# I Relativistic relative velocity

1. Mar 17, 2017

### davidge

I'm interested in deriving the relativistic relative velocity of two particles moving near the speed of light. It turns out to be (with $c = 1$) $$\frac{V^{(1)} - V^{(2)}}{1 - V^{(1)}V^{(2)}}$$

How should I approach to this problem? Maybe I should not think of particles at all, but instead simply to consider the two reference frames as two set of coordinate values? So

Frame 1: $(x^0,x^1,...)$; Frame 2: $(y^0,y^1,...)$.
Then the velocities are $\partial x^i / \partial t$ and $\partial y^i / \partial t'$.
$t \equiv x^0$, $t' \equiv y^0$

Now, $dy^\alpha = \Lambda^\alpha{}_{\beta}dx^\beta$. The solution for the $\Lambda$ are easy for the case when the displacement $dx^i$ are null in one of the frames, but this does not help in solving for the situation in question.

Last edited: Mar 17, 2017
2. Mar 17, 2017

### robphy

First, check that expression. Is the square root there?

One could derive it from a Lorentz Transformation, probably with the aid of a spacetime diagram.

Using rapidity is quicker and more geometrical.

One can also use vector-algebraic methods with dot products.

3. Mar 17, 2017

### davidge

Could you give me a hint ?

Oh, there's no sqrt in the expression. Thanks.

4. Mar 17, 2017

### robphy

One way to start off is like this:
$\hat t_{1}=\cosh(\theta_1-\theta_0) \hat t_0 + \sinh(\theta_1-\theta_0)\hat t_{0,\perp}=\cosh(\theta_1-\theta_0)\left( \hat t_0 + \tanh(\theta_1-\theta_0)\hat t_{0,\perp}\right)$
$\hat t_{2}=\cosh(\theta_2-\theta_0) \hat t_0 + \sinh(\theta_2-\theta_0)\hat t_{0,\perp}$
then what you seek is in here
$\hat t_{2}=\cosh(\theta_2-\theta_1) \hat t_1 + \sinh(\theta_2-\theta_1)\hat t_{1,\perp}$
..to be expressed in terms of the preceding.
[You could simplify the notation by writing $\theta_{21}\equiv\theta_{2}-\theta_{1}$].

One might not use the trig functions from the start... it can be deduced later, if desired.
So, $\hat t_1=\gamma_{10}(\hat t_0 + \beta_{10} \hat t_{0,\perp})$.... but that already gives a hint.

$\hat t_1=A\hat t_0 + B\hat t_{0,\perp}$ and deduce that $\hat t_1=A(\hat t_0 + V\hat t_{0,\perp})$ is a useful form when interpreted physically.

5. Mar 17, 2017

### vanhees71

This formula is only true for particles moving along the same direction in the given reference frame. The relative velocity in general is defined as the velocity of one particle in the restframe of the other particle (of course the latter must be massive if the definition should make sense). For details, see p. 16 of

http://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf

6. Mar 17, 2017

### davidge

In this case, would we end up with $\Lambda^0{}_{0} = \gamma$, $\Lambda^i{}_{0} = v^i \gamma$? (Where $v^i$ denotes the i-th component of the velocity)

7. Mar 17, 2017

### SiennaTheGr8

The method here is relatively straightforward, but you need to use notation in a way that won't confuse you. Subscripts can help.

Say we have two inertial observers in "standard configuration," so that the "primed" observer is moving in the "unprimed" observer's positive x-direction. Then the Lorentz transformations are:

$\mathrm{d}x^{\prime} = \dfrac{\mathrm{d}x - \beta_{rel} \, c \, \mathrm{d}t}{\sqrt{1 - \beta_{rel}^2}} \qquad \mathrm{d}x = \dfrac{\mathrm{d}x^{\prime} + \beta_{rel} \, c \, \mathrm{d}t^{\prime}}{\sqrt{1 - \beta_{rel}^2}} \\ \mathrm{d}y^{\prime} = \mathrm{d}y \\ \mathrm{d}z^{\prime} = \mathrm{d}z \\ \\ c\, \mathrm{d}t^{\prime} = \dfrac{c \, \mathrm{d}t - \beta_{rel} \, \mathrm{d}x}{\sqrt{1 - \beta_{rel}^2}} \qquad c\, \mathrm{d}t = \dfrac{c \, \mathrm{d}t^{\prime} + \beta_{rel} \, \mathrm{d}x^{\prime}}{\sqrt{1 - \beta_{rel}^2}}$
where $\beta_{rel}$ is the (normalized) relative speed of the observers ($v_{rel}/c$). Note that $\beta_{rel} < 1$ (because observers cannot travel at the speed of light).

Now say there's something else that's moving with velocity $\vec{\beta} = (\beta_x, \beta_y, \beta_z)$ according to the "unprimed" observer, and $\vec{\beta} \, ^{\prime} = (\beta_x^{\prime}, \beta_y^{\prime}, \beta_z^{\prime})$ according to the "primed" observer (no "rel" subscript now). It could even be something traveling at the speed of light.

Our goal is to find the transformation formulas for these primed and unprimed velocity components. Specifically, you're looking for the transformation of the $\beta_x$-component, since the relative motion of our observers is along their coincident $x$-axes.

Well, we just use the fact that $\beta_x = \frac{\mathrm{d}x}{c \, \mathrm{d}t}$, etc.:

$\beta_x^{\prime} = \dfrac{\mathrm{d}x^\prime}{c \, \mathrm{d}t^\prime} = \dfrac{\frac{\mathrm{d}x - \beta_{rel} \, c \, \mathrm{d}t}{\sqrt{1 - \beta_{rel}^2}}}{\frac{c \, \mathrm{d}t - \beta_{rel} \, \mathrm{d}x}{\sqrt{1 - \beta_{rel}^2}}} = \dfrac{\mathrm{d}x - \beta_{rel} \, c \, \mathrm{d}t}{c \, \mathrm{d}t - \beta_{rel} \, \mathrm{d}x}$

Divide top and bottom by $c \, \mathrm{d}t$, and we have:

$\beta_x^{\prime} = \dfrac{\beta_x - \beta_{rel}}{1 - \beta_x \beta_{rel}}$.

The same procedure generates the inverse transformation:

$\beta_x = \dfrac{\beta_x^\prime + \beta_{rel}}{1 + \beta_x^\prime \beta_{rel}}$.

Since these are vector components, they can be positive or negative. They are not speeds! The only speed here is $\beta_{rel}$.

If you follow the same procedure for the primed and unprimed $y$- (or $z$-) components, you'll find that $\beta_y^\prime \neq \beta_y$. So even the velocity-components perpendicular to the observers' axis of relative motion get transformed. That's not so in Galilean relativity!

In the special case that $\beta_y = \beta_y^\prime = \beta_z = \beta_z^\prime = 0$, the $x$-component is the only relevant one, and we can say:

$\vec{\beta} \, ^{\prime} = \left( \dfrac{\beta_x - \beta_{rel}}{1 - \beta_x \beta_{rel}} \right) \hat{\imath}^\prime \qquad \vec{\beta} = \left( \dfrac{\beta_x^\prime + \beta_{rel}}{1 + \beta_x^\prime \beta_{rel}} \right) \hat{\imath}$,

where $\hat{\imath}$ is the basis vector $(1, 0, 0)$.

That's the familiar formula you were looking for. I kept the vector notation because there is sometimes confusion over what's a speed, what's a velocity, and what's a velocity-component. The value of what's in the parentheses can be positive or negative, since it's really the $x$-component of the velocity vector. Its absolute value is the speed (the magnitude of the velocity). So $\left| \beta_x \right| = \| \vec{\beta} \| = \beta$ and $\left| \beta_x^\prime \right| = \| \vec{\beta} \, ^\prime \| = \beta^\prime$.

Note that if $\beta = 1$ here, then so does $\beta^\prime$ (and vice versa). And always remember that $\beta_{rel} < 1$.

[Edited because I got confused myself a bit at the end there!]

Last edited: Mar 17, 2017
8. Mar 17, 2017

### davidge

Thank you SiennaTheGr8. This was very helpful. I don't understand just why the transformation for the "spatial part", $dx'$, is given by your expression above. Since $dx'\ ^i = \Lambda^i{}_{j}dx^j$, if the $dx^j$ are not equal to zero, I thought we would have a different expression for $dx'$ than the one that you wrote down.

9. Mar 17, 2017

### SiennaTheGr8

You mean at the very top of my post? That's just the standard Lorentz transformation for space and time coordinates (rather, the differentials thereof).

By the way, I did edit my post just now because I'd made an error at the end.

10. Mar 17, 2017

### pervect

Staff Emeritus
First, one needs some basic formula from special relativity. I'd suggest Bondi's K-calculus as one of the easiest approaches. This can be found in Bondi's book, "Relativity and common sense".

The key result from Bondi's analysis is a relationship between the doppler shift factor k, and the velocity v. I'll give the result initially without proof:

$$\beta = \frac{k^2 - 1}{k^2 + 1} \quad k = \sqrt{\frac{1+\beta}{1-\beta}}$$

Here I've given both $\beta=v/c$ as a function of the doppler shift factor k, and k as a function of $\beta$. Wiki has the later formula, in their article on "Relativistic doppler effect" <link>.

If you send out a 1 hz signal, i.e. one flash of light per second, from observer 1, then observer 2 will receive one flash every 2 seconds if k=2. This happens at a velocity v = 3/5c, or $\beta = .6$.

A third observer, moving at the same velocity relative to the second observer as the second has relative to the first, must receive one flash every four seconds. Basically, the doppler k-factors must multiply.

It remains only to do some algebra. If you substitute

$$\beta_1 = \frac{k_1^2 -1}{k_1^2 + 1} \quad \beta_2 = \frac{k_2^2 - 1}{k_2^2 + 1}$$

into
$$\beta_T = \frac{\beta_1 + \beta_2}{1 + \beta_1 \beta_2}$$

you get

$$\beta_T = \frac{(k1\,k2)^2 -1}{ (k1\,k2)^2 +1 }$$

which shows that this formula makes the doppler factor k's multiplicative, which is the correct behavior.

I'm torn as to whether or not to present a proof of the relationship between k and $\beta$. The proof isn't that hard, but I think to properly understand it would take a more detailed description than I have the patience to write. Since Bondi (and other authors) have already done a much better write-up than I could do in a short post, I think it might be best if I defer the details to another source, with possibly a post if this turns out to be hard to track down. If you can't get a hold of Bondi's original book, try googling for "bondi k-calculus". It's not really calculus, by the way - just algebra.

The point I'm hoping to get across is the following basic logic. Velocity can be computed knowing the doppler shift factor k, and vica-versa. The doppler shift k factors must multiply. This gives us the "velocity addition rule".

11. Mar 17, 2017

### robphy

Bondi's method is great for physical motivation.
I never liked the Lorentz-transformation approach involving dividing "differentials".

Underlying Bondi's multiplicative result that Doppler factors multlply $k_{20}=k_{21}k_{10}$
is that since $k=e^\theta$, we have the additivity of rapidities $\theta_{20}=\theta_{21}+\theta_{10}$
so that the relative-rapidity is $\theta_{21}=\theta_{20}-\theta_{10}$ or simply $\theta_{rel}=\theta_{2}-\theta_{1}$.

Since physics prefers velocities (slopes) over rapidities (angles),
this suggests $v_{rel}=\tanh{\theta_{rel}}=\tanh\left(\theta_{2}-\theta_{1}\right) \equiv \frac{\tanh \theta_{2}-\tanh\theta_{1}}{1-\tanh\theta_2\tanh\theta_1}=\frac{v_2-v_1}{1-v_2v_1}$,
where we used a mathematical identity.
Indeed many of the formulas in special relativity are hyperbolic-trigonometric identities written in terms of $\tanh\theta$
(as if you took all of the trig formulas from ordinary geometry and wrote them in terms of the slope $m$... example: write $\cos\theta$ in terms of $\tan\theta$.).
Once you realize that... a lot of the mathematical mystery goes away [if you accept the model of Minkowski spacetime]...
if not, at least you can be guided by your ordinary trigonometric intuition rather than hunting around for how to do algebra with $\gamma$ and $v$.

It might be worth stating that
the Doppler factor $k$ is an eigenvalue of the Lorentz boost... so, it is arguably more important mathematically than the velocity.
And writing a composition of boosts in the eigenbasis [light-cone coordinates] immediately leads to Bondi's multiplicative result.

$k=e^\theta=\cosh\theta+\sinh\theta=\cosh\theta(1+\tanh\theta)=\gamma(1+v)$
$1/k=e^{-\theta}=\cosh\theta-\sinh\theta=\cosh\theta(1-\tanh\theta)=\gamma(1-v)$
By multiplication, $1=\gamma^2(1-v^2)$.
By division, $k^2=e^{2\theta}=\frac{1+v}{1-v}$.

Half-the-difference is $\frac{1}{2}(k-k^{-1})=\sinh\theta=\cosh\theta \tanh\theta=\gamma v$.
Half-the-sum is $\frac{1}{2}(k+k^{-1})=\cosh\theta=\gamma$.
The ratio is $v=\tanh\theta=\frac{k-k^{-1}}{k+k^{-1}}=\frac{k^2-1}{k^2+1}$.

Last edited: Mar 17, 2017
12. Mar 17, 2017

### davidge

Yes. Can you show me the derivation of that expression for $dx'$?

Thanks Pervect and Robphy. I will carefully read your posts and look on the internet for the Bondi's analysis, but I think it's easier (for me) to derive the expressions from the Lorentz transformation method.

13. Mar 17, 2017

### SiennaTheGr8

I wrote:

To be clear, in the aforementioned special case we have:

$\beta^\prime = \left| \beta_x^\prime \right| = \left| \dfrac{\beta_x - \beta_{rel}}{1 - \beta_x \beta_{rel}} \right| = \dfrac{\left| \beta_x - \beta_{rel} \right|}{\left| 1 - \beta_x \beta_{rel} \right|} \qquad \beta = \left| \beta_x \right| = \left| \dfrac{\beta_x^\prime + \beta_{rel}}{1 + \beta_x^\prime \beta_{rel}} \right| = \dfrac{\left| \beta_x^\prime + \beta_{rel} \right|}{\left| 1 + \beta_x^\prime \beta_{rel} \right|}$.

But perhaps it's worth emphasizing that the absolute value of a sum (or difference) of two terms does not generally equal the sum (or difference) of the absolute values of the terms. So we can't just isolate $\left| \beta_x \right|$ and $\left| \beta_x^\prime \right|$ on the right sides of those expressions and sub in $\beta$ and $\beta^\prime$ for them.

In other words, in the special case that $\beta_y = \beta_y^\prime = \beta_z = \beta_z^\prime = 0$, it is still NOT generally true that:

$\beta^\prime = \dfrac{\beta - \beta_{rel}}{1 - \beta \beta_{rel}} \qquad \beta = \dfrac{\beta^\prime + \beta_{rel}}{1 + \beta^\prime \beta_{rel}}$.

These often-seen formulas only hold if $\beta_x^\prime$ isn't negative. Either way, though, $\left| \beta_x \right| = \beta$ and $\left| \beta_x^\prime \right| = \beta^\prime$, and the sign of the $x$-component of course specifies the vector's direction in either frame.

Hope I've got that all right.

14. Mar 17, 2017

### SiennaTheGr8

@davidge, are you already comfortable with the Lorentz transformation for coordinates? Namely:

$x^{\prime} = \dfrac{x - \beta ct}{\sqrt{1 - \beta^2}} \qquad x = \dfrac{x^{\prime} + \beta ct^{\prime}}{\sqrt{1 - \beta^2}} \\ y^{\prime} = y \\ z^{\prime} = z \\ ct^{\prime} = \dfrac{ct - \beta x}{\sqrt{1 - \beta^2}} \qquad ct = \dfrac{ct^{\prime} + \beta x^{\prime}}{\sqrt{1 - \beta^2}}$

If you are, then use those formulas to derive the transformation for $\Delta x = x_2 - x_1$ (etc.). The differentials are just infinitesimal changes in the coordinates, so you can then just swap the deltas for d's.

To save you the trouble: the differential transformation looks identical to the coordinate transformation, except with the d's tossed in.

15. Mar 17, 2017

### robphy

By introducing "speed" instead of just keeping the [signed] x-component $\beta_x$,
it seems your equations are unnecessarily complicated and restrictive.

There is no restriction that you use the relative-"speed"... if you do, it's best to make that explicit in the formula with $|\beta|$.
Using "speed" in the Lorentz transformations is like saying that a rotation can only involve a positive angle.

By using "speed", you end up claiming...
Those often-seen formulas are fine
if you stick with x-components of velocity and x-components of relative-velocity, as is usually done.

16. Mar 17, 2017

### davidge

yea

Well, I know how to derive them only for the case where the spatial displacements $\bf{dx}$ are equal to zero in one of the frames:

Let's suppose a particle is seen stationary in the unprimed frame, but has velocity of magnitude $v'$ in the primed frame. So all $dx^i$ are equal to zero, and we have the transformation equations:
$$dx'^{{\ }i} = \Lambda^i{}_{0}dt \\ dt' = \Lambda^0{}_{0}dt$$
The velocity i-th component is $dx'^{{\ }i} / dt'$. So, dividing the above, we get
$$v'^{{\ }i} \Lambda^0{}_{0} = \Lambda^i{}_{0}$$
$$\Lambda^\alpha{}_{\sigma}\Lambda^\beta{}_{\kappa}\eta_{\alpha \beta} = \eta_{\sigma \kappa}$$
Therefore
$$-1 = -(\Lambda^0{}_{0})^2 + \sum_i (\Lambda^i{}_{0})^2\\$$
The solution is
$$\Lambda^0{}_{0} = \frac{1}{\sqrt{1-v'^2}} \\ \Lambda^i{}_{0} = \frac{v'^{{\ }i}}{\sqrt{1-v'^2}}$$
And we can write the $dx'^{{\ }i}$ as
$$dx'^{{\ }i} = \frac{v'^{{\ }i}}{\sqrt{1-v'^2}}dt$$

We can see that this last expression is different from the one that you wrote for the $dx'$. Probably it's because I've assumed that the particle is at rest in one of the frames. I dont know how to derive it for the case where the particle is moving in both frames, because in that case we would have a long summ in the transformation equations, not just one term as in above.

Last edited: Mar 17, 2017
17. Mar 17, 2017

### SiennaTheGr8

@robphy

Thanks for the response.

I find the full vector equations clearer, more illuminating, and certainly less restrictive (perpendicular components!).

As opposed to a third $x$-component, do you mean?

I may be mistaken, but I believe that there is an implied restriction that $w$ be a "relative speed" in the often-seen $v = (u - w) / (1 - uw/c^2)$. The inverse equation is $u = (v + w) / (1 + vw/c^2)$, so: if "$v$ guy" subtracts $w$, and "$u$ guy" adds it, doesn't it have to be an unsigned magnitude? Additionally, we have the requirement that $w \neq c$, but no such requirement of $v$ and $u$.

As for being explicit about speed vs. velocity vs. velocity-components, I agree, and that's actually why I prefer the full vector equations!

But what else could the $v$ (or $\beta$) in the Lorentz transformations be, other than relative speed?

"As is usually done"—I don't think that most presentations are particularly clear about what's a speed, what's a velocity, and what's a velocity-component. That's one reason why I prefer the full vectorial approach.

18. Mar 17, 2017

### SiennaTheGr8

19. Mar 17, 2017

### davidge

The problem with that article is that they use a different approach in their derivation. I would like to see one that uses the same approach and notation that I used in my previous post.

20. Mar 17, 2017

### robphy

From a text search on that page, all references to "speed" are for "speed of light"... not to any particle or frame.
On the other hand, "velocity" is used throughout.

agreed... but one doesn't have to overcomplicate it.
A good consistent notation might suffice, as in the wikipedia article you suggested.

relative-velocity, where $v$ can have any sign (or $v_x$ to be explicit).
If you want speed, then you probably should be explicit with $|\vec v|$.
Can $x$ and $t$ take arbitrary signs?

if you write down a rotation transformation in euclidean space,
is the transformation parameter "the angle" or "the magnitude of the angle"?

If you write the Lorentz Transformation as a matrix,
you can compose transformations and see the group properties.
With magnitudes, I think you lose that ability.

For rotations in the plane, if i compose a rotation by 30-degrees and then by (-30)-degrees, I should get the identity.
In "slope form" (where the parameter is the slope $m=\tan\theta$ instead of angle), these correspond to $m_A=\tan 30$ and $m_B=-\tan(30)$.
If I use magnitudes, I don't think you get that... unless you say, "for clockwise rotations, you have to use this other form of the transformations".
The same argument applies to the Lorentz Transformations.

Implicit in this discussion is that their axes have the same handedness... in a boost, we're not looking at each other in a mirror of some kind.
We agree on the general direction of "large x".
Technically speaking, we agree that the positive x-axis (in spacetime) of one observer projects onto the positive x-axis of the other. (As vectors in spacetime [in a spacetime diagram], their x-axes are not parallel.)