# What are the Lorentz Transformation equations when the relative velocity of an entity is in an arbitrary direction?

The Lorentz Transformation equations are generally presented as a transformation of spacetime between two observers, when they each move at a velocity $$v$$ along the $$x$$ axis relative to each other. My question is, what are the equations if the relative motion between the observers is not strictly along the $$x$$ axis. For example, if an observer moves at $$0.4c$$ along the $$x$$ axis and at $$0.3c$$ along the $$y$$ axis (which gives it a total relative velocity of $$0.5c$$ at an angle of about $$36.9^\circ$$ from the $$x$$ axis). Intuition says that one would simply solve for the $$x'$$ equation for $$0.3c$$ and $$0.4c$$ to get $$y'$$ and $$x'$$ respectively (essentially, solving for each coordinate individually as one would for $$x'$$), and solve for $$t'$$ using the total relative velocity and the distance from the origin using the square root of $$x^2$$ and $$y^2$$. However, this seems like it could be wrong through the exponential nature of the addition of velocities (as the sum of the velocities ($$v+u$$) approaches (or exceeds) $$c$$, the added velocity ($$(v+u)/(1+vu/c^2)$$) differs more and more from said sum), as each component of the velocity is farther from $$c$$ than the total velocity of the observer. So, what are the Lorentz Transformations if the relative velocity between the observers is in an arbitrary direction?

• In many of my answers I use a more "General matrix Lorentz transformation" for example in : Is it a typo in David Tong's derivation of spin-orbit interaction?, see Figure-01 and the more general matrix Lorentz transformation in equations (02),(03). Jun 28, 2020 at 8:07
• For the composition of two 2+1-Lorentz boosts along two perpedicular directions see in: General matrix Lorentz transformation. Jun 28, 2020 at 8:07
• This issue is usually avoided by saying that the reference frame is always choosen so that the velocity is parallel to the x-axis. However, I guess this doesn't help you... Frobenius comments above seem to be what you are looking for. Jun 28, 2020 at 10:30
• Your result $\,w=0.5c\,$ for the addition of $\,v=0.4c\,$ along the $\,x-$axis with $\,u=0.3c\,$ along the $\,y-$axis is incorrect. Instead of this use $\,\gamma_w=\gamma_u\gamma_v\,$ to find the correct one $\,w=0.48539c$. Jun 28, 2020 at 11:01

Let $$S'$$ move at velocity $$\vec v$$ w.r.t. $$S$$. In $$S$$, denote the spacial position vector of an event by $$\vec r$$ and time by $$t$$: we want to know how $$\vec r$$ and $$t$$ transform as we switch from $$S$$ to $$S'$$.

We can decompose $$\vec r$$ into two components: one parallel to $$\vec v$$ and the other perpendicular, i.e. $$\vec r=\vec r_{\|}+\vec r_{\perp}$$, where $$\vec r_{\|}=(\vec v\cdot\vec r)\frac{\vec v}{v^2}$$ We already know how the position transforms when it is parallel to the velocity. Also, the transformation of time also depends only on the component of position parallel to the velocity. Why? You've already seen it: if one frame is moving w.r.t. the other along the $$x$$ direction only, then the $$t'$$ expression contains $$t$$ and $$x$$ only, not $$y$$ or $$z$$. Also, in the movement in $$x$$ direction case, $$y'=y, z'=z$$, so you know the perpendicular dimensions remain unaffected. i.e. the perpendicular component of position should remain unaffected: $$t'=\gamma\bigg[t-\frac{\vec v\cdot \vec r_{\|}}{c^2}\bigg] \\\vec r'_{\|}=\gamma(\vec r_{\|}-\vec vt) \\\vec r'_{\perp}=\vec r_{\perp}$$

Adding $$\vec r'_{\|}$$ and $$\vec r'_{\perp}$$, $$\vec r'=\vec r'_{\|}+\vec r'_{\perp}=\gamma(\vec r_{\|}-\vec vt)+\vec r_{\perp}=\gamma(\vec r_{\|}-\vec vt)+\vec r-\vec r_{\|}$$

Substituting the value of $$\vec r_{\|}$$, you get $$\vec r'=\gamma(\vec r-\vec vt)+\frac{\gamma^2}{c^2(1+\gamma)}\vec v\times(\vec v\times\vec r)$$

How does velocity addition work? Suppose a particle has velocity $$\vec u$$ in $$S$$ and $$\vec u'$$ in $$S'$$. From the transformation equations of $$\vec r'$$ and $$\vec t$$, you have $$d\vec r'=\gamma(d\vec r-\vec vdt)+\frac{\gamma^2}{c^2(1+\gamma)}\vec v\times(\vec v\times d\vec r) \\dt'=\gamma[dt-d\vec r\cdot\vec v/c^2]=\gamma dt[1-\vec u\cdot\vec v/c^2]$$

Dividing $$d\vec r'$$ by $$dt'$$, $$\vec u'=\frac{\vec u-\vec v}{1-\vec u\cdot\vec v/c^2}+\frac{\gamma}{c^2(1+\gamma)}\frac{\vec v\times(\vec v\times\vec u)}{1-\vec v\cdot\vec u/c^2}$$

But we want $$\vec u$$ in terms of $$\vec u'$$ and $$\vec v$$. For that, switch between primed and unprimed coordinates and let $$\vec v\to -\vec v$$ (imagine the relative motion of the frames and convince yourself that this is so!)

$$\vec u=\frac{\vec u'+\vec v}{1+\vec u'\cdot\vec v/c^2}+\frac{\gamma}{c^2(1+\gamma)}\frac{\vec v\times(\vec v\times\vec u')}{1+\vec v\cdot\vec u'/c^2}$$

If you want to get transformation of the perpendicular and parallel components of the velocity separately, then instead of forming the differential of $$\vec r'$$, do so for $$\vec r'_{\|}$$ and $$\vec r'_{\perp}$$, and divide $$d\vec r'_{\|}$$ (or $$d\vec r'_{\perp}$$) by $$dt'$$.

Let $$\boldsymbol{v}$$ be the velocity vector ascribed to $$\Sigma'$$ in $$\Sigma$$.

If $$(\boldsymbol{r},t) = (x,y,z,t)$$ are the coordinates of an event E in $$\Sigma$$ and $$(\boldsymbol{r'},t') = (x',y',z',t')$$ the coordinates of E in $$\Sigma'$$, it is useful to define a symmetric space operator 𝐿 acting on the vectors position of Σ characterized by the eigenvalues

\begin{align} & L\,\boldsymbol{r} = \boldsymbol{r} & &\text{if } \boldsymbol{r} \perp \boldsymbol{v} & && \\ & L\,\boldsymbol{r} = \gamma\boldsymbol{r}& &\text{if }\boldsymbol{r} \parallel \boldsymbol{v}& \text{where } \gamma =\frac{1}{\sqrt{1-v^2/c^2}}&& \end{align} In explicit form we have

$$$$L\,\boldsymbol{r} =\boldsymbol{r} + (\gamma -1)\; (\boldsymbol{r}\cdot\boldsymbol{v})\;\frac{\boldsymbol{v}}{v^2}$$$$

The vector form of Lorentz transformations answers the question $$$$\begin{cases} \boldsymbol{r'} &= L\,\boldsymbol{r} - \gamma\,\boldsymbol{v} \,t \\ ct' &= \gamma\left(ct - \frac{1}{c}\,\boldsymbol{v}\cdot\boldsymbol{r} \right) \end{cases} \mspace{100mu} \text{where } \gamma = \frac{1}{\sqrt{1-v^2/c^2}}$$$$

For an spacetime interval we have $$$$\begin{cases} \Delta\boldsymbol{r'} &= L\,\Delta\boldsymbol{r} - \gamma \boldsymbol{v}\Delta t \\ \Delta t' &= \gamma\left(\Delta t - \frac{\boldsymbol{v}\cdot\Delta \boldsymbol{r}}{c^2} \right) \end{cases}$$$$ and by division we obtain the law of speed composition $$$$\boldsymbol{u'} = \frac{ L\,\Delta\boldsymbol{r} - \gamma \boldsymbol{v}\Delta t } { \gamma\left(\Delta t - \frac{\boldsymbol{v}\cdot\Delta\boldsymbol{r}}{c^2} \right)} = \frac{ L\,\boldsymbol{u} - \gamma \boldsymbol{v} } { \gamma\left(1 - \frac{\boldsymbol{v}\cdot\boldsymbol {u}}{c^2} \right)} = \frac{ \gamma \boldsymbol{u}_\parallel + \boldsymbol{u}_\perp - \gamma \boldsymbol{v} } { \gamma\left(1 - \frac{\boldsymbol{v}\cdot\boldsymbol{u}}{c^2} \right)}$$$$ For $$\boldsymbol{u}_\perp=0$$ (that is if $$\boldsymbol{u}\parallel\boldsymbol{v}$$) the law takes the well-known form $$$$\boldsymbol{u'} = \frac{\boldsymbol{u} -\boldsymbol{v} } { 1 - \frac{\boldsymbol{v}\cdot\boldsymbol{u}}{c^2}} \qquad \qquad \qquad \boldsymbol{u} = \frac{\boldsymbol{u'} +\boldsymbol{v} } { 1 + \frac{\boldsymbol{v}\cdot\boldsymbol{u'}}{c^2}}$$$$

• Good answer. To help the OP, have you any idea how to add relativistically two not collinear velocities ? Jun 29, 2020 at 22:17
• Well done (for the addition of velocities). Jun 30, 2020 at 14:02

A vector $$𝐀$$ decomposes relative to another vector $$𝐁 ≠ 𝟎$$ into the components $$𝐀 = \frac{𝐁·𝐀𝐁}{B^2} + \frac{𝐁×𝐀×𝐁}{B^2},$$ (with $$B = \sqrt{𝐁·𝐁}$$) that are, respectively, collinear with and perpendicular to $$𝐁$$.

So, for $$𝐫 = (x,y,z)$$ and $$𝐫' = (x',y',z')$$, where the boost velocity $$𝐯 ≠ 𝟎$$ is in an arbitrary direction, replace $$vx$$ by $$𝐯·𝐫$$, $$x$$ by $$𝐯·𝐫𝐯/v^2$$ and $$(y,z)$$ by $$𝐯×𝐫×𝐯/v^2$$: $$t' = \frac{t - 𝐯·𝐫/c^2}{\sqrt{1 - (v/c)^2}}, \hspace 1em 𝐫' = \frac{1}{\sqrt{1 - (v/c)^2}}\left(\frac{𝐯·𝐫𝐯}{v^2} - 𝐯t\right) + \frac{𝐯×𝐫×𝐯}{v^2},$$ where $$v = \sqrt{𝐯·𝐯}$$.

Alternatively, undoing the decomposition of $$𝐫$$, you can write $$𝐯×𝐫×𝐯 = v^2𝐫 - 𝐯·𝐫𝐯$$ and substitute back in to get: $$𝐫' = 𝐫 - \frac{𝐯t}{\sqrt{1 - (v/c)^2}} + \left(\frac{1}{\sqrt{1 - (v/c)^2}} - 1\right)\frac{𝐯·𝐫𝐯}{v^2}.$$ Use the identity: $$\frac{1}{\sqrt{1 - x}} - 1 = \frac{1 - \sqrt{1 - x}}{\sqrt{1 - x}} = \frac{1}{\sqrt{1 - x}}\frac{1 - (1 - x)}{1 + \sqrt{1 - x}} = \frac{1}{\sqrt{1 - x}}\frac{x}{1 + \sqrt{1 - x}}$$ to rewrite this as: $$𝐫' = 𝐫 - \frac{𝐯t}{\sqrt{1 - (v/c)^2}} + \frac{1}{c^2}\frac{𝐯}{\sqrt{1 - (v/c)^2}}\frac{𝐯·𝐫}{1 + \sqrt{1 - (v/c)^2}}.$$ This version weaves back in the case $$𝐯 = 𝟎$$.