Appearance of an angle of inclination on a horizontal rod moving upwards after a Lorentz Transformation [closed]

If i have in one inertial reference frame called $S':(x',y',z',t')$ a rod with lenght $l$ in the plane $z'=0$ and parallel with the $x'$ axis, moving with a velocity $u$ in the direction of the $y'$ axis, so i think therefore that the extreme points of the rod can be given in $S'$ by $$P'_1:(x'_1 , y'_1 , z'_1 , t'_1) = (x'_1 , ut'_1 ,z'_1 , t'_1)$$ $$P'_2:(x'_2 , y'_2 , z'_2 , t'_2) =(x'_1 + l , ut'_1 , z'_1 , t'_1)$$

Given that there is a inertial reference frame $S:(x,y,z,t)$ ; in relation to which the reference frame $S'$ is moving with velocity $v$ in the $x$ direction, and such that in the time $t=0$ both reference frames coincided, and after that, the $z'$ axis continued parallel to the $z$ axis as did the $y'$ axis parallel to $y$ axis and the $x'$ axis continued in the same direction as the $x$ axis.

So to get the points $P'_1$ and $P'_2$ in the frame $S$ i tried to do the lorentz inverse transform in each point as it is shown bellow:

$$P'_1 \Rightarrow P_{1} \begin{cases} x_1=\gamma (x'_1+vt'_1) \\ y_1 = y'_1 =ut'_1\\ z_1 = z'_1\\ t_1 = \gamma(t'_1+\frac{v}{c^2}x'_1) \end{cases}$$

$$P'_2 \Rightarrow P_{2} \begin{cases} x_2=\gamma ((x'_1+l)+vt'_2) = \gamma ((x'_1+l)+vt'_1)\\ y_2 = y'_2 =ut'_2=ut'_1\\ z_2 = z'_2 = z'_1\\ t_2 = \gamma(t'_2+\frac{v}{c^2}(x'_1+l)) = \gamma(t'_1+\frac{v}{c^2}(x'_1+l)) \end{cases}$$

But then if $t'_2 = t'_1$ there would be no $y$ axis inclination as the problem sugests, what i am doing wrong ?

• Please note that we don't answer homework or worked example type questions. Please see this Meta post on asking homework/exercise questions and this Meta post for "check my work" problems. – John Rennie Nov 16 '17 at 6:29
• But it says that for homework type of problems, as long as it is targeted at a specific conceptual physics question it is alright to ask, and in this post targets the conceptual problem of how do angles transform under Lorentz transformations, in a given occurrence. If you had read it is not a "check my work" type of question because i actually didn't come to any conclusion, i just have laid the foundations of how i think the problem should have been solved, and came to a conclusion that contradicts what was expected demonstrating a misunderstanding of the method of solution. – Felipe Dilho Nov 16 '17 at 6:59
• Take a look in my answer herein :General matrix Lorentz transformation. May be answers your question too. – Frobenius Nov 16 '17 at 17:24
• @FelipeDilho You're right that it's fine to ask about some physics concept that happens to be involved in the problem you're solving. But I don't see that here. All I see you asking is "what am I doing wrong?" That being said, I think there may be a way to make this question fine for the site with a small edit. I'll think about what that might be. – David Z Nov 16 '17 at 22:25

Hint : We have here 3 frames and 2 Lorentz Transformations. (1) Frame $$\:\mathrm S\left(x,y,z\right)\:$$ (2) frame $$\:\mathrm S'\left(x',y',z'\right)\:$$ moving with respect to $$\:\mathrm S\left(x,y,z\right)\:$$ with velocity $$\:\mathbf{v}=\left(v,0,0\right)\:$$ along the common $$\:x-,x'-$$axis (3) frame $$\:\mathrm S''\left(x'',y'',z''\right)\:$$, the rest frame of the rod, moving with respect to $$\:\mathrm S'\left(x',y',z'\right)\:$$ with velocity $$\:\mathbf{u}=\left(0,u,0\right)\:$$ along the common $$\:y'-,y''-$$axis.

I suggest to take a look in my answer therein : General matrix Lorentz transformation

EDIT

We have here two 2+1-Lorentz transformations. So the axes $$\:z,z',z''\:$$ are ignored.

From Figure 01 :

Lorentz Transformation from $$\:\mathrm{S}\equiv \{xy\eta, \eta=ct\}\:$$ to $$\:\mathrm{S'}\equiv \{x'y'\eta', \eta'=ct'\}\:$$ $$$$\begin{bmatrix} x'\\ y'\\ \eta' \end{bmatrix} = \begin{bmatrix} \hphantom{-}\cosh\zeta & 0 & \boldsymbol{-}\sinh\zeta \\ 0 & 1 & 0 \\ \boldsymbol{-}\sinh\zeta & 0 & \hphantom{-}\cosh\zeta \\ \end{bmatrix} \begin{bmatrix} x\\ y\\ \eta \end{bmatrix} \,, \quad \tanh\zeta=\dfrac{v}{c} \tag{01}$$$$ or $$$$\mathbf{X'}=\mathrm{L'}\mathbf{X}\,, \qquad \mathrm{L'}= \begin{bmatrix} \hphantom{-}\cosh\zeta & 0 & \boldsymbol{-}\sinh\zeta \\ 0 & 1 & 0 \\ \boldsymbol{-}\sinh\zeta & 0 & \hphantom{-}\cosh\zeta \\ \end{bmatrix} \tag{01"}$$$$

From Figure 02:

Lorentz Transformation from $$\:\mathrm{S'}\equiv \{x'y'\eta', \eta'=ct'\}\:$$ to $$\:\mathrm{S''}\equiv \{x''y''\eta'', \eta''=ct''\}\:$$ $$$$\begin{bmatrix} x''\\ y''\\ \eta'' \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 &\hphantom{-}\cosh\xi & \boldsymbol{-}\sinh\xi \\ 0 & \boldsymbol{-}\sinh\xi & \hphantom{-}\cosh\xi \\ \end{bmatrix} \begin{bmatrix} x'\\ y'\\ \eta' \end{bmatrix} \,, \quad \tanh\xi=\dfrac{u}{c} \tag{02}$$$$ or $$$$\mathbf{X''}=\mathrm{L''}\mathbf{X'}\,, \qquad \mathrm{L''}= \begin{bmatrix} 1 & 0 & 0 \\ 0 &\hphantom{-}\cosh\xi & \boldsymbol{-}\sinh\xi \\ 0 & \boldsymbol{-}\sinh\xi & \hphantom{-}\cosh\xi \\ \end{bmatrix} \tag{02"}$$$$ Note that because of the Standard Configurations the matrices $$\:\mathrm{L'}, \mathrm{L''}\:$$ are real symmetric.

From equations (01) and (02) we have $$$$\mathbf{X''}=\mathrm{L''}\mathbf{X'}=\mathrm{L''}\mathrm{L'}\mathbf{X}\Longrightarrow \mathbf{X''}=\Lambda\mathbf{X} \tag{03}$$$$ where $$\:\Lambda\:$$ the composition of the two Lorentz Transformations $$\:\mathrm{L'}, \mathrm{L''}\:$$ $$$$\Lambda=\mathrm{L''}\mathrm{L'}= \begin{bmatrix} 1 & 0 & 0 \\ 0 &\hphantom{-}\cosh\xi & \boldsymbol{-}\sinh\xi \\ 0 & \boldsymbol{-}\sinh\xi & \hphantom{-}\cosh\xi \\ \end{bmatrix} \begin{bmatrix} \hphantom{-}\cosh\zeta & 0 & \boldsymbol{-}\sinh\zeta \\ 0 & 1 & 0 \\ \boldsymbol{-}\sinh\zeta & 0 & \hphantom{-}\cosh\zeta \\ \end{bmatrix} \tag{04}$$$$ that is $$$$\Lambda= \begin{bmatrix} \hphantom{-}\cosh\zeta & 0 & \boldsymbol{-}\sinh\zeta \\ \hphantom{-}\sinh\zeta\sinh\xi &\hphantom{-}\cosh\xi & \boldsymbol{-}\cosh\zeta\sinh\xi \\ \boldsymbol{-}\sinh\zeta\cosh\xi & \boldsymbol{-}\sinh\xi & \hphantom{-}\cosh\zeta\cosh\xi \\ \end{bmatrix} \tag{04"}$$$$

In the following it's necessary to have the inverse transformation $$\:\Lambda^{\boldsymbol{-}1}\:$$ which could be determined from : $$$$\Lambda^{\boldsymbol{-}1}=\mathrm{L\!'}^{\boldsymbol{-}1}\mathrm{L\!''}^{\boldsymbol{-}1} \tag{05}$$$$ For the inverse transformations $$\:\mathrm{L\!'}^{\boldsymbol{-}1},\mathrm{L\!''}^{\boldsymbol{-}1}\:$$ we have respectively $$$$\mathrm{L'}\left(\zeta\right)= \begin{bmatrix} \hphantom{-}\cosh\zeta & 0 & \boldsymbol{-}\sinh\zeta \\ 0 & 1 & 0 \\ \boldsymbol{-}\sinh\zeta & 0 & \hphantom{-}\cosh\zeta \\ \end{bmatrix} \Longrightarrow \mathrm{L\!'}^{\boldsymbol{-}1}=\mathrm{L'}\left(\boldsymbol{-}\zeta\right)= \begin{bmatrix} \cosh\zeta & 0 & \sinh\zeta \\ 0 & 1 & 0 \\ \sinh\zeta & 0 & \cosh\zeta \\ \end{bmatrix} \tag{06}$$$$ $$$$\mathrm{L''}\left(\xi\right)= \begin{bmatrix} 1 & 0 & 0 \\ 0 &\hphantom{-}\cosh\xi & \boldsymbol{-}\sinh\xi \\ 0 & \boldsymbol{-}\sinh\xi & \hphantom{-}\cosh\xi \\ \end{bmatrix} \Longrightarrow \mathrm{L\!''}^{\boldsymbol{-}1}=\mathrm{L''}\left(\boldsymbol{-}\xi\right)= \begin{bmatrix} 1 & 0 & 0 \\ 0 &\cosh\xi & \sinh\xi \\ 0 & \sinh\xi & \cosh\xi \\ \end{bmatrix} \tag{07}$$$$ so $$$$\!\!\!\!\!\! \Lambda^{\boldsymbol{-}1} \!=\! \begin{bmatrix} \cosh\zeta & 0 & \sinh\zeta \\ 0 & 1 & 0 \\ \sinh\zeta & 0 & \cosh\zeta \\ \end{bmatrix} \!\!\! \begin{bmatrix} 1 & 0 & 0 \\ 0 &\cosh\xi & \sinh\xi \\ 0 & \sinh\xi & \cosh\xi \\ \end{bmatrix} \!=\! \begin{bmatrix} \cosh\zeta & \sinh\zeta\sinh\xi & \sinh\zeta\cosh\xi \\ 0 &\hphantom{-}\cosh\xi & \sinh\xi \\ \sinh\zeta & \cosh\zeta\sinh\xi & \cosh\zeta\cosh\xi \\ \end{bmatrix}\!\!\!\! \tag{08}$$$$

We need also the velocity vector, let $$\:\mathbf{w}$$, of the translational motion of the frame $$\:\mathrm S''\left(x'',y''\right)\:$$ with respect to the frame $$\:\mathrm S\left(x,y\right)$$. This vector is the relativistic sum of two orthogonal velocity vectors $$\:\mathbf{v}=\left(\upsilon\,,0\right),\mathbf{u}=\left(0\,,u\right)$$(1)
$$$$\mathbf{w}=\mathbf{v}+\dfrac{\mathbf{u}}{\gamma_{\upsilon}}=\left[\upsilon\,,\left(\!1\!-\!\frac{\upsilon^{2}}{c^{2}}\right)^{\!\!\frac12}\!\!u\right]\,,\quad \gamma_{\upsilon} = \left(\!1\!-\!\frac{\upsilon^{2}}{c^{2}}\right)^{\!\!\boldsymbol{-}\frac12}=\cosh\zeta \tag{09}$$$$ not to be confused with the relativistic sum of two collinear velocity vectors pointing to the same direction $$$$w \ne \dfrac{\upsilon\!+\!u}{1+\dfrac{\upsilon u}{c^{2}}} \tag{10}$$$$ From (09) we have \begin{align} \dfrac{w_{x}}{c} & = \dfrac{\upsilon}{\:\:c\:\:}=\tanh\zeta \tag{11.1}\\ \dfrac{w_{y}}{c} & = \dfrac{u}{\gamma_{\upsilon}c}= \dfrac{\tanh\xi}{\cosh\zeta} \tag{11.2}\\ \left(\dfrac{w}{c}\right)^{2} & = \left(\dfrac{w_{x}}{c}\right)^{2}+\left(\dfrac{w_{y}}{c}\right)^{2}=1-\left(\dfrac{1}{\cosh\zeta\cosh\xi}\right)^{2}=\dfrac{\gamma^{2}_{w}\!-\!1}{\gamma^{2}_{w}} \tag{11.3}\\ \gamma_{w} & = \left(\!1\!-\!\frac{w^{2}}{c^{2}}\right)^{\boldsymbol{-}\frac12}=\cosh\zeta\cosh\xi=\gamma_{\upsilon}\gamma_{u} \tag{11.4} \end{align}

Now, let a rod of length $$\:\boldsymbol{\ell}\:$$ at rest in frame $$\:\mathrm S''\:$$ parallel to the $$\:x''\!\!-$$axis, see Figure 03. The end points of the rod are observed in $$\:\mathrm S''\:$$ simultaneously at a given time moment $$\:t''\!\!-$$ so we have two events $$\:\mathrm 1''\:$$ and $$\:\mathrm 2''\:$$ separated by the space-time vector $$$$\Delta\mathbf{X''}= \begin{bmatrix} \Delta x''\\ \Delta y''\\ \Delta \eta'' \end{bmatrix} = \begin{bmatrix} x''_{2}-x''_{1}\\ y''_{2}-y''_{1}\\ c\left(t''_{2}-t''_{1}\right) \end{bmatrix} = \begin{bmatrix} \boldsymbol{\ell}\\ 0\\ 0 \end{bmatrix} \tag{12}$$$$

In frame $$\:\mathrm S\:$$ the two events are observed as two events $$\:\mathrm 1\:$$ and $$\:\mathrm 2\:$$ separated by the space-time vector $$$$\Delta\mathbf{X}= \begin{bmatrix} \Delta x\\ \Delta y\\ \Delta \eta \end{bmatrix} = \begin{bmatrix} x_{2}-x_{1}\\ y_{2}-y_{1}\\ c\left(t_{2}-t_{1}\right) \end{bmatrix} \tag{13}$$$$ which is connected with $$\:\Delta\mathbf{X''}\:$$ by the $$\:\Lambda^{\boldsymbol{-}1}\:$$ transformation $$$$\Delta\mathbf{X}=\Lambda^{\boldsymbol{-}1}\Delta\mathbf{X''} \tag{14}$$$$ From equations (08) and (12) we have $$$$\begin{bmatrix} \Delta x\\ \Delta y\\ \Delta \eta \end{bmatrix} = \begin{bmatrix} \cosh\zeta & \sinh\zeta\sinh\xi & \sinh\zeta\cosh\xi \\ 0 &\hphantom{-}\cosh\xi & \sinh\xi \\ \sinh\zeta & \cosh\zeta\sinh\xi & \cosh\zeta\cosh\xi \end{bmatrix} \begin{bmatrix} \boldsymbol{\ell}\\ 0\\ 0 \end{bmatrix} \tag{15}$$$$ so $$$$\Delta x=\boldsymbol{\ell} \cosh\zeta \,, \quad \Delta y=0\,, \quad \Delta \eta= c\Delta t=\boldsymbol{\ell}\sinh\zeta \tag{16}$$$$

Equation (16) is represented in Figure 04. But this view is not a snapshot of the rod at a given time moment $$\:t\:$$ in $$\:\mathrm S\left(x,y\right)$$. To the contrary, it's a montage of snapshots of the points along the rod, each point shown at its own time moment. So, the end $$\:\mathrm 1\:$$ of the rod is shown on its position at the time moment $$\:t_{1}\:$$, the end $$\:\mathrm 2\:$$ of the rod is shown on its position at the time moment $$\:t_{2}\:$$ and an intermediate point $$\:\mathrm 3\:$$ with coordinate $$\:x_{3}\in \left[x_{1},x_{2}\right]\:$$ at the time moment $$\:t_{3}\:$$ $$$$t_{3}=t_{1}+\dfrac{x_{3}-x_{1}}{x_{2}-x_{1}}\left(t_{2}-t_{1}\right) \tag{17}$$$$ To observe the rod at a given time moment $$\:t\:$$ in $$\:\mathrm S\left(x,y\right)$$, for example at $$\:t_{1}\:$$, we must return the end point $$\:\mathrm 2\:$$ from its position at time $$\:t_{2}\:$$ to its position at time $$\:t_{1}\:$$ since $$\:\mathrm 1\:$$ is already shown at time $$\:t_{1}$$.

This is done in Figure 05. Since the rod is in translational motion with velocity vector $$\:\mathbf{w}\:$$ in frame $$\:\mathrm S\left(x,y\right)$$, the two positions of $$\:\mathrm 2\:$$ at time moments $$\:t_{1},t_{2}$$ are separated by the vector $$\:\mathbf{w}\left(t_{2}-t_{1}\right)$$ and from equations (11.1),(11.2),(16) we have $$$$\mathbf{w}\Delta t=\!\dfrac{\mathbf{w}}{c}\Delta \eta\!=\left(\dfrac{w_{x}}{c},\dfrac{w_{y}}{c}\right)\boldsymbol{\ell}\sinh\!\zeta= \left(\tanh\!\zeta,\dfrac{ \tanh\!\xi}{\cosh\!\zeta}\right)\boldsymbol{\ell}\sinh\!\zeta \tag{18}$$$$ so $$$$\mathbf{w}\Delta t= \left(\boldsymbol{\ell}\sinh\!\zeta\tanh\!\zeta,\boldsymbol{\ell}\tanh\!\zeta\tanh\!\xi\right) \tag{19}$$$$ From details in Figure 05 about the lengths of line segments $$$$\tan\!\theta=\dfrac{\boldsymbol{\ell}\!\tanh\!\zeta\!\tanh\!\xi}{\dfrac{\boldsymbol{\ell}}{\cosh\!\zeta}}=\sinh\!\zeta\tanh\!\xi \tag{20}$$$$ but $$$$\sinh\!\zeta=\underbrace{\cosh\!\zeta}_{\gamma_{\upsilon}}\underbrace{\tanh\!\zeta}_{\upsilon/c}=\dfrac{\gamma_{\upsilon}\upsilon}{c}\,, \quad \tanh\!\xi=\dfrac{u}{c} \tag{21}$$$$ so $$$$\boxed{\:\tan\!\theta=\dfrac{\gamma_{\upsilon}\upsilon u}{c^{2}}=\left(1-\dfrac{\upsilon^{2}}{c^{2}}\right)^{\boldsymbol{-}\frac12}\dfrac{\upsilon u}{c^{2}}\:} \tag{22}$$$$ For the length $$\:\boldsymbol{\ell}_{\mathrm S}\:$$ of the inclined rod in frame $$\: \mathrm S\:$$ we have from the orthogonal triangle in Figure 05 $$$$\boldsymbol{\ell}_{\mathrm S}^{2}=\left(\dfrac{\boldsymbol{\ell}}{\cosh\!\zeta}\right)^{\!\!2}+\left(\boldsymbol{\ell}\!\tanh\!\zeta\!\tanh\!\xi\vphantom{\dfrac12}\right)^{\!\!2}=\left[1 -\dfrac{\upsilon^{2}}{c^{2}}\left(1-\dfrac{u^{2}}{c^{2}}\right)\right]\boldsymbol{\ell}^{2} \tag{23}$$$$ so $$$$\boxed{\:\boldsymbol{\ell}_{\mathrm S}=\sqrt{1 -\dfrac{\upsilon^{2}}{c^{2}}\left(1-\dfrac{u^{2}}{c^{2}}\right)}\:\boldsymbol{\ell}\:} \tag{24}$$$$

From (22) and (24) we have verifications of the following two special cases :

1. If $$\:\color{blue}{u=0}\:$$ then $$\:\theta\!=\! 0\:$$, the rod is moving parallel to the $$\:x-$$axis with speed $$\:\upsilon\:$$ , contracted from $$\:\boldsymbol{\ell}\:$$ to $$\:\boldsymbol{\ell}/\cosh\!\zeta$$.

2. If $$\:\color{blue}{\upsilon\!=\! 0}\:$$ then $$\:\theta\!=\! 0\:$$, the rod is parallel to the $$\:x-$$axis with unchanged length $$\:\boldsymbol{\ell}$$ moving in the direction of the $$\:y-$$axis with speed $$\:u$$.

Numerical Examples : $$$$\begin{Bmatrix} \upsilon/c=0.60\\ u/c=0.40 \end{Bmatrix} \quad =\!=\!=\!=\!\Longrightarrow\quad \begin{Bmatrix} \tan\!\theta=0.30 \\ \theta=16.70^{\,\rm o} \\ \boldsymbol{\ell}_{\mathrm S}=0.835\, \boldsymbol{\ell} \end{Bmatrix} \tag{25}$$$$

$$$$\begin{Bmatrix} \upsilon/c=0.80\\ u/c=0.60 \end{Bmatrix} \quad =\!=\!=\!=\!\Longrightarrow\quad \begin{Bmatrix} \tan\!\theta=0.80 \\ \theta=38.66^{\,\rm o} \\ \boldsymbol{\ell}_{\mathrm S}=0.768\, \boldsymbol{\ell} \end{Bmatrix} \tag{26}$$$$

$$$$\begin{Bmatrix} \upsilon/c\:\longrightarrow\:1\\ u/c\:\longrightarrow\:1 \end{Bmatrix} \quad =\!=\!=\!=\!\Longrightarrow\quad \begin{Bmatrix} \tan\!\theta\:\longrightarrow\:+\infty \\ \hphantom{\:\tan\theta}\theta\:\longrightarrow\:90.00^{\,\rm o} \\ \boldsymbol{\ell}_{\mathrm S}\:\longrightarrow\:\boldsymbol{\ell} \end{Bmatrix} \text{!!!!!!} \tag{27}$$$$

(1) We can derive equations (11.1),(11.2) for the components of the velocity vector $$\:\mathbf{w}\:$$ using the Lorentz transformation $$\:\mathrm{L\!'}^{\boldsymbol{-}1}\:$$ between frames $$\:\mathrm S',\mathrm S\:$$ instead of equation (09)(the last is given without proof and comes from on a general 3+1-Lorentz Transformation)

So, suppose that the origin $$\:\mathrm O''\:$$ of frame $$\:\mathrm S''\:$$ is a particle moving in frame $$\:\mathrm S'\:$$ with velocity vector $$$$\mathbf{u}=\left(\dfrac{\mathrm d x'}{\mathrm d t'},\dfrac{\mathrm d y'}{\mathrm d t'}\right)=\left(0,u\right) \tag{ft-01}$$$$ From equation (06) for the Lorentz transformation $$\:\mathrm{L\!'}^{\boldsymbol{-}1}\:$$ $$$$\begin{bmatrix} \mathrm d x\\ \mathrm d y\\ \mathrm d\eta \end{bmatrix} = \begin{bmatrix} \cosh\zeta & 0 & \sinh\zeta \\ 0 & 1 & 0 \\ \sinh\zeta & 0 & \cosh\zeta \\ \end{bmatrix} \begin{bmatrix} \mathrm d x'\\ \mathrm d y'\\ \mathrm d\eta' \end{bmatrix} \tag{ft-02}$$$$ \begin{align} \mathrm dx & =\cosh\!\zeta \,\mathrm d x'+\sinh\!\zeta \,\mathrm d\eta' \tag{ft-03a}\\ \mathrm dy & =\mathrm dy' \tag{ft-03b}\\ \mathrm d\eta & = \sinh\!\zeta \,\mathrm dx'+\cosh\!\zeta \,\mathrm d\eta' \tag{ft-03c} \end{align} dividing (ft-03a) and (ft-03b) by (ft-03c) \begin{align} \dfrac{\mathrm dx}{\mathrm d\eta} & =\dfrac{\cosh\!\zeta \,\mathrm d x'+\sinh\!\zeta \,\mathrm d\eta'}{\sinh\!\zeta \,\mathrm dx'+\cosh\!\zeta \,\mathrm d\eta'}=\dfrac{\cosh\!\zeta \, \left(\dfrac{\mathrm dx'}{\mathrm d\eta'}\right)+\sinh\!\zeta}{\sinh\!\zeta \, \left(\dfrac{\mathrm dx'}{\mathrm d\eta'}\right)+\cosh\!\zeta} \tag{ft-04a}\\ \dfrac{\mathrm dy}{\mathrm d\eta} & =\dfrac{\mathrm dy'}{\sinh\!\zeta \,\mathrm dx'+\cosh\!\zeta \,\mathrm d\eta'}=\dfrac{\left(\dfrac{\mathrm dy'}{\mathrm d\eta'}\right)}{\sinh\!\zeta \, \left(\dfrac{\mathrm dx'}{\mathrm d\eta'}\right)+\cosh\!\zeta} \tag{ft-04b} \end{align} From (ft-01) having in mind that $$\eta=ct,\eta'=ct'$$ $$$$\dfrac{\mathrm dx'}{\mathrm d\eta'}=0\,,\qquad \dfrac{\mathrm dy'}{\mathrm d\eta'}=\dfrac{u}{c}=\tanh\!\xi \tag{ft-05}$$$$ Inserting expressions (ft-05) in (ft-04a,b) we have finally \begin{align} \dfrac{w_{x}}{c} & =\dfrac{\mathrm dx}{\mathrm d\eta} =\tanh\zeta=\dfrac{\upsilon}{\:\:c\:\:} \tag{11.1}\\ \dfrac{w_{y}}{c} & = \dfrac{\mathrm dy}{\mathrm d\eta}=\dfrac{\tanh\xi}{\cosh\zeta}=\dfrac{u}{\gamma_{\upsilon}c} \tag{11.2} \end{align}

------Figure 06 : The Length------ Figure 07 : The Angle------