Friday, March 25, 2016

Basic Physics of Nuclear Medicine

I’m cheap and I’m proud of it; I love free stuff. Intermediate Physics for Medicine and Biology isn’t free. Russ Hobbie and I appreciate our readers’ willingness to spend their money to purchase our book. Thank you! But what if you want more? What if—heaven forbid—you find our book is not totally clear, complete, or comprehensive? In IPMB we cite many references at the end of each chapter, so you have many sources of additional information. But often these sources cost money or may be difficult to obtain. Is there anywhere you can go online for free to augment IPMB?

A screenshot of the wikibook Basic Physics of Nuclear Medicine.
A screenshot of the wikibook
Basic Physics of Nuclear Medicine.
One option is the wikibook Basic Physics of Nuclear Medicine. This book covers much of the same material as in the last half of IPMB. It analyzes in depth nuclear medicine (our Chapter 17), but it also covers the interaction of radiation with tissue (our Chapter 15), Fourier methods and tomography (our Chapters 11 and 12), detectors and x-ray imaging systems (our Chapter 16), ultrasound (our Chapter 13), and even a little magnetic resonance imaging (our Chapter 18).

Some of my favorite parts of the wikibook are not covered in IPMB:
What are the advantages of IPMB? For one thing, IPMB has a large collection of homework problems, more extensive than in Basic Physics of Nuclear Medicine. Also, I think our book has a better focus on using mathematical modeling to illustrate medical and biological physics concepts. Moreover, the entire first half of IPMB—about biomechanics, biothermodynamics, diffusion, bioelectricity, biomagnetism, and feedback—is absent from Basic Physics of Nuclear Medicine. Finally, and most importantly, Basic Physics of Nuclear Medicine doesn’t have a blog with weekly updates.

If you are looking for a free, easily accessible online textbook to use as a supplement (please, not a replacement!) for Intermediate Physics for Medicine and Biology, consider Basic Physics of Nuclear Medicine. It’s worth every penny.

Friday, March 18, 2016

Phineas Gage: Neuroscience’s Most Famous Patient

When Russ Hobbie and I discuss transcranial magnetic stimulation in Intermediate Physics for Medicine and Biology, we write that “because TMS is noninvasive and nearly painless, it can be used to study learning and plasticity (changes in brain organization over time).” When I worked with Mark Hallett and Leo Cohen at the National Institutes of Health, they were using TMS to study plasticity in patients who had undergone amputations or spinal cord injuries.

A photograph of Phineas Gage.
Phineas Gage.
How much can the brain reorganize and rehabilitate after an injury? We gain insight into this question by examining the amazing case of Phineas Gage. Recently, science writer Sam Kean published the article “Phineas Gage, Neuroscience’s Most Famous Patient” in the online magazine Slate. Let me quote Kean’s opening lines.
On Sept. 13, 1848, at around 4:30 p.m., the time of day when the mind might start wandering, a railroad foreman named Phineas Gage filled a drill hole with gunpowder and turned his head to check on his men. It was the last normal moment of his life….

The Rutland and Burlington Railroad had hired Gage’s crew that fall to clear away some tough black rock near Cavendish, Vermont, and it considered Gage the best foreman around. Among other tasks, a foreman sprinkled gunpowder into blasting holes, and then tamped the powder down, gently, with an iron rod. This completed, an assistant poured in sand or clay, which got tamped down hard to confine the bang to a tiny space. Gage had specially commissioned his tamping iron from a blacksmith. Sleek like a javelin, it weighed 13¼ pounds and stretched 3 feet 7 inches long. (Gage stood 5-foot-6.) At its widest, the rod had a diameter of 1¼ inches, although the last foot—the part Gage held near his head when tamping—tapered to a point.

Gage’s crew members were loading some busted rock onto a cart, and they apparently distracted him. Accounts differ about what happened after Gage turned his head. One says Gage tried to tamp the gunpowder down with his head still turned, and scraped his iron against the side of the hole, creating a spark. Another says Gage’s assistant (perhaps also distracted) failed to pour the sand in, and when Gage turned back, he smashed the rod down hard, thinking he was packing inert material. Regardless, a spark shot out somewhere in the dark cavity, igniting the gunpowder, and the tamping iron rocketed upward.

The iron entered Gage’s head point-first, striking below the left cheekbone. It destroyed an upper molar, passed behind his left eye, and tore into the underbelly of his brain’s left frontal lobe. It then plowed through the top of his skull, exiting near the midline, just behind where his hairline started. After parabola-ing upward—one report claimed it whistled as it flew—the rod landed 25 yards away and stuck upright in the dirt, mumblety-peg-style. Witnesses described it as streaked with red and greasy to the touch, from fatty brain tissue.
Gage survived after his rod destroyed much of his frontal lobe. He eventually recovered much neural function, but his personality changed; Gage “was no longer Gage”. At least, so goes the traditional story as told in many neuroscience textbooks. Kean argues that these personality changes were not as dramatic as claimed, and were temporary. Years after the accident, Gage enjoyed fairly good health and lived a nearly normal life. His brain recovered. Kean writes
Modern neuroscientific knowledge makes the idea of Gage’s recovery all the more plausible. Neuroscientists once believed that brain lesions caused permanent deficits: Once lost, a faculty never returned. More and more, though, they recognize that the adult brain can relearn lost skills. This ability to change, called brain plasticity, remains somewhat mysterious, and it happens achingly slowly. But the bottom line is that the brain can recover lost functions in certain circumstances.
If transcranial magnetic stimulation had been developed in the first half of the nineteenth century (and why not? Faraday discovered electromagnetic induction 17 years before Gage’s accident), perhaps neuroscientists would have had the tool they needed to monitor and map Gage’s brain during his recovery. Magnetic stimulation—a classic application of physics to medicine—has taught us much about how the brain can change and heal. This knowledge might have implications for how we treat all sorts of brain injuries, from concussion to stroke to dementia to rods shot through our head. As Kean concludes, “If even Phineas Gage bounced back—that’s a powerful message of hope.”

Friday, March 11, 2016

Mass Attenuation Coefficient and Areal Density

I don’t like the mass attenuation coefficient; it bugs me and it has weird units. Yet researchers studying the attenuation of x-rays in materials usually quote the mass attenuation coefficient rather than the linear attenuation coefficient in their publications.

As x-rays pass through a material, they fall off exponentially as exp(−μL), where L is the distance (m) and μ is the linear attenuation coefficient (m−1). But often researchers multiply and divide by the density ρ, so the exponential becomes exp(−(μ/ρ)(ρL)), where μ/ρ is the mass attenuation coefficient (m2/kg) and ρL is the areal density (kg/m2).

In Chapter 15 of Intermediate Physics of Medicine and Biology, Russ Hobbie and I explain some of the advantages of using μ/ρ.
The mass attenuation coefficient has the advantage of being independent of the density of the target material, which is particularly useful if the target is a gas. It has an additional advantage if Compton scattering is the dominant interaction. If σtot = C, then μatten/ρ = CNA/A [Z is the atomic number, A the mass number, NA is Avogadro’s number, and σC is the Compton cross section]. Since Z/A is nearly 1/2 for all elements except hydrogen, this quantity changes very little throughout the periodic table. This constancy is not true for the photoelectric effect or pair production. Figure 15.10 plots the mass attenuation coefficient vs energy for three substances spanning the periodic table. It is nearly independent of Z around 1 MeV where Compton scattering is dominant. The K and L absorption edges can be seen for lead; for the lighter elements they are below 10 keV. Figure 15.11 shows the contributions to μatten/ρ for air from the photoelectric effect, incoherent scattering, and pair production. Tables of mass attenuation coefficients are provided by the National Institute of Standards and Technology (NIST) at http://www.nist.gov/pml/data/xcom/index.cfm.
Let me offer an example where it makes sense to consider the mass attenuation coefficient.

Imagine you have a large box of area S. You measure a mass M of the fluid pentane and poor it into the box. Then you place a source of x-rays under the box, directed upwards. You measure the intensity of the radiation incident on the underside of the box to be Io, and then move your detector to above the box and measure the intensity of radiation that passes through the pentane to be I. Finally, use your ruler to measure the thickness of the pentane layer, L.

You now have enough data to determine both the linear attenuation coefficient and the mass attenuation coefficient of pentane. For the linear attenuation coefficient, use the relationship I = Io exp(−μL) and solve for μ = ln(Io/I)/L. You can also calculate the density ρ = M/(SL). If you want the mass attenuation coefficient, you can now easily determine it: S ln(Io/I)/M. You can also calculate the areal density: M/S.

Next you perform the same experiment on neopentane. You use the same box with area S and measure out the same mass M of fluid. You find that Io/I is unchanged, but L is about 6% larger. You conclude the linear attenuation coefficient and the density both decrease by 6%, but the mass attenuation coefficient and the areal density are unchanged.

Why is Io/I the same for both fluids? Pentane and neopentane are isomers. They have exactly the same chemical formula, C5H12, but they have different structures. Pentane is an unbranched hydrocarbon and neopentane has a central carbon bonded to four other carbon atoms. Because the mass M of both substances is the same, the number of the atoms is the same in each case. The x-ray attenuation only depends on the number of atoms and the type of atoms, but not how those atoms are arranged (the density). This is one advantage of the mass attenuation coefficient: it depends only on the atoms and not their arrangement.

You can calculate the mass attenuation coefficient and the areal density without knowing L. If for some reason L were difficult to measure, you could still determine the mass attenuation coefficient even if you could not calculate the linear attenuation coefficient.

In a gas, the number of molecules is fixed but the density depends on the pressure and temperature. The mass attenuation coefficient does not change with the pressure and temperature. Again, it just depends on the atoms and not their distribution.

Water has a density of 1 g/cm3. If you express the mass attenuation coefficient in cm2/g and the linear attenuation coefficient in cm, then the mass attenuation coefficient and the linear attenuation coefficient have the same numerical value. Most tissue has a density close to that of water, so this trick works well for tissue too.

Given these advantages, have I started liking the mass attenuation coefficient? No, I still think it’s weird. But I can tolerate it a little better now.

Friday, March 4, 2016

Welcome Home Scott Kelly

A photograph of Scott Kelly, when he returned to earth after a year on the space station.
Scott Kelly, when he returned to earth
after a year on the space station.
This week astronaut Scott Kelly returned to Earth after nearly a year on the International Space Station. One goal of his mission was to determine how astronauts would function during long trips in space. I suspect we will learn a lot from Kelly about life in a weightless environment. But one of the biggest risks during a mission to Mars would be radiation exposure, and we may not learn much about that from trips to the space station.

In space, the major source of radiation is cosmic rays, consisting mostly of high energy (GeV) protons. Most of these particles are absorbed by our atmosphere and never reach Earth, or are deflected by Earth’s magnetic field. The space station orbits above the atmosphere but within range of the geomagnetic field, so Kelly was partially shielded from cosmic rays. He probably experienced a dose of about 150 mSv. This is much larger than the annual background dose on the surface of the earth. According to Chapter 16 of Intermediate Physics for Medicine and Biology, we all are exposed to about 3 mSv per year.

A photograph of Scott and Mark Kelly.
Scott and Mark Kelly.
Is 150 mSv in one year dangerous? This dose is below the threshold for acute radiation sickness. It would, however, increase your chances of developing cancer. A rule of thumb is that the excess relative risk of cancer is about 5% per Sv. This does not mean Kelly has a 0.75% chance of getting cancer (5%/Sv times 0.15 Sv). Instead, it means that Scott Kelly has a 0.75% higher chance of getting cancer than his brother Mark Kelly, who remained on Earth. This is a significant increase in risk, but may be acceptable if your goal in life is to be an astronaut. The Kelly twins are both 52 years old, and the excess relative risk goes down with age, so the extra risk of Scott Kelly contracting cancer is probably less than 0.5%.

NASA’s goal is to send astronauts to Mars. Such a mission would require venturing beyond the range of Earth’s geomagnetic field, increasing the exposure to cosmic rays. Data obtained by the Mars rover Curiosity indicate that a one-year interplanetary trip would result in an exposure of 660 mSv. This would be four times Kelly's exposure in the space station. 660 mSv would be unlikely to cause serious acute radiation sickness, but would increase the cancer risk. NASA would have to either shield the astronauts from cosmic rays (not easy given their high energy) or accept the increased risk. I’m guessing they will accept the risk.