Particles collisions at speeds greater than c :surprised

[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?

 

[Meir Achuz]

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 ?

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
$$u_{rel}=\frac{u_1+u_2}{1+u_1u_2/c^2}$$.

 

[Doc Al]

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 ?

No. You must compute the relative velocity relativistically:

Relativistic addition of velocities:
$$V_{a/c} = \frac{V_{a/b} + V_{b/c}}{1 + (V_{a/b} V_{b/c})/c^2}$$

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

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

 

[jambaugh]

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?

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"

Regards,
James Baugh

 

[rfwebster]

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

 

[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.

 

[robphy]

Concerning "velocity addition"...
It's probably best to drop the word "addition" and instead use "composition".
(For relativistic pedagogical purposes, it might be good to somehow begin this refinement at the introductory level... long before the topic of special relativity is discussed.)

 

[neutrino]

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 ?

Just about a couple of hours ago, I was reading this very example in an SR textbook.

As observed from the laboratory, two particles move in opposite directions along the x axis, each with speed 3c/4. Call their coordinates x1, x2 so that the distance between them is
s $\equiv$ x1-x2. The rate of change of s is called the closing speed (as observed in the laboratory). Evidently |ds/dt| = 3c/2. No paradox is involved, because s is not the distance between the particles in the rest frame of either. By contrast, the speed of one of the particles measured by an observer sitting on the other certainly is bounded by c