# The Physics of Tides

## Introduction

Tides turn out to be very interesting. The main patterns are:

• A twice-daily variation
• A difference between the first and second tide of a day
• The spring–neap cycle
• The annual variation

As to how this works at a particular point on the coastline is determined by many factors including the orientation of the coastline, the size of the body of water around it and the characteristics of the sea bed. Although two high and low tides per day are usual, other combinations are not uncommon, as shown below. Of course, other local features like tidal bores also occur.

So what causes tides?

## Rotating Reference Frames

An inertial reference frame is one in which a body with no net force acting on it is not accelerating. This is Newton’s First Law. Newton’s Second Law states for a force, F, causes a body of mass, m to accelerate with acceleration, a :

$\vec{F} = m\vec{a}$

A non-inertial reference frame is one which is accelerating. Consider the following diagram:

So there are two reference frames. Frame A is inertial and the position of a particle at time t is given by $\vec{x_A}$.

Frame B is non-inertial (or accelerating) and the position of the particle at time, t is given by $\vec{x_B}$.

The origins of Frame B from Frame A is given by $\vec{X_{AB}}$.

We know in Frame A, Newton’s Second Law holds so lets see what happens in Frame B.

Let the coordinate axis in B be represented by unit vectors uj with j any of { 1, 2, 3 } for the three coordinate axes, such that:

$\vec{x_B} = \sum\limits_{j=1}^3 x_j.\vec{u_j}$

From Frame A, the equation becomes:

$\vec{x_A} = \vec {X_{AB}} + \sum\limits_{j=1}^3 x_j.\vec{u_j}$

Now, take a derivative with respect to time:

$\frac{d\vec{x_A}} {dt} = \frac{d\vec {X_{AB}}} {dt} + \sum\limits_{j=1}^3 \frac{dx_j} {dt}.\vec{u_j} + \sum\limits_{j=1}^3 x_j.\frac{d\vec{u_j}} {dt}$

So substituting velocity into this equation we get:

$\vec{v_A} = \vec{v_{AB}} + \vec{v_B} + \sum\limits_{j=1}^3 x_j.\frac{d\vec{u_j}} {dt}$

What this equation represents is that the velocity observed in Frame A is the velocity observed in Frame B plus 2 other terms; the first is the relative velocity at which the origin of the two reference frames are moving and the second represents the contribution to velocity due to the rotation of Frame B.

So lets differentiate again with respect to time to get accelerations:

$\frac{d^2\vec{x_A}} {dt^2} = \frac{d^2\vec {X_{AB}}} {dt^2} + \frac{d^2\vec{x_B}} {dt^2} + \sum\limits_{j=1}^3 \frac{d^2x_j} {dt^2}.\vec{u_j} + \sum\limits_{j=1}^3 \frac{dx_j} {dt}.\frac{d\vec{u_j}} {dt} + \sum\limits_{j=1}^3 \frac{dx_j} {dt}.\frac{d\vec{u_j}} {dt} + \sum\limits_{j=1}^3 x_j.\frac{d^2\vec{u_j}} {dt^2}$

So, on the right the second differential of position is the acceleration in Frame A. On the right, the first term is the acceleration of the origin of Frame B from Frame A. The second term is the acceleration of the particle as observed in Frame B.

$\vec{a_A} = \vec {a_{AB}} + \vec{a_B} + 2.\sum\limits_{j=1}^3 \frac{dx_j} {dt}.\frac{d\vec{u_j}} {dt} + \sum\limits_{j=1}^3 x_j.\frac{d^2\vec{u_j}} {dt^2}$

So this equation relates the acceleration observed in Frame A to the acceleration observed in Frame B with these 3 extra terms. The equation can be easily turned into a force equation by using Newton’s Second Law.

$\vec{F_A} = \vec {F_B} + \vec{F_{AB}} + 2.m.\sum\limits_{j=1}^3 \frac{dx_j} {dt}.\frac{d\vec{u_j}} {dt} + m.\sum\limits_{j=1}^3 x_j.\frac{d^2\vec{u_j}} {dt^2}$

These 3 additional terms are often called fictitious, pseudo or inertial forces. If we make Frame B the primary frame we are interested in then:

$F_B = F_A + F_{Pseudo}$

and the Pseudo Forces can be equated to:

$F_{Pseudo} = -\vec{F_{AB}} - 2.m.\sum\limits_{j=1}^3 \frac{dx_j} {dt}.\frac{d\vec{u_j}} {dt} - m.\sum\limits_{j=1}^3 x_j.\frac{d^2\vec{u_j}} {dt^2}$

The next step is to apply this general result to a B frame of reference of the Earth’s surface.

### A Frame of Reference on the Earth’ Surface

Let $\vec{\omega}$ represent the rotation of the Earth about its axis such that its magnitude is given by:

$\omega = \frac {2.\pi} {1 day} = 0.00007 / sec$

The first time derivative is given by:

$\sum\limits_{j=1}^3 \frac{d.\vec{u_j}} {dt} = \sum\limits_{j=1}^3 \vec{\omega}.\vec{u_j}$

The second time derivative is given by:

$\sum\limits_{j=1}^3 \frac{d^2.\vec{u_j}} {dt^2} = \sum\limits_{j=1}^3 \frac{d.\vec{\omega}} {dt}\times\vec{u_j} + \sum\limits_{j=1}^3 \vec{\omega}\times\frac{d.\vec{u_j}} {dt} = \sum\limits_{j=1}^3 \frac{d.\vec{\omega}} {dt}\times\vec{u_j} + \sum\limits_{j=1}^3 \vec{\omega}\times(\vec{\omega}\times\vec{u_j})$

Where “x” is the vector cross product operator. Substituting this into the earlier equation for acceleration gives:

$\vec{a_A} = \vec {a_{AB}} + \vec{a_B} + 2.\vec{\omega} \times \sum\limits_{j=1}^3 \frac{dx_j} {dt}.\vec{u_j} + \sum\limits_{j=1}^3 x_j.\frac{d.\vec{\omega}} {dt}\times\vec{u_j} + \sum\limits_{j=1}^3 x_j.\vec{\omega}\times(\vec{\omega}\times\vec{u_j})$

Next setting the translation acceleration $\vec {a_{AB}} = 0$ and collecting the terms up:

$\vec{a_A} = \vec{a_B} + 2.\vec{\omega} \times \vec{v_B} + \frac{d.\vec{\omega}} {dt}\times\vec{x_B} + \vec{\omega}\times(\vec{\omega}\times\vec{x_B})$

Lastly, putting this result back into the equation for Pseudo Forces gives:

$F_{Pseudo} = -2.m.\vec{\omega} \times \vec{v_B} -m.\frac{d.\vec{\omega}} {dt}\times\vec{x_B} -m.\vec{\omega}\times(\vec{\omega}\times\vec{x_B})$

The first term is what is called the Coriolis Force, the second the Euler Force and the third term the Centrifugal Force.

### Pseudo Forces

So what are these confusing Pseudo Forces?

So to start off they do not arise in inertial (or non-accelerating frames of reference). When the frame of reference is accelerating (or non-inertial) then these forces are a consequence of being in an accelerating frame of reference. This occurs in everyday life, for example, in a car going around a bend. If the bend is to the left and the speed of the car remains constant then the driver will feel a force pushing him or her to the right. This is called a centrifugal force and is one of the Pseudo Forces. The car seat pushes back on the driver to the left. This is the centripetal force and has the same magnitude but opposite sign to the centrifugal force. The driver therefore has no net force on him or her and follows the line of the corner inside the car.

From the perspective of an observer standing outside the car watching it drive around the corner, the observer sees the car and driver follow a corner to the left. So a net force to the left must be applied (the centripetal force) to make the car change direction. From the standing observer’s perspective, in an inertial frame of reference, the centripetal force exists but the centrifugal force does not.

#### The Euler Force

The Euler Force is given by:

$F_{Euler} = -m.\frac{d.\vec{\omega}} {dt}\times\vec{x_B}$

The first thing to note is that it contains $\frac{d.\vec{\omega}} {dt}$ which is zero for a reference frame spinning at a constant speed (like the Earth), so it plays no part in tides.

An example of the Euler Force would be sitting on a carousel, initially at rest, that starts moving. The force is proportional to the angular acceleration and the distance from the carousel’s centre.

To work out the direction of the force we know the angular acceleration $\frac{d.\vec{\omega}} {dt}$ is vertically upwards and that the position vector $\vec{x_B}$ is outwards from the carousel’s centre. The vector cross product direction is derived from the right hand rule, so would point forwards, in the direction of motion. The force formula, however, has a minus sign which reverses the direction to point backwards.

So back on the carousel, the Euler Force pushes the person back into the seat. As the ride ends, and the carousel decelerates the Euler Force acts to lift the person out of the seat. So it behaves in this case like the real (non-pseudo force) acting on a passenger in a car, initially at rest, that accelerates pushing the passenger back into the seat. When the car decelerates to a stop, the passenger is lifted out of the seat.

#### The Centrifugal Force

The Centrifugal Force is given by:

$F_{Centrifugal} = -m.\vec{\omega}\times(\vec{\omega}\times\vec{x_B})$

Taking the mass away to give the magnitude of the acceleration (where R is the Earth’s radius):

$\omega^2.\rho = \omega^2.R.\cos(\lambda)$

So at the poles, $\cos(\lambda) = 0$ so there is no centrifugal acceleration. At the equator $\cos(\lambda) = 1$ so the centrifugal acceleration becomes:

$\omega^2.R = (\frac{2.pi()} {24.60.60})^2.6,378,000 = 0.034m/s^2$

Note that this is quite small compared to the force of gravity (about 1/3 of 1% of the force of gravity). Note also that the force is independent of the motion of a particle on the Earth’s surface (whether stationary or moving).

The direction of the Centrifugal Force is given by following the right hand rule for the two vector cross products. Firstly, $\vec{\omega}\times\vec{x_B})$ is a vector pointing downwards. Performing the cross product of $\vec{\omega}$ with the previous result is a vector pointing inwards towards the axis of rotation. The minus sign in the equation for Centrifugal force reverses the direction of the force to point away from the axis of rotation.

Resolving the centripetal acceleration into a component parallel to the centre of the Earth gives: $\omega^2.R.\cos^2(\lambda)$. So that gravity on the Earth’s surface is effectively reduced by this factor.

There is also a component parallel to the Earth’s surface in the north south direction of: $\omega^2.R.\cos(\lambda).\sin(\lambda)$

This parallel component is zero at the equator and poles and acts elsewhere in the direction of the equator. Its maximum is at:

$\frac{d.\cos(\lambda).\sin(\lambda)} {d\lambda} = 0$

Which is at +/-45 degrees latitude.

So one of the theories is that this force, small that it is, acting towards the equator is responsible over the lifetime of the Earth for producing the oblate nature of the Earth. That is, it is not quite spherical, but fatter at the equator than at the poles.

#### The Coriolis Force

The Coriolis Force is given by:

$F_{Coriolis} = -2.m.\vec{\omega} \times \vec{v_B}$

So the first thing to note is that the Coriolis Force is proportional to a particle’s mass and to its speed in Frame B, so for stationary particles the force is zero. It has an acceleration magnitude of:

$2.\omega.v = 2.\frac{2.pi()} {24.60.60} = 0.00015.v.m/s^2$

The next thing to note is that the direction of the force is given by the vector cross product of angular velocity and velocity by the right hand rule:

• If the velocity is parallel to the axis of rotation then the cross product and therefore the Coriolis Force is zero.
• If a ball is dropped at the equator, the cross product points west. So the Coriolis Force’s negative sign reverses the direction, and the Coriolis Force on the ball is to the east (in the direction of rotation of the Earth).
• For a ball thrown vertically upwards the Coriolis Force is to west (opposite to the direction of rotation of the Earth).

A major impact of the Coriolis Force is on the atmosphere and oceans. In the atmosphere, when an area of low pressure develops, air would flow towards the area of low pressure from surrounding areas of high pressure. On the revolving Earth, these flows of air are subject to the Coriolis Force, perpendicular to the direction of motion of high to low pressure air flow.

In the northern hemisphere, air flowing towards an area of low pressure (blue arrows) is subject to the Coriolis Force (red arrows) and is deflected as shown in the diagram. This leads to an anticlockwise rotating mass of air, or anticyclone.

In the Southern Hemisphere, the forces act in the opposite direction and lead to clockwise rotating air masses, or cyclones.

In fact, the existence of the Coriolis Force was used as evidence that the Earth rotates; rather than the sun, moon and stars rotating about the Earth. Faucault’s Pendulum is a good example of this, where a swinging pendulum can be seen to precess indicating a force other than just gravity is acting on it.

## Orbiting Reference Frames

So, lets park rotating reference frames for a while and look at the case of an orbiting reference frame.

Lets keep things simple and assume a particle is following orbiting a central point. We know the general equation for acceleration is given by:

$\frac{d^2\vec{x_A}} {dt^2} = \frac{d^2\vec {X_{AB}}} {dt^2} + \frac{d^2\vec{x_B}} {dt^2} + \sum\limits_{j=1}^3 \frac{d^2x_j} {dt^2}.\vec{u_j} + \sum\limits_{j=1}^3 \frac{dx_j} {dt}.\frac{d\vec{u_j}} {dt} + \sum\limits_{j=1}^3 \frac{dx_j} {dt}.\frac{d\vec{u_j}} {dt} + \sum\limits_{j=1}^3 x_j.\frac{d^2\vec{u_j}} {dt^2}$

The rotational components are zero (by definition as there is no rotation is Fame B in this example), so the equation simplifies to:

$\frac{d^2\vec{x_A}} {dt^2} = \frac{d^2\vec {X_{AB}}} {dt^2} + \frac{d^2\vec{x_B}} {dt^2}$

Now if the origin of frame A is at the centre of the circle orbited by the particle then:

$\vec{X_{AB}} = R.(\cos(\omega.t), \sin(\omega.t))$

$\vec{V_{AB}} = \frac{d\vec{X_{AB}}} {dt} = \vec{\omega} \times \vec{X_{AB}}$

And the acceleration is given by (remembering that $\omega$ is constant with respect to time:

$\vec{a_{AB}} = \frac{d\vec{V_{AB}}} {dt} = \vec{\omega} \times (\vec{\omega} \times \vec{X_{AB}})$

Lastly, turning this into a Pseudo Force in the orbiting reference frame B:

$F_{Pseudo} = -m.\vec{\omega} \times (\vec{\omega} \times \vec{X_{AB}})$

So we can see that the format is the same as the format of the centrifugal force in the rotating Earth example, and that the magnitude of the acceleration is:

$\omega^2.R = (\frac{2.pi()} {365.24.60.60})^2.149,600,000,000 = 0.0059m/s^2$

The direction of the force using the right hand rule again, gives the force in the orbital plane, pointing outwards from the centre of the orbit.

## The Earth’s Motion around the Sun

In the above diagram we can see Earth orbiting the sun in an anticlockwise direction. Lets assume a circular (rather than elliptic) orbit for now. More on that later. The rotation of the Earth is anticlockwise and inclined at 23.5 degrees to the sun / Earth orbital plane.

In addition, of course, the moon orbits the Earth. The plane of the moon’s orbit is inclined at about 5 degrees to the earth / sun orbital plane.

Technically, the moon does not orbit the Earth, but rather, both the moon and the Earth both orbit their centre of mass (or barycentre). Because the Earth is much more massive than the moon, the barycentre is located within the Earth so the effect on the Earth is for it to appear to wobble, as it orbits the sun, as shown below:

### Pseudo Forces

Thinking about the major Pseudo Forces involved from the perspective of a non-inertial reference frame on the Earth’s surface (Frame B) in relation to an inertial reference frame in deep space (Frame A). The Earth’s rotation contains the following components:

• Centrifugal Force. Acts outwards from the Earth’s axis of rotation $F_{Centrifugal} = -m.\vec{\omega}\times(\vec{\omega}\times\vec{x_B})$. Has an acceleration magnitude of $0.034m/s^2$
• Coriolis Force. Is speed dependent so will be quite small for slow moving things like water currents, but has an acceleration magnitude of $0.00015.v.m/s^2$
• Euler Force. Assuming a constant rate of rotation this will be zero.

The Earth’s orbit around the sun generates the following components.

• A Centrifugal Force like component. Acts outwards from a line joining the Sun and Earth, in the Sun / Earth plane. Has an acceleration magnitude of $\omega^2.R = (\frac{2.pi()} {365.24.60.60})^2.149,600,000,000 = 0.0059m/s^2$

The Earth’s orbit around the moon (or rather the Earth / moon barycentre) generates the following components:

• A Centrifugal Force like component. Acts outwards from a line joining the Earth and moon, in the Earth / moon plane. Has an acceleration magnitude of $\omega^2.R = (\frac{2.pi()} {27.24.60.60})^2.4,600,000 = 0.000033m/s^2$

### Gravity

Now lets look at gravity. The acceleration due to gravity of a mass M at distance R is given by:

$a_g = \frac{G.M} {R^2}$ where G = 6.674×10−11 m3⋅kg−1⋅s−2

So, plugging the numbers for the mass (5.97e24 kg) and radius of the Earth (6.371e6 m) into this equation gives the acceleration on the surface of the Earth due to gravity = 9.82 ms−2

Plugging the numbers for the mass of the sun (1.99e30 kg) and the distance between the Earth and the sun (1.50e11 m) gives an acceleration on the surface of the Earth towards the sun of 0.0059 ms−2

Hang on you are thinking, I’ve seen that number before! Yep, it’s the value of the Pseudo Centrifugal Force on the Earth due to the Earth orbiting the sun. This makes complete sense that these forces are of equal size (gravity acts towards the sun; the Pseudo Force acts away from the Sun). So in the reference frame of the Earth, the Sun’s gravitational pull is offset by the Pseudo force with no net resultant force so we never move any nearer to the sun.

Plugging the numbers in for the mass of the moon (7.35e22 kg) and the distance between the Earth and the moon (3.84e8 m) gives an acceleration on the surface of the Earth towards the moon of 0.000033 ms−2. Bingo, this matches the centrifugal like Pseudo Force calculated above, but acting in the opposite direction so as to cancel out.

## Tide Generating Forces

So now we know a little about the forces acting on the surface of the Earth we can start to think about how tides might be created. Firstly, the Earth’s gravity and Pseudo Forces produced by the Earth’s rotation are clearly not involved in producing the observed tidal patterns simply because these forces are constant with respect to time and therefore would not lead to the observed periodic properties. Although the forces change over the surface of the Earth, they are constant at any point on the surface.

So this leaves the impact of Sun and the Moon. To avoid having to deal with a 3-body problem, lets take the impact of the Sun and the Moon separately, starting with the Sun.

So lets look more carefully at the forces acting at different points on the Earth’s surface. As already stated we can discard the Earth’s own gravitational force and the Pseudo Forces produced by the Earth’s rotation and concentrate on the gravitational force towards the Sun, $F_{gr}$ and the Pseudo Force produced by the Earth’s orbit around the Sun, $F_{in}$.

At points A and B, you can see that there is a net downward force towards the centre of the Earth. At point Z, where the Sun is overhead, the force of gravity will be a little higher than elsewhere on Earth because the point is nearer the Sun so will be a little larger than the Pseudo Force acting away from the Sun, resulting in a net force towards the Sun. At point N, the force of gravity from the Sun will be a little less than elsewhere on Earth due to this point being further from Sun, so the net force at this point will be pointing away from the Sun.

The above diagram shows the direction of the net force of gravity and pseudo force due to the Sun. The thing to note is that there is a tendency to create a tidal force at the point nearest the Sun, and also at the point furthest from the Sun. As the Earth spins about its axis, so the points affected by these maximal tidal forces rotate, and a point on the Earth’s surface will have these maximal tidal forces twice per day.

This is about as far as a static analysis of tides can usefully go. The next step is to consider a dynamic analysis, by thinking about the net, tide producing forces at a point on the Earth as periodic, for example, as the Earth rotates about the Sun, or as the Moon rotates about the Earth.

## Harmonic Analysis

The starting point here is that the tide at a given point rises and falls in a periodic way and that this can be expressed as a Fourier series of periodic terms related to astronomical conditions. So, a general equation for the height, h of the tide at time, t, would be:

$h = H_0 + A.\cos(a.t + \alpha) + B.\cos(b.t + \beta) + C.\cos(c.t + \gamma) +...$

where $H_0$ is the mean tide level. The A, B, C, etc are the amplitudes of each component, the a, b, c, etc are the speed of the component and the $\alpha, \beta, \gamma$, etc are the phase offset of each component.

The more accurate you want the analysis to be the more components you can add in. Doodson did the pioneering work on this which was published in 1921. He identified 6 orbital attributes (3 for Sun and 3 for the Moon) that described their elliptical orbits and constructed 388 periodic components from this.

By measuring the height of the tide using a tidal gauge at regular intervals over an extended period of time, it is possible to calculate the amplitudes (A, B, C, etc) and phase offsets ($\alpha, \beta, \gamma$, etc) of each component and then use the above equation to predict forward in time the tide height. This is the basis of tide tables.

The main components can be understood in terms of their frequencies: semi-diurnal, diurnal, longer periods, and shorter periods.

### Semi-Diurnal Components

The most significant component is due to the Moon and, by convention, is called $M_2$. “M” for Moon and “2” for approximately twice per day. A lunar day is 24hrs and 50minutes (not 24hrs) as the Earth / Moon combination are orbiting the Sun so the Moon needs to travel a bit more than 24hrs to get to the same spot relative to the Earth (which has moved over the 24hrs). So $M_2$ has a period of 12hrs 25mins and explains why, for example, high tides in tide tables are usually separated by 12hrs 25mins.

The next most significant component is due to the Sun, $S_2$, and the radiation difference between night and day leading to different atmospheric pressure. By definition this occurs at a period of 12hrs.

Then comes, $N_2$, which accounts for the Moon’s elliptical (rather than circular) orbit.

There are potentially many more components all with slightly different periods, but their amplitudes, and therefore their contribution, is smaller.

### Diurnal Components

The most important components are $K_1$ and $O_1$, that both relate to the moon but have slightly different periods.

Again there are potentially many more components, this time with a period around 24hrs.

### Longer Periods

These relate to things like the monthly Moon orbit and the annual orbit of the Earth about the Sun. These components are usually small.

### Shorter Periods

The effect of shallow water causes interaction with the bottom to generate harmonics of the larger components already discussed. For example, $M_4$ with a period of 6hrs 13mins, $M_6$ and $S_4$ with a period of 6hrs.

## Putting it all Together

The US use 37 components and publish data for these components. For example, for The Battery in New York, so you could use this to calculate your own Tide Tables.

Obviously, these calculations are pretty simple for a computer to do now, but back in the days before computers, mechanical computers were built to do this.

Lord Kelvin had one built in 1872 to use 26 components. It was used by Doodson to calculate Tide Tables for the Normandy Landings in the Second World War.

The landings had to be at low water as the beeches had been booby-trapped, and low tide had to be first light so the troops could cross the channel under the cover of darkness, so the Moon had to rise late.

A note was sent to Doodson with 11 harmonics.

Its unclear where these came from, but crunching the numbers Doodson worked out that 5th-7th June would be the most suitable dates for D-Day.