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Dimensional Analysis Page 1 Biochemistry 102a Fall Term 2002 Background Handout: Dimensional Analysis 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 Physics and Chemistry. 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 times 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 mathematics 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 dimensions 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 length [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 time 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 dimension 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 seconds 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 temperature [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 (lengths, times, masses, charges, etc… ) like the SI system (mass has units of kilograms, 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 angles, 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 frequency or the rate 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 second 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 angle, 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.
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