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I have learned that if a medium can transmit longitudinal waves and transverse waves, then the longitudinal wave will travel faster.

Why is this the case?

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  • 2
    $\begingroup$ I don't think this statement is even true. There are many types of waves in nature. $\endgroup$ – Ben Crowell Oct 10 '13 at 1:03
  • $\begingroup$ Im certain that EM waves travel in air faster than Sound waves. $\endgroup$ – Ubaid Hassan Apr 13 at 22:54
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Pretty much what "Ben" and "akhmeteli" said. In general, you can't say which will travel faster without specifying the system. I actually have an easily-visualizable discretized (rather than continuous) explanation of why this is the case. The utility of a discrete example is that it has a finite-dimensional space of modes and a discrete spectrum which can be easily computed and visualized.

Consider a chain of $\kappa$ particles connected by springs, where the unstressed springs have an equilibrium rest length of $L$. The chain is affixed to the ceiling and allowed to come to rest hanging down loosely. At rest, any small disturbances applied to the chain link positions admits a finite-dimensional vector representation in the eigenmode basis of the system. To that end, the eigenmodes of the system are computed and plotted. For example, here is the case of $\kappa=55$ and $L=0$ (picture is big, 1333 x 7261):

http://www.flickr.com/photos/104348204@N05/10180403584/sizes/o/

The eigenfrequencies are noted at the top of each mode, and they are sorted in descending order of frequency. The red node is attached to the ceiling, and the blue ones are free to move. The black lines indicate displacement vectors. Note that each mode is doubly degenerate, and the degenerate subspace admits a decomposition into a transverse and longitudinal component; the longitudinal ones are kind of hard to see because the displacement vectors point up and down the chain and all blob together, but if you rotate the displacement vector of the transverse one by 90 degrees, that's what the longitudinal one would look like if there was a better way of plotting it. The degeneracy pairing is apparent in the eigenspectrum:

http://www.flickr.com/photos/104348204@N05/10180521124/sizes/o/

So in this case, the longitudinal waves and transverse waves will propagate at the same speeds through the chain.

The scenario changes when $L\neq 0$. Here are the eigenmodes of the system for $L=100$ (units are a bit arbitrary):

http://www.flickr.com/photos/104348204@N05/10180600473/sizes/o/

And here is the corresponding eigenspectrum:

http://www.flickr.com/photos/104348204@N05/10180532024/sizes/o/

Something dramatic happens here. In this case the separation into longitudinal and transverse modes is preserved, but the double-degeneracy is lost. The longitudinal waves predominate at high frequency. The density of states shows a clear discontinuity, and the eigenspectrum looks like it's the sum of two distributions. In this type of chain, longitudinal disturbances propagate faster than transverse ones. What is the dependence on equilibrium spring rest length? The following photo demonstrates the eigenspectra of identical systems with $\kappa=100$ and various values of $L$:

http://www.flickr.com/photos/104348204@N05/10180551864/sizes/l/

For completeness, here is a decomposition of the case of $\kappa=100,L=50$ into longitudinal and transverse mode branches:

(http://www.flickr.com/photos/104348204@N05/10180571834/sizes/o/)

Note that the longitudinal branch generally has higher frequencies than the transverse branch.

All the previous examples had exactly the same spring constant between each link, but varied in their spring rest length.

So, now I've convinced you that in a dangling chain, longitudinal disturbances usually propagate faster than transverse ones, and the longer the rest length of the connecting springs is, the greater the discrepancy. Is there an intuitive explanation of why this is the case?

The answer is yes. For a long rest length spring, when it is at rest in the chain, a displacement sideways will tug the spring less than a displacement up and down (because of the angles involved). Thus the motion is "looser" in the transverse direction, and you'd expect things to vibrate slower in that manner. The longer the rest length is, the greater this discrepancy.

So how does this connect to the propagation velocities of sound in solids and other continuum materials? The transverse modes of a continuum material are like the sideways motions of the dangling chains, and the longitudinal modes are like the up-and-down motions of the chains. As "akhmetali" said, the shear modulus of materials is often less than the compressibility, so the longitudinal branch will predominate in terms of velocity.

One final note: I have neatly skipped over the reasoning as to why a higher average eigenspectrum frequency of a branch will mean than a displacement at one end of a chain will propagate to the other end at a higher speed when the disturbance is longitudinal. But with a little generalization of the thinking involved in Fourier analysis, you can represent the disturbance in the vector representation, and a little thinking will convince you of why the propagation speed will be faster in the longitudinal case.

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For example, in solids, the velocities of transverse and the longitudinal waves depend on the shear modulus and compressive modulus, respectively, and shear modulus is less than compression modulus (the velocities also depend on the density, which is the same for both types of waves).

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  • $\begingroup$ "shear modulus is less than compression modulus ". Always? $\endgroup$ – Keith McClary Oct 2 '15 at 4:26
  • $\begingroup$ @KeithMcClary: I don't know, maybe there are some exceptions, but typically - see en.wikipedia.org/wiki/… $\endgroup$ – akhmeteli Oct 2 '15 at 6:55
  • $\begingroup$ It seems to be stated here, p.265 . It would be nice to have a simpler "why" answer. $\endgroup$ – Keith McClary Oct 2 '15 at 16:40
  • $\begingroup$ @KeithMcClary: I don't know what level of that "why" answer you seek. My take is rather superficial: the expressions for speeds of longitudinal and transverse waves (say, in an isotropic elastic medium) are pretty well-known (they are derived, say, from the equations of the theory of elasticity), the relations between the moduli in those expressions are pretty well-known. I am afraid I cannot offer a deeper / more intuitive explanation. $\endgroup$ – akhmeteli Oct 2 '15 at 17:22
  • $\begingroup$ According to Mark Barton's answer on Quora the "P-wave modulus" (for pure compression in one direction only) is greater than the shear modulus (regardless of the bulk modulus). Maybe there is an intuitive reason for that. I will think about it for a few more months. $\endgroup$ – Keith McClary Jan 7 '16 at 6:34
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A little bit of completion to akhmeteli's answer:

In homogeneous and isotropic materials, we have that the stress tensor is $$\sigma_{ij}=2\mu \epsilon_{ij}+\lambda\delta_{ij}\epsilon_{kk}$$ The momentum equation is: $$\varrho\frac{\partial^2 \vec{u}}{\partial t^2}=\text{Div}(\sigma)$$ Where $$(\text{Div}(\sigma))_i=\partial_j\sigma_{ij}$$ $$=\partial_j(2\mu \epsilon_{ij}+\lambda\delta_{ij}\epsilon_{kk})$$ $$=\partial_j(\mu (\partial_i u_j+\partial_j u_i)+\lambda\delta_{ij}\partial_ku_k)$$ $$=\mu \partial_i \partial_j u_j+\mu \partial_j \partial_j u_i+\lambda\partial_j\delta_{ij}\partial_ku_k$$ $$=\mu \partial_i \partial_j u_j+\mu \partial_j \partial_j u_i+\lambda\partial_i\partial_ku_k$$ $$=\mu \partial_i \partial_j u_j+\mu \partial_j \partial_j u_i+\lambda\partial_i\partial_ju_j$$ $$=(\mu+\lambda) \partial_i \partial_j u_j+\mu \partial_j \partial_j u_i$$ $$=(\mu+\lambda) (\text{grad}(\text{div}(\vec{u})))_i+\mu (\Delta \vec{u})_i$$ So the momentum equation is: $$\varrho \frac{\partial^2 \vec{u}}{\partial t^2}=(\mu+\lambda) \text{grad}(\text{div}(\vec{u}))+\mu \Delta \vec{u}$$ If we substitute in the planewave $$\vec{u}=\vec{a}e^{-i \omega t}e^{i \vec{k}\vec{r}}$$ After simplification, we will get that $$\varrho \vec{a}\omega^2=\mu(\vec{k}\vec{k})\vec{a}+(\mu+\lambda)\vec{k}(\vec{k}\vec{a})$$ If we let now $\vec{k}=k\vec{n}$ with $|\vec{n}|=1$, and we divide by $k^2$, we get that $$\varrho \vec{a}\frac{\omega^2}{k^2}=\mu(\vec{n}\vec{n})\vec{a}+(\mu+\lambda)\vec{n}(\vec{n}\vec{a})$$ $$\varrho \vec{a}\frac{\omega^2}{k^2}=\mu\vec{a}+(\mu+\lambda)\vec{n}(\vec{n}\vec{a})$$ The $\vec{n}(\vec{n}\vec{a})$ is the projection of $\vec{a}$ to $\vec{n}$, so if we let $P$ be the projection matrix to $\vec{n}$, we can write it as: $$\varrho \vec{a}\frac{\omega^2}{k^2}=\mu\vec{a}+(\mu+\lambda)P\vec{a}$$ Now let $Q$ be another projector, with $Q+P=I$. If we factor out $\vec{a}$ we get that: $$\varrho \vec{a}\frac{\omega^2}{k^2}=(\mu I+(\mu+\lambda)P)\vec{a}$$ $$\varrho \vec{a}\frac{\omega^2}{k^2}=(\mu (P+Q)+(\mu+\lambda)P)\vec{a}$$ And using that $\frac{\omega^2}{k^2}$ is the phase velocity squared: $$\varrho \vec{a}V^2=(\mu Q+(2\mu+\lambda)P)\vec{a}$$ And finally: $$V^2 \vec{a}=\left(\frac{\mu}{\varrho} Q+\frac{2\mu+\lambda}{\varrho}P\right)\vec{a}$$ Now, we can notice that the matrix on the right side is in it's projector decomposition, so it's eigenvalues are the constants before the projectors: $\frac{\mu}{\varrho}$ and $\frac{2\mu+\lambda}{\varrho}$. Because $P$ projects to the direction of it's propagation, we can conclude that $\frac{2\mu+\lambda}{\varrho}$ is the square speed of the longtudinal wave, and because the $Q$ projects to the plane perpendicular to it, $\frac{\mu}{\varrho}$ will be the square speed of the transverse wave.

Because $\mu$ and $\varrho$ are always positive, and $\lambda$ is positive for almost all materials (see Lamé parameters), we can conclude that $2\mu+\lambda \geq \mu$, so $c_{\text{longitudinal}}^2\geq c_{\text{transverse}}^2$, and finally $$c_{\text{longitudinal}}\geq c_{\text{transverse}}$$

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  • $\begingroup$ You could amend that positive definiteness of the stiffness requires $\lambda>-2\mu/3$, but $c_{\rm long}=c_{\rm trans}$ requires $\lambda=-\mu$. Hence isotropic materials that have equal longitudinal and transversal propagation speeds cannot exist. $\endgroup$ – Rainer Glüge Jun 6 at 8:23
  • $\begingroup$ @RainerGlüge Thank you, I didn't know it! $\endgroup$ – Botond Jun 6 at 10:54

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