0
$\begingroup$

First of all, these answers (How to derive the law of velocity transformation using chain rule?, The chain rule and velocity transformation in relativity, and other from a quick search on this site.) didn't help me at all. $$ * * * $$

Consider the frame of reference $S$ and another frame $S'$. Both inertial ones and in standard configuration (which means that this question deals only with 1D motion on $x-axis$ direction, and the simplest form of Lorentz Transformation will be used).

Now, consider the following points, which may be disconnected but I will explain and merge then together in the final item:

I) Coordinate Transformations

Consider, for example, the change of coordinates from polar coordinates to cartesian. The mathematical fact is that the coordinates functions which relates older coordinates to the new ones are:

$$x(r, \theta)$$

$$y(r, \theta)$$

And the inverse:

$$r(x,y)$$

$$\theta(x,y)$$

These are examples of coordinate transformation equations which relates the cartesian system with polar system of coordinates. Then, following the same spirit, for both coordinates cartesian systems $S$ and $S'$ we will have, undoubtly:

$$x'(t, x)$$

$$t'(t, x)$$

And the inverse:

$$x(t', x')$$

$$t(t', x')$$

It is wrong to assume a priori that the transformations are in the form as:

$$x'(t(t'), x(t')) \tag{1} $$

$$ t'(t(t'), x(t')) $$

And the inverse:

$$x(t'(t), x'(t)) \tag{2}$$

$$t(t'(t), x'(t))$$

II) The Chain Rule, the "differential" (total differential), directional derivative and total derivative

The chain rule from one single-variable calculus is undoubtly:

If $f(y(x))$ then:

$$\frac{df}{dx} = \frac{df}{dy}\frac{dy}{dx} \tag{3}$$

The chain rule from multivariable calculus is:

If $f(x,y)$ and $x(u,v) , y(u,v)$ then:

$$\frac{\partial f}{\partial u} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial u} \tag{4} $$

$$ \frac{\partial f}{\partial v} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial v}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial v} \tag{5}$$

The total differential of a function is:

If $f(x,y)$, then:

$$\Delta f=\frac{\partial f}{\partial x}.\Delta x + \frac{\partial f}{\partial y}\Delta y$$

$$df=: \lim_{\Delta t \to 0} \frac{\partial f}{\partial x}.\Delta x + \frac{\partial f}{\partial y}\Delta y = \frac{\partial f}{\partial x}.dx + \frac{\partial f}{\partial y}dy \tag{6}$$

Of course if we divide by $\Delta t$ the both sides and take the limit procedure we will have the total derivative:

$$\frac{\Delta f}{\Delta t}=\frac{\partial f}{\partial x}.\frac{\Delta x}{\Delta t} + \frac{\partial f}{\partial y}\frac{\Delta y}{\Delta t}$$

$$\lim_{\Delta t \to 0}\frac{\Delta f}{\Delta t}= \lim_{\Delta f \to 0} \frac{\partial f}{\partial x}.\frac{\Delta x}{\Delta t} + \frac{\partial f}{\partial y}\frac{\Delta y}{\Delta t} = \frac{\partial f}{\partial x}.\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{d y}{d t} := \frac{df}{dt} \tag{7}$$

Note that the quotient by $\Delta t$ if a valid procedure because this quantity is a number, not a infinitesiamal quantity like $dt$. So, simply saying that "we can divide $dx$ by $dt$" leads you to a ill-defined procedure.

The directional derivative is defined as:

If $f(x,y)$ and also, $\alpha(t) = (x(t),y(t))$ a curve parametrized bt $t$, then the derivative of $f$ along the curve $\alpha(t)$ in the direction of $\vec{t}$ is:

$$\frac{d}{dt}(f\circ \alpha)(t) \equiv \frac{df(x(t),y(t))}{dt} =\langle \vec{\nabla}f, \vec{t} \rangle =: \sum_{i=1}^{2}\frac{\partial f}{\partial x^{i}}\frac{dx^{i}}{dt} \tag{8}$$

where $i=\{1,2\}$ and $\{x^{1},x^{2}\} = \{x,y\}$

Of course that in the level of multivariable calculus we can say that $(8)$ is "is the same as" $(7)$.

III) Vectors and coordinate transformations

A vector (more precisely components of contravariant vector fields) are the quantities which under a change of coordinates(i.e. $x^{j} \to x'^{i}(x^{j})$, transforms as:

$$V'^{i} = \frac{\partial x'^{i}(x^{j})}{\partial x^{j} } V^{j} \tag{9}$$

IV) Velocity components

Consider then the $S$ frame and $S'$. The mean velocity is a quantity which are build upon the concept of displacement $\Delta x^{i} = \mid x^{i}(t)-x^{i}(t_{0})\mid$ and time interval $\Delta t = t- t_{0}$ , forming the quotient (for instance using the $S$ frame):

$$ \bar{v^{i}} = \frac{x^{i}(t)-x^{i}(t_{0})}{t- t_{0}} = \frac{\Delta x^{i}}{\Delta t} \tag{10}$$

The components of velocity vector are then the derivative of $(10)$:

$$ v^{i} \equiv \frac{dx^{i}(t)}{dt} =: \lim_{\Delta t \to 0} \frac{\Delta x^{i}}{\Delta t} \tag{11}$$

And is a completely non-sense to write something like:

$$ v^{i} \equiv \frac{dx^{i}(t)}{dt'} =: \lim_{\Delta t \to 0} \frac{\Delta x^{i}}{\Delta t'} \tag{12}$$

Because an observer $O$ at rest with respect to $S$, constructs the velocity ONLY WITH $S$ coordinates i.e. $(t,x)$. Because this observer measures the physical systems with non-primed quantities. Conversely,for an observer $O'$ in $S'$, he or she constructs the velocity $v'$ with coordinates $(t',x')$.

Then we have:

The components of velocity vector of $S$:

$$ v^{i} \equiv \frac{dx^{i}(t)}{dt} =: \lim_{\Delta t \to 0} \frac{\Delta x^{i}}{\Delta t} \tag{13}$$

The components of velocity vector of $S'$:

$$ v'^{i} \equiv \frac{dx'^{i}(t)}{dt'} =: \lim_{\Delta t' \to 0} \frac{\Delta x'^{i}}{\Delta t'} \tag{14}$$

V) Lorentz Transformations

Consider the following transformations between $S$ and $S'$ CARTESIAN SYSTEMS $(t,x,y,z)$ and $(t',x',y',z')$, AND IN STANDARD CONFIGURATION:

$$\displaystyle \begin{cases} x' = \gamma(x-vt)\\y'=y\\z'=z\\t'=\gamma \Big(t-\displaystyle \frac{v}{c^{2}}x\Big) \end{cases} \tag{15}$$

If we take the differential of $dt'$ we will have:

$$dt' = \gamma \Big( dt - \frac{v}{c^2}dx \Big) = \gamma \Big( 1 - \frac{v}{c^2}\frac{dx}{dt}\Big) dt = \gamma \Big( 1 - \frac{v}{c^2}u^{x}\Big) dt\tag{16}$$

So, by $(16)$ we can define:

$$ \frac{\mathrm{dt'}}{\mathrm{dt}} =: \gamma \Big( 1 - \frac{v}{c^2}u^{x}\Big) \tag{17}$$

and

$$ \frac{\mathrm{dt}}{\mathrm{dt'}} =: \frac{1}{\gamma \Big( 1 - \frac{v}{c^2}u^{x}\Big)} \tag{18}$$

$(17)$ and $(18)$ have the limit procedure of $(7)$ assumed.

VI) Velocity Transformations

These are the velocity transformations between $S$ and $S'$ $(S \to S')$:

$$ \begin{cases} u'^{x} = \frac{u^{x} - V}{1 - \frac{V}{c^{2}}u^{x}} \\ u'^{y} = \frac{u^{y}}{\gamma \Big(1 - \frac{V}{c^{2}}u^{x}\Big)} \\ u'^{z} = \frac{u^{z}}{\gamma \Big(1 - \frac{V}{c^{2}}u^{x}\Big)} \\ \end{cases} \tag{19}$$

$$ * * * $$

The Question:

Consider then the Lorentz transformations from $S \to S'$, i.e. the set $(15)$.

It is absolute to say that $x'(t,x,y,z)$, $y'(t,x,y,z)$, $z'(t,x,y,z)$ and $t'(t,x,y,z)$. If we aim to construct the velocity (for instance in $x'$ coordinate) components in $S'$ then we have to assume a curve $\alpha'(t')$ and then take the directional derivative (or total derivative) for the coordinates $x'(t(t'),x(t'),y(t'),z(t'))$, $y'(t(t'),x(t'),y(t'),z(t'))$, $z'(t(t'),x(t'),y(t'),z(t'))$ and $t'(t(t'),x(t'),y(t'),z(t'))$ :

$$ \frac{dx'}{dt'} =\frac{\partial x'}{\partial t}\frac{dt'}{dt} + \frac{\partial x'}{\partial x}\frac{dx}{dt'} + \frac{\partial x'}{\partial y}\frac{dy}{dt} + \frac{\partial x'}{\partial z}\frac{dz}{dt} = \frac{\partial x'}{\partial t}\frac{dt'}{dt} + \frac{\partial x'}{\partial x}\frac{dx}{dt'} + 0 + 0 = \frac{\partial}{\partial t}[\gamma(x-vt)]\frac{dt'}{dt} + \frac{\partial}{\partial x}[\gamma(x-vt)]\frac{dx}{dt'} = \gamma(u^{x} - v) + \gamma u^{x} \neq u'^{x} \tag{20} $$

Conversely, consider the coordinate transformation $S \to S'$, i.e. the set $(15)$, nothing special.Then the vector components (for instance $u'^{x}$) transforms as $(9)$. Then we have:

$$u'^{x}=\frac{dx'}{dt'} = \frac{\partial x'(t,x,y,z)}{\partial x^{j}}\frac{dx^{j}}{dt} = \frac{\partial x'(t,x,y,z)}{\partial x^{j}} u^{j}$$

Then,

$$ \frac{dx'}{dt'} = \frac{\partial x'(t,x,y,z)}{\partial x^{j}}\frac{dx^{j}}{dt} = \frac{\partial x'(t,x,y,z)}{\partial t}\frac{dt}{dt} + \frac{\partial x'(t,x,y,z)}{\partial x}\frac{dx}{dt} + \frac{\partial x'(t,x,y,z)}{\partial y}\frac{dy}{dt} + \frac{\partial x'(t,x,y,z)}{\partial z}\frac{dz}{dt} = \frac{\partial x'(t,x,y,z)}{\partial t}\frac{dt}{dt} + \frac{\partial x'(t,x,y,z)}{\partial x}\frac{dx}{dt} + 0 + 0 = \gamma(u^{x} -v) + \gamma u^{x} \neq u'^{x} \tag{21} $$

Of course that $(20)$ and $(21)$ aren't the velocity transformations in $x$ direction.

But, consider the following equation:

$$\frac{dx'}{dt'} = \frac{dx'}{dt} \frac{dt}{dt'} \tag{22}$$

By this expression we reach the right velocity transformations in $x$ direction:

$$ \frac{dx'}{dt} \frac{dt}{dt'} = \frac{d}{dt}[\gamma (x-vt)]\frac{1}{\gamma \Big( 1 - \frac{v}{c^2}u^{x}\Big)} = (u^{x}-v)\frac{1}{\Big( 1 - \frac{v}{c^2}u^{x}\Big)} = u'^{x} \tag{23}$$

Now, expression $(23)$ is the correct $x$ transformation for velocity. But this is the chain rule $(3)$ and this is valid JUST IF WE ASSUME THAT $x'(t(t'))$, and we know that $x'$ do not depends only on $t$! This leads to a mathematical mistake of some kind which I didn't grasp yet. Also, physically didn't make much sense for me a primed quantity $x'$ varying with respect a non-primed $t$ like in $x'(t(t'))$. Also, it's seems so strange that a simply application of the change of coordinates formula $(9)$ in $(21)$ didn't work, it seems that the factor $1/ \gamma(1-v/c^{2}u^{x})$ is missing, and this is strange because perform a vector component transformation like $(9)$ are the most general way to see this problem, as this didn't reach the right results!

So my question is what is the relationship between, single-variable, multivariable chain rule, vector component transformation, and equation $(22)$ ?

$\endgroup$
  • $\begingroup$ I have two comments. 1. There are many formulas that are just basic facts about calculus. They make the post unnecessarily long and may make potential answeres lose interest before getting to the end. 2. It's custom practice to typeset emphasised sentences with a cursive font rather than with caps lock. $\endgroup$ – MannyC Apr 14 at 21:32
  • $\begingroup$ I appreciate your comments. But this was the best way which I found to speak my mind properly. $\endgroup$ – M.N.Raia Apr 14 at 21:35
  • $\begingroup$ Eq 1 must be correct, because x' can be a function of t' only $\endgroup$ – Wolphram jonny Apr 14 at 21:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.