We are referring to the second equality in equation (9.24) in section 9.1.3 Boundary Conditions: Reflection and Transmission.

$$ \frac{\lambda_1}{\lambda_2} = \frac{k_2}{k_1} = \frac{v_1}{v_2}. \tag{9.24}$$

Shouldn't this be $ \frac{k_2}{k_1} = \frac{v_1}{v_2} \frac{\omega_2}{\omega_1}$?

Namely, using the substitution $\omega = 2\pi \nu = kv$.

Context. We are considering the boundary conditions at the knot at which two strings are tied to each other. $k_1,\lambda_1$ and $v_1$ are the parameters for the first string. We look at the incident, reflected and transmitted waves.

Supplement. Introduction to Electrodynamics, Fourth Edition, Griffiths

  • 1
    $\begingroup$ Have you checked the errata on Dr Griffiths web page? Note also that likely many of us do not have the 4th edition, so you may want to include more of the text than the single equation. $\endgroup$
    – Kyle Kanos
    Jun 15, 2015 at 12:47

1 Answer 1


This is not a typo. In this situation both waves must have the same frequency, $\omega_1=\omega_2$, so the extra factor is simply unity. The equality of both frequencies is due to the fact that, at the interface itself, the waves must remain in step for all time, which can only happen if they oscillate at the same frequency. (In contrast, the wavelengths can differ, because neither medium can "see" what happens in the interior of the other one. The wavelength - i.e. the wavevector $k$ - happens internally to the medium; the interface only sees the frequency and the transverse component of $\mathbf k$.)

  • $\begingroup$ Of course, they have the same frequency, indeed. The guy swinging the string at $z=-\infty$ governs both $\omega_1$ and $\omega_2$. $\endgroup$ Jun 15, 2015 at 12:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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