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.

___________________________________

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

And Now for Something Completely Different: Rotation Heat Pump!

Heat pumps for space heating are all very similar: Refrigerant evaporates, pressure is increased by a scroll compressor, refrigerant condenses, pressure is reduced in an expansion value. *yawn*

The question is:

Can a compression heat pump be built in a completely different way?

Austrian start-up ECOP did it: They  invented the so-called Rotation Heat Pump.

It does not have a classical compressor, and the ‘refrigerant’ does not undergo a phase transition. A pressure gradient is created by centrifugal forces: The whole system rotates, including the high-pressure (heat sink) and low-pressure (source) heat exchanger. The low pressure part of the system is positioned closer to the center of the rotation axis, and heat sink and heat source are connected at the axis (using heating water). The system rotates at up to 1800 rounds per minute.

A mixture of noble gases is used in a Joule (Brayton) process, driven in a cycle by a ventilator. Gas is compressed and thus heated up; then it is cooled at constant pressure and energy is released to the heat sink. After expanding the gas, it is heated up again at low pressure by the heat source.

In the textbook Joule cycle, a turbine and a compressor share a common axis: The energy released by the turbine is used to drive the compressor. This is essential, as compression and expansion energies are of the same order of magnitude, and both are considerably larger than the net energy difference – the actual input energy.

In contrast to that, a classical compression heat pump uses a refrigerant that is condensed while releasing heat and then evaporated again at low pressure. There is no mini-turbine to reduce the pressure but only an expansion valve, as there is not much energy to gain.

This explains why the Rotation Heat Pumps absolutely have to have compression efficiencies of nearly 100%, compared to, say, 85% efficiency of a scroll compressor in heat pump used for space heating:

Some numbers for a Joule process (from this German ECOP paper): On expansion of the gas 1200kW are gained, but 1300kW are needed for compression – if there would be no losses at all. So the net input power is 100kW. But if the efficiency of the compression is reduced from 100% to 80% about 1600kW are needed and thus a net input power of 500kW – five times the power compared to the ideal compressor! The coefficient of performance would plummet from 10 to 2,3.

I believe these challenging requirements are why Rotation Heat Pumps are ‘large’ and built for industrial processes. In addition to the high COP, this heat pump is also very versatile: Since there are no phase transitions, you can pick your favorite corner of the thermodynamic state diagram at will: This heat pump works for very different combinations temperatures of the hot target and the cold source.

Re-Visiting Carnot’s Theorem

The proof by contradiction used in physics textbooks is one of those arguments that appear surprising, then self-evident, then deceptive in its simplicity. You – or maybe only: I – cannot resist turning it over and over in your head again, viewing it from different angles.

tl;dr: I just wanted to introduce the time-honored tradition of ASCII text art images to illustrate Carnot’s Theorem, but this post got out of hand when I mulled about how to  refute an erroneous counter-argument. As there are still research papers being written about Carnot’s efficiency I feel vindicated for writing a really long post though.

Carnot‘s arguments prove that there is a maximum efficiency of a thermodynamic heat engine – a machine that turns heat into mechanical energy. He gives the maximum value by evaluating one specific, idealized process, and then proves that a machine with higher efficiency would give rise to a paradox. The engine uses part of the heat available in a large, hot reservoir of heat and turns it into mechanical work and waste heat – the latter dumped to a colder ‘environment’ in a 4-step process. (Note that while our modern reformulation of the proof by contradiction refers to the Second Law of Thermodynamics, Carnot’s initial version was based on the caloric theory.)

The efficiency of such an engine η – mechanical energy per cycle over input heat energy – only depends on the two temperatures (More details and references here):

\eta_\text{carnot} = \frac {T_1-T_2}{T_1}

These are absolute temperatures in Kelvin; this universal efficiency can be used to define what we mean by absolute temperature.

I am going to use ‘nice’ numbers. To make ηcarnot equal to 1/2, the hot temperature
T1 = 273° = 546 K, and the colder ‘environment’ has T2 = 0°C = 273 K.

If this machine is run in reverse, it uses mechanical input energy to ‘pump’ energy from the cold environment to the hot reservoir – it is a heat pump using the ambient reservoir as a heat source. The Coefficient of Performance (COP, ε) of the heat pump is heat output over mechanical input, the inverse of the efficiency of the corresponding engine. εcarnot is 2 for the temperatures given above.

If we combine two such perfect machines – an engine and a heat pump, both connected to the hot space and to the cold environment, their effects cancel out: The mechanical energy released by the engine drives the heat pump which ‘pumps back’ the same amount of energy.

In the ASCII images energies are translated to arrows, and the number of parallel arrows indicates the amount of energy per cycle (or power). For each device, the number or arrows flowing in and out is the same; energy is always conserved. I am viewing this from the heat pump’s perspective, so I call the cold environment the source, and the hot environment room.

Neither of the heat reservoirs are heated or cooled in this ideal case as the same amount of energy flows from and to each of the heat reservoirs:

|----------------------------------------------------------|
|         Hot room at temperature T_1 = 273°C = 546 K      |
|----------------------------------------------------------|
           | | | |                         | | | |
           v v v v                         ^ ^ ^ ^
           | | | |                         | | | |
       |------------|                 |---------------|
       |   Engine   |->->->->->->->->-|   Heat pump   |
       |  Eta = 1/2 |->->->->->->->->-| COP=2 Eta=1/2 |
       |------------|                 |---------------|
             | |                             | |
             v v                             ^ ^
             | |                             | |
|----------------------------------------------------------| 
|        Cold source at temperature T_2 = 0°C = 273 K      | 
|----------------------------------------------------------|

If either of the two machines works less than perfectly and in tandem with a perfect machine, anything is still fine:

If the engine is far less than perfect and has an efficiency of only 1/4 – while the heat pump still works perfectly – more of the engine’s heat energy input is now converted to waste heat and diverted to the environment:

|----------------------------------------------------------|
|         Hot room at temperature T_1 = 273°C = 546 K      |
|----------------------------------------------------------|
           | | | |                           | |  
           v v v v                           ^ ^  
           | | | |                           | |  
       |------------|                 |---------------|
       |   Engine   |->->->->->->->->-|   Heat pump   |
       |  Eta = 1/4 |                 | COP=2 Eta=1/2 |
       |------------|                 |---------------|
            | | |                             |
            v v v                             ^
            | | |                             |
|----------------------------------------------------------| 
|        Cold source at temperature T_2 = 0°C = 273 K      | 
|----------------------------------------------------------|

Now two net units of energy flow from the hot room to the environment (summing up the arrows to and from the devices):

|----------------------------------------------------------|
|         Hot room at temperature T_1 = 273°C = 546 K      |
|----------------------------------------------------------|
                              | |                                
                              v v                                
                              | | 
                     |------------------|
                     |   Combination:   |
                     | Eta=1/4 COP=1/2  |
                     |------------------|                            
                              | |                              
                              v v                              
                              | |                             
|----------------------------------------------------------| 
|        Cold source at temperature T_2 = 0°C = 273 K      | 
|----------------------------------------------------------|

Using a real-live heat pump with a COP of 3/2 (< 2) together with a perfect engine …

|----------------------------------------------------------|
|         Hot room at temperature T_1 = 273°C = 546 K      |
|----------------------------------------------------------|
          | | | |                             | | | 
          v v v v                             ^ ^ ^ 
          | | | |                             | | |
       |------------|                 |-----------------|
       |   Engine   |->->->->->->->->-|    Heat pump    |
       |  Eta = 1/2 |->->->->->->->->-|     COP=3/2     |
       |------------|                 |-----------------|
            | |                                 |
            v v                                 ^
            | |                                 |
|----------------------------------------------------------| 
|        Cold source at temperature T_2 = 0°C = 273 K      | 
|----------------------------------------------------------|

… causes again a non-paradoxical net flow of one unit of energy from the room to the environment.

In the most extreme case  a poor heat pump (not worth this name) with a COP of 1 just translates mechanical energy into heat energy 1:1. This is a resistive heating element, a heating rod, and net heat fortunately flows from hot to cold without paradoxes:

|----------------------------------------------------------|
|         Hot room at temperature T_1 = 273°C = 546 K      |
|----------------------------------------------------------|
            | |                                |   
            v v                                ^   
            | |                                |   
       |------------|                 |-----------------|
       |   Engine   |->->->->->->->->-|   'Heat pump'   |
       |  Eta = 1/2 |                 |     COP = 1     |
       |------------|                 |-----------------|
             |                                 
             v                                 
             |                                 
|----------------------------------------------------------| 
|        Cold source at temperature T_2 = 0°C = 273 K      | 
|----------------------------------------------------------|

The textbook paradox in encountered, when an ideal heat pump is combined with an allegedly better-than-possible engine, e.g. one with an efficiency:

ηengine = 2/3 (> ηcarnot = 1/2)

|----------------------------------------------------------|
|         Hot room at temperature T_1 = 273°C = 546 K      |
|----------------------------------------------------------|
           | | |                           | | | |
           v v v                           ^ ^ ^ ^
           | | |                           | | | |
       |------------|                 |---------------|
       |   Engine   |->->->->->->->->-|   Heat pump   |
       |  Eta = 2/3 |->->->->->->->->-| COP=2 Eta=1/2 |
       |------------|                 |---------------|
             |                               | |
             v                               ^ ^
             |                               | |
|----------------------------------------------------------| 
|        Cold source at temperature T_2 = 0°C = 273 K      | 
|----------------------------------------------------------|

The net effect / heat flow is then:

|----------------------------------------------------------|
|        Hot room at temperature T_1 = 273°C = 546 K       | 
|----------------------------------------------------------| 
                             | 
                             ^ 
                             | 
                   |------------------| 
                   |   Combination:   | 
                   | Eta=3/2; COP=1/2 | 
                   |------------------| 
                             | 
                             ^ 
                             | 
|----------------------------------------------------------| 
|       Cold source at temperature T_2 = 0°C = 273 K       | 
|----------------------------------------------------------|

One unit of heat would flow from the environment to the room, from the colder to the warmer body without any other change being made to the system. The combination of these machines would violate the Second Law of Thermodynamics; it is a Perpetuum Mobile of the Second Kind.

If the heat pump has a higher COP than the inverse of the perfect engine’s efficiency, a similar paradox arises, and again one unit of heat flows in the forbidden direction:

|----------------------------------------------------------|
|         Hot room at temperature T_1 = 273°C = 546 K      |
|----------------------------------------------------------|
            | |                             | | |
            v v                             ^ ^ ^
            | |                             | | |
       |------------|                 |---------------|
       |   Engine   |->->->->->->->->-|   Heat pump   |
       |  Eta = 1/2 |                 |    COP = 3    |
       |------------|                 |---------------|
             |                               | |
             v                               ^ ^
             |                               | |
|----------------------------------------------------------| 
|        Cold source at temperature T_2 = 0°C = 273 K      | 
|----------------------------------------------------------|

A weird question: Can’t we circumvent the paradoxes if we pair the impossible superior devices with poorer ones (of the reverse type)?

|----------------------------------------------------------|
|         Hot room at temperature T_1 = 273°C = 546 K      |
|----------------------------------------------------------|
           | | |                             | |  
           v v v                             ^ ^  
           | | |                             | |  
       |------------|                 |---------------|
       |   Engine   |->->->->->->->->-|   Heat pump   |
       |  Eta = 2/3 |->->->->->->->->-|    COP = 1    |
       |------------|                 |---------------|
             |                                
             v                                
             |                                
|----------------------------------------------------------| 
|        Cold source at temperature T_2 = 0°C = 273 K      | 
|----------------------------------------------------------

Indeed: If the COP of the heat pump (= 1) is smaller than the inverse of the (impossible) engine’s efficiency (3/2), there will be no apparent violation of the Second Law – one unit of net heat flows from hot to cold.

An engine with low efficiency 1/4 would ‘fix’ the second paradox involving the better-than-perfect heat pump:

|----------------------------------------------------------|
|         Hot room at temperature T_1 = 273°C = 546 K      |
|----------------------------------------------------------|
           | | | |                          | | |
           v v v v                          ^ ^ ^
           | | | |                          | | |
       |------------|                 |---------------|
       |   Engine   |->->->->->->->->-|   Heat pump   |
       |  Eta = 1/4 |                 |     COP=3     |
       |------------|                 |---------------|
            | | |                            | |
            v v v                            ^ ^
            | | |                            | |
|----------------------------------------------------------| 
|        Cold source at temperature T_2 = 0°C = 273 K      | 
|----------------------------------------------------------|

But we cannot combine heat pumps and engines at will, just to circumvent the paradox – one counter-example is sufficient: Any realistic engine combined with any realistic heat pump – plus all combinations of those machines with ‘worse’ ones – have to result in net flow from hot to cold …

The Second Law identifies such ‘sets’ of engines and heat pumps that will all work together nicely. It’s easier to see this when all examples are condensed into one formula:

The heat extracted in total from the hot room – Q1 –  is the difference of heat used by the engine and heat delivered by the heat pump, both of which are defined in relation to the same mechanical work W:

Q_1 = W\left (\frac{1}{\eta_\text{engine}}-\varepsilon_\text{heatpump}\right)

This is also automatically equal to Qas another quick calculation shows or by just considering that energy is conserved: Some heat goes into the combination of the two machines, part of it – W – flows internally from the engine to the heat pump. But no part of the input Q1 can be lost, so the output of the combined machine has to match the input. Energy ‘losses’ such as energy due to friction will flow to either of the heat reservoirs: If an engine is less-then-perfect, more heat will be wasted to the environment; and if the heat pump is less-than-perfect a greater part of mechanical energy will be translated to heat only 1:1. You might be even lucky: Some part of heat generated by friction might end up in the hot room.

As Q1 has to be > 0 according to the Second Low, the performance numbers have to related by this inequality:

\frac{1}{\eta_\text{engine}}\geq\varepsilon_\text{heatpump}

The equal sign is true if the effects of the two machines just cancel each other.

If we start from a combination of two perfect machines (ηengine = 1/2 = 1/εheatpump) and increase either ηengine or εheatpump, this condition would be violated and heat would flow from cold to hot without efforts.

But also an engine with efficiency = 1 would work happily with the worst heat pump with COP = 1. No paradox would arise at first glance  – as 1/1 >= 1:

|----------------------------------------------------------|
|         Hot room at temperature T_1 = 273°C = 546 K      |
|----------------------------------------------------------|
             |                                |   
             v                                ^   
             |                                |   
       |------------|                 |-----------------|
       |   Engine   |->->->->->->->->-|   'Heat pump'   |
       |   Eta = 1  |                 |      COP=1      |
       |------------|                 |-----------------|
                                               
                                               
                                               
|----------------------------------------------------------| 
|        Cold source at temperature T_2 = 0°C = 273 K      | 
|----------------------------------------------------------|

What’s wrong here?

Because of conservation of energy ε is always greater equal 1; so the set of valid combinations of machines all consistent with each other is defined by:

\frac{1}{\eta_\text{engine}}\geq\varepsilon_\text{heatpump}\geq1

… for all efficiencies η and COPs / ε of machines in a valid set. The combination η = ε = 1 is still not ruled out immediately.

But if the alleged best engine (in a ‘set’) would have an efficiency of 1, then the alleged best heat pump would have an Coefficient of Performance of only 1 – and this is actually the only heat pump possible as ε has to be both lower equal and greater equal than 1. It cannot get better without creating paradoxes!

If one real-live heat pump is found that is just slightly better than a heating rod – say
ε = 1,1 – then performance numbers for the set of consisent, non-paradoxical machines need to fulfill:

\eta_\text{engine}\leq\eta_\text{best engine}

and

\varepsilon_\text{heatpump}\leq\varepsilon_\text{best heatpump}

… in addition to the inequality relating η and ε.

If ε = 1,1 is a candidate for the best heat pump, a set of valid machines would comprise:

  • All heat pumps with ε between 1 and 1,1 (as per limits on ε)
  • All engines with η between 0 and 0,9 (as per inequality following the Second Law plus limit on η).

Consistent sets of machines are thus given by a stronger condition – by adding a limit for both efficiency and COP ‘in between’:

\frac{1}{\eta_\text{engine}}\geq\text{Some Number}\geq\varepsilon_\text{heatpump}\geq1

Carnot has designed a hypothetical ideal heat pump that could have a COP of εcarnot = 1/ηcarnot. It is a limiting case of a reversible machine, but feasible on principle. εcarnot  is thus a valid upper limit for heat pumps, a candidate for Some Number. In order to make this inequality true for all sets of machines (ideal ones plus all worse ones) then 1/ηcarnot = εcarnot also constitutes a limit for engines:

\frac{1}{\eta_\text{engine}}\geq\frac{1}{\eta_\text{carnot}}\geq\varepsilon_\text{heatpump}\geq1

So in order to rule out all paradoxes, Some Number in Between has to be provided for each set of machines. But what defines a set? As machines of totally different making have to work with each other without violating this equality, this number can only be a function of the only parameters characterizing the system – the two temperatures

Carnot’s efficiency is only a function of the temperatures. His hypothetical process is reversible, the machine can work either as a heat pump or an engine. If we could come up with a better process for a reversible heat pump (ε > εcarnot), the machine run in reverse would be an engine with η less than ηcarnot, whereas a ‘better’ engine would lower the upper bound for heat pumps.

If you have found one truly reversible process, both η and ε associated with it are necessarily the upper bounds of performance of the respective machines, so you cannot push Some Number in one direction or the other, and the efficiencies of all reversible engines have to be equal – and thus equal to ηcarnot. The ‘resistive heater’ with ε = 1 is the iconic irreversible device. It will not turn into a perfect engine with η = 1 when ‘run in reverse’.

The seemingly odd thing is that 1/ηcarnot appears like a lower bound for ε at first glance if you just declare ηcarnot an upper bound for corresponding engines and take the inverse, while in practice and according to common sense it is the maximum value for all heat pumps, including irreversible ones. (As a rule of thumb a typical heat pump for space heating has a COP only 50% of 1/ηcarnot.)

But this ‘contradiction’ is yet another way of stating that there is one universal performance indicator of all reversible machines making use of two heat reservoirs: The COP of a hypothetical ‘superior’ reversible heat pump would be at least 1/ηcarnot  … as good as Carnot’s reversible machine, maybe better. But the same is true for the hypothetical superior engine with an efficiency of at least ηcarnot. So the performance numbers of all reversible machines (all in one set, characterized by the two temperatures) have to be exactly the same.

Steam pump / Verkehrt laufende Dampfmaschine

Historical piston compressor (from the time when engines with pistons looked like the ones in textbooks), installed 1878 in the salt mine of Bex, Switzerland. 1943 it was still in operation. Such machines used in salt processing were considered the first heat pumps.

An Efficiency Greater Than 1?

No, my next project is not building a Perpetuum Mobile.

Sometimes I mull upon definitions of performance indicators. It seems straight-forward that the efficiency of a wood log or oil burner is smaller than 1 – if combustion is not perfect you will never be able to turn the caloric value into heat, due to various losses and incomplete combustion.

Our solar panels have an ‘efficiency’ or power ratio of about 16,5%. So 16.5% of solar energy are converted to electrical energy which does not seem a lot. However, that number is meaningless without adding economic context as solar energy is free. Higher efficiency would allow for much smaller panels. If efficiency were only 1% and panels were incredibly cheap and I had ample roof spaces I might not care though.

The coefficient of performance of a heat pump is 4-5 which sometimes leaves you with this weird feeling of using odd definitions. Electrical power is ‘multiplied’ by a factor always greater than one. Is that based on crackpottery?

Heat pump.

Our heat pump. (5 connections: 2x heat source – brine, 3x heating water hot water / heating water supply, joint return).

Actually, we are cheating here when considering the ‘input’ – in contrast to the way we view photovoltaic panels: If 1 kW of electrical power is magically converted to 4 kW of heating power, the remaining 3 kW are provided by a cold or lukewarm heat source. Since those are (economically) free, they don’t count. But you might still wonder, why the number is so much higher than 1.

My favorite answer:

There is an absolute minimum temperature, and our typical refrigerators and heat pumps operate well above it.

The efficiency of thermodynamic machines is most often explained by starting with an ideal process using an ideal substance – using a perfect gas as a refrigerant that runs in a closed circuit. (For more details see pointers in the Further Reading section below). The gas would be expanded at a low temperature. This low temperature is constant as heat is transferred from the heat source to the gas. At a higher temperature the gas is compressed and releases heat. The heat released is the sum of the heat taken in at lower temperatures plus the electrical energy fed in to the compressor – so there is no violation of energy conservation. In order to ‘jump’ from the lower to the higher temperature, the gas is compressed – by a compressor run on electrical power – without exchanging heat with the environment. This process is repeating itself again and again, and with every cycle the same heat energy is released at the higher temperature.

In defining the coefficient of performance the energy from the heat source is omitted, in contrast to the electrical energy:

COP = \frac {\text{Heat released at higher temperature per cycle}}{\text{Electrical energy fed into the compressor per cycle}}

The efficiency of a heat pump is the inverse of the efficiency of an ideal engine – the same machine, running in reverse. The engine has an efficiency lower than 1 as expected. Just as the ambient energy fed into the heat pump is ‘free’, the related heat released by the engine to the environment is useless and thus not included in the engine’s ‘output’.

100 1870 (Voitsberg steam power plant)

One of Austria’s last coal power plants – Kraftwerk Voitsberg, retired in 2006 (Florian Probst, Wikimedia). Thermodynamically, this is like ‘a heat pump running in reverse. That’s why I don’t like when a heat pump is said to ‘work like a refrigerator, just in reverse’ (Hinting at: The useful heat provided by the heat pump is equivalent to the waste heat of the refrigerator). If you run the cycle backwards, a heat pump would become sort of a steam power plant.

The calculation (see below) results in a simple expression as the efficiency only depends on temperatures. Naming the higher temperature (heating water) T1 and the temperature of the heat source (‘environment’, our water tank for example) T….

COP = \frac {T_1}{T_1-T_2}

The important thing here is that temperatures have to be calculated in absolute values: 0°C is equal to 273,15 Kelvin, so for a typical heat pump and floor loops the nominator is about 307 K (35°C) whereas the denominator is the difference between both temperature levels – 35°C and 0°C, so 35 K. Thus the theoretical COP is as high as 8,8!

Two silly examples:

  • Would the heat pump operate close to absolute zero, say, trying to pump heat from 5 K to 40 K, the COP would only be
    40 / 35 = 1,14.
  • On the other hand, using the sun as a heat source (6000 K) the COP would be
    6035 / 35 = 172.

So, as heat pump owners we are lucky to live in an environment rather hot compared to absolute zero, on a planet where temperatures don’t vary that much in different places, compared to how far away we are from absolute zero.

__________________________

Further reading:

Richard Feynman has often used unusual approaches and new perspectives when explaining the basics in his legendary Physics Lectures. He introduces (potential) energy at the very beginning of the course drawing on Carnot’s argument, even before he defines force, acceleration, velocity etc. (!) In deriving the efficiency of an ideal thermodynamic engine many chapters later he pictured a funny machine made from rubber bands, but otherwise he follows the classical arguments:

Chapter 44 of Feynman’s Physics Lectures Vol 1, The Laws of Thermodynamics.

For an ideal gas heat energies and mechanical energies are calculated for the four steps of Carnot’s ideal process – based on the Ideal Gas Law. The result is the much more universal efficiency given above. There can’t be any better machine as combining an ideal engine with an ideal heat pump / refrigerator (the same type of machine running in reverse) would violate the second law of thermodynamics – stated as a principle: Heat cannot flow from a colder to a warmer body and be turned into mechanical energy, with the remaining system staying the same.

KarnoyiCikl

Pressure over Volume for Carnot’s process, when using the machine as an engine (running it counter-clockwise it describes a heat pump): AB: Expansion at constant high temperature, BC: Expansion without heat exchange (cooling), CD: Compression at constant low temperature, DA: Compression without heat exhange (gas heats up). (Image: Kara98, Wikimedia).

Feynman stated several times in his lectures that he does not want to teach history of physics or downplayed the importance of learning about history of science a bit (though it seems he was well versed in – as e.g. his efforts to follow Newton’s geometrical prove of Kepler’s Laws showed). For historical background of the evolution of Carnot’s ideas and his legacy see the the definitive resource on classical thermodynamics and its history – Peter Mander’s blog carnotcycle.wordpress.com:

What had puzzled me is once why we accidentally latched onto such a universal law, using just the Ideal Gas Law.The reason is that the Gas Law has the absolute temperature already included. Historically, it did take quite a while until pressure, volume and temperature had been combined in a single equation – see Peter Mander’s excellent article on the historical background of this equation.

Having explained Carnot’s Cycle and efficiency, every course in thermodynamics reveals a deeper explanation: The efficiency of an ideal engine could actually be used as a starting point defining the new scale of temperature.

Temperature scale according to Kelvin (William Thomson)

Carnot engines with different efficiencies due to different lower temperatures. If one of the temperatures is declared the reference temperature, the other can be determined by / defined by the efficiency of the ideal machine (Image: Olivier Cleynen, Wikimedia.)

However, according to the following paper, Carnot did not rigorously prove that his ideal cycle would be the optimum one. But it can be done, applying variational principles – optimizing the process for maximum work done or maximum efficiency:

Carnot Theory: Derivation and Extension, paper by Liqiu Wang

A Sublime Transition

Don’t expect anything philosophical or career-change-related. I am talking about water and its phase transition to ice because …

…the fact that a process so common and important as water freezing is not fully resolved and understood, is astonishing.

(Source)

There are more spectacular ways of triggering this transition than just letting a tank of water cool down slowly: Following last winter’s viral trend, fearless mavericks turned boiling water vapor into snow flakes. Simply sublime desublimation?

Here is an elegant demo of Boiling water freezing in midair in the cold:

The science experiment took its toll: About 50 hobbyist scientists scalded themselves, ignoring the empirical rule about spraying any kind of liquid and wind direction:

“I accidentally threw all the BOILING water against the wind and burnt myself.”

Can it really be desublimation of water vapor? The reverse of this process, sublimation, is well known to science fiction fans:

Special effects supervisor Alex Weldon was charged with devising a way to realistically recreate the look of pools of steaming milky water that had been at the location. He concocted similar liquid with evaporated milk and white poster paint, mixed with water and poured into the set’s pools. Steam bubbling to the top was created with dry ice and steam machines, passed into the water via hidden tubing.

(Source: Star Trek online encyclopedia Memory Alpha on planet Vulcan.)

Dry ice is solid carbon dioxide, and it is the combination of temperature and atmospheric pressure on planet earth that allow for the sublimation of CO2. The phase diagram shows that at an air pressure of 1 bar and room temperature (about 293 K = 20°C) only solid and gaseous CO2 can exist:

Carbon dioxide p-T phase diagramIf a chunk of dry ice is taken out of the refrigerator and thrown onto the disco’s dance floor it will heat up a bit, and cross the line between the solid and gas areas in the diagram.

Sublimation of dry ice (Wikimedia, public domain)On the contrary, the phase diagram of water shows that at 1 bar (= 100 kPa) the direct transition from vapor to ice is the is not an option. Following the red horizontal 1-bar-line you need to cross the green realm of the liquid phase:

Phase diagram of water (Wikimedia, User cmglee)You would need to do the experiment in an atmosphere less than 1/100 as dense to sublimate ice or desublimate vapor.

But experiments show that the green area seems to be traversed in the fraction of a second – and boiling water seems to cool down much faster than colder water!

It seems paradoxical as more heat energy need to be removed from boiling water (or vapor!) to cool it down to 0°C. The heat of vaporization is about 2.300 kJ/kg whereas the specific heat of water is only 4 kJ/kgK.

I believe that the sudden freezing  is due to the much more efficient heat transfer between the ambient air and vapor / tiny droplets versus the smaller heat flow from larger droplets to the air.

Mixing water vapor with air will provide for the best exposure of the wildly shaking water molecules to the slower air molecules. If not-yet-vaporized water droplets are thrown into the air, I blame the faster freezing on water’s surface tension decreasing with increasing temperature:

Temperature dependence surface tension of waterSurface tension indicates the work it takes to create or maintain a surface between different phases or substances. The internal pressure inside a water droplet is proportional to surface tension and inverse proportional to its radius. This follows from the work against air pressure needed to increase the size of a droplet. Assuming that droplets of different sizes will be created with similar internal pressures, the average size of droplets will be smaller for higher temperatures.

A cup of water at 90°C will be dispersed into a larger number of smaller droplets and thus a bigger surface exposed to air than a cup at 70°C. The liquid with the lower surface tension will evaporate more quickly.

One more twist: If droplets are created in mid air, as precipitates from condensation or desublimation, it takes work to create their surfaces – proportional to surface tension and area. On the other hand, you gain energy from  these processes – proportional to volume. If the surface tension is lower but the area is larger the total volume is the same – and thus the net effect in terms of energy balance might be the same. But arguments based on energy balance only don’t take into account the dynamic nature of this process, far off thermodynamic equilibrium: The theoretical energy gain can only be cashed in (within the time frame we are interested in it) if condensation or freezing or desublimation is actually initiated – which in turn depends in the shape and area of the surface and on nuclei for droplets.

Heat transfer is of course more efficient for a larger temperature differences between air and water; perhaps that’s why the trend started in Siberia:

I have for sure not discussed any phenomenon involved here. Even hot water kept in a vessel can cool down and freeze faster than initially cooler water: This is called the Mpemba effect, a phenomenon known to our ancestors and rediscovered by the scientific community in the 1960s – after a curious African student refused to believe that his teachers called his observations on making ice cream ‘impossible’. The effect is surprisingly difficult to explain!

In 2013 an Mpemba effect contest had been held and the paper quoted at the top of this post was the winner (out of 22.000 submissions!). Physical chemist Nikola Bregovic emphasizes the impact of heat transfer and convection: Hot water is cooled faster due to more efficient heat transfer to the environment. Stirring the liquid will disturb convective flows inside the vessel and can prevent the Mpemba effect.

The  effect could also be due to different spontaneous freezing temperatures of supercooled water. Ice crystals can start to grow instantly at a temperature below the theoretical freezing point:

Various parameters and processes – such as living organisms in the water or heating water to higher temperatures before! –  might destroy or create nucleation sites for ice crystals. Supercooling of vapor might also allow for a jump over the green liquid area in the phase diagram, and thus for deposition of ice from vapor even at normal pressures.

Quoting Bregovic again:

I did not expect to find that water could behave in such a different manner under so similar conditions. Once again this small, simple molecule amazes and intrigues us with it’s magic.
~
Ice in our underground water tank, growing at the top layer of heat exchanger tubes. These are only covered with water if a bulk of ice underneath will make the water level rise.

Pumped Heat from the Tunnel

The idea to use a reservoir of water as a heat pump’s heat source is not new. But now and then somebody dares to do it again in a more spectacular way. Provided governmental agencies give you permit, lakes or underground aquifers could be used.

Today a (German) press release about a European research project called Sinfonia caught my attention. The cities of Innsbruck (AT) and Bolzano (IT) plan to reduce energy demands by 40-50% and increase the percentage of renewable energies used for heating and electrical power by 30%. The results should serve as best practices applicable to other cities.

Diverse activities are planned, such as improving insulation, installing solar thermal collectors or photovoltaic panels, and developing ways to renovate even buildings that are subject to monument protection. The latter is quite a challenge in European cities as laws do typically not allow for installing anything that impacts the view of historical rooftops or the structure of facades.

In a smart grid infrastructure, energy demands and supply should be managed, for example by cooling down refrigerators to -30°C if too much electrical energy is available – effectively storing energy in the ice.

My favorite: Heat pumps should utilize water from a very special source – drain water from the tunnel underneath Brenner pass (called tunnel water or mountain water in German).

This source would provide water flowing at 200-300 liters per second at a  temperature of 22°C, resulting in about 10 Megawatt of heating power.

I want to cross-check these numbers:

Assuming low-temperature floor heating loops, heat pumps would need to operate between 22°C ‘input’ temperature and about 40°C ‘output’ temperature.

The maximum theoretical efficiency is limited by principles of thermodynamics: This is Carnot’s Coefficient of Performance, which is:

Thot / (Thot – Tcold)

There are absolute temperatures in Kelvin, so 273 K needs to be added to the temperatures in °C.

Thus the COP is about:

COP = (40 + 273) / (40 + 273 – 22 – 273) = 313 / 18 ~ 17

Carnot’s perfect circular process does not include any phase change – as the evaporation and condensation in a real heat pump – and there are different sources of energy loss.

Thus a real-live heat pump shows a much lower COP. But there is a simple rule of thumb based on experience that is surprisingly accurate: Divide Carnot’s COP by 2 to calculate the realistic COP.

So these heat pumps would operate at a COP of 8,5 which is still very high. The temperature of the tunnel water is expected to be rather constant – as ground water – so COPs will also be high in winter.

Standard ‘geothermal’ brine-water heat pumps show COPs of about 4 – 4,5 when operating between the standard temperatures (values used in standardized tests) of 0°C brine temperature and 35° heating water temperature (B0/W35). As we discussed the meaning of ‘brine’ in the comments recently: In relation to heat pumps this term always refers to a solution of glycol-based frost protection in water.

Water-water heat pumps utilizing ground water with a temperature of about 10°C show a COP of 5,5 to 6 (W10/W35).

I have picked 40° rather than 35° in my estimate, accounting for losses in a district heating system attached to the gigantic heat pumps. Had I calculated with 35° my numbers for the tunnel water heat pumps would even be higher. So the whole exercise is more of an order-of-magnitude check.

The mass flow of about 300 kg (= 300 liters) per second can be converted into power retrieved by the heat pump: by multiplying it with the specific heat of water (4,19 kJ/kg) and with the temperature drop of the water caused by passing the heat pump’s evaporator unit.

The temperature difference in the brine circuit connecting the heat source to the heat pump is about 5 K for standard ‘small’ heat pumps. Water-water heat pumps might also use and additional brine circuit and thus an additional heat exchanger to transfer heat from the source water to the heat pump. I use those 5 K nonetheless as it is not a constant anyway – mentally insert error bars of several 10% here.

The drain water ‘carries’ approximately:

300 kg/s * 4,19 kJ/kgK * 5K ~ 6285 kJ/s = 6285 kW ~ 6,3 MW

The COP represents the factor the electrical energy feed into the heat pump is multiplied with to yield the heating power.

[Heating power] = COP * [Electrical power]

Given a COP of 8 an electrical power of 1 MW would result in 8 MW of heating power, delivered to floor heaters for example. However, the remaining 7 MW then need to be retrieved from the heat source, and the heat source really needs to be able to deliver them:

[Heating power] = [Power retrieved from source] + [Electrical power]

Thus:

[Power to be retrieved from source] = [Heating power] * (COP – 1)/COP

Aiming at 10 MW heating power output using a heat pump with a COP of 8, the drain water would need to deliver:

[Required power from drain water] = 10 MW * 7/8 = 8,8 MW

This is higher than the calculated power of 6,3 MW but the temperature drop I used was just an estimate and 7 K would also as well be OK – let alone all the other assumptions for operating temperatures and COP. So the numbers from the press release are self-consistent.

Brenner basetunnel portalBrenner base tunnel, portal in Austria. The tunnel has been subject to endless political debates and completion seems now to be scheduled for 2025 (Image by Wikimedia user B.Zsolt) I guess the drain water will be available earlier though.

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.)

Gravity Wells Potential Plus Kinetic Energy - Circle-Ellipse-Parabola-Hyperbola

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.