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Four vectors and lorentz transformations

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1. Consider a four vector [tex]x^{\mu}[/tex], that is timelike (i.e [tex]x^{2}>0[/tex]. show that it is always possible to find a frame where the coordinates of x are of the form [tex](x^{0'},0)[/tex]. Determine the lorentz transformation relating the initial frame to this particular frame

3. I figured that assuming that the starting 4 vector could be of the form [tex](x^{0},0,0,x^{3})[/tex] then the resulting answer would be [tex]x^{'\mu}=\Lambda_{\nu}^{\mu}x^{\mu}
[/tex]
 

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  • #2
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1. Consider a four vector [tex]x^{\mu}[/tex], that is timelike (i.e [tex]x^{2}>0[/tex]. show that it is always possible to find a frame where the coordinates of x are of the form [tex](x^{0'},0)[/tex]. Determine the lorentz transformation relating the initial frame to this particular frame

3. I figured that assuming that the starting 4 vector could be of the form [tex](x^{0},0,0,x^{3})[/tex] then the resulting answer would be [tex]x^{'\mu}=\Lambda_{\nu}^{\mu}x^{\mu}
[/tex]
It would probably be easier to work backwards starting with [itex](x^{0'},0)[/itex].
 
  • #3
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And then find the Lorentz transformation that would give me x?

What I'm also curious about is, how do I deal with x=y=0 in the Lorentz transformation matrix? Would I just put 0 for each component? If so, I dont know how to write a transformation that only acts on t but leaves z' the same as z...
 
  • #4
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And then find the Lorentz transformation that would give me x?

What I'm also curious about is, how do I deal with x=y=0 in the Lorentz transformation matrix? Would I just put 0 for each component? If so, I dont know how to write a transformation that only acts on t but leaves z' the same as z...
Just use the inverse lorentz transformation matrix on it in terms of the lambdas. See what that gives.
 
  • #5
vanhees71
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I'd think more intuitively. Suppose your time-like vector is the time-like momentum of a (massive!) particle. Then, how do you have to change the inertial reference frame to make the spatial part of the four-momentum (i.e., its usual three-momentum vector) vanish? Then it's easy to write down the prove for the corresponding Lorentz-transformation matrix.

PS: I cannot stress it often enough in this forum: The concise notation of the Lorentz matrices in the index notation is crucial, i.e., you should type
Code:
{\Lambda^{\mu}}_{\nu}
resulting in [itex]{\Lambda^{\mu}}_{\nu}[/itex] with properly ordered upper and lower indices!
 
  • #6
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[tex]x'_{\mu}=(\Lambda^{-1})_{\mu}^{\alpha}x_{\alpha}[/tex]

I'm trying to figure out in latex how to write it properly, as I'm aware of what it's meant to look like anyways.
 
  • #7
BruceW
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vanhees' post has the way to write the ordered indices (in latex). Anyway, you are getting to the answer. I'm guessing your plan is that since you know the Lorentz transform, you can then just write down the inverse? Also, I'm guessing that for this problem, you are only allowed to use proper orthochronous linear transformations. The problem doesn't specify this, but it would be a weird question if it allowed any kind of Poincare transform.
 
  • #8
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Yeah I tried adding an underscore but it came out funny.. Oh well....

Yes I am assuming I can only use orthochronous lorentz transforms.

I don't understand the latter part of what you said though. When you say the inverse do you mean the inverse of [tex](\Lambda^{-1})_{\mu}^{\alpha}[/tex] ?
 
  • #9
vanhees71
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Sure, the intuitive idea to think about I suggested in my previous posting, automatically shows that a proper orthochronous Lorentz transform is sufficient to achieve the goal. It's really obvious!
 
  • #10
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[tex]\Lambda_{\alpha}^{\mu}[/tex]
 
  • #11
BruceW
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What I meant was: if you know the form of the proper orthochronous Lorentz transform going from the vector ##(x_0,0)## to another vector, then the answer for the problem is the inverse of this transform. But as vanhees is saying, it might actually be easier to just think of how to transform to the ##(x_0,0)## vector, from another vector. (using proper orthochronous transform).

edit: p.s. 'proper orthochronous Lorentz transform' is probably what most people simply call 'Lorentz transform'. Even though they are technically not the same thing.
 
  • #12
vela
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I figured that assuming that the starting 4 vector could be of the form [tex](x^{0},0,0,x^{3})[/tex]
You can't assume this. The problem tells you only that ##x^\mu## is timelike.
 
  • #13
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I've been told in the question to assume this....
 
  • #14
BruceW
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it will be easier to write it like ##(x^0,x^1)## if you are allowed, since the matrix will be smaller ;) But you can write it like ##(x^0,0,0,x^3)## if that makes more sense to you.
 
  • #15
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I don't know how to make the spatial part of the transformation vanish.

I'm wondering whether the matrix would be made of cosh and sinh because then I guess I could have the spatial part vanish provided the angle was right...?
 
Last edited:
  • #16
BruceW
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it is more simple than you are thinking. write down the usual Lorentz transform matrix between two reference frames with some velocity ##\beta##. And now just choose ##\beta## to get the specific transform you want.
 
  • #17
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I don't know how to make the spatial part of the transformation vanish.
That's why I'm suggesting using the inverse transformation. Have you learned the form of the Lorentz Transformation when the relative velocity vector for the two frames of reference is not in the same direction as one of their spatial coordinate axes? If not, just use the standard configuration inverse Lorentz Transformation to start with.

Ax=γ(Ax'+βAt')

At=γ(At'+βAx')

In your system, Ax' is equal to zero.

Chet
 
  • #18
vanhees71
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Again my suggestion: Suppose your time-like vector is the four-momentum of a particle. Setting [itex]c=1[/itex] it's given by [itex]p=(p^0,\vec{p})[/itex]. The three-velocity of the particle is given by [itex]\vec{v}=\vec{p}/p^0[/itex]. Now, with which Lorentz transform can you go over to a reference frame, where the particle is at rest? It's really very simple!

Then you write down the appropriate proper orthochronous Lorentz-trafo matrix as an ansatz and prove that it really fulfills the task to bring the particle to rest in the new reference frame, i.e.,
[tex]p'=(p'{}^0,0,0,0).[/tex]
 
  • #19
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Continuing from where I left off in #17, if Ax'=0, you get

Ax=βγAt'

At=γAt'
 
  • #20
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Hey,
Well I managed to solve it a few days ago :-)

Was dead simple... I was massively overthinking it...
 
  • #21
BruceW
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hehe, yeah I thought so :)
 

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