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Let me say that particle A hits particle B and two particles come out - C and D;

In system S I can write: $$p_A^μ+p_B^μ=p_C^μ+p_D^μ;\tag{1}$$ here $p_N^μ$ is the 4-momentum.

Using the Lorentz transformation I want to prove that energy and momentum are also conserved in frame S'. I rewrite $(1)$ like that: $$p_A^μ+p_B^μ-p_C^μ-p_D^μ=0; (2)$$

Now I write something similar for the system S', except I do not know yet whether it's equal to zero: $$p_A^{'μ}+p_B^{'μ}-p_C^{'μ}-p_D^{'μ}=C;(3)$$

My goal is to find that $C=0$;

I know that for Lorentz transformations this holds true: $$p^{'μ}=Λ_ν^{μ}p^ν ;(4)$$

So if I put (4) into (3) , I get $$Λ_ν^{μ}p_A^{ν}+Λ_ν^{μ}p_B^{ν}-Λ_ν^{μ}p_C^{ν}-Λ_ν^{μ}p_D^{ν}=C;(5)$$

Now, this will be my question, if I consider each particle's transformation $Λ_ν^{μ}$ to be the same, I can bring out the common factor $Λ_ν^{μ}(p_A^{ν}+p_B^{ν}-p_C^{ν}-p_D^{ν})$ (6) and inside the parentheses I have the same equation (2), thus $C=0$ and 4-momentum is conserved.

My questions are: 1) Why can I consider that $Λ_ν^{μ}$ is the same for every particle's transformation?

2) Also, is my method of proving the 4-momentum conservation alright, or am I doing something ineffectively?

Henrikas
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  • Because $\Lambda^\mu_\nu$ depends on the relative velocity between the frames (and not the particles). 2) It's alright, but you could also have noted that for any frame $S'$ obtained by Lorentz-boosting the frame $S$, you would get $C=\Lambda^\mu_\nu 0=0$.
    • – Demosthene Mar 23 '15 at 20:11
  • @Demosthene Why write this as a comment? – Bill N Mar 23 '15 at 20:22
  • the sum of time-like four momentums is always time-like. This is proved with the Schwartz inequality – diffeomorphism Mar 23 '15 at 20:27
  • you use the Schwartz inequality to prove that time-like four vectors are a group with the vector addition operation – diffeomorphism Mar 23 '15 at 20:28
  • What level of answer are you expecting? I don't really understand your confusion. – innisfree Mar 23 '15 at 20:38
  • I just wanted to make sure I'm doing everything correctly. As far as I understand user Demosthene gave the reason I was asking for, so I guess I understand it now.

    My confusion was, that I didn't fully understand the exact reason why $Λ^μ_ν$ is the same for every particle, even though it had to be.

     – Henrikas Mar 23 '15 at 20:56
  • Great, well maybe somebody will add something, but if not, did you know that you can answer your own question? – innisfree Mar 23 '15 at 21:21
  • I will do that, alright – Henrikas Mar 23 '15 at 23:01
  • Henrikas: Re-writing your eq. (1) a little more careful than you did, I get instead: $$p_A^{\mu}+p_B^{\mu}-p_C^{\mu}-p_D^{\mu} = 0^{\mu}\tag{2~},$$ and consequently your eq. (6) evaluates rather more consistently: $$\Lambda^{\nu}{\mu}~(p_A^{\mu}+p_B^{\mu}-p_C^{\mu}-p_D^{\mu}) = \Lambda^{\nu}{\mu}~0^{\mu} = 0^{\nu}. \tag{6~ }$$"am I doing something ineffectively?" -- Yes: you could obviously leave off the "primes" in your eqs. (3) and (4) [... contd.] – user12262 May 10 '15 at 08:46
  • Noting that "$\mu$" and "$\nu$" are distinctive reference base indices, corresponding to the two reference systems under consideration, you could write more effectively: $$p_A^{\nu} + p_B^{\nu} - p_C^{\nu} - p_D^{\nu} = C^{\nu} \tag{3~},$$ with the goal of proving "$C^{\nu} = 0^{\nu}$"; and $$p^{\nu} = \Lambda^{\nu}_{\mu}~p^{\mu}. \tag{4~}$$ – user12262 May 10 '15 at 08:51