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## Main Question or Discussion Point

Two particles of masses m1 and m2 are travelling at v1 and v2, and, collide. after the collision, their speeds become v3 and v4 (every speed w.r.t. ground frame). now, if i watch this collision in a moving reference frame with velocity 'v' (in the opposite direction to the particles)

Laws used:

Conservation of momentum (classical and relativistic)

conservation of total energy(relativistic only)

notations:

'c' = speed of light in vacuum.

bi = sqrt(1-vi^2/c^2) => b1 = sqrt(1-v1^2/c^2)

vi' = speed vi as seen in the moving frame of reference.

classical: vi' = vi + v

relativistic: vi' = (vi+v)/(1+(vi*v/c^2))

classical collision:

momentum conservation:

gnd frame:

m1*v1+m2*v2 = m1*v3+m2*v4.

knowing v3, i can uniquely find v4.

moving frame:

m1*v1'+m2*v2' = m1*v3'+m2*v4'

knowing v3', i can uniquely find v4'. then, transform it to gnd frame (v4), and these two things match up. no problem here.

now, coming to relativistic collision:

gnd frame:

momentum conservation:

m1*v1/b1 + m2*v2/b2 = m1'*v3/b3 + m2'*v4/b4. yes, the rest masses change due to collision. so, even if we know v3 (we choose it arbitrarily), we cannot solve this equation of 3 unknowns.

so, we go for TOTAL energy conservation.

m1*c^2/b1 + m2*c^2/b2 = m1'*c^2/b3 + m2'*c^2/b4.

i.e., m1/b1 + m2/b2 = m1'/b3 + m2'/b4.

now, we have 2 equations and 4 unknowns. (remember, v3 is also unknown)

for perfectly elastic collision, m1'=m1, m2'=m2. so, 2 equations, 2 unknowns. these are trancedental equations. solved in both the frames of reference, and then, converting moving frame speeds to ground frame speeds, answers match up.

for perfectly inelastic collision, m1' and m2' are not there, it is only M (both the bodies are attached together and travel as one). also, M>m1+m2, and v3=v4. so, 2 equations, 2 unknowns, simple equations. solving them in both frames, transferring moving frame speeds to ground frame, the results match up.

now, for partially inelastic collisions:

for classical:

we have coefficient of restitution. this is a measure of loss of kinetic energy in the collision. (actually, this is the total energy. no potential energy, and, of course, no mass energy)

e = (v3-v4) / (v1-v2)

in moving frame:

e = (v3'-v4')/(v1'-v2') = (v3+v -(v4+v))/(v1+v -(v2+v)) = e in gnd frame.

so, it is a unique number.

in relativistic collision:

we are already conserving the total energy. so, wot other property can we have for such collisions

Laws used:

Conservation of momentum (classical and relativistic)

conservation of total energy(relativistic only)

notations:

'c' = speed of light in vacuum.

bi = sqrt(1-vi^2/c^2) => b1 = sqrt(1-v1^2/c^2)

vi' = speed vi as seen in the moving frame of reference.

classical: vi' = vi + v

relativistic: vi' = (vi+v)/(1+(vi*v/c^2))

classical collision:

momentum conservation:

gnd frame:

m1*v1+m2*v2 = m1*v3+m2*v4.

knowing v3, i can uniquely find v4.

moving frame:

m1*v1'+m2*v2' = m1*v3'+m2*v4'

knowing v3', i can uniquely find v4'. then, transform it to gnd frame (v4), and these two things match up. no problem here.

now, coming to relativistic collision:

gnd frame:

momentum conservation:

m1*v1/b1 + m2*v2/b2 = m1'*v3/b3 + m2'*v4/b4. yes, the rest masses change due to collision. so, even if we know v3 (we choose it arbitrarily), we cannot solve this equation of 3 unknowns.

so, we go for TOTAL energy conservation.

m1*c^2/b1 + m2*c^2/b2 = m1'*c^2/b3 + m2'*c^2/b4.

i.e., m1/b1 + m2/b2 = m1'/b3 + m2'/b4.

now, we have 2 equations and 4 unknowns. (remember, v3 is also unknown)

for perfectly elastic collision, m1'=m1, m2'=m2. so, 2 equations, 2 unknowns. these are trancedental equations. solved in both the frames of reference, and then, converting moving frame speeds to ground frame speeds, answers match up.

for perfectly inelastic collision, m1' and m2' are not there, it is only M (both the bodies are attached together and travel as one). also, M>m1+m2, and v3=v4. so, 2 equations, 2 unknowns, simple equations. solving them in both frames, transferring moving frame speeds to ground frame, the results match up.

now, for partially inelastic collisions:

for classical:

we have coefficient of restitution. this is a measure of loss of kinetic energy in the collision. (actually, this is the total energy. no potential energy, and, of course, no mass energy)

e = (v3-v4) / (v1-v2)

in moving frame:

e = (v3'-v4')/(v1'-v2') = (v3+v -(v4+v))/(v1+v -(v2+v)) = e in gnd frame.

so, it is a unique number.

in relativistic collision:

we are already conserving the total energy. so, wot other property can we have for such collisions