Chapter 2, Problem 10P

To determine

Relation between the time measured in moving frame and time measured in frame at rest with the help of equations of the spherical wavefronts.

Answer to Problem 10P

Relation between the time measured in moving frame and time measured in frame at rest with the help of equations of the spherical wavefronts is $t\prime =\frac{\left(t-\frac{vx}{c^2}\right)}{\sqrt{1-\frac{v^2}{c^2}}}$.

Explanation of Solution

Let’s a frame Kʹ moving with a uniform velocity v along the x-axis with respect to a frame K in rest. The points of frame K coincide with the points of the frame Kʹ at $t=0$  and at this time xʹ-axis is parallel to x-axis. Thus, from theory of relativity,

$x\prime =\alpha (x-vt)$        (I)

Where, $\alpha$ is constant, which will be found latter.

Frame Kʹ is moving along the x-axis, therefore y and z coordinates will remain same. Thus,

$y\prime =y$ and

$z\prime =z$        (II)

Write an equation to relate time (tʹ) measured in frame Kʹ with the time and space coordinates of the frame K, tʹ surely depends on t, x, y, and z linearly due to homogeneity but due to symmetry z and y will not affect tʹ. Thus,

$t\prime =\gamma t+\beta x$        (III)

Where, $\beta$ and $\gamma$ are constants, which will be found latter

Write the equation for the spherical wavefronts in frame K. Thus,

$x^2+y^2+z^2=c^2{t}^2$        (IV)

Write the equation for the spherical wavefronts in frame Kʹ. Thus,

$x{\prime }^2+y{\prime }^2+z{\prime }^2=c^{2}t{\prime }^2$        (V)

Substitute the values of the xʹ, yʹ, zʹ, and tʹ in equation (V) from equations (I), (II), and (III). Thus,

$\begin{array}{l}{\alpha }^2(x-vt){}^2+y^2+z^2=c^2{\left(\gamma t+\beta x\right)}^2\\ {\alpha }^2(x^2+v^2{t}^2-2xvt)+y^2+z^2=c^2\left({\gamma }^2{t}^2+{\beta }^2{x}^2+2\beta x\gamma t\right)\\ {\alpha }^2{x}^2-c^2{\beta }^2{x}^2+y^2+z^2=c^2{\gamma }^2{t}^2-v^2{t}^2{\alpha }^2+2\beta x\gamma t{c}^2\text{+}2xvt{\alpha }^2\\ x^2\left({\alpha }^2-c^2{\beta }^2\right)+y^2+z^2=t^2\left(c^2{\gamma }^2-v^2{\alpha }^2\right)+2xt(\beta \gamma c^2+v{\alpha }^2)\end{array}$

Compare the above equation with equation (IV), thus,

$\begin{array}{c}{\alpha }^2-c^2{\beta }^2=1\\ c^2{\gamma }^2-v^2{\alpha }^2=c^2\\ 2xt(\beta \gamma c^2+v{\alpha }^2)=0\\ \beta \gamma c^2+v{\alpha }^2=0\end{array}$        (VI)

From above equation,

${\alpha }^2=1+c^2{\beta }^2$

Substitute $\beta \gamma c^2+v{\alpha }^2=0$ in equation $c^2{\gamma }^2-v^2{\alpha }^2=c^2$. Thus,

$\begin{array}{c}{c}^2{\gamma }^2-v^2{\alpha }^2=c^2\\ c^2{\gamma }^2-v\left(v{\alpha }^2\right)=c^2\\ c^2{\gamma }^2-v\left(-\beta \gamma c^2\right)=c^2\\ {\gamma }^2+v\beta \gamma =1\\ \gamma \left(\gamma +v\beta \right)=1\\ \beta =\frac{1}{v}\left(\frac{1}{\gamma }-\gamma \right)\end{array}$        (VII)

Substitute ${\alpha }^2=1+c^2{\beta }^2$ in equation $\beta \gamma c^2+v{\alpha }^2=0$. Thus,

$\begin{array}{l}\beta \gamma c^2+v\left(1+c^2{\beta }^2\right)=0\\ \beta \gamma c^2+v{c}^2{\beta }^2+v=0\\ \beta c^2\left(\gamma +\beta v\right)=-v\end{array}$        (VIII)

Substitute $\beta =\frac{1}{v}\left(\frac{1}{\gamma }-\gamma \right)$ from equation (VII) in equation (VIII). Thus,

$\begin{array}{l}\frac{1}{v}\left(\frac{1}{\gamma }-\gamma \right)c^2\left(\gamma +\frac{1}{v}\left(\frac{1}{\gamma }-\gamma \right)v\right)=-v\\ \left(\frac{1}{\gamma }-\gamma \right)\left(\gamma +\left(\frac{1}{\gamma }-\gamma \right)\right)=-\frac{v^2}{c^2}\\ \left(1-{\gamma }^2+\frac{1}{{\gamma }^2}+{\gamma }^2-2\right)=-\frac{v^2}{c^2}\\ \left(\frac{1}{{\gamma }^2}-1\right)=-\frac{v^2}{c^2}\end{array}$

Simplify the above equation, thus,

$\begin{array}{l}\frac{1}{{\gamma }^2}=1-\frac{v^2}{c^2}\\ {\gamma }^2=\frac{1}{1-\frac{v^2}{c^2}}\end{array}$        (IX)

Substitute $\left(\frac{1}{{\gamma }^2}-1\right)=-\frac{v^2}{c^2}$ in equation (VII). Thus,

$\begin{array}{l}\beta =\frac{1}{v}\left(\frac{1}{\gamma }-\gamma \right)\\ =\frac{\gamma }{v}\left(\frac{1}{{\gamma }^2}-1\right)\\ =\frac{\gamma }{v}\left(-\frac{v^2}{c^2}\right)\\ =-\frac{\gamma v}{c^2}\end{array}$        (X)

Substitute value from equation (X) in equation $\beta \gamma c^2+v{\alpha }^2=0$. Thus,

$\begin{array}{l}\left(-\frac{\gamma v}{c^2}\right)\gamma c^2+v{\alpha }^2=0\\ -{\gamma }^{2}v+v{\alpha }^2=0\\ {\alpha }^2={\gamma }^2\\ \alpha =\gamma \end{array}$        (XI)

Choose positive sign of the root on substituting all theses values of the $\alpha ,\beta ,$ and $\gamma$ in equation (I) and (III). Thus,

$x\prime =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}(x-vt)$

And

$\begin{array}{l}t\prime =\gamma t+\left(-\frac{\gamma v}{c^2}\right)x\\ =\gamma \left(t-\frac{vx}{c^2}\right)\\ \text{=}\frac{\left(t-\frac{vx}{c^2}\right)}{\sqrt{1-\frac{v^2}{c^2}}}\end{array}$

Conclusion:

Therefore, relation between the time measured in moving frame and time measured in frame at rest with the help of equations of the spherical wavefronts is $t\prime =\frac{\left(t-\frac{vx}{c^2}\right)}{\sqrt{1-\frac{v^2}{c^2}}}$.