (i) to (viii):

$$\begin{gathered} \mathrm{c}\overline{\mathrm{t}}=\mathrm{\gamma }(\mathrm{ct}-\mathrm{\beta X}) \\ \mathrm{c}\overline{\mathrm{t}}=\mathrm{\gamma }\left[\mathrm{ct}-\mathrm{\beta }(\cos \mathrm{\varphi x}+\sin \mathrm{\varphi y})\right]....\left(\mathrm{ix}\right) \end{gathered}$$

Determination of Lorentz transformation matrix

To get the Lorenz matrix, we have to rotate from to by using the equation (1.29) with negative and putting the respectives values from the above equations. We get

Therefore,

$\begin{array}{rcl}\overline{\mathrm{x}}& =& \mathrm{cos\varphi }\overline{\mathrm{X}}-\mathrm{sin\varphi }\overline{\mathrm{Y}}\\ & =& \mathrm{\gamma cos\varphi }\left(\cos \mathrm{\varphi x}+\sin \mathrm{\varphi y}-\mathrm{\beta ct}\right)-\mathrm{sin\varphi }\left(-\sin \mathrm{\varphi x}+\cos \mathrm{\varphi y}\right)\\ & =& \left({\mathrm{\gamma cos}}^2\mathrm{\varphi }+{\sin }^2\mathrm{\varphi }\right)\mathrm{x}+(\mathrm{\gamma }-1)\mathrm{sin\varphi cos\varphi y}-\mathrm{\gamma \beta cos\varphi }\left(\mathrm{ct}\right)....\left(\mathrm{x}\right)\end{array}$

And,

$$\begin{gathered} \overline{\mathrm{y}}=\mathrm{sin\varphi }\overline{\mathrm{X}}+\mathrm{cos\varphi }\overline{\mathrm{Y}} \\ =\mathrm{\gamma sin\varphi }[\cos \mathrm{\varphi x}+\sin \mathrm{\varphi y}-\mathrm{\beta ct}]+\mathrm{cos\varphi }[-\sin \mathrm{\varphi x}+\cos \mathrm{\varphi y}] \\ =({\mathrm{\gamma sin}}^2\mathrm{\varphi }+{\cos }^2\mathrm{\varphi })\mathrm{y}+(\mathrm{\gamma }-1)\mathrm{sin\varphi cos\varphi x}-\mathrm{\gamma \beta sin\varphi }\left(\mathrm{ct}\right)........\left(\mathrm{xi}\right) \end{gathered}$$

By convention (x) and (xi) into matrix form:

$\left(\begin{array}{c}\mathrm{c}\overline{\mathrm{t}}\\ \overline{\mathrm{x}}\\ \overline{\mathrm{y}}\\ \overline{\mathrm{z}}\end{array}\right)=\left[\begin{array}{cccc}\mathrm{\gamma }& -\mathrm{\gamma \beta cos}\mathrm{\varphi }& -\mathrm{\gamma \beta sin}\mathrm{\varphi }& 0\\ -\mathrm{\gamma \beta cos}\mathrm{\varphi }& \left({\mathrm{\gamma cos}}^2\mathrm{\varphi }+{\sin }^2\mathrm{\varphi }\right)& \left(\mathrm{\gamma }-1\right)\mathrm{sin\varphi cos\varphi }& 0\\ -\mathrm{\gamma \beta sin}\mathrm{\varphi }& \left(\mathrm{\gamma }-1\right)\mathrm{sin\varphi cos\varphi }& \left({\mathrm{\gamma sin}}^2\mathrm{\varphi }+{\cos }^2\mathrm{\varphi }\right)& 0\\ 0& 0& 0& 1\end{array}\right]\left(\begin{array}{c}\mathrm{ct}\\ \mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right)$