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Energymomentum for a point particle and 4vectors 
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#37
Mar808, 01:57 PM

P: 3,967

In practice you would sum all the x components and then transform that quantity. The final transformed momentum of the system is obtained from [tex] p ' = \sqrt{ (\Sigma p_x ' )^2 + (\Sigma p_y ' )^2 + (\Sigma p_z ' )^2} [/tex] 


#38
Mar808, 02:10 PM

P: 321




#39
Mar808, 02:10 PM

P: 3,967

When [tex]\vec{v_1}=\vec{v_2}[/tex] the term [tex]\vec{p_1} \vec{p_2}[/tex] is zero which leaves just [tex]E^2[/tex]. It is this quantity, the energy squared of the sytem in the rest frame, that is invariant of the [tex]E_1E_2\vec{p_1} \vec{p_2}[/tex] expression. [tex]E_1E_2\vec{p_1} \vec{p_2}[/tex] gives the "rest energy" of the system and not the rest mass. The only time this expression gives the rest mass is when all the particles are stationary in the rest frame. 


#40
Mar808, 03:16 PM

P: 321

I don't think this one is right :[tex]\vec{p_1} \vec{p_2}=\gamma^2 m_1m_2v_1^2[/tex] 


#41
Mar808, 03:22 PM

P: 3,967

where [tex]\Sigma \textbf{p}^2[/tex] goes to zero, because the momenta are summed before they are squared. [tex] [p_1 + (p_1)]^2 = 0[/tex] 


#42
Mar808, 04:40 PM

P: 321

I don't think this is right either, we were taliking about a system of two particles , so [tex]\vec{p_1}+\vec{p_2}=\gamma (m_1m_2) \vec{v_1}[/tex]. 


#43
Mar808, 07:39 PM

P: 3,967

Equal masses and opposite velocities is the simplest case and that was what I assumed we were talking about. Also remember that by definition the total momentum in the centre of mass rest frame is zero so if [itex] u_1 = u_2 [/itex] then [itex]m_1[/itex] must equal [itex]m_2[/itex]. 


#44
Mar908, 11:27 AM

P: 321




#45
Mar908, 11:28 AM

P: 321




#46
Mar908, 11:41 AM

P: 3,967




#47
Mar908, 01:01 PM

P: 3,967

[tex] m_1+m_2+{2m_1m_2(1u_1u_2) \over \sqrt{1u_1^2} \sqrt{1u_2^2}[/tex] which reduces to [tex] {4m_1^2 \over (1u_1^2)[/tex] when [itex]m_2=m_1[/itex] and [itex]v_2 = v_1[/itex] 


#48
Mar908, 04:43 PM

P: 321

Why is so difficult for you to accept when you make a mistake? 


#49
Mar908, 05:17 PM

P: 3,967

P.S. No one else has said I have made a mistake ;) 


#50
Mar908, 05:20 PM

P: 321

[tex]\vec{p}=\gamma(v)*m_0 \vec{v}[/tex] [tex]m_0[/tex] is the invariant mass. In two dimensions: [tex]p_x= \gamma(v)*m_0 v_x[/tex] [tex]p_y= \gamma(v)*m_0 v_y[/tex] In frame S' , moving with speed [tex]u[/tex] along the aligned x axes: [tex]\vec{p'}=\gamma(v')*m_0 \vec{v'}[/tex] [tex]p'_x= \gamma(v')*m_0 v'_x[/tex] [tex]p'_y= \gamma(v')*m_0 v'_y[/tex] where : [tex]v'^2=v'_x^2+v'_y^2[/tex] [tex]v'_x=\frac{v_x+u}{1+v_xu/c^2}[/tex] [tex]v'_y=\frac{v_y \sqrt(1u^2/c^2)}{1+v_xu/c^2}[/tex] [tex]\gamma(v')=\gamma(v) \gamma(u) (1+v_xu/c^2)[/tex] so: [tex]\gamma(v')v'_x=\gamma(v) \gamma(u) (v_x+u)[/tex] [tex]\gamma(v')v'_y=\gamma(v) v_y[/tex] so, indeed: [tex]p'_y=p_y[/tex] [tex]p'_x=m_0 \gamma(v')v'_x=\gamma(v) \gamma(u) (m_0v_x+m_0u)= \gamma(u)(p_x+\gamma(v)m_0u)=\gamma(u)(p_x+\frac {uE}{c^2})[/tex] 


#51
Mar908, 05:23 PM

P: 321




#52
Mar908, 05:26 PM

P: 3,967




#53
Mar908, 05:36 PM

P: 3,967

You do not seem to accept my definition of the rest frame of a system of particles. If we have a cloud of particles all with random masses and velocities, how would you define the rest frame of that system? If you were watching this cloud of particles and the cloud as a whole was drifting away from you (ignore any expansion) then would you still claim to be in the rest frame of the cloud of particles? If the total momentum of the cloud is not zero by your measurements, do you accept that the cloud as a whole will drift away from you? 


#54
Mar908, 05:44 PM

P: 321




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