Hi everyone. Welcome back. I'd like to do an example. It's a classical historical example of how dimensional analysis was used to best US intelligence. What we're going to do is called the Trinity Test example. The Trinity Test is a code name for the first detonation of a nuclear bomb by the US government. It happened on July 16th, 1945. They were testing as part of the Manhattan Project, if you know a little bit about that, the US was testing this idea of nuclear weapons during World War II and the Trinity test is named after the Trinity site. This is in New Mexico, in the Southwest portion of the US. The US had this new bomb and they wanted to show their power, their military might. They wanted to tell people they had it so they can intimidate their enemies, so they released data about the explosion, but they showed a video, which we'll watch in the second. Then they only showed some screenshots. They wanted to show people, but of course they held back how big the bomb was, how much energy there was. It was very highly classified information. Let's watch the video now on this bomb from July 16th, 1945. So we saw the bomb, right? You can imagine, you'd like to know how big it was, but that's not really. You don't quite know. The fear of the unknown is there. Well, there's a British physicist that saw the video and his name was Sir, he was knighted, G.I Taylor. So he's a British physicist who took the screenshots from that and they also were released and they looked something like this where they had these spear shaped blasts pictures and in particular it had the distance of which in meters so you could give an estimate of how big the radius was and it had the time. It had these stills of pictures. Using this information, he was able to find how big the bomb blast was, how much energy the bomb released using dimensional analysis. It's really an amazing application of dimensional analysis to find something that was classified information. So what do we do? What does any good scientists do whenever they're trying to get something? They collect data. Since the distance was shown in a bunch of pictures as well as the time. We create a table with our time and radius from all the still shots. So our time is measured in seconds and our radius is measured in meters. From this table of data, we can graph the scatter plot to visualize the data and we see what looks like, if you were to try to think of what kind of curve this would look like, maybe it looks like a square root function or something like that, some sort of root function. But in particular, the picture that comes to mind here might be a power function. So the relationship between time and rate is something that could be a power functionalist from visualizing the data. But really if you think about it from a scientific perspective, we know that radius is going to depend on time and of course, the energy of the blast and some other properties in particular, like the density of air. You can imagine the more dense the air is then the more resistance there is to the bomb blast, then the less dense, then less resistance. So if you think about it, the radius is a multivariable function. We're going to do time, energy, and we'll use density and the common letter we use in density is a Greek letter. This is Rho. This is a Greek letter R-H-O, not R-O-W, Rho, Greek letter Rho for density. So the idea is, can we, from just this information find the energy? It turns out that using dimensional analysis you absolutely can. Let's write our formula again here just so we have it on this slide. We have radius as a function of time, energy, and air density. So let's think about what our units are going to be for this equation so we can line them up. The units of radius, this is a distance or a length, so it's going to be meters. The units of time, fundamental units, we're going to use seconds and the units of energy, now this is a derived dimension here. So energy, what is energy? It's like a force per distance. Better go into your physics book a little bit. What did Newton say a force was? Force is mass times acceleration, still times distance. Now I can start putting in units. Units for mass, kilograms, units for acceleration. This is distance, so meters per second squared, and then the units for distance again, put another meters in there as well. Let's combine things up and we get kilogram per mass squared and then we'll put s to the minus two, since it was in the denominator. We're going to use negative exponents instead of fractions. Last but not least, density. This is another derived dimension, so what the units for. Density, is what is the amount of things there are over your volume. This is what is your mass over some given space? This is going to be kilograms per cubic meters, and when we rewrite that using negative integer exponents, we have kilograms times meters to the negative three. Okay, so there's our units. Now, with our units in hand, let's go ahead and compare dimensions. What would the dimensions be of this thing? Whatever the dimension is for the radius, we use square brackets to denote dimensions, this would be the dimension of time. But remember, you're allowed to have integer powers, so some power a, which we don't know. It has to also match the energy piece, which will be some other integer exponent b, which we don't know. Then the density for air, which we'll call it rho, looks like a p. The point is A, B, and C, these are integer exponents, we don't know what they are. Once we have our dimensions set to be equal to each other, we're now going to use our units. Units of radius is meters, our unit for time is seconds, but now it's raised to the a. Our unit for energy, is kilograms times mass squared times second to the minus two, and the entire expression raised to the b. Then our unit for density, we saw was kilograms times m to the minus three and that expression is raised to our unknown variable c. Let's simplify this a little bit, so we're going to combine terms. So I have a second raised to the a, and if I bring this two in I get a minus 2_b. Same variables when you multiply them, you add the exponents. Then let's see what else we have. We have kilograms. So there is a b that's going to be on top of the kilograms and then there's a plus c from the third expression for density. Last but not least, we have mass. Mass, is going to be 2_b, remember, the b is [inaudible] mass as well. Then we have minus three c from our expression for density. Now, this looks a little bit intimidating, like what do we do from here? What's nice about this is, this looks like one equation, but it's really three in disguise. What we can do here is we can compare coefficients. This is exactly, which I told you did. So we can compare coefficients. Remember, when there's no coefficients exponent, when there's no exponent written, it's a one. So [inaudible] to be equal, the one on the m on the left side, has equal to 2_b minus 3_c on the right side, that gives us our first equation, one equals 2_b minus 3_c. Now, there's no s and no kg, but you can think about it like it's really s is zero and kg to zero, just giving us one, so that gives us our other two equations. We get zero is a minus 2_b and zero equals b plus c. Three equations, three unknowns, we can solve for this. We can solve this using basic algebra. Take third equation, might be the easiest. You can do this a bunch of different ways, but let's just say third equation gives us b equals negative c. So if I plug that into the first equation, I get one equals 2_b, and then plus 3_b, placing negative c with b, which of course gives us that one is 5_b. Solving for b gives us that b is one-fifth. Going back and then solving into c that tells us that c is negative b, negative one-fifth. Use b now as positive one-fifth to go get a from our second equation. So you get a equals 2_b, which implies that a is two-fifths over b. So we are finding the exponents that will equate the dimensions for our equation. All right, so remember what we have so far. We have r is now equal to t as the dimension, so t to the a, which was two-fifth, we have e to the b, b was one-fifth, and then c was our exponent on density, which was negative one-fifth. So as dimensions, this is now my equation. We can turn this into a function. This says that r is going to equal to some constant, t to the two-fifths, e to the one-fifth as variables. Then we'll write our negative exponent downstairs, Rho to the one fifth. K is some constant, some real number here. What do we have? What do we know? Let's pause for a minute and see where we are at here. The radius, this is known at least in terms of its relation to time. I have the table, I have the scatter plot. These two things are related to each other. K is some constant, I don't know that. Energy is the thing I'm after, I don't have that either, and Rho is my density. Now, Rho looks like a variable, but remember, this is just the density of air. This is a known quantity, so we can go look this up in some table or we can grab it, but the density of air, I'll put it over here, so this is density. Now of course, this will vary a little bit depending on where you are on Earth and stuff like that, but the density of air is approximately 1.25 kg per cubic meter, so times m minus 3. So 1.25. That number is known. Now here's sort brilliance of G. I. Taylor to find the coefficient, there's two unknowns at this point. The k, the number in front, the constant, and the energy, and I don't know either one. So what he did was he conducted a series of experiments with small blasts where you can control the energy. Think about like lighting fireworks off or something like that. Like you know the energy, the energy is known and you can measure the density of air and you have r and t. So using experimental evidence, he was able to guesstimate or find using science and everything, k. So he was able to get k as a constant. So this is experimental data and the number that he found is 1.033. So this is from experimental data. Now we're starting to get there. So our function that we're trying to find here is 1.033, t to the two fifths times E to the one fifth divided by Rho is 1.25 to the one fifth. This coefficient in front. Who cares? It's some number, 1.033 divided by 1.25 to the one fifth, and as promised, it'll certainly look like a power function. We have exponents with our variables and exponents are constants and variables and the base. So now here's comes the next step, and this is what we know how to do. I want to start to use the relationship, I haven't really used that yet between r and t to find this. So let's do that now. So what do we do with power functions when we want to play around them? We take the logs of both sides. So let's do that. We have the natural log of r, would be equal to the natural log of this entire expression. So I'm going to take the natural log of the entire expression. To save some room we're not going to rewrite the whole thing over, but it will all go here. Whatever you do to one side, you must do the other side. I'm also going to use properties of logarithms now. I really have three things multiplied by each other. I have the big constant in front times t to the two fifth, times E to the one fifth. Log of a product is the sum of the logs. So this becomes the natural log of 1.033 divided by 1.25 to the one fifth. That's just some constant, the calculator can figure out what that is. We're going to add it to the log of t to the two fifths. Mostly we're going to use the property logarithms. So two fifths, I'm just going to write this in front just to save myself from having to write the equation again but normally it's t to the two fifths, but we'll bring it out. Hopefully we have done this enough, you're okay with me doing this in one step and the same idea, I would take log of E to the one fifth. The one fifth would fall out in front and I would have natural log of E. So I have some constant, I know it's scary-looking, but it's some constant C plus two fifths log of t, plus one fifth log of E. Maybe I'll write this one more time so we can see it. Natural log of r is some constant C, whatever this is, who cares? Plus two fifths, log of t plus one fifth log of E. Stare that for a minute. Where's this going? Hopefully see what this is going. Let's think about this in terms of like a line. Natural log of r, think about that as your y value. We have some constant, that's fine and we have natural log of t, this used to be like two fifths x plus one fifth log of E. Let's think of that as our b. The constant in front is just another constant. We can deal with that later but what I'm looking at here, if you think of C and b is kind of one big constant, it's a line. This is a line, this is linear but the only difference about this line versus a regular scatter plot is that the line now due to our dimensional analysis has sloped two fifths. So this is linear with our slope to be two fifths. Wherever we go to the data and fit our line, our curve or best fit, our linear regression, I'm going to set the line to be two fifths. Let's go over to Desmos and do that now. So the first thing we're going to do is we're going to take our table from time and radius and we're going to convert them to get ready for a log-log plot. We're going to take the natural log of the time and the natural log of the radius. So you can see I just let Excel do all this for me using equals LN, and I drag the formula down, and I get a nice table of new values for my log-log plot. This is the scatter plot for time versus radius of the original table, not the second table. I then copy and paste this table into Desmos. Once you copy and paste the table of your logs of your time and radius you get what looks like at least from visual inspection to be a pretty good line. So now we want to start to evaluate this line. So remember where we left off, hope you're keeping track still. We have the natural log of the radius. So that's going to be my y value. So we do y1. So y_1 squiggle, so let's introduce what we have here. So we have the natural log of this big constant. Now I'm just going to type it in, Desmos is a calculator, 1.033 divided by 1.25 raised to the one-fifth. Some calculator, who cares. But that's some number, and then we'll close that up. That's my constant, plus, remember what else I had here, I had two-fifths to the natural log of the time, that's my x1 variable, and plus some b, plus my shift, the other part of my constant. This b is representing the one-fifth natural log of b. My goal is to get this b. You can see that in the blink of an eye Desmos puts the line of best fit through these points and it calculates b for me. In particular it gives me that b is 3.60704. So remember the equation, compare this what we have so far. This b, we're going to go back to our PowerPoint software. We're going to use this and equate it to our unknown. So let's go ahead and do that. From all our work in Desmos doing linear regression we get that one-fifth natural log of e is equal to 3.60704. This is using linear regression to get this equation. Remember the ultimate goal of this thing was to find the energy, this classified information, how big this very first nuclear bomb was. Well now you have very simple expression in terms of e multiply both sides by 5. I guess we could do this. Multiply by 5, you get 18.0352. Then take e to both sides, take the exponential to both sides, and you get that the energy is 6.8012359.23. So it's probably more than we want. But remember the units that we're dealing with in energy is like kilograms per meter squared times meters squared times seconds to the negative 2. It's more common to think about this in terms of joules or even kilotons as it's an explosion. So if you convert the units, so again the units are important. So kilograms, meters squared, seconds to the negative 2. This was the answer. This is the right right in terms of the calculation. You just found the classified information. But just to show you where we're at so far. Let's round this off because we don't want that many decimals. We're about 6.8 times 10_13, and we put this in terms of joules. Or if you just want to talk about explosions, it's about 19.2 kilotons. So this was the estimate that GI Taylor found using dimensional analysis and published in a paper much to the chagrin and much to dismay of the US National Intelligence. So once the paper got published the US government was pressured to reveal what it actually was. When they revealed the true number their measurements gave 21 kilotons. So using dimensional analysis, the estimate 19.2 kilotons was extremely close to the true number and that's mildly amazing. This is just a nice historical example of where dimensional analysis was used to find some classified information. But it lend some insight to how you can use data and the tools of regression to find unknown variables. This process, this model is how in physics they are able to measure the energy of a blast of a supernova that happened a long time ago and they can solve for the other variables. So maybe you know the radius, maybe you know the energy or the density, but you don't know time. So you can go and find how long ago something occurred in the history of the universe. So when you hear these things about space and explosions and you wonder how are they doing it, it is through these sort analysis calculations. Really a nice example here of using dimensional analysis. Great historical significance. I hope you enjoy this lecture. This is always a fun one for me to do. Thanks and I'll see you next time.