Relationship between mass preserving four-fources and proper acceleration

Question

Background

I studied Rindler's book on Relativity. Relevant information from this book is available on-line here:

http://www.scholarpedia.org/article/Special_relativity:_mechanics

Given that we work with four-vectors $\mathbf{R} = \left(\vec{r},ct\right)$ where the four-velocity is defined as

$$ \tag{1} \mathbf{U}=\frac{d\mathbf{R}}{d\tau}=\gamma\frac{d\mathbf{R}}{dt}=\gamma\left(u\right)\left(\vec{u},c\right), $$

$\vec{u}$ is the spacial "three-velocity" and the four-acceleration is defined as

$$ \tag{2} \mathbf{A}=\gamma\frac{d\mathbf{U}}{d\tau}=\gamma\frac{d}{dt}\left(\gamma\vec{u},\gamma c\right)=\gamma\left(\dot{\gamma}\vec{u}+\gamma\vec{a},\dot{\gamma}c\right), $$

we can define the four-force as

$$ \tag{3} \mathbf{F}=\frac{d}{ {d\tau } } \mathbf{P}=\frac{d}{ {d\tau}}(m_{0}\mathbf{U})=m_{0}\mathbf{A}+\frac{ {dm}_{0} }{ {d\tau} }\mathbf{U}. $$ The four-force $\mathbf{F}$ is said to be mass-preserving if $dm_0/d\tau$ equals zero, i.e. the force leaves the resting mass of the particle unchanged. Moreover

$$ \tag{4} \mathbf{F}=\frac{d}{d\tau}\mathbf{P}=\gamma(u)\frac{d}{ {dt} }( \vec{p},mc)=\gamma (u)\left(\vec{f},\frac{1}{c}\frac{dE}{dt}\right). $$

We note that $\mathbf{A}\cdot\mathbf{U}=0$ and therefore for a mass preserving four-force $\mathbf{F}\cdot\mathbf{U}=0$ from Eq. (3) or from Eq. (4)

$$ \tag{5} \mathbf{F}\cdot\mathbf{U}=\gamma^2(u)\left(\frac{dE}{dt}-\vec{f}\cdot\vec{u}\right)=0 $$

which means that

$$ \frac{dE}{dt}=\vec{f}\cdot\vec{u} $$

and

$$ \tag{6} \mathbf{F}=\gamma(u)\left(\vec{f},\vec{f}\cdot\vec{u}/c\right). $$

We also have the relationship

$$ \tag{7} \gamma(u)\vec{f}=m_0\frac{d^2\vec{r}}{d\tau^2}. $$

Moreover we have that the proper acceleration $\alpha$ is defined as

$$ \mathbf{A}\cdot\mathbf{A}=\mathbf{A}^2=-\alpha^2. $$

I studied the questions formulated in

Why proper acceleration is $du/dt$ and not $du/d\tau$?

Relativistic factor between coordinate acceleration and proper acceleration

and realized that the spatial vectors can be broken down into a parallel component and an orthogonal component, i.e. $\vec{v}=\vec{v}_\parallel+\vec{v}_\perp$ or $\vec{a}=\vec{a}_\parallel+\vec{a}_\perp$ where one makes a distinction between the component of the velocity that is parallel with the acceleration and vice versa.

Question

Now it can be shown that for a force $\vec{f}$ acting upon a particle traveling with a velocity vector $\vec{u}$ where there is an angle $\theta$ between $\vec{f}$ and $\vec{u}$ the relationship between the proper acceleration of the particle and the force $\vec{f}$ is

$$ \tag{8} \vec{f}^2=m_0^2\frac{\gamma_\parallel^2}{\gamma^2}\alpha^2 $$

where

$$ \tag{9} \gamma_\parallel=\frac{1}{\sqrt{1-u_\parallel^2/c^2}}\;\;\;\text{and}\;\;\;\gamma=\frac{1}{\sqrt{1-u^2/c^2}}. $$

Now how can I derive or prove equation (8)? I have tried every feasible combination of Eq. (1) to (7) without success. I've also tried the more basic variants of the Lorentz transformation

https://en.wikipedia.org/wiki/Lorentz_transformation

but it gets very confusing when there are two components to consider when assessing the relativistic effects. Any ideas?

Answer 1

It's quite easy to get confused by differences in notation: in this post of mine I defined $v_\parallel$ and $v_\perp$ as the velocity components parallel and perpendicular to the coordinate acceleration $\vec{a}$. In your question however, you're interested in the angle between the velocity and the force vector $\vec{f}$, and that's a different angle: $\vec{f}$ and $\vec{a}$ are not parallel! So I'll try to be as careful as I can to avoid further confusion.

I'll denote the coordinate velocity as $\vec{v}$, and the spacial part of the four-velocity vector as $\vec{U}$, thus $$ \vec{U} = \frac{\text{d}\vec{x}}{\text{d}\tau}=\gamma\vec{v}, $$ and $$ \gamma\vec{f} = m_0\frac{\text{d}\vec{U}}{\text{d}\tau}. $$ The acceleration four-vector $\boldsymbol{A}$ is $$ \boldsymbol{A} = \left(c\frac{\text{d}\gamma}{\text{d}\tau},\frac{\text{d}\vec{U}}{\text{d}\tau} \right), $$ and its scalar product is equal to $\boldsymbol{A}\bullet\boldsymbol{A} = A_0^2 - (\vec{A})^2 =- \alpha^2$, with $\alpha$ the proper acceleration. Thus $$ \alpha^2 = \left(\frac{\text{d}\vec{U}}{\text{d}\tau}\right)^2 - c^2\left(\frac{\text{d}\gamma}{\text{d}\tau}\right)^2 = m_0^{-2}\gamma^2 f^2 - c^2\left(\frac{\text{d}\gamma}{\text{d}\tau}\right)^2. $$ Now, $$ \frac{\text{d}\gamma}{\text{d}\tau} = \gamma^4\frac{\vec{v}\cdot\vec{a}}{c^2}, $$ where $\vec{a} = \text{d}\vec{v}/\text{d}t$, and $$ \frac{\text{d}\vec{U}}{\text{d}\tau} = m_0^{-1}\gamma \vec{f} = \frac{\text{d}\gamma}{\text{d}\tau}\vec{v} + \gamma\frac{\text{d}\vec{v}}{\text{d}\tau} = \gamma^4\left(\frac{\vec{v}\cdot\vec{a}}{c^2}\right)\vec{v} + \gamma^2\vec{a}. $$ As you can see, $\vec{f}$ and $\vec{a}$ will in general not be parallel. If we take the inner product of $\vec{f}$ and $\vec{v}$ and divide by $c^2$, we get $$ \begin{align} \frac{m_0^{-1}}{c^2}\gamma \left(\vec{f}\cdot\vec{v}\right) = \frac{m_0^{-1}}{c^2}\gamma fv\cos\theta &= \gamma^4\left(\frac{\vec{v}\cdot\vec{a}}{c^2}\right)\frac{v^2}{c^2} + \gamma^2\left(\frac{\vec{v}\cdot\vec{a}}{c^2}\right)\\ &= \gamma^4\frac{\vec{v}\cdot\vec{a}}{c^2}\\ &= \frac{\text{d}\gamma}{\text{d}\tau}. \end{align} $$ In other words, $$ \alpha^2 = m_0^{-2}\gamma^2 f^2\left(1 - \frac{v^2}{c^2}\cos^2\theta\right)^2. $$

Answer 2

Take the scalar product of $(1)$ and $(4)$ to get $$\mathbf{F\cdot U}= \gamma^2(u)\left(\frac{dE}{dt} - \vec f\cdot \vec u\right)$$

In the proper frame this equals $\frac{dE}{dt}$ the rate of change of the rest energy so that for a rest-mass preserving force where $\frac{dE}{dt}=0$: $\mathbf{F\cdot U}= 0$. The four-force in this case can now be written as

$$\mathbf{F} = \gamma(u)(\vec f, \vec f\cdot\vec u/c)$$ giving $$\mathbf{F}^2 = \gamma^2(u)f^2(1 - (u\cos\theta)^2/c^2) = m_0^2\alpha^2$$ $$f^2 = m_0^2\alpha^2\frac{(1 - (u\cos\theta)^2/c^2)}{\gamma^2(u)}=m_0^2\alpha^2\frac{\gamma^2_{||}}{\gamma^2} $$

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Rindler: Special, General and Cosmological; page 124, equations (6.43), (6.44)