# LOG#010. Relativistic velocities.

In our daily experience, we live in a “non-relativistic” world with a very high degree of accuracy. Thus, if you see a train departing from you ( you are at rest relative to it) with speed $V$ (in the positive direction of the x-axis), you  move with relative velocity $V-u$ respect to the train if you run in pursuit of it with speed $u$, or maybe you can also run with relative speed $V+u$ if you run away from it in the opposite direction of motion.

However, light behaviour is diferent to material bodies. Light, a.k.a. electromagnetic waves, is weird.  I hope you have realized it from previous posts. We will see what happen with an SR analogue gedanken experiment of the previously mentioned “non-relativistic” train(S’-frame)-track(S-frame) experiment and that we have seen lot of times in our ordinary experience. We will discover that velocities close to the speed of light add in a different way, but we recover the classical result ( like the above) in the limit of low velocities ( or equivalently, in the limit $c\rightarrow \infty$).

Problem to be solved:

In the S’-frame, an object (or particle) moves at constant velocity $\vec{v}=\mathbf{v}=(v,0,0)$ relative to the S-frame. In the S’-frame, the object/particle moves with velocity

$\vec{u}\,'=\mathbf{u}'=(u'_x,u'_y,u'_z)=\left( \dfrac{dx'}{dt'},\dfrac{dy'}{dt'},\dfrac{dz'}{dt'}\right)$

The question is: what is the velocity in the S-frame

$\vec{u}=\mathbf{u}=(u_x,u_y,u_z)=\left( \dfrac{dx}{dt},\dfrac{dy}{dt},\dfrac{dz}{dt}\right)$

CAUTION ( important note): v is constant (a relative velocity from one frame into another). u is not constant, since it is a vector describing the motion of a particle in some frame.

From the definiton of velocity, and the Lorentz transformation for a parallel motion, we have

$u_x=\dfrac{dx}{dt}=\dfrac{\dfrac{dx}{dt'}}{\dfrac{dt}{dt'}}=\dfrac{\dfrac{d}{dt'}\left[\gamma (x'+\beta ct')\right]}{\dfrac{d}{dt'}\left[\gamma (t'+\frac{\beta }{c}x')\right]}=\dfrac{\dfrac{dx'}{dt'}+\beta c \dfrac{dt'}{dt'}}{\dfrac{dt'}{dt'}+\dfrac{\beta}{c}\dfrac{dx'}{dt'}}$

and thus we get the addition law of velocities in the direction of motion

$\boxed{u_x=\dfrac{u'_x+v}{1+\dfrac{u'_x v}{c^2}}}$

We can also calculate the transformation of the transverse components to the velocity in the sense of motion. We only calculate the component $u_y$ since the remaining one would be identical but labelled with other letter(the z-component indeed):

$u_y=\dfrac{dy}{dt}=\dfrac{\dfrac{dy}{dt'}}{\dfrac{dt}{dt'}}=\dfrac{\dfrac{dy'}{dt'}}{\gamma \left( 1+\dfrac{u'_x v}{c^2}\right)}$

$\boxed{u_y=\dfrac{u'_y}{\gamma \left( 1+\dfrac{u'_x v}{c^2}\right)}}$

Therefore, the transverse velocity also changes in that way! There is an alternative deduction of the above formula using space and time coordinates. We will proceed in two important cases only.

The first case is when the motion happens with parallel relative velocity. Suppose two inertial frames S and S’. S is moving relative to S’ with velocity $V$ along the X-axis. Moreover, suppose an object that is moving parallel to OX, with velocity $v$. Imagine two “frozen pictures” of the object at two different times according to S, e.g., fix two times $t_1$ and $t_2$. The two events have coordinates of space and time given by

$E_1(t_1,x_1,y_1,z_1)$ and $E_2(t_2,x_2,y_2,z_2)$

But we do know that $t_2=t_1+\Delta t$ and so,

$(t_2,x_2,y_2,z_2)=(t_1+\Delta t,x_1+v\Delta t,y_1,z_1)$

What does the S’-frame observe? Using the Lorentz boost with speed $-V$, we get

$(t'_1,x'_1,y'_1,z'_1) = (\gamma (V)(t_1+Vx_1/c^2),\gamma (V) (x_1+Vt_1),y_1,z_1)$

and

$(t'_2,x'_2,y'_2,z'_2)=(\gamma (V)(t_2+Vx_2/c^2),\gamma (V)(x_2+Vt_2),y_2,z_2)$

Therefore, the velocity of the object according to the S’-frame will be:

$v'=\dfrac{\Delta x'}{ \Delta t'}=\dfrac{x'_2-x'_1}{t'_2-t'_1}=\dfrac{x_2+Vt_2-(x_1+Vt_1)}{t_2+(V/c^2)x_2-(t_1+(V/c^2)x_1)}$

and in this way, we obtain, dividing by $t_2-t_1$

$v'=\dfrac{\dfrac{(x_2-x_1)+V(t_2-t_1)}{(t_2-t_1)}}{\dfrac{(t_2-t_1)+(V/c^2)(x_2-x_1)}{(t_2-t_1)}}$

or equivalently

$\boxed{v'=\dfrac{v+V}{1+\dfrac{V \cdot v}{c^2}}}$

as before.

The second case is when in the velocities between the frames are orthogonal (perpendicular) can be also calculated in this way. Suppose that some object is moving in the orthogonal ( perpendicular) direction to the OX axis. For instance, we can suppose it moves along the y-axis (OY axis) with velocity v measured in the S-frame. We proceed in the same fashion that the previous calculation. We take two “imaginary pictures” of the body at $t_1$ and $t_2=t_1+\Delta t$. We write the coordinates of space and time of this object as

$E_1(t_1,x_1,y_1,z_1)$ and $E_2(t_2,x_2,y_2,z_2)$

But we do know that $t_2=t_1+\Delta t$ and so,

$(t_2,x_2,y_2,z_2)=(t_1+\Delta t,x_1,y_1+v\Delta t,z_1)$

We make the corresponding Lorentz boost on those coordinates

$(t'_1,x'_1,y'_1,z'_1) = (\gamma (V)(t_1+Vx_1/c^2),\gamma (V) (x_1+Vt_1),y_1,z_1)$

$(t'_2,x'_2,y'_2,z'_2)=(\gamma (V)(t_2+Vx_2/c^2),\gamma (V)(x_2+Vt_2),y_2,z_2)$

and now, the two components of this motion in the S’-frame will be given by ( note than our two set of coordinates have $x_2=x_1$ and $y_2\neq y_1$ in this particular case):

$v'_{x'}=\dfrac{\Delta x'}{\Delta t'}=\dfrac{x_2+Vt_2-(x_2+Vt_1)}{t_2+(V/c^2)x_2-(t_1+(V/c^2)x_1)}=V$

$v'_{y'}=\dfrac{\Delta y'}{\Delta t'}=\dfrac{y'_2-y'_1}{\gamma (V)(t_1+(V/c^2)x_2-(t_1+(V/c^2)x_1))}=\dfrac{\dfrac{(y_2-y_1)}{(t_2-t_1)}}{\gamma (V)}=\dfrac{v}{\gamma (V)}$

and so, in summary, in the orthogonal relative motion we have

$\boxed{ v'_{x'}=V \;\; v'_{y'}=\dfrac{v}{\gamma (V)}}$

Indeed, these two cases are particular cases of the general transformations we got before.

The complete transformation of the velocity components and their inverses (obtained with the simple rule of mapping v into -v, and primed variables into unprimed variables)  can be summarized  by these formulae:

$\boxed{\mbox{SR: Adding velocity(I)}\begin{cases}u_x=\dfrac{u'_x+v}{1+\dfrac{u'_x v}{c^2}} \; \; u_y=\dfrac{u'_y}{\gamma \left( 1+\dfrac{u'_x v}{c^2}\right)}\; \; u_z=\dfrac{u'_z}{\gamma \left( 1+\dfrac{u'_x v}{c^2}\right)}\\ \; \\ u'_x=\dfrac{u_x-v}{1-\dfrac{u_x v}{c^2}} \; \; u'_y=\dfrac{u_y}{\gamma \left( 1-\dfrac{u_x v}{c^2}\right)}\; \; u'_z=\dfrac{u_z}{\gamma \left( 1-\dfrac{u_x v}{c^2}\right)}\\ \; \\ \mathbf{u}=(u_x,u_y,u_z)\;\; \mathbf{u}'=(u'_x,u'_y,u'_z)\;\; \gamma =\dfrac{1}{\sqrt{1-\beta ^2}}\;\; \beta =\dfrac{v}{c}\end{cases}}$

Of course, these transformations are valid in the case of a parallel relative motion between S and S’. What are the transformations in the case of non-parallel motion? Suppose that

$\vec{\beta} =\dfrac{\mathbf{v}}{c}=(\beta_x,\beta_y,\beta_z)$

and $\mathbf{u}=(u_x,u_y,u_z), \; \mathbf{u}'=(u'_x,u'_y,u'_z)$ as before. Then, using the most general Lorentz transformations

$\mathbf{u}'=\dfrac{d\mathbf{r}'}{dt'}=\dfrac{d\mathbf{r}+(\gamma - 1)\dfrac{\left(\vec{\beta}\cdot \mathbf{u}\right)\vec{\beta}}{\beta^2}-\gamma \vec{\beta}cdt}{\gamma dt - \dfrac{1}{c}\gamma \vec{\beta}\cdot d\mathbf{r}}$

then, using the same trick as above

$\mathbf{u}'=\dfrac{d\mathbf{r}'}{dt'}=\dfrac{\dfrac{d\mathbf{r}'}{dt}}{\dfrac{dt'}{dt}}$

$\mathbf{u}'=\dfrac{\dfrac{1}{\gamma}\mathbf{u}+\left(1-\dfrac{1}{\gamma}\right)\dfrac{\left(\vec{\beta}\cdot \mathbf{u}\right)\vec{\beta}}{\beta^2}-\vec{\beta} c}{1-\dfrac{1}{c}\vec{\beta} \cdot \mathbf{u}}$

We have got the following transformations (we apply the same recipe to obtain the inverse transformations, also included in the box below):

$\boxed{\mbox{SR: Adding velocity(II)}\begin{cases} \mathbf{u}'=\dfrac{1}{1-\dfrac{\mathbf{u}\cdot\mathbf{v}}{c^2}}\left[\dfrac{\mathbf{u}}{\gamma}+\left[\left(1-\dfrac{1}{\gamma}\right)\dfrac{\mathbf{u}\cdot\mathbf{v}}{v^2}-1\right]\mathbf{v}\right]\\ \;\\ \mathbf{u}=\dfrac{1}{1+\dfrac{\mathbf{u}'\cdot\mathbf{v}}{c^2}}\left[\dfrac{\mathbf{u}'}{\gamma}-\left[-\left(1-\dfrac{1}{\gamma}\right)\dfrac{\mathbf{u}'\cdot\mathbf{v}}{v^2}-1\right]\mathbf{v}\right]\end{cases}}$

We observe that these equations are a non-linear addition of velocities. Equivalently, they can be rewritten as follows after some elementary algebra using a mathematical structure called gyrovector (or gyrovector addition):

$\boxed{\mbox{gyrovector law}\begin{cases}\mathbf{u}'\equiv -\mathbf{v}\biguplus_{REL}\mathbf{u}=\dfrac{1}{1-\dfrac{\mathbf{u}\cdot \mathbf{v}}{c^2}}\left[-\mathbf{v}+\dfrac{\mathbf{u}}{\gamma_{\mathbf{v}}}+\dfrac{1}{c^2}\left(\dfrac{\gamma_{\mathbf{v}}}{\gamma_{\mathbf{v}}+1}\right)\left(\mathbf{v}\cdot \mathbf{u}\right)\mathbf{v}\right]\\ \; \\ \mathbf{u}\equiv \mathbf{v}\biguplus_{REL}\mathbf{u}'=\dfrac{1}{1+\dfrac{\mathbf{u}'\cdot \mathbf{v}}{c^2}}\left[\mathbf{v}+\dfrac{\mathbf{u}'}{\gamma_{\mathbf{v}}}+\dfrac{1}{c^2}\left(\dfrac{\gamma_{\mathbf{v}}}{\gamma_{\mathbf{v}}+1}\right)\left(\mathbf{v}\cdot \mathbf{u}'\right)\mathbf{v}\right]\end{cases}}$

These 2 cases can be seen as particular examples in the addition rule of velocities as a “gyrovector sum”, the nonlinear addition rule given by the formula:

$\mathbf{u}\biguplus_{REL}\mathbf{v}=\dfrac{1}{1+\dfrac{\mathbf{u}\cdot \mathbf{v}}{c^2}}\left[\mathbf{u}+\dfrac{\mathbf{v}}{\gamma_{\mathbf{u}}}+\dfrac{1}{c^2}\left(\dfrac{\gamma_{\mathbf{u}}}{\gamma_{\mathbf{u}}+1}\right)\left(\mathbf{u}\cdot \mathbf{v}\right)\mathbf{u}\right]$

This formula is usually written in a more intuitive expression with the following arguments. Suppose some object moves with velocity $\mathbf{v}$ in some inertial frame S. S is moving itself with relative velocity $\mathbf{V}$ respect to another frame S’. In the S’-frame, the velocity is given by:

$\boxed{\mathbf{v}'\equiv \mathbf{V}\biguplus_{REL}\mathbf{v}=\dfrac{\mathbf{v}_\parallel+\gamma^{-1}(V)\mathbf{v}_\perp + \mathbf{V}}{1+\dfrac{\mathbf{V}\cdot \mathbf{v}}{c^2}}}$

and where we have defined the projections of $\mathbf{v}$ in the direction parallel and orthogonal to $\mathbf{V}$. They are given by:

$\boxed{\mathbf{v}_\parallel =\dfrac{(\mathbf{V}\cdot \mathbf{u})\mathbf{V}}{V^2}\;\;\;\;\;\; V^2=\vert \mathbf{V}\vert^2\;\;\;\;\;\mathbf{v}_\perp =\mathbf{v}-\mathbf{v}_\parallel}$

We will talk about gyrovectors more in a future post. They have a curious mathematical structure and geometry, and they are not well known  by physicists since they are not in the basic curriculum and background of SR courses. Of course, the non-associative composition rule for velocities is not a standard formula you can find in books about relativity, so I will write it here:

$\boxed{\mathbf{u}\boxplus\mathbf{v}=\dfrac{\mathbf{u}+\mathbf{v}}{1+\dfrac{\mathbf{u}\cdot \mathbf{v}}{c^2}}+\dfrac{\gamma_{\mathbf{u}}}{c^2(\gamma_{\mathbf{u}}+1)}\dfrac{\mathbf{u}\times\left(\mathbf{u}\times\mathbf{v}\right)}{1+\dfrac{\mathbf{u}\cdot \mathbf{v}}{c^2}}}$

and where we used the previous formula for $\mathbf{u}\biguplus_{REL}\mathbf{v}$ and after some algebra we used the known relationship for the cross product of three vectors, two being the same,

$u\times(u\times v)=(u\cdot v)u-(u\cdot u)v$

Let’s go back to the beginning. Suppose now we imagine a train (our S’-frame) travelling at velocity $v$. Suppose that Special Relativity matters now. We are in the tracks as observers “in relative rest” with respect to the train ( we are the S-frame) and suppose that we take into account the SR corrections above to the addition of velocities. Inside the train some object is being thrown with velocity $u'_x$ in the direction of motion. What is the velocity $u_x$ in the S-frame? That is, what is the velocity we observe in the tracks? In this simple example, we use the easier addition rule of velocities ( named addition rule SR(I) above). Firstly, we note some expected features from the mathematical structure of the relativistic addition rule of velocities (valid propterties as well in the general case (II) with a suitable generalization):

1st. For low velocities, i.e., if $u'_x/c<<1$ and/or $\beta=\dfrac{v}{c}<<1$, the result approaches the nonrelativistic “ordinary life” experience: $u_x=u'_x+v$.

2nd. For positive velocities $u'_x>0$ and a positive relative velocity between frames $v>0$, the addition of velocities is generally $u_x, i.e., we get a velocity smaller that in the non-relativistic (ordinary or “common” experience) limit.

Now, some easy numerical examples to see what is going on bewteen the train (S’) and the track (S) where we are:

Example 1. Moderate velocity case. We have, e.g., velocities $v=u'_x=30 km\cdot s^{-1}=10^{-4}c$. This gives, using (I):

$u_x=\dfrac{2 \cdot 10^{-4}c}{1+10^{-8}}\approx 60 km\cdot s^{-1} - 0.6 mm\cdot s^{-1}$

Then, the deviation with respect to the non-relativistic value ( 60km/s) is negligible for all the practical purposes! This typical velocity, 30km/s, is about the typical velocities in 20th and early 21st century space flight. So, our astronauts can not note/observe relativistic effects. The addition theorem in SR is not practical in current space travel (20th/early 21st century).

Example 2. Case velocities are “close enough” to the speed of light. E.g.: One quarter and one half of the speed of light. In the first case,

$v=u'_x=0.25c$ and then $u_x=\dfrac{0.25c+0.25c}{1+0.0625}=\dfrac{0.5c}{1.0625}\approx 0.47c$

In the seconde case, we write $v=u'_x=0.5c$. This provides

$u_x=\dfrac{0.5c+0.5c}{1+0.25}=\dfrac{c}{1.25}=0.8c$

Thus, we observe that a higher velocities, e.g., those in particle accelerators, some processes in the Universe, etc, the relativistic effects of the non-linear addition of velocities can NOT be neglected. The effect is important and becomes increasingly important when the velocity increases itself ( you can note how large the SR effect is if you compare the 0.25c and 0.5c examples above).

Example 3. Speed of light case. Extreme case: we are trying to exceed the velocity of light. Suppose now, that the train could move with relative velocity equal to c. The object is thrown with relative speed $u'_x=c$ and $u'_y=u'_z=0$. What we do see on the track. Naively, ordinary life would suggest the answer 2c, but we do know that velocities transform non-linearly, so, we plug the values in the formula to get the answer:

$u_x=\dfrac{c+c}{1+1}=c$ and $u_y=u_z=0$. Therefore, if a trian is travelling at the speed of light, and inside the train an object is thrown forward at c, we DO NOT observe/measure a 2c velocity, we observe/measure it has velocity $c$!!!! if we stand at rest on the track. Amazing!

Suppose we try to do it in a “transverse way”. That is, suppose that the velocities are now $v=c$, $u'_x=0$ and the transverse speed is not $u'_y=c$. This case results in the numbers:

$u_x=c$ and $u_y=\dfrac{c}{\gamma}=0$ since $\gamma \rightarrow \infty$ and thus $u=c$.

Therefore, if the train travels at c, an inside of the train an object is launched at right angles to the direction of motion at the velocity c, the object itself is measured/seen to have velocity c measured from a rest observer placed beside the tracks. Amazing, surprise again!

Example 4. Case: Superluminal relative motion. Suppose, somehow, the relative motion between the two frames provides $v=2c$ (even you can plug $v=nc$ with $n>1$ if you wish). Suppose the object is measured to have the extremal limit speed $u'_x=c$ (imagine we consider a light beam/flash, for instance). Again, using the addition law we would get:

$u_x=\dfrac{3c}{1+2}=c$ and $u_x=\dfrac{n+1}{1+n}c=c$

Even if the inertial frames move at superluminal velocities relative to each other, a light beam would remain c in the S-frame if SR holds! Surprise, again!

In this way, we can conclude one of the most important conclusions of special relativity ( something that it is ignored by many Sci-fi writers, and that we would like to be able to overcome somehow if we have to master the interstellar travel/interstellar communications as Sci-fi fans, or as an interstellar civilization, you should get some trick to avoid/”live with” it.):

$\boxed{\mbox{The speed of light can never be exceeded by adding velocities in SR.}}$

If SR holds, the velocity (or speed) of light is the maximum speed attainable in the Universe. You can like it or hate it, but if SR is true, you can not avoid this conclusion.

There is another special case of motion important in practical applications: two dimensional motion. I mean, imagine that in the S’-frame, an object has the velocity $\mathbf{u}'=(u'_x,u'_y,0)$. The velocity subtends an angle $\theta '$ with the x’-axis. See the figure below:

What is the angle $\theta$ in the frame that we observe between $\mathbf{u}$ and the x-axis? For the S-frame we find:

$\tan \theta= \dfrac{u_y}{u_x}=\dfrac{\dfrac{u'_y}{\gamma \left( 1+\dfrac{u'_x v}{c^2}\right)}}{\dfrac{u'_x+v}{1+\dfrac{u'_x v}{c^2}}}$

and after some easy algebraic manipulations we get the important result

$\boxed{\tan \theta =\dfrac{1}{\gamma}\dfrac{u'_y}{u'_x+v}}$

We observe that, according to this last equation, with teh exception of $v=0$, $\tan \theta$ is smaller in the S frame than $\tan \theta '=\dfrac{u'_y}{u'_x}$ in the S’-frame. In the non-relativist limit, we recover the result that our ordinary intuition and experience provides ($\gamma \rightarrow 1$):

$\theta_{nonrel}=\dfrac{u'_y}{u'_x+v}$

It is logical. In the nonrelativistic limit we do know that $u'_y=u_y$ and $u_x=u'_x+v$, so the result agrees with our experience in the low velocity realm.