particle accelerator

[Brian W (Brian)]

Hi all, in my naive and simplistic world particle accelerators accelerate particles in opposite directions at very fast speeds until at just the right speed and time they collide and some really really neat stuff happens.

Here's where I need some educating; these particles are moving at close to the speed of light (?)... this means that when they collide they have a closing speed greater than the speed of light(?)... if the maxim that for every action there is an equal and opposite reaction applies in this case does that mean that what comes out of this collision is travelling at a speed greater than the speed of light?

Brian

[sjastro]

Originally Posted by Brian W

Hi all, in my naive and simplistic world particle accelerators accelerate particles in opposite directions at very fast speeds until at just the right speed and time they collide and some really really neat stuff happens.

Here's where I need some educating; these particles are moving at close to the speed of light (?)... this means that when they collide they have a closing speed greater than the speed of light(?)... if the maxim that for every action there is an equal and opposite reaction applies in this case does that mean that what comes out of this collision is travelling at a speed greater than the speed of light?

Brian

Brian,

You may have heard of the Lorentz transformations for time and distance in Special Relativity. These can be used to derive how velocities add up in Special Relativity.

They add up according to the equation:
u=(u'+v)/(1+(u'v/c^2))

u is the velocity relative to a centre of mass frame.
u' is the velocity of a particle.
v is the velocity of the frame of reference of the particle moving in the opposite direction.

Suppose the particles are moving towards each other at 0.99c.
The inituitive calculation is that the speed in the centre of mass frame should be u=0.99c+0.99c= 1.98c.

However it is found to be u=(0.99c+0.99c)/(1+(0.99c*0.99c)/c^2))
= 0.999949497c

Regards

Steven

[Robh (Rob)]

Thanks Brian for that interesting question and Steven for that illuminating answer.

I note that if u' and v are much less than c then
then the equation reduces pretty much to u = u' + v.

Regards, Rob

[snas (Stuart)]

Thanks for the comments Steve and Rob. Just wondering, at what speed does the equation change from the intuitive equation u = u' + v to the special relativity equation u=(u'+v)/(1+(u'v/c^2))

Stuart

[sjastro]

Originally Posted by snas

Thanks for the comments Steve and Rob. Just wondering, at what speed does the equation change from the intuitive equation u = u' + v to the special relativity equation u=(u'+v)/(1+(u'v/c^2))

Stuart

Hello Stuart,

As Rob points out it is based on the u'v/c^2 term. It depends on how accurate you want to be.

If the 2 particles approach each other at say 100km/hr or 0.28 km/sec, uv'/c^2 becomes (0.28*0.28)/300000^2 = 0.000000000000871. For all intents and purposes the uv'/c^2 term is zero and we use u=u'+v.

Even for particles travelling towards each other at 0.1c or 30000 km/sec, the uv'/c^2 term is small.

(0.1c*0.1c)/c^2 = 0.01.
u=u'+v is still a good approximation.

u=u'+v
= 0.2c

u=(u'+v)/(1+(u'v/c^2))
=0.2c/(1.01)
=0.198c

Regards

Steven