Acceleration 4 vector question

[jason12345]

This is question 5.8 from Rindler's Relativity: Special, General, Cosmological book, with A, U the 4 acceleration and velocity; a, u the 3 acceleration and velocity. It has 3 parts which I think I've answered correctly below, but need help with the last part.

So could you have a look and comment on it?

a) Prove that, in any inertial frame where a is orthogonal to u, $$\textbf{A}= \gamma^{2} (\textbf{a},0)$$, and conversely.

my answer:

$$\dot{\gamma}=\gamma^{3}/c^{2}\textbf{a}\cdot\textbf{u}$$

$$\textbf{a}\cdot\textbf{u} = 0$$

$$\textbf{A}= \gamma d\textbf{U}/dt = \gamma d/dt(\gamma\textbf{u},\gamma c) = \gamma(\dot{\gamma}\textbf{u} + \gamma\textbf{a}, \dot{\gamma}c )$$

and so

$$\textbf{A}= \gamma^{2} (\textbf{a},0)$$ --- (1)

which proves the first part.

Given (1), then

$$\dot{\gamma}c = 0$$ and so

$$\dot{\gamma}=\gamma^{3}/c^{2}\textbf{a}\cdot\textbf{u} = 0$$

which is only possible for $$\textbf{a}\cdot\textbf{u} = 0$$

Hence the converse is also true.

b) Deduce that for any instantaneous motion it is possible to find an inertial frame S_|_ in which a, u are orthogonal to one another.

my answer:

In s_|_ , $$\textbf{A}= \gamma^{2} (\textbf{a},0)$$

I need to show $$\dot{\gamma}c$$ in A can transform to 0 in S_|_:

$$\gamma_{v}(\gamma_{u}\dot{\gamma_{u}} - v(\gamma_{u}\mathbf{a_{x}} + \dot{\gamma_{u}}\mathtbf{u_{x}})/c^{2}) = 0$$

$$v = \gamma_{u}\dot{\gamma_{u}}c^{2} / (\gamma_{u}\mathbf{a_{x}} + \dot{\gamma_{u}}\mathbf{u_{x}})$$

$$v = \gamma_{u}^{4}\mathbf{a}\cdot\mathbf{u} / (\gamma_{u}\mathbf{a_{x}} + \dot{\gamma_{u}}\mathbf{u_{x}})$$--- (2)

Hence there exists a v and therefore s_|_.

c) Moreover, prove that there is a whole class of such frames, all moving relative to S_|_ in directions orthogonal to a.

My answer:

The stationary frame can be rotated giving a new a_x, u_x in (2) and hence another v and s_|_. But I can't show they're moving relative to S_|_ in directions orthogonal to a.

Thanks for your help.

 

[mark726]

I appreciate the effort you have put into answering the first two parts of the question. It seems like you have a good understanding of the concepts involved.

For the last part, you are correct that there is a whole class of frames that satisfy the conditions of being orthogonal to a and moving relative to S_|_ in directions orthogonal to a. This is because rotations in the plane orthogonal to a do not affect the orthogonality between a and u.

To show this, we can consider a rotation matrix R that rotates the frame S_|_ by an angle θ in the plane orthogonal to a. This rotation matrix can be written as:

R = \begin{bmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix}

Using this rotation matrix, we can transform the velocities a and u into the new frame S'_|_:

\mathbf{a}' = R\mathbf{a} = \begin{bmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} a_x \\ a_y \\ a_z \end{bmatrix} = \begin{bmatrix} a_x\cos\theta - a_y\sin\theta \\ a_x\sin\theta + a_y\cos\theta \\ a_z \end{bmatrix}

Similarly,

\mathbf{u}' = R\mathbf{u} = \begin{bmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} u_x \\ u_y \\ u_z \end{bmatrix} = \begin{bmatrix} u_x\cos\theta - u_y\sin\theta \\ u_x\sin\theta + u_y\cos\theta \\ u_z \end{bmatrix}

Since we are only interested in the plane orthogonal to a, we can ignore the z-components of a and u. This means that the dot product between a and u in the new frame S'_|_ is given