Unable to solve an approximation

[Kidiz]

Homework Statement

I'm trying to understand an approximation Griffiths does (in his solutions' manual - exercise 9.18-b) and I'm not quite getting it.
Let
$$k = \omega \sqrt{\dfrac{\epsilon \mu}{2}} [\sqrt{ 1 + (\dfrac{\sigma}{\epsilon \omega}})^2-1]^{1/2}$$

He says that, because $\sigma >> \omega \epsilon$, we have:$$k = \sqrt{\dfrac {\mu \sigma \omega}{2}}$$

The Attempt at a Solution

[/B]
Essentially, my attempt was to say that $(\dfrac{\sigma}{\epsilon \omega})^2 \approx \sigma ^2$. After that, and since sigma >> 1, I'd say that $\sqrt {1 + \sigma ^2} \approx \sigma$, and for the same reasoning $\sigma -1 \approx \sigma$, which would make everything inside the straight brakets be $\sqrt{\sigma}$. After all this, $k$ would be $\omega \sqrt {\dfrac{\epsilon \mu \sigma}{2}}$. Now, sadly, that's no way near the solution I'm supposed to arrive.

 

[mfb]

Essentially, my attempt was to say that $(\dfrac{\sigma}{\epsilon \omega})^2 \approx \sigma ^2$.

That would correspond to $\epsilon \omega \approx 1$. The approximation gives $(\dfrac{\sigma}{\epsilon \omega})^2 \approx 1$ which allows to simplify the expression.
You can use the same approximation for the first ω in the equation and then ignore some prefactor.

 

[Kidiz]

That would correspond to $\epsilon \omega \approx 1$. The approximation gives $(\dfrac{\sigma}{\epsilon \omega})^2 \approx 1$ which allows to simplify the expression.
You can use the same approximation for the first ω in the equation and then ignore some prefactor.

I don't quite understand what you're saying. You're saying that $(\dfrac{\sigma}{\epsilon \omega})^2 \approx 1$ solves my problem? If so, I don't understand how, as it would eliminate my $\omega$.

 

[rock.freak667]

If σ >> ωε then (σ/ωε) >>1.

And so (σ/ωε)² +1 ≈ (σ/ωε)²

You'd then use your same reasoning however carrying the σ/ωε expression instead of just the σ.

 

[mfb]

Oh sorry, I read $\approx$ instead of >>. See the post above this how to do the approximation then.

 

[Kidiz]

If σ >> ωε then (σ/ωε) >>1.

And so (σ/ωε)² +1 ≈ (σ/ωε)²

You'd then use your same reasoning however carrying the σ/ωε expression instead of just the σ.

Thank you! This works wonderfully. I wonder if there's a text somewhere that explains how to do this type of approximations or if one just picks them up as one goes.

Oh sorry, I read $\approx$ instead of >>. See the post above this how to do the approximation then.

That was my bad mfb, sorry. Initially, I had $\approx$, but then edited and by mistake made it $>>$, and then saw the error and edited again.