- #1

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I mean let v[itex]_{α}[/itex] be a four velocity vector.

Is there a form of Lorentz tfm matrix such that:

v[itex]^{'}[/itex][itex]_{α}[/itex] = [itex]\Lambda^{β}[/itex][itex]_{α}[/itex]v[itex]_{β}[/itex] ?

- Thread starter ssamsymn
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- #1

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I mean let v[itex]_{α}[/itex] be a four velocity vector.

Is there a form of Lorentz tfm matrix such that:

v[itex]^{'}[/itex][itex]_{α}[/itex] = [itex]\Lambda^{β}[/itex][itex]_{α}[/itex]v[itex]_{β}[/itex] ?

- #2

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##v'_{\alpha} = \Lambda_{\alpha}^{~\beta} v_{\beta} ##

- #3

BruceW

Homework Helper

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- #4

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- #5

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I was trying to ask is there a matrix which includes all the "four vector information" in itself and we can act it on salt

[itex]\vec{v}[/itex] 's as (c, [itex]\vec{v}[/itex] ) (not a four velocity but still)

Maybe it seems meaningless but I was confused with all the γ factors because I couldn't find out which of the velocities should be used in them.

I tried to find L in the form of a matrix equation:

V' = L(not the Lorentz tfm matrix) V

and V vectors have 4 components.

Thanks again anyway.

- #6

Fredrik

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$$\Lambda=\frac{1}{\sqrt{1-v^2}}\begin{pmatrix}1 & -v\\ -v & 1\end{pmatrix},$$ where ##v\in(-1,1)## is the velocity of the new coordinate system in the old. This is of course just a Lorentz transformation. I don't understand what you want when you say "not the Lorentz tfm matrix".

- #7

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[tex]u^{\mu}=\gamma \begin{bmatrix}{1 \\ \vec{v}}\end{bmatrix}[/tex]

with [itex]\gamma=1/\sqrt{1-v^2}[/itex].

Of course you get the three-velocity as

[tex]\vec{v}=\vec{u}/u^0.[/tex]

Then, by definition you define the three-velocity in the new frame as

[tex]\vec{v}'=\vec{u}'/u'^0[/tex]

with

[tex]u'^{\mu}={\Lambda^{\mu}}_{\nu} u^{\nu},[/tex]

where [itex]{\Lambda^{\mu}}_{\nu}[/itex] is an arbitrary O(1,3) matrix.

- #8

BruceW

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[tex]u'^{\mu}=\gamma' \begin{bmatrix}{1 \\ \vec{v}'}\end{bmatrix} [/tex]

So the velocity of the object (or whatever), with respect to the old frame is unprimed. And the velocity of the object according to the new frame is primed. (when I say primed, I mean ' this thing).

And also, the unprimed gamma corresponds to the unprimed velocity and the primed gamma corresponds to the primed velocity.

The reason that this form of the 4-velocity is used, is because it is a four-vector. i.e. it has an invariant scalar product, and you can use a Lorentz matrix on it to boost it. And it has a lot of nice properties. On the other hand, the column matrix:

[tex]\begin{bmatrix}{1 \\ \vec{v}}\end{bmatrix}[/tex]

Does not have all the nice properties of a 4-vector. If you wanted an equation for this thing, then you could rearrange the equation for the four-vector as:

[tex]\begin{bmatrix}{1 \\ \vec{v}'}\end{bmatrix} = \frac{\gamma}{\gamma'} {\Lambda^{\mu}}_{\nu} \begin{bmatrix}{1 \\ \vec{v}}\end{bmatrix}[/tex]

But people don't usually write stuff like this, because the physics is really contained in the equation for the four-vector. So the 'proper' four-vector is the thing that you should try to get used to thinking about.

Also, as vanhees says, if you want to then talk about the new 3-velocity, then you can do this:

[tex]\vec{v}'=\vec{u}'/u'^0[/tex]

Which is a much nicer way, because here we can clearly see how it relates to the four-vector, which is the physically important quantity.

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