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Energymomentum for a point particle and 4vectors 
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#55
Mar908, 06:22 PM

P: 321

[tex] p_z' = p_z [/tex], I had to do the calculations myself, it wasn't obvious. This leads to: [tex]E'^2(p'_xc)^2=E^2(p_xc)^2[/tex] and ultimately to: [tex]E'^2(\vec{p'}c)^2=E^2(\vec{p}c)^2[/tex] Now I am very happy. 


#56
Mar908, 06:39 PM

P: 3,967

Like me, you are cursed with having to prove everything to yourself from first principles. 


#57
Mar908, 06:40 PM

P: 321

Please check your formula for [tex]p'_x[/tex], it is wrong, so our results do not agree :) 


#58
Mar908, 07:20 PM

P: 3,967

> [tex]p_x ' = \gamma(u_x)\gamma(v)m_o (v+u_x) = \gamma(v)p_x+\gamma(u_x)\gamma(v)m_o v [/tex] > [tex]p_x ' = \gamma(u_x)\gamma(v)m_o u_x+m_ov) = \gamma(v)(p_x+\gamma(u_x)m_o v) [/tex] You have used v for the particle velocities in the rest frame and u for velocity of frame S relative to S' while I have done the opposite. That is why the end results look different. Swap the u's and v's and they are the same. 


#59
Mar908, 07:53 PM

P: 321

[tex]p'_x[/tex] contains [tex]\gamma(u_x)[/tex] This can't be right, you shouldn't have any vector components in the [tex]\gamma[/tex] expression, you should only have the vector norm. 


#60
Mar1008, 08:39 AM

P: 321




#61
Mar1008, 04:00 PM

P: 3,967

You are right, that I am wrong. You see, I can accept when I have made a mistake :P
I would be interested to see how you got to: [tex]\gamma(v')=\gamma(v) \gamma(u) (1+v_xu/c^2)[/tex] from: [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] 


#62
Mar1008, 04:25 PM

P: 321




#63
Mar1008, 05:36 PM

P: 3,967

I can get a similar answer to yours except for uncertainty about the signs due to imaginary roots. I notice your answer differs from Dr Greg's transformation in sign too. Dr Greg has [tex] p_x ' = \gamma_u \left(p_x  \frac{u E}{c^2} \right) [/tex] while you appear to have [tex] p_x ' = \gamma_u \left(p_x + \frac{u E}{c^2} \right) [/tex] 


#64
Mar1008, 05:55 PM

P: 321

[tex]v'_x=\frac{v_x+u}{1+v_xu/c^2}[/tex] If you use instead: [tex]v'_x=\frac{v_xu}{1+v_xu/c^2}[/tex] you will get dr.Greg's form. 


#65
Mar1108, 01:05 PM

Sci Advisor
PF Gold
P: 1,848

Yes, [itex]\sum E[/itex] is conserved in collisions. Yes, [itex]\sum \textbf{p}[/itex] is conserved. Therefore [itex](\Sigma E)^2c^2\\sum \textbf{p}\^2[/itex] is conserved, in a closed system. It is [itex]\sum E^2  c^2 \sum \\textbf{p}\^2[/itex] (i.e. [itex]c^4 \sum m^2[/itex]) that is not conserved, as your example shows: it's 2 before and 0 after. 


#66
Mar1108, 01:07 PM

Sci Advisor
PF Gold
P: 1,848

After a bit of extra reading over the weekend, I should point out some extra terms and conditions.
It's already been pointed out that the technique we've been using applies only to closed systems, i.e. where there's no interaction with the outside world. Note that hitting an enclosing wall counts as interaction and thus invalidates this approach unless you include the walls as part of your system of particles. It also only applies (in the form expressed so far) where each particle moves freely (inertially) between collisions. The only interaction allowed is at events (single points in spacetime); interaction over a distance (e.g. between charged particles) is forbidden. However, interaction over a distance can be included in the system by means of a potential. This means that the [itex]\sum E[/itex] term has to include potential energy as well as each particle's [itex]\sqrt{\\textbf{p}c\^2 + m^2 c^4}[/itex] energy. Unfortunately, that's the point where my knowledge of S.R. falls over but I believe it works. Finally, note that photons count as particles just as much as massive particles, and [itex]E^2 = \\textbf{p}c\^2 + m^2 c^4[/itex] still works for photons (with m = 0). 


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