This question may have a simple solution. I did a change of variables and I'd like to check it; what do you recommend?. Altough I believed there's a way to compare term-by-term doing

expr1 == expr2

I couldn't come with an answer (I expected a Boolean output like True or False). Also I've tried Replace All replacing literally and simplifying but it's not helping me.

I'm writing below what I did:

  1. I have equations:

a) $$e_{1eq} = \frac{1}{\sqrt{2}}\sqrt{\frac{\frac{t_{e,1}}{t_{a,2}}-\frac{t_{e,1}}{t_{a,1}}}{(q+1) \left(\frac{m_1 q}{\alpha m_2 (q+1)}+1\right) \left(\frac{m_1 q^2 f_{2 p}^2 t_{e,1}}{\alpha f_1^2 m_2 (q+1)^2 t_{e,2}}+1\right)}}$$

which can be written as:

e1eq =Sqrt[(-(te1/ ta1) + te1/ta2)/((1 + q) (1 + (m1 q)/(m2 (1 + q) \[Alpha])) (1 + ( f2p^2 m1 q^2 te1)/(f1^2 m2 (1 + q)^2 te2 \[Alpha])))]/Sqrt[2]

b) $$\Delta eq = \sqrt{2} \sqrt{-\frac{\alpha ^2 m_2^2 t_{a,1} \left(\frac{m_1 q}{\alpha m_2 (q+1)}+1\right) \left(\frac{m_1 q^2 f_{2 p}^2 t_{e,1}}{\alpha m_2 (q+1) t_{e,2}}+f_1^2 (q+1)\right)}{M_*^2 q^2 \left(1-\frac{t_{a,1}}{t_{a,2}}\right) t_{e,1}}}$$


Deltaeq=Sqrt[2] Sqrt[-(( m2^2 ta1 (1 + (m1 q)/(m2 (1 + q) \[Alpha])) (f1^2 (1 + q) + ( f2p^2 m1 q^2 te1)/(m2 (1 + q) te2 \[Alpha])) \[Alpha]^2)/( M^2 q^2 (1 - ta1/ta2) te1))]

where (Aclaration!) the application of the expressions is restricted to the case where (i) $t_{a,1}/t_{a,2}>1$ for $t_{a,1}, t_{a,2} >0$, (ii) $t_{a,1}<0$ and $t_{a,2}>0$ or (iii) $t_{a,1}/t_{a,2}<1$ for $t_{a,1},t_{a,2} <0$. In other words, $\Delta_{eq}$ is has $\mathrm{Im}{(\Delta_{eq})} \equiv 0$

I recognize common terms for $e_{1eq}$ and $\Delta_{eq}$ and calculated (by hand) that

$$e_{1,eq} = \frac{\alpha m_2 |f_1|}{q M_* \Delta_{eq}}$$

which is an expression I find more friendly to use. How would you check an algebra calculation?

By the way, I checked on this question but

e1eq /. {Sqrt[2] Sqrt[-(( m2^2 ta1 (1 + (m1 q)/(m2 (1 + q) \[Alpha])) (f1^2 (1 + q) + ( f2p^2 m1 q^2 te1)/(m2 (1 + q) te2 \[Alpha])) \[Alpha]^2)/( M^2 q^2 (1 - ta1/ta2) te1))] -> Subscript[\[CapitalDelta], eq]}

won't do the trick.


1 Answer 1


Assume that all parameters are positive.

Then consider the case ta1 < ta2. Then e1eq as well as Deltaeq will have of the form:

+I times some-real-positive-number

On the other hand. The quotient:

enter image description here

must have the form:

-I times some-real-positive-number

Therefore e1eq and the quotient can not be equivalent, at least if on always takes the plus branch of the square root.

  • $\begingroup$ I've edited my question to manage the exception of possible negative arguments in the square root, which in this case have non physical answers. Thanks for noting it! @DanielHuber $\endgroup$
    – nuwe
    Nov 5, 2021 at 18:13

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