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


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

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


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

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.


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

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


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

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


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

  • 1
    $\begingroup$ Do you have a reference for eq (8)? I can't find it in your link. $\endgroup$
    – Pulsar
    Commented Dec 12, 2013 at 20:47
  • $\begingroup$ Other than you find a similar form of it in Exercise (6.26) in Rindler: $\endgroup$
    – Anodinium
    Commented Dec 12, 2013 at 21:19
  • $\begingroup$ Deduce that if a 3-force $\vec{f}$ acts on a particle with constant rest mass $m_0$ moving at velocity $\vec{u}$ making an angle $\theta$ with $\vec{f}$, then $\endgroup$
    – Anodinium
    Commented Dec 12, 2013 at 21:22
  • $\begingroup$ $\alpha=m_0^{-1}f\gamma(u)[1-(u^2/c^2)cos^2\theta]^{1/2}$ $\endgroup$
    – Anodinium
    Commented Dec 12, 2013 at 21:24

2 Answers 2


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, $$ \alpha^2 = m_0^{-2}\gamma^2 f^2\left(1 - \frac{v^2}{c^2}\cos^2\theta\right)^2. $$

  • $\begingroup$ Thank you! I need some time to digest this though. I'm trying to get a better understanding in how the $\gamma$ looks like in the transformations when velocity and acceleration are not parallel. I was about to write Lorentz transformations, but perhaps Poincarè transformations are what should be preferred. I was thinking that there could be something qualitative to say about the perpendicular component of the acceleration when looking at it relativistically... $\endgroup$
    – Anodinium
    Commented Dec 13, 2013 at 0:34
  • $\begingroup$ I'm saying this because Rindler (the top link) keeps referring to the basic Lorentz transformations in the derivation of Equations (3) - (7). Look for example at Equation (78) in the first link in my prior post. $\endgroup$
    – Anodinium
    Commented Dec 13, 2013 at 0:38

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

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


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