You are not going to like this page. Aged 11-16 you were told a bunch of science. You probably believed it all (why wouldn’t you?). Then aged 16-18 you were (or might be) told that same science but with added complexity. Often accompanied by the words ‘remember what you learned last year? That wasn’t quite right…’ You may have thought ‘why did they lie to me?’. You might even have thought ‘well, I bet this is wrong too, but at least if I do a degree in a science subject, they’ll finally tell me the truth’.
Sorry to break it to you… there are lots of simplifications at undergraduate level too.
But this isn’t a shortcoming in the education system; the ability to make simplifications is a vital skill in a physical scientist. Science doesn’t really consist of ‘revealing the truth’, so much as constructing models that can be tested against experiment, and incrementally improved. (Unless you’re Einstein and you just revolutionise everything all in one go.) Your science education at school consisted of accessing incrementally better models.
Simplifying a model to make the maths easier is often called making an ‘approximation’. The word ‘approximation’ suggests a use to make estimates, but they are more important than that. Lots of situations, in even seemingly simple situations, are not directly solvable. Then an approximation can sometimes make it easier to solve (or even possible at all).
An approximation we have already seen…
In Section 04: Differential Equations we investigated motion under freefall according to the differential equation
[latex]mg-kv^2=m\frac{dv}{dt}[/latex]
We were quite proud at the time, because we had generalised the simple case of zero drag to the case where drag is present, and given by [latex]D=kv^2[/latex]. In fact, we then recognised that our model of the motion was still simplified in certain ways:
- It assumed that [latex]kv^2[/latex] is a good representation of the drag, and that is likely to be true only if the object is falling through a fluid of very low viscosity. Terms in the drag such as a linear term, [latex]cv[/latex], will become significant if that is not the case.
- It assumed that buoyancy forces, [latex]F_B, were negligible, which in turn requires that the object has a density much greater than that of the medium it is falling through
So our model of the motion is a simplified version of this equation:
[latex]mg-(kv^2+cv+other terms)-F_B=m\frac{dv}{dt}[/latex]
And as such, the equation:
[latex]mg-kv^2=m\frac{dv}{dt}[/latex]
is itself an approximation.
And yet if we don’t want to solve this equation, either numerically (as we did in Section 04) or exactly (which we didn’t do, but can be done), we can simplify it further by assuming that there is no drag or buoyancy forces. Then the equation is the familiar:
[latex]mg=m\frac{dv}{dt}[/latex]
Which means that the accelerating force, the weight [latex]mg[/latex], is equal to mass x acceleration. And this in turn leads to the solution that the acceleration is equal to [latex]g[/latex], which is the familiar case of freefall without drag. When the assumptions in the bullets above are reasonable, we get freefall very nearly at constant [latex]g[/latex], such as when Galileo (maybe) dropped cannonballs off the leaning tower of Pisa. When the assumptions are not reasonable, say when dropping a feather rather than a cannonball, then the solution given by the approximation does not describe the observed behaviour well, and either an exact solution or a better approximation is required.
We mention this just because it is interesting that freefall with no drag is an approximation to the equation [latex]mg-kv^2=m\frac{dv}{dt}[/latex], which in turn is an approximation to the general situation of a complicated drag force and buoyancy present. So in that case, how do you know which level of approximation is valid and sensible? It is a good question, and one with no hard and fast answers. It can take experience as a scientist to be able to judge, which may be one reason why students are not that keen on approximations! One answer could be related to the uncertainty of any measurements you are taking: if your approximation is causing more error than your measurements, then maybe your approximation is too crude?
Powers of [latex]x[/latex] when [latex]x[/latex] is small
One kind of approximation occurs when we have terms in powers of [latex]x[/latex], and [latex]x[/latex] is very small, so that the large powers can be ignored. For example, if [latex]x=0.01[/latex], then [latex]x^2=0.0001[/latex] and [latex]x^3=0.000001[/latex]. Depending on the situation and the level of accuracy required, one or more of these terms might be ignored due to their small size compared with the [latex]x[/latex] term.
As an example, we will ask
“By how much does the energy stored in a spring increase when the extension increases by, say, 5 %?”
We will solve it exactly, and then solve it via an approximation. It will help us see an example of the question raised above: “how do you when an approximation is justified?” The working out is done in the video below.
VIDEO
The approximate solution as worked out in the video is reproduced here. In summary:
The energy stored by a spring at two extensions, [latex]x[/latex] and [latex]x+\Delta x[/latex], is
[latex]E=\frac{1}{2}kx^2; E+\Delta E=\frac{1}{2}k(x+\Delta x)^2[/latex]
Expanding the bracket gives:
[latex]E+\Delta E=\frac{1}{2}k(x^2+2x\Delta x+(\Delta x)^2)[/latex]
Subtracting [latex]E[/latex] from the left-hand side and [latex]\frac{1}{2}kx^2[/latex] (which is equal to [latex]E[/latex]) from the right hand side gives:
[latex]\Delta E=\frac{1}{2}k(x^2+2x\Delta x+(\Delta x)^2)-\frac{1}{2}kx^2[/latex]
Which simplifies (there’s a [latex]\frac{1}{2}kx^2[/latex] and a [latex]-{1}{2}kx^2[/latex], and the [latex]{1}{2}[/latex] applies to the whole bracket…) to
[latex]\Delta E=kx\Delta x+\frac{1}{2}k(\Delta x)^2)[/latex]
Dividing through by [latex]E=\frac{1}{2}kx^2[latex] gives us:
[latex]\frac{\Delta E}{E}=\frac{kx\Delta x+\frac{1}{2}k(\Delta x)^2)}{\frac{1}{2}kx^2}[/latex]
Or:
[latex]\frac{\Delta E}{E}=2\frac{\Delta x}{x}+(\frac{\Delta x}{x})^2[/latex]
[latex]\frac{\Delta x}{x}[/latex] is the fractional change in length, which we know is 5 %, or 0.05. So we have:
[latex]\frac{\Delta E}{E}=2 \times 0.05+0.05^2[/latex]
And the answer to our question ‘By how much does the energy in a spring increase when the extension increases by 5 %?’ is 10.25 %
Of the answer 10.25 %, 10 % comes from the first term and 0.25 % from the second. The first is 40 x larger than the second. So unless a high level of accuracy is required, just calculating the first term is likely to be sufficient. And in such situations, we could have settled for:
So that the approximate solution would have been 10 %. Easy to calculate, and not very different from the ‘correct’ answer of 10.25 %.
One step further – the binomial expansion
In Section 07, we saw that a power series is a series in which each of the terms is a different power of an unknown, say , multiplied by a coefficient. That is,
where the coefficients are labelled so that their number (in the subscript) is equal to the power of it belongs to.
One example of a power series is called the binomial expansion. I’ll begin discussion of the binomial expansion through a very personal example! I remember very clearly my first interview for an undergraduate physics course, aged 18. It went averagely, shall I say, and I was called back for a second interview later that evening. On entering the study of the interviewer, his first words were not “good evening” or “welcome”, but
“what is the square root of 1.01?”
I was rather flustered. And it took me a while to realise this was a question on the binomial expansion. And that ‘while’ was quite an awkward silence!
If you don’t know what the ‘binomial expansion’ means, the following video is for you. If you are fully conversant in it, maybe skip this video?
VIDEO
After the aside in the video, to introduce the binomial expansion, let’s return to the question of . Well, because 1.01 is equal to 1 + 0.01, and taking a square root is the same as raising to the power of ½, this means that .
That expression can be expanded using the binomial theorem, as described in the video:
And because , the value of the terms in successive powers of probably drops off pretty quickly, so that we don’t need to evaluate many terms. Which, in an interview situation, without pen, paper or calculator, is quite helpful… So, using and , gives the first two terms as:
And that is easily computable as 1.005. Which was the answer I gave, as the idea of computing more terms under stress was not appealing… (That was the approach anticipated by the interviewer, which was nice! But the rest of the interview didn’t go so well, and it was only after four interviews that the university accepted me!)
In case I had managed to compute more terms, what good would it have done me? One more term would have given
Notice the extra term makes a very small difference to our previous estimate of 1.005. In fact, a difference of just 0.001 %. And it is only 6 parts per million out from the true value of (I didn’t reason all that out in the interview!). Further terms make even less difference. So in this case, depending on the accuracy required, it is often justifiable to stop the expansion at .
Let’s see a more useful example (it pertains to the world at large rather than the torture of undergraduate candidates) that can be approximated using the binomial expansion.
“By what percentage is the acceleration due to gravity less, due to altitude, on top of Mount Everest, as compared to at sea level?”
To answer that, we will note that the acceleration, due to gravity, at a distance from the centre of a planet of mass , is given fairly well by
where is the constant of proportionality needed to make the numbers work out, called Newton’s gravitational constant. Let’s pretend the Earth is completely spherical, and we’ll assign the following values:
- Height of Everest above sea level,
- Radius of Earth,
Therefore the height of Everest as a fraction of Earth’s radius is
We met the relationship above for indirectly in Section 08. It isn’t obvious necessarily that this problem is related to the binomial expansion, but it is! The video below works through the problem.
VIDEO
The “order” of expansions
You may well hear people say things like “To first order, A is a good approximation of B…” What this means is that if we find a series expansion A, and take just the linear term (the term in ) and any constant term, that is a decent estimate of B. The estimate to second order would include the quadratic term of the expansion. And so on…
So, let’s summarise the video. To first order, the answer to the question “by what percentage is the acceleration due to gravity less, due to altitude, on top of Mount Everest?” is 0.00278 %.
Notice that we don’t need to work out , as all we need are the fractional or percentage deviations from it.
The exact solution for the difference in between Everest and sea level (based on the model ) could be found instead, from the difference:
This has a value of , corresponding to a difference of 0.00276 %. That means we can be quite impressed with the result of our first order approximation (0.00278 %)!! For many purposes the tiny increase in accuracy from working out further order terms in the expansion is not worth the extra effort.
Another step further: Taylor expansions
We saw in Section 07 that a power series is a series in which each of the terms is a different power of an unknown, say , multiplied by a coefficient.
That is,
is the ‘coefficient’ of , that is, the number that needs to be multiplied by, to make the whole thing work. There is a piece of mathematics called Taylor’s theorem, which is very useful. It says that ANY function can be expressed as an infinite series of powers of , as above, provided it meets certain conditions. We aren’t going to worry about those conditions – we are trying to give an insight into what is happening; for more mathematical rigour, an undergraduate maths book might be a good starting point!
Here’s a video to give you a sense of what Taylor’s theorem means. Afterwards, we’ll apply it to the context of this page – that is, approximations!
VIDEO
Just like we saw with the binomial expansion, sometimes we are justified in neglecting all but a few of the terms in a Taylor series expansion. One area in science this occurs over and over again is in what is called the harmonic oscillator.
First, we’ll remind ourselves of what a truly harmonic oscillator is! (We met them in Section 09: Conservation Laws).
Then we’ll find that any mass vibrating in stable equilibrium acts as a harmonic oscillator if the vibrations are small enough. The smaller the vibrations, the better the approximation of a real oscillator to a harmonic oscillator…
So, first! Truly harmonic oscillators!
We saw in Section 09 the behaviour of oscillating springs. When there is a restoring force proportional to the extension of the spring (in other words, to the displacement from equilibrium), as in , then the energy, , stored in the elastic potential spring is proportional to the square of that extension. And in fact, .
During the oscillation of a mass on a spring, energy is continually transformed from kinetic energy to elastic potential energy on the way out from the equilibrium point to the motion’s extremity… And then back from potential to kinetic on the way back in toward the centre… Since the total energy of the oscillation is constant (assuming the ideal case of no friction of other effects that would dissipate energy out of the system), and since the potential energy is a U-shaped parabola (), then the kinetic energy is an upturned parabola. In other words, with reference to the diagram below, at each displacement, x, the potential energy (shown in blue) plus the kinetic energy (red) gives the constant total energy in yellow.
What’s so special about springs? Well, in one sense, nothing. But in another sense, they are everywhere! Because anything that oscillates due to a restoring force towards an equilibrium point can be modelled as a mass on a spring.
But what if the restoring force is not proportional to the extension? That is, what if does not hold? Then oscillations may well still occur, and the potential energy curve will still have some form of U-shape. But it will not be truly parabolic, according to .
And next: oscillators that are not truly harmonic
Atoms within a molecule is an example of a real-life system that is a little like a spring. Atoms within molecules have an ‘equilibrium separation’, , the distance apart at which they are neither attracted to, nor repelled from, each other. If two atoms within a molecule stray further than , they are subject to an attractive force. If they move closer than , they repel each other. In molecules, the atoms are not continually at the equilibrium separation – they oscillate about it due to a restoring force, like a mass on a spring!
Except that the restoring force is not proportional to the displacement from equilibrium, and the potential energy curve is not a parabola. In fact it is not even symmetrical, because the attractive and repulsive forces behave differently. Here is a picture of what is often called the ‘Lennard-Jones potential’ for such a situation. In the diagram, is the equilibrium separation.
Notice we called it a potential, rather than a potential energy. The potential at a point in the field in the electrical potential energy per unit charge, of an object in the field. The reason for doing this is that the potential energy depends on the charge of the particle in the field. By inventing ‘potential’ as potential energy per unit charge, we create a quantity that is a property of the electric field itself, which will then affect an object in it, but doesn’t depend on that object. But you can just imagine it as potential energy per unit charge.
When dealing with gravitational fields, rather than electric fields, there is an analogous quantity called the gravitational potential of the field. This is the potential energy per unit mass of an object in the field, as opposed to potential energy per unit charge. That is because gravitational fields act on masses, whereas electric fields act on charges.
The video below explains visually why the harmonic oscillator is a good approximation to potential energy functions that are not given by .
VIDEO
So, to summarise the video, the bottom of the L-J potential energy curve can be approximated by the familiar parabola. The approximation becomes less good as we get further from the equilibrium separation. So small oscillations are approximated quite well by a potential energy function . But the larger the oscillation (the greater the energy of the particle), the less good the approximation.
And why is that so important? Why would we want to make that approximation anyway? Well, the harmonic oscillator is a system for which the maths is relatively straightforward and well-known. For more complicated potential energy functions, it is often a good idea to approximate it as a harmonic oscillator, and then make adjustments to the model for situations where the oscillations are too large for the model to work well.
Back to power series, with the harmonic oscillator!
We have gained a conceptual understanding that the harmonic oscillator can be an approximation to more realistic potential energy functions, provided the oscillations are small enough.
Now we’ll describe semi-mathematically why the harmonic oscillator potential is the right potential! That is, why a parabola, rather than some other U-shape? We know a parabola works for a truly harmonic potential. But why is it the best approximation for any potential with a minimum that constrains movement (in the form of oscillations) to a particular region? If you don’t want to engage with maths in this section, that’s ok – you can stick with the visual explanation we provided in the video above.
The Lennard-Jones potential energy function was just one example. This paragraph is about more general behaviour, and it may be worth reading it step-by-step in conjunction with the diagram below. Imagine an arbitrary potential energy function that has a local minimum at a point (there may or may not be other minima). Imagine also that the particle has a total energy that is only slightly larger than . Then the particle is constrained to move only in regions such that , as shown as the shaded region and arrows in the diagram below.
The Taylor expansion of the potential energy function around the equilibrium point , is:
The first of these terms is simply the constant value . However, we can shift the potential energy values by any constant value without altering the science (much like in gravitational potential energy calculations we measure height, and therefore energy, from any convenient height, because it is changes in the height – and energy – that count).
So let’s set the zero of the potential energy to be zero at the point , so that .
The second term (the linear one) in the expansion includes the first derivative, , evaluated at the point around which we are expanding.
However, the point corresponds to a potential minimum. A minimum has zero gradient, and so the derivative is zero. Therefore the linear term is zero.
The third term is , which involves the second derivative. Unlike the linear term, there is no reason for this term to be zero in general.
If we assume that the particle energy, , is only just larger than (so that the region around that the particle in constrained to explore is small) then it will be a reasonable approximation to only expand the potential energy to second order (just like we did for the expansions earlier in this page. In this case, the Taylor expansion for the potential energy function becomes
Therefore,
And if we set (don’t worry – that’s just a label to simplify things; no science has actually happened in there!), then we get:
Thus almost any particle which is moving in the vicinity of a potential minimum can be effectively described by a potential energy function of this form. But this is nothing other than the potential energy of a spring, centred on , because is just the distance from the equilibrium point.
So, at risk of labouring the point, because I am aware we have said this before… Almost any oscillating mechanical system will behave like a vibrating spring (harmonic oscillator), so long as the oscillations are small enough. Therefore, the harmonic oscillator is one of the most important mechanical systems in all of science. If it’s that important for oscillations, then, you might not be surprised to find out that we’ll revisit it in Section 14: Oscillations and Waves!
Link to Section 09: Conservation Laws and Springs – COMING SOON
Return here to find out more about simple harmonic motion in the context of a mass vibrating on a spring, and how energy conservation applies to this kind of motion.
Link to Section 13: Circular motion and phasors
Where we describe simple harmonic motion as a 1D projection of 2D circular motion.
Link to Section 14: Oscillations and Waves – COMING SOON
This is a logical step after this page. We concentrated here on the potential energy function for a perfect freely oscillating system with no energy loss, In section 14 we look in more detail at oscillators that are driven by external forces, and have energy losses.
Link to Home Page
And you can always return here…