- #1
ChrisJ
- 70
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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}##
##p \cdot p = m^2##
##E^2 = m^2 + |\vec{p}|^2##
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,
[tex]
p_w = p_u - p_v \\
p_w \cdot p_w = (p_u - p_v) \cdot (p_u - p_v) \\
m_w^2 = m_u^2 + m_v^2 - 2 ( p_u \cdot p_v ) \\
p_u \cdot p_v = \frac{m_u^2+m_v^2-m_w^2}{2}
[/tex]
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,
[tex]
p_u \cdot p_v = E_v(Ev + E_w) - \vec{p_v}(\vec{p_v}+\vec{p_w}) \\
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
[/tex]
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 :)
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,
[tex]
p_w = p_u - p_v \\
p_w \cdot p_w = (p_u - p_v) \cdot (p_u - p_v) \\
m_w^2 = m_u^2 + m_v^2 - 2 ( p_u \cdot p_v ) \\
p_u \cdot p_v = \frac{m_u^2+m_v^2-m_w^2}{2}
[/tex]
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,
[tex]
p_u \cdot p_v = E_v(Ev + E_w) - \vec{p_v}(\vec{p_v}+\vec{p_w}) \\
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
[/tex]
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 :)
Last edited: