# I Invariant mass plots for resonance 'particles'

1. Jul 6, 2018

### IAN 25

The interaction p + π- → n + π- + π + may proceed by the creation of an intermediate 'particle' or resonance called a rho. This can be detected as a peak in the plot of invariant rest mass energy of the emergent pions versus frequency of pions observed. My question is quite simply, invariant rest mass is 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/ c2 for both, so how can you have a range of their rest mass energies?

2. Jul 6, 2018

### Orodruin

Staff Emeritus
You are confusing the invariant mass of the system with the sum of the invariant masses. The invariant mass of the system is the square of the total 4-momentum, which is also the total energy in the CoM frame. This is not equal to the sum of the invariant masses of the pions.

3. Jul 6, 2018

### Orodruin

Staff Emeritus
As a curiosity, a system of two photons have a non-zero invariant mass (unless they are collinear). The Higgs was discovered as a peak in the photon-photon invariant mass spectrum at 125 GeV.

4. Jul 6, 2018

### IAN 25

Thank you. Okay, I understand what the invariant mass of the system is - but if it is the sum of the invariant masses of the (two?) pions that is plotted, why is there a continuous range?

5. Jul 6, 2018

### Orodruin

Staff Emeritus
Based on the next part of your post, you don’t.

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).

6. Jul 6, 2018

### IAN 25

To me invariant mass means invariant, whether its a sum for several particles or not; the masses have discrete values and

E2 -(pc)2 = m2c4 for each particle (or for a sum of particles)

which as I understand it, is a scalar invariant - which has the same value in all inertial frames, does it not?
So, how can you have a continuous range of energies on this plot, unless its (rest mass + kinetic ) energies. That I could understand. The Ek of the pions getting bigger until the threshold for the mass of the Rho is reached.

7. Jul 6, 2018

### Orodruin

Staff Emeritus
Again. It is not the invariant masses of the particles added together. It is the invariant mass of the system, i.e., $E^2 - p^2$ for the system as a whole (in reasonable units where c=1).

8. Jul 6, 2018

### IAN 25

I still don't see why this gives a range of rest mass energies. That's the point I am trying to understand - the continuous range?

9. Jul 6, 2018

### Orodruin

Staff Emeritus
Because:
$$(E_1 + E_2)^2 - (\vec p_1 + \vec p_2)^2 = E_1^2 - \vec p_1^2 + E_2^2 - \vec p_2^2 + 2( E_1 E_2 - \vec p_1 \cdot \vec p_2) = m_1^2 + m_2^2 + 2( E_1 E_2 - \vec p_1 \cdot \vec p_2) \neq (m_1 + m_2)^2.$$

10. Jul 6, 2018

### Staff: Mentor

If the pions are at rest relative to each other, the invariant mass of the system is simply the sum of their rest masses. 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.

11. Jul 7, 2018

### IAN 25

Got it! I can follow the algebra but the two worded sentences of physics (mfb) add clarity. Thank you both.

12. Jul 7, 2018

### Orodruin

Staff Emeritus
Just to add some more food for thought.

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.

13. Jul 9, 2018

### IAN 25

Yes that is an interesting example. In you original expression, it is clearly the term 2(E1 E2 - p1. p2 which provides the range of kinetic energies of the pions observed either side of the peak in the laboratory frame.

14. Jul 9, 2018

### IAN 25

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

15. Jul 10, 2018

### IAN 25

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 (E1 + E2 )2 - (p1 + p2)2 will not change as a result will it? Since, the momentum term is zero. Or is the rest mass energy available in the laboratory frame bigger as a result? If so is it because of the 2(E1 E2 - p1 . p2) term, which appears in the invariant rest mass expression in the latter frame?

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. Jul 10, 2018

### Orodruin

Staff Emeritus
Yes it will. You said it yourself, the $E_i$ increase because there is more kinetic energy.

17. Jul 10, 2018

### IAN 25

Of course, yes. Right , thanks.