Showing posts with label new homework problem. Show all posts
Showing posts with label new homework problem. Show all posts

Friday, December 29, 2023

Special Relativity in IPMB

Electricity and Magnetism, by Edward Purcell, superimposed on Intermediate Physics for Medicine and Biology.
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 (TMc2, 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 TMc2.

(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.

Friday, May 26, 2023

Terminal Speed of Microorganisms

A Paramecium aurelia seen through an optical microscope
A Paramecium aurelia seen through an optical microscope.
Source: Wikipedia (http://en.wikipedia.org/wiki/Image:Paramecium.jpg)

Homework Problem 28 at the end of Chapter 2 in Intermediate Physics for Medicine and Biology asks the reader to calculate the terminal speed of an animal falling in air. Although this problem provides insight, it includes a questionable assumption. Russ and I tell the student to “assume that the frictional force is proportional to the surface area of the animal.” If, however, the animal falls at low Reynolds number, this assumption is not valid. Instead, the drag force is given by Stokes’ law, which is proportional to the radius, not the surface area (radius squared). The new homework problem given below asks the reader to calculate the terminal speed for a microorganism falling through water at low Reynolds number.

Section 2.8

Problem 28 ½. Calculate the terminal speed, V, of a paramecium sinking in water. Assume that the organism is spherical with radius R, and that the Reynolds number is small so that the drag force is given by Stokes’ law. Include the effect of buoyancy. Let the paramecium’s radius be 100 microns and its specific gravity be 1.05. Verify that its Reynolds number is small.
The reader will first need to get the density ρ and viscosity η of water, which are ρ = 1000 kg/m3 and η = 0.001 kg/(m s). The specific gravity is not defined in IPMB, but it’s the density divided by the density of water, implying that the density of the paramecium is 1050 kg/m3. Finally, Stokes’ law is given in IPMB as Eq. 4.17, Fdrag = –6πRηV.

I’ll let you do your own calculation. I calculate the terminal speed to be about 1 mm/s, so it takes about a fifth of a second to sink one body diameter. The Reynolds number is 0.1, which is small, but not exceptionally small.

You should find that the terminal speed increases as the radius squared, in contrast to a drag force proportional to the surface area for which the terminal speed increases in proportion to the radius. Bigger organisms sink faster. The dependence of terminal speed on size is even more dramatic for aquatic microorganisms than for mammals falling in air. To paraphrase Haldane’s quip, “a bacterium is killed, a diatom is broken, a paramecium splashes,” except the speeds are small enough that none of the “wee little beasties” are really killed (the terminal speed is not terminal...get it?) and splashing is a high Reynolds number phenomenon.

Buoyancy is not negligible for aquatic animals. The effective density of a paramecium in air would be about 1000 kg/m3, but in water its effective density drops to a mere 50 kg/m3. Microorganisms are made mostly of water, so they are almost neutrally buoyant. In this homework problem, the effect of gravity is reduced to only five percent of what it would be if buoyancy were ignored.

A paramecium is a good enough swimmer that it can swim upward against gravity if it wants to. Its surface is covered with cilia that beat together like a Roman galley to produce the swimming motion (ramming speed!).

Whenever discussing terminal speed, one should remember that we assume the fluid is initially at rest. In fact, almost any volume of water will have currents moving at speeds greater than 1 mm/s, caused by tides, gravity, thermal convection, wind driven waves, or the wake of a fish swimming by. A paramecium would drift along with these currents. To observe the motion described in this new homework problem, one must be careful to avoid any bulk movement of water.

If you watched a paramecium sink in still water, would you notice any Brownian motion? You can calculate the root-mean-squared thermal speed with Eq. 4.12 in IPMB, using the mass of the paramecium as four micrograms and a temperature of 20° C. You get approximately 0.002 mm/s. That is less than 1% of the terminal speed, so you wouldn’t notice any random Brownian motion unless you measured extraordinarily carefully.

Friday, February 24, 2023

A Simple Mathematical Function Representing the Intracellular Action Potential

In Problem 14 of Chapter 7 in Intermediate Physics for Medicine and Biology, Russ Hobbie and I chose a strange-looking function to represent the intracellular potential along a nerve axon, vi(x). It’s zero everywhere except in the range −a < x < a, where it’s
 

 
Why this function? Well, it has several nice properties, which I’ll leave for you to explore in this new homework problem.
Section 7.4

Problem 14 ¼. For the intracellular potential, vi(x), given in Problem 14
(a) show that vi(x) is an even function,
(b) evaluate vi(x) at x = 0,
(c) show that vi(x) and dvi(x)/dx are continuous at x = 0, a/2 and a, and
(d) plot vi(x), dvi(x)/dx, and d2vi(x)/dx2 as functions of x, over the range −2a < x < 2a.
This representation of vi(x) has a shape like that of an action potential. Other functions also have a similar shape, such as a Gaussian. But our function is nice because it’s non-zero over only a finite region (−a < x < a) and it’s represented by a simple, low-order polynomial rather than a special function. An even simpler function for vi(x) would be triangular waveform, like that shown in Figure 7.4 of IPMB. However, that function has a discontinuous derivative and therefore its second derivative is infinite at discrete points (delta functions), making it tricky (but not too tricky) to deal with when calculating the extracellular potential (Eq. 7.21). Our function in Problem 14 ¼ has a discontinuous but finite second derivative.

The main disadvantage of the function in Problem 14 ¼ is that the depolarization phase of the “action potential” has the same shape as the repolarization phase. In a real nerve, the upstroke is usually briefer than the downstroke. The next new homework problem asks you to design a new function vi(x) that does not suffer from this limitation.
Section 7.4

Problem 14 ½. Design a piecewise continuous mathematical function for the intracellular potential along a nerve axon, vi(x), having the following properties. 
(a) vi(x) is zero outside the region −a < x < 2a
(b) vi(x) and its derivative dvi(x)/dx are continuous. 
(c) vi(x) is maximum and equal to one at x = 0. 
(d) vi(x) can be represented by a polynomial bi + ci x + di x2, where i refers to four regions: 
        i = 1,    −a < x < −a/2 
        i = 2, −a/2 < x < 0 
        i = 3,      0 < x < a
        i = 4,      a < x < 2a.
Finally, here’s another function that I’m particularly fond of.
Section 7.4

Problem 14 ¾. Consider a function that is zero everywhere except in the region −a < x < 2a, where it is

(a) Plot vi(x) versus x over the region −a < x < 2a,
(b) Show that vi(x) and its derivative are each continuous. 
(c) Calculate the maximum value of vi(x).
Simple functions like those described in this post rarely capture the full behavior of biological phenomena. Instead, they are “toy models” that build insight. They are valuable tools when describing biological phenomena mathematically.

Friday, December 23, 2022

Think Before You Calculate!

I encourage students to build their qualitative problem solving skills by recasting equations in dimensionless variables, analyzing the limiting behavior of mathematical expressions, and sketching plots showing how functions behave. “Think Before You Calculate!” is my mantra. But how, specifically, do you do this? Let me show you an example.

A plot of the solution to the logistic equation.
Fig. 2.16 from IPMB. A plot of the solution
of the logistic equation when y0 = 0.1,
y = 1.0, b0 = 0.0667. Exponential
growth with the same values of
y0 and b0 is also shown.
In Section 2.10 of Intermediate Physics for Medicine and Biology, Russ Hobbie and I discuss the logistic model.
Sometimes a growing population will level off at some constant value. Other times the population will grow and then crash. One model that exhibits leveling off is the logistic model, described by the differential equation

dy/dt = b0 y (1 – y/y) ,                           (2.28)

where b0 and y are constants….

If the initial value of y is y0, the solution of Eq. 2.28 is

y(t) = 1 / [1/y + (1/y0 – 1/y) eb0t] .    (2.29)
Below is a new homework problem, analyzing the logistic equation in a way to build insight. Consider it an early Christmas present. Santa won’t give you the answer, so you need to solve the problem yourself to gain anything from this post.
Section 2.10

Problem 36 ½. Consider the logistic model.

(a) Write Eq. 2.28 in terms of dimensionless variables Y and T, where Y = y/y and T = b0t.

(b) Express the solution Eq. 2.29 in terms of Y, T, and Y0 = y0/y.

(c) Verify that your solution in part (b) obeys the differential equation you derive in part (a).

(d) Verify that your solution in part (b) is equal to Y0 at T = 0.

(e) In a plot of Y(T) versus T, which of the three constants (y, y0, and b0) affect the qualitative shape of the solution, and which just scale the Y and T axes? 

(f) Verify that your solution in part (b) approaches 1 as T goes to infinity.

(g) Find an expression for the slope of the curve Y = Y(T). What is the slope at time T = 0? For what value of Y0 is the initial slope largest? For what values of Y0 is the slope small?

(h) The plot in Fig. 2.16 compares the solution of logistic equation with the exponential Y = Y0 eT. The figure gives the impression that the exponential is a good approximation to the logistic curve at small times. Do the two curves have the same value at T = 0? Do the two curves have the same slope at T = 0?

(i) Sketch plots of Y versus T for Y0 = 0.0001, 0.001, 0.01, and 0.1.

(j) Rewrite the solution from part (b), Y = Y(T), using the constant T0, where T0 = ln[(1−Y0)/Y0]. Show that varying Y0 is equivalent to shifting the solution along the T axis. What value of Y0 corresponds to T0 = 0?

(k) How does the logistic curve behave if Y0 > 1? Sketch a plot of Y versus T for Y0 =1.5.

(l) How does the logistic curve behave if Y0 < 0? Sketch a plot of Y versus T for Y0 = –0.5.

(m) Plot Y versus T for Y0 = 0.1 on semilog graph paper.

If you solve this new homework problem and want to compare you solution to mine, email me at roth@oakland.edu and I’ll send you my solution. 

The Logistic Equation, MIT OpenCourseWare

https://www.youtube.com/watch?v=TCkLSYxx21c&t=69s

Friday, August 19, 2022

The Loudest Sound

In Table 13.1 of Intermediate Physics for Medicine and Biology, Russ Hobbie and I list the approximate intensity levels of various sounds, in decibels. The minimum perceptible sound is 0 dB, a typical office has a sound level of 50 dB, a jack hammer is 100 dB, and the loudest sound listed is a rocket launch pad at 170 dB.

Can there be even louder sounds? Yes, there can! This new homework problem lets you calculate the loudest possible sound.
Section 13.4

Problem 17 ½. Let us calculate the loudest possible sound in air. 
(a) Use Eq. 13.29 to calculate the intensity of a sound in W m−2, using 428 Pa s m−1 for the acoustic impedance of air and one atmosphere (1.01 × 105 Pa) for the pressure. This pressure is the largest that can exist for a sinusoidally varying sound wave, as an even louder sound would create a minimum pressure below zero (less than a vacuum). 
(b) Use the result from part (a) to calculate the intensity in decibels using Eq. 13.34.
For those of you who don’t have a copy of IPMB at your side, here are the two equations you need

                I = ½ p2/Z                                         (13.29)

                Intensity level = 10 log10(I/I0)         (13.34)

where I is the intensity, Z is the acoustic impedance, p is the pressure, I0 is the minimum perceptible intensity (10−12 W m−2), and log10 is the common logarithm.

I’ll let you do the calculation, but you should find that the loudest sound is about 191 dB. Is this really an upper limit? No, you could have a peak pressure larger than one atmosphere, but in that case you wouldn’t be dealing with a traditional sound wave (with pressure ranging symmetrically above and below the ambient pressure) but more of a nonlinear acoustic shock wave.

Krakatoa, by Simon Winchester, superimposed on Intermediate Physics for Medicine and Biology.
Krakatoa,
by Simon Winchester.
Has there ever been a sound that loud? Or, more interestingly, what is the loudest sound ever heard on earth? That’s hard to say for sure, but one possibility is the 1883 eruption of the Krakatoa volcano. We know this sound was loud, because people heard it so far from where the eruption occurred.

Simon Winchester tells this story in his fascinating book Krakatoa: The Day the World Exploded, August 27. 1883. Krakatoa is an island that is now part of Indonesia. When it erupted, people on the island of Rodriguez in the western Indian Ocean, nearly 3000 miles from Krakatoa, could hear it. Winchester writes
In August 1883 the chief of police on Rodriguez was a man named James Wallis, and in his official report… for the month he noted:
On Sunday the 26th the weather was stormy, with heavy rain and squalls; the wind was from SE, blowing with a force of 7 to 10, Beaufort scale. Several times during the night (26th–27th) reports were heard coming from the eastward, like the distant roar of heavy guns. These reports continued at intervals of between three and four hours, until 3 pm on the 27th, and the last two were heard in the directions of Oyster Bay and Port Mathurie [sic].
This was not the roar of heavy guns, however. It was the sound of Krakatoa—busily destroying itself fully 2,968 miles away to the east. By hearing it that night and day, and by noting it down as any good public servant should, Chief Wallis was unknowingly making for himself two quite separate entries in the record books of the future. For Rodriguez Island was the place furthest from Krakatoa where its eruptions could be clearly heard. And the 2,968-mile span that separates Krakatoa and Rodriguez remains to this day the most prodigious distance recorded between the place where unamplified and electrically unenhanced natural sound was heard and the place where that same sound originated.
Winchester concludes
The sound that was generated by the explosion of Krakatoa was enormous, almost certainly the greatest sound ever experienced by man on the face of the earth. No manmade explosion, certainly, can begin to rival the sound of Krakatoa—not even those made at the height of the Cold War’s atomic testing years.

No one knows how many decibels Krakatoa’s eruption caused on the island itself. The sound was almost certainly in the nonlinear regime, and probably had an intensity of over 200 dB.

An Interview with Simon Winchester. 

https://www.youtube.com/watch?v=MGNLJb1m2fg

 

Friday, June 17, 2022

Limping

In Chapter 1 of Intermediate Physics for Medicine and Biology, Russ Hobbie and I discuss biomechanics. One of our most important examples is the force on the hip.
The forces in the hip joint can be several times a person’s weight, and the use of a cane can be very effective in reducing them.
Indeed, a cane is very useful, as Russ and I show in Section 1.8 of IPMB. But what if you don’t have a cane handy, or if you prefer not to use one? You limp. In this post, we examine the biomechanics of limping.

When you limp, you lean toward the injured side to reduce the forces on the hip. The reader can analyze limping in this new homework problem.
Section 1.8

Problem 11 ½. The left side of the illustration below analyzes normal walking and reproduces Figures 1.11 and 1.12. The right side shows what happens when you walk with a limp. By leaning toward the injured side you reduce the distance between the hip joint and the body’s center of gravity, and your leg is more vertical than in the normal case.
Pertinent features of the anatomy of the leg: normal (left) and limping (right), based on figures 1.11 and 1.12 in Intermediate Physics for Medicine and Biology.
Pertinent features of the anatomy of the leg: normal (left) and limping (right).

(a) Reproduce the analysis of Section 1.7 to calculate of the forces on the hip during normal walking using the illustration on the left. Begin by making a free-body diagram of the forces acting on the leg like in Figure 1.13. Then solve the three equations for equilibrium: one for the vertical forces, one for the horizontal forces, and one for the torques. Verify that the magnitude of the force on the hip joint is 2.4 times the weight of the body. 
(b) Reanalyze the forces on the hip when limping. Use the geometry and data shown in the illustration on the right. Assume that any information missing from the diagram is the same as for the case of normal walking; For example, the abductor muscle makes an angle of 70° with the horizontal for both the normal and limping cases. Draw a free-body diagram and determine the magnitude of the force on the hip joint in terms of the weight of the body.

I won’t solve the entire problem for you, but I’ll tell you this: limping reduces the force on the hip from 2.4 times the body weight in the normal case to 1.2 times the body weight in the case of limping. No wonder we limp! 

The main reason for the lower force when limping is the smaller moment arm. If we calculate torques about the head of the femur, then in the normal case the moment arm for the force that the ground exerts on the foot is 18 – 7 = 11 cm. When limping, this moment arm reduces to 9 – 7 = 2 cm. The moment arm for the abductor muscles (the gluteus minimus and gluteus medius) is the same in the two cases. Therefore, rotational equilibrium can be satisfied with a small muscle force when limping, although a large muscle force is required normally. The torque is a critical concept for understanding biomechanics.

What do you do if both hips are injured? When walking, you first lean to one side and then the other; you waddle. This reduces the forces on the hip, but results in a lot of swinging from side to side as you walk.

If you are having trouble solving this new homework problem, contact me and I’ll send you the solution. 

Enjoy!


Friday, June 10, 2022

The Genetics of Cystic Fibrosis

In Appendix H of Intermediate Physics for Medicine and Biology, Russ Hobbie and I briefly mention the severe lung disease cystic fibrosis. Analyzing this disease provides an opportunity to examine the prevalence of a genetic disorder. I’ll do this by creating a new homework problem.
Appendix H

Problem 5 ½. About 1 in every 2500 people is born with cystic fibrosis, an autosomal recessive disorder. What is the probability of the gene responsible for cystic fibrosis in the population? What fraction of the population are carriers of the disease?

To answer these questions, first we must know that an “autosomal recessive disorder” is one in which you only get the disease if you have two copies of a recessive gene. To a first approximation, there are often two variants (or alleles) of a gene governing a particular protein: dominant (A) and recessive (a). In order to have cystic fibrosis, you must have two copies of the recessive allele (aa). If you have only one copy (Aa), you are healthy but are a carrier for the disease: your children could potentially get the disease if your mate is also a carrier. If you have no copies of the recessive allele (AA) then you’re healthy and your children will also be healthy.

Let’s assume the probability of the dominant allele is p, and the probability of the recessive allele is q. Since we assume there are only two possibilities, we know that p + q = 1. Our goal is to find q, the probability of the gene responsible for cystic fibrosis in the population.

When two people mate, they each pass on to their offspring one of their two copies of the gene. The probability that both parents are dominant (AA), so the child is normal, is p2. The probability that both parents are recessive (aa), so the child has the disease, is q2. There are two ways for the child to be a carrier: A from dad and a from mom, or a from dad and A from mom. So, the probability of a child being a carrier (Aa) is 2pq. There are only three possibilities or genotypes: AA, Aa, and aa. The sum of their probabilities must equal one: p2 + 2pq + q2 = 1. But this expression is equivalent to (p + q)2 = 1, and we already knew that p + q = 1, so the result isn’t surprising.

The only people that suffer from cystic fibrosis have the genotype aa, so q2 is equal to the fraction of people with the disease. The problem states that this fraction is 1/2500 (0.04%). So, q is the square root of 1/2500, or 1/50 (2%; wasn’t that nice of me to make the fraction be the reciprocal of a perfect square?). One out of every fifty copies of the gene governing cystic fibrosis is defective (that is, it is the recessive version that can potentially lead to the disease). If q is 1/50, then p is 49/50 (98%). The fraction of carriers is 2pq, or 3.92%. The only reason this result is not exactly 4% is that we don’t count someone with the disease (aa) as a carrier, even though they could pass the disease to their children (a carrier by definition has the genotype Aa). If we are rounding off our result to the nearest percent, then 1 out of every 25 people (4% of the population) are carriers.

This calculation is based on several assumptions: no natural selection, no inbreeding, and no selection of embryos based on genetic testing. Cystic fibrosis is such a severe disease that often victims don’t survive long enough to have children (modern medicine is making this less true). The untreated disease is so lethal that one wonders why natural selection didn’t eliminate it from our gene pool long ago. One possible reason is that carriers of cystic fibrosis might be better able to resist other diseases—such as cholera, typhoid fever, or tuberculosis—than are normal people. 

A nice discussion of cystic fibrosis.
https://www.youtube.com/watch?v=QfjIGXNey3g
 

 
This video from the Cystic Fibrosis Foundation suggests that gene editing may cure cystic fibrosis. https://www.youtube.com/watch?v=nWj7Be6PSS4

Friday, April 1, 2022

Diffusion with a Buffer

The Mathematics of Diffusion, by John Crank, superimposed on Intermediate Physics for Medicine and Biology.
The Mathematics of Diffusion,
by John Crank.
Homework Problem 27 in Chapter 4 of Intermediate Physics for Medicine and Biology examines diffusion in the presence of a buffer. The problem shows that the buffer slows diffusion and introduces the idea of an effective diffusion constant. I like this problem but I admit it’s rather long-winded. Recently, when thumbing through John Crank’s book The Mathematics of Diffusion (doesn’t everyone thumb through The Mathematics of Diffusion on occasion?), I found an easier way to present the same basic idea. Below is a simplified version of Problem 27.

Section 4.8

Problem 27½. Calcium ions with concentration C diffuse inside cells. Assume that this free calcium is in instantaneous local equilibrium with calcium of concentration S that is bound to an immobile buffer, such that

S = RC ,          (1)

where R is a dimensionless constant. Calcium released from the buffer acts as a source term in the diffusion equation

   C/∂t = D2C/∂x2 − ∂S/∂t .        (2)

(a) Explain in words why ∂S/∂t is the correct source term. Be sure to address why there is a minus sign.

(b) Substitute Eq. (1) into Eq. (2), derive an equation of the form

C/∂t = Deff2C/∂x2 ,         (3)

and obtain an expression for the effective diffusion constant Deff.

(c) If R is much greater than one, describe the physical affect the buffer has on diffusion.

(d) Show that this problem corresponds to the case of Problem 27 when [B] is much greater than [CaB]. Explain physically what this means.

To do part (d), you will need to look at the problem in IPMB

The bottom line: an immobile buffer hinders diffusion. The stronger the buffer (the larger the value of R), the slower the calcium diffuses. The beauty of the homework problem is that it illustrates this property with only a little mathematics.

Enjoy!

Friday, February 4, 2022

Does a Nerve Axon Have an Inductance?

When I was measuring the magnetic field of a nerve axon in graduate school, I wondered if I should worry about a nerve’s inductance. Put another way, I asked if the electric field induced by the axon’s changing magnetic field is large enough to affect the propagation of the action potential.

Here is a new homework problem that will take you through the analysis that John Wikswo and I published in our paper “The Magnetic Field of a Single Axon” (Biophysical Journal, Volume 48, Pages 93–109, 1985). Not only does it answer the question about induction, but also it provides practice in back-of-the-envelope estimation. To learn more about biomagnetism and magnetic induction, see Chapter 8 of Intermediate Physics for Medicine and Biology.
Section 8.6

Problem 29½. Consider an action potential propagating down a nerve axon. An electric field E, having a rise time T and extended over a length L, is associated with the upstroke of the action potential.

(a) Use Ohm’s law to relate E to the current density J and the electrical conductivity σ
(b) Use Ampere’s law (Eq. 8.24, but ignore the displacement current) to estimate the magnetic field B from J and the permeability of free space, μ0. To estimate the derivative, replace the curl operator with 1/L
(c) Use Faraday’s law (Eq. 8.22, ignoring the minus sign) to estimate the induced electric field E* from B. Replace the time derivative by 1/T
(d) Write your result as the dimensionless ratio E*/E
(e) Use σ = 0.1 S/m, μ0 = 4 π × 10-7 T m/A, L = 10 mm, and T = 1 ms, to calculate E*/E
(f) Check that the units in your calculation in part (e) are consistent with E*/E being dimensionless. 
(g) Draw a picture of the axon showing E, J, B, E*, and L
(h) What does your result in part (e) imply about the need to consider inductance when analyzing action potential propagation along a nerve axon.

For those of you who don’t have IPMB handy, Equation 8.24 (Ampere’s law, ignoring the displacement current) is

∇×B = μ0 J

and Eq. 8.22 (Faraday’s law) is

∇×E = −∂B/∂t .

I’ll leave it to you to solve this problem. However, I’ll show you my picture for part (g).

Also, for part (e) I get a small value, on the order of ten parts per billion (10-8). The induction of a nerve axon is negligible. We don't need an inductor when modeling a nerve axon.

Friday, January 28, 2022

How Far Can Bacteria Coast?

Random Walks in Biology, by Howard Berg, superimposed on Intermediate Physics for Medicine and Biology.
Random Walks in Biology,
by Howard Berg.
In last week’s blog post, I told you about the recent death of Howard Berg, author of Random Walks in Biology. This week, I present a new homework problem based on a topic from Berg’s book. When discussing the Reynolds number, a dimensionless number from fluid dynamics that is small when viscosity dominates inertia, Berg writes
The Reynolds number of the fish is very large, that of the bacterium is very small. The fish propels itself by accelerating water, the bacterium by using viscous shear. The fish knows a great deal about inertia, the bacterium knows nothing. In short, the two live in very different hydrodynamic worlds.

To make this point clear, it is instructive to compute the distance that the bacterium can coast when it stops swimming.
Here is the new homework problem, which asks the student to compute the distance the bacterium can coast.
Section 1.20

Problem 54. When a bacterium stops swimming, it will coast to a stop. Let us calculate how long this coasting takes, and how far it will go.

(a) Write a differential equation governing the speed, v, of the bacterium. Use Newton’s second law with the force given by Stokes law. Be careful about minus signs.

(b) Solve this differential equation to determine the speed as a function of time.

(c) Write the time constant, τ, governing the decay of the speed in terms of the bacterium’s mass, m, its radius, a, and the fluid viscosity, η.

(d) Calculate the mass of the bacterium assuming it has the density of water and it is a sphere with a radius of one micron.

(e) Calculate the time constant of the decay of the speed, for swimming in water having a viscosity of 0.001 Pa s.

(f) Integrate the speed over time to determine how far the bacterium will coast, assuming its initial speed is 20 microns per second.
I won’t solve all the intermediate steps for you; after all, it’s your homework problem. However, below is what Berg has to say about the final result.
A cell moving at an initial velocity of 2 × 10-3 cm/sec coasts 4 × 10-10 cm = 0.04 , a distance small compared with the diameter of a hydrogen atom! Note that the bacterium is still subject to Brownian movement, so it does not actually stop. The drift goes to zero, not the diffusion.

Berg didn’t calculate the deceleration of the bacterium. If the speed drops from 20 microns per second to zero in one time constant, I calculate the acceleration to be be about 91 m/s2, or nearly 10g. This is similar to the maximum allowed acceleration of a plane flying in the Red Bull Air Race. That poor bacterium.

Friday, August 27, 2021

Can Induced Electric Fields Explain Biological Effects of Power-Line Magnetic Fields?

Sometimes proponents of pseudoscience embrace nonsense, but other times they propose plausible-sounding ideas that are wrong because the numbers don’t add up. For example, suppose you are discussing with your friend about the biological effects of power-line magnetic fields. Your friend might say something like this:
“You keep claiming that magnetic fields don’t have any biological effects. But suppose it’s not the magnetic field itself, but the electric field induced by the changing magnetic field that causes the effect. We know an electric field can stimulate nerves. Perhaps power-line effects operate like transcranial magnetic stimulation, by inducing electric fields.”
Well, there’s nothing absurd about this hypothesis. Transcranial magnetic stimulation does work by creating electric fields in the body via electromagnetic induction, and these electric fields can stimulate nerves. The qualitative idea is reasonable. But does it work quantitatively? If you do the calculation, the answer is no. The electric field induced by a power line is less than the endogenous electric field associated with the electrocardiogram. You don’t have to perform a difficult, detailed calculation to show this. A back-of-the-envelope estimation suffices. Below is a new homework problem showing you how to make such an estimate.
Section 9.10

Problem 36 ½
. Estimate the electric field induced in the body by a power-line magnetic field, and compare it to the endogenous electric field in the body associated with the electrocardiogram.

(a) Use Eq. 8.25 to estimate the induced electric field, E. The magnetic field in the vicinity of a power line can be as strong as 5 μT (Possible Health Effects of Exposure to Residential Electric and Magnetic Fields, 1997, Page 32), and it oscillates at 60 Hz. The radius, a, of the current loop in our body is difficult to estimate, but take it as fairly large (say, half a meter) to ensure you do not underestimate the induced electric field.

(b) Estimate the endogenous electric field in the torso from the electrocardiogram, using Figures 7.19 and 7.23.

(c) Compare the electric fields found in parts (a) and (b). Which is larger? Explain how an induced electric field could have an effect if it is smaller than the electric fields already existing in the body.
Let’s go through the solution to this new problem. First, part (a). The amplitude of the magnetic field is 0.000005 T. The field oscillates with a period of 1/60 Hz, or about 0.017 s. The peak rate of change will occur during only a fraction of this period, and a reasonable approximation is to divide the period by 2π, so the time over which the magnetic field changes is 0.0027 s. Thus, the rate of change dB/dt is 0.000005 T/0.0027 s, or about 0.002 T/s. Now use Eq. 8.25, E = a/2 dB/dt (ignore the minus sign in the equation, which merely indicates the phase), with a = 0.5 m, to get E = 0.0005 V/m.

Now part (b). Figure 7.23 indicates that the QRS complex in the electrocardiogram has a magnitude of about ΔV = 0.001 V (one millivolt). Figure 7.19 shows that the distance between leads is on the order of Δr = 0.5 m. The magnitude of the electric field is approximately ΔV/Δr = 0.002 V/m.

In part (c) you compare the electric field induced by a power line, 0.0005 V/m, to the electric field in the body caused by the electrocardiogram, 0.002 V/m. The field produced by the ECG is four times larger. So, how can the induced electric field have a biological effect if we are constantly exposed to larger electric fields produced by our own body? I don’t know. It seems to me that would be difficult.

Hart and Gandhi (1998) Phys. Med. Biol., 43:3083–3099, superimposed on Intermediate Physics for Medicine and Biology.
Hart and Gandhi (1998)
Phys. Med. Biol.,
43:3083–3099.
But wait! Our calculation in part (b) is really rough. Perhaps we should do a more detailed calculation. Rodney Hart and Om Gandhi did just that (“Comparison of cardiac-induced endogenous fields and power frequency induced exogenous fields in an anatomical model of the human body,”  Physics in Medicine & Biology, Volume 43, Pages 3083–3099, 1998). They found that during the QRS complex the endogenous electric field varied throughout the body, but it is usually larger than what we estimated. It’s giant in the heart itself, about 3 V/m. All through the torso it’s more than ten times what we found; for instance, in the intestines it’s 0.04 V/m. Even in the brain the field strength (0.014 V/m) is seven times larger than our estimate (0.002 V/m).

Moreover, the heart isn’t the only source of endogenous fields (although it’s the strongest). The brain, peripheral nerves, skeletal muscle, and the gut all produce electric fields. In addition, our calculation of the induced electric field is evaluated at the edge of the body, where the current loop is largest. Deeper within the torso, the field will be less. Finally, our value of 5 μT is extreme. Magnetic fields associated with power lines are usually about one tenth of this. In other words, in all our estimates we took values that favor the induced electric field over the endogenous electric field, and the endogenous electric field is still four times larger.

What do we conclude? The qualitative mechanism proposed by your friend is not ridiculous, but it doesn’t work when you do the calculation. The induced electric field would be swamped by the endogenous electric field.

The moral of the story is that proposed mechanisms must work both qualitatively and quantitatively. Doing the math is not an optional step to refine your hypothesis and make it more precise. You have to do at least an approximate calculation to decide if your idea is reasonable. That’s why Russ Hobbie and I emphasize solving toy problems and estimation in Intermediate Physics for Medicine and Biology. Without estimating how big effects are, you may go around saying things that sound reasonable but just aren’t true.

Friday, August 20, 2021

The Central Slice Theorem: An Example

The central slice theorem is key to understanding tomography. In Intermediate Physics for Medicine and Biology, Russ Hobbie and I ask the reader to prove the central slice theorem in a homework problem. Proofs are useful for their generality, but I often understand a theorem better by working an example. In this post, I present a new homework problem that guides you through every step needed to verify the central slice theorem. This example contains a lot of math, but once you get past the calculation details you will find it provides much insight.

The central slice theorem states that taking a one-dimensional Fourier transform of a projection is equivalent to taking the two-dimensional Fourier transform and evaluating it along one direction in frequency space. Our “object” will be a mathematical function (representing, say, the x-ray attenuation coefficient as a function of position). Here is a summary of the process, cast as a homework problem.

Section 12.4 

Problem 21½
. Verify the central slice theorem for the object

(a) Calculate the projection of the object using Eq. 12.29,

Then take a one-dimensional Fourier transform of the projection using Eq. 11.59,
 
(b) Calculate the two-dimensional Fourier transform of the object using Eq. 12.11a,
Then transform (kx,ky ) to (θ,k) by converting from Cartesian to polar coordinates in frequency space.
(c) Compare your answers to parts (a) and (b). Are they the same?


I’ll outline the solution to this problem, and leave it to the reader to fill in the missing steps. 

 
Fig. 12.12 from Intermediate Physics for Medicine and Biology, showing how to do a projection.
Fig. 12.12 from IPMB, showing how to do a projection.

The Projection 

Figure 12.12 shows that the projection is an integral of the object along various lines in the direction θ, as a function of displacement perpendicular to each line, x'. The integral becomes


Note that you must replace x and y by the rotated coordinates x' and y'


You can verify that x2 + y2= x'2 + y'2.

After some algebra, you’re left with integrals involving eby'2 (Gaussian integrals) such as those analyzed in Appendix K of IPMB. The three you’ll need are


The resulting projection is


Think of the projection as a function of x', with the angle θ being a parameter.

 

The One-Dimensional Fourier Transform

The next step is to evaluate the one-dimensional Fourier transform of the projection

The variable k is the spatial frequency. This integral isn’t as difficult as it appears. The trick is to complete the square of the exponent


Then make a variable substitution u = x' + ik2b. Finally, use those Gaussian integrals again. You get


This is our big result: the one-dimensional Fourier transform of the projection. Our next goal is to show that it’s equal to the two-dimensional Fourier transform of the object evaluated in the direction θ.

Two-Dimensional Fourier Transform

To calculate the two-dimensional Fourier transform, we must evaluate the double integral


The variables kx and ky are again spatial frequencies, and they make up a two-dimensional domain we call frequency space.

You can separate this double integral into the product of an integral over x and an integral over y. Solving these requires—guess what—a lot of algebra, completing the square, and Gaussian integrals. But the process is straightforward, and you get


Select One Direction in Frequency Space

If we want to focus on one direction in frequency space, we must convert to polar coordinates: kx = k cosθ and ky = k sinθ. The result is 

This is exactly the result we found before! In other words, we can take the one-dimensional Fourier transform of the projection, or the two-dimensional Fourier transform of the object evaluated in the direction θ in frequency space, and we get the same result. The central slice theorem works.

I admit, the steps I left out involve a lot of calculations, and not everyone enjoys math (why not?!). But in the end you verify the central slice theorem for a specific example. I hope this helps clarify the process, and provides insight into what the central slice theorem is telling us.

Friday, July 16, 2021

The Bragg Peak (Continued)

In last weeks post, I discussed the Bragg peak: protons passing through tissue lose most of their energy near the end of their path. In Chapter 16 of Intermediate Physics for Medicine and Biology, Russ Hobbie and I present a homework problem in which the student calculates the stopping power (energy lost per distance traveled), S, as a function of depth, x, given a relationship between stopping power and energy, T. This problem is a toy model illustrating the physical origin of the Bragg peak. Often its helpful to have two such exercises; one to assign as homework and one to work in class (or put on an exam). Heres a new homework problem similar to the one in IPMB, but with a different assumption about how stopping power depends on energy.

Section 16.10

Problem 31 ½. Assume the stopping power of a particle, S = −dT/dx, as a function of kinetic energy, T, is S = So eT/To
(a) What are the units of So and To
(b) If the initial kinetic energy at x = 0 is Ti, calculate T(x). 
(c) Determine the range R of the particle as a function of To, So, and Ti
(d) Plot S(x) vs. x. Does this plot contain a Bragg peak? 
(e) Discuss the implications of the shape of S(x) for radiation treatment using this particle.

The answer to part (d) is difficult, because your conclusion is different depending on the relative magnitude of Ti and To. You might consider adding a part (f)

(f) Plot T(x), S(x), and R(Ti) for Ti >> To and for Ti << To.

The case Ti >> To has a conspicuous Bragg peak; the case Ti << To doesnt. 

The homework problem in IPMB is more realistic than this new one, because Fig. 15.17 indicates that the stopping power decreases as 1/T (assumed in the original problem) rather than exponentially (assumed in the new problem). This changes the particles behavior, particularly at low energies (near the end of its range, in the Bragg peak). Nevertheless, having multiple versions of the problem is useful. 

The answer to part (e) is given in IPMB.

Protons are also used to treat tumors... Their advantage is the increase of stopping power at low energies. It is possible to make them come to rest in the tissue to be destroyed, with an enhanced dose relative to intervening tissue and almost no dose distally.

 Enjoy!

Friday, June 25, 2021

Cerenkov Luminescence Imaging: Physics Principles and Potential Applications in Biomedical Sciences

When a particle travels faster than the speed of light, it emits Cerenkov radiation. This phenomenon has resulted in new medical imaging applications, as described in a 2017 review paper by Esther Ciarrocchi and Nicola Belcari (Cerenkov Luminescence Imaging: Physics Principles and Potential Applications in Biomedical Sciences, EJNMMI Physics, Volume 4, Article 14). This is an open access article, so you can read it for free.

Russ Hobbie and I don’t discuss Cerenkov Luminescence Imaging in Intermediate Physics for Medicine and Biology, but you can learn a lot about it using the physics we do discuss. For example, can particles  travel faster than the speed of light? They can’t travel faster than the speed of light in a vacuum, but they can travel faster than the speed of light in a material such as water or tissue where light is slowed and the medium has an index of refraction. Below is a new homework problem, in which we consider electrons emitted in tissue by beta decay of the isotope iodine-131, used in many medical applications.
Problem 9 ¼. The end point kinetic energy (see Fig. 17.8) for beta decay of 131I is 606 keV, and tissue has an index of refraction of 1.4. Do any of the emitted electrons have a speed faster than the speed of light in the tissue? To determine this speed, use Eq. 14.1. Because the electrons move near the speed of light, to determine their speed as a function of their kinetic energy use a result from special relativity, Eq. 17.1.

For those who don’t have IPMB at your side (shame on you!), Eq. 14.1 is cn = c/n, where cn is the speed of light in the medium, c is the speed of light in a vacuum (3 × 108 m/s), and n is the index of refraction, and Eq. 17.1 is T + mc2 = mc2/√(1 − v2/c2), where v is the speed of the particle, T is its kinetic energy, and mc2 is the rest mass of an electron expressed as energy (511 keV).

If you solved this problem correctly, you found that some of the more energetic electrons emitted during beta decay of 131I do travel faster than the speed of light in tissue.

Cerenkov radiation is emitted at an angle θ with respect to the direction that the particle is moving. This distribution of light is characteristic of a shock wave, and is similar to the distribution of sound in a sonic boom made by a plane when it flies faster than the speed of sound. The new problem below requires the reader to calculate θ.

Problem 9 ½. The drawing below shows a particle moving to the right faster than the speed of light in the medium. The position of the particle at several instants is indicated by the purple dots. The location of light emitted by the particle at each position is shown by the black circles. The light adds to form a conical wave front, shown by the green lines. 
(a) Use the red right triangle to calculate the angle θ as a function of the particle speed, v, and the index of refraction, n
(b) Compute the value of θ for the fastest electrons emitted by beta decay of 131I in tissue.

The number of photons emitted tends to be greatest at short wavelengths, so Cerenkov radiation often has a blue tinge. However, readers of IPMB learned in Chapter 14 that the spectrum of radiation can look different when viewed as a function of frequency (or energy) rather than as a function of wavelength. Below is a new problem to explore this effect.

Problem 9 ¾. The number of photons dN emitted with a wavelength between λ and λ + is approximately dN = C/λ2, where C is a constant.
(a) Sketch a plot of dN/ versus λ. Don’t worry about the scale of the axes (in other words, don't worry about the value of C); just make the plot qualitatively correct. 
(b) Use methods similar to those introduced in Section 14.8 to determine the number of photons emitted with an energy between E and E + dE. Don’t worry about constant factors, just determine how dN/dE varies with E
(c) Sketch a plot of dN/dE versus E. Again, just make the plot qualitatively correct.

If you solved part (c) correctly, you should have drawn a plot with a flat line, because dN/dE is independent of E. Of course, there must be some limits to this result, otherwise the particle would emit an infinite amount of energy when integrated over all photon energies. See Ciarrocchi and Belcari’s review for an explanation.

Perhaps the most interesting part of Ciarrocchi and Belcari’s article is their discussion of biomedical applications. You can use Cerenkov radiation to image beta emitters like 131I, positron emitters like 18F used in positron emission tomography, and high-energy protons required for proton therapy.

To learn more about Cerenkov radiation, watch this video by Don Lincoln. Enjoy!

How does Cerenkov radiation work?

https://www.youtube.com/watch?v=Yjx0BSXa0Ks