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## Imaging the Galaxy’s supermassive blackhole – part 1

Last week, I blogged about the theoretical arguments for the Galaxy harbouring a supermassive black hole at its centre, and here I blogged about the observational evidence. The work done by the UCLA and MPE teams, discussed here, has led to a determination that the central black hole has a mass of between 4.4 and 4.5 million solar  masses. I am going to take the upper end  of this range, just for convenience.

An artist’s impression of Sgr A*, showing the central supermassive black hole and the accretion disk which surrounds it.

## The size of the event horizon

In this blog here I showed that the radius of a blackhole’s event horizon can be calculated by using the equation for the escape velocity $v_{esc}$ when that velocity is equal to the speed of light $c$. That is

$v_{esc} = c = \sqrt{ \frac{2GM }{ R } }$

where $M$ is the mass of the blackhole, $G$ is the universal gravitational constant, and $R$ is the size of the object, which in this case is the radius of the event horizon (also known as the Swarzchild radius $R_{s}$). So, we can write

$R_{s} = \frac{ 2GM }{ c^{2} }$

Putting in a mass of 4.5 million solar masses, we find

$R_{s} = 1.33 \times 10^{10} \text{ metres}$

Converting this to AUs, we find the radius of the event horizon is 0.09 AUs, much smaller than the radius of Mercury’s orbit, which is about 0.3 AUs.

At the distance of the Galactic centre, 8 kpc, this would subtend an angle of
$\theta = 6.17 \times 10^{-9} \text{ degrees}$ (remember to double $R_{s}$ to get the diameter of the event horizon). This is the same as

$\boxed{ \theta = 22.22 \text{ micro arc seconds} }$

Converting this to radians, we get

$\theta ( \text{in radians}) = 1.08 \times 10^{-10}$

In fact, we do not need to resolve the event horizon itself, but rather the “shadow” of the event horizon, which is about four times the size, so we need to resolve an angle of

$\theta ( \text{in radians}) \approx 4 \times 10^{-10}$

## The resolution of a telescope

There is a very simple formula for the resolving power of a telescope, it is given by

$\theta( \text{in radians}) = \frac{ 1.22 \lambda }{ D }$

where $D$ is the diameter of the telescope and $\lambda$ is the wavelength of the observation. Let us work out the diameter of a telescope necessary to resolve an object with an angular size of $50 \times 10^{-4} \text{ radians }$ at various wavelengths.

For visible light, assuming $\lambda = 550 \text{ nanometres}$

$D = \frac{ 1.22 \times 550 \times 10^{-9} }{ 4 \times 10^{-10 } }, \boxed{ D = 1.68 \text { km} }$

There is no visible light telescope this large, nor will there ever be. At the moment, visible-light interferometry is still not technically feasible over this kind of a baseline, so imaging the event horizon of the Galaxy’s supermassive blackhole is not currently possible at visible wavelengths.

$D = \frac{ 1.22 \times 21 \times 10^{-2} }{ 4 \times 10^{-10 } }, \boxed{ D = 640,000 \text { km} }$

This is more than the distance to the Moon (which is about 400,000 km away). So, until we have a radio dish in space, we cannot resolve the supermassive blackhole at 21cm either.

For millimetre waves, we have

$D = \frac{ 1.22 \times 1 \times 10^{-3} }{ 4 \times 10^{-10 } }, \boxed{ D = 3,100 \text { km} }$

which is feasible with very long baseline interferometry (VLBI). So, with current technology, imaging the event horizon of the Milky Way’s supermassive blackhole is only feasible at millimetre wavelengths. Millimetre waves lie in a niche between visible light and radio waves. They are long enough that we can do VLBI, but they are short enough that the baseline to image the supermassive black hole’s event horizon is small enough to be possible with telescope on the Earth.

Next week I will talk about a project to do just that!

## The direction of the angular velocity vector

In this blog, I introduced the idea of angular velocity, which is the rotational equivalent of linear velocity. The angular velocity $\omega$ is usually measured in radians per second, where radians are the more natural measurement of an angle than the more familiar degrees. But, just as linear velocity is a vector and therefore has a direction, so too does angular velocity. So, what is the direction of the angular velocity vector?

## The relationship between linear velocity and angular velocity

As we saw in the blog where I introduced the concept of angular velocity, it can be defined as simply the angle $\theta$ moved per unit time, or

$\omega = \frac{ \theta }{ t }$

which of course leads to it being usually measured in radians per second. However, we can also write the angular velocity in terms of the linear velocity. To see how to do this let us remind ourselves of the definition of a radian, the more natural unit for measuring an angle. As I introduced it in this blog, measuring an angle in radians just means dividing the length of the arc $l$ by the radius of the circle $r$.

An angle measured in radians is simply the length of the arc $l$ divided by the radius $r$

We can write that the angle measured in radians is

$\theta \text{ (in radians) } = \frac{ l }{ r }$

But, the linear velocity $v$ is just defined as distance divided by time, so we can write

$v = \frac{ l }{ t }$

Re-writing $l$ in terms of $\theta \text{ and } t$ we can write

$v = \frac{ \theta r }{ t }$

and so we can write the angle $\theta$ as

$\theta = \frac{ v t }{ r }$

Using this for $\theta$ we can write the angular velocity $\omega$ as

$\omega = \frac{ v t }{ r } \cdot \frac{ 1 }{ t }, \text{ so } \boxed{ \omega = \frac{ v }{ r } }$

## The direction of the angular velocity vector $\vec{ \omega }$

Writing this in terms of vectors, and remembering that division of vectors is not defined, we instead write that

$\boxed{ \vec{ \omega } = \frac{ \vec{ r} \times \vec{ v } }{ | \vec{ r } |^{2} } }$

where $\vec{ r } \times \vec{ v }$ is the vector product (or cross-product), as I discussed in this blog here.

The direction of the radius vector $\vec{ r }$ is away from the centre of the circle, and the direction of the linear velocity $\vec{ v }$ for an object moving anti-clockwise is in the direction shown in the diagram below, tangential to the circle so at right angles to the radial vector $\vec{ r }$.

The direction of the radius vector $\vec{ r }$ is away from the centre of the circle, the direction of the linear velocity $\vec{ v }$ for an object moving anti-clockwise is as shown, at right angles to the radius vector.

To find the direction of $\vec{ \omega }$, we can use the right-hand rule, as shown in the figure below.

The right-hand rule for determining the direction of the result of the vector product

In our example here, our first-finger is in the direction of the radial vector $\vec{ r }$, and our second-finger is in the direction of the linear velocity $\vec{ v }$, leading to the angular velocity $\vec{ \omega }$ (represented by the thumb) being outwards, or towards us, as shown in the figure below.

The direction of the angular velocity vector $\vec{ \omega}$ is perpendicular to the plane of rotation of the object.

Another way to find this direction is to wrap the fingers of the right hand in the direction of the rotation, the thumb will then show the direction of the angular velocity vector.

The direction of the angular velocity can also be found as shown in this figure.

## Derivation of the centripetal force

I have mentioned a few times in previous blogs that an object moving in a circle at a constant speed does so because of a force acting towards the centre. We call this force the centripetal force. The force is given by the equation

$F = \frac{ mv^{2} }{ r }$

where $m$ is the mass of the object moving in the circle, $v$ is its speed, and $r$ is the radius of the circle. More correctly, remembering that force is a vector, it should be written

$\boxed{ \vec{F} = - \frac{ mv^{2} }{ |\,\vec{r}\,| } \hat{r} }$

where $|\,\vec{r}\,|$ is the magnitude (size) of the radial vector $\vec{r}$, and $\hat{r}$ is the unit vector in the direction of $\vec{r}$.

But, from where does this formula come?

## The acceleration of an object moving in a circle

The acceleration of any object is defined at the change in its velocity divided by the change in time. Both acceleration and velocity are vectors, so mathematically we can write this as

$\vec{a} = \frac{ \Delta \vec{v} }{ \Delta t }$

We are going to consider an object moving at a constant speed in a circle, as illustrated in the diagram below. In a time $\Delta t$ the object has moved through an angle $\theta$, and its initial velocity $v_{1}$ has changed to $v_{2}$, where the only change in the velocity is its direction. Remember, the velocity is tangential to the radius, so makes a right angle with the radius vector $\vec{r}$. The direction of the radius vector is, by definition, radially outwards.

We will consider an object moving in a circle with a constant speed. It moves through an angle $\theta$ in time $t$.

In my blog about vectors I mentioned that, to combine vectors which have different directions, we need to split the vectors into components, add the components and then recombine the resultant components. The components need to be at right-angles to each other, and usually (but not always) we choose the x and y-directions when the vectors are in two dimensions.

To find the acceleration of our object in this example, we want to find the change or difference in the velocity, that is $\Delta \vec{ v } = \vec{ v_{2} } - \vec{ v_{1} }$. We start by splitting the two vectors into their x and y-components.

$\vec{ v_{1} } = v_{1}\hat{x} + v_{1}\hat{y} \; \; \text{ and } \; \; \vec{ v_{2} } = v_{2}\hat{x} + v_{2}\hat{y}$

Looking at our diagram, we can write

$\vec{ v_{1} } = 0\hat{x} + v\hat{y} \; \; \text{ and } \; \; \vec{ v_{2} } = v\sin(\theta)\hat{x} + v\cos(\theta)\hat{y}$

where $v$ is the speed, the size of the vectors $\vec{ v_{1} } \text{ and } \vec{ v_{2} }$.

The change in the velocity in the x-direction, which we will call $(\Delta v)\hat{x}$ is then just

$(\Delta v)\hat{x} = v\sin(\theta) - 0 = v\sin(\theta)$

Similarly, the change in the velocity in the y-direction, which we will call $(\Delta v)\hat{y}$ is given by

$(\Delta v)\hat{y} = v\cos(\theta) - v = v(\cos(\theta) - 1)$

To correctly calculate the acceleration, we need to find the change in velocity with time as the time interval tends to zero. This means the angle $\theta$ tends towards zero also, and when $\theta$ is very small (and expressed in radians) we can write

$\cos(\theta) \rightarrow 1 \; \; \text{ and } \sin(\theta) \rightarrow \theta$

So we can then write

$(\Delta v)\hat{x} \rightarrow v\theta \; \; \text{ and } (\Delta v)\hat{y} \rightarrow v(1 - 1) =0$

The overall change in velocity, $\Delta \vec{v}$ is then just the change in velocity in the x-direction, $\Delta \vec{v} = v\theta$. The direction of the change in velocity is in the positive x-direction, which as $\theta \rightarrow 0$ is along the radial vector towards the centre of the circle (that is, in the $- \vec{r}$ direction).

However, we can do a substation for the angle $\theta$. Remember, the arc-length, which we shall call $l$ is related to the angle $\theta$ via our definition of the radian (see this blog here), $\theta = l/r$, so we can write

$\Delta \vec{v} = \frac{ vl }{ |\,\vec{r}\,| }\hat{r} \; \text{ (Equation 1)}$

The acceleration $\vec{a} = \Delta \vec{v} / \Delta t$. But the speed $v$ and time $t$ are related via $v = l/t$, so we can write that $l = vt$. Substituting this into equation 1 above gives

$\frac{ vl }{ r } = \frac{ v^{2}t }{ r } \; \text{ (Equation 2)}$

and so the acceleration $\vec{a}$ is given by

$\vec{a} = \frac{ \Delta \vec{v} }{ t } = \frac{ v^{2}t }{ |\,\vec{r}\,| t } = \boxed{ - \frac{ v^{2} }{ |\,\vec{r}\,| } \hat{r} }$

where we have added the minus sign to remind us that the change in velocity, and hence the acceleration, is directed towards the centre of the circle.

If you prefer to think of vectors pictorially, then the direction of $\Delta\vec{v} = \vec{v_{2}} - \vec{v_{1}}$ can be seen from this diagram:

This shows the direction of $\Delta \vec{v} = \vec{v_{2}} - \vec{v_{1}}$, and it is along the radius vector towards the centre of the circle, in the opposite direction to the radius vector $\vec{r}$.

The centripetal force is found by remembering that $\vec{F} = m\vec{a}$ (Newton’s 2nd law), so finally we can write that the centripetal force is

$\boxed{ \vec{F} = - \frac{ mv^{2} }{ |\,\vec{r}\,| } \hat{r} }$

## Safely into the harbour

In yesterday’s blog, I discussed how mathematicians and physicists measure angles in radians rather than the more familiar degrees. I finished the blog by mentioning a problem to do with water level in a harbour. Let me state the problem in more detail, and then show how we go about solving it.

## The tide is high

Suppose the height in metres of the water in a harbour can be described by the equation

$y = 6\sin(\frac{\pi}{4}t) + 8\cos(\frac{\pi}{4}t) + 11$

where $t$ is the time in hours. We will assume that the time $t=0$ corresponds to midnight. We want to find the following

1. The maximum height of the water
2. The minimum height of the water
3. The time when the first maximum water height occurs
4. The time when the first minimum water height occurs
5. If a boat needs 2 metres of water to be able to use the harbour, how many hours a day can it use the harbour?

As I showed yesterday, the period of this up and down motion of the water level is given by $\frac{2\pi}{\omega}$, so in this example above the period is $\frac{2\pi}{\pi/4}=8\; \text {hours}$. In reality, of course, if the up and down motion of the level of water is due to the tides, then the period has to be 12 hours, as there are two high and two low tides each day. But, let’s assume, just for variety, that the period can differ from 12 hours, so in this example it will be 8 hours.

Notice that the equation describing the level of the water $y$ is a combination of sines and cosines. How do we go about solving this?

## Combining sines and cosines

The trick to being able to solve this equation is to use the compound angle formula. A few pages of algebra will show that

$\sin(A+B) = \sin(A)\cos(B) + \cos(A)\sin(B)$.

We will use that to rewrite our formula for the water level. We can write

$\sin( \frac{\pi}{4}t + \alpha ) = \sin( \frac{\pi}{4}t ) \cos( \alpha ) + \cos( \frac{\pi}{4}t ) \sin( \alpha )$

Multiplying all the terms by $R$ we get

$R \sin( \frac{\pi}{4}t + \alpha ) = R \sin( \frac{\pi}{4}t ) \cos(\alpha) + R \cos( \frac{\pi}{4}t ) \sin( \alpha )$

and, re-ordering this we get

$R \sin( \frac{\pi}{4}t + \alpha ) = R \cos( \alpha) \sin( \frac{\pi}{4}t ) + R \sin( \alpha ) \cos( \frac{\pi}{4}t )$

By writing it in this order we can see that in our original equation $6=R \cos( \alpha )$ and $8=R \sin( \alpha )$. If we square these two equations, and remember that $\sin^{2}( \alpha ) + \cos^{2}( \alpha ) = 1$ for any value of $\alpha$, then $36=R^{2} \cos^{2}( \alpha )$ and $64 = R^{2} \sin^{2}( \alpha )$, so adding these we get $36 + 64 = 100 = R^{2} (\cos^{2} ( \alpha ) + \sin^{2} ( \alpha ) ) = R^{2} (1)$, so $R = \sqrt{100} = 10$.

In order to calculate the phase angle $\alpha$ we not that $\frac{ R \sin ( \alpha ) }{ R \cos ( \alpha ) } = \tan ( \alpha ) = \frac{8}{6} = \frac{4}{3}$. So $\alpha = \tan^{-1} (4/3) = 0.93 \; \text{radians}$. (This is about $53^{\circ}$).

So, now we have an equation for $y$ which is entirely in terms of $\sin$, we have

$y = 10 \sin( \frac{\pi}{4}t + 0.93 ) + 11$.

Plots of $y = \sin( \frac{\pi}{4}t)$ (black curve) and $y=\sin(\frac{\pi}{4}t + 0.93)$ (green curve). The blue dashed line is where y=-9. These plots do not include the +11 constant, so of course +11-9=2 metres.

## The maximum and minimum heights of water in the harbour

Sine is a function which varies between +1 and -1, so the maximum value of $10 \sin ( \frac{\pi}{4}t + 0.93 ) = 10$. This means the maximum height of the water in the harbour will be $10 + 11 = 21 \; \text{metres}$.

The minimum height will be when $10 \sin ( \frac{\pi}{4}t + 0.93 ) = -10$, and so the minimum height will be $-10 + 11 = 1 \; \text{metre}$.

## At what time does the water level first reach its maximum value?

We have shown that the maximum value happens when $\sin( \frac{\pi}{4}t + 0.93 ) = 1$. Let us call $(\frac{\pi}{4}t + 0.93 )$ a new angle $\theta$. We know that $\sin( \theta ) = 1 \; \text{when} \; \theta = \pi/2$. But $\theta = ( \frac{\pi}{4}t + 0.93 ) \; \text{so} \; \frac{\pi}{4}t = ( \theta - 0.93 ) = ( \pi/2 - 0.93 ) = 0.64 \; \text{radians}$. This leads to $t = (0.64)(4)/\pi = 0.81 \; \text{hours} = 49 \; \text{minutes}$ after midnight.

## At what time does the water level first reach its minimum value?

This happens when $\sin( \frac{\pi}{4}t + 0.93 ) = \sin ( \theta ) = -1$. The first time this happens is when $\theta = 3\pi/2, \; \text{so} \; \frac{\pi}{4}t = ( \theta - 0.93 ) = ( 3\pi/2 - 0.93 ) = 3.78 \; \text{radians}$. This leads to $t = (3.74)(4)/\pi = 4.8 \; \text{hours} = 4 \; \text{hours and} \; 49 \; \text{minutes}$ after midnight. Note, this is half a period (4 hours) after the maximum, as one would expect.

## How many hours each day can we use the harbour?

If we can only use the harbour when the water level is above 2 metres, we need to find the time when the water level first drops below 2 metres (remember it goes down as low as 1 metre). At midnight, the water level is $y = 10 \sin(0.93) + 11 = 10(0.8)+11=19 \; \text{metres}$.

We need to find $y = 10 \sin( \frac{\pi}{4}t + 0.93 ) + 11 = 2$. Re-arranging, this is $10 \sin( \frac{\pi}{4}t + 0.93 ) = 2 - 11 = -9, \; \text{so} \; \sin( \theta ) = - 9/10$.

This value is shown by the horizontal dashed blue line on the zoomed-in sine functions plotted above. This gives a value of $\theta = \sin^{-1}( -9/10 ) = - 1.12 \; \text{radians}$. But, this answer is no good for us, as it gives a negative time, as shown by the first red dot to the left of the y-axis. We have to find the next solution, which will be when at a time 1 period later, so $-1.12 + 2\pi = 5.16 \; \text{radians}$. So $( \frac{\pi}{4}t + 0.93 ) = 5.16$, the 3rd red dot on the plot. This value leads to $t = 5.39 \; \text{hours}$.

If we compare this time to the time of the first minimum, $4.8 \; \text{hours}$, we can see that is is after it. What has gone wrong? What we have found is the time when the water level comes back up to 2m after being as low as 1 metre when the time was $4.8 \; \text{hours}$. The difference between $5.39 \; \text{and} \; 4.8 = 0.59 \; \text{hours}$. From the symmetry of the sine function, the water will drop below the 2 metre level at the same time difference before the minimum, i.e. at a time of $4.8 - 0.59 = 4.21 \; \text{hours}$. This is the 2nd red dot on the plot. So, the time the water is below 2 metres during each cycle is the time between the 2nd and 3rd red dots, i.e. the part shaded in yellow. That is $2 \times 0.59 = 1.18 \; \text{hours which is} \; 1 \; \text{hour and} \; 10.8 \; \text{minutes}.$

Because the period of our sine function is 8 hours, there are $24/8 = 3$ full cycles in each 24 hour day. During each cycle the water level is below 2 metres for $1.18 \; \text{hours}$, the total number of hours each day for which we cannot use the harbour is $1.18 \times 3 = 3.54 \; \text{hours} \; = 3 \; \text{hours and} \; 32.4 \; \text{minutes}$.

We were asked for the number of hours each day we can use the harbour, so this will be $24 - 3.54 = 20.46 \; \text{hours} \; = 20 \; \text{hours and} \; 27.6 \; \text{minutes}$ each day.

## Why are there $360^{\circ}$ in a circle?

I would imagine all of you reading this know that when we learn about angles, we learn that we measure angles in degrees, and that there are $360^{\circ}$ in a full circle. Why this strange number?

It was the Babylionians who chose to have $360^{\circ}$ in a full circle. When you think about it, they could have chosen any figure, say 400 or 10 or 1,000. Why did they choose 360? Well, it turns out that 360 was a special number to the Babylonians, or to be more correct 12 and 30 were, and 360 is 12 multiplied by 30.

12 was special to the Babylonians because there are 12 and a bit cycles of the Moon in a year. It is from the time between e.g. each new Moon (or full Moon) that we get our idea of months. 30 was special to the Babylonians because this was, roughly speaking, the number of days in each lunar cycle.

So, if 12 and 30 were special, then the number produced by multiplying them together, 360, was even more special. That’s why we are stuck with the strange number for the degrees in a circle.

## Radians – more natural units

Mathematicians, physicists and a lot of other scientists most often use a different unit to measure angles, a radian. It is a more natural unit than degrees. A radian is very simply defined. If we have a circle with a radius $r$, and we draw an angle $\theta$ as shown in the diagram below, then we will call the length of the arc produced $l$.

The definition of the radian is quite simply that if $\theta$ is measured in radians then

$\boxed { \theta = \frac{l}{r} }$

## Converting between degrees and radians

Converting between these two ways of measuring an angle is quite simple. Remember that the circumference of a circle, the distance all the way around, is $2\pi r$. But this is just the length $l$ of our arc when $\theta = 360^{\circ}$. So when $\theta=360^{\circ}$ this is equivalent to $\theta = \frac{2\pi r}{r} = 2 \pi$ radians.

This table shows the conversion from degrees to radians for some commonly used angles.

 Degrees Radians 0 0 30 $\frac{\pi}{6}$ 45 $\frac{\pi}{4}$ 60 $\frac{\pi}{3}$ 90 $\frac{\pi}{2}$ 180 $\pi$ 270 $\frac{3\pi}{2}$ 360 $2 \pi$

## The sine function

The sine function, $\sin(\theta)$, can be multiplied by a factor to produce a regularly varying function of any amplitude. For example, $4 \sin(\theta)$ would look like this (remember we are expressing $\theta$ in radians, not degrees).

The sine function is very useful in many areas of mathematics and physics.

## Time varying waves

Suppose we have a wave travelling out from a stone which has been dropped into the water in a smooth pond. The ripples (waves) will travel outwards from where the stone was dropped. If we look at the height of the water at some point on the surface of the pond, then we find that the height $y$ of the water will vary with time $t$ as $y=A + R\sin (\omega t + \phi)$. In this expression $A$ is the mean (average) height of the water (the level of the water before we dropped the stone), $R$ is the size of the wave (the peak height), $\omega$ is something called the angular velocity, $t$ is time (in seconds), and $\phi$ is something called the phase angle.

### Angular velocity

Linear velocity, the velocity of something moving in a straight line in, say, the x-direction is just defined as $v=\frac{x}{t}$. If an object is moving in a circle, we can define a similar concept called the angular velocity, which is not the linear distance divided by time, but rather the angular distance divided by time. But, the angular distance is just $\theta$ (when expressed in radians), so the angular velocity $\omega = \frac{ \theta }{t}$ and is measured in radians per second. Re-arranging this, we can see that $\theta = \omega t$, so in other words $\omega t$ has the same units as radians, they multiply together to measure an angle. This means that $\sin{\theta}$ is equivalent to $\sin(\omega t)$, enabling us to consider the variation of a wave as a function of time.

When expressed in radians, the angle for one complete cycle of a sine curve is just given by $\theta = 2\pi$. Using our relationship between $\theta$ and $\omega t$ we can say that the time taken for one complete cycle, which we called the period $T$ is just given by $T = \frac{2\pi}{\omega}$. So, depending on the value of the angular velocity $\omega$, this determines how many seconds it takes for the sine function to complete one cycle.

### The phase angle

The phase angle, $\phi$ above, is needed as our zero time may not be when the expression $R \sin( \omega t)$ is zero. So, it just allows us to correct for when we start timing, just in case this is not at the moment when our sine function has zero value.

### Putting this together

Below are plots of two sine curves, the first (the blue curve) is given by $y = 4 \sin ( \frac{\pi}{4} t)$. The second one (the green curve) is of $y = 4 \sin ( \frac{\pi}{4} t + 0.5 )$ (remember the phase angle $\phi$ is expressed in radians, and one radian is approximately $57^{\circ}$.

Because $\omega=\frac{\pi}{4}$, this means the period of the wave will be $T = \frac{2\pi}{\omega} = \frac{2\pi}{\pi/4} = 8s$. You can see that the blue curve does indeed complete one cycle in 8 seconds. The green curve has the same value of $\omega$, so it has the same period. However, it is shifted in time due to the phase angle $\phi$. So, at a time of $t=0$, the green curve is not zero, and we say that it is ahead or leads the blue curve as it reaches its first peak before the blue curve does. The number of seconds it leads the blue curve can be found by considering that $\phi = \omega t$ so $t = \frac{\phi}{\omega}$, which in this example is $t = \frac{1/2}{\pi/4} = \frac{2}{\pi} \approx 0.63 s$.

In tomorrow’s blog I will discuss an exam question I have been going over with my students, where the height of water in metres in a harbour is given by e.g. $y = 6\sin(\frac{\pi}{4}t) + 8\cos(\frac{\pi}{4}t) + 11$ (where $t$ is expressed in hours). I will discuss how we can work out the maximum and minimum heights of the water, when e.g. the first maximum occurs, and how many hours a day we can use the harbour if we need a minimum of e.g. 2 metres of water in the harbour.