# Using Lorentz transformations

The question states that two particles in a high-energy accelerator experiment approach each other head-on with a relative speed of $0.890 \, c$. Both particles travel at the same speed on a straight line as measured and seen in the laboratory.

I am supposed to find the speed of each particle, as measured in the laboratory.

I have the derived Lorentz Transformations for proper and relative velocities respectively;

\begin{align} v'_x &= \frac{v_x - u}{1 - (v_x u / c^2) } \\ v_x &= \frac{v'_x+u}{1 + (v'_x u / c^2)} \end{align}

where $u$ is the speed of the moving object measured in the rest frame of the observer, and $c$ is the speed of light.

Since I am only given the numerical value of $u$, and I am looking for proper velocity $v'_x$, will I have to substitute the equation $v_x$ into the equation for $v'_x$ to solve for $v'_x$, or is there an easier way to solve the problem?

• I'm having trouble understanding the meaning of $89%c$. Is it the closing speed of two particles or the speed of one particle from a reference frame of the other? – Gonenc Aug 12 '15 at 21:40
• I assume that $0.890c$ is the relative speed as measured in the frame of either particle. Otherwise the answer is sort of trivial. @mnmakrets can you confirm this? – gleedadswell Aug 12 '15 at 22:11
• Yes that's what I thought so as well @gleedadswell – mnmakrets Aug 12 '15 at 22:31

There is a much easier way to solve the problem. Quick derivation: if something is moving past you at speed $u = \alpha~c$ then in a reference frame travelling at the speed $\beta ~c$ in that direction, the Lorentz boost puts its trajectory as:$$\gamma\begin{bmatrix}1&-\beta\\-\beta&1\end{bmatrix}\begin{bmatrix}c t\\ut\end{bmatrix} = \gamma \begin{bmatrix}c t - \beta \,ut\\ut - \beta\,ct\end{bmatrix}$$and substituting in you find a speed $c~\frac{\alpha - \beta}{1 - \alpha \beta}.$ But suppose we define the hyperbolic sine and cosine as $\sinh x = (e^x - e^{-x})/2,$ $\cosh x = (e^x + e^{-x})/2.$ Then we find that the products of these various things are\begin{align} \sinh a\,\sinh b =& \frac 14 \left(e^{a + b} - e^{a - b} - e^{-a + b} + e^{-a - b} \right),\\ \sinh a\,\cosh b =& \frac 14 \left(e^{a + b} + e^{a - b} - e^{-a + b} - e^{-a - b} \right),\\ \cosh a\,\sinh b =& \frac 14 \left(e^{a + b} - e^{a - b} + e^{-a + b} - e^{-a - b} \right),\text{ and}\\ \cosh a\,\cosh b =& \frac 14 \left(e^{a + b} + e^{a - b} + e^{-a + b} + e^{-a - b} \right). \end{align}This yields something very much like the trigonometric "angle sum rules" when you try to cancel out the $e^{\pm a \mp b}$ terms, \begin{align}\sinh(a + b) =& \sinh a\,\cosh b + \cosh a\,\sinh b \\ \cosh(a + b) =& \cosh a\,\cosh b + \sinh a\,\sinh b \end{align}For $a - b$, simply replace both $+$ signs with $-$ signs on the right hand side, as $\sinh$ is odd and $\cosh$ is even.

Now rewrite each of the above expressions as $\alpha = \sinh a / \cosh a,\; \beta = \sinh b / \cosh b$. It turns out that this function $\tanh x$ is totally invertible, so you can always do this. Then multiply both top and bottom of the fraction by $\cosh a \, \cosh b$ to find: $$c \frac{\alpha - \beta}{1 - \alpha\beta} = c~\frac{\sinh a\,\cosh b - \cosh a\,\sinh b}{\cosh a\,\cosh b - \sinh a\,\sinh b}= c~\frac{\sinh(a - b)}{\cosh(a - b)} = c \tanh(a - b)$$So when you see something moving at a speed $u = c~\tanh a$ and you boost into a reference frame with speed $c~\tanh b$ relative to where you started, you find that the new velocity is just $u' = c~\tanh(b - a).$

These numbers $w = \tanh^{-1}(u/c)$ are usually called rapidities. In special relativity, rapidities are linear. So for example, if you want to know what "constant acceleration" looks like in the relativistic limit, it is $v = c~\tanh(a t / c).$

So, find the rapidity, divide it by 2, boost into the frame with that rapidity, and you should find that both of these particles are coming towards you with the same rapidity.

Assuming that $0.890c$ is the approach speed of one particle in the frame of reference of the other, here's my suggestion.

First of all, you've misinterpreted variables. If the convention of the symbols is typical then $u$ is not the numerical value you have been given. $u$ refers to the speed of one frame of reference relative to another frame of reference. The $0.890c$ is either $v_x$ or $v_x'$, depending on how you decide to define your frames of reference.

Let's arbitrarily call the lab frame S and the frame of particle 1 S'. So we know the following:

$u = v_1$ (as usual $u$ refers to the speed of S' relative to S and we have attached S' to particle 1)

$v_2' = 0.890c$

There is at least one more crucial thing that you know. Writing it down may clarify things for you. You'll then need to substitute all of these into one of your Lorentz transformation equations and re-solve for the variable you are looking for. [Hint: it will be much easier to solve if you work in terms of $\beta = v_x/c$]

\begin{equation} v_x = \frac{v'_x+u}{1 + (v'_x u / c^2)} \tag{01} \end{equation}

Two Particles : Particle $\rm{P}$ and $\rm{P}^{\prime}$.

$\:v_x\:$ : velocity of $\rm{P}$ in laboratory (real number, not the non-negative norm).

$\:v_x^{\prime}\:$ : velocity of $\rm{P}^{\prime}$ in laboratory (real number, not the non-negative norm).

$\:u =\alpha\cdot c, (\alpha=\pm0.89) \:$ : velocity of $\rm{P}$ as seen by $\rm{P}^{\prime}$. By far the most convenient way for calculations is using four velocities. In lab reference frame components of four velocities are:

$U_1=(\gamma,\gamma v_{lab},0,0)$

$U_2=(\gamma,-\gamma v_{lab},0,0)$

Where $v_{lab}$ is speed of particles as measured in lab, $\gamma=(1-v_{lab})^{-1/2}$. Now in rest frame of first particle those 4 velocities have components:

$U_1=(1,0,0,0)$

$U_2=(\gamma_2,-\gamma _2 v_{rel},0,0)$

Where $v_{rel}$ is relative velocity of particles and $\gamma_2$ is defined in an obvious way.

Now you can calculate scalar product if those four velocities in both frames. You know that results will be equal, and resulting equation can be easily solved.

As a rule of thumb: performing Lorentz transformations is almost always extremely cumbersome. The most convenient way is always to look for some invariants.