# LOG#006. Lorentz Transformations(II).

$\boxed{ \begin{cases} x'_0=ct'=\gamma (ct - \mathbf{\beta} \cdot \mathbf{r}) \\ \mathbf{r'}=\mathbf{r}+(\gamma -1) \dfrac{(\mathbf{\beta}\cdot \mathbf{r})\mathbf{\beta}}{\beta^2} -\gamma \beta ct \\ \gamma = \dfrac{1}{\sqrt{1-\beta^2}}= \dfrac{1}{\sqrt{1-\beta_x^2-\beta_y^2-\beta_z^2}} \end{cases}}$

$\boxed{\left( \begin{array}{c} ct' \\ x' \\ y' \\ z' \end{array} \right) = \begin{pmatrix} \gamma & -\gamma \beta_x & -\gamma \beta_y & -\gamma \beta_z \\ -\gamma \beta_x & 1+(\gamma -1)\dfrac{\beta_{x}^{2}}{\beta^2} & (\gamma -1)\dfrac{\beta_x \beta_y}{\beta^2} & (\gamma -1)\dfrac{\beta_x \beta_z}{\beta^2} \\ -\gamma \beta_y & (\gamma -1)\dfrac{\beta_y \beta_x}{\beta^2} & 1+(\gamma -1)\dfrac{\beta_{y}^{2}}{\beta^2} & (\gamma -1)\dfrac{\beta_y \beta_z}{\beta^2} \\ -\gamma \beta_z & (\gamma -1)\dfrac{\beta_z \beta_x}{\beta^2} & (\gamma -1)\dfrac{\beta_z \beta_y}{\beta^2} & 1+(\gamma -1)\dfrac{\beta_{z}^{2}}{\beta^2} \end{pmatrix} \left( \begin{array}{c} ct\\ x\\ y\\ z\end{array}\right)}$

These equations define the most general (direct) Lorentz transformations  and we see they are not those in the previous post! I mean, they are not the one with the relative velocity in the direction of one particular axis, as we derived in the previous log. We will derive these equations. How can we derive them?

The most general Lorentz transformation involves the following scenario (a full D=3+1 motion):

1st) The space-time coordinates of an event E are described by one observer (and frame) A at rest at the origin of his own frame S. The observer B is at rest at the origin in a second frame S’. S and S’ have parallel axes.

2nd) The origin of the S and S’ frames coincide at t=t’=0.

3rd) B moves relative to A with a velocity  3d vector (space-like) given by $\mathbf{v}=(v_x,v_y,v_z)$.

4th) The position vector of the event in the S-frame is $\mathbf{r}=(x,y,z)$. It is decomposed into  “horizontal/vertical” or parallel/orthogonal pieces as follows

$\mathbf{r}=\mathbf{r}_\parallel + \mathbf{r}_ \perp$

The following transformation is suitable for the S’-frame, defining $\beta=\mathbf{v}/c=(v_x/c,v_y/c,v_z/c)$:

$ct'=x_0=\gamma(ct-\beta r_\parallel)=\gamma (ct - \mathbf{\beta} \cdot \mathbf{r})$

$\mathbf{r'}_\parallel = \gamma (\mathbf{r}_\parallel - \mathbf{\beta}ct )$

$\mathbf{r'}_\perp=\mathbf{r}_\perp$

where the dot represents scalar product. Using the elementary knowledge and application of the scalar product with projections of vectors, we calculate the projection of the position vector onto the velocity vector in any frame with the scalar product of the position vector with a normalized velocity vector, $\mathbf{\hat{v}}=\mathbf{v}/v$ . In the S’-frame we will get the projection $\mathbf{\hat{v}}\cdot \mathbf{r}$. Therefore,

$\mathbf{r}_\parallel = (\hat{\mathbf{v}}\cdot \mathbf{r})\hat{\mathbf{v}}$

and thus, the component of the position vector with respect to the parallel direction to the velocity will be:

$\mathbf{r}_\parallel = (\hat{\mathbf{v}}\cdot \mathbf{r})\hat{\mathbf{v}}= \dfrac{(\mathbf{v}\cdot \mathbf{r})\mathbf{v}}{v^2}$

or

$\mathbf{r}_\parallel = \dfrac{(\mathbf{\beta}\cdot \mathbf{r})\mathbf{\beta}}{\beta^2}$

Then, since $\mathbf{r}_\perp = \mathbf{r}-\mathbf{r}_\parallel$, we have

$\mathbf{r}_\perp = \mathbf{r}- \dfrac{(\mathbf{\beta}\cdot \mathbf{r})\mathbf{\beta}}{\beta^2}$

Finally, we put together the vertical/horizontal (orthogonal/parallel) pieces of the general Lorentz transformations:

$\mathbf{r'} =\mathbf{r'}_\parallel + \mathbf{r'}_ \perp = \gamma \left( \dfrac{(\mathbf{\beta}\cdot \mathbf{r})\mathbf{\beta}}{\beta^2} -\beta ct \right)+\mathbf{r}- \dfrac{(\mathbf{\beta}\cdot \mathbf{r})\mathbf{\beta}}{\beta^2}$

Then, the general 4D=3d+1 Lorentz transformation (GLT) from S to S’ are defined through the equations:

$\boxed{GLT(S\rightarrow S') \begin{cases} x'_0=ct'=\gamma (ct - \mathbf{\beta} \cdot \mathbf{r}) \\ \mathbf{r'}=\mathbf{r}+(\gamma -1) \dfrac{(\mathbf{\beta}\cdot \mathbf{r})\mathbf{\beta}}{\beta^2} -\gamma \beta ct \\ \gamma = \dfrac{1}{\sqrt{1-\beta^2}}= \dfrac{1}{\sqrt{1-\beta_x^2-\beta_y^2-\beta_z^2}} \end{cases}}$

Q.E.D.

The inverse GLT (IGLT) will be:

$\boxed{IGLT(S'\rightarrow S) \begin{cases} x_0=ct=\gamma (ct' + \mathbf{\beta} \cdot \mathbf{r}') \\ \mathbf{r}=\mathbf{r}'+(\gamma -1) \dfrac{(\mathbf{\beta}\cdot \mathbf{r}')\mathbf{\beta}}{\beta^2} +\gamma \beta ct' \\ \gamma = \dfrac{1}{\sqrt{1-\beta^2}}= \dfrac{1}{\sqrt{1-\beta_x^2-\beta_y^2-\beta_z^2}} \end{cases}}$

Indeed, these transformations allow a trivial generalization to D=d+1, i.e., these transformations are generalized to d-spatial dimensions simply allowing a d-space velocity and beta parameter, while time remains 1d. Indeed, the Lorentz transformations form a group. A group is a mathematical gadget with certain “nice features” that physicists and mathematicians love. You can imagine the Lorentz group in D=d+1 dimensions as a generalization of the rotation group  called Lorentz group. The Lorentz group involves rotations around the spatial axes plus the so-called “boosts”, transformations involving mixing of space and time coordinates. Indeed, the Lorentz transformations involving relative motion along one particular axis IS a (Lorentz) boost! That is, the simplest Lorentz transformations like the one in the previous posts are “boosts”.

With the above transformations, the GLT can be easily written in components:

$\boxed{\left( \begin{array}{c} ct' \\ x' \\ y' \\ z' \end{array} \right) = \begin{pmatrix} \gamma & -\gamma \beta_x & -\gamma \beta_y & -\gamma \beta_z \\ -\gamma \beta_x & 1+(\gamma -1)\dfrac{\beta_{x}^{2}}{\beta^2} & (\gamma -1)\dfrac{\beta_x \beta_y}{\beta^2} & (\gamma -1)\dfrac{\beta_x \beta_z}{\beta^2} \\ -\gamma \beta_y & (\gamma -1)\dfrac{\beta_y \beta_x}{\beta^2} & 1+(\gamma -1)\dfrac{\beta_{y}^{2}}{\beta^2} & (\gamma -1)\dfrac{\beta_y \beta_z}{\beta^2} \\ -\gamma \beta_z & (\gamma -1)\dfrac{\beta_z \beta_x}{\beta^2} & (\gamma -1)\dfrac{\beta_z \beta_y}{\beta^2} & 1+(\gamma -1)\dfrac{\beta_{z}^{2}}{\beta^2} \end{pmatrix} \left( \begin{array}{c} ct\\ x\\ y\\ z\end{array}\right)}$

Q.E.D.

These transformations can be written in a symbolic way using matrix notation as $\mathbb{X}'=\mathbb{L}\mathbb{X}$ or using tensor calculus:

$x^{\mu'}=\Lambda^{\mu'}_{\;\nu} x^\nu$

The inverse GLT (IGLT) will be in component way:

$\boxed{\left( \begin{array}{c} ct \\ x \\ y \\ z \end{array} \right) = \begin{pmatrix} \gamma & \gamma \beta_x & \gamma \beta_y & \gamma \beta_z \\ \gamma \beta_x & 1+(\gamma -1)\dfrac{\beta_{x}^{2}}{\beta^2} & (\gamma -1)\dfrac{\beta_x \beta_y}{\beta^2} & (\gamma -1)\dfrac{\beta_x \beta_z}{\beta^2} \\ \gamma \beta_y & (\gamma -1)\dfrac{\beta_y \beta_x}{\beta^2} & 1+(\gamma -1)\dfrac{\beta_{y}^{2}}{\beta^2} & (\gamma -1)\dfrac{\beta_y \beta_z}{\beta^2} \\ \gamma \beta_z & (\gamma -1)\dfrac{\beta_z \beta_x}{\beta^2} & (\gamma -1)\dfrac{\beta_z \beta_y}{\beta^2} & 1+(\gamma -1)\dfrac{\beta_{z}^{2}}{\beta^2} \end{pmatrix} \left( \begin{array}{c} ct'\\ x'\\ y'\\ z'\end{array}\right)}$

and they can be written as $\mathbb{X}=\mathbb{L}^{-1}\mathbb{X'}$, or using tensor notation

$x^\rho=(\Lambda^{-1})^\rho_{\;\mu'} x^{\mu'}$

in such a way that

$x^{\mu'} = \Lambda^{\mu}_{\; \nu} x^\nu \rightarrow (\Lambda^{-1})^{\rho}_{\; \mu'}x^{\mu'} = (\Lambda^{-1})^{\rho}_{\;\mu'}(\Lambda)^{\mu'}_{\;\nu} x^{\nu} = x^{\rho} = \delta ^{\rho}_{\; \nu}x^\nu$

Thus, $(\Lambda^{-1})^{\rho}_{\;\mu'}(\Lambda)^{\mu'}_{\;\nu} = \delta ^{\rho}_{\; \nu}$

or equivalently $\mathbb{L}^{-1}\mathbb{L}=\mathbb{L}\mathbb{L}^{-1}=\mathbb{I}$.

$\delta ^{\rho}_{\; \nu}$ is the “unity” tensor, also called Kronecker delta, meaning that its components are 1 if $\rho = \nu$ and 0 otherwise (if $\rho \neq \nu$). The Kronecker delta is therefore the “unit” tensor with two indexes.

NOTATIONAL CAUTION: Be aware, some books and people use to assume you know when you need the matrix $\mathbb{L}$ or its inverse $\mathbb{L}^{-1}$. Thus, you will often read and see this

$x^{\mu'}=\Lambda^{\mu'}_{\;\nu} x^\nu \rightarrow x^\nu= \Lambda^\nu_{\;\mu'} x^{\mu'}$

where certain abuse of language since it implies that

$\Lambda^\nu_{\;\mu'} = (\Lambda^{-1})^{\mu'}_{\;\nu}$

and because we have  to be mathematically consistent, the following  relationship is required to hold

$\Lambda^\nu_{\;\mu'}\Lambda^{\mu'}_{\;\nu}=1$

or more precisely, taking care with the so-called free indexes

$\Lambda^\rho_{\;\mu'} \Lambda^{\mu'}_{\;\sigma}=\delta^\rho_\sigma$

as before.