Finding the Energy of an Outgoing Particle in a Momentum Conservation Problem

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SUMMARY

The discussion focuses on calculating the energy of an outgoing particle in a momentum conservation problem involving particle decay. Specifically, it derives the energy of particle v, denoted as ##E_v##, using the formula ##E_v = \frac{m_u^2+m_v^2-m_w^2}{2m_u}##, where ##m_u##, ##m_v##, and ##m_w## are the masses of particles u, v, and w, respectively. The analysis employs conservation of four-momentum and the Minkowski metric, confirming that in the rest frame of particle u, the energy and momentum components are conserved. The final solution is reached by recognizing that in this frame, the four-momentum simplifies to ##p_u = (m_u, 0)##.

PREREQUISITES
  • Understanding of four-momentum in particle physics
  • Familiarity with the Minkowski metric
  • Knowledge of energy-momentum conservation principles
  • Basic algebra and vector operations
NEXT STEPS
  • Study the derivation of four-momentum conservation in particle decay scenarios
  • Learn about the implications of the Minkowski metric in relativistic physics
  • Explore examples of energy calculations in particle collisions using natural units
  • Investigate the role of rest frames in simplifying particle physics problems
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This discussion is beneficial for physics students, particularly those studying particle physics, as well as educators and researchers involved in teaching or exploring concepts related to momentum conservation and particle decay processes.

ChrisJ
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This problem assumes working in natural units where ##c=1##, and using the Minkowski metric where the time component is positive and the space ones negative (as I know the opposite convention is just as commonly used).

EDIT: I had intended to display 4-vectors as bold and the 3-vectors with the arrow but forgot that it won't parse the Tex properly. So I left the 3-vectors with the arrows, and if it has no arrow and has no component subscript (i.e. x/y/z) assume its a 4-vector.

1. Homework Statement

Particle u of mass ##m_u## has total energy ##E_u## in the the frame of the lab. It decays into two new particles, particle v of mass ##m_v## and particle w of mass ##m_w##.

Show that in particle u's rest frame, the energy of outgoing particle v is given by ##E_v = \frac{m_u^2+m_v^2-m_w^2}{2m_u}##

Homework Equations


##p \cdot p = m^2##
##E^2 = m^2 + |\vec{p}|^2##

The Attempt at a Solution


Never done conservation of four momentum problem before, but I assume that it is the case that ##p_u = p_v + p_w## and with the energy components being conserved, and the 3-momentum components being conserved, so that ##p_u = (E_v + E_w, \vec{p_v} + \vec{p_w})## ?

So far I have done,

<br /> p_w = p_u - p_v \\<br /> p_w \cdot p_w = (p_u - p_v) \cdot (p_u - p_v) \\<br /> m_w^2 = m_u^2 + m_v^2 - 2 ( p_u \cdot p_v ) \\<br /> p_u \cdot p_v = \frac{m_u^2+m_v^2-m_w^2}{2}<br />

Which implies that ##p_u \cdot p_v = E_v m_u ## but this is where I come unstuck and can't get that result.

I have that ##p_u = (E_v + E_w, \vec{p_v} + \vec{p_w}) = (E_u , \vec{p_u}) ##
and that ##p_v = (E_v , \vec{p_v}) ##

Doing the dot product I get,
<br /> p_u \cdot p_v = E_v(Ev + E_w) - \vec{p_v}(\vec{p_v}+\vec{p_w}) \\<br /> p_u \cdot p_v = E_v^2+E_v E_w - p_{v_x}^2 - p_{w_x}^2 - p_{v_y}^2 - p_{w_y}^2 - p_{v_z}^2 - p_{w_z}^2<br />

I am not sure if this is the correct way to go about it, but I can't seem to progress from here, not sure if it is because I have gone down the wrong path or if its just a algebra/connection issue. Any help/advice is much appreciated, thanks :)
 
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ChrisJ said:
<br /> p_w = p_u - p_v \\<br /> p_w \cdot p_w = (p_u - p_v) \cdot (p_u - p_v) \\<br /> m_w^2 = m_u^2 + m_v^2 - 2 ( p_u \cdot p_v ) \\<br /> p_u \cdot p_v = \frac{m_u^2+m_v^2-m_w^2}{2}<br />
Nice

Which implies that ##p_u \cdot p_v = E_v m_u ## but this is where I come unstuck##
This follows easily if you keep in mind that you are in the rest frame of particle u.
 
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TSny said:
NiceThis follows easily if you keep in mind that you are in the rest frame of particle u.

Thanks for reminding about that, managed to get it now, since ##p_u = (E_u, 0)=(m_u,0)##, Cheers :)
 

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