3
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Is there anything on RHS of these equations? $\endgroup$
    – Paras
    Jul 7, 2020 at 12:12
  • 1
    $\begingroup$ @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

7
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ 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
  • 10
    $\begingroup$ @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
7
$\begingroup$

The book simply use the well-known equivalence $\:A^2=B^2\iff A=\pm B$.

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

$\endgroup$
2
  • 1
    $\begingroup$ 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
6
$\begingroup$

$$ 4\omega^2 = 2\Big(\frac{eB}{m}\Big)^2+4\omega_0^2 \pm\ 2\sqrt{\Big[\Big(\frac{eB}{m}\Big)^2+2\omega_0^2\Big]^2-4\omega_0^4}\\ = 2\Big(\frac{eB}{m}\Big)^2+4\omega_0^2 \pm\ 2\sqrt{\Big(\frac{eB}{m}\Big)^4+ 4\Big(\frac{eB}{m}\Big)^2 \omega_0^2}\\ = 2\Big(\frac{eB}{m}\Big)^2+4\omega_0^2 \pm\ 2\Big(\frac{eB}{m}\Big)\sqrt{\Big(\frac{eB}{m}\Big)^2+ 4 \omega_0^2}\\ = \left[ \frac{eB}{m} \pm \sqrt{\Big(\frac{eB}{m}\Big)^2+ 4 \omega_0^2} \right]^2\\ $$ so it gives you the "nice" formula for the roots of $\omega^2 \pm \frac{eB\omega}{m} - \omega_0^2$

$\endgroup$
5
  • $\begingroup$ How much simpler would it be to factorise that quartic using difference of two squares first, though? $\endgroup$ Jul 7, 2020 at 12:33
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.