Question

Define proper acceleration in the obvious way:

${\mathit{\alpha }}^{\mathbf{\mu }}\mathbf{=}\frac{\mathbf{d}{\mathbf{\eta }}^{\mathbf{\mu }}}{\mathbf{d}\mathbf{\tau }}\mathbf{=}\frac{{\mathbf{d}}^{\mathbf{2}}{\mathbf{x}}^{\mathbf{\mu }}}{\mathbf{d}{\mathbf{\tau }}^{\mathbf{2}}}$

(a) Find${\mathit{\alpha }}^{\mathbf{0}}$and α in terms of u and a (the ordinary acceleration).

(b) Express${\mathit{\alpha }}_{\mathbf{\mu }}{\mathit{\alpha }}^{\mathbf{\mu }}$in terms of u and a.

(c) Show that${\mathit{\eta }}^{\mathbf{\mu }}{\mathit{\alpha }}_{\mathbf{\mu }}\mathbf{=}\mathbf{0}$.

(d) Write the Minkowski version of Newton’s second law, in terms of${\mathit{\alpha }}_{\mathbf{\mu }}$. Evaluate the invariant product${\mathit{K}}^{\mathbf{\mu }}{\mathit{\eta }}_{\mathbf{\mu }}$.

Short answer

(a)The value of${\alpha }^0$ and $\alpha$are$\frac{ua}{c{\left(1-{\displaystyle \frac{u^2}{c^2}}\right)}^2}$ and$\frac{1}{\sqrt{1-{\displaystyle \frac{u^2}{c^2}}}}\times \left[a+\frac{u\left(ua\right)}{\left(1-{\displaystyle \frac{u^2}{c^2}}\right)}\frac{1}{c^2}\right]$ respectively.

(b)The value of${\alpha }_{\mu }{\alpha }^{\mu }$ is $\left[a^2{\left(1-\frac{u^2}{c^2}\right)}^2+\frac{{\left(ua\right)}^2}{c^2}\left(1-\frac{u^2}{c^2}\right)\right]$.

(c)The value of${\eta }^{\mu }{\eta }_{\mu }$ is zero.

(d) The value of the invariant product is $m{\alpha }^{\mu }{\eta }_{\mu }$.

Step-by-step solution

Expression for the proper acceleration and ordinary acceleration:

Write the expression for the proper acceleration.

${\mathit{\alpha }}^{\mathbf{\mu }}\mathbf{=}\frac{\mathbf{d}{\mathbf{\eta }}^{\mathbf{\mu }}}{\mathbf{d}\mathbf{\tau }}$ …… (1)

Here,${\alpha }^{\mu }$ is the proper acceleration and$\frac{d{\eta }^{\mu }}{d\mathcal{b}}$ is the change in the proper velocity with respect to the time.

Write the expression for the ordinary acceleration.

$$\begin{gathered} {\mathit{\alpha }}^{\mathbf{0}}\mathbf{=}\frac{\mathbf{d}{\mathbf{\eta }}^{\mathbf{0}}}{\mathbf{d}\mathbf{\tau }} \\ {\mathit{\alpha }}^{\mathbf{0}}\mathbf{=}\frac{\mathbf{d}{\mathbf{\eta }}^{\mathbf{0}}}{\mathbf{d}\mathbf{\tau }}\frac{\mathbf{d}\mathbf{t}}{\mathbf{d}\mathbf{\tau }} \end{gathered}$$ …… (2)

Here, ${\alpha }^0$is the proper acceleration and $\frac{d{\eta }^0}{d\mathcal{b}}$is the change in the ordinary velocity with respect to the time and$\frac{dt}{d\mathcal{b}}$ is the time function.

Determine the value of α0 and α (angular acceleration):

(a)

Write the expression for the ordinary velocity.

${\eta }^0=\frac{c}{\sqrt{1-{\displaystyle \frac{u^2}{c^2}}}}$ …… (3)

Here, c is the velocity of light and u is the linear relative velocity.

Write the expression for the time function.

$\frac{dt}{d\mathcal{b}}=\frac{1}{\sqrt{1-{\displaystyle \frac{u^2}{c^2}}}}$ …… (4)

Write the expression for the angular acceleration.

$\alpha =\frac{d\eta }{dt}\frac{dt}{d\mathcal{b}}$ …… (5)

Here,$\frac{d\eta }{d\mathcal{b}}$is the change in the angular velocity.

Substitute localid="1654064044527" $\frac{c}{\sqrt{1-{\displaystyle \frac{u^2}{c^2}}}}$for ${\eta }^0$and $\frac{1}{\sqrt{1-{\displaystyle \frac{u^2}{c^2}}}}$for $\frac{dt}{d\mathcal{b}}$in equation (2).

$$\begin{gathered} {\alpha }^0=\frac{d}{d\tau }\left(\frac{c}{\sqrt{1-\frac{u^2}{c^2}}}\right)\times \frac{1}{\sqrt{1-\frac{u^2}{c^2}}} \\ {\alpha }^0=\frac{1}{\sqrt{1-\frac{u^2}{c^2}}}\times c\frac{d}{d\tau }\left({\left(\sqrt{1-\frac{u^2}{c^2}}\right)}^{-\frac{1}{2}}\right) \\ {\alpha }^0=\frac{1}{\sqrt{1-\frac{u^2}{c^2}}}\times \left(c\left(\frac{-1}{2}\right){\left(1-\frac{u^2}{c^2}\right)}^{-\frac{3}{2}}\left(\frac{-2ua}{c^2}\right)\right) \end{gathered}$$

Here, a is the linear acceleration.

Solve as further,

$$\begin{gathered} {\alpha }^0=\frac{1}{\sqrt{1-\frac{u^2}{c^2}}}\times \left(\frac{c\left(\frac{ua}{c^2}\right)}{{\left(1-\frac{u^2}{c^2}\right)}^{-\frac{3}{2}}}\right) \\ {\alpha }^0=\frac{ua}{c{\left(1-\frac{u^2}{c^2}\right)}^2} \end{gathered}$$

Substitute $\eta =\frac{u}{\sqrt{1-\frac{u^2}{c^2}}}$and $\frac{dt}{d\tau }=\frac{1}{\sqrt{1-\frac{u^2}{c^2}}}$in equation (5).

$$\begin{gathered} \alpha =\frac{d}{dt}\left(\frac{u}{\sqrt{1-\frac{u^2}{c^2}}}\right)\times \frac{1}{\sqrt{1-\frac{u^2}{c^2}}} \\ \alpha =\frac{1}{\sqrt{1-\frac{u^2}{c^2}}}\times \left[{\left(1-\frac{u^2}{c^2}\right)}^{\frac{1}{2}}a+u\left(-\frac{1}{2}\right){\left(1-\frac{u^2}{c^2}\right)}^{-\frac{3}{2}}\left(-\frac{2ua}{c^2}\right)\right] \\ \alpha =\frac{1}{\sqrt{1-\frac{u^2}{c^2}}}\times \left[a+\frac{u\left(ua\right)}{\left(1-\frac{u^2}{c^2}\right)}\frac{1}{c^2}\right] \end{gathered}$$

Therefore, the value of ${\alpha }^0$and$\alpha$ are$\frac{ua}{c{\left(1-\frac{u^2}{c^2}\right)}^2}$ and $\frac{1}{\sqrt{1-\frac{u^2}{c^2}}}\times \left[a+\frac{u\left(ua\right)}{\left(1-\frac{u^2}{c^2}\right)}\frac{1}{c^2}\right]$respectively.

Determine the value of αμαμ :

(b)

Write the general expression for the product of invariant acceleration$\left({\alpha }_{\mu }\right)$and proper acceleration $\left({\alpha }^{\mu }\right)$.

${\alpha }_{\mu }{\alpha }^{\mu }=-{\left({\alpha }^0\right)}^2+{\alpha }^2$

Substitute${\alpha }^0=\frac{ua}{c{\left(1-{\displaystyle \frac{u^2}{c^2}}\right)}^2}$and$\alpha =\frac{1}{\sqrt{1-{\displaystyle \frac{u^2}{c^2}}}}\times \left[a+\frac{u\left(ua\right)}{\left(1-{\displaystyle \frac{u^2}{c^2}}\right)}\frac{1}{c^2}\right]$in the above expression.

$$\begin{gathered} {\alpha }_{\mu }{\alpha }^{\mu }=-{\left(\frac{ua}{c{\left(1-{\displaystyle \frac{u^2}{c^2}}\right)}^2}\right)}^2+{\left(\frac{1}{\sqrt{1-{\displaystyle \frac{u^2}{c^2}}}}\times \left[a+\frac{u\left(ua\right)}{\left(1-{\displaystyle \frac{u^2}{c^2}}\right)}\frac{1}{c^2}\right]\right)}^2 \\ {\alpha }_{\mu }{\alpha }^{\mu }=-\left(\frac{u^2{a}^2}{c^2{\left(1-{\displaystyle \frac{u^2}{c^2}}\right)}^4}\right)+\frac{1}{\left(1-{\displaystyle \frac{u^2}{c^2}}\right)}\times \left[{\left(a+\frac{u\left(ua\right)}{\left(1-{\displaystyle \frac{u^2}{c^2}}\right)}\frac{1}{c^2}\right)}^2\right] \\ {\alpha }_{\mu }{\alpha }^{\mu }=-\left(\frac{u^2{a}^2}{c^2{\left(1-{\displaystyle \frac{u^2}{c^2}}\right)}^4}\right)+{\left[\frac{1}{{\left(1-{\displaystyle \frac{u^2}{c^2}}\right)}^2}\times \left[a\left(1-\frac{u^2}{c^2}\right)+\left(\frac{u\left(ua\right)}{c^2-u^2}\right)\left(\frac{c^2-u^2}{c^2}\right)\right]\right]}^2 \end{gathered}$$

Solve as further,

$$\begin{gathered} {\alpha }_{\mu }{\alpha }^{\mu }=-\left(\frac{u^2{a}^2}{c^2{\left(1-{\displaystyle \frac{u^2}{c^2}}\right)}^4}\right)+\frac{1}{{\left(1-{\displaystyle \frac{u^2}{c^2}}\right)}^4}{\left[a\left(1-\frac{u^2}{c^2}\right)+\frac{1}{c^2}u\left(ua\right)\right]}^2 \\ {\alpha }_u{\alpha }^{\mu }=\left(\frac{1}{{\left(1-{\displaystyle \frac{u^2}{c^2}}\right)}^4}\right)\left[a^2{\left(1-\frac{u^2}{c^2}\right)}^2+\frac{{\left(ua\right)}^2}{c^2}\left(1-\frac{u^2}{c^2}\right)\right] \\ {\alpha }_{\mu }{\alpha }^{\mu }=\frac{1}{{\left(1-{\displaystyle \frac{u^2}{c^2}}\right)}^2}\left[a^2+\frac{{\left(ua\right)}^2}{c^2-u^2}\right] \end{gathered}$$

Therefore, the value of${\alpha }_{\mu }{\alpha }^{\mu }$ is $\left[a^2{\left(1-\frac{u^2}{c^2}\right)}^2+\frac{{\left(ua\right)}^2}{c^2}\left(1-\frac{u^2}{c^2}\right)\right]$.

Determine the value of ημημ:

(c)

Write the general expression for the product of invariant velocity$\left({\eta }_{\mu }\right)$ and proper velocity $\left({\eta }^{\mu }\right)$.

${\eta }^{\mu }{\eta }_{\mu }=-c^2$

Differentiate the above expression with respect to time (t).

$\frac{d}{dt}\left({\eta }^{\mu }{\eta }_{\mu }\right)={\eta }^{\mu }{\eta }_{\mu }+{\eta }^{\mu }{\eta }_{\mu }$ …… (6)

Substitute$-c^2$ for${\eta }^{\mu }{\eta }_{\mu }$ in the above expression

$\frac{d}{dt}\left({\eta }^{\mu }{\eta }_{\mu }\right)=\frac{d}{dt}\left(-c^2\right)$

It is known that the derivative of a constant is zero. Then,

$\frac{d}{dt}\left({\eta }^{\mu }{\eta }_{\mu }\right)=0$

Substitute 0 for$\frac{d}{dt}\left({\eta }^{\mu }{\eta }_{\mu }\right)$ in equation (6) as,

$$\begin{gathered} 0={\eta }^{\mu }{\eta }_{\mu }+{\eta }^{\mu }{\eta }_{\mu } \\ 0=2{\eta }^{\mu }{\eta }_{\mu } \\ {\eta }^{\mu }{\eta }_{\mu }=0 \end{gathered}$$

Therefore, the value of${\eta }^{\mu }{\eta }_{\mu }$ is zero.

Determine the value of the invariant product Kμημ using the Minkowski version of Newton’s second law:

(d)

Write the expression for the proper force using Newton’s second law.

$k^{\mu }=\frac{d{p}^{\mu }}{dt}$ …… (7)

Here,$\frac{d{p}^{\mu }}{dt}$ is the change of proper momentum and$k^{\mu }$ is the proper force.

Write the expression for the proper momentum.

$p^{\mu }=m{\eta }^{\mu }$

Substitute$p^{\mu }=m{\eta }^{\mu }$ in equation (7).

$$\begin{gathered} k^{\mu }=\frac{d}{dt}\left(m{\eta }^{\mu }\right) \\ {k}^{\mu }=m\frac{d}{dt}\left({\eta }^{\mu }\right) \\ {k}^{\mu }=m{\alpha }^{\mu } \end{gathered}$$

Here,${\alpha }^{\mu }$ is the proper acceleration, and m is the mass.

Substitute$k^{\mu }=m{\alpha }^{\mu }$ in the invariant product $K^{\mu }{\eta }_{\mu }$.

$K^{\mu }{\eta }_{\mu }=m{\alpha }^{\mu }{\eta }_{\mu }$

Therefore, the value of the invariant product is $m{\alpha }^{\mu }{\eta }_{\mu }$.