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{..everything that follows is in the domain of relativistic kinematics..}

Say a particle A collides with a particle B at rest and produces particles C and D.

What exactly is the definition of "threshold energy" for a reaction ?

Is it the energy that A should have so that so that the heavier of C and D is produced at rest?

Given the masses of all the particles, how does one calculate the threshold energy that A must have to cause this reaction ?

I intuitively feel that a threshold energy scenario would mean that C and D are moving along the same direction in which A was coming in. I think it is an unnecessary "waste" of energy for C and D to develop momentum in the two transverse directions of A's motion. But I can't prove this in general.

I would like to know how this situation is analyzed.

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2 Answers 2

Luboos has already explained that even at threshold the products are produced at rest only in the Center of Momentum frame of the initial system, but this is a nice problem to demonstrate the power of calculating with relativistic invariants.

To find the threshold is a naive way, we could calculate the velocity of the center of momentum frame as a function of the beam energy, then boost both beam and target into that frame and calculate the total energy. It is not a particularly difficult task.

But we can do better than that.

The total four-momentum vector of the products at threshold, in the CoM from is

$$ \mathbf{P} = \mathbf{p}_C + \mathbf{p}_D = (m_C,\vec{0}) + (m_D,\vec{0}) $$

and squaring that we obtain the (relativisticly invariant) mass of the system at threshold:

$$ M_f^2 = (m_C + m_D, \vec{0})^2 = m_C^2 + 2 m_C m_D + m_D^2 $$

We can calculate the same quantity for the precursors in the lab frame of reference:

$$ \mathbf{p}_A = (E_A,\vec{p}_A) \approx (p_A, \vec{p}_A) $$ $$ \mathbf{p}_B = (E_B,\vec{p}_B) = (m_B, \vec{0}) $$

where I have assumed that the beam is fully relativistic (hey, I spent a lot of time at an electron accelerator facility). This makes the mass squared of the initial system

$$ M_i^2 = (p_A + m_B, \vec{p}_A)^2 = (p_A^2 + 2 p_A m_B + m_B^2) - p_A^2 $$

which we simplify and set equal to the mass squared of the final products to find the beam momentum needed for threshold production:

$$ p_{A,\text{threshold}} = \frac{ (m_C + m_D)^2 - m_B^2}{2 m_B} $$

One half of this calculation was done in the CoM frame and the other half in the lab frame, but because I only compared Lorentz invariant quantities I was able to combine them with impunity.

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Hi, why is it that we can calculate mass invariance? I thought the formula was (E/c,p) for energy momentum 4 vector, if E is rest mass then wouldn't it be mc? – inya Nov 24 at 0:08
You don't care what frame you used when computing invariant quantities because they are invariant (not dependent on the frame), and I have used $c=1$ units. Feel free to re-inert the $c$s in a consistent way; if you go with the form you've just used than my masses should be multiplied by $c$ ( $m \implies mc$) and my energies divided by $c$ ($E \implies E/c$) and my momenta left alone. This procedure won't reveal any new physics, but will make the equations suitable for use with SI units. Note that the final result won't change beyond a overall factor of $c$ on the RHS. – dmckee Nov 24 at 2:21

Dear Anirbit, the threshold energy is the minimum energy for which the production of a particular collection of particles in the final state may occur.

It is easy to see that this minimum energy occurs when all the particles - if there are many - of the final state are moving with the same velocity, i.e. if they have the same values of $\vec p / E$. Equivalently, there exists an inertial system in which all the final particles are rest.

However, as you correctly pointed out, it doesn't mean that this inertial system where the final particles are at rest has to agree with a particular inertial frame of the initial state. If one considers head-on collisions of two equally heavy particles going in the opposite direction, then it's true: the threshold energy means that the final particles are produced at rest.

However, if one considers e.g. collisions with a fixed target - a moving proton with a $\vec p=0$ proton - it is not the case. Instead, the final velocity will be going in the same direction as the initial moving proton. That's dictated by the momentum conservation. See Wikipedia for a particular calculation of the threshold energy.

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I was looking precisely for a "proof" of this idea that you said - that at "threshold" all the final particles are at rest in the COM frame. Can you kindly help prove that? – user6818 May 31 '11 at 18:03
@Anirbit: If there were moving they would have non-zero kinetic energy, but threshold is defined as the point where there is only enough energy to generate the particles with none left over. – dmckee May 31 '11 at 22:43

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