I am reading "Introduction to Electrodynamics" [Griffiths] and in section 9.1.1, there is an explanation for why a stretched string supports wave motion. It begins as follows:

It identifies 2 points, $z$ and $z + \Delta z$ in in the direction that the wave is propagating, and states the net transverse force on the segment between $z$ and $z + \Delta z$ is:

$\Delta F = T (sin\Theta' - sin\Theta)$

so far so good. Then, provided that the distortion of the string is not too great, tan can replace sine because the angles will be small so:

$\Delta F \cong T (tan\Theta' - tan\Theta) = T(\frac{\partial f}{\partial z} - \frac{\partial f}{\partial z})$

where each of the above partial derivatives are taken at points $(z + \Delta z)$ and $z$ respectively.

The next step I am unclear on the reasoning for. It says:

$T(\frac{\partial f}{\partial z} - \frac{\partial f}{\partial z}) \cong T \frac{\partial ^2 f}{\partial z^2} \Delta z$

I suppose I follow this in the sense that both the functions are going to be close to 0, but I feel that there's a more nuanced reasoning for this approximation that I'm not following.

Or perhaps it is that simple and I'm just more accustomed to a more easily geometrically visible reason for an approximation?


2 Answers 2


I'm going to notate the partials with subscripts and restore the points of evaluation. The step you're having trouble with is then written $$T(f_z(z+\Delta z)-f_z(z))\approx Tf_{zz}(z)\Delta z.$$

This is just from the definition of the derivative! When you compare a function (here $f_z$) at two nearby points (here separated by $\Delta z$), you approximate the derivative of the function: $$f_{zz}(z)\approx\frac{f_z(z+\Delta z)-f_z(z)}{\Delta z},$$ with the approximation becoming better as $\Delta z\to0.$ Multiply the above line by $\Delta z,$ then plug it in to your expression.

There's no need for $f_z(z)$ or $f_z(z+\Delta z)$ to be near zero for this step. This step is justified by

  1. smallness of your choice of $\Delta z,$
  2. smoothness of (the displacement of) the string, i.e. that $f$ is $C^2$

This is simply using

$$\frac{\frac{\partial f}{\partial z}\big{|}_{z+\Delta z}-\frac{\partial f}{\partial z}\big{|}_{z}}{\Delta z}\approx\frac{\partial^2 f}{\partial z^2}$$ which follows from the definition of the derivative. For simplicity call $g = f'$ and the formal definition of the derivative is$$ \lim_{\Delta z\rightarrow 0}\frac{g(z+\Delta z)-g(z)}{\Delta z}=\frac{\partial g}{\partial z}=\frac{\partial^2 f}{\partial z^2}$$ and so Griffiths removed the limit and called the formula approximate (where the approximation is better and better the smaller $\Delta z$ is). I assume later he will take $\Delta z \rightarrow 0$.


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.