# 3.8.1: Dynamic theory of tides


So far, the earth was schematised as if it were completely covered with water, with the earth rapidly turning through the slowly varying tide. For the semi-diurnal tide this is equivalent to a propagating tidal wave covering the entire circumference of the earth in one day. In reality, continents prevent the development of the tidal ellipsoid and the land masses do not move through the tide but move the water masses along with them. Also the limited water depth of the oceans prevents the development of an equilibrium tide. This can be realised when considering the propagation properties of the tidal wave. The tidal wave is a long wave because the wavelength $$L \gg h$$ and has a small amplitude (order 1 m on an average ocean water depth of 4000 m). Therefore and if friction can be neglected, the wave propagation speed follows from (see Sect. 3.5.2):

$c = \sqrt{gh}\label{eq3.8.1.1}$

where

 $$c$$ wave propagation speed $$m/s$$ $$g$$ gravitation acceleration $$m/s^2$$ $$h$$ water depth $$m$$

Note that this phase velocity can also be derived directly on the basis of continuity and by balancing acceleration and pressure gradient, see Intermezzo 3.5.

## Intermezzo 3.5 Tidal propagation in the open oceans

The tide can be described by the shallow water equations (shallow water approximation to the Navier-Stokes equations). For the tidal propagation in open oceans we can simplify these equations by neglecting advection, friction, horizontal diffusion and short wave effects. We also assume a small tidal amplitude and bottom slope. The resulting momentum equation is a balance between acceleration (inertia) and the pressure gradient:

$\dfrac{\partial u}{\partial t} = -g \dfrac{\partial \eta}{\partial x}\label{eq3.8.1.2}$

where $$u$$ is the horizontal velocity and $$\eta$$ the oscillatory surface elevation. We further have the vertically integrated continuity equation:

$\dfrac{\partial \eta}{\partial t} = -h \dfrac{\partial u}{\partial x}\label{eq3.8.1.3}$

where h is the water depth. Eqs. $$\ref{eq3.8.1.2}$$ and $$\ref{eq3.8.1.3}$$ together describe the tidal propagation.

Taking $$\tfrac{\partial}{\partial t}$$ of Eq. $$\ref{eq3.8.1.3}$$ and subtracting $$h \tfrac{\partial}{\partial x}$$ of Eq. $$\ref{eq3.8.1.2}$$ gives the classical wave equation:

$\dfrac{\partial^2 \eta}{\partial t^2} = gh \dfrac{\partial^2 \eta}{\partial x^2}\label{eq3.8.1.4}$

Substitution of a progressive sine (or equivalently: cosine) wave $$\eta = a\sin (\omega t - kx)$$ into Eq. $$\ref{eq3.8.1.4}$$ yields:

$\omega^2 = ghk^2$

and thus (only taking the positive solution for $$c$$):

$c = \dfrac{\omega}{k} = \sqrt{gh}$

The velocity is in phase with the surface elevation:

$u = \dfrac{gak}{\omega} \sin (\omega t - kx)$

Evidently, for progressive waves the maximum velocities are found under the crest of the wave. The velocity amplitude is:

$\hat{u} = \dfrac{gak}{\omega} = \dfrac{a\omega}{kh} = a \sqrt{\dfrac{g}{h}}\label{eq3.8.1.8}$

Compare this expression with the velocity amplitude according to linear wave theory in the shallow water approximation (Eq. 5.4.1.2). As you can see it is identical.

For the development of the equilibrium tide, the semi-diurnal tide should cover half of the circumference of the earth in one period of 12 hours and 25 minutes. The circumference at the equator is $$2\pi R = 2\pi \times 6.37 \times 10^3\ km = 4.00 \times 10^4 \ km$$ and therefore the required propagation speed is

$c = 4.00 \times 10^4 \ km /(2 \times 12.42h) = 447\ m/s.$

This means that the tidal wave requires a water depth of 20 km – much more than the depth of the oceans – to travel fast enough (use Eq. $$\ref{eq3.8.1.1}$$). The travel distance reduces towards the poles and therefore the required propagation speed and water depth as well. The result is that the water depth becomes less of a limitation at higher latitudes.

Only in the Southern Hemisphere at the latitude of about 65°S, an equilibrium tide can more or less exist. To the south of Africa, South America and Australia the earth is circled by an uninterrupted band of water such that the tidal wave can travel around the earth. Besides, at this latitude the water depth is not so much of a limitation for the propagation velocity so that the original ellipsoid can develop.

From around 65°S the tidal wave propagates to the north into the Atlantic, Indian and Pacific Oceans. Because it takes time for the tidal wave to progress through the oceans the tidal constituents at a particular location away from the area of tidal generation (around 65°S) lag behind the theoretical constituents from equilibrium theory. The further the location is away from the South Pole, the longer is the time shift between the celestial event and its appearance in the form of the tide.

Assuming an average depth in the ocean of 4000 m, the propagation speed of the tidal wave is about $$200\ m/s$$ and for $$T = 12.42\ h$$ the wavelength $$L = 9000\ km$$. At a speed of $$200\ m/s$$ (or $$720\ km/h$$) and if travelling in a more less straight line, it takes the tidal wave less than a day to travel from 65°S to say, Scotland. To reach the Netherlands, the tidal wave needs to cross the shallow North Sea basin. The latter takes another day or so, since the propagation speed strongly decreases in the shallow North Sea. This means that in the Netherlands spring and neap tide occur about two days after the corresponding moon configurations (see Fig. 3.22).

On its way, the tidal wave is distorted by local differences in water depth and – due to land masses – by restriction of the width or reflection. The period does not change, only the length of the wave changes via changes in propagation speed. The latter can be illustrated as follows. For the long tidal wave the wave celerity is proportional to the square root of the water depth. When the tidal waves gets into shallower water the celerity will decrease which results in a concentration of energy and thus an increase in tidal amplitude (comparable to the shoaling effect of wind waves in shallow water; see Sect. 5.2.2). In shallower water along the ocean coasts and shallow seas like the North Sea we have for instance a wavelength of $$L = 450\ km$$ (with $$h = 10\ m$$ and $$c = 10\ m/s$$). Depending on the geometry of a bay, sea or ocean basin resonance may occur, leading to amplification of the tidal amplitude at the coast (standing waves or – in combination with Coriolis – rotary standing waves, see Sect. 3.8.2).

In deep water tidal current velocities are very small. For an ocean depth of $$4000\ m$$ and a tidal amplitude of for instance $$0.25\ m$$ we get with Eq. $$\ref{eq3.8.1.8}$$ a tidal velocity amplitude of $$1.2\ cm/s$$. In shallower water this is quite different: tidal amplitude and tidal current velocities are larger. If $$a = 1\ m$$ and $$h = 10\ m$$ the tidal current velocity is $$1\ m/s$$, hence two orders of magnitude larger.

This page titled 3.8.1: Dynamic theory of tides is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Judith Bosboom & Marcel J.F. Stive (TU Delft Open) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.