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I am reading this paper. There are two known algebra equations (Eq. (2.9) in the paper) $$(e_++p_+)\frac{v_+^2}{1-v_+^2}+p_+=(e_-+p_-)\frac{v_-^2}{1-v_-^2}+p_-, \tag{1}$$ $$(e_++p_+)\frac{v_+}{1-v_+^2}=(e_-+p_-)\frac{v_-}{1-v_-^2}. \tag{2} $$ Then it is mentioned that these two relations can be rearranges as $$v_+v_-=\frac{p_+-p_-}{e_+-e_-}, \tag{3}$$ $$\frac{v_+}{v_-}=\frac{e_-+p_+}{e_+ + p_-}.\tag{4}$$

Although Eqs. (1), (2) are algebra equations, I found it very complicated (and failed) to rederive Eqs. (3), (4) from (1), (2).

Since Eqs. (3), (4) look quite clean, I would guess there may be some clever way to derive them. Does anybody have any insights? And further, can Mathematica solve such algebra equations?

Wein Eld
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1 Answers1

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There may be better ways of solving this problem but here goes. I assume that no division by zero occurs at any step. Let $$x = (e_{+}+p_{+})\frac{v_{+}}{1-v_{+}^{2}} = (e_{-}+p_{-})\frac{v_{-}}{1-v_{-}^{2}}\tag{A.1}.$$ Using $(1)$ in the question we get $$v_{+}-v_{-} = \frac{p_{-}-p_{+}}{x}\tag{A.2}.$$ From $(\text{A.1})$ we get $$1-v_{+}^{2} = \frac{(e_{+}+p_{+})v_{+}}{x},\ 1-v_{-}^{2} = \frac{(e_{-}+p_{-})v_{-}}{x}\tag{A.3}.$$ Subtracting the first equation from the second and using $(\text{A.2})$ we have $$(v_{+}+v_{-})\frac{p_{-}-p_{+}}{x} = \frac{e_{-}+p_{-}}{x}v_{-}-\frac{e_{+}+p_{+}}{x}v_{+}\tag{A.4}.$$ Simplifying the above we obtain $(4)$ in the question. Let $v_{+}/v_{-} = \alpha$. Then using $(\text{A.1})$ we note that $$\frac{(e_{+}+p_{+})\alpha}{1-\alpha v_{+}v_{-}} = \frac{e_{-}+p_{-}}{1-\dfrac{v_{+}v_{-}}{\alpha}}.$$ After simplifying and substituting the value of $\alpha$ from $(4)$ in the question we obtain $(3)$.

Karthik Kannan
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