Is the speed of one beam in the LHC nearly twice that of the other?

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Hi all,
I've learned that the LHC has just started. There are two beams of protons which travel clockwise and anti-clockwise at nearly the speed of light. Then the two beams are to colide into each other. My question is whether the speed of one beam compared to the other is at nearly 2 c ?
 
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Yes the closing speed is 2c, if you were traveling with one bunch of particles you would measure the other bunch approaching you at c (although what you would measure this with - I'm not sure)

This doesn't violate relativity in any way, no THING is traveling at 2c, just the rate of decrease of the distance between them.
 
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The speed of one beam wrt other is 'c'. NOT 2c
 
astromandi said:
The speed of one beam wrt other is 'c'. NOT 2c

Just before collision, the coordinate distance in the frame of the Earth between bunches of particles traveling in opposite directions decreases at the rate of (nearly) 2c. This is what mgb_phys meant by "closing speed."
 
George Jones said:
Just before collision, the coordinate distance in the frame of the Earth between bunches of particles traveling in opposite directions decreases at the rate of (nearly) 2c. This is what mgb_phys meant by "closing speed."

So if one man stood on one beam, he would see the second moving at 2c ?. In other way, one beam would travel at 2c if the other beam was considered the frame?
 
If the speed of one beam is u and the speed of the other v (both relative to the ground), then the speed of one of the beams relative to the other is

\frac{u+v}{1+uv/c^2}

For example, if u=v=0.99c (exactly), then the result is (0.99+0.99)c/(1+0.99*0.99)=0.99994949750012... c.
 
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