# Spheres in a Space with Trillions of Dimensions

I don’t venture into speculative science writing – this is just about classical statistical mechanics; actually about a special mathematical aspect. It was one of the things I found particularly intriguing in my first encounters with statistical mechanics and thermodynamics a long time ago – a curious feature of volumes.

I was mulling upon how to ‘briefly motivate’ the calculation below in a comprehensible way, a task I might have failed at years ago already, when I tried to use illustrations and metaphors (Here and here). When introducing the ‘kinetic theory’ in thermodynamics often the pressure of an ideal gas is calculated first, by considering averages over momenta transferred from particles hitting the wall of a container. This is rather easy to understand but still sort of an intermediate view – between phenomenological thermodynamics that does not explain the microscopic origin of properties like energy, and ‘true’ statistical mechanics. The latter makes use of a phase space with with dimensions the number of particles. One cubic meter of gas contains ~1025 molecules. Each possible state of the system is depicted as a point in so-called phase space: A point in this abstract space represents one possible system state. For each (point-like) particle 6 numbers are added to a gigantic vector – 3 for its position and 3 for its momentum (mass times velocity), so the space has ~6 x 1025 dimensions. Thermodynamic properties are averages taken over the state of one system watched for a long time or over a lot of ‘comparable’ systems starting from different initial conditions. At the heart of statistical mechanics are distributions functions that describe how a set of systems described by such gigantic vectors evolves. This function is like a density of an incompressible fluid in hydrodynamics. I resorted to using the metaphor of a jelly in hyperspace before.

Taking averages means to multiply the ‘mechanical’ property by the density function and integrate it over the space where these functions live. The volume of interest is a  generalized N-ball defined as the volume within a generalized sphere. A ‘sphere’ is the surface of all points in a certain distance (‘radius’ R) from an origin

$x_1^2 + x_2^2 + ... + x_ {N}^2 = R^2$

($x_n$ being the co-ordinates in phase space and assuming that all co-ordinates of the origin are zero). Why a sphere? Because states are ordered or defined by energy, and larger energy means a greater ‘radius’ in phase space. It’s all about rounded surfaces enclosing each other. The simplest example for this is the ellipse of the phase diagram of the harmonic oscillator – more energy means a larger amplitude and a larger maximum velocity.

And here is finally the curious fact I actually want to talk about: Nearly all the volume of an N-ball with so many dimensions is concentrated in an extremely thin shell beneath its surface. Then an integral over a thin shell can be extended over the full volume of the sphere without adding much, while making integration simpler.

This can be seen immediately from plotting the volume of a sphere over radius: The volume of an N-ball is always equal to some numerical factor, times the radius to the power of the number of dimensions. In three dimensions the volume is the traditional, honest volume proportional to r3, in two dimensions the ‘ball’ is a circle, and its ‘volume’ is its area. In a realistic thermodynamic system, the volume is then proportional to rN with a very large N.

The power function rN turn more and more into an L-shaped function with increasing exponent N. The volume increases enormously just by adding a small additional layer to the ball. In order to compare the function for different exponents, both ‘radius’ and ‘volume’ are shown in relation to the respective maximum value, R and RN.

The interesting layer ‘with all the volume’ is certainly much smaller than the radius R, but of course it must not be too small to contain something. How thick the substantial shell has to be can be found by investigating the volume in more detail – using a ‘trick’ that is needed often in statistical mechanics: Taylor expanding in the exponent.

A function can be replaced by its tangent if it is sufficiently ‘straight’ at this point. Mathematically it means: If dx is added to the argument x, then the function at the new target is f(x + dx), which can be approximated by f(x) + [the slope df/dx] * dx. The next – higher-order term would be proportional to the curvature, the second derivation – then the function is replaced by a 2nd order polynomial. Joseph Nebus has recently published a more comprehensible and detailed post about how this works.

So the first terms of this so-called Taylor expansion are:

$f(x + dx) = f(x) + dx{\frac{df}{dx}} + {\frac{dx^2}{2}}{\frac{d^2f}{dx^2}} + ...$

If dx is small higher-order terms can be neglected.

In the curious case of the ball in hyperspace we are interested in the ‘remaining volume’ V(r – dr). This should be small compared to V(r) = arN (a being the uninteresting constant numerical factor) after we remove a layer of thickness dr with the substantial ‘bulk of the volume’.

However, trying to expand the volume V(r – dr) = a(r – dr)N, we get:

$V(r - dr) = V(r) - adrNr^{N-1} + a{\frac{dr^2}{2}}N(N-1)r^{N-2} + ...$
$= ar^N(1 - N{\frac{dr}{r}} + {\frac{N(N-1)}{2}}({\frac{dr}{r}})^2) + ...$

But this is not exactly what we want: It is finally not an expansion, a polynomial, in (the small) ratio of dr/r, but in Ndr/r, and N is enormous.

So here’s the trick: 1) Apply the definition of the natural logarithm ln:

$V(r - dr) = ae^{N\ln(r - dr)} = ae^{N\ln(r(1 - {\frac{dr}{r}}))}$
$= ae^{N(\ln(r) + ln(1 - {\frac{dr}{r}}))}$
$= ar^Ne^{\ln(1 - {\frac{dr}{r}}))} = V(r)e^{N(\ln(1 - {\frac{dr}{r}}))}$

2) Spot a function that can be safely expanded in the exponent: The natural logarithm of 1 plus something small, dr/r. So we can expand near 1: The derivative of ln(x) is 1/x (thus equal to 1/1 near x=1) and ln(1) = 0. So ln(1 – x) is about -x for small x:

$V(r - dr) = V(r)e^{N(0 - 1{\frac{dr}{r})}} \simeq V(r)e^{-N{\frac{dr}{r}}}$

3) Re-arrange fractions …

$V(r - dr) = V(r)e^{-\frac{dr}{(\frac{r}{N})}}$

This is now the remaining volume, after the thin layer dr has been removed. It is small in comparison with V(r) if the exponential function is small, thus if ${\frac{dr}{(\frac{r}{N})}}$ is large or if:

$dr \gg \frac{r}{N}$

Summarizing: The volume of the N-dimensional hyperball is contained mainly in a shell dr below the surface if the following inequalities hold:

${\frac{r}{N}} \ll dr \ll r$

The second one is needed to state that the shell is thin – and allow for expansion in the exponent, the first one is needed to make the shell thick enough so that it contains something.

This might help to ‘visualize’ a closely related non-intuitive fact about large numbers, like eN: If you multiply such a number by a factor ‘it does not get that much bigger’ in a sense – even if the factor is itself a large number:

Assuming N is about 1025  then its natural logarithm is about 58 and…

$Ne^N = e^{\ln(N)+N} = e^{58+10^{25}}$

… 58 can be neglected compared to N itself. So a multiplicative factor becomes something to be neglected in a sum!

I used a plain number – base e – deliberately as I am obsessed with units. ‘r’ in phase space would be associated with a unit incorporating lots of lengths and momenta. Note that I use the term ‘dimensions’ in two slightly different, but related ways here: One is the mathematical dimension of (an abstract) space, the other is about cross-checking the physical units in case a ‘number’ is something that can be measured – like meters. The co-ordinate  numbers in the vector refer to measurable physical quantities. Applying the definition of the logarithm just to rN would result in dimensionless number N side-by-side with something that has dimensions of a logarithm of the unit.

Using r – a number with dimensions of length – as base, it has to be expressed as a plain number, a multiple of the unit length $R_0$ (like ‘1 meter’). So comparing the original volume of the ball $a{(\frac{r}{R_0})}^N$ to one a factor of N bigger …

$aN{(\frac{r}{R_0})}^N = ae^{\ln{(N)} + N\ln{(\frac{r}{R_0})}}$

… then ln(N) can be neglected as long as $\frac{r}{R_0}$ is not extreeeemely tiny. Using the same argument as for base e above, we are on the safe side (and can neglect factors) if r is of about the same order of magnitude as the ‘unit length’ $R_0$. The argument about negligible factors is an argument about plain numbers – and those ‘don’t exist’ in the real world as one could always decide to measure the ‘radius’ in a units of, say, 10-30 ‘meters’, which would make the original absolute number small and thus the additional factor non-negligible. One might save the argument by saying that we would always use units that sort of match the typical dimensions (size) of a system.

Saying everything in another way: If the volume of a hyperball ~rN is multiplied by a factor, this corresponds to multiplying the radius r by a factor very, very close to 1 – the Nth root of the factor for the volume. Only because the number of dimensions is so large, the volume is increased so much by such a small increase in radius.

As the ‘bulk of the volume’ is contained in a thin shell, the total volume is about the product of the surface area and the thickness of the shell dr. The N-ball is bounded by a ‘sphere’ with one dimension less than the ball. Increasing the volume by a factor means that the surface area and/or the thickness have to be increased by factors so that the product of these factors yield the volume increase factor. dr scales with r, and does thus not change much – the two inequalities derived above do still hold. Most of the volume factor ‘goes into’ the factor for increasing the surface. ‘The surface becomes the volume’.

This was long-winded. My excuse: Also Richard Feynman took great pleasure in explaining the same phenomenon in different ways. In his lectures you can hear him speak to himself when he says something along the lines of: Now let’s see if we really understood this – let’s try to derive it in another way…

And above all, he says (in a lecture that is more about math than about physics)

Now you may ask, “What is mathematics doing in a physics lecture?” We have several possible excuses: first, of course, mathematics is an important tool, but that would only excuse us for giving the formula in two minutes. On the other hand, in theoretical physics we discover that all our laws can be written in mathematical form; and that this has a certain simplicity and beauty about it. So, ultimately, in order to understand nature it may be necessary to have a deeper understanding of mathematical relationships. But the real reason is that the subject is enjoyable, and although we humans cut nature up in different ways, and we have different courses in different departments, such compartmentalization is really artificial, and we should take our intellectual pleasures where we find them.

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Further reading / sources: Any theoretical physics textbook on classical thermodynamics / statistical mechanics. I am just re-reading mine.

# You Never Know

… when obscure knowledge comes in handy!

You can dismantle an old gutter without efforts, and without any special tools:

Just by gently setting it into twisted motion, effectively applying ~1Hz torsion waves that would lead to fatigue break within a few minutes.

I knew my stint in steel research in the 1990s would finally be good for something.

If you want to create a meme from this and tag it with Work Smart Not Harder, don’t forget to give me proper credits.

# Lest We Forget the Pioneer: Ottokar Tumlirz and His Early Demo of the Coriolis Effect

Two years ago I wrote an article about The Myth of the Toilet Flush, comparing the angular rotation caused by the earth’s rotation to the typical rotation in experiments with garden hoses that make it easy to observe the Coriolis effect. There are several orders of magnitude in difference, and the effect can only be observed in an experiment done extremely carefully, not in the bathtub sink or toilet flush.

Now two awesome science geeks have finally done such a careful experimenteven a time-synchronized one, observing vortices on either hemisphere!

The effect has been demonstrated in a similarly careful experiment in 1908. It had been done on the Northern hemisphere only, but if it can attributed it to the Coriolis effect by ruling out other disturbances, the different senses of rotations are straight-forward.

Austrian physicist Ottokar Tumlirz had published a German  paper called “New physical evidence on the axis of rotation of the earth”. I had created this ugly sketch of his setup:

Rough sketch based on the abstract of Tumlirz’ paper, not showing the vessel containing these components [*]

A cylindrical vessel (not shown in my drawing) is filled with water, and two glass plates are placed into it. The bottom plate has a hole, as well as the vessel. Both holes are connected by a glass tube that has many small holes. The space between the two plates is filled with water and water slowly flows out – from the bulk of the vessel through the the tiny holes into the tube. These radial red lines are bent very slightly due to the Coriolis force, and the Tumlirz added a die to make them visible. He took a photo 24 hours after starting the experiment, and the water must not flow out faster than 1 mm per minute.

Ernst Mach has given an account of Tumlirz’ experiment, quoted in an article titled Inventors I Have Met – anecdotes by a physicist approached by ‘outsider scientists’, once called paradoxers, today often called crackpots. I learned about Ernst Mach’s article from the reference and re-print of the article on this history of physics website.

Mach refers to Tumlirz’ experiment as an example of an idea that seems to belong in the same category at first glance, but is actually correct:

To be sure, Professor Tumlirz has recently performed an experiment which, while externally similar to this, is correct. By this experiment the rotation of the earth can be imitated, if the utmost care is taken, by the direction of the current of water flowing axially out of a cylindrical vessel. Further details are to be found in an article by Tumlirz in the Sitzungsberichte der Wiener Akademie, Vol. 117, 1908. I happened to know the origin of the thought that gave rise to this invention. Tumlirz noticed that the water flowing somewhat unsymmetrically in a glass funnel assumed a swift rotation in the neck of the funnel so that it formed a whirl of air in the axis of the flowing jet. This put it in his mind to increase the slight angular velocity of the water at rest with reference to the earth, by contraction in the axis.

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Comment on the German abstract: It seems one line or sentence got lost or mangled when processing the original as this does not make sense: so bendet sich das Wasser zwischen den beiden Glasscheiben [here something is missing] nach dem Rohrchen durch die kleinen Öffnungen.

I have not managed to find the full version of the old paper and the figures and photos online. I would be grateful for pointers.

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Update added August 2016: C. Schiller quotes this historical experiment in vol. 1 of his free physics textbook Motion Mountain (p. 135):

Only in 1962, after several attempts by other researchers, Asher Shapiro was the first to verify that the Coriolis effect has a tiny influence on the direction of the vortex flowing out of the bathtub.

Ref: A. H. SHAPIRO, Bath-tub vortex, Nature 196, pp. 1080-1081, 1962

# All Kinds of Turbines

Dave asked an interesting question, commenting on the heat-from-the-tunnel project:

Has anyone considered the fact that the water can be used to first drive turbines and then distributed to supply the input source for the heat pumps?

I am a water turbine fan, and every time I spot a small hydro power plant on a hiking map, I have to find it.

Pelton turbine. The small regional utility has several of them, the flow rate is typically a few 100 liters per second. The NSA should find an image of myself in the reflections.

This does not mean I have developed intuition for the numbers, so I have to do some cross-checks.

You can harvest either kinetic or potential energy from a flowing river in a hydro power plant. Harvesting kinetic energy could be done by something like the ‘under-water version’ of a wind turbine:

Tidal stream generator, rotor raised (Wikimedia user Fundy)

The tunnel produces a flow of 300 liters per second but this information is not yet sufficient for estimating mechanical power.

The kinetic energy of a mass $m$ moving at velocity $v$ is:  $\frac{mv^{2}}{2}$. From the mean velocity in a flow of water we could calculate the energy carried by flow by replacing $m$ in this expression by mass flow.

If 300 liters per second flow through a pipe with an area of 1 m2, the flow velocity is equal to  0,3 m3/s divided by this area, thus 0,3 m/s. This translates to a kinetic energy of:

$\frac{300 ^{.} 0,3^{2}}{2}$ W = 13,5 W

… only, just enough for a small light bulb.

If the cross-section of the pipe would be ten times smaller, the power would be 100 times larger – 1,35 kW.

(Edit: This is just speculating about typical sizes of the natural pipe determined by rocks or whatever. You cannot create energy out of nothing as increasing velocity by a sort of funnel would decrease pressure. I was rather thinking of a river bed open to ambient air – and ambient pressure – than a closed pipe.)

On the other hand, if that water would be allowed to ‘fall’, we could harvest potential energy:

Also this mill wheel is utilizing potential energy from the height difference of a few meters. (Critically inspected by The Chief Engineer, photo by elkement)

This is how commercial hydro power plants work, including those located at rivers in seemingly flat lowlands.

The potential energy of a point mass at height $h$ is $mgh$, $g$ being the acceleration due to gravity (~ 10m/s2). Assuming a usable height of 10m, 300kg/s would result in about

300 . 10 . 10 W = 30kW – quite a difference!

Of course there are huge error bars here but the modest output of kinetic energy is typical for the topography of planet earth.

Mass flow has to be conserved, and it enters both expressions as a factor. If I am interested in comparing potential and kinetic energies relative to each other, it is sufficient to compare $\frac{v^{2}}{2}$ to $gh$.

Cross-checking this for a flow of water we know more about:

The Danube flows at about 3-10 m/s, so

$\frac{v_{Danube}^{2}}{2}$ = 4,5 – 50m2/s2

But we cannot extract all that energy: The flow of water would come to a halt at the turbine – where should the water go then? For the same reasons there is a theoretical maximum percentage of wind power that turbines can harvest, even if perfectly frictionless.

In addition, such a turbine would need to be much smaller than the cross-section of the river. Mass flow needs to be conserved: when part of the water slows down, it gets spread over a larger cross-section.

So the realistic $\frac{v_{Danube}^{2}}{2}$ will be smaller.

I have stumbled upon an Austrian startup offering floating turbines, designed for operations in larger rivers and delivering about 70kW at 3,3m/s flow velocity (Images on the German site). This is small compared to the overall kinetic energy of the Danube of about several MW, calculated from 2.000m3/s (mass flow near Vienna) and about 3m/s.

The first hydro power plant at the Danube in Austria, built in 1959 – an icon of post World War II reconstruction (Wikimedia). The plant is currently modernised, the rated power will be increased by 5% to 250MW. Utilized difference in height: 10m.

So the whole kinetic energy – that cannot be extracted anyway – is still small compared to the rated power of typical power plants which are several 100MW!

If the water of the Danube ‘falls’ about 10m then

$gh_{Danube}$ ~ 100

… which is much larger than realistic values of $\frac{v_{Danube}^{2}}{2}$! Typical usable kinetic energies are lower than typical potential energies.

So if tunnel drain water should drive a turbine, the usable height is crucial. But expected powers are rather low compared to the heat power to be gained (several MW) so this is probably not economically feasible.

I was curious about the largest power plants on earth: Currently the Chinese Three Gorges Dam delivers 22GW. I have heard about plans in Sweden to build a plant that could deliver 50GW – a pumped hydro storage plant utilizing a 50km tunnel between two large lakes, with a difference in altitude of 44m (See the mentions here or here.)

Three Gorges Dam in China (Wikimedia user Filnko)

# Grim Reaper Does a Back-of-the-Envelope Calculation

I have a secondary super-villain identity. People on Google+ called me:
Elke the Ripper or Master of the Scythe.

[FAQ] No, I don’t lost a bet. We don’t have a lawn-mower by choice. Yes, we tried the alternatives including a reel lawn-mower. Yes, I really enjoy doing this.

It is utterly exhausting – there is no other outdoor activity in summer that leaves me with the feeling of really having achieved something!

So I was curious if Grim Reaper the Physicist can express this level of exhaustion in numbers.

Just holding a scythe with arms stretched out would not count as ‘work’. Yet I believe that in this case it is the acceleration required to bring the scythe to proper speed that matters; so I will focus on work in terms of physics.

In order to keep this simple, I assume that the weight of the scythe is a few kilos (say: 5kg) concentrated at the end of a weightless pole of 1,5m length. All the kinetic energy is concentrated in this ‘point mass’.

But how fast does a blade need to move in order to cut grass? Or from experience: How fast do I move the scythe?

One sweep with the scythe takes a fraction of second – probably 0,5s. The blade traverses an arc of about 2m.

Thus the average speed is: 2m / 0,5s = 4m/s

However, using this speed in further calculations does not make much sense: The scythe has two handles that allow for exerting a torque – the energy goes into acceleration of the scythe.

If an object with mass m is accelerated from a velocity of zero to a peak velocity vmax the kinetic energy acquired is calculated from the maximum velocity: m vmax2 / 2. How exactly the velocity has changed with time does not matter – this is just conservation of energy.

But what is the peak velocity?

For comparison: How fast do lawn-mower blades spin?

This page says: at 3600 revolutions per minute when not under load, dropping to about 3000 when under load. How fast would I have to move the scythe to achieve the same?

Velocity of a rotating body is angular velocity times radius. Angular velocity is 2Pi – a full circle – times the frequency, that is revolutions per time. The radius is the length of the pole that I use as a simplified model.

So the scythe on par with a lawn-mower would need to move at:
2Pi * (3000 rev./minute) / (60 seconds/minute) * 1,5m = 471m/s

This would result in the following energy per arc swept. I use only SI units, so the resulting energy is in Joule:

Energy needed to accelerate to 314m/s: 5kg * (471m/s)2 / 2 = 555.000J = 555kJ

I am assuming that this energy is just consumed (dissipated) to cut the grass; the grass brings the scythe to halt, and it is decelerated to 0m/s again.

1 kilocalorie is 4,18kJ, so this amounts to about 133kcal (!!)

That sounds way too much already: Googling typical energy consumptions for various activities I learn that easy work in the garden needs about 100-150kcal kilocalories per half an hour!

If scything were that ‘efficient’ I would put into practice what we always joke about: Offer outdoor management trainings to stressed out IT managers who want to connect with their true selves again through hard work and/or work-out most efficiently. So they would pay us for the option to scythe our grass.

But before I crank down the hypothetical velocity again, I calculate the energy demand per half an hour:

I feel exhausted after half an hour of scything. I pause a few seconds before the next – say 10s – on average. In reality it is probably more like:

scythe…1s…scythe…1s…scythe…1s….scythe…1s….scythe…longer break, gasping for air, sharpen the scythe.

I assume a break of 9,5s on average to make the calculation simpler. So this is 1 arc swept per 10 seconds, 6 arcs per minute, and 180 per half an hour. After half on hour I need to take longer break.

So using that lawn-mower-style speed this would result in:

Energy per half an hour if I were a lawn-mower: 133kJcal * 180 = 23.940kcal

… about five times the daily energy demands of a human being!

Velocity enters the equation quadratically. Assuming now that my peak scything speed is really only a tenth of the speed of a lawn-mower, 47m/2, which is still about 10 times my average speed calculated the beginning, this would result in one hundredth the energy.

A bit more realistic energy per half an hour of scything is then: 239kcal

Just for comparison – to get a feeling for those numbers: Average acceleration is maximum velocity over time. Thus 47m/s would result in:

Average acceleration: (47m/s) / (0,5s)  =  94m/s2

A fast car accelerates to 100km/h within 3 seconds, at (100/3,6)m/s / 3s = 9m/s2

So my assumed scythe’s acceleration is about 10 times a Ferrari’s!

Now I would need a high-speed camera, determine speed exactly and find a way to calculate actual energy needed for cutting.

Is there some conclusion?

This was just playful guesswork but the general line of reasoning and cross-checking orders of magnitude outlined here is not much different from when I try to get my simulations of our heat pump system right – based on unknown parameters, such as the effect of radiation, the heat conduction of ground, and the impact of convection in the water tank. The art is not so much in gettting numbers exactly right but in determining which parameters matter at all and how sensitive the solution is to a variation of those. In this case it would be crucial to determine peak speed more exactly.

In physics you can say the same thing in different ways – choosing one way over the other can make the problem less complex. As in this case, using total energy is often easier than trying to figure out the evolution of forces or torques with time.

The two images above were taken in early spring – when the ‘lawn’ / meadow was actually still growing significantly. Since we do not water it relentless Pannonian sun already started to turn it into a mixture of green and brown patches.

This is how the lawn looks now, one week after latest scything. This is not intended to be beautiful – I wanted to add a realistic picture as I had been asked about the ‘quality’ compared to a lawn-mower. Result: Good enough for me!

# Non-Linear Art. (Should Actually Be: Random Thoughts on Fluid Dynamics)

In my favorite ancient classical mechanics textbook I found an unexpected statement. I think 1960s textbooks weren’t expected to be garnished with geek humor or philosophical references as much as seems to be the default today – therefore Feynman’s books were so refreshing.

Natural phenomena featured by visual artists are typically those described by non-linear differential equations . Those equations allow for the playful interactions of clouds and water waves of ever changing shapes.

So fluid dynamics is more appealing to the artist than boring electromagnetic waves.

Is there an easy way to explain this without too much math? Most likely not but I try anyway.

I try to zoom in on a small piece of material, an incredibly small cube of water in a flow at a certain point of time. I imagine this cube as decorated by color. This cube will change its shape quickly and turn into some irregular shape – there are forces pulling and pushing – e.g. gravity.

This transformation is governed by two principles:

• First, mass cannot vanish. This is classical physics, no need to consider the generation of new particles from the energy of collisions. Mass is conserved locally, that is if some material suddenly shows up at some point in space, it had to have been travelling to that point from adjacent places.
• Second, Newton’s law is at play: Forces are equal to a change momentum. If we know the force acting at time t and point (x,y,z), we know how much momentum will change in a short period of time.

Typically any course in classical mechanics starts from point particles such as cannon balls or planets – masses that happen to be concentrated in a single point in space. Knowing the force at a point of time at the position of the ball we know the acceleration and we can calculate the velocity in the next moment of time.

This also holds for our colored little cube of fluid – but we usually don’t follow decorated lumps of mass individually. The behavior of the fluid is described perfectly if we know the mass density and the velocity at any point of time and space. Think little arrows attached to each point in space, probably changing with time, too.

Digesting that difference between a particle’s trajectory and an anonymous velocity field is a big conceptual leap in my point of view. Sometimes I wonder if it would be better to not learn about the point approach in the first place because it is so hard to unlearn later. Point particle mechanics is included as a special case in fluid mechanics – the flowing cannon ball is represented by a field that has a non-zero value only at positions equivalent to the trajectory. Using the field-style description we would say that part of the cannon ball vanishes behind it and re-appears “before” it, along the trajectory.

Pushing the cube also moves it to another place where the velocity field differs. Properties of that very decorated little cube can change at the spot where it is – this is called an explicit dependence on time. But it can also change indirectly because parts of it are moved with the flow. It changes with time due to moving in space over a certain distance. That distance is again governed by the velocity – distance is velocity times period of time.

Thus for one spatial dimension the change of velocity dv associated with dt elapsed is also related to a spatial shift dx = vdt. Starting from a mean velocity of our decorated cube v(x,t) we end up with v(x + vdt, t+dt) after dt has elapsed and the cube has been moved by vdt. For the cannon ball we could have described this simply as v(t + dt) as v was not a field.

And this is where non-linearity sneaks in: The indirect contribution via moving with the flow, also called convective acceleration, is quadratic in v – the spatial change of v is multiplied by v again. If you then allow for friction you get even more nasty non-linearities in the parts of the Navier-Stokes equations describing the forces.

My point here is that even if we neglect dissipation (describing what is called dry water tongue-in-cheek) there is already non-linearity. The canonical example for wavy motions – water waves – is actually rather difficult to describe due to that, and you need to resort to considering small fluctuations of the water surface even if you start from the simplest assumptions.

# Mastering Geometry is a Lost Art

I am trying to learn Quantum Field Theory the hard way: Alone and from text books. But there is something harder than the abstract math of advanced quantum physics:

You can aim at comprehending ancient texts on physics.

If you are an accomplished physicist, chemist or engineer – try to understand Sadi Carnot’s reasoning that was later called the effective discovery of the Second Law of Thermodynamics.

At Carnotcycle’s excellent blog on classical thermodynamics you can delve into thinking about well-known modern concepts in a new – or better: in an old – way. I found this article on the dawn of entropy a difficult ready, even though we can recognize some familiar symbols and concepts such as circular processes, and despite or because of the fact I was at the time of reading this article a heavy consumer of engineering thermodynamics textbooks. You have to translate now unused notions such as heat received and the expansive power into their modern counterparts. It is like reading a text in a foreign language by deciphering every single word instead of having developed a feeling for a language.

Stephen Hawking once published an anthology of the original works of the scientific giants of the past millennium: Corpernicus, Galieo, Kepler, Newton and Einstein: On the Shoulders of Giants. So just in case you googled for Hawkins – don’t expect your typical Hawking pop-sci bestseller with lost of artistic illustrations. This book is humbling. I found the so-called geometrical proofs most difficult and unfamiliar to follow. Actually, it is my difficulties in (not) taming that Pesky Triangle that motivated me to reflect on geometrical proofs.

I am used to proofs stacked upon proofs until you get to the real thing. In analysis lectures you get used to starting by proving that 1+1=2 (literally) until you learn about derivatives and slopes. However, Newton and his processor giants talk geometry all the way! I have learned a different language. Einstein is most familiar in the way he tackles problems though his physics is on principle the most non-intuitive.

This amazon.com review is titled Now We Know why Geometry is Called the Queen of the Sciences and the reviewer perfectly nails it:

It is simply astounding how much mileage Copernicus, Galileo, Kepler, Newton, and Einstein got out of ordinary Euclidean geometry. In fact, it could be argued that Newton (along with Leibnitz) were forced to invent the calculus, otherwise they too presumably would have remained content to stick to Euclidean geometry.

Science writer Margaret Wertheim gives an account of a 20th century giant trying to recapture Isaac Newton’s original discovery of the law of gravitation in her book Physics on the Fringe (The main topic of the book are outsider physicists’ theories, I have blogged about the book at length here.).

This giant was Richard Feynman.

Today the gravitational force, gravitational potential and related acceleration objects in the gravitational fields are presented by means of calculus: The potential is equivalent to a rubber membrane model – the steeper the membrane, the higher the force. (However, this is not a geometrical proof – this is an illustration of underlying calculus.)

Model of the gravitational potential. An object trapped in these wells moves along similar trajectories as bodies in a gravitational field. Depending on initial conditions (initial position and velocity) you end up with elliptical, parabolic or hyperbolic orbits. (Wikimedia, Invent2HelpAll)

(Today) you start from the equation of motion for a object under the action of a force that weakens with the inverse square of the distance between two massive objects, and out pops Kepler’s law about elliptical orbits. It takes some pages of derivation, and you need to recognize conic sections in formulas – but nothing too difficult for an undergraduate student of science.

Newton actually had to invent calculus together with tinkering with the law of gravitation. In order to convince his peers he needed to use the geometrical language and the mental framework common back then. He uses all kinds of intricate theorems about triangles and intersecting lines (;-)) in order to say what we say today using the concise shortcuts of derivatives and differentials.

Wertheim states:

Feynman wasn’t doing this to advance the state of physics. He was doing it to experience the pleasure of building a law of the universe from scratch.

Feynman said to his students:

“For your entertainment and interest I want you to ride in a buggy for its elegance instead of a fancy automobile.”

But he underestimated the daunting nature of this task:

In the preparatory notes Feynman made for his lecture, he wrote: “Simple things have simple demonstrations.” Then, tellingly, he crossed out the second “simple” and replaced it with “elementary.” For it turns out there is nothing simple about Newton’s proof. Although it uses only rudimentary mathematical tools, it is a masterpiece of intricacy. So arcane is Newton’s proof that Feynman could not understand it.

Given the headache that even Corpernicus’ original proofs in the Shoulders of Giants gave me I can attest to:

… in the age of calculus, physicists no longer learn much Euclidean geometry, which, like stonemasonry, has become something of a dying art.

Richard Feynman has finally made up his own version of a geometrical proof to fully master Newton’s ideas, and Feynman’s version covered hundred typewritten pages, according to Wertheim.

Everybody who indulges gleefully in wooden technical prose and takes pride in plowing through mathematical ideas can relate to this:

For a man who would soon be granted the highest honor in science, it was a DIY triumph whose only value was the pride and joy that derive from being able to say, “I did it!”

Richard Feynman gave a lecture on the motion of the planets in 1964, that has later been called his Lost Lecture. In this lecture he presented his version of the geometrical proof which was simpler than Newton’s.

The proof presented in the lecture have been turned in a series of videos by Youtube user Gary Rubinstein. Feynman’s original lecture was 40 minutes long and confusing, according to Rubinstein – who turned it into 8 chunks of videos, 10 minutes each.

The rest of the post is concerned with what I believe that social media experts call curating. I am just trying to give an overview of the episodes of this video lecture. So my summaries do most likely not make a lot of sense if you don’t watch the videos. But even if you don’t watch the videos you might get an impression of what a geometrical proof actually is.

In Part I (embedded also below) Kepler’s laws are briefly introduced. The characteristic properties of an ellipse are shown – in the way used by gardeners to creating an elliptical with a cord and a pencil. An ellipse can also be created within a circle by starting from a random point, connecting it to the circumference and creating the perpendicular bisector:

Part II starts with emphasizing that the bisector is actually a tangent to the ellipse (this will become an important ingredient in the proof later). Then Rubinstein switches to physics and shows how a planet effectively ‘falls into the sun’ according to Newton, that is a deviation due to gravity is superimposed to its otherwise straight-lined motion.

Part III shows in detail why the triangles swept out by the radius vector need to stay the same. The way Newton defined the size of the force in terms of parallelogram attached to the otherwise undisturbed path (no inverse square law yet mentioned!) gives rise to constant areas of the triangles – no matter what the size of the force is!

In Part IV the inverse square law in introduced – the changing force is associated with one side of the parallelogram denoting the deviation from motion without force. Feynman has now introduced the velocity as distance over time which is equal to size of the tangential line segments over the areas of the triangles. He created a separate ‘velocity polygon’ of segments denoting velocities. Both polygons – for distances and for velocities – look elliptical at first glance, though the velocity polygon seems more circular (We will learn later that it has to be a circle).

In Part V Rubinstein expounds that the geometrical equivalent of the change in velocity being proportional to 1 over radius squared times time elapsed with time elapsed being equivalent to the size of the triangles (I silently translate back to dv = dt times acceleration). Now Feynman said that he was confused by Newton’s proof of the resulting polygon being an ellipse – and he proposed a different proof:
Newton started from what Rubinstein calls the sun ‘pulsing’ at the same intervals, that is: replacing the smooth path by a polygon, resulting in triangles of equal size swept out by the radius vector but in a changing velocity.  Feynman divided the spatial trajectory into parts to which triangles of varying area e are attached. These triangles are made up of radius vectors all at the same angles to each other. On trying to relate these triangles to each other by scaling them he needs to consider that the area of a triangle scales with the square of its height. This also holds for non-similar triangles having one angle in common.

Part VI: Since ‘Feynman’s triangles’ have one angle in common, their respective areas scale with the squares of the heights of their equivalent isosceles triangles, thus basically the distance of the planet to the sun. The force is proportional to one over distance squared, and time is proportional to distance squared (as per the scaling law for these triangles). Thus the change in velocity – being the product of both – is constant! This is what Rubinstein calls Feynman’s big insight. But not only are the changes in velocity constant, but also the angles between adjacent line segments denoting those changes. Thus the changes in velocities make up for a regular polygon (which seems to turn into a circle in the limiting case).

Part VII: The point used to build up the velocity polygon by attaching the velocity line segments to it is not the center of the polygon. If you draw connections from the center to the endpoints the angle corresponds to the angle the planet has travelled in space. The animations of the continuous motion of the planet in space – travelling along its elliptical orbit is put side-by-side with the corresponding velocity diagram. Then Feynman relates the two diagrams, actually merges them, in order to track down the position of the planet using the clues given by the velocity diagram.

In Part VIII (embedded also below) Rubinstein finally shows why the planet traverses an elliptical orbit. The way the position of the planet has finally found in Part VII is equivalent to the insights into the properties of an ellipse found at the beginning of this tutorial. The planet needs be on the ‘ray’, the direction determined by the velocity diagram. But it also needs to be on the perpendicular bisector of the velocity segment – as force cause a change in velocity perpendicular to the previous velocity segment and the velocity needs to correspond to a tangent to the path.