I'm trying to understand solution of the following problem from Landau, Lifshitz, Classical Theory Of Fields: enter image description here

(ending skipped).

What I see when I "write out the expression for $w^iw_i$", using definition $w^i=\frac{du^i}{ds}=\frac{du^i}{cdt\sqrt{1-v^2/c^2}}$: $$\frac1{c^2-v^2}\left(\left(\frac{d}{dt}\frac1{\sqrt{1-v^2/c^2}}\right)^2-\left(\frac{d}{dt}\frac{v}{c\sqrt{1-v^2/c^2}}\right)^2\right)=-\frac{w^2}{c^4}$$

I then take the derivatives and arrive at the following result: $$-\frac{c^2\dot v^2}{(c^2-v^2)^3}=-\frac{w^2}{c^4}$$

I can get rid of the squares, use some guessing to determine correct signs, and then solve the resulting equation, getting correct results (according to the book). But the original solution in the book looks way simpler and must be better justified than mine.

So, the question: how did LL get their bottom-left equation? Did they "write out the expression for $w^iw_i$" in some other way than I did? How should I do here instead?

  • $\begingroup$ More on acceleration in special relativity. $\endgroup$
    – Qmechanic
    Commented Sep 24, 2013 at 18:15
  • $\begingroup$ See also this previous answer $\endgroup$
    – Trimok
    Commented Sep 24, 2013 at 18:50
  • $\begingroup$ @Trimok unfortunately that answer doesn't really answer my question. The question is how to derive the equation given in the book, not how to define proper acceleration. $\endgroup$
    – Ruslan
    Commented Sep 26, 2013 at 13:11
  • $\begingroup$ @Ruslan : I made an answer. $\endgroup$
    – Trimok
    Commented Sep 26, 2013 at 15:15

2 Answers 2


I work in units $c=1$, I use the notation $\gamma = \frac{1}{\sqrt{1-v^2}}$.

The relation between the proper time $\tau$ and the time $t$ is :

$$dt = \gamma ~ d \tau\tag{1}$$

In the particle-comoving inertial frame $F$, the 4-velocity of the particle is :

$$U^\mu = (1,0,0,0)\tag{2}$$

We define the acceleration by $A^\mu = \large \frac{dU^\mu}{d\tau}$, we note that, because $U^\mu U_\mu = 1$, then $A^\mu U_\mu = 0$, so the acceleration has only spatial component that we choose on the axis $x$ :

$$A^\mu = (0,w,0,0)\tag{3}$$

Now, in the stationnary inertial frame $F'$, the $4$ velocity is :

$$U'^\mu = \gamma (1,v,0,0)\tag{4}$$ You may check this by noticing that $U^\mu$ transforms like a vector in a Lorentz transformation, so we have simply :

$$\begin{pmatrix} U'^0\\U'^1\end{pmatrix} = \gamma \begin{pmatrix} 1&v\\v&1\end{pmatrix}\begin{pmatrix} U^0\\U^1\end{pmatrix}= \gamma \begin{pmatrix} 1&v\\v&1\end{pmatrix}\begin{pmatrix} 1\\0\end{pmatrix} \tag{5}$$

Now, looking at the acceleration $A'^\mu = \large \frac{dU'^\mu}{d\tau}$, in the stationary frame $F'$, we may notice that $A$ transforms also as a vector in a Lorentz transformation, so we have :

$$\begin{pmatrix} A'^0\\A'^1\end{pmatrix} = \gamma \begin{pmatrix} 1&v\\v&1\end{pmatrix}\begin{pmatrix} A^0\\A^1\end{pmatrix}= \gamma \begin{pmatrix} 1&v\\v&1\end{pmatrix}\begin{pmatrix} 0\\w\end{pmatrix} \tag{6}$$

So, we have : $$A'^1 = \frac{dU'^1}{d\tau}= \frac{d(\gamma v)}{d \tau} = \gamma w \tag{7}$$

That is :

$$\frac{d(\gamma v)}{dt} = w \tag{8}$$

See also this lecture of Susskind

  • $\begingroup$ Thanks, this is quite a clear derivation. But I still fail to see, how could this be derived starting from "expression for $A^\mu A_\mu$", mentioned in the book? $\endgroup$
    – Ruslan
    Commented Sep 26, 2013 at 16:53
  • $\begingroup$ $A^\mu$ transforms as a vector under Lorentz transformation, so you have an invariant ; $A^\mu A_\mu = A'^\mu A'_\mu$. If you want, you may start with this equation (instead of using equation $(6)$), and the fact that $A'^\mu U'_\mu = 0$, to find $A'^1$, and you get equation $(7)$ $\endgroup$
    – Trimok
    Commented Sep 26, 2013 at 17:36
  • $\begingroup$ Well, I was hoping to get insight into book's logic, that's why my tries to stick to $A^\mu A_\mu$. Anyway, I accept your answer since it still solves the problem, although not quite in a way I was looking for. $\endgroup$
    – Ruslan
    Commented Sep 26, 2013 at 17:46
  • $\begingroup$ Thanks, I think that my previous comment corresponds to the logic of the book, that is, using the invariant $A^\mu A_\mu $, instead of using the Lorentz transform $(6)$for the acceleration. $\endgroup$
    – Trimok
    Commented Sep 26, 2013 at 17:59
  • 1
    $\begingroup$ OK, thanks for your answer. I was completely stuck for more than a week trying to understand the solution. $\endgroup$
    – Ruslan
    Commented Sep 26, 2013 at 18:05

Landau's hint goes as follows:

Four-velocity is $$ u^{i} = \frac{dx^{i}}{ds} = \gamma_{v}(1,\frac{1}{c}\mathbf{v}), $$ where $$ \gamma_{v} = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}. $$ We have used $$ ds = \frac{cdt}{\gamma_{v}}. $$ To proceed, we note that $$ \dot{\gamma}_{v} = \frac{\gamma^{3}_{v}}{c^2}\mathbf{v}\cdot\dot{\mathbf{v}}, $$ therefore $$ w^{i} = \frac{dw^{i}}{ds} = \frac{\gamma_{v}}{c}\left\{\dot{\gamma}_{v}(1,\frac{1}{c}\mathbf{v})+\gamma_{v}(0,\frac{1}{c}\dot{\mathbf{v}})\right\} = \frac{\gamma^{2}_{v}}{c^2}(\frac{\gamma^{2}_{v}}{c}(\mathbf{v}\cdot\dot{\mathbf{v}}),\frac{\gamma^{2}_{v}}{c^2}(\mathbf{v}\cdot\dot{\mathbf{v}})\mathbf{v}+\dot{\mathbf{v}}). $$ Considering only one-dimensional motion (along $x$), using a initial reference system in which the particle with $v=0$ (but $\dot{v}=w\ne0$) at this moment, because $\gamma_{v}=1$, we have $$ w^{i} = (0,\frac{w}{c^2},0,0). $$ If the current velocity is not vanishing, we have $$ w^{i} = \frac{\gamma^{2}_{v}}{c^2}(\frac{\gamma^{2}_{v}}{c}v\dot{v},\frac{\gamma^{2}_{v}}{c^2}v^2\dot{v}+\dot{v},0,0) = \frac{\gamma^{4}_{v}}{c^2}(\frac{v}{c}\dot{v},\dot{v},0,0). $$ The scalar $w^{i}w_{i}$ is $$ w^{i}w_{i} = -\frac{\gamma^{6}_{v}}{c^4}\dot{v}^{2}. $$ Landau hints that it should be a constant under Lorentz transformation, i.e., $$ -\frac{\gamma^{6}_{v}}{c^4}\dot{v}^{2} = -\frac{w^2}{c^4}, $$ therefore $$ \gamma^{3}_{v}\dot{v} = w, $$ which is equivalent to $$ \frac{d}{dt}(\gamma_{v}v) = w. $$ Solving this differential equation with initial condition, we reach Landau's solution.


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