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I was solving a normal-mode problem and got a different result for the quadratic equation. The book provides a simpler solution than mine so I suspect I am the one who's wrong. Let's check it out.

Let us start from the following determinant

$$ \begin{vmatrix} \omega_o^2-\omega^2 & \frac{-ieB\omega}{m} \\ \frac{ieB\omega}{m} & \omega_o^2-\omega^2 \\ \end{vmatrix}= (\omega_o^2-\omega^2)^2-\Big(\frac{eB\omega}{m}\Big)^2=\omega^4-\Big[\Big(\frac{eB}{m}\Big)^2+2\omega_0^2\Big]\omega^2+\omega_0^4 $$

OK so far.

From here on I proceeded as follows; I looked for the roots, i.e. $\omega^4-\Big[\Big(\frac{eB}{m}\Big)^2+2\omega_0^2\Big]\omega^2+\omega_0^4=0$

$$\omega^2= \frac 1 2 \Big[\Big(\frac{eB}{m}\Big)^2+2\omega_0^2 \pm\ \sqrt{\Big[\Big(\frac{eB}{m}\Big)^2+2\omega_0^2\Big]^2-4\omega_0^4}\Big]$$

This leads to pretty ugly roots for $\omega$.

However, the book states that $(\omega_o^2-\omega^2)^2-\Big(\frac{eB\omega}{m}\Big)^2$ leads to $\omega^2 \pm \frac{eB\omega}{m} - \omega_o^2$. This leads to good looking roots for $\omega$.

My struggle is that I do not see how to show that's indeed the case.

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  • $\begingroup$ Is there anything on RHS of these equations? $\endgroup$
    – Paras
    Jul 7, 2020 at 12:12
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    $\begingroup$ @Paras note that firstly we compute the determinant and then we look for the roots of such computation (i.e. we equate such computation to zero). $\endgroup$
    – JD_PM
    Jul 7, 2020 at 12:21
  • $\begingroup$ You appear to have made a mistake by absorbing the constant $\left(\tfrac{eB\omega}{m}\right)^2$ into the $\omega^2$ term. $\endgroup$
    – J.G.
    Jul 8, 2020 at 6:49

3 Answers 3

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Factorise $(\omega_0^2 - \omega^2)^2 - \left({\dfrac {eBw} m}\right)^2$ by difference of two squares and you get:

$\left({\omega_0^2 - \omega^2 - \left({\dfrac {eBw} m}\right) }\right) \left({\omega_0^2 - \omega^2 + \left({\dfrac {eBw} m}\right) }\right)$

which gets you practically there.

Note that the equation you are left with is itself a quadratic which has not yet been solved.

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    $\begingroup$ Ahhh so the key was to use $(a^2-b^2)=(a+b)(a-b)$; neat! $\endgroup$
    – JD_PM
    Jul 7, 2020 at 12:39
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    $\begingroup$ @JD_PM: One of the keys to success in physics is not to forget the mathematical tools you learned in high school, even after you've acquired a bunch of shiny new university-level mathematical tools to play with. $\endgroup$ Jul 7, 2020 at 21:53
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$$ 4\omega^2 = 2\Big(\frac{eB}{m}\Big)^2+4\omega_0^2 \pm\ 2\sqrt{\Big[\Big(\frac{eB}{m}\Big)^2+2\omega_0^2\Big]^2-4\omega_0^4}\\ = 2\Big(\frac{eB}{m}\Big)^2+4\omega_0^2 \pm\ 2\sqrt{\Big(\frac{eB}{m}\Big)^4+ 4\Big(\frac{eB}{m}\Big)^2 \omega_0^2}\\ = 2\Big(\frac{eB}{m}\Big)^2+4\omega_0^2 \pm\ 2\Big(\frac{eB}{m}\Big)\sqrt{\Big(\frac{eB}{m}\Big)^2+ 4 \omega_0^2}\\ = \left[ \frac{eB}{m} \pm \sqrt{\Big(\frac{eB}{m}\Big)^2+ 4 \omega_0^2} \right]^2\\ $$ so it gives you the "nice" formula for the roots of $\omega^2 \pm \frac{eB\omega}{m} - \omega_0^2$

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  • $\begingroup$ How much simpler would it be to factorise that quartic using difference of two squares first, though? $\endgroup$ Jul 7, 2020 at 12:33
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    $\begingroup$ it is simpler, but I wanted to prove directly that the even using the other way, you could obtain the nicer formula for the roots anyway, since the OP couldn't see it $\endgroup$
    – Exodd
    Jul 7, 2020 at 12:36
  • $\begingroup$ Fair comment. Fair play for multiplying through by 4 first, by the way, I didn't actually think of that. It does make it easier to handle. $\endgroup$ Jul 7, 2020 at 12:38
  • $\begingroup$ @Exodd Your answer taught me how to get the desired answer out of mine; +1 $\endgroup$
    – JD_PM
    Jul 7, 2020 at 12:41
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    $\begingroup$ In many cases I find that when a question asks "how does X lead to Y", people will often answer "here is how you find Y without using X" which might solve the broader problem but is not very useful to explaining what the question actually asked. So thank you for posting this Exodd. $\endgroup$
    – David Z
    Jul 7, 2020 at 20:59
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The book simply use the well-known equivalence $\:A^2=B^2\iff A=\pm B$.

Namely, in the present case: $$(\omega_o^2-\omega^2)^2=\Bigl(\frac{eB\omega}{m}\Bigr)^{\!2}\iff \omega_o^2-\omega^2=\pm\frac{eB\omega}{m}\iff \omega^2 \pm\frac{eB\omega}{m}-\omega_o^2=0$$

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    $\begingroup$ Neater than my analysis. Nice one. $\endgroup$ Jul 7, 2020 at 16:36
  • $\begingroup$ @PrimeMover: Thank you for your kind appreciation! $\endgroup$
    – Bernard
    Jul 7, 2020 at 18:11

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