Monday, 13 July 2015

special relativity - General matrix Lorentz transformation


I just finished an introduction course into theory of relativity and am trying to find the general matrix Lorentz transformation. I have already looked into this question, but I could not make much out of it.


Basically, we know that for one space vector relating a frame S and S': [ctx]=[γγβγβγ][ctx] This I simplify to x=L1x. My thinking therefore is that if S' moves from S in two space coordinates (x and y), then I can use first move in x and then in y, such that x=L1L2x, where in L1 I keep the y coordinate fixed, and in L2 I keep the x coordinate fixed. Writing this out would be: [ctxy]=[γxγxβx0γxβxγx0001][γy0γyβy010γyβy0γy][ctxy]


[ctxy]=[γxγyγxβxγxγyβyγxβxγyγxγxβxγyβyγyβy0γy][ctxy] To me all of this looks quite neat, but when I try to apply it to velocity addition, I get false results. As far as I could make it out, the zero in the last 3x3 matrix is wrong (,which won't disappear neither if I add a z coordinate as well... ).


I am therefore hoping that someone can indicate to me where I am doing something wrong when trying to create a more general lorentz matrix equation. I have found on wikipedia a big matrix equation, but because it starts talking about rotations and so on, plus it doesnt show how the components are put together, I dismissed it for now.


(In case this is correct and my sense that I got it wrong is false due to the velocity addition method I apply, do let me know, and I can elaborate on that method as well. )




Answer



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From Figure 01 :


Lorentz Transformation from S{xyη,η=ct} to S1{x1y1η1,η1=ct1} [x1y1η1]=[coshζ0sinhζ010sinhζ0coshζ][xyη],tanhζ=uc or X1=L1X,L1=[coshζ0sinhζ010sinhζ0coshζ]


enter image description here


From Figure 02:


Lorentz Transformation from S1{x1y1η1,η1=ct1} to S2{x2y2η2,η2=ct2} [x2y2η2]=[1000coshξsinhξ0sinhξcoshξ][x1y1η1],tanhξ=wc or X2=L2X1,L2=[1000coshξsinhξ0sinhξcoshξ] Note that because of the Standard Configurations the matrices L1,L2 are real symmetric.


From equations (01) and (02) we have X2=L2X1=L2L1XX2=ΛX where Λ the composition of the two Lorentz Transformations L1,L2 Λ=L2L1=[1000coshξsinhξ0sinhξcoshξ][coshζ0sinhζ010sinhζ0coshζ] that is Λ=[coshζ0sinhζsinhζsinhξcoshξcoshζsinhξsinhζcoshξsinhξcoshζcoshξ]


The Lorentz Transformation matrix Λ is not symmetric, so the systems S,S2 are not in the Standard configuration. But it could be written as Λ=RL where L is the symmetric Lorentz Transformation matrix from S to an intermediate system S2 in Standard configuration to it and co-moving with S2, while R is a purely spatial transformation from S2 to S2.


enter image description here



Now it's up to you to find the Lorentz Transformation matrix L first and then to prove that R is R=[cosϕsinϕ0sinϕcosϕ0001],wheretanϕ=sinhζsinhξcoshζ+coshξ,ϕ(π2,+π2)11111 representing a plane rotation from S2 to S2, see Figure 03.




EDIT


The Lorentz Transformation matrix L, from S to the intermediate system S2 in Standard Configuration to it, is : L(υ)=[1+(γυ1)n2x(γυ1)nxnyγυυxc(γυ1)nynx1+(γυ1)n2yγυυycγυυxcγυυycγυ] In (07) υ=(υx,υy)n=(nx,ny)=υ where \:\boldsymbol{\upsilon}\: is the velocity vector of the origin \:\mathrm{O'}_{\!\!2}\left(\equiv \mathrm{O}_{2}\right)\: with respect to \:\mathrm{S}, \:\mathbf{n}\: the unit vector along \:\boldsymbol{\upsilon}\: and \:\gamma_{\upsilon}\: the corresponding \:\gamma-factor.


The velocity vector \:\boldsymbol{\upsilon}\: could be expressed in terms of the rapidities \:\zeta,\xi\: and so we could express the matrix \:\mathrm{L}\: as function of them. To begin with this we first note that the velocity vector \:\boldsymbol{\upsilon}\: is the relativistic sum of two orthogonal velocity vectors \:\mathbf{u}=\left(u\,,0\right),\mathbf{w}=\left(0\,,w\right) \begin{equation} \boldsymbol{\upsilon}=\mathbf{u}+\dfrac{\mathbf{w}}{\gamma_{\!u}}=\left[u\,,\left(\!1\!-\!\frac{u^{2}}{c^{2}}\right)^{\!\!\frac12}\!\!w\right]\,,\quad \gamma_{u} = \left(\!1\!-\!\frac{u^{2}}{c^{2}}\right)^{\!\!-\frac12} \tag{09} \end{equation} not to be confused with the relativistic sum of two collinear velocity vectors pointing to the same direction \begin{equation} \upsilon \ne \dfrac{u\!+\!w}{1+\dfrac{uw}{c^{2}}} \tag{10} \end{equation} From (09) we have \begin{align} \dfrac{\upsilon_{x}}{c} & = \dfrac{u}{\:\:c\:\:}=\tanh\zeta \tag{11.1}\\ \dfrac{\upsilon_{y}}{c} & = \dfrac{w}{\gamma_{u}c}= \dfrac{\tanh\xi}{\cosh\zeta} \tag{11.2}\\ \left(\dfrac{\upsilon}{c}\right)^{2} & = \left(\dfrac{\upsilon_{x}}{c}\right)^{2}+\left(\dfrac{\upsilon_{y}}{c}\right)^{2}=1-\left(\dfrac{1}{\cosh\zeta\cosh\xi}\right)^{2}=\dfrac{\gamma^{2}_{\upsilon}\!-\!1}{\gamma^{2}_{\upsilon}} \tag{11.3}\\ \gamma_{\upsilon} & = \left(\!1\!-\!\frac{\upsilon^{2}}{c^{2}}\right)^{-\frac12}=\cosh\zeta\cosh\xi \tag{11.4} \end{align} and \begin{align} \dfrac{\gamma_{\!\upsilon}\upsilon_{x}}{c} & = \sinh\zeta \cosh\xi \tag{12.1}\\ \dfrac{\gamma_{\!\upsilon}\upsilon_{y}}{c} & = \sinh\xi \tag{12.2}\\ 1\!+\!\left(\gamma_{\!\upsilon}\!-\!1\right)\!\mathrm{n}^{2}_{x} & = 1\!+\!\left(\gamma_{\!\upsilon}\!-\!1\right)\dfrac{\left(\dfrac{\upsilon_{x}}{c}\right)^{2}}{\left(\dfrac{\upsilon}{c}\right)^{2}}=1\!+\!\dfrac{\gamma^{2}_{\!\upsilon}}{1\!+\!\gamma_{\!\upsilon}}\tanh^{2}\!\zeta=1\!+\!\dfrac{\sinh^{2}\!\zeta\cosh^{2}\!\xi}{1\!+\!\cosh\!\zeta\cosh\!\xi} \tag{12.3}\\ 1\!+\!\left(\gamma_{\!\upsilon}\!-\!1\right)\!\mathrm{n}^{2}_{y} & = 1\!+\!\left(\gamma_{\!\upsilon}\!-\!1\right)\dfrac{\left(\dfrac{\upsilon_{y}}{c}\right)^{2}}{\left(\dfrac{\upsilon}{c}\right)^{2}}=1\!+\!\dfrac{\gamma^{2}_{\!\upsilon}}{1\!+\!\gamma_{\!\upsilon}}\dfrac{\tanh^{2}\!\xi}{\cosh^{2}\!\zeta}=1\!+\!\dfrac{\sinh^{2}\!\xi}{1\!+\!\cosh\!\zeta\cosh\!\xi} \tag{12.4}\\ \left(\gamma_{\!\upsilon}\!-\!1\right)\!\mathrm{n}_{x}\mathrm{n}_{y} & =\left(\gamma_{\!\upsilon}\!-\!1\right)\dfrac{\left(\dfrac{\upsilon_{x}}{c}\right)\!\!\left(\dfrac{\upsilon_{y}}{c}\right)}{\left(\dfrac{\upsilon}{c}\right)^{2}}=\dfrac{\gamma^{2}_{\!\upsilon}}{1\!+\!\gamma_{\!\upsilon}}\dfrac{\tanh\!\zeta\tanh\!\xi}{\cosh\!\zeta}=\dfrac{\sinh\!\zeta\sinh\!\xi\cosh\!\xi}{1\!+\!\cosh\!\zeta\cosh\!\xi} \tag{12.5} \end{align} So the matrix \:\mathrm{L}\left(\boldsymbol{\upsilon} \right)\: of equation (07) as function of the rapidities \:\zeta,\xi\: is \begin{equation} \mathrm{L}\left(\boldsymbol{\upsilon} \right)= \begin{bmatrix} 1\!+\!\dfrac{\sinh^{2}\!\zeta\cosh^{2}\!\xi}{1\!+\!\cosh\!\zeta\cosh\!\xi} & \dfrac{\sinh\!\zeta\sinh\!\xi\cosh\!\xi}{1\!+\!\cosh\!\zeta\cosh\!\xi} & \!-\sinh\zeta \cosh\xi \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ \dfrac{\sinh\!\zeta\sinh\!\xi\cosh\!\xi}{1\!+\!\cosh\!\zeta\cosh\!\xi} & 1\!+\!\dfrac{\sinh^{2}\!\xi}{1\!+\!\cosh\!\zeta\cosh\!\xi} & \!-\sinh\xi \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ \!-\sinh\zeta \cosh\xi & \!-\sinh\xi & \cosh\zeta\cosh\xi \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}} \end{bmatrix} \tag{13} \end{equation} Now, in order to determine the spatial transformation \:\mathrm{R}\: we have from (05) \begin{equation} \mathrm{R}=\Lambda\cdot\mathrm{L}^{-1} \tag{14} \end{equation} For \:\mathrm{L}^{-1}\: equation (07) yields \begin{equation} \mathrm{L}^{-1}=\mathrm{L}\left(\!-\!\boldsymbol{\upsilon} \right)= \begin{bmatrix} 1\!+\!\left(\gamma_{\!\upsilon}\!-\!1\right)\!\mathrm{n}^{2}_{x} & \left(\gamma_{\!\upsilon}\!-\!1\right)\!\mathrm{n}_{x}\mathrm{n}_{y} & \dfrac{\gamma_{\!\upsilon}\upsilon_{x}}{c} \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ \left(\gamma_{\!\upsilon}\!-\!1\right)\!\mathrm{n}_{y}\mathrm{n}_{x} & 1\!+\!\left(\gamma_{\!\upsilon}\!-\!1\right)\!\mathrm{n}^{2}_{y} & \dfrac{\gamma_{\!\upsilon}\upsilon_{y}}{c} \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ \dfrac{\gamma_{\!\upsilon}\upsilon_{x}}{c} & \dfrac{\gamma_{\!\upsilon}\upsilon_{y}}{c} & \gamma_{\!\upsilon} \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}} \end{bmatrix} \tag{15} \end{equation} and from (13) \begin{equation} \mathrm{L}^{-1}= \begin{bmatrix} 1\!+\!\dfrac{\sinh^{2}\!\zeta\cosh^{2}\!\xi}{1\!+\!\cosh\!\zeta\cosh\!\xi} & \dfrac{\sinh\!\zeta\sinh\!\xi\cosh\!\xi}{1\!+\!\cosh\!\zeta\cosh\!\xi} & \sinh\!\zeta \cosh\!\xi \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ \dfrac{\sinh\!\zeta\sinh\!\xi\cosh\!\xi}{1\!+\!\cosh\!\zeta\cosh\!\xi} & 1\!+\!\dfrac{\sinh^{2}\!\xi}{1\!+\!\cosh\!\zeta\cosh\!\xi} & \sinh\!\xi \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ \sinh\!\zeta \cosh\!\xi & \sinh\!\xi & \cosh\!\zeta\cosh\!\xi \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}} \end{bmatrix} \tag{16} \end{equation} So \begin{equation} \mathrm{R}= \begin{bmatrix} \hphantom{-}\cosh\zeta & 0 & -\sinh\zeta \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ \hphantom{-}\sinh\zeta\sinh\xi &\hphantom{-}\cosh\xi & -\cosh\zeta\sinh\xi\vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}} \\ -\sinh\zeta\cosh\xi & -\sinh\xi & \hphantom{-}\cosh\zeta\cosh\xi \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ \end{bmatrix} \begin{bmatrix} 1\!+\!\dfrac{\sinh^{2}\!\zeta\cosh^{2}\!\xi}{1\!+\!\cosh\!\zeta\cosh\!\xi} & \dfrac{\sinh\!\zeta\sinh\!\xi\cosh\!\xi}{1\!+\!\cosh\!\zeta\cosh\!\xi} & \sinh\!\zeta \cosh\!\xi \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ \dfrac{\sinh\!\zeta\sinh\!\xi\cosh\!\xi}{1\!+\!\cosh\!\zeta\cosh\!\xi} & 1\!+\!\dfrac{\sinh^{2}\!\xi}{1\!+\!\cosh\!\zeta\cosh\!\xi} & \sinh\!\xi \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ \sinh\!\zeta \cosh\!\xi & \sinh\!\xi & \cosh\!\zeta\cosh\!\xi \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}} \end{bmatrix} \tag{17} \end{equation} Above matrix multiplication ends up to the following expression \begin{equation} \mathrm{R}= \begin{bmatrix} \dfrac{\cosh\!\zeta\!+\!\cosh\!\xi}{1\!+\!\cosh\!\zeta\cosh\!\xi} &\!- \dfrac{\sinh\!\zeta\sinh\!\xi}{1\!+\!\cosh\!\zeta\cosh\!\xi} & \hphantom{-} 0 \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ \dfrac{\sinh\!\zeta\sinh\!\xi}{1\!+\!\cosh\!\zeta\cosh\!\xi} & \hphantom{\!-} \dfrac{\cosh\!\zeta\!+\!\cosh\!\xi}{1\!+\!\cosh\!\zeta\cosh\!\xi} & \hphantom{-} 0 \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ 0 & 0 & \hphantom{-} 1 \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}} \end{bmatrix} \tag{18} \end{equation} But \begin{equation} \left(\dfrac{\cosh\!\zeta\!+\!\cosh\!\xi}{1\!+\!\cosh\!\zeta\cosh\!\xi}\right)^{2}+\left(\dfrac{\sinh\!\zeta\sinh\!\xi}{1\!+\!\cosh\!\zeta\cosh\!\xi}\right)^{2}=1 \tag{19} \end{equation} so we can define \begin{equation} \cos\phi \stackrel{def}{\equiv}\dfrac{\cosh\!\zeta\!+\!\cosh\!\xi}{1\!+\!\cosh\!\zeta\cosh\!\xi}\,, \qquad \sin\phi =\dfrac{\sinh\!\zeta\sinh\!\xi}{1\!+\!\cosh\!\zeta\cosh\!\xi}\,, \qquad \phi \in \left(-\tfrac{\pi}{2},+\tfrac{\pi}{2}\right) \tag{20} \end{equation} and finally \begin{equation} \mathrm{R}= \begin{bmatrix} \cos\phi & -\sin\phi & 0 \\ \sin\phi &\hphantom{-}\cos\phi & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} \tag{21} \end{equation} proving that \:\mathrm{R}\: is a rotation, see Figure 03.


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