Presentation: On theories

This presentation was made to explore what a theory is, how it is developed, what makes a good theory as well as related topics such as the concepts of accuracy and precision, the scientific method, Karl Popper’s requirements of a theory, Occam’s razor, the ideas of  provability v. unfalsifiability etc.

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Launching Physics Capsule, the online magazine

Long back I was part of an online science update project called The Scientific Papers. Those of you who have followed me there know it ran pretty successfully for three years before I decided to shut it down. Until now, my only presence online has been here, on my personal website; but starting June the 24th, that is about to change.

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Abraham Lincoln’s unitary method problem

They say Lincoln had very little or no formal education. The former US president himself is known to have spoken few good things about his childhood schooling.

However, sometime back in June last year two mathematics professors in Illinois confirmed the discovery and authenticity of two papers believed to have been part of Linoln’s cyphering book — that is a 19th century term for “workbook”, in case you are wondering.

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On projecting an octahedron onto ellipses

I recently came across an interesting animation that was shared with a rather poor choice of words: “Dots that are moving around in a circle! Connect them and you’ll get a moving octahedron…”

I find that this statement is faulty for reasons I will explain below; but first, the animation in question:

I will try to limit the mathematics to the necessities; generally, neither is this a two-dimensional figure, nor are those circles. I think the animation, in its present form, merely benefits from a rather tricky choice of perspective.

Before we actually talk about this, I hope to clear up a few things (read, prove a couple of lemma) regarding the construction and structure of an octahedron.

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Dark energies, Higgs boson seesaws and the tiny Grand Unified Scale

ONE OF THE BIGGEST PROBLEMS in contemporary theoretical physics is the dark matter-dark energy duo. Back when Einstein forced himself to tailor his theory of general relativity to a static universe, he had come up with what was termed the cosmological constant; it was, for all practical purposes, a sort of correction parameter which made sure his famous field equation worked in a static universe. Once turned off (i.e. the Lambda in the equation below (the upside-down V) once set to zero) one term is knocked off the equation once again returning it to a state of satisfying a changing universe.


Quantum fluctuations and the cosmological constant

The entire elaboration that was the cosmological constant has since died and been resurrected. Physicists are now toying with the idea that the value-equivalents of temporary fluctuations of energy in space (as a result of Heisenberg’s uncertainty principle) called quantum vacuum fluctuations is nothing but the cosmological constant.

Click here to read an interesting paper introducing the cosmological constant pedagogically. Or you can read Steven Weinberg’s slightly denser paper instead.

A lot of hope was stacked on the Higgs Boson prior to its discovery, and if anything, that has all be trebled after the fact. Now, the latest attribution to the so-called God particle, is the mysterious behaviour of dark matter (no, not the comic,) and dark energy and, in turn, the cosmological constant itself.

Ride the Higgs boson seesaw?

In their PRL paper, “A Higgs seesaw mechanism as a source for dark energy,” physicists Lawrence Kraus and James Dent propose using the seesaw mechanism (which was originally used to understand neutrino masses which are a mere millionths of, say, particles in the Lepton family) to understand how the Higgs particle can create an energy density in the Grand Unified Scale (about 16 orders smaller than a proton) at which the three known energies are believed to tie to form a unification à la Einstein’s Grand Unification Theory — quite aptly shortened to GUT.

Promises of a minuscule order

The discovery of the Higgs boson has provided a much needed injection of fact into what was otherwise turning out to appear like a lot of wild conjecture.

Dark matter is what occupies a majority of the mass on any given galaxy, but can only be detected indirectly (by its gravitational influence) since it does not interact with light (things such as light shining off it do not occur.) Such an energy, called dark energy for obvious reasons, appears to have a magnitude of electroweak physics’ GUT order.

This webpage explains the neutrino seesaw mechanism in (once again) a pedagogical manner better than I care to elaborate here. It is a fairly typical but more fluid treatment by one, Robert Klauber.

Kraus and Dent’s theory of Higgs coupling along the lines of the Neutrino seesaw explanation arrives at precisely this order of magnitude of energy contribution, as it were, to a quantum vacuum. The duo calls this an “extension of the standard model” — another promising, if not fully proven, foundation on which to explain their ideas.

The pith of this paper is that by operating on a factual entity (the Higgs boson,) using, analogously, a previously tested approach (the neutrino seesaw,) treated on the platform of a well-accepted idea (the standard model,) we end up with energies of exactly the same minuscule order of magnitude as the one we found hard to explain convincingly.

Perhaps this paper is more of a nudge in the right direction than a revelation itself, but sometimes in physics nudges are exactly what we need.

 Cover image: Rosette nebula via Flickr/bobfamiliar 

Does laser eye-surgery last indefinitely?

Laser eye surgery is without doubt an amazing procedure, giving people the opportunity to throw away their glasses and contact lenses. Laser eye surgery has actual been around a lot longer than most people realise and it was first carried out over 20 years ago, admittedly not with the same impressive results as we expect today.

In the early days, laser eye surgery was far more invasive than it is today, leaving people with extensive corneal scarring and variable results. Laser eye surgery risks were also far higher and they also tended to be more serious.

Today however, laser eye surgery is minimally invasive and the results are extremely impressive. Around 95% of people having the procedure will end up with 20:20 vision and almost 100% will have driving standard of vision or better. Laser eye surgery is the most common elective surgery available and is far more popular than even the most common cosmetic surgery procedures.

For those deciding whether or not to have laser eye surgery then one of the most commonly asked questions is just how long the visual results are likely to last. This article is going to outline the most important points you need to know about this aspect of laser eye surgery:

Laser re-treatments

Around 6% of all laser eye surgery procedures will have to be repeated with a ‘top up’ treatment. This is normally done for free by the clinic that carried out the original procedure and it may only be required for one eye. To illustrate how laser re-enhancements work, it is best to use an example.

Imagine your prescription was -7.00 Dioptres prior to your surgery and then after surgery your prescription was completely eliminated to zero, meaning you had perfect vision. Then over the course of the next year your prescription slowly regressed to -1.00.

This is obviously only a fraction of what your prescription was before you had the surgery but it is enough to affect your vision. The laser is then re-applied to re-correct this remaining prescription.

Long sighted prescriptions

Long sighted prescriptions (plus e.g. +5.00) are more likely to regress than short sighted (minus e.g. -3.00) prescriptions.

Higher prescriptions

The higher the prescription the more likely you are to need a laser re-enhancement. The reason for this is that the more something has changed, the more likely it is to return back to its original state.

Reading glasses as you get older

Regardless of whether you have had laser eye surgery or not, you are still going to need reading glasses as you reach your mid 40’s. This is related to the lens in your eye that slowly loses its power which is a condition called presbyopia.

Eye diseases

As we get older so the chances of us getting age related conditions such as cataracts increases. Cataracts typically start affecting us aged around 70 years old and they can cause a change in our prescription. So even though you may have had laser eye surgery previously the cataracts could mean you end up needing glasses as you get older.

So, as you can see, having laser eye surgery does not guarantee that you will be free from glasses/contact lenses for life but for most people they will at least have perfect distance vision for a prolonged period of time.

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Tim-HarwoodTim Harwood is an optometrist with over 8 years in practice with a specialist interest in both laser eye surgery and contact lenses. He has worked both in the United Kingdom as well as Australia and has worked for both multiple and independent opticians. Tim also provides useful information on his own website,

[/sws_blue_box]  Cover image: Flickr/Peretz Partensky 

Integration 101 – Session 3: integration and area, and the fundamental theorem of calculus

Last session we saw two important aspects under integration: methods of solution and the steps to solve a definite integral. In this session, we shall be seeing why definite integrals can be solved this way. In other words, we’ll see why it works.

This is mathematics, so it isn’t like somebody decided you could subtract integrals of limit points and whatever you get is the right answer. And then we didn’t follow all of that for centuries to satisfy one man’s fancy. There is a reason it works (or proof, if you want to call it that,) and we will be understanding that today.

First of all, if you have not been here for sessions 1 and 2, my advice is that you head right back and go through them before continuing this session. You can read session 1 here and you’ll find session 2 here. Go on, I’ll wait.

The weird thing about areas

If you remember last session well, I asked you to find the area of a weird looking object. My point then was to show you basically that integration of something can tell you the area of that thing.

[sws_red_box box_size="100%"] Now, this is obviously not always true — a little common sense will tell us that. For instance, integration of velocity won’t give you the area of anything. What I intend to say is that if the area of something can be measured, such as an apple, integration will most likely give you its value. [/sws_red_box]

As always, to understand how this works, let’s reduce this real-world problem to a mathematical model. (That’s just a fancy way of saying let’s use a wooden plank and a toy car to understand the effects of toppling over instead of trying it with real cars, real mountain roads and real humans.

Here’s our mathematical model:

Graph 1

Ugh, not very interesting, is it? Well, what model were you expecting?

Now let’s zoom in to this part:

Graph 2
Graph 2

I hope you recognise where the region shown in graph 2 is present in graph 1. Got it? Good, now I’ve been a very kind soul and scribbled a few things on this graph:

1. I’ve marked two points: x = 2, y=0 and x = 4, y=4

2. I’ve marked another point, A, in red

3. I’ve marked the curve between (2,0) and (4,4) as C

4. I’ve shaded the area under curve C in light grey

5. I’ve pointed out to the graph’s equation in case you can’t find it anywhere

6. I’ve spoken to dolphins undersea

Now, once again, this is the equation for our graph:

Let’s start with the absolute basics. Firstly, whether you say y = something or f(x) = something, it doesn’t matter because y=f(x), they’re just two ways of saying the same thing. Secondly, the graph comes by plotting points by giving values to x, not y. You always give values to the thing on the right hand and see what happens to the thing on the left. Therefore, if I were to say x = 2, I get… y=0. Now, if x=3, y=1; and if x=4 then y=4. Notice the parts I’ve written in bold: these are the points I’ve marked on the graph, (2,0) and (4,4)

So what’s the fuss about all these points? Let’s do a little exercise. Let’s integrate eq. 1 with respect to x. I mean, that’s what we’re here for isn’t it? It’s been seven paragraphs into this third integration session and we still have not integrated anything? Well, try it yourself and see what you’ll get.

But wait! We’re geniuses, so let’s go one step further and try to make this thing definite. Sounds like fun.

Now how exactly do you do that? By finding the limits, of course. Yes, but how? Just ask yourself these simple questions:

1. What’s our superstar variable?  Answer — x

2. What’s out curve? Answer — c

3. What’s the lowest value of our variable in our curve? Answer — 2

4. What’s the highest value of our variable in our curve? Answer — 4

So, there. For an equation (like eq. 1) given for curve C in variable x, the upper limit is 4 and lower limit is 2. How simple was that?

Good, now integrate eq. 1 with limits and you’ll get eq. 2 below. (I’ve shown the intermediate steps in case you’re lost.)

Aren’t we feeling creative? Now let’s do something else. Let’s join points (2,0) and (4,4) and make a triangle like I’ve shown in the figure below. I’ve now even given the points (vertices) names as M, N and O.

Graph 3
Graph 3

Great, now let’s find the area of that triangle. Yes, I see your stare. Do we have no better work? Right now, no, so let’s find the area of the triangle. We know it’s a right angled triangle, so the area is given by,

From graph 3, our base stretches from x=2 all the way to x=4 (i.e. the line MO) which is 4 – 2 = 2 units in length. Next, out height is the line NO, which stretches from y=4 to 4=0 putting its length at 4 – 0 = 4 units but you already knew all that.

So our area is,

Well, well, well, will you look at that! In both equations 2 and 3, our answers are roughly similar. Granted, eq. 3 is very slightly (2/3 times) greater, but that is because of the extra region as I’ve highlighted in grey in graph 3.

So what does that tell us? It tells us that integration of a curve gives you the area of graph under it. Beautiful!

[sws_red_box box_size="100%"] Please note: for the precise minded, if the slight difference arising due to the extra region in graph 3 does not satisfy you, there are more advanced proofs. We’ll discuss that once again in session 4.[/sws_red_box] [sws_red_box box_size="100%"]Please note this too: the above integration-area equivalence is not always true. For instance, in a line integral, the integration restricts itself entirely to the line. In particular this integration-area equivalence arises when the lower limits of our region (the shaded part in graph 1) has a limit at the x-axis. What if it goes beyond? Or into the third dimension? :-) [/sws_red_box]

Alright, here’s an exercise for you, and this is why I marked point A. Just like we’ve done above for curve C: MN, try to prove the same result for curve D: AMN. Hint: make two triangles with their only common vertex at M.

The fundamental theory of calculus

We’re here at last. Let us quickly sum up a couple of important things we have learnt since session 1. We know how these work, and the fundamental theorem of calculus will tell us why.

Firstly, we know integration can definitively be done over an a range of values with a fixed beginning and end called limits; secondly, we know this gives the area under graph in most general circumstances; thirdly, we know that integrating a function and then subtracting the integral at specific situations corresponding to its two limits will give us the definite value of that integral.

Alright, so here is what the fundamental theory of calculus says:

Huh? We’ve seen that before haven’t we? Let me explain the notation: the capital F stands for integral. While eq. * above is the standard notation, here is another way to write the same thing:

There, you have definitely seen eq. 4 before! We discussed it in session two when learning how to solve definite integrals, but we just learnt it like everything else then, we never called it with a fashionable name like, ‘the fundamental theorem of calculus.’

So now you might be wondering why this is held in such high regard. All it does is state what we already know, right, so what’s so fundamental about it?

In short, this theorem says our method of integration is correct. (Well, that’s reassuring.) But it also has other deep implications such as the area thing we saw above. A second fundamental theorem of calculus is even more fundamental (and rather obvious) and it simply says how differentiation and integration and interchangeable, like this below. (Note that x and t are variables, a is a constant.)

Since you probably are waiting for proof of this theorem, let me give you some good news and some bad news: the good news is, we’re not going to prove this theorem now. The bad news is, we’ll prove it after we learn yet another theorem called the mean value theorem in integration, or, as Doofenshmirtz would say, “Behold! The ‘for every definite integral there exists a rectangle with the same area and width -inator,’ or the FEDITERWSAW-inator.”

[sws_red_box box_size="100%"] If you don’t know who Dr Heinz Doofenshmirtz is, it’s time to abandon integration and turn on the Disney Channel. Seriously. [/sws_red_box]

It’s question time

After sessions 1 and 2, I received a few emails carrying doubts regarding stuff we have talked about so far. I’m going to take time today in session 3 to clarify a couple of those because it may help others too to be free and fully clarified when learning the ultra-important things coming up in sessions 4 and 5. (See contents at the bottom of this article.)

Josh Fraser asks, “What really is the exponent (e) in math? I’ve seen that in so many places.”

First of all, John, that’s not the exponent. If I say 2 is raised to the power 5, then 5 is the exponent, another name for power. The letter e you see used everywhere, which you’re referring to, is actually called Euler’s number after the Swiss mathematics superstar, Leonhard Euler (say leon-hard-oiler.) This e is just a mathematical constant like pi and is approximately equal to 2.7182. It is defined like this:

Why is it important? The Euler number (sometimes called Napier’s constant) is its own derivative and integral. In other words, almost always, no matter what you do with e — throw it off a cliff, run it over with a car, integrate it, drown it, differentiate it, light it on fire — it remains just as it is. Kind of adamant, eh? And, needless to say, such properties get us maths and physics nerds terribly interested.

Shania Hubbard asks, “How do I integrate the cube of cosine?”

How indeed! I was overjoyed when I saw this question because I remember being asked this in an examination and not being able to answer it till the very last moment when it struck me like a light bulb but it was too late. Anyway, here’s the question the way it’s supposed to look:

First try it yourself before looking at the answer below.

This one is solved by substitution; in fact, it is a good example of how substitution can sometimes get messy. Now if I directly substitute something for cos x, say u, then deriving it would go, 2 times the square of sin x, which is nowhere in the function. A little more effort will reveal this:

Now substitute u in place of sin x, so,

Substituting this in eq. 5 we get,

Now integrate eq. 6 as usual,

Now just substitute back u = sin x in eq. 7 and you have your answer:

Now try it for the cube of sin, tan and other trigonometric functions. Then try it for the fourth power and see how you can handle it.

Dave Moreno asks, “I’m going back to my college days with your Integration 101 series and I’m enjoying it more than I did in college. But I remember we were taught a formula method for substitution whereas you explained it in today’s session [session 2] without any formula. I actually understood and enjoyed it, but how different is it from the formula?”

Well, big question, Dave. First of all, congratulations on deciding to re-visit college calculus; there are not many who are brave enough to do that. But, you’re right, the substitution method can be taught with a formula. And it looks like this:

See how you made that “Ugh! What is that spelschtick? I’m gettin’ oatta heer!” look on your face? That’s why I didn’t mention this in session 2. In reality, you do not need to know this to perform substitution because this exists just to generalise the substitution rule and prove that it works. (What you see above is sufficient proof.)

When you’re working out mathematics, functions rarely show up in the convenient f'(g(x))g'(x) form as in eq. 8, so identifying them in this manner becomes simpler when you forget about this rule/formula hullabaloo and just do it the way we saw in session 2 (and even in Shania’s question above.) Trust me, forget this formula and you’ll fare better.

I hope that has covered most doubts. Other questions I received were in some way or other related to these three (especially the last two) so I figured if I answered these publicly, it should help others solve their own problems, if any.

It’s integration table time

We’ll look at ten hyperbolic functions today — yes, there are exactly ten; and no, we still have not let go of trigonometric functions.  Here’s a look at the future: there are 32 more. Just the trigonometric ones, that is. And then there are about thrice as many left and then — I think I’ll stop now.


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Note how the left hand sides of v16 – v25 are very similar to those of v6-v15 except that v16-v25 are hyperbolic.

Also note there is another way to represent hyperbolic functions: arc-. So sinh is arcsin, cosh is arccos, csch is arccsc and so on. I usually prefer the arcsec, arccot… representation but stuck to the more popular method in the table above.

So, in today’s session we’ve covered the fundamental theorem of calculus (which we will prove in session 4) and seen an introduction to the integration-area equivalence (which we’ll further understand after session 4.) In other words, sessions 3 and 4 are very closely linked and you might want to treat this as an introduction to session 4 which goes into the beginning of the dark depths of maths that have made students shiver for centuries.

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Here is the plan for session 4:

1. Reimann sums, mean value theorem in integrals, and hence proof of the fundamental theorem of calculus

2. Properties of definite integrals

3. Integration by parts


Until next time, enjoy calculus… muhahahaha! [vhb]

 Cover image: Flickr/functoruser