A New Homework Problem on Modeling

Russ Hobbie and I like to use the homework problems in Intermediate Physics for Medicine and Biology to illustrate modeling. But rarely does a problem encompass the entire process of constructing and analyzing a mathematical model. Gene Surdutovich and I try to do better in the 6th edition of IPMB (due out in a few months). Here is a homework problem that is not in any edition of IPMB, but that requires the student to analyze a model in its entirety. At least, that’s the goal. 

Section 11.1

Problem 4½. Consider a variable that changes discretely (in steps, not continuously). You measure it in consecutive steps and get 

step	data
1	3.2184
2	4.0680
3	2.2896
4	4.6490
5	0.3875
6	1.3951

Your hypothesis is that this data can be obtained from a logistic map. Develop a mathematical model. Quantify it, analyze it, determine any unknown parameters, and decide if the data support your hypothesis.

The central question to be answered by this problem is: can this data be explained by a logistic map? You can’t definitively answer this question, but you can assess if the data support this hypothesis.

First you have to quantify what a logistic map is. The first equation in Section 10.9 of IPMB states that the logistic map is represented by the equation 

The counter j indicates which data point is which, and y_j is the value of the j^th data point. There are two parameters: a and y_∞.

We will want to use least squares to determine these parameters. Least squares is simpler if the parameters enter the model linearly. Ours don’t, but we can define b = a/y_∞ and then write the logistic map as 

Now we are in a position to apply the method of linear least squares, as described in Section 11.1 of IPMB. Define the quantity Q as 

Q represents the average of the squares of the differences between the data and the model. (Although we have six data points, the sum goes from one to five. We can’t use j = 6 because then we would need a seventh point for y_j+1). Our goal is to minimize Q, thereby obtaining the best fit to the data. The variables we can vary to minimize Q are a and b. To find the minimum, we set ∂Q/∂a = ∂Q/∂b = 0. Setting ∂Q/∂a = 0 gives

which reduces to

Similarly, setting ∂Q/∂b = 0 reduces to

Next, you need to calculate all these sums. I find it easiest to construct a table like that below

j	yj	yj2	yj3	yj4	yj yj+1	yj yj+12
1	3.2184	10.3581	33.3365	107.2902	13.0925	42.1367
2	4.0680	16.5486	67.3198	273.8570	9.3141	37.8897
3	2.2896	5.2423	12.0027	27.4814	10.6444	24.3713
4	4.6490	21.6132	100.4798	467.1305	1.8015	8.3751
5	0.3875	0.1502	0.0582	0.0226	0.5406	0.2095
6	1.3951
sum		53.9124	213.1970	875.7817	35.3931	112.9823

The two equations for a and b become

53.9124 a – 213.1970 b = 35.3931

213.1970 a – 875.7817 b = 112.9823

You can solve these two linear equations for the two unknowns. You will find 

a = 3.920 and b = 0.8253 

which means that 

a = 3.92 and y_∞ = 4.75

These are exactly the parameters I used to construct the original data. If you calculate Q, you will get zero (expect, perhaps, for some round-off error) because I didn’t add any noise to the data. The model explains the data well.

The two parameters have different interpretations. The parameter y_∞ merely scales the size of the data. Dividing y_j by y_∞ transforms the data so it lies in the range between zero and one. The parameter a, however, cannot be scaled away. It’s a fundamental parameter characteristic of the logistic map (Eq. 10.39 in IPMB). Moreover, a = 3.92 is well into the range for which the logistic map results are chaotic.

Other than for practice, why create this new problem? It requires the student to go through the entire modeling procedure. Translating the hypothesis (the logistic map) into quantitative form, identifying the unknown parameters (a and y_∞), using least squares to evaluate the parameters from the data, and examining the quality of the fit to determine if the calculation supports the hypothesis. You are getting about as close to modeling as you can hope for with a simple homework problem. 

The Five Step Method: Math Modelling, with Jason Bramburger  

https://www.youtube.com/watch?v=zw9Y4t-Nh3E 

A Toy Model for Radiation Damage

Sometimes a toy model (a simple model that strips away all the detail to expose the underlying mechanisms more clearly) can be useful. Today I present a new homework problem that contains a toy model for understanding equivalent dose.

Section 16.12

Problem 34 ½. Consider two scenarios.

Scenario 1: N* particles are distributed evenly in a volume V*, so the concentration is C* = N*/V*.

Scenario 2: The volume V* is divided into two noninteracting regions of volume V₁ and V₂, where V* = V₁ + V₂. All N* particles are placed in V₂. Therefore, the concentration of particles in V₁ is C₁ = 0, and the concentration in V₂ is C₂ = N*/V₂ = N*/V*[(V₁ + V₂)/V₂] = C*[(V₁ + V₂)/V₂].

Now, examine two cases about how, in a local region, cellular damage, D, relates to the concentration C.

Case 1: Damage is proportional to the concentration. In other words, D = αC, where α is a constant of proportionality.

Case 2: Damage is proportional to the square of the concentration. In other words, D = βC², where β is another constant of proportionality.

For both cases and both scenarios (a total of four different situations), average the damage over the entire volume V* to get D. Find how D is related to C*.

Stop! To get the most out of this blog post, stop reading and solve this homework problem yourself...

...Okay, so you solved it and now you’re back. Help me explain it to that fellow who didn’t bother to solve it for himself.

Case 1 (damage proportional to concentration)

Scenario 1: The concentration is uniform throughout V*. Averaging the local relation D = αC over V* simply gives D = αC*. The average relationship is the same as the local relationship.

Scenario 2: Locally, D₁ = 0 because all the particles are in V₂ so C₁ = 0. Moreover, D₂ = α C₂ = αC*[(V₁ + V₂)/V₂]. Now, average the damage over the volume V*. You get D = [V₁/(V₁ + V₂)] (0) + [V₂/(V₁ + V₂)] αC*[(V₁ + V₂)/V₂]. But all those complicated factors cancel out, and you get simply D = αC*. This is the same result as in scenario 1. The average damage is proportional to C*.

Case 2 (damage proportional to concentration squared)

Scenario 1: Again, the concentration is uniform throughout V*. So you just get D = βC*². All that matters is the average concentration, C*.

Scenario 2: Locally, D₁ = 0 and D₂ = βC₂² = βC*²[(V₁ + V₂)/V₂]². Now average over the volume V*. You get D = βC*²[(V₁ + V₂)/V₂]. If V₂ is much less than V₁, then D is much greater than βC*². It is as if the average damage is supercharged by the concentration being, well, concentrated. In this scenario, the average damage depends on both C* and the ratio V₁/V₂.

This is all interesting, but what does it mean? It means that if you deposit energy locally, then the concentration (or "dose") alone may not tell the whole story. It depends on how the damage depends on the concentration. What is an example of when the damage would be proportional to the square of the concentration? Suppose we are talking about damage to DNA. The concentration might refer to the number of “breaks” in the DNA strand caused by radiation. Now suppose further that DNA has a repair mechanism that can fix breaks as long as they are far apart. That is, as long as they are isolated. But if you get two breaks near each other, then the repair mechanism is overwhelmed and doesn’t work. So, you need two “breaks” close together or you get no damage (in the jargon of radiobiology, you need double-strand breaks instead of just single-strand breaks). The concentration squared tells you something about having two events happen at the same place. You need a “break” to happen at some target spot along the DNA (proportional to the concentration) and then you need another “break” to happen nearby (again, proportional to the concentration), so the probability of getting two breaks near the target spot is proportional to the concentration squared.

Now let’s compare x-rays and alpha particles. Suppose you irradiate tissue so that the energy deposited in the tissue is the same for both. Then, the “dose” (energy per unit mass, analogous to C) is the same in both cases. But the alpha particles (scenario 2) deposit all their energy along a few thin tracks, whereas x-rays (scenario 1) deposit their energy all over the place randomly. You might say: well, for alpha particles the energy has a high density along the path, but everywhere else there is nothing, so on average those effects balance out. That’s true if damage is proportional to concentration (case 1 above). But if damage is proportional to concentration squared (case 2), it’s not true. The average damage caused by alpha particles is more extensive than for x-rays, even if the energy deposited into the tissue (the dose) is the same. The “equivalent dose” (another term for “damage”) is higher for the alpha particles than for the x-rays.

Intermediate Physics for
Medicine and Biology.

In Section 16.12 of Intermediate Physics for Medicine and Biology, Russ Hobbie and I introduce the concept of equivalent dose. To find the equivalent dose, the dose is multiplied by a dimensionless weighting factor (in the jargon, called the “relative biological effectiveness”), which is one for x-rays and twenty for alpha particles. The equivalent dose even has its own unit, the sievert (as opposed to the gray, the unit of the dose). Both the sievert and the gray are abbreviations for joules per kilogram, but the sievert includes the weighting factor. Alpha particles just do more damage than x-rays for a given dose. This is because alpha particles deposit their energy in a smaller volume, and damage depends on DNA being hit twice close together. In other words, damage depends on the concentration squared. In our toy model, the weighting factor is (V₁ + V₂)/V₂.

Our whole story about DNA repair mechanisms is reasonable and most likely true. But any other mechanism that results in the damage depending on the concentration (or dose) squared would give the same behavior. This result is not limited to DNA repair processes.

In general, case 1 (damage proportional to concentration) and case 2 (damage proportional to the square of the concentration) are not mutually exclusive. For instance, instead of DNA repair mechanisms being perfect for single-strand breaks and being useless for double-strand breaks, perhaps they are 90% effective for single-strand breaks and only 10% effective for double-strand breaks. In Section 16.9 of IPMB, Russ and I show that cell survival curves typically have two terms, one proportional to the dose and one proportional to the dose squared. At low doses the linear term dominates, but at high doses the quadratic one does.

The goal of toy models is to provide insight. I hope that even though the model in this new homework problem is oversimplified and artificial, it helps you get an intuitive feel for the equivalent dose.

A Toy Model for Straggling

One of the homework problems in Intermediate Physics for Medicine and Biology (Problem 31 in Chapter 16) introduces a toy model for the Bragg peak. I won’t review that entire problem, but students derive an equation for the stopping power, S, (the energy per unit distance deposited in tissue by a high energy ion) as a function of the depth below the tissue surface, x

where S₀ is the ion’s stopping power at the surface (x = 0) and R is the ion’s range. At a glance you can see how the Bragg peak arises—the denominator goes to zero at x = R so the stopping power goes to infinity. That, in fact, is why proton therapy for cancer is becoming so popular: Energy is deposited primarily at one spot well below the tissue surface where a tumor is located, with only a small dose to upstream healthy tissue. 

One topic that comes up when discussing the Bragg peak is straggling. The idea is that the range is not a single parameter. Instead, protons have a distribution of ranges. When preparing the 6th edition of Intermediate Physics for Medicine and Biology, I thought I would try to develop a toy model in a new homework problem to illustrate straggling. 

Section 16.10 

Problem 31 ½. Consider a beam of protons incident on a tissue. Assume the stopping power S for a single proton as a function of depth x below the tissue surface is

Furthermore assume that instead of all the protons having the same range R, the protons have a uniform distribution of ranges between R – δ/2 and R + δ/2, and no protons have a range outside this interval. Calculate the average stopping power by integrating S(x) over this distribution of ranges. 

This calculation is a little more challenging than I had expected. We have to consider three possibilities for x. 

x < R — δ/2

In this case, all of the protons contribute so the average stopping power is

We need to solve the integral 

First, let

With a little analysis, you can show that

So the integral becomes

This new integral I can look up in my integral table

Finally, after a bit of algebra, I get

Well, that was a lot of work and the result is not very pretty. And we are not even done yet! We still have the other two cases. 

 R — δ/2 <  x <  R + δ/2

In this case, if the range is less than x there is no contribution to the stopping power, but if the range is greater than x there is. So, we must solve the integral

I’m not going to go through all those calculations again (I’ll leave it to you, dear reader, to check). The result is 

x   >  R + δ/2

This is the easy case. None of the protons make it to x, so the stopping power is zero. 

Well, I can’t look at these functions and tell what the plot will look like. All I can do is ask Mr. Mathematica to make the plot (he’s much smarter than I am). Here’s what he said: 

The peak of the “pure” (single value for the range) curve (the red one) goes to infinity at x = R, and is zero for any x greater than R. As you begin averaging, you start getting some stopping power past the original range, out to R + δ/2. To me the most interesting thing is that for x = R – δ/2, the stopping power is larger than for the pure case. The curves all overlap for x  > R + δ/2 (of course, they are all zero), and for fairly small values x (in these cases, about x <  0.5) the curves are all nearly equal (indistinguishable in the plot). Even a small value of δ (in this case, for a spread of ranges equal to one tenth the pure range), the peak of the stopping power curve is suppressed. 

The curves for straggling that you see in most textbooks are much smoother, but that’s because I suspect they assume a smoother distribution of range values, such as a normal distribution. In this example, I wanted something simple enough to get an analytical solution, so I took a uniform distribution over a width δ . 

Will this new homework problem make it into the 6th edition? I’m not sure. It’s definitely a candidate. However, the value of toy models is that they illustrate the physical phenomenon and describe it in simple equations. I found the equations in this example to be complicated and not illuminating. There is still some value, but if you are not gaining a lot of insight from your toy model, it may not be worth doing. I’ll leave the decision of including it in the 6th edition to my new coauthor, Gene Surdutovich. After all, he’s the expert in the interaction of ions with tissue.

Dipole-Dipole Interaction

One strength of Intermediate Physics for Medicine and Biology is its many homework problems. The problems stress (but perhaps not enough) the ability to make general arguments about how some quantity will depend on a variable. Often getting a calculation exactly right is not as important as just knowing how something varies with something else. For instance, you could spend all day learning how to compute the volume and surface area of complicated objects, but it’s still useful simply to know that volume goes as size cubed and surface area as size squared. Below is a new homework problem that emphasizes the ability to determine a functional form.

Section 6.7

Problem 20½. Consider an electric dipole p a distance r from a small dielectric object. Calculate how the energy of interaction between the dipole and the induced dipole in the dielectric varies with r. Will the dipole be attracted to or repelled from the dielectric? Use the following facts:

1. The energy U of a dipole in an electric field E is U = – p · E,

2. The net dipole induced in a dielectric, p', is proportional to the electric field the dielectric experiences,

3. The electric potential produced by a dipole is given by Eq. 7.30.

Let’s take a closer look at these three facts.

1. When discussing magnetic resonance imaging in Chapter 18 of IPMB, we give the energy U of a magnetic dipole μ in a magnetic field B as U = – μ · B (Eq. 18.3). An analogous relationship holds for an electric dipole in an electric field. The energy is lowest when the dipole and the electric field are in the same direction, and varies as the cosine of the angle between them. I suggest treating the original dipole p as producing the electric field E, and the induced dipole p' as interacting it. 

2. Section 6.7 of IPMB discusses how an electric field polarizes a dielectric. The net dipole p' induced in the dielectric object will depend on the electric field and the objects shape and volume. I don’t want you to have to worry about the details, so the problem simply says that the net dipole is proportional to the electric field. You might get worried and say “wait, the electric field in the dielectric is not uniform!” That’s why I said the dielectric object is small. Assume that it’s small enough compared to the distance to the dipole that the electric field is approximately uniform over the volume of the dielectric. 

3. What is the electric field produced by a dipole? Russ Hobbie and I don’t actually calculate that, but we do give an equation for a dipole’s electrical potential, which falls off as one over the square of the distance. (It may look like the cube of the distance in Eq. 7.13, but there’s a factor of distance in the numerator that cancels one factor of distance cubed in the denominator, so it’s an inverse square falloff.) The electric field is the negative gradient of the potential. Calculating the electric field can be complicated in the general case. I suggest you assume the dipole p points toward the dielectric. Fortunately, the functional dependence of the energy on the distance r does not depend on the dipole direction.

I won’t work out all theentire solution here. When all is said and done, the energy falls off as 1/r⁶, and the dipole is attracted to the dielectric. It doesn’t matter if the dipole originally pointed toward the dielectric or away from it, the force is always attractive.

This result is significant for a couple reasons. First, van der Waals interactions are important in biology. Two dielectrics attract each other with an energy that falls as 1/r⁶. Why is there any interaction at all between two dielectrics? Because random thermal motion can create a fluctuating dipole in one dielectric, which then induces a dipole in a nearby dielectric, causing them to be attracted. These van der Waals forces play a role in how biomolecules interact, such as during protein folding.

From Photon to Neuron:
Light, Imaging, Vision.

Second, there is a technique to determine the separation between two molecules called fluorescence resonance energy transfer (FRET). The fluorescence of two molecules, the donor and the acceptor, is affected by their dipole-dipole interaction. Because this energy falls off as the sixth power of the distance between them, FRET is very sensitive to distance. You can use this technique as a spectroscopic ruler. It’s not exactly the same as in the problem above, because both the donor and acceptor have permanent dipole moments, instead of one being a dielectric in which a dipole moment is induced. But nevertheless, the 1/r⁶ argument still holds, as long as the dipoles aren’t too close together. You can learn more about FRET in Philip Nelson’s book From Photon to Neuron: Light, Imaging, Vision.

The Cyclotron Resonance Hypothesis

Want a sneak peek at one of the new homework problems tentatively included in the 6th edition of Intermediate Physics for Medicine and Biology? Today I present a problem related to the flawed “cyclotron resonance hypothesis.” A lot of nonsense has been written about the idea of extremely low frequency electromagnetic fields influencing biology and medicine, and one of the proposed mechanisms for such effects is cyclotron resonance. 

In Section 8.1 of the 5th edition of IPMB, Russ Hobbie and I discuss the cyclotron.

One important application of magnetic forces in medicine is the cyclotron. Many hospitals have a cyclotron for the production of radiopharmaceuticals, especially for generating positron-emitting nuclei for use in Positron EmissionTomography (PET) imaging (see Chap. 17).

Consider a particle of charge q and mass m, moving with speed v in a direction perpendicular to a magnetic field B. The magnetic force will bend the path of the particle into a circle. Newton’s second law states that the mass times the centripetal acceleration, v²/r, is equal to the magnetic force

      mv²/r = qvB.      (8.5)

The speed is equal to [the] circumference of the circle, 2πr, divided by the period of the orbit, T. Substituting this expression for v into Eq. (8.5) and simplifying, we find

       T = 2π m/(qB).   (8.6)

In a cyclotron particles orbit at the cyclotron frequency, f = 1/T. Because the magnetic force is perpendicular to the motion, it does not increase the particles’ speed or energy. To do that, the particles are subjected periodically to an electric field that changes direction with the cyclotron frequency so that it is always accelerating, not decelerating, the particles. This would be difficult if not for the fortuitous disappearance of both v and r from Eq. (8.6), so that the cyclotron frequency only depends on the charge-to-mass ratio of the particles and the magnetic field, but not on their energy.

This analysis of cyclotron motion works great in a vacuum. The trouble begins when you apply the cyclotron concept to ions in the conducting fluids of the body. The proposed hypothesis says that when an ion is moving about in the presence of the earth’s magnetic field, the resulting magnetic force causes it to orbit about the magnetic field lines, with an orbital period equal to the reciprocal of the cyclotron frequency. If any electric field is present at that same frequency, it could interact with the ion, increasing its energy or causing it to cross the cell membrane.

Below is a draft of the new homework problem, which I hope debunks this erroneous hypothesis.

Section 9.1

Problem 7. One mechanism for how organisms are influenced by extremely low frequency electric fields is the cyclotron resonance hypothesis. 

(a) The strength of the earth's magnetic field is about 5 × 10^–5 T. A calcium ion has a mass of 6.7 × 10^–26 kg and a charge of 3.2 × 10^–19 C. Calculate the cyclotron frequency of the calcium ion. If an electric field exists in the tissue at that frequency, the calcium ion will be in resonance with the cyclotron frequency, which could magnify any biological effect. 

(b) This mechanism seems to provide a way for an extremely low frequency electric field to interact with calcium ions, and calcium influences many cellular processes. But consider this hypothesis in more detail. Use Eq. 4.12 to calculate the root-mean-square speed of a calcium ion at body temperature. Use this speed in Eq. 8.5 to calculate the radius of the orbit. Compare this to the size of a typical cell. 

(c) Now make a similar analysis, but assume the radius of the calcium ion orbit is about the size of a cell (since it would have difficulty crossing the cell membrane). Then use this radius in Eq. 8.5 to determine the speed of the calcium ion. If this is the root-mean-square speed, what is the body temperature? 

(d) Finally, compare the period of the orbit to the time between collisions of the calcium ion with a water molecule. What does this imply for the orbit?

This analysis should convince you that the cyclotron resonance hypothesis is unlikely to be correct. Although the frequency is reasonable, the orbital radius will be huge unless the ions are traveling extraordinarily slowly. Collisions with water molecules will completely disrupt the orbit.

For those who don't have the 5th edition of IPMB handy, Eq. 4.12 says the root-mean-square speed is equal to the square root of 3 times Boltzmann's constant times the absolute temperature divided by the mass of the particle. 

I won’t give away the solution to this problem (once the 6th edition of IPMB is out, instructors can get the solution manual for free by emailing me at roth@oakland.edu). But here are some order-of-magnitude results. The cyclotron frequency is tens of hertz. The root-mean-square (thermal) speed of calcium at body temperature is hundreds of meters per second. The resulting orbital radius is about a meter. That is bigger than the body, and vastly bigger than a cell. To fit the orbit inside a cell, the speed would have to be much slower, on the order of a thousandth of a meter per second, which corresponds to a temperature of about a few nanokelvins. The orbital period is a couple hundredths of a second, but the time between collisions of the ion with a water molecule is one the order of 10^–13 seconds, so there are many billions of collisions per orbit. Any circular motion will be destroyed by collisions long before anything like an orbit is established. I’m sorry, but the hypothesis is rubbish.

Are Electromagnetic Fields
Making Me Ill?

If you want to learn more about how extremely low frequency electric fields interact with tissue, see my book Are Electromagnetic Fields Making Me Ill?

Finally, for you folks who are really on the ball, you may be wondering why this homework problem is listed as being in Chapter 9 when the discussion of the cyclotron is in Chapter 8 of the 5th edition of IPMB. (In this post I changed the equation numbers in the homework problem to match the 5th edition, so you would have them.) Hmm.. is there a new chapter in the 6th edition? More on that later…

 To be fair, I should let my late friend Abraham Liboff tell you his side of the story. In this video, Abe explains how he proposed the cyclotron resonance hypothesis. I liked Abe, but I didn’t like his hypothesis.

https://www.youtube.com/watch?v=YL-wqJ-PMAQ&list=PLCO-VktC6wofkMeEeZknT9Y4WhMnP76Ee&index=6

The Song of the Dodo

The Song of the Dodo,
by David Quammem.

One of my favorite science writers is David Quammen. I’ve discussed several of his books in this blog before, such as Breathless, Spillover, and The Tangled Tree. A copy of one of his earlier books—The Song of the Dodo: Island Biogeography in an Age of Extinctions—has sat on my bookshelf for a while, but only recently have I had a chance to read it. I shouldn’t have waited so long. It’s my favorite.

Quammen is not surprised that the central idea of biology, natural selection, was proposed by two scientists who studied islands: Charles Darwin and the Galapagos, and Alfred Russell Wallace and the Malay Archipelago. The book begins by telling Wallace’s story. Quammen calls him “the man who knew islands.” Wallace was the founder of the science of biogeography: the study of how species are distributed throughout the world. For example, Wallace’s line lies between two islands in Indonesia that are only 20 miles apart: Bali (with plants and animals similar to those native to Asia) and Lombok (with flora and fauna more like that found in Australia). Because islands are so isolated, they are excellent laboratories for studying speciation (the creation of new species through evolution) and extinction (the disappearance of existing species).

Quammen is the best writer about evolution since Stephen Jay Gould. I would say that Gould was better at penning essays and Quammen is better at authoring books. Much of The Song of the Dodo deals with the history of science. I would rank it up there with my favorite history of science books: The Making of the Atomic Bomb by Richard Rhodes, The Eighth Day of Creation by Horace Freeland Judson, and The Maxwellians by Bruce Hunt.

Yet, The Song of the Dodo is more than just a history. It’s also an amazing travelogue. Quammen doesn’t merely write about islands. He visits them, crawling through rugged jungles to see firsthand animals such as the Komodo Dragon (a giant man-eating lizard), the Madagascan Indri (a type of lemur), and the Thylacine (a marsupial also known as the Tasmanian tiger). A few parts of The Song of the Dodo are one comic sidekick away from sounding like a travel book Tony Horwitz might have written. Quammen talks with renowned scientists and takes part in their research. He reminds me of George Plimpton, sampling different fields of science instead of trying out various sports.

Although I consider myself a big Quammen fan, he does have one habit that bugs me. He hates math and assumes his readers hate it too. In fact, if Quammen’s wife Betsy wanted to get rid of her husband, she would only need to open Intermediate Physics for Medicine and Biology to a random page and flash its many mathematical equations in front of his face. It would put him into shock, and he probably wouldn’t last the hour. In his book, Quammen only presents one equation and apologizes profusely for it. It’s a power law relationship

S = c Aⁿ .

This is the same equation that Russ Hobbie and I analyze in Chapter 2 of IPMB, when discussing log-log plots and scaling. How do you determine the dimensionless exponent n for a particular case? As is my wont, I’ll show you in a new homework problem.

Section 2.11

Problem 40½. In island biogeography, the number of species on an island, S, is related to the area of the island, A, by the species-area relationship: S = c Aⁿ, where c and n are constants. Philip Darlington counted the number of reptile and amphibian species from several islands in the Antilles. He found that when the island area increased by a factor of ten, the number of species doubled. Determine the value of n.

Let me explain to mathaphobes like Quammen how to solve the problem. Assume that on one island there are S₀ species and the area is A₀. On another island, there are 2S₀ species and an area of 10A₀. Put these values into the power law to find S₀ = cA₀ⁿ and 2S₀ = c(10A₀)ⁿ. Now divide the second equation by the first (c, S₀, and A₀ all cancel) to find 2 = 10ⁿ. Take the logarithm of both sides, so log(2) = log(10ⁿ), or using a property of logarithms log(2) = n log(10). So n = log(2)/log(10) = 0.3. Note that n is positive, as it should be since increasing the area increases the number of species.

When I finished the main text of The Song of the Dodo, I thumbed through the glossary and found an entry for logarithm. “Aww,” I thought, “Quammen was only joking; he likes math after all.” Then I read his definition: “logarithm. A mathematical thing. Never mind.”

About halfway through, the book makes a remarkable leap from island biogeography—interesting for its history and relevance to exotic tropical isles—to mainland ecology, relevant to critical conservation efforts. Natural habitats on the continents are being broken up into patches, a process called fragmentation. The expansion of towns and farms creates small natural reserves surrounded by inhospitable homes and fields. The few remaining native regions tend to be small and isolated, making them similar to islands. A small natural reserve cannot support the species diversity that a large continent can (S = c Aⁿ). Extinctions inevitably follow.

The Song of the Dodo also provides insight into how science is done. For instance, the species-area relationship was derived by Robert MacArthur and Edward Wilson. While it’s a valuable contribution to island biogeography, scientists disagree on its applicability to fragmented continents, and in particular they argue about its relevance to applied conservation. Is a single large reserve better than several small ones? In the 1970s a scientific battle raged, with Jared Diamond supporting a narrow interpretation of the species-area relationship and Dan Simberloff advocating for a more nuanced and less dogmatic view. As in any science, the key is to get data to test your hypothesis. Thomas Lovejoy performed an experiment in the Amazon to test the species-area relationship. Parts of the rainforest were being cleared for agriculture or other uses, but the Brazilian government insisted on preserving some of the native habitat. Lovejoy obtained permission to create many different protected rainforest reserves, each a different size. His team monitored the reserves before and after they became isolated from adjacent lands, and tracked the number of species supported in each of these “islands” over time. While the results are complicated, there is a correlation between species diversity and reserve size. Area matters.

One theme that runs through the story is extinction. If you read the book, you better have your hanky ready when you reach the part where Quammen imagines the death of the last Dodo bird. Conservation efforts are featured throughout the text, such as the quest to save the Mauritius kestrel.  

 

The Song of the Dodo concludes with a mix of optimism and pessimism. Near the end of the book, when writing about his trip to Aru (an island in eastern Indonesia) to observe a rare Bird of Paradise, Quammen writes

The sad, dire things that have happened elsewhere, in so many parts of the world—biological imperialism, massive habitat destruction, fragmentation, inbreeding depression, loss of adaptability, decline of wild populations to unviable population levels, ecosystem decay, trophic cascades, extinction, extinction, extinction—haven’t yet happened here. Probably they soon will. Meanwhile, though, there’s still time. If time is hope, there’s still hope.

An interview with David Quammen, by www.authorsroad.com

https://www.youtube.com/watch?v=Quq7PNH1zWM

Can the Microwave Auditory Effect Be ‘Weaponized’

“Can the Microwave Auditory
Effect be Weaponized?”

I was recently reading Ken Foster, David Garrett, and Marvin Ziskin’s paper “Can the Microwave Auditory Effect Be Weaponized?” (Frontiers in Public Health, Volume 9, 2021). It analyzed if microwave weapons could be used to “attack” diplomats and thereby cause the Havana syndrome. While I am interested in the Havana syndrome (I discussed it in my book Are Electromagnetic Fields Making Me Ill?), today I merely want to better understand Foster et al.’s proposed mechanism by which an electromagnetic wave can induce an acoustic wave in tissue.

As is my wont, I will present this mechanism as a homework problem at a level you might find in Intermediate Physics for Medicine and Biology. I’ll assign the problem to Chapter 13 about Sound and Ultrasound, although it draws from several chapters.

Forster et al. represent the wave as decaying exponentially as it enters the tissue, with a skin depth λ. To keep things simple and to focus of the mechanism rather than the details, I’ll assume the energy in the electromagnetic wave is absorbed uniformly in a thin layer of thickness λ, ignoring the exponential behavior.

Section 13.4

Problem 17 ½. Assume an electromagnetic wave of intensity I₀ (W/m²) with area A (m²) and duration τ (s) is incident on tissue. Furthermore, assume all its energy is absorbed in a depth λ (m).

(a) Derive an expression for the energy E (J) dissipated in the tissue.

(b) Derive an expression for the tissue’s increase in temperature ΔT (°C), E = C ΔT, where C (J/°C) is the heat capacity. Then express C in terms of the specific heat capacity c (J/°C kg), the density ρ (kg/m³), and the volume where the energy was deposited V (m³). (For a discussion of the heat capacity, see Sec. 3.11).

(c) Derive an expression for the fractional increase in volume, ΔV/V, caused by the increase in temperature, ΔV/V = αΔT, where α (1/°C) is the tissue’s coefficient of thermal expansion.

(d) Derive an expression for the change in pressure, ΔP (Pa), caused by this fractional change in volume, ΔP = B ΔV/V, where B (Pa) is the tissue’s bulk modulus. (For a discussion of the bulk modulus, see Sec. 1.14).

(e) You expression in part d should contain a factor Bα/cρ. Show that this factor is dimensionless. It is called the Grüneisen parameter.

(f) Assume α = 0.0003 1/°C, B = 2 × 10⁹ Pa, c = 4200 J/kg °C, and ρ = 1000 kg/m³. Evaluate the Grüneisen parameter. Calculate the change in pressure ΔP if the intensity is 10 W/m², the skin depth is 1 mm, and the duration is 1 μs.

I won’t solve the entire problem for you, but the answer for part d is

                            ΔP = I₀ (τ/λ) [Bα/cρ] .

I should stress that this calculation is approximate. I ignored the exponential falloff. Some of the incident energy could be reflected rather than absorbed. It is unclear if I should use the linear coefficient of thermal expansion or the volume coefficient. The tissue may be heterogeneous. You can probably identify other approximations I’ve made. 

Interestingly, the pressure induced in the tissue varies inversely with the skin depth, which is not what I intuitively expected. As the skin depth gets smaller, the energy is dumped into a smaller volume, which means the temperature increase within this smaller volume is larger. The pressure increase is proportional to the temperature increase, so a thinner skin depth means a larger pressure.

You might be thinking: wait a minute. Heat diffuses. Do we know if the heat would diffuse away before it could change the pressure? The diffusion constant of heat (the thermal diffusivity) D for tissue is about 10^-7 m²/s. From Chapter 4 in IPMB, the time to diffuse a distance λ is λ²/D. For λ = 1 mm, this diffusion time is 10 s. For pulses much shorter than this, we can ignore thermal diffusion. 

Perhaps you’re wondering how big the temperature rise is? For the parameters given, it’s really small: ΔT  = 2 × 10^–9 °C. This means the fractional change in volume is around 10^–12. It’s not a big effect.

The Grüneisen parameter is a dimensionless number. I’m used to thinking of such numbers as being the ratio of two quantities with the same units. For instance, the Reynolds number is the ratio of an inertial force to a viscous force, and the Péclet number is the ratio of transport by drift to transport by diffusion. I’m having trouble interpreting the Grüneisen parameter in this way. Perhaps it has something to do with the ratio of thermal energy to elastic energy, but the details are not obvious, at least not to me.

What does this all have to do with the Havana syndrome? Not much, I suspect. First, we don’t know if the Havana syndrome is caused by microwaves. As far as I know, no one has ever observed microwaves associated with one of these “attacks” (perhaps the government has but they keep the information classified). This means we don’t know what intensity, frequency (and thus, skin depth), and pulse duration to assume. We also don’t know what pressure would be required to explain the “victim’s” symptoms. 

In part f of the problem, I used for the intensity the upper limit allowed for a cell phone, the skin depth corresponding approximately to a microwave frequency of about ten gigahertz, and a pulse duration of one microsecond. The resulting pressure of 0.0014 Pa is much weaker than is used during medical ultrasound imaging, which is known to be safe. The acoustic pressure would have to increase dramatically to pose a hazard, which implies very large microwave intensities.

Are Electromagnetic Fields
Making Me Ill?

That such a large intensity electromagnetic wave could be present without being noticeable seems farfetched to me. Perhaps very low pressures could have harmful effects, but I doubt it. I think I’ll stick with my conclusion from Are Electromagnetic Fields Making Me Ill?

Microwave weapons and the Havana Syndrome: I am skeptical about microwave weapons, but so little evidence exists that I want to throw up my hands in despair. My guess: the cause is psychogenic. But if anyone detects microwaves during an attack, I will reconsider.

Special Relativity in IPMB

Electricity and Magnetism,
by Edward Purcell.

In Intermediate Physics for Medicine and Biology, Russ Hobbie and I rarely discuss special relativity. We briefly mention that magnetism is a consequence of relativity in Chapter 8 (Biomagnetism) but we don’t develop our study of magnetic fields from this point of view. (If you want to see magnetism analyzed in this way, I suggest looking at the textbook Electricity and Magnetism, by Edward Purcell, which is Volume 2 of the Berkeley Physics Course). We use the relationship between the energy and momentum of a photon, E = pc, in Chapter 15 (Interaction of Photons and Charged Particles with Matter) when analyzing Compton scattering and pair production. And we use Einstein’s famous equation E = mc², relating a particles energy to its rest mass, when calculating the binding energy of nuclei in Chapter 17 (Nuclear Physics and Nuclear Medicine).

The most relativisticish equation we present is in Chapter 15 when analyzing how charged particles (such as protons, electrons, or alpha particles) lose energy when passing through tissue at relativistic speeds. We write

The stopping powers are plotted vs particle speed in the form β = v/c. At low energies (β ≪ 1) β is related to kinetic energy by 

For larger values of β, the relativistically correct expression

was used to convert Fig. 15.17 to 15.18. 

Fig. 15.17

 

Fig. 15.18

Here’s a new homework problem examining the relationship between a particle’s speed and kinetic energy when its speed is near the speed of light.

Section 15.11

Problem 41 ½. A charged particle’s kinetic energy, T, is related to its mass M and its speed, v. We often express speed in terms of the parameter β = v/c, where c is the speed of light.

(a) At low energies (T ≪ Mc², or equivalently β ≪ 1), show that Eq. 15.47 is consistent with the familiar expression from classical mechanics, T = ½ mv².

(b) Show that Equation 15.48 (the relativistically correct relationship between β and T) reduces to Eq. 15.47 when T ≪ Mc².

(c) Plot β versus T/Mc², both in a linear plot (0 < T/Mc² < 3) and in a log-log plot (0.0001 < T/Mc² < 100).

(d) Take a few data points from Fig. 15.17 for a proton, replot them in Fig. 15.18, where the dependent variable is β, not T. See how well they match. Be sure to adjust for the different units for Stopping Power in the two plots.