Consider a frame $S$ and $S'$ which is coincides at $t=0$ and then $S'$ starts moving with velocity $v$ in $+x$ direction.
By Lorentz transformation equation,
\begin{align} x'&=\gamma(x-vt) \\ t'&=\gamma\left(t-\frac{vx}{c^2}\right) \\ x&=\gamma(x'+vt') \\ t&=\gamma\left(t'+\frac{vx'}{c^2}\right)\end{align} where $\gamma=\sqrt{1-\frac{v^2}{c^2}}$ .
Consider a body $A$ having coordinates $(x,y,z,t)$ in $S$ frame and $(x',y',z',t')$ in $S'$ frame.
In time $dt$, coordinates become $(x+dx,y+dy,z+dz,t+dt)$ in $S$ frame.
In $S'$ frame, in interval $dt'$, coordinate become $(x'+dx',y'+dy',z'+dz',t'+dt')$

In books the derivation for velocity transformation equations are given as

Dividing the numerator and denominator by $dt$, we get
$\displaystyle u'_x=\frac{\frac{dx}{dt}-v}{1-\frac{v(dx/dt)}{c^2}}$
$\displaystyle u'_x=\frac{u_x-v}{1-\frac{vu_x}{c^2}}$

I have tried the derivation using chain rule of differentiation,
As $x'=x'(x,t)$
$\frac{dx'}{dt'}=\frac{\partial x'}{\partial x}\Big|_t\frac{dx}{dt'}+\frac{\partial x'}{\partial t}\Big|_x\frac{dt}{dt'}\tag{1}$
As the transformation equations are invertible so,
So, $\frac{dx}{dt'}=\frac{\partial x}{\partial x'}\Big |_{t'}\frac{dx'}{dt'}+\frac{\partial x}{\partial t'}\Big |_{x'}\frac{dt'}{dt'}\tag{2}$

$\frac{dt}{dt'}=\frac{\partial t}{\partial x'}\Big|_{t'}\frac{dx'}{dt'}+\frac{\partial t}{\partial t'}\Big |_{x'}\frac{dt'}{dt'}\tag{3}$

Plugging $(2)$ and $(3)$ in $(1)$,
$\frac{dx'}{dt'}=\frac{\partial x'}{\partial x}\Big|_t\Big(\frac{\partial x}{\partial x'}\Big |_{t'}\frac{dx'}{dt'}+\frac{\partial x}{\partial t'}\Big |_{x'}\Big)+\frac{\partial x'}{\partial t}\Big|_{x}\Big(\frac{\partial t}{\partial x'}\Big|_{t'}\frac{dx'}{dt'}+\frac{\partial t}{\partial t'}\Big |_{x'}\Big)$

I have calculated all the above partial derivatives from Lorentz transformation. And I get
$\implies 1-1=0$
$\implies 0=0$

I have two doubts,
i) If I use chain rule for differentiation then why I am not able to get transformation equation for velocity. Why I get $0=0$ instead of getting an expression for $u'_x$? Have I made some mistake or there is some other way to derive it using chain rule?
ii) The way used in books is that they treated derivative as fraction. They write expression for $dx'$ and $dt'$ and divide them. But isn't derivative a sort of operation. In elementary calculus courses, we have been told that $\frac{dx}{dt}$ is actually $\frac{d}{dt}(x)$. It is not like dividing $dx$ and $dt$. I get very confused how derivative as a fraction is justified.

Please help!

  • $\begingroup$ If you put x=.. and t=.. in first two equations you obtain x‘=x‘ and t‘=t‘ but this is trivial. so you can’t do it like this $\endgroup$
    – Eli
    May 13, 2021 at 7:43
  • $\begingroup$ But then how to prove the velocity transformation equations using chain rule? $\endgroup$
    – Iti
    May 13, 2021 at 7:52
  • $\begingroup$ I give you a hint, I'am not sure if it works. $\;x',t'\:$ are not explicit functions of $\:x,t\:$ as in Lorentz transformation. They refer to the space-time coordinates on the worldline of motion of the particle. I think you must consider the parametric equation $\:x(\tau),t(\tau)\:$ in the rest frame $\:S\:$ with parameter the proper time $\:\tau$. Similarly you have the parametric equation $\:x'(\tau),t'(\tau)\:$ in the moving frame $\:S'$. Then use the chain rule with differentials with respect to $\:\tau$. $\endgroup$
    – Frobenius
    May 13, 2021 at 8:39
  • $\begingroup$ @Forbenius, how can I parametrize the curve with $\tau$. You said that the parameter $\tau$ is the proper time $\tau$. That means $\tau=t$ and $x$ becomes $x(t)$. I am not able to understand what exactly you are saying. Are you referring proper time to be the time elapsed in that frame in which the body appears to be at rest (frame also move with body's velocity)? Can you please explain in detail. $\endgroup$
    – Iti
    May 13, 2021 at 9:12
  • 1
    $\begingroup$ $\:\tau\:$ is a Lorentz invariant parameter. $$\gamma^{-1}(u)\mathrm dt =\mathrm d\tau=\gamma^{-1}(u')\mathrm dt'$$ If you don't know and/or you don't have experience about the proper time don't try my hint. $\endgroup$
    – Frobenius
    May 13, 2021 at 9:40

3 Answers 3


The particle is moving in the 1+1-dimensional Minkowski space-time on a (world-line) curve which could be represented as function of the parameter $\,\tau$, the proper time. More precisely $\,s=c\,\tau\,$ is the relativistic $''$arc length$''$, a scalar invariant under Lorentz transformations. It corresponds to the arc length (natural) parameter $\,s\,$ of Euclidean curves, a scalar invariant under space transformations.

So, for the parametric representation of the curve we have \begin{align} \texttt{in system } \mathrm S & : \mathbf X\left(\tau\right)\boldsymbol{=}\left[x\left(\tau\right),t\left(\tau\right)\right] \tag{01a}\label{01a}\\ \texttt{in system } \mathrm S'\! & : \mathbf X'\!\left(\tau\right)\boldsymbol{=}\left[x'\!\left(\tau\right),t'\!\left(\tau\right)\right] \tag{01b}\label{01b} \end{align} The space-time coordinates are related by a Lorentz boost transformation with velocity $\,\upsilon \in \left(-c,c\right)\,$ along the common $\,x,x'-$axis in differential form \begin{align} \mathrm dx' & \boldsymbol{=} \gamma_v\left(\mathrm dx\boldsymbol{-}\upsilon \mathrm dt\right) \tag{02a}\label{02a}\\ \mathrm dt' & \boldsymbol{=}\gamma_v\left(\mathrm dt\boldsymbol{-}\dfrac{\upsilon}{c^2} \mathrm dx\right) \tag{02b}\label{02b}\\ \gamma_v & \boldsymbol{=}\left(1\boldsymbol{-}\dfrac{\upsilon^2}{c^2}\right)^{\boldsymbol{-}\frac12} \tag{02c}\label{02c} \end{align} Now by the chain rule and differentiation with respect to $\,\tau\,$ we have \begin{align} u'_x& \boldsymbol{=}\dfrac{\mathrm dx'}{\mathrm dt'} \boldsymbol{=}\dfrac{\dfrac{\mathrm dx'}{\mathrm d\tau}}{\dfrac{\mathrm dt'}{\mathrm d\tau}} \boldsymbol{=}\dfrac{\dfrac{\partial x'}{\partial x}\dfrac{\mathrm dx}{\mathrm d\tau}\boldsymbol{+}\dfrac{\partial x'}{\partial t}\dfrac{\mathrm dt}{\mathrm d\tau}}{\dfrac{\partial t'}{\partial x}\dfrac{\mathrm dx}{\mathrm d\tau}\boldsymbol{+}\dfrac{\partial t'}{\partial t}\dfrac{\mathrm dt}{\mathrm d\tau}}\boldsymbol{=}\dfrac{\gamma_v\dfrac{\mathrm dx}{\mathrm d\tau}\boldsymbol{+}\left(\boldsymbol{-}\gamma_v\upsilon\right)\dfrac{\mathrm dt}{\mathrm d\tau}}{\left(\boldsymbol{-}\gamma_v\dfrac{\upsilon}{c^2} \right)\dfrac{\mathrm dx}{\mathrm d\tau}\boldsymbol{+}\gamma_v\dfrac{\mathrm dt}{\mathrm d\tau}} \nonumber\\ & \boldsymbol{=}\dfrac{\gamma_v\left(\dfrac{\mathrm dx}{\mathrm d\tau}/\dfrac{\mathrm dt}{\mathrm d\tau}\right)\boldsymbol{-}\gamma_v\upsilon}{\left(\boldsymbol{-}\gamma_v\dfrac{\upsilon}{c^2} \right)\left(\dfrac{\mathrm dx}{\mathrm d\tau}/\dfrac{\mathrm dt}{\mathrm d\tau}\right)\boldsymbol{+}\gamma_v} \boldsymbol{=}\dfrac{\gamma_v\dfrac{\mathrm dx}{\mathrm dt}\boldsymbol{-}\gamma_v\upsilon}{\left(\boldsymbol{-}\gamma_v\dfrac{\upsilon}{c^2} \right)\dfrac{\mathrm dx}{\mathrm dt}\boldsymbol{+}\gamma_v} \nonumber\\ & \boldsymbol{=}\dfrac{u_x\boldsymbol{-}\upsilon}{\boldsymbol{-}\dfrac{\upsilon}{c^2} u_x\boldsymbol{+}1} \tag{03}\label{03} \end{align} that is \begin{equation} u'_x \boldsymbol{=}\dfrac{u_x\boldsymbol{-}\upsilon}{1\boldsymbol{-}\dfrac{\upsilon u_x}{c^2} } \tag{04}\label{04} \end{equation}



An other way :

\begin{align} \require{cancel} u'_x & \boldsymbol{=}\dfrac{\mathrm dx'}{\mathrm dt'} \boldsymbol{=}\dfrac{\dfrac{\partial x'}{\partial x}\mathrm dx\boldsymbol{+}\dfrac{\partial x'}{\partial t}\mathrm dt}{\dfrac{\partial t'}{\partial x}\mathrm dx\boldsymbol{+}\dfrac{\partial t'}{\partial t}\mathrm dt}\boldsymbol{=}\dfrac{\left(\dfrac{\partial x'}{\partial x}\dfrac{\mathrm dx}{\mathrm dt}\boldsymbol{+}\dfrac{\partial x'}{\partial t}\right)\cancel{\mathrm dt}}{\left(\dfrac{\partial t'}{\partial x}\dfrac{\mathrm dx}{\mathrm dt}\boldsymbol{+}\dfrac{\partial t'}{\partial t}\right)\cancel{\mathrm dt}} \nonumber\\ & \boldsymbol{=}\dfrac{\gamma_v\dfrac{\mathrm dx}{\mathrm dt}\boldsymbol{-}\gamma_v\upsilon}{\left(\boldsymbol{-}\gamma_v\dfrac{\upsilon}{c^2} \right)\dfrac{\mathrm dx}{\mathrm dt}\boldsymbol{+}\gamma_v} \boldsymbol{=}\dfrac{u_x\boldsymbol{-}\upsilon}{\boldsymbol{-}\dfrac{\upsilon}{c^2} u_x\boldsymbol{+}1} \tag{A-01}\label{A-01} \end{align} that is \begin{equation} u'_x \boldsymbol{=}\dfrac{u_x\boldsymbol{-}\upsilon}{1\boldsymbol{-}\dfrac{\upsilon u_x}{c^2} } \tag{A-02}\label{A-02} \end{equation}

Any way we use is equivalent to the division of the Lorentz equations \eqref{02a},\eqref{02b} side by side so I don't think that there exists any sense of the chain rule use. It doesn't give to us a better way or something new.

  • $\begingroup$ I have understood the proof. But I have difficulty in understanding the initial part (invariance of arc length under space transformations). Isn't arc length also gets transformed in relativistic kinematics (length contraction), so how this remain invariant. I have studied very little differential geometry, in that there is a concept "unit speed reparameterization of curves". Is it somewhat related to the unit speed reparameterization"? $\endgroup$
    – Iti
    May 13, 2021 at 15:10
  • 1
    $\begingroup$ Proper time $\:\tau\:$ is that of a clock attached to the particle. It's invariant in the following sense : Suppose that the particle is moving from a point A to a point B infinitesimaly close to A spending the infinitesimal time $\,\mathrm dt\,$ as seen from observer $\,\mathrm S$, time $\,\mathrm dt'\,$ as seen from observer $\,\mathrm S'$ and $\,\mathrm d\tau\,$ as seen from an observer on the particle.... $\endgroup$
    – Frobenius
    May 13, 2021 at 15:35
  • 1
    $\begingroup$ ....If you ask observer $\,\mathrm S\,$ what is the time period that the clock on the particle is counting he/she will respond with $\sqrt{1-u^2_x/c^2}\mathrm dt=\mathrm d\tau $ while if you ask observer $\,\mathrm S'\,$ what is the time period that the clock on the particle is counting he/she will respond with $\sqrt{1-u'^2_x/c^2}\mathrm dt'=\mathrm d\tau $. That is any inertial observer will agree that that the particle-time (proper time) is $\mathrm d\tau$ $\endgroup$
    – Frobenius
    May 13, 2021 at 15:36
  • $\begingroup$ Thanks for the reply, so from the discussion i have made the following conclusion: In S frame the particle has certain trajectory $x(t)$. Proper time (time measured at constant position) $(\tau)$ is the smallest time interval measured in the frame of the particle. So we can write $t=\gamma\tau$. So, $t(\tau)$, thus $x(\tau)$. Hence, $x'(\tau)$. Is my conclusion correct? I have last doubt that without choosing $\tau$ why we have difficulty in deriving the velocity transformation equations using chain rule? $\endgroup$
    – Iti
    May 13, 2021 at 15:56
  • 1
    $\begingroup$ Precisely !!! We take advantage of the invariance of $\,\tau\,$ to represent the curve in any inertial frame with the same parameter. If we choose an other non-invariant scalar parameter, say $\,\lambda\,$ in $\,\rm S$, in system $\,\rm S'\,$ would be $\,\lambda'\,$ and to use the chain rule we must find the functional relation $\,\lambda'=f(\lambda)$. By the way may be there exists another way to derive the velocity transformation equations using chain rule which we miss. $\endgroup$
    – Frobenius
    May 13, 2021 at 21:55

Let the velocity of $S′$ in relation to $S$ in x-direction be $v_0$, let $\gamma_0=\gamma(v_0)=\frac{1}{\sqrt{1-v_0^2/c^2}}$ and let a position and velocity in spacetime be $$x^\mu=(ct,x,y,z)^T,\,\,\,\,\,u^\mu=(c,\vec v)^T=(c,v_x,v_y,v_z)^T.$$ If you want to solve it using differentials (which is formally and mathematically not the best way) you can first calculate the lorentztransformation of the spacetime coordinates $x$: $$\begin{pmatrix}ct'\\x'\\y'\\z'\end{pmatrix}\equiv x'^\mu=\Lambda^\mu\,_\nu x^\nu=\begin{pmatrix} c\gamma_0(t-\frac{v_0}{c^2}x)\\ \gamma_0(x-v_0t)\\ y\\ z\end{pmatrix}.\,\,\,\,\,\,\,\,\,(1)$$ Thus we get for the velocity $$\gamma'\begin{pmatrix}c\\ v'_x\\v'_y\\v'_z\end{pmatrix}\equiv u'^\mu=\frac{dx'^\mu}{d\tau}=\gamma'\frac{dx'^\mu}{dt'}\stackrel{(1)}{=}\gamma'\frac{d}{d(\gamma_0(t-\frac{v_0}{c^2}x))}\begin{pmatrix} c\gamma_0(t-\frac{v_0}{c^2}x)\\ \gamma_0(x-v_0t)\\ y\\ z\end{pmatrix}=\gamma'\begin{pmatrix}c\\\frac{\gamma_0(dx-v_0dt)}{\gamma_0(dt-\frac{v_0}{c^2}dx)}\\ \frac{dy}{\gamma_0(dt-\frac{v_0}{c^2}dx)}\\ \frac{dz}{{\gamma_0(dt-\frac{v_0}{c^2}dx)}}\end{pmatrix},$$ where $\tau$ is the proper time. Dividing the numerators and denominators by $dt$ and simplifying, the space components become $$\gamma'\begin{pmatrix}v'_x\\v'_y\\v'_z\end{pmatrix}=\gamma'\begin{pmatrix}\frac{v_x-v_0}{1-\frac{v_0}{c^2}v_x}\\\frac{v_y}{\gamma_0(1-\frac{v_0}{c^2}v_x)}\\\frac{v_z}{\gamma_0(1-\frac{v_0}{c^2}v_x)}\end{pmatrix}$$ yielding the same result as in my other formally correcter answer: $$\vec v'(\vec v)=\frac{1}{1-\frac{v_0}{c^2}v_x}\begin{pmatrix}v_x-v_0\\ v_y/\gamma_0\\ v_z/\gamma_0\end{pmatrix}.$$


Let the velocity of $S'$ in relation to $S$ in x-direction be $v_0$, let $\gamma_0=\gamma(v_0)=\frac{1}{\sqrt{1-v_0^2/c^2}}$ and let the velocity in a System be $$(u^\mu)=\gamma(c,v_x,v_y,v_z)^T.$$ The velocity $u$ transforms as follows:$$u'^\mu=\Lambda^\mu\,_\nu u^\nu=\begin{pmatrix} \gamma_0 & -\gamma_0v_0/c & 0&0\\ -\gamma_0v_0/c & \gamma_0 & 0&0\\ 0&0&1&0\\ 0&0&0&1 \end{pmatrix}(u^\mu)$$ $$\Leftrightarrow \gamma'\begin{pmatrix}c\\ v_x'\\ v_y'\\ v_z' \end{pmatrix}=\gamma\begin{pmatrix} \gamma_0(c-v_0v_x/c)\\ \gamma_0(-v_0+v_x)\\ v_y\\ v_z\end{pmatrix}$$ From the first component we get $$\gamma'=\gamma\gamma_0(1-\frac{v_0v_x}{c^2}).\,\,\,\,\,\,(1)$$ For the second component we get $$v_x'=\frac{\gamma}{\gamma'}\gamma_0(v_x-v_0)\stackrel{(1)}{=}\frac{v_x-v_0}{1-\frac{v_0v_x}{c^2}}.$$ For the third component we get $$v_y'=\frac{\gamma}{\gamma'}v_y\stackrel{(1)}{=}\frac{v_y/\gamma_0}{1-\frac{v_0v_x}{c^2}}.$$ For the fourth component we get $$v_z'=\frac{\gamma}{\gamma'}v_z\stackrel{(1)}{=}\frac{v_z/\gamma_0}{1-\frac{v_0v_x}{c^2}}.$$

  • $\begingroup$ I don't believe this answers the question. The OP wants to use the chain rule. $\endgroup$
    – garyp
    May 13, 2021 at 10:54
  • $\begingroup$ @garyp, yes I basically want to understand how to prove the transformation equations of velocities using chain rule. $\endgroup$
    – Iti
    May 13, 2021 at 11:01

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