How can transverse waves on a string carry longitudinal momentum? In general, if a wave carries energy density $u$ with velocity $v$, it also carries momentum density $u/v$. I've seen this explicitly shown for electromagnetic waves and (longitudinal) sound waves.
However, I'm having trouble seeing how the momentum density of a transverse wave on a string can be anything besides zero. The string elements only ever move up or down, so they can't have longitudinal momentum. And if you compute the force  on any small piece of string, its net horizontal component is zero. These suggest the momentum density and flow of momentum density are both zero.
I realize that accounting for higher-order effects, such as making the wave not purely transverse, or having nonuniform stretching of the string, can produce a longitudinal momentum. But if these effects are included, the waves won't satisfy the ideal wave equation. Maybe we have to account for funny nonlinear effects to get the right answer here, but we don't have to do this for other kinds of waves. So I don't think this approach is right, unless transverse string waves are somehow unique.
How do you find the momentum density of a transverse wave on a string? What approximations, if any, need to be removed?
 A: You are absolutely right in everything you said. The momentum is non zero only if the wave has a longitudinal mode, which is in fact the realistic case. Moreover when this is the case, the wave equation is not that simple. Let me try show this.
Longitudinal Mode
Let us assume that when in equilibrium the string, of density $\mu$, is along with the $x\equiv x_1$ axis and has tension $T_0$. The general displacement of the string is
$$\vec \xi(x_1,t)=\xi_1(x_1,t)\vec e_1+\xi_2(x_1,t)\vec e_2\equiv\xi_i(x_1,t)\vec e_i.$$
A small section $ds$ of the string is acted by a force
$$d\vec T=\frac{\partial\vec T}{dx_1}dx_1=\mu dx_1\frac{\partial^2\vec\xi}{dt^2},$$
by Newton's second law. The magnitude of $\vec T$ is $T_0$ plus an increment proportional to the stretched amount 
$$\frac{ds-dx_1}{dx_1}=\frac{ds}{dx_1}-1.$$
If the string has cross-sectional area $A$ and Young modulus $Y$, the increment in tension is
$$AY\left(\frac{ds}{dx_1}-1\right).$$ 
Hence
$$\vec T=\left[T_0+AY\left(\frac{ds}{dx_1}-1\right)\right]\frac{d\vec s}{ds}$$ 
where $d\vec s$ is directed along $ds$. We have
$$d\vec s=(d\xi_1+dx_1)\vec e_1+d\xi_2\vec e_2,$$
$$ds=\sqrt{\left(1+\frac{\partial\xi_1}{\partial x_1}\right)^2+\left(\frac{\partial\xi_2}{\partial x_1}\right)^2}dx_1$$
As you can see, when we plugging $\vec T$ back into the equation of motion we get three non linear and coupled partial differential equation (Though isn't it?). To simplify, let us assume small displacements, i.e.
$$\frac{\partial\xi_1}{\partial x_1}\approx\frac{\partial\xi_2}{\partial x_1}\ll 1.$$
The goal now is to expand $\vec T$ up to first order in $\frac{d\xi_i}{dx_1}$. First notice that
$$\frac{ds}{dx_1}-1=\frac{\partial\xi_1}{\partial x_1}+O((\partial\xi_1/\partial x_1)^2).$$
Then
\begin{align}
\vec T&\approx\frac{\left(T_0+AY\frac{\partial\xi_1}{\partial x_1}\right)\left(\vec e_1+\frac{\partial\xi_i}{\partial x_i}\vec e_i\right)}{\sqrt{\left(1+\frac{\partial\xi_1}{\partial x_1}\right)^2+\left(\frac{\partial\xi_2}{\partial x_2}\right)^2}},\\
&\approx\left(1-\frac{\partial\xi_1}{\partial x_1}\right)\left(T_0+AY\frac{\partial\xi_1}{\partial x_1}\right)\left(\vec e_1+\frac{\partial\xi_i}{\partial x_i}\vec e_i\right),\\
&\approx\left(T_0+AY\frac{\partial\xi_1}{\partial x_1}\right)\vec e_1+T_0\frac{\partial\xi_2}{\partial x_1}\vec e_2.
\end{align}
Plugging this back into the equation of motion we get two wave equation,
$$\frac{\partial^2\xi_i}{\partial t^2}=c_i^2\frac{\partial^2\xi_i}{\partial x_1^2},$$
whose speeds are
$$c_1=\sqrt{AY/\mu},\quad c_2=\sqrt{T_0/\mu}.$$
Notice that if the Young (elastic) coefficient of the string is neglected, then the longitudinal mode disappear. Also notice that the speed of the longitudinal wave is in general greater than the speed of the transverse wave since typical values of the Young modulus is in general large, $Y\sim 10^{9}\, Pa$. For a string of area $A\sim 10^{-4}\, m^2$ we get
$$\frac{c_1}{c_2}\sim\sqrt{\frac{10^5}{T_0}}.$$
Longitudinal Momentum
In this post it is computed the potential energy density of a string (remember $\xi_2$ is the transverse displacement),
$$U=\frac{T_0dx}{2}\left(\frac{\partial \xi_2}{\partial x_1}\right)^2.$$
Then the "force density" in the longitudinal direction is
$$f_1=-\frac{\partial U}{\partial x_1}=-T_0\frac{\partial \xi_2}{\partial x_1}\frac{\partial^2 \xi_2}{\partial x_1^2}.$$
Let us call $p_1$ the "momentum density" in the longitudinal direction. Then
$$\frac{dp_1}{dt}=-T_0\frac{\partial \xi_2}{\partial x_1}\frac{\partial^2 \xi_2}{\partial x_1^2}=-\mu\frac{\partial \xi_2}{\partial x_1}\frac{\partial^2\xi_2}{\partial t^2},$$
where we used the wave equation. Integrating by parts (in time) we finally get the momentum density
$$p_1=-\mu\frac{\partial \xi_2}{\partial x_1}\frac{\partial\xi_2}{\partial t}.$$
A: This perennial problem is due to a failure to distinguish between Newtonian momentum (the conserved quantity obtained via Noethers theorem from the invariance of the system under a simultaneous translation of the string and any waves on it) and and pseudomomentum (the conserved quantity obtained via Noethers theorem from an  invariance of the system under the translation of the  waves, while the string itself is not translated) Pseudomomentum is given by $-\rho \partial_x y \partial_t y$. It is conserved only if the density of the string is independent of $x$. The consservation of pseudomentum can be derived from the wave equation which requires no knowledge of eleastic constants such as Young's modulus. 
Any real disturbance of the string will also excite longitudinal waves that travel at a speed that depends on Young's modulus. This is  explained in  the paper of McIntyre mentioned above. It is  also discussed by Peierls in his "surprises in theoretical physics" book undeer the heading "what is the momentum of a phonon." It turns out that pseudomentum is more useful than actual Newtonian momentum, as it is changes in pseudomentum that correspond to forces.
See my paper "Phonons and Forces: Momentum versus Pseudomomentum in Moving Fluids,"  arXiv:cond-mat/0012316.
A: A fake derivation
We can rather easily compute a horizontal velocity for the string fi we assume that the total velocity vector is everywhere normal to the string (this assumption is not always valid, see below). The following picture then illustrates the computation:

Take two infinitesimally separated points $x$ and $x+\mathrm{d}x$ and let the wave motion be $\varphi(x,t)$. The vertical/transverse velocity is $v_\text{vert} = \partial_t \varphi(x,t)$, and the horizontal component is $v_\text{hor} = -v_\text{vert}\tan(\vartheta)$, where $\vartheta$ is the angle between the normal and the vertical, and the minus sign is because if we measure $\vartheta$ in the usual counterclockwise direction then the horizonal velocity points to $-x$ for small $\vartheta$. Now $\tan(\vartheta)$ is $\frac{\varphi(x+\mathrm{d}x) - \varphi(x)}{\mathrm{d}x} = \partial_x\varphi(x)$ , so we get
$$ v_\text{hor} = -\partial_t\varphi\partial_x\varphi$$
and if you plug in the sinusoidal solution and take the time average you get exactly the same result as for longitudinal waves. However, you might protext - the transverse wave equation was derived assuming no longitudinal motion, and this computation just blatantly assumes something different.
A Lagrangian derivation
Oddly enough, the result of the above computation is the correct momentum for a pure transverse wave. The Lagrangian of a transverse wave is
$$ L = \frac{1}{2}\rho (\partial_t\varphi)^2 - \frac{1}{2}\tau(\partial_x\varphi)^2$$
and translation invariance gives us a momentum density
$$ T_{xt} = \partial_x L \partial_t \varphi = - \rho\partial_x\varphi\partial_t\varphi$$
which is conserved by Noether's theorem.
The actual answer
In reality, there are no purely transverse waves on a string, there will always be secondary longitudinal waves generated when trying to excite it purely transversely. The "true" momentum of a realistic "transverse" wave is rather half of the theoretical prediction, i.e. $\frac{1}{2}\rho\partial_t\varphi\partial_x\varphi$, for more on this see "The missing wave momentum mystery"[pdf link] by Rowland and Pask.
