<p>Remember that a wave when it passes moves every particle up and down (with no horizontal component) so in deep water tsunamis have very little affect on anything.  In addition, the wavelength/amplitude of these waves is huge so the water rises very gently and it is hardly noticed in deep water. In fact, the safest place to be when a tsunami is coming (beside high ground) is out at sea in deep water.  </p>

CC BY-SA 3.0
2015-07-04T19:34:50.383Z
I thought that particles would move in a circular/ellipsoidal fashion like this https://upload.wikimedia.org/wikipedia/commons/8/8b/Deep_water_wave_after_three_periods.png
reddit
CC BY-SA 3.0
2015-07-04T21:22:50.350Z
I think the confusion is how the wave is generated.  A sinusoidal wave passing through water is not going to have a net horizontal movement (e.g., when a wave from a tsunami passes through water).  However, waves generated by friction from wind blowing on the surface do generate some net horizontal movement.  But your question relates to the affect of life in the oceans as a tsunami wave passes by.  I tell my students that descriptions of tsunami's passing on the ocean under boats are rarely felt because the wavelength/amplitude is so huge.  Basically life is unaffected.
see you
CC BY-SA 3.0
2015-07-05T05:49:52.523Z
@MarcDefant But if it is sinusoidal it makes me think even more strongly that the path should be circular/ellipsoidal (does not mean there is net forward transport). Wouldnt biology follow that orbital path as well?
reddit
CC BY-SA 3.0
2015-07-05T05:52:21.653Z
Interesting question and instructive answer! From what I understand of linear (sinusoidal) waves, the water particles may have a small non-zero *instantaneous* horizontal velocity (as part of the circular motion reddit referred to), but the *net* or *average* velocity is zero, as Marc Defant pointed out.
user4624937
CC BY-SA 3.0
2015-07-05T07:16:09.690Z
As noted in other comments, this answer is wrong. Particle motion in a transverse wave follows an elliptical pattern, and I have no reason to think that a tsunami would be different. It is true that the *net* horizontal motion will be close to zero (but then again, so will the net vertical motion), but this does not mean there is no affect on the water column or the seabed. I imagine that there *is* little effect because, as noted, in deep water tsunamis have low amplitude - but I'm not expert enough to say this for sure.
Semidiurnal Simon
CC BY-SA 3.0
2015-07-05T23:34:42.807Z
The bit about wave motion in this answer is wrong. Particle motion in a horizontally moving transverse wave is entirely vertical (so Simon W's remark is incorrect too). But ocean waves (including tsunamis) are not transverse waves. For one thing, [fluids do not support transverse waves](http://physics.stackexchange.com/questions/164491/why-cant-transverse-waves-travel-through-a-liquid). Another clue to the sub-circular motion is in [the withdrawal of the water](http://earthscience.stackexchange.com/questions/420/why-does-the-shoreline-sometimes-recede-prior-to-a-tsunami) before a tsunami.
Matt Hall
CC BY-SA 3.0
2015-07-05T23:37:53.263Z
Sorry for another comment but I ran out of characters. I just wanted to add that the wavelength of tsunamis in deep water is indeed large, but the amplitude is not. I could edit the answer I guess, but given the other problem (see my other comment) I think I'll let Marc do it.
Matt Hall
CC BY-SA 3.0
2015-07-06T15:08:14.080Z
-1 This answer is wrong. In an Eulerian framework, water particle has both horizontal and vertical non-zero velocity that averages to zero over the period of the wave. In a Lagrangian framework (following the wave particle), orbits are not closed and there is net movement of mass in the direction of the wave (Stokes drift). Vertical velocity averages to zero in a Lagrangian framework, but there exists a vertical pseudo-velocity due to divergence in the horizontal component of Stokes drift.
milancurcic
CC BY-SA 3.0
2015-07-04T13:49:53.883Z
2015-07-04T13:49:53.883Z
see you
<p>The answer to your question, based on linear theory, is no. The short answer is that the slope of these waves is very small, and the displacement of the particle paths is proportional to (ak) (times, perhaps, a depth dependent term) with a the wave amplitude and k its wavenumber. Now let's make this answer rigorous. </p>

<p>Recall, for irrotational inviscid 2 dimensional water waves the governing equation is</p>

<p>$$\nabla^2 \phi = 0$$</p>

<p>where $\nabla^2 \equiv \partial_{xx}+\partial_{zz}$ and $\phi$ is the velocity potential. We require particles on the free surface to remain there, and for the pressure to be continuous across the air-water interface, which gives us (assuming the atmospheric pressure to be 0) our surface boundary conditions </p>

<p>$$\left.\eta_t+\phi_x\eta_x = \phi_z\right|_{z=\eta}; \quad \quad \left. \phi_t+\frac{1}{2}(\nabla \phi)^2 +gz=0\right|_{z=\eta},$$</p>

<p>where $\eta(x,t)$ is the free surface displacement. Finally, we require the bottom to be impermeable, hence</p>

<p>$$\phi_z = 0 \quad @ z= -h$$</p>

<p>for $h$ the depth of the water. These are a closed set of equations. Note, the governing equation, ie Laplace's equation, is linear and hence simple to solve. However, the free surface conditions are nonlinear, and more significantly, evaluated at a moving interface. This last condition makes these equations very difficult to solve. </p>

<p>Therefore, to make progress, we generally take asymptotic expansions of these equations. Here, we'll just linearize the equations about some small parameter $\epsilon = ak$ for characteristic wave amplitude $a$ and wavenumber $k$. We will also consider periodic monochromatic waves, so that one can show that the following $\eta,\phi$ satisfy the governing equations to first order in $\epsilon$: </p>

<p>$$\eta = a \cos \theta; \quad \phi = a\omega \sin \theta \frac{\sinh k(z+h)}{\sinh kh},$$</p>

<p>where $\omega$ is the angular frequency given by the dispersion relationship</p>

<p>$$\omega = \sqrt{gk \tanh kh},$$
and $\theta\equiv kx-\omega t$. </p>

<p>To answer your question, we would like to know the behavior of a particle with position $(\xi+x_o,\zeta+z_o)$ originally located at some position $(x_o,z_o)$.This is the Lagrangian description of the fluid. Recall, we can relate the Eulerian and Lagrangian descriptions of a fluid through the velocity field, that is </p>

<p>$$\phi_x = \frac{d \xi}{d t}; \quad \phi_z = \frac{d\zeta}{dt}.$$</p>

<p>Substituting in the relations for $\phi$ found above, and then integrating in time, we find </p>

<p>$$\xi =-a\frac{\cosh k(z_o+h)}{\sinh kh} \sin \theta_o; \quad \quad \zeta = a\frac{\sinh k(z_o+h)}{\sinh kh} \cos \theta_o,$$
where $\theta_o =kx_o - \omega t$. We note that the particle displacements from the origin scales with the wave amplitude $a$. </p>

<p>Now, we can eliminate $\theta_o$ by squaring the expressions for the particle position to find </p>

<p>$$\frac{\xi^2}{\left(a\frac{\cosh k(z_o+h)}{\sinh kh} \right)^2 } +\frac{\zeta^2}{\left(a\frac{\sinh k(z_o+h)}{\sinh kh} \right)^2 }= 1$$</p>

<p>which describes an ellipse. I've attached a figure from Kundu and Cohen which illustrates this. <img src="https://i.sstatic.net/cdq5k.png" alt="enter image description here"></p>

<p>Now we can finally talk about some physics, and answer your question. For deep water waves, when $kh \ll 1$, the above relation reduces to the equation for a circle, hence the particles have a circular orbit which decays with depth (with an e-folding scale that goes like the wave number). For shallow water waves, i.e.  $kh$ small, our equation describes an ellipse with a semi major axis that is much much larger than the semi minor axis, hence there's is nearly horizontal motion that doesn't decay much with depth. </p>

<p>Tsunamis have wavenumbers that are large enough so that even in the middle of the ocean they fall into the shallow water regime, meaning that the particle trajectories are approximately very flat ellipses, or horizontal lines, and this does not decay with depth. However, their deviations from their origin will be very small, as their slope $ak$ is very very small, e.g. $\mathcal{O}(1^{-5})$ for numbers I looked up online. Therefore, it's very unlikely that this will have much of an effect on a passive tracer.</p>

<p>Now, the story is very different as the wave approaches shore and steepens. But this is a different story for a different day. </p>

<p>I hope this helps. Feel free to ask any other questions,</p>

<p>Nick</p>

CC BY-SA 3.0
2015-07-05T22:40:08.990Z
2015-07-06T16:44:30.757Z
2015-07-06T16:44:30.757Z
Nick P
<p>Is everything in the water column going to be affected?</p>

<ocean><waves><tsunami>
How does a tsunami affect biology and other matter in its path while in deep water
reddit
# How does a tsunami affect biology and other matter in its path while in deep water

(Asked by: username_0 on 2015-07-04)

Is everything in the water column going to be affected?

**Tags:** ocean, waves, tsunami


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## 2 answers

### Answer by username_6 on 2015-07-05. Score: 7 | Accepted answer

The answer to your question, based on linear theory, is no. The short answer is that the slope of these waves is very small, and the displacement of the particle paths is proportional to (ak) (times, perhaps, a depth dependent term) with a the wave amplitude and k its wavenumber. Now let's make this answer rigorous.

Recall, for irrotational inviscid 2 dimensional water waves the governing equation is

$$\nabla^2 \phi = 0$$

where $\nabla^2 \equiv \partial\_{xx}+\partial\_{zz}$ and $\phi$ is the velocity potential. We require particles on the free surface to remain there, and for the pressure to be continuous across the air-water interface, which gives us (assuming the atmospheric pressure to be 0) our surface boundary conditions

$$\left.\eta\_t+\phi\_x\eta\_x = \phi\_z\right|\_{z=\eta}; \quad \quad \left. \phi\_t+\frac{1}{2}(\nabla \phi)^2 +gz=0\right|\_{z=\eta},$$

where $\eta(x,t)$ is the free surface displacement. Finally, we require the bottom to be impermeable, hence

$$\phi\_z = 0 \quad @ z= -h$$

for $h$ the depth of the water. These are a closed set of equations. Note, the governing equation, ie Laplace's equation, is linear and hence simple to solve. However, the free surface conditions are nonlinear, and more significantly, evaluated at a moving interface. This last condition makes these equations very difficult to solve.

Therefore, to make progress, we generally take asymptotic expansions of these equations. Here, we'll just linearize the equations about some small parameter $\epsilon = ak$ for characteristic wave amplitude $a$ and wavenumber $k$. We will also consider periodic monochromatic waves, so that one can show that the following $\eta,\phi$ satisfy the governing equations to first order in $\epsilon$:

$$\eta = a \cos \theta; \quad \phi = a\omega \sin \theta \frac{\sinh k(z+h)}{\sinh kh},$$

where $\omega$ is the angular frequency given by the dispersion relationship

$$\omega = \sqrt{gk \tanh kh},$$
and $\theta\equiv kx-\omega t$.

To answer your question, we would like to know the behavior of a particle with position $(\xi+x\_o,\zeta+z\_o)$ originally located at some position $(x\_o,z\_o)$.This is the Lagrangian description of the fluid. Recall, we can relate the Eulerian and Lagrangian descriptions of a fluid through the velocity field, that is

$$\phi\_x = \frac{d \xi}{d t}; \quad \phi\_z = \frac{d\zeta}{dt}.$$

Substituting in the relations for $\phi$ found above, and then integrating in time, we find

$$\xi =-a\frac{\cosh k(z\_o+h)}{\sinh kh} \sin \theta\_o; \quad \quad \zeta = a\frac{\sinh k(z\_o+h)}{\sinh kh} \cos \theta\_o,$$
where $\theta\_o =kx\_o - \omega t$. We note that the particle displacements from the origin scales with the wave amplitude $a$.

Now, we can eliminate $\theta\_o$ by squaring the expressions for the particle position to find

$$\frac{\xi^2}{\left(a\frac{\cosh k(z\_o+h)}{\sinh kh} \right)^2 } +\frac{\zeta^2}{\left(a\frac{\sinh k(z\_o+h)}{\sinh kh} \right)^2 }= 1$$

which describes an ellipse. I've attached a figure from Kundu and Cohen which illustrates this. ![enter image description here](https://i.sstatic.net/cdq5k.png)

Now we can finally talk about some physics, and answer your question. For deep water waves, when $kh \ll 1$, the above relation reduces to the equation for a circle, hence the particles have a circular orbit which decays with depth (with an e-folding scale that goes like the wave number). For shallow water waves, i.e. $kh$ small, our equation describes an ellipse with a semi major axis that is much much larger than the semi minor axis, hence there's is nearly horizontal motion that doesn't decay much with depth.

Tsunamis have wavenumbers that are large enough so that even in the middle of the ocean they fall into the shallow water regime, meaning that the particle trajectories are approximately very flat ellipses, or horizontal lines, and this does not decay with depth. However, their deviations from their origin will be very small, as their slope $ak$ is very very small, e.g. $\mathcal{O}(1^{-5})$ for numbers I looked up online. Therefore, it's very unlikely that this will have much of an effect on a passive tracer.

Now, the story is very different as the wave approaches shore and steepens. But this is a different story for a different day.

I hope this helps. Feel free to ask any other questions,

username_6


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### Answer by username_1 on 2015-07-04. Score: 1

Remember that a wave when it passes moves every particle up and down (with no horizontal component) so in deep water tsunamis have very little affect on anything. In addition, the wavelength/amplitude of these waves is huge so the water rises very gently and it is hardly noticed in deep water. In fact, the safest place to be when a tsunami is coming (beside high ground) is out at sea in deep water.

> **Comments:**
> **username_0**: I thought that particles would move in a circular/ellipsoidal fashion like this https://upload.wikimedia.org/wikipedia/commons/8/8b/Deep_water_wave_after_three_periods.png  
> **username_1**: I think the confusion is how the wave is generated.  A sinusoidal wave passing through water is not going to have a net horizontal movement (e.g., when a wave from a tsunami passes through water).  However, waves generated by friction from wind blowing on the surface do generate some net horizontal movement.  But your question relates to the affect of life in the oceans as a tsunami wave passes by.  I tell my students that descriptions of tsunami's passing on the ocean under boats are rarely felt because the wavelength/amplitude is so huge.  Basically life is unaffected.  
> **username_0**: @MarcDefant But if it is sinusoidal it makes me think even more strongly that the path should be circular/ellipsoidal (does not mean there is net forward transport). Wouldnt biology follow that orbital path as well?  
> **username_2**: Interesting question and instructive answer! From what I understand of linear (sinusoidal) waves, the water particles may have a small non-zero *instantaneous* horizontal velocity (as part of the circular motion username_0 referred to), but the *net* or *average* velocity is zero, as Marc Defant pointed out.  
> **username_3**: As noted in other comments, this answer is wrong. Particle motion in a transverse wave follows an elliptical pattern, and I have no reason to think that a tsunami would be different. It is true that the *net* horizontal motion will be close to zero (but then again, so will the net vertical motion), but this does not mean there is no affect on the water column or the seabed. I imagine that there *is* little effect because, as noted, in deep water tsunamis have low amplitude - but I'm not expert enough to say this for sure.  
> **username_4**: The bit about wave motion in this answer is wrong. Particle motion in a horizontally moving transverse wave is entirely vertical (so username_3 W's remark is incorrect too). But ocean waves (including tsunamis) are not transverse waves. For one thing, [fluids do not support transverse waves](http://physics.stackexchange.com/questions/164491/why-cant-transverse-waves-travel-through-a-liquid). Another clue to the sub-circular motion is in [the withdrawal of the water](http://earthscience.stackexchange.com/questions/420/why-does-the-shoreline-sometimes-recede-prior-to-a-tsunami) before a tsunami.  
> **username_4**: Sorry for another comment but I ran out of characters. I just wanted to add that the wavelength of tsunamis in deep water is indeed large, but the amplitude is not. I could edit the answer I guess, but given the other problem (see my other comment) I think I'll let Marc do it.  
> **username_5**: -1 This answer is wrong. In an Eulerian framework, water particle has both horizontal and vertical non-zero velocity that averages to zero over the period of the wave. In a Lagrangian framework (following the wave particle), orbits are not closed and there is net movement of mass in the direction of the wave (Stokes drift). Vertical velocity averages to zero in a Lagrangian framework, but there exists a vertical pseudo-velocity due to divergence in the horizontal component of Stokes drift.  

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