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Biochemistry 102a Fall Term 2002 Background Handout: Page 1

The attached handout by Tom Pologruto is background reading; I intend it to refresh your memory about (and give a bit more insight into) things that you should already know from studying and . I’m not assigning the problems, but they would be useful to do on your own. Dimensional Analysis

Thomas Pologruto Harvard University Department of Biophysics [email protected] www.fas.harvard.edu/~tpologr

Introduction

Answering questions in chemistry, physics, or biology requires two things: physical interpretation of the problem and mathematical manipulation of the associated equations. While knowing the math and being able to compute the desired quantities in a given problem is a very important thing to master, often the equation or the quantities you are deriving lose their physical meaning when only interpreted as solutions of mathematical equations. The physical, common sense interpretation of equations is what is really important in developing an understanding of the material and in gaining the necessary mental tools to tackle more complex problems later on in life. It is precisely this intuition for physical problems that often gets lost when complex is used.

I want to convince you that physical intuition can help you understand a problem (and thus arrive at the correct answer) more than complex mathematics. You have to know the math to get a detailed answer, but you need to understand the problem and what it is asking before you can even write a single equation down. One of the easiest and most powerful ways to look at a problem and get a “feel” for what it means is by doing Dimensional Analysis.

We will see that equations that represent real-life situations have to be consistent in the following way: all of the must be arranged such that the equation and its solutions make physical sense. If I ask you how long a piece of rope is, and you respond “1 liter,” that is obviously an inappropriate answer. Rope is measured in feet or meters or any other unit that has dimensions of [L]†. Liters measure volume and have dimensions of length cubed [L]3. As the old adage from algebra goes “you can’t mix apples and oranges,” you also cannot equate quantities that have different dimensions‡.

Thus, when you are asked a question, either in the lab, in the ER, or on an exam, making sure that the answer you give makes physical sense is very important. Imagine the confused look on a patient’s face if you said to take five-hundred millimeters of penicillin three times a day, instead of five-hundred milligrams! While this seems really ridiculous, it often happens that on

† The notation for “has dimensions of X” is [X]. Thus, we will see that things with dimensions of length are written as [L], with dimensions of are written as [T], and so on.

‡ You can equate two quantities with different units (like feet and meters), since they are related by some (dimensionless) conversion factor, and because both feet and meters represent the of length. This is an important distinction. Units are specific to a given situation, whereas dimensions are universal. I can measure volume in units of liters, gallons, cubic yards, or quarts, but all of these units has dimensions of volume, [L]3.

1 exams, in homework problems, or in lab, students lose the dimensions of the problem when they do the math, and the answers they derive are as absurd as the penicillin dosage being given in millimeters.

The point of these exercises is to familiarize you with dimensional analysis by going through some examples and giving you some problems to do on your own. You will see how you can develop intuition for a problem simply by analyzing the dimensions given, even in cases where you have never even seen (or understand!) the governing equation. I will then introduce the idea of a dimensionless quantity, which is extremely useful in seeing how the solutions of complex differential equations should be approached. Finally, I will briefly discuss unit conversions, a topic that is very important in chemistry as well as for back-of-the-envelope calculations.

Dimensions and Units

We associate a dimension with nearly every physical quantity we observe§. Dimensions are what give meaning to the numbers we derive using mathematical equations. “5 “ has no meaning until we place a dimension after it; so we write “5 “ meters or to be correct. This annotation is not optional in science. Every number must have dimensions associated with it to be correct. In general, this also provides a way for you to check that your answer at least makes sense.

Mass [M], time [T], length [L], charge [Q], moles [O], and [K] are fundamental dimensions, and other quantities, like force [F] = [M][L][T]-2 , energy [E] = [M][L]2 [T]-2, volume [L]3 , or pressure [F] [L]-2, can be expressed as some combination of these fundamental dimensions, as seen here. A list of dimensions, some useful dimensional combinations, the most common units for a given dimension, and equivalent representations of the same dimension are given in Table 1.

Units refer to a specific set of dimensions (, times, , charges, etc… ) like the SI system ( has units of , time has units of seconds, and length has units of meters) or the CGS system (mass has units of grams, time has units of seconds, and length has units of centimeters). Making this distinction between units and dimensions is important. While an equation has to be dimensionally correct, it does not necessarily have to be unit-wise correct, since the conversion between units simply involves multiplication by a constant, dimensionless, conversion factor. Some of the common units for a given dimension are listed in Table 1. We will discuss conversion units in more detail later.

Using Dimensional Analysis

Now that we know what dimensions are, lets see how useful they can be. We will begin with a simple example from high school physics. We all know Newton said that force equals

§ Somethings are inherently dimensionless, like , geometric constants (like π), the probability of an event occurring, and the number of objects (as in the number of particles in a box, or the number of chickens in a yard). We will comment on this later in the paper.

2 mass times acceleration. In fact, this is the common definition of a force. But could we interpret this in another way based on dimensional analysis?

Referring to Table 1, we see that a force has dimensions of [M][L][T]-2, which we could regroup as follows: {[M][L][T]-1}{[T]-1}. Now, again referring to Table 1, we know that the first term {[M][L][T]-1} has dimensions of momentum (mass • velocity), while the latter quantity {[T]-1 } has the dimensions of or the of change of something. Therefore, we could “guess” that momentum is what is changing in time when a force is applied, just based on dimensional analysis. This turns out to be exactly correct, without the need for any additional constants. Newton’s law can then be written as follows:

dp Force = F = ma = (1) dt where I have used the notation from differential calculus to express the instantaneous time rate of change of the momentum as dp/dt, and the bold letters refer to the vector nature of these quantities. In general, you can cancel dimensions that appear in both the numerator and denominator or on opposite sides of an equal sign, since the manipulation of dimensions is analogous to algebraic manipulations.

In a classic example of dimensional analysis, you are asked to look at a pendulum swinging at low amplitude, and to determine its angular acceleration as a function of the displacement , θ, the length of the rope, L, and the gravitational acceleration, g.

The angular acceleration is given as d2θ/dt2 which has dimensions of [T]-2. The reason for this assignment of dimensions is that angles are dimensionless, while the d2/dt2 operator has dimensions of [T]-2, and thus the entire quantity has the dimensions of [T]-2. The angular acceleration can be determined via dimensional analysis by looking at what combination of g, θ, and L have the same dimensions as d2θ/dt2 .

Playing around with these quantities a little, we see that θ has no dimensions, so you might expect that the angular acceleration is simply proportional to θ (or more precisely, -θ, since the force opposes the displacement, like in a spring). The only combination of g and L that yields dimensions of [T]-2 is g • L-1, which has dimensions of [L][T]-2[L]-1 = [T]-2 after canceling the [L] terms. Thus, simply based on dimensional analysis, we would suspect naively that:

d 2θ g = −θ (2) dt 2 L

d 2θ g The actual answer is = −sinθ , where for small θ, we exactly recover our naïve dt 2 L expectation given in (2). It is remarkable that we are able to get so close to the correct answer without writing down any equation of motion, and we actually predicted precisely the behavior in the small θ limit (when sinθ ~ θ). Dimensional analysis will never reliable predict geometrical things like sines and cosines, or directions which are implied by the physics, but rather it shows

3 that for a given set of constraints (in this example, g and L), only certain combinations of quantities is permissible in the final answer.

This is precisely why dimensional analysis is so important. Say you were solving the pendulum differential equations during an exam, and you worked for three blue-book pages to yield that the proper equation was:

d 2θ g = −θ (3) dt 2 L2

This answer seems simple and reasonable at first sight, since only the quantities you were asked to solve in terms of appear in the solution. However, you would know immediately that something was wrong. The dimensions do not work out. The left hand side of (3) has dimensions of [T]-2 while the right-hand side has dimensions of [T]-2[L]-1.

You can either look at the dimensions before or after you solve the equations, but make sure that they are consistent all the way through. In the case of the pendulum, the incorrect but reasonable answer (3) for the would have incorrect dimensions, and we would be prescribing millimeters of penicillin. This prescription does not seem so ridiculous now that we see how easily dimensions can get lost while doing mathematics (If you don’t believe me, try solving the pendulum problem explicitly. If you need help getting started, just ask me!).

Problem 1.1 Einstein’s Photons The wavelength of a photon is inversely proportional to its frequency. Using Table 1 and dimensional analysis, what are the dimensions of this conversion factor? What do you think it represents? The energy of a photon is related directly to its frequency. What are the dimensions of this conversion factor? Is there a fundamental constant with these dimensions?

Problem 1.2 Hamiltonian Dynamics The kinetic energy of a particle can be written as:

1 2 E = mv , where vx is the velocity of the particle in the x-direction and m is the particle’s Kinetic 2 x 2 -2 mass. We see that Ekinetic has the proper dimensions for energy (i.e. [M][L] [T] or the same dimensions as mc2 as I always remember it), ignoring the constant ½ factor. Using the dimensions for momentum given in Table 1, re-write this energy in terms of the particle’s momentum and mass. The answer you get is actually how we will see kinetic energy appearing when we look at the Schrodinger Equation.

As we have demonstrated in the case of the pendulum, mathematical operations like differentiation can introduce added dimensions into equations. For instance, in our pendulum, the operator d2/dt2 has dimensions of [T]-2, which was something critical to realize in order to understand the problem. Generally, applying the rules for mathematical dimensional analysis can help you solve and remember differential equations, even if you have never seen them before, simply because if these equations represent physical reality, they must be dimensionally consistent.

4 Suppose we have some particles diffusing in , like ink or small biological molecules, and we want to know what the concentration, C(x,t), of these particles is as a function of time and position. If the molecules can only move up and down the x-axis, the governing differential equation is:

dC(x,t) d 2C(x,t) = [X ] (4) dt dx 2 and it is called the Diffusion Equation. The [X] represents a constant that is necessary to include so that the dimensions of this equation are consistent; namely, that the dimensions of the left- hand side are the same as the right-hand side. We can rewrite this entire equation in terms of dimensions, noting that the operators d/dt and d2/dx2 have dimensions of [T]-1 and [L]-2, respectively, as:

[T]-1[C] = [X] [L]-2[C] (5) where we have left the dimensions of concentration in the simplified form, [C], since this same quantity appears on both sides of the equation, and simply cancels out. We can solve for [X] to yield that it must have dimensions of:

[X] = [L]2[T]-1 (6a) which we call the Diffusion Coefficient (D) of the system, and for small molecules like ink diffusing in water , D equals 10-5 cm2 sec-1. Thus, the correct dynamical equation for how the concentration of some substance (ink, enzymes, etc…) changes in space (1D) and time due to diffusion is:

dC(x,t) d 2C(x,t) = D (6b) dt dx 2

On an exam you might remember the form of an equation, but be unsure where to place the constant(s). All you need to remember is dimensional analysis in order to determine the placement of constants, since you are given the value of the constants on exams, along with their appropriate dimensions (like D = 10-5 cm2 sec-1) . Suppose you had not read about dimensional analysis, then you might have written equation (6b) as:

dC(x,t) d 2C(x,t) D = (7) dt dx 2 which is obviously incorrect because each side of the equation has different dimensions.

As a final example of using dimensional analysis, we will see that the probability, P, that a particle is located somewhere between x1 and x2 is given as:

5 x P = ∫ 2ψdx (8) x1

where ψ is the wavefunction for the particle. Since probability has no dimensions, that means that all of the dimensions on the right-hand side of (8) must cancel exactly. Here, there is only ψ and the integral, which, referring to Table 1, has dimensions of [L], since it is an single integral. The dimensions of ψ must then be [L]-1 so that the right-hand side of (8) is dimensionless. Moreover, we can interpret ψ as the probability per unit distance that a particle is located at some position on the x-axis, and ψ is known as a probability function.

Problem 1.3 Probability Fun In three-dimensions, the probability to find a particle somewhere is written as:

P = ∫∫∫ ψdxdydz (9) where all the quantities are the same as described previously. What must the dimensions of ψ be in this case? Can you give a physical interpretation of this ψ? Problem 1.4 Schrodinger Equation The time-independent Schrodinger Equation for a free particle can be written as:

− h 2 d 2ψ = Eψ (10) 2m dx 2 where m is the particle’s mass, h is the over 2π, E is the energy of the particle, and Ψ is the particle’s wavefunction. Using dimensional analysis, re-write this equation in terms of only dimensions, referring to Table 1 as necessary. After cancellation, what must - h2 d2/dx2 represent? What does i h d/dx represent? (Hint: Remember that −1 = i and look at Problem 1.2). You just made a really deep connection between classical and quantum mechanics.

Problem 1.5 Maxwell Equations Maxwell derived one of the most remarkable things in all of science from his set of 4 equations. He was able to show that the Electric and Magnetic fields satisfy a Wave Equation given as:

d 2 E 1 d 2 E = (11) 2 µ ε 2 dt 0 0 dx where E represents the Electric field, and 1/(µ0 ε0) is a constant that fell out of the Maxwell Equations. What are the dimensions of this constant? What is a possible physical interpretation 1 of the quantity ? If you get this last part right, you will have found the constant that most µ ε 0 0 of special relativity theory hinges upon, based simply on dimensional analysis.

6 Dimensionless Quantities

As we have seen with probability , sometimes lack of dimensions is as useful as anything else. Here we will look at some dimensionless quantities, and see what we can learn from these important things.

One of the most difficult subjects in all of science is , since the governing equation (termed the Navier-Stokes Equation) is a very messy, non-linear thing**. The derivation of this equation is straightforward†† but it solution is very particular to the geometry of the situation, the size and of the particles moving in the fluid, as well as to the density and the of the fluid itself.

Let’s see if we can analyze this problem without doing any of the messy mathematics. We will take an example of a sphere moving through a liquid with viscosity η (in dimensions [M][T]-1[L]-1 ) and mass density in dimensions of ([M][L]-3). The sphere has a radius [L], moving at speed [L][T]-1. Is there some way to arrange these four quantities so that the combination is completely dimensionless?

Sometimes it is useful to group things that are similar. For instance, the particle’s size, speed, and the fluid density all are related to inertia, in the sense that they are like the size, speed, and mass of a free particle, respectively. The only other term is the viscosity term. Taking the ratio of the product of the inertial terms to the viscous term seems reasonable and yields:

R’ = [L] [L][T]-1 [M][L]-3 / [M][T]-1[L]-1 (12a) R’ = 1 (12b)

Wow! All of the dimensions completely canceled out, and we have a dimensionless quantity, R’. Now, a guy named Reynolds had already discovered this 100 years ago, but that is OK. This number is called the of the system, and it is very important in determining when turbulence sets in (usually when R’ ≥ 2000) and for telling the strength of inertial versus viscous forces in a fluid.

The big idea behind dimensionless numbers is that any combination of the parameters involved yields the same resulting physics if the different combinations are numerically equivalent. This is why you can build model airplanes that operate in wind tunnels to mimic real planes in the air. Engineers match the Reynolds number by adjusting the length of the model plane, the air velocity, and the air density so that the Reynolds number for the airplane in the sky and for the model in the wind tunnel are identical, and then all of the physics is the same. All of the properties of a system can be determined by these dimensionless quantities, which are often called scaling parameters.

Another set of useful dimensionless quantities arises in statistical mechanics, particularly when we are looking at Boltzmann . In these statistics, the probability for a system to

** Actually, it is so difficult to solve that its solution is worth one-million dollars! See the Clay Mathematics Institute at www.claymath.com for details. †† For a derivation, you can see The Feynman Lectures on Physics, Volume II.

7 be in a state with energy E, P(E), is given as:

− E P(E) ∝ e kBT (13) where kBT is the unit of thermal energy, given as the product of the and the absolute temperature. The dimensions for this combination are energy (see Table 1). Now, it turns out that when you exponentiate something, the thing in the exponent must be dimension- less, since there is nothing physical about e[L] (or e raised to any other dimension for that matter).

Thus, whenever you see something in an exponential (or in a trigonometric function or logarithm), it must be dimensionless. Here, since we know probabilities are dimensionless, the right hand side must also be dimensionless, indicating that eX is also without dimension, for any X. In equation (13), E/kT is dimensionless, so that it can be placed in the exponent. Table 1 lists the rules (and exceptions) for dimensionless quantity manipulation.

Remembering these rules for where dimensionless quantities must exist can help you tremendously when doing complicated problems in statistical mechanics, , and kinetics, to name only a few places. The problems below will demonstrate the power of knowing when a certain equation necessitates a dimensionless quantity, and what the meaning of those quantities is.

Problem 1.6 Bacterial Growth The equation that governs how bacteria grow in a culture dish is one of the oldest in all of calculus. Its solution has the form:

= +kt N(t) N 0e (14)

where N(t) is the number of bacteria in the culture at time t, N0 is the number of bacteria initially in the dish, t is time, and k is an undetermined constant. What are the dimensions of k? What does a large versus a small k indicate physically? (Hint: Sometimes graphing can help you visualize these types of limits.)

8 Problem 1.7 Feynman’s Quantum Theory Richard Feynman formulated quantum theory from first principles based on a Lagrangian equation of motion (the more common theory is based on a Hamiltonian equation of a motion). He said that the fundamental thing to look at was the Action of a particle (denoted S), which is related to the difference between the particle’s kinetic and potential energy, integrated over all time. Thus, the action has dimensions of [E][T]. What fundamental constant, which is quantum in character, has [E][T] as its dimensions? As it turns out, the key quantity in Feynman’s theory is:

iS e X (15)

where S is the action, i is the imaginary number, and X is an unknown constant. What are the dimensions of X? Compare the dimensions of X with the dimensions of the answer from the first part of this question. Can you draw any conclusions about the possible identity of X?

Problem 1.8 Diffusion Revisited We have already looked at the Diffusion Equation (4), and now I will give you its solution, assuming we start from a localized ink drop. The solution is:

− x2 C(x,t) ∝ Ae 2Dt (16)

where C(x,t) is the concentration of ink at position x and time t, and D is the diffusion coefficient. What are the dimensions of the thing we are exponentiating, (-x2/2Dt) ? What about the dimensions of A?

Unit Conversion

The last thing we will look at is how to convert between different units of the same dimension. This is equivalent to converting between feet and meters or gallons and liters. You have all probably done this before, so I will be brief. The basic idea is that you get so many of one unit for so many of another unit. For example, 2.54 cm equals one inch. So if we have 5 inches of rope and we want to know how many cm of rope that equals, all we need to do is arrange the conversion factor so that the units we are looking for do not cancel out, while the other units do. Mathematically, we would say that:

5 inches x 2.54 cm/1 inches = 12.7 cm (17a) where we have canceled out the extraneous units. For the inverse problem of finding out how many inches is 5 cm, we would have the following:

5 cm x 1 inch/ 2.54 cm ≈ 1.96 inches (17b)

9 where we simply inverted the dimensionless conversion factor [2.54 cm/1 inches]. That is pretty much it, but for long conversions, be sure to take your time, as you can get lost in these manipulations very easily.

The choice of units in a given situation is arbitrary, but making the proper choice can lead to a simplification of the answers and the underlying equations. For instance, when talking about atomic systems, we would not want to use SI units, since the SI unit of distance, the meter, is very large compared to the size of atoms and molecules. Rather, the use of angstroms (which are 10-10 meters) is more appropriate. Choosing proper units can make your life much easier, and conversely, the choice of improper units can make calculations a nightmare.

Sometimes, scientists will even “make-up” units to simplify their theories or equations. They define a set of units where some constant equals one (like the or the Planck constant), and therefore when you write out equations, this constant no longer appears explicitly. We know it must still be there for the dimensions to be correct, but it is “hidden”. This is very confusing to people, and when these conventions are made, authors are (mostly) diligent about stating them.

Problem 1.9 Spectroscopy Language Chemists have some confusing nomenclature as well. In spectroscopy, the quantity you are looking at is the photons absorbed by molecules as they vibrate, rotate, or shift around. You then sort these absorbed photons according to how energetic they are. You can then compute a spectrum by plotting the number of photons absorbed with a certain energy (called the absorbance) versus the actual photon energy. This histogram can reveal much useful information about the molecule of interest.

Spectroscopists have expressed the energy of the absorbed photons in terms of the photon’s , which equals 2π times the inverse of the wavelength of the photon, with dimensions of [L]-1. We know that this is not the dimension of energy, but it is a rather convenient unit to use in order to represent the energy of a photon. The dimensions of the photon wavenumber (often in units of cm-1) must be simply related to the dimensions of energy. What are the dimensions of this conversion dimension? Look at the list of fundamental constant dimensions in Table 1. What combination of these has the same dimensions as our conversion factor? (Hint: Look over your answer to Problem 1.1)

Conclusion

I hope this paper has demonstrated the power of dimensional analysis, since you have been able to analyze and think about problems from fluid mechanics, quantum mechanics, diffusion theory, and quantum field theory without ever solving an equation. These last few problems will allow you to get a better feel for the power of this formalism, since you will be predicting new laws of nature based solely on your physical intuition developed from dimensional analysis.

10 Problem 1.10 Fluctuations and Responses We saw that D governs the diffusion of a particle caused by the thermal fluctuations of the fluid in which it is immersed. The frictional coefficient, µ, is related to the between the fluid and the particle as it moves. Since the diffusive motion and the associated friction are both caused by the thermal agitation of the fluid, the relationship between these two quantities and the thermal energy is called the fluctuation- response theorem (or sometimes simply an Einstein Relation). Using dimensional analysis, what combination of D and µ yield an energy? Since D and µ are both related to the available thermal energy, what might be a good way to write this characteristic energy? If you get this correct, you will arrive at a simple and powerful result that is exactly true, without the need to add any constants, for a wide spectrum of systems.

Problem 1.11 Fluids Revisited We looked briefly at fluid mechanics when we talked about Reynolds number, and we saw that a fluid is governed by two quantities, its mass density ρ and its viscosity η. Using the dimensions from Table 1, is there any combination of these two quantities that equals a force? What is that relationship? What do you think the force represents in light of the Reynolds number?

11 Table 1. Basic Dimensional Equivalents & Quantity Dimension Definitions Common Units

Fundamental Dimensions

Mass [M] Kilograms (kg) Length [L] Meters (m), Feet Time [T] Seconds (sec) Moles [O] 6.022 x 1023 objects per Moles (mol) Temperature [K] Kelvin (K), Celsius, Fahrenheit Charge [Q] Coulombs, ESUs

Geometric Dimensions

Perimeter [L] Circumference m (A) [L]2 Length • Width m2 Volume (V) [L]3 Area • Distance m3

Linear Mechanical Dimensions

Velocity (S) [L][T]-1 Distance/Time m•sec-1 Acceleration [L][T]-2 Force/Mass; Velocity/Time m•sec-2 Momentum [M][L][T]-1 Mass • Velocity kg•m•sec-1 (P) Force [F] [M][L][T]-2; Mass • Acceleration Newtons (N), Pounds Energy [E] [M][L]2[T]-2; Force•Distance Joules (J) , Kcals

Rotational Mechanical Dimensions

Torque [F][L] [E]; Same dimensions as Energy. N•m, foot-pounds Angular Velocity [T]-1 dθ/dt; Same dimension as Frequency. •sec-1 Angular Acceleration [T]-2 dθ2/dt2; Radians•sec-2 Angular Momentum [P][L] of Inertia•Angular Velocity ≈ I• dθ/dt kg•m2•sec-1

Wave Dimensions

Frequency (ν) [T]-1 Identical to Rate of Change Hertz (Hz) Angular Frequency (ω) [T]-1 2π Radians per Second Radians (Rads)•sec-1 Wavelength (λ) [L] Usually of a Wave Nanometers (nm)

12

Wavenumber (k) [L]-1 Usually of a Wave; 2π /Wavelength Centimeters-1 (cm-1)

Electromagnetic Dimensions

Electric Field [F][Q]-1 Force/Charge; Volt•Distance N•Coulomb-1 (N•C-1) Potential (V) [E][Q]-1 Energy/Charge Volt (V) = [J•C-1] Current [Q][T]-1 Charge/Second Ampere (A) = [C•sec-1] Density Dimensions

-3 -3 ρMass [M][L] Mass per unit Volume Grams (g)•cm -3 -3 ρNumber [L] Number of Objects per unit Volume cm -3 ρMolar [O][ L] Number of Moles per unit Volume Molar = [mol•Liter-1 (l-1)] -1 ρMolal [O][M] Number of Moles per unit Mass Molal = [mol•kg-1] Statistical Mechanics and Hydrodynamic Dimensions

Diffusion Constant (D) [L]2[T]-1 Diffusion Coefficient cm2•sec-1 Frictional Constant (µ) [M][T]-1 Viscous Force F = µ • Velocity kg•sec-1 Viscosity (η) [M][T]-1[L]-1 Viscosity of a fluid; [F][T][L]-2 Poise = [kg•sec-1•m-1] Dynamic Viscosity (υ) [L]2[T]-1 Viscosity/Mass Density; Same dimensions as Diffusion Coefficient cm2•sec-1 Reynolds Number (R’) Dimensionless Scaling Term in Fluid Mechanics; Describes onset of turbulence; Ratio of inertial to viscous terms.

Thermodynamic Dimensions

-1 -1 Entropy (S) [E][K] Has Same Dimensions as kB. J•K Pressure [F][L]-2 Force per unit Area; ATMs, Pascals = Energy per unit Volume [N•m2]

Useful Combinations of Dimensions

S•K [E] Entropy • Temperature J kB•K [E] Thermal Energy per particle J; picoN•nm

13 R•K [E][O]-1 Thermal Energy per Mole J•Mol-1 hν [E] Energy of a photon of frequency ν •Volts hω [E] “ “ angular frequency ω Electron•Volts

Mathematical Operator Dimensions

N N x [L] This is true for any x, not just lengths d/dx [L]-1 First derivative with respect to distance d2/dx2 [L]-2 Second derivative with respect to distance d/dt [T]-1 First derivative with respect to time d2/dt2 [T]-2 Second derivative with respect to time. d/dn [N]-1 First Derivative with respect to n with dimensions of [N]-1. d2/dn2 [N]-2 Second Derivative with respect to n with dimensions of [N]-2. ∫dx [L] Integral with respect to x. ∫dt [T] Integral with respect to t. ∫dn [N] Integral with respect to N with dimensions of [N]. ∫∫dxdy [L]2 Double Integral with respect to x &y. ∫∫∫dxdydz [L]3 Double Integral with respect to x,y, & z. ex Dimensionless ex implies x is dimensionless. ln(x) Dimensionless ln(x) implies x is dimensionless‡‡. sin(x) or cos(x) Dimensionless Any trigonometric function is dimensionless and implies x is also.

Dimensionless Quantities

Angles Dimensionless Radians or degrees are dimensionless. Numbers Dimensionless The number of molecules (or anything else) is dimensionless. Probabilities Dimensionless The probability of an event occurring is the ratio of 2 numbers. Constants Dimensionless e, π, ½ , 1,000,000, etc…

Fundamental Constant Dimensions

-1 23 -1 NA [O] Avogadro’s Number 6.022 x 10 Mol h [M][L]2 [T]-1 [E][T]; Planck Constant 6.626 x 10-34 J•sec 1.5 x 10-27 g cm2 sec-1 h [M][L]2 [T]-1 [E][T]; Planck Constant/ 2π 1.054 x 10-34 J•sec e [Q] Fundamental Charge 1.602 x 10-19 C c [L][T]-1 Speed of Light 2.977 x 1010 cm•sec-1

‡‡ Remember the properties of logarithms: log(A/B) = log(A) - log(B) and log(AB) = log(A) + log(B), where the forms to the right of the equal sign violate this rule. However, the combination on the right does not violate the rule, and you only see logs in these appropriate combinations. Please be careful when dealing with logarithms for this reason.

14 2 -2 -1 -1 -23 -1 kB [M][L] [T] [K] [E][K] ; Boltzmann Constant 1.380 x 10 J•K R [M][L]2 [T]-2 [K]-1[O]-1 [E][K]-1 [O]-1; Gas Constant 8.314 J•K-1•Mol-1 -2 -7 -2 µ0 [M][L][Q] Permeability of Free Space π x 10 N•A 2 2 -3 -1 -12 -1 ε0 [Q] [T] [L] [M] Permittivity of a Vacuum 8.854 x 10 F•m 2 -2 9 2 -2 1/4πε0 [N][L] [Q] 8.987 x 10 N•m •C F [Q][O]-1 Faraday Constant 96485 C•mol-1 -31 me [M] Mass of an electron 9.109 x 10 kg -27 mp [M] Mass of a 1.672 x 10 kg -27 mn [M] Mass of a neutron 1.674 x 10 kg -10 a0 [L] Bohr Radius .529 x 10 m ATM [F][L]-2 Atmosphere 101325 pascals G [L]3 [M]-1 [T]-2 6.7 x 10-8 cm3 g-1 sec- 2 ------N.B. The first time a unit is used, it is written explicitly, and a shorter definition is given, which is used throughout the rest of the table. For example, Meters are written as m, Joules as J, etc… after they are introduced at the beginning of the table.

15