It's a Wonderful Line: Data Analysis Using Graphs
Few things in science are as daunting as a pile of unreduced data. "Here's a bunch of numbers. Draw a conclusion." Talk about a plunge into the icy cold water! Fortunately, we have graphing as a powerful way to help us dry off, visualize the data and translate it into mathematical relationships.
Example 1: A particle in a magnetic field
We know that if a charged particle like a proton with some momentum p enters a magnetic field, its path will be bent into a circle or a section of a circle. We ought to be able to infer the magnitude of p from the radius of curvature of the path R and from the magnetic field B. But how?
Below we have hypothetical data from an experiment in which particles of various known momenta are injected into a 10 T magnetic field. The data table is to the left. On the right is a graph of that data.
The data points in the graph form a pretty good straight line. It's not perfectsome variation in the data is to be expected, but it is clearly a line which intercepts the axes at the origin. Straight lines take the form
Radius R (cm) Momentum p (MeV/c) 54 168 67 195 90 279 99 292 117 348 122 368 154 458 175 526 186 563 197 595 201 598 y = mx + bbut here momentum p in MeV/c takes the place of y and radius R in cm takes the place of x. We already said that the y-intercept, b, is equal to zero. So we get p = mRwhere m is the slope of the line. Let's draw a line of best fit and find the slope:
The slope m equals the "rise" divided by the "run."
m = 450/150 = 3.0 MeV/c-cmThus p = (3.0 MeV/c-cm)RGreat! We have derived a mathematical relationship from data! (They say every time a mathematical relationship is derived somewhere a physicist gets accelerator time. Or is that only an old story?) However . . . What about B? This data is for a steady B = 10 T. So let's now look at data for momentum p as a function of B, with R held constant:
Clearly, this is also a straight-line relationship. So if p is linear in B and, as we saw previously, also linear in R, then we can say
B (T) P (MeV/c) 2 50 4 136 5 144 7 201 10 305 11 322 12 354 14 439 15 460 16 486 p = kBRwhere k is some constant. But wait! We know that when B = 10 T p = (3.0 MeV/c-cm)RThen we can say kB = 3.0 MeV/c-cmand thus
k(10 T) = 3.0 MeV/c-cm
k = 0.30 MeV/c-cm-T
p = (0.30 MeV/c-cm-T)BRUsing two graphs, we've established a relationshipa sturdy mathematical equationbetween p, B, and R. This equation can be found in textbooks and is derivable from the Lorentz force, though our units here are a bit odd outside the world of particle physics.
The key to our success has been using the equation for a straight line. However, not all relationships show a straight-line relationship. What do we do then?