Chapter 38, Problem 1P

In a laboratory frame of reference, an observer notes that Newton’s second law is valid. Assume forces and masses are measured to be the same in any reference frame for speeds small compared with the speed of light. (a) Show that Newton’s second law is also valid for an observer moving at a constant speed, small compared with the speed of light, relative to the laboratory frame. (b) Show that Newton’s second law is not valid in a reference frame moving past the laboratory frame with a constant acceleration.

To determine

To show: The Newton’s second law is valid for an observer moving at a constant speed of light, relative to the laboratory frame.

Answer to Problem 1P

The Newton’s second law is valid for an observer moving at a constant speed of light, relative to the laboratory frame.

Explanation of Solution

Assume $v$ is the constant speed of the frame.

The Galilean coordinate transformation is,

$x^{\prime }=x-vt$

To find the velocity, take a time derivative $\frac{d}{d{t}^{\prime }}$, $dt=d{t}^{\prime }$ and $v$ is constant.

$u^{\prime }=\frac{d{x}^{\prime }}{d{t}^{\prime }}$

Substitute $x-vt$ for $x^{\prime }$ in above equation.

$\begin{array}{c}{u}^{\prime }=\frac{d\left(x-vt\right)}{d{t}^{\prime }}\\ =\frac{dx}{d{t}^{\prime }}-v\frac{d\left(t\right)}{d{t}^{\prime }}\\ =\frac{dx}{dt}-v\frac{d\left(t\right)}{dt}\\ =u-v\end{array}$

To find the acceleration, take another time derivative $\frac{d}{d{t}^{\prime }}$ and $dt=d{t}^{\prime }$.

$a^{\prime }=\frac{d{u}^{\prime }}{d{t}^{\prime }}$

Substitute $u-v$ for $u^{\prime }$ in above equation,

$\begin{array}{c}{a}^{\prime }=\frac{d}{d{t}^{\prime }}\left(u-v\right)\\ a^{\prime }=\frac{du}{d{t}^{\prime }}-\frac{dv}{d{t}^{\prime }}\\ a^{\prime }=\frac{du}{d{t}^{\prime }}\\ a=\frac{du}{dt}\end{array}$

It is shown from the above equation that the accelerations are identical.

The Newton’s second law is the same.

$\begin{array}{c}{F}^{\prime }=m{a}^{\prime }\\ F=ma\end{array}$

Thus, the Newton’s second law is valid for an observer moving at a constant speed of light, relative to the laboratory frame.

Conclusion:

Therefore, the Newton’s second law is valid for an observer moving at a constant speed of light, relative to the laboratory frame.

To determine

The Newton’s second law is not valid in a reference frame moving past the laboratory frame with a constant acceleration.

Answer to Problem 1P

The Newton’s second law is not valid in a reference frame moving past the laboratory frame with a constant acceleration.

Explanation of Solution

Assume $a_0$ is the constant acceleration of the frame and at time $t=t^{\prime }=0$ the two frames are together, then at some arbitrary time $t$ later, the distance between the two frames is $\frac{1}{2}{a}_0{t}^2$.

The Galilean coordinate transformation is,

$x^{\prime }=x-\frac{1}{2}{a}_0{t}^2$

To find the velocity, take a time derivative $\frac{d}{d{t}^{\prime }}$, $dt=d{t}^{\prime }$ and $v$ is constant.

$u^{\prime }=\frac{d{x}^{\prime }}{d{t}^{\prime }}$

Substitute $x-\frac{1}{2}{a}_0{t}^2$ for $x^{\prime }$ in above equation,

$\begin{array}{c}{u}^{\prime }=\frac{d}{d{t}^{\prime }}\left(x-\frac{1}{2}{a}_0{t}^2\right)\\ =\frac{dx}{d{t}^{\prime }}-a_{0}t\frac{dt}{d{t}^{\prime }}\\ =u-a_{0}t\end{array}$

To find the acceleration, take another time derivative $\frac{d}{d{t}^{\prime }}$ and $dt=d{t}^{\prime }$.

$a^{\prime }=\frac{d{u}^{\prime }}{d{t}^{\prime }}$

Substitute $u-a_{0}t$ for $u^{\prime }$ in above equation,

$\begin{array}{c}{a}^{\prime }=\frac{d}{d{t}^{\prime }}\left(u-a_{0}t\right)\\ =\frac{du}{d{t}^{\prime }}-a_0\frac{dt}{d{t}^{\prime }}\\ =a-a_0\end{array}$

It is shown from the above equation that the accelerations are not identical.

The Newton’s second does not have same value in two different frames,

$F^{\prime }=m{a}^{\prime }$

Substitute $a-a_0$ for $a^{\prime }$ in above equation,

$\begin{array}{c}{F}^{\prime }=m\left(a-a_0\right)\\ =ma-m{a}^{\prime }\\ =F-m{a}^{\prime }\end{array}$

Conclusion:

Therefore, Newton’s second law is not valid in a reference frame moving past the laboratory frame with a constant acceleration.