Collid'em

One of the reactions you may expect on a hadron accelerator is

p + p p + p + p + p . p + p \; \rightarrow \; p + p + p + \overline{p}.

(Check it for yourself by the conservation of baryon, electron and muon numbers, strangeness, charge and anything else you can think of. Just remember that upon collision, the resulting rest mass may be greater than the rest mass before the collision [why?]). Suppose one of the protons (in the lab rest frame) collides with the second proton which was standing still. The lower bound to the energy of the incident proton for this reaction to happen is

E = N m . E = Nm.

Here we have taken c c to be 1 1 , m m the mass of the proton and N N is an integer number. What is N N ?


The answer is 7.

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1 solution

In the lab frame (taking c = 1 c = 1 ), the four-momentum vector L μ L^\mu may be written as L μ = ( E + m , p ) L^\mu = (E+m, \overrightarrow{p}) , where p \overrightarrow{p} is the momentum of the moving proton. If we contract this four-vector we get

L μ L μ = ( E + m ) 2 p 2 = E 2 p 2 + 2 E m + m 2 . L^\mu L_\mu = (E+m)^2 - \left |\overrightarrow{p} \right |^2 = E^2 - \left |\overrightarrow{p} \right |^2 + 2Em + m^2.

But according to Einstein's energy equation, E 2 p 2 = m 2 E^2 - \left |\overrightarrow{p} \right |^2 = m^2 . Therefore one may rewrite the above contraction as

L μ L μ = 2 m 2 + 2 E m . L^\mu L_\mu = 2m^2 + 2Em.

Now, there exists a frame (called center of momentum frame) where the total momentum of the system is null. The existence of such frame is easy to prove in this case (which I encourage you to do), but it also exists for any other system. In this case, the four-momentum vector in the center of momentum frame M μ M^\mu after the collision (one may evaluate the four-momentum before or after the collision and it must be the same) is M μ = ( 4 m , 0 ) M^\mu = (4m, \overrightarrow{0}) since there are 4 4 particles with mass m m . As the contraction of the four-momentum vector is a scalar, it must be the same, no matter what frame we're using:

L μ L μ = M μ M μ 2 m 2 + 2 E m = 16 m 2 L^\mu L_\mu = M^\mu M_\mu \;\;\Rightarrow \;\; 2m^2 + 2Em = 16m^2

E = 7 m . \therefore E = 7m.

I spent 22 years working at Fermi National Accelerator Laboratory (a.k.a., Fermilab or FNAL) as a computer programmer with most of that time at the CDF experiment and this is the clearest exposition of this topic that I have seen. Thank you.

You might want to expand the explanation of the derivation of the E 2 p 2 = m 2 E^2-\left|\vec{p}\right|^2=m^2 equation, for the benefit of others. That part, I did understand..

See also Review of formulas for relativistic motion Example 4.

A Former Brilliant Member - 2 years, 3 months ago

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