# Conservation of relavistic energy

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1. Jan 30, 2015

### Samuelriesterer

Problem statement, equations, and work done:

A particle called a Kaon is moving at 0.8c through a detector when it decays into two pions.

Kaon particle: mass = 493.7 MeV/c^2
Pion+: mass = 139.6 MeV/c^2
Pion0: mass = 135.0 MeV/c^2

1) Apply conservation of momentum/energy to determine the momentum of each pion in the kaon frame of reference (where K is at rest).Use the fact that the momentum of the original kaon in its own frame is zero and don't forget to use the relativistic momentum energy conservation equation. Leave the momentum in MeV/c units.

Since the momentum of the kaon is 0 because it is at rest in its own frame, the magnitudes of the momentum of the pions should be equal.

$E_k = m_k c^2 = 493.7 \frac{MeV}{c^2} = (m_{\pi +} + m_{\pi 0}) c^2 + (p_{\pi +} + p_{\pi 0})c$

$493.7 \frac{MeV}{c} = (139.6 + 135.0) \frac{MeV}{c} + 2|p|$

$219.1 \frac{MeV}{c} = 2|p|$

$p = 109.55 \frac{MeV}{c}$

Solving for velocities:

$p{\pi +} = 109.55 \frac{MeV}{c} = \gamma_{\pi +} m_{\pi +} v_{\pi +} = \gamma_{\pi +} 139.6 \frac{MeV}{c^2} v_{\pi +}$

From this:

$v_{\pi +} = .61733 (c), and thus : \gamma_{\pi +} = 1.27113$

And:

$v_{\pi 0} = .63012 (c), and thus: \gamma_{\pi 0} = 1.28783$

From this there is no Conservation of energy:

$E_k = E_{\pi +} + E_{\pi 0} = 493.7 \frac{MeV}{c^2} = \gamma_{\pi +} m_{\pi +} + \gamma_{\pi 0} m_{\pi 0} = doesn't = 493.7!$

(I left units off)

2)Now transform the momentum-energy vectors for each pion and the kaon to the detector frame. Assume that the pion+ is moving in the direction the Kaon was moving in the detector frame.

3) Check to see that the momentum-energy conservation law is followed in the detector frame.

4) If the pions are moving off perpendicular to the Kaon's motion in the detector frame, calculate the momentum energy vectors for the pions in the detector frame and again check to see that momentum-energy is conserved.

2. Jan 30, 2015

### DEvens

What's that? What is the relativistic formula for kinetic energy?

3. Jan 30, 2015

### PeroK

I would do this for a general decay of a particle of mass M into two particles of masses m_1 and m_2, then plug the numbers into the final equation. You're getting bogged down in numbers. At the level you're at, you should be working more algebraically and not plugging in awkward values at the first opportunity!

4. Jan 30, 2015

### Samuelriesterer

That is E = mc^2 + pc which is total energy = rest energy + KE energy which equals the combined energy totals of the lions. The first part is how my teacher should us, but I can't bring it around to the conservation of energy

5. Jan 30, 2015

### DEvens

That's because kinetic energy is not equal to pc for a particle with mass.

6. Jan 30, 2015

### Staff: Mentor

I second this recommendation.

I also recommend that you always keep m together with c2 as (mc2), and p together with c as (pc). Then the only unit you need to use is MeV. That is, for the kaon, don't think m = 493.7 MeV/c2, but rather (mc2) = 493.7 MeV.

Finally, you should never need to calculate any velocities in a problem like this, unless the problem statement explicitly asks for it. Use conservation of energy, conservation of momentum, and the general relationship between energy, mass and momentum: E2 = (pc)2 + (mc2)2.

7. Jan 30, 2015

### Samuelriesterer

I wonder what my fantastic teacher was doing then.

I only was trying to calculate the velocities to show it does not fit into conservation of energy.

Can anybody point me in the right direction, I'm only trying to understand this all.

8. Jan 30, 2015

### PeroK

Your starting point is the three equations:

$E_1 + E_2 = Mc^2, \ \ E_1^2 = p^2c^2 + m_1^2c^4, \ \ E_2^2 = p^2c^2 + m_2^2c^4$

Can you explain these - at least to yourself?

There are now common ways to work with these sorts of equations: common algebraic tricks. In this case I recommend trying to get an equation for the energy of either pion. Once you have the energy, you can get the momentum from that.

9. Jan 30, 2015

### DEvens

I just did.

Kinetic energy for a particle with mass is (gamma - 1) mc^2, total energy (mass plus kinetic) is gamma mc^2.
Momentum is gamma m v.

Alternatively, as suggested by jtbell, E^2 = (pc)^2 + (mc^2)^2
You have E = pc + mc^2, which is wrong.

Your very first equation is wrong. As I have now told you for the third time.

Last edited: Jan 30, 2015
10. Jan 30, 2015

### Samuelriesterer

That equation is an approximation that our teacher said we could use in this situation.

Yes I understand those, thank you. Isn't that 3 equations with 4 unknowns? Don't I need: p = p1 + p2 = 0?

11. Jan 30, 2015

### PeroK

I've used p1 = p2 = p, as the momentum of the two pions must be equal.

12. Jan 30, 2015

### DEvens

Ok, I think I have discovered the fault.

13. Jan 30, 2015

### Samuelriesterer

Oh I see that now. Thanks.

14. Jan 30, 2015

### Samuelriesterer

I just solved the system of equations using the approximation and got 109.55 for the momentum. I also got the conservation of energy with E1 = 249.15 and E2 = 244.55, so that works.

Now to convert E1 and E2 to the second frame, what equation do I use? If it involves gamma, then what gamma is it and how to calculate it.

Thanks guys.

15. Jan 30, 2015

### PeroK

The mometum should be nearly double what you've calculated. The energy looks correct. If you have the energy and the mass of a particle, then you can calculate its momentum.

16. Jan 30, 2015

### Samuelriesterer

Well if E1 = m1 + p then p = E1 - m1 = 249.15 - 139.6 = 109.55.

Leaving the c and units out.

17. Jan 30, 2015

### PeroK

Energy isn't the sum of mass and momentum. It's $E^2 = p^2 + m^2$

18. Jan 30, 2015

### Samuelriesterer

I see now. So was it OK to use that approximation to solve the system of equations? Even though I got a different p in that system. Would that throw the values of E off too?

19. Jan 30, 2015

### PeroK

I haven't seen how you solved the equations. You seem to have got the right energies.

If you look at that approximation mathematically, it can only be valid where one of the terms is very small: where E is approx p or E is approx m. That's not the situation here.

20. Jan 30, 2015

### Samuelriesterer

Basically it was your 3 equations without the squares. <shrugs> when I get the time I'll do the system with the squares and see how it comes out.

Can you point me in the right direction to get the energy and momentum in the second frame?