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][ct′x′] 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][ct′x′y′]
[ctxy]=[γxγyγxβxγxγyβyγxβxγyγxγxβxγyβyγyβy0γy][ct′x′y′] 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
From Figure 01 :
Lorentz Transformation from S≡{xyη,η=ct} to S1≡{x1y1η1,η1=ct1} [x1y1η1]=[−coshζ0−sinhζ010−sinhζ0−coshζ][xyη],tanhζ=uc or X1=L1X,L1=[−coshζ0−sinhζ010−sinhζ0−coshζ]
From Figure 02:
Lorentz Transformation from S1≡{x1y1η1,η1=ct1} to S2≡{x2y2η2,η2=ct2} [x2y2η2]=[1000−coshξ−sinhξ0−sinhξ−coshξ][x1y1η1],tanhξ=wc or X2=L2X1,L2=[1000−coshξ−sinhξ0−sinhξ−coshξ] Note that because of the Standard Configurations the matrices L1,L2 are real symmetric.
From equations (01) and (02) we have X2=L2X1=L2L1X⟹X2=ΛX where Λ the composition of the two Lorentz Transformations L1,L2 Λ=L2L1=[1000−coshξ−sinhξ0−sinhξ−coshξ][−coshζ0−sinhζ010−sinhζ0−coshζ] that is Λ=[−coshζ0−sinhζ−sinhζsinhξ−coshξ−coshζsinhξ−sinhζcoshξ−sinhξ−coshζcoshξ]
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=R⋅L where L is the symmetric Lorentz Transformation matrix from S to an intermediate system S′2 in Standard configuration to it and co-moving with S2, while R is a purely spatial transformation from S′2 to S2.
Now it's up to you to find the Lorentz Transformation matrix L first and then to prove that R is R=[cosϕ−sinϕ0sinϕ−cosϕ0001],wheretanϕ=sinhζsinhξcoshζ+coshξ,ϕ∈(−π2,+π2)11111 representing a plane rotation from S′2 to S2, see Figure 03.
EDIT
The Lorentz Transformation matrix L, from S to the intermediate system S′2 in Standard Configuration to it, is : L(υ)=[1+(γυ−1)n2x(γυ−1)nxny−γυυxc(γυ−1)nynx1+(γυ−1)n2y−γυυyc−γυυxc−γυυycγυ] In (07) υ=(υx,υy)n=(nx,ny)=υ‖υ‖=υυγυ=(1−υ2c2)−12=1√1−υ2c2 where υ is the velocity vector of the origin O′2(≡O2) with respect to S, n the unit vector along υ and γυ the corresponding γ−factor.
The velocity vector υ could be expressed in terms of the rapidities ζ,ξ and so we could express the matrix L as function of them. To begin with this we first note that the velocity vector υ is the relativistic sum of two orthogonal velocity vectors u=(u,0),w=(0,w) υ=u+wγu=[u,(1−u2c2)12w],γu=(1−u2c2)−12 not to be confused with the relativistic sum of two collinear velocity vectors pointing to the same direction υ≠u+w1+uwc2 From (09) we have υxc=uc=tanhζυyc=wγuc=tanhξcoshζ(υc)2=(υxc)2+(υyc)2=1−(1coshζcoshξ)2=γ2υ−1γ2υγυ=(1−υ2c2)−12=coshζcoshξ and γυυxc=sinhζcoshξγυυyc=sinhξ1+(γυ−1)n2x=1+(γυ−1)(υxc)2(υc)2=1+γ2υ1+γυtanh2ζ=1+sinh2ζcosh2ξ1+coshζcoshξ1+(γυ−1)n2y=1+(γυ−1)(υyc)2(υc)2=1+γ2υ1+γυtanh2ξcosh2ζ=1+sinh2ξ1+coshζcoshξ(γυ−1)nxny=(γυ−1)(υxc)(υyc)(υc)2=γ2υ1+γυtanhζtanhξcoshζ=sinhζsinhξcoshξ1+coshζcoshξ So the matrix L(υ) of equation (07) as function of the rapidities ζ,ξ is L(υ)=[1+sinh2ζcosh2ξ1+coshζcoshξsinhζsinhξcoshξ1+coshζcoshξ−sinhζcoshξsinhζsinhξcoshξ1+coshζcoshξ1+sinh2ξ1+coshζcoshξ−sinhξ−sinhζcoshξ−sinhξcoshζcoshξ] Now, in order to determine the spatial transformation R we have from (05) R=Λ⋅L−1 For L−1 equation (07) yields L−1=L(−υ)=[1+(γυ−1)n2x(γυ−1)nxnyγυυxc(γυ−1)nynx1+(γυ−1)n2yγυυycγυυxcγυυycγυ] and from (13) L−1=[1+sinh2ζcosh2ξ1+coshζcoshξsinhζsinhξcoshξ1+coshζcoshξsinhζcoshξsinhζsinhξcoshξ1+coshζcoshξ1+sinh2ξ1+coshζcoshξsinhξsinhζcoshξsinhξcoshζcoshξ] So R=[−coshζ0−sinhζ−sinhζsinhξ−coshξ−coshζsinhξ−sinhζcoshξ−sinhξ−coshζcoshξ][1+sinh2ζcosh2ξ1+coshζcoshξsinhζsinhξcoshξ1+coshζcoshξsinhζcoshξsinhζsinhξcoshξ1+coshζcoshξ1+sinh2ξ1+coshζcoshξsinhξsinhζcoshξsinhξcoshζcoshξ] Above matrix multiplication ends up to the following expression R=[coshζ+coshξ1+coshζcoshξ−sinhζsinhξ1+coshζcoshξ−0sinhζsinhξ1+coshζcoshξ−coshζ+coshξ1+coshζcoshξ−000−1] But (coshζ+coshξ1+coshζcoshξ)2+(sinhζsinhξ1+coshζcoshξ)2=1 so we can define cosϕdef≡coshζ+coshξ1+coshζcoshξ,sinϕ=sinhζsinhξ1+coshζcoshξ,ϕ∈(−π2,+π2) and finally R=[cosϕ−sinϕ0sinϕ−cosϕ0001] proving that R is a rotation, see Figure 03.
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