# Lorentz transformation for a system composed of a 3 dimensional velocity

I'm working on learning general relativity with the help of math papers and original Einstein's general relativity paper. I only have a pre-calculus math knowledge but I learned on my own differentials, integrals and the basis of matrix operations. I'm working on tensors and they are quite complicated. I'm trying to get the complete Lorentz tensor for the equation $ds^2=g_{uv}dx_udx_v=g^{\sigma\tau}d\xi_\sigma d\xi_\tau$ where $g_{uv}$ is the identity matrix with the sign correction. I'm using Einstein notation from 1 to 4.

$g_{uv}= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 &-1 \end{pmatrix}$

Want I want to bring with this is that I'm looking for the tensor (if it's the proper name assigned to that object) that will convert the $(x_1,x_2,x_3,x_4)$ or $(x,y,z,t)$ to the $(\xi_1,\xi_2,\xi_3,\xi_4)$ or $(\xi,\eta,\zeta,\tau)$.

In special relativity, the basic lorentz transformation that we could easily define would be :

$\xi=(x-vt)\beta$

$\eta=y$

$\zeta=z$

$\tau=(t-\frac {vx}{c^2})\beta$

Where $\beta$ is the usual Lorentz factor. It's limited to a motion on the x axis. I'm struggling on finding a way to extend the motion to the y and z axis to include it in the lorentz transformation, then adding it to a matrix that would work with the upper $ds^2$ equation.

If we consider the "basic" Lorentz transformation with only 1 dimensional velocity, we can form a matrix from the transformation, and the product of the transformation matrix times the vector will give the other system of coordinates. By reversing the process we can get the inverse operation out of it. It is also possible to prove the Lorentz factor this way.

If there is a way to figure out the complete Lorentz transformation for this, I will make a big step foward in my understandings of the tensor. I'm not perfectly at ease with those operations yet, but each step I made here ,I believe, is correct. I think this is general relativity applied to a special case with no acceleration. I would really like to have some external opinion about this.

Note by Samuel Hatin
6 years, 9 months ago

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To start with, if you are working with $g_{uv} \; = \; \left( \begin{array}{cccc}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \end{array}\right)$ your coordinates need to be $(x^1,x^2,x^3,x^4) = (x,y,z,ct)$, and your standard Lorentz boost is $\begin{array}{rcl} \xi^1 & = & \gamma(x^1 - \tfrac{v}{c}x^4) \\ \xi^2 & = & x^2 \\ \xi^3 & = & x^3 \\ \xi^4 & = & \gamma(-\tfrac{v}{c}x^1 + x^4) \end{array}$ or $\left( \begin{array}{c}\xi^1 \\ \xi^2 \\ \xi^3 \\ \xi^4\end{array}\right) \; = \; \left( \begin{array}{cccc}\gamma & 0 & 0 & -\tfrac{\gamma v}{c} \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -\tfrac{\gamma v}{c} & 0 & 0 & \gamma \end{array}\right)\left(\begin{array}{c} x^1\\x^2\\x^3\\x^4\end{array}\right)$ where $\gamma = (1 - \tfrac{v^2}{c^2})^{-\frac12}$.

In addition to these Lorentz boosts, Special Relativity permits Galilean transformations, which include rotations of the spatial coordinates which leave the time coordinate unchanged. Coordinate changes for frames would then use the formula $\left( \begin{array}{c}\xi^1 \\ \xi^2 \\ \xi^3 \\ \xi^4\end{array}\right) \; = \; \left( \begin{array}{cccc} & & & 0 \\ & A & & 0 \\ & & & 0 \\ 0 & 0 & 0 & 1 \end{array}\right)\left(\begin{array}{c} x^1\\x^2\\x^3\\x^4\end{array}\right)$ where $A$ is a $3\times3$ rotation matrix, so that $AA^T = A^TA = I$ and $\mathrm{det} A = 1$.

Suppose that you want to compare two frames, where one is moving with velocity $(u,v,w)$ with respect to the other. Apply a Galilean transformation to make the direction of motion along the $x^1$-axis, apply a standard Lorentz boost and invert the Galilean transformation.

To be specific, find a rotation matrix $A$ such that $A\left(\begin{array}{c}u\\v\\w\end{array}\right) \; = \; V\left(\begin{array}{c}1\\0\\0\end{array}\right)$ where $V \,=\, \sqrt{u^2+v^2+w^2}$. Of course $V$ is the speed of one frame relative to the other. To do this we need to choose the first row of $A$ to be $(\tfrac{u}{V},\tfrac{v}{V},\tfrac{w}{V})$, with the three rows forming a right-handed orthonormal triad. Consider the Lorentz transformation $\left( \begin{array}{c}\xi^1 \\ \xi^2 \\ \xi^3 \\ \xi^4\end{array}\right) \; = \; L\left(\begin{array}{c} x^1\\x^2\\x^3\\x^4\end{array}\right)$ where $L \; = \; \left( \begin{array}{cccc} & & & 0 \\ & A^T & & 0 \\ & & & 0 \\ 0 & 0 & 0 & 1 \end{array}\right) \left( \begin{array}{cccc}\gamma & 0 & 0 & -\tfrac{\gamma V}{c} \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -\tfrac{\gamma V}{c} & 0 & 0 & \gamma \end{array}\right) \left( \begin{array}{cccc} & & & 0 \\ & A & & 0 \\ & & & 0 \\ 0 & 0 & 0 & 1 \end{array}\right)$ where $\gamma \,=\, \big(1 - \tfrac{V^2}{c^2}\big)^{-\frac12}$.

The origin of the $(x^1,x^2,x^3,x^4)$ frame has timeline written as $\left(\begin{array}{c}0\\0\\0\\ct\end{array}\right)$ in that frame. Thus, in the $(\xi^1,\xi^2,\xi^3,\xi^4)$ frame, the coordinates of the timeline of the origin of the other frame are $L\left(\begin{array}{c}0\\0\\0\\ct\end{array}\right) \; = \; \left( \begin{array}{cccc} & & & 0 \\ & A^T & & 0 \\ & & & 0 \\ 0 & 0 & 0 & 1 \end{array}\right) \left( \begin{array}{cccc}\gamma & 0 & 0 & -\tfrac{\gamma V}{c} \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -\tfrac{\gamma V}{c} & 0 & 0 & \gamma \end{array}\right)\left(\begin{array}{c}0\\0\\0\\ct\end{array}\right) \; = \; \left( \begin{array}{cccc} & & & 0 \\ & A^T & & 0 \\ & & & 0 \\ 0 & 0 & 0 & 1 \end{array}\right) \left(\begin{array}{c} -\gamma Vt \\ 0 \\ 0 \\ \gamma ct\end{array}\right)$ and hence $L\left(\begin{array}{c}0\\0\\0\\ct\end{array}\right) \; = \; \left( \begin{array}{c} -\gamma ut\\-\gamma vt\\-\gamma wt \\ \gamma ct\end{array}\right) \; = \; \left(\begin{array}{c} -u\tau\\-v\tau\\-w\tau\\c\tau\end{array}\right)$ and hence the $(\xi^1,\xi^2,\xi^3,\xi^4)$ frame is indeed moving with speed $(u,v,w)$ with respect to the other frame. You will find that spatial measurements in directions perpendicular to that velocity are unchanged, as should be.

- 6 years, 9 months ago

Thank you very much for your answer, I never used rotation matrix but by the name I guess what the use of it. This is quite clever and I will look more into it!

- 6 years, 9 months ago