I'm currently reading a PDE book, and it's deriving the wave equation.

The book begins by stating that the net transverse tension force in a small segment is:

$\sum T = T\sin(\theta_2)-T\sin(\theta_1)$

Which makes sense. However, it goes on to say that

$\sum T = T\sin(\theta_2)-T\sin(\theta_1) \cong T(u_x(x_1+dx,t)-u_x(x_1,t))$

What I don't understand is that this implies $\sin(\theta_2)=u_x(x_2)$.

I can see how this could be true for small $x$, as I know $\lim_{x\to0} \sin(x)=x$, but what am I missing here? Maybe it's something to do with the $\cong$ sign.

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  • $\begingroup$ Is it because over a small change in $x$, $dx$, the change in $θ, dθ$ must also be very small, showing that the approximation $\sin(\theta_0) = \frac{\partial u}{\partial x}\bigg|_{x=x_0}$ is valid? $\endgroup$
    – Jon Snow
    Dec 24 '15 at 2:58

When $x$ changes by a very small amount, the curve you've designated as $dm$ is almost a straight line. In that limit, the two angles $\theta_1$ and $\theta_2$ are very small. Therefore, $\tan\theta_1 \approx \sin\theta_1 \approx \theta_1$ and similarly for $\theta_2$.

Once we agree with the approximation $\theta_i \approx \tan\theta_i$, ($i = 1, 2$), we use the fact that the tangent at a point is just the derivative. Therefore, $\theta_i \approx u_x(x_i, t)$.

  • $\begingroup$ I don't think the angles $\theta_1$ and $\theta_2$ have to be small even if $dm$ were a straight line. I do understand what you're saying though, but maybe it's because $\Delta\theta$ is very small, rather than the individual $\theta$s? $\endgroup$
    – Jon Snow
    Dec 24 '15 at 4:35

From the figure we see that $\frac{\Delta u}{\Delta x}\approx \tan \theta(x,t)$ which is exact in the limit where $\Delta x\to 0$:

$$\tan \theta(x,t) = \lim_{\Delta x \to 0} \frac{\Delta u}{\Delta x} = \frac{\partial u(x,t)}{\partial x}$$

I here use a notation where $\theta_1 = \theta(x_1,t)$ and $\theta_2 = \theta(x_2,t)$. Using the relation $\sin z = \frac{\tan z}{\sqrt{1+\tan^2 z}}$ it follows that

$$\sin(\theta(x,t)) = \frac{\frac{\partial u(x,t)}{\partial x}}{\sqrt{1+\left(\frac{\partial u(x,t)}{\partial x}\right)^2}}$$

Now as long as the vibrations are small enough in the sense that $\frac{\partial u(x,t)}{\partial x} \ll 1$, or equivalently $|\theta(x,t)| \ll 1$, we have $\sqrt{1+\left(\frac{\partial u(x,t)}{\partial x}\right)^2} \approx 1$ so

$$\sin(\theta(x,t)) \simeq \frac{\partial u(x,t)}{\partial x}$$

This leads to the desired result

$$T_{\rm transverse} = T\sin(\theta(x+\Delta x,t)) - T\sin(\theta(x,t)) \approx T\left(\frac{\partial u(x +\Delta x,t)}{\partial x} - \frac{\partial u(x,t)}{\partial x}\right) \approx T \Delta x \frac{\partial^2 u}{\partial x^2}$$

To finish the derivation we use Newton's second law $T_{\rm transverse} = \Delta m \frac{\partial^2 u}{\partial t^2}$ where $$\Delta m = \rho\sqrt{(\Delta x)^2 + (\Delta u)^2} \approx \rho \sqrt{1 + \left(\frac{\Delta u}{\Delta x}\right)^2} \Delta x \approx \rho \Delta x$$

where we have used the same approximation, $\frac{\partial u(x,t)}{\partial x} \ll 1$, as above. Putting it togeather it follows that $$\frac{\partial^2 u}{\partial t^2} \approx v^2 \frac{\partial^2 u}{\partial x^2}$$ where $v = \sqrt{\frac{T}{\rho}}$. A more detailed (and general) derivation can be found here.


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