# Lorentz Transformation

Can Lorentz Transformation be applied directly to a four velocity vector?

I mean let v$_{α}$ be a four velocity vector.

Is there a form of Lorentz tfm matrix such that:

v$^{'}$$_{α}$ = $\Lambda^{β}$$_{α}$v$_{β}$ ?

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dextercioby
Homework Helper
Yes, why not ? As long as the velocity in the transformation is the relative velocity between frames, thus independent of the spatial components of the v 4-vector. And we usually write

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

BruceW
Homework Helper
yeah man, as wikipedia says: "The transformation matrix is universal for all four-vectors, not just 4-dimensional spacetime coordinates" on this page https://en.wikipedia.org/wiki/Lorentz_transformation about halfway down, under the subsection heading "transformation of other physical quantities"

I didn't encounter the matrix anywhere, they always use the addition of the velocities, so I wasn't sure. I think I should construct on my own that matrix. Thank you

Correction:
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
$\vec{v}$ 's as (c, $\vec{v}$ ) (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.

Fredrik
Staff Emeritus
Gold Member
In units such that c=1,
$$\Lambda=\frac{1}{\sqrt{1-v^2}}\begin{pmatrix}1 & -v\\ -v & 1\end{pmatrix},$$ 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".

vanhees71
Gold Member
2019 Award
Of course, you cannot simply transform a non-covariant object. The right way to "add" velocities in the sense of three-vector velocities is to first go over to the proper velocity
$$u^{\mu}=\gamma \begin{bmatrix}{1 \\ \vec{v}}\end{bmatrix}$$
with $\gamma=1/\sqrt{1-v^2}$.

Of course you get the three-velocity as
$$\vec{v}=\vec{u}/u^0.$$
Then, by definition you define the three-velocity in the new frame as
$$\vec{v}'=\vec{u}'/u'^0$$
with
$$u'^{\mu}={\Lambda^{\mu}}_{\nu} u^{\nu},$$
where ${\Lambda^{\mu}}_{\nu}$ is an arbitrary O(1,3) matrix.

BruceW
Homework Helper
I think vanhees has explained it pretty well. From the equations, we get:
$$u'^{\mu}=\gamma' \begin{bmatrix}{1 \\ \vec{v}'}\end{bmatrix}$$
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:
$$\begin{bmatrix}{1 \\ \vec{v}}\end{bmatrix}$$
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:
$$\begin{bmatrix}{1 \\ \vec{v}'}\end{bmatrix} = \frac{\gamma}{\gamma'} {\Lambda^{\mu}}_{\nu} \begin{bmatrix}{1 \\ \vec{v}}\end{bmatrix}$$
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:
$$\vec{v}'=\vec{u}'/u'^0$$
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.