- #1

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*invariant*. So, how can you have a spectrum of rest mass energies? The rest mass of the pi plus and pi minus are well known to be 139.6 Mev/ c

^{2}for both, so how can you have a range of their rest mass energies?

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- #4

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Based on the next part of your post, you don’t.Okay, I understand what the invariant mass of the system is

It is not the sum of the invariant masses of the pions. It is the invariant mass of the system of pions (in this case the system consists of three pions).but if it is the sum of the invariant masses of the (two?) pions that is plotted, why is there a continuous range?

- #6

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E

which as I understand it, is a scalar invariant - which has the

So, how can you have a continuous range of energies on this plot, unless its (rest mass + kinetic ) energies. That I could understand. The E

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$$

(E_1 + E_2)^2 - (\vec p_1 + \vec p_2)^2 = E_1^2 - \vec p_1^2 + E_2^2 - \vec p_2^2 + 2( E_1 E_2 - \vec p_1 \cdot \vec p_2)

= m_1^2 + m_2^2 + 2( E_1 E_2 - \vec p_1 \cdot \vec p_2) \neq (m_1 + m_2)^2.

$$

- #10

mfb

Mentor

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You can always consider the rest frame of one of the pions where the expression from my previous post takes the form

$$

M^2 = m_1^2 + m_2^2 + 2 m_1\sqrt{m_2^2 + p_2^2} \geq (m_1+m_2)^2,

$$

with equality holding only when ##p_2 = 0##, i.e., when the pions are at relative rest.

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- #14

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I did read an article featuring this in the UCL Physics department Annual Departmental Review in the year concerned.

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If the pions move relative to each other, the invariant mass of the system (=the total energy in the center of mass frame) is higher.

If they move relative to each other, I can see there will be more kinetic energy in the C.M. frame. However, the sum of their momenta will be zero in that frame. So, the expression (E

I am still a bit confused as to how you can have more energy in the C.M. frame if the momenta sum to zero in that frame.

- #16

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Yes it will. You said it yourself, the ##E_i## increase because there is more kinetic energy.So, the expression (E1 + E2 )2 - (p1 + p2)2 will not change as a result will it?

- #17

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Of course, yes. Right , thanks.

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