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?
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