# Particles collisions at speeds greater than c :surprised

1. Jun 14, 2007

### rfwebster

i recently read that it was possible to acelerate an electron, using a synchrotron to 3/4 c !
Does this imply that if two electrn beams were to be accelerated in this way that the resultant velocity of a collision would be 3/2 c ?
If so what would this b like, i.e. there be lots of mass created from the energy of the beam or would there be a geet massive explosion?

2. Jun 14, 2007

### Meir Achuz

No. Special Relativity shows that the relative velocity between two particles can never exceed c. The law of addition of parallel velocities in SR is
$$u_{rel}=\frac{u_1+u_2}{1+u_1u_2/c^2}$$.

3. Jun 14, 2007

### Staff: Mentor

No. You must compute the relative velocity relativistically:

$$V_{a/c} = \frac{V_{a/b} + V_{b/c}}{1 + (V_{a/b} V_{b/c})/c^2}$$

Which gives you a relative speed of about 24/25 c.

(Oops... Meir beat me to it!)

4. Jun 14, 2007

### jambaugh

This question should go in the relativity section.
The answer is no. Relativistic velocities don't add that way in SR.
I know it is counter intuitive but remember when you say "velocity of a collision" you are asking how fast one particle appears in the other's frame of reference. Think of the velocity as a direction in space-time, i.e. the speed is slope of a line instead of as vector. In Euclidean space it will be the angles which add. In Minkowski space-time it will be pseudo-angles which add.

$$v_1 = \tanh(\phi_1),\quad v_2 = \tanh(\phi_2),\quad v_{12}=\tanh(\phi_1+\phi_2)$$
(velocities expressed as multiples of c.)
So use your calculator to figure:
$$v_{12} = \tanh\left( \tanh^{-1}(v_1) + \tanh^{-1}(v_2) \right)$$

Try this with various velocities and you'll get some idea of how velocity addition behaves. In particular your example of v1=v2=3/4c gives:

$$\tanh^{-1}(0.75)=0.97295507452765665255267637172159$$
$$\tanh(2\times 0.97295507452765665255267637172159)=0.96$$
so their relativistic speed, each w.r.t. the other is 96% of c.

However if you want to consider momentum and energy instead of velocity then Yes their momenta will add up (you can add them in our frame the center of mass frame) and you can get arbitrarily large collision energies in principle. To get really interesting effects you use a beam of electrons hitting a beam of positrons so that their rest energies also go into the mix (and so the net charge, lepton number, and other conserved quantities cancel).

See: http://en.wikipedia.org/wiki/Large_Electron_Positron" [Broken]

Regards,
James Baugh

Last edited by a moderator: May 2, 2017
5. Jun 14, 2007

### rfwebster

thanks all, i think i need to read up alot orelativity in all forms

6. Jun 14, 2007

### MeJennifer

It becomes interesting when we add velocities that are not parallel. In such cases the velocity addition formula is neither associative nor commutative.

Velocity addition together with Thomas precession form a group, sometimes called a gyrogroup.

7. Jun 14, 2007