# The speed of light, objects moving against each other and the effects of a clash

Physics is a hobby interest of mine, so I may be asking a senseless question, but there is something I cannot get my head around when it comes to relative frames, the constant speed of light (as a top-limit for any speed) and the following situation:

Imagine two spaceships, capable of reaching speed close to $c$, flying in opposite directions and passing each other by. According to the special relativity, we will calculate the speeds this way: $$V_{\rm total} = \frac{u+v}{1+\frac{uv}{c^2}}$$ where ${u}$ and $v$ would be the speeds of each spaceship, both close to $c$.

According to this, I perfectly understand that each ship will "see" the other moving with near the speed of light, not faster. An observer in a separate reference frame will see each ship moving with a speed near that of the light too, and it is OK.

Now, let's assume the ships crash. Would the effect of the crash be the same as if only one of them was moving, as opposed to both speeding against each other? I am thinking there would not be much difference, as if the first ship is stationary, it will still experience the other crashing at him with the speed near that of the light. The external observer would also confirm this.

If this is so, would there be any reason to accelerate particles near the speed of light in opposite directions as it is being done in CERN? I assume (may be wrongly) that the above observation for the spaceships can be applied to particles also. So, would it be enough if the particles are just accelerated in one direction to "bomb" a stationary object, in order for them to get scattered? Am I missing something that would apply to particles but not to larger objects (like ships) or even something more-fundamental?

In the center of mass frame, if two particle beams have the same energy, using energy-momentum 4-vectors we get:

$s =(P_1 + P_2)\cdot(P_1 + P_2) = (E + E, \vec{p}-\vec{p})\cdot(E + E, \vec{p}-\vec{p}) = (2E, 0)\cdot(2E,0) = 4E^2$

Therefore the $E_{CM} = \sqrt{s} = 2E$

For a fixed target ($E_b$ = Energy of the beam and $m_t$ = mass of the target):

$s =(P_1 + P_2)\cdot(P_1 + P_2) = (E_b + m_t, \vec{p_b})\cdot(E_b + m_t, \vec{p_b}) = E_b^2 + m_t^2 + 2E_bm_t - p_b^2 = m_b^2 + m_t^2 + 2E_bm_t$

Therefore, if the masses are negligible compared to the beam energy:

$E_{CM} = \sqrt{s} = \sqrt{2E_bm_t}$

We can see that for a fixed target the energy in the center of a mass frame is significantly lower than the case with two moving targets. The same applies to the spaceship in your particular situation.

• Great +1! I like the mathematical approach, it explains it. I am obviously missing the effects on the amount of energy, which will be released in the clash. Thanks. – Ivaylo Slavov Apr 11 '14 at 11:07

I'm not particularly confident with experimental Physics, nevertheless I will try to answer to your interesting question.

Not every scattering experiment in Particle Physics needs the acceleration of particles in opposite directions, there are a lot of experiment (for example Rutherford scattering) in which a fixed target is used.

However in doing so the all energy available for the scattering phenomenon is due to the moving particles.

Loosely speaking if we are able to accelerate particles only up to the velocity $v_{max}$ the maximum energy will be that corresponding to the value $v_{max}$ of the velocity, while if we accelerate particles in opposite directions up to $v_{max}$, the maximum energy at our disposal will be that corresponding to the velocity $V_{max}$ calculated composing the opposite velocities of the particles.

Thus accelerating particles in opposite directions enhances the energy at our disposal.

• Nice to learn there are experiments with static particles. Somewhat I used to think the only way used to scatter a particle was to crash accelerated particles against each other. +1! – Ivaylo Slavov Apr 11 '14 at 11:06