Musings on my faith going into Holy Week 1/n

As I stand on the precipice of entering Holy Week, the holiest, most important time of the year for Christians that the rest of the world kind of (mercifully) ignores because it has only managed to co-opt the Easter Egg and candy part of things, which is literally the least important part, I have been reflecting, as I ought, on what my faith means. A kind of all compassing musing on what it is I believe, why I bother to believe it, and going all the way to "What do I call myself, since 'Christian Scientist' is something other than what I am?" I'm going to try to write as much of it as I can on this blog, because I feel it is important, but being musings I can't promise they will be thesis like. They may ramble a bit. Some may be long and some may be short. If you come here for physics posts, sorry not sorry for the theological interlude.

Holy Week, particularly in the liturgical tradition, throws sharp relief on a lot of doctrinal points that Christians tend to go 'yeah, yeah I know' at and non-Christians think we are crazy for believing. It can also bring up, if you run in the right circles, friendly debates about atonement vs. redemption theology, the sufficiency of Christ's sacrifice, and even the purpose of baptism, getting into the paedobaptism vs. believer baptism debate. The practice of Holy Week is designed to remind us, in case Lent did not, that we are broken, and that Christ died to heal that brokenness, and rose again to usher in the coming of wholeness.

That we are broken is something of which I have no doubt. I don't see how anyone can disagree with it. As my father observed, "The doctrine of total depravity has never lacked for outside proof"[ETA: This is apparently a quotation from G.K. Chesterton]. That Christ died to heal that brokenness I also have no doubt, though this is where a lot of the people I know think I've jumped the shark, so to speak. A fair number of my peers (and superiors and inferiors, I have no doubt) think that my faith is odd, nutty, a bit of a relic or even 'something [I'll] outgrow'. I have no problem with the ones who think the first two, I can understand, though not agree with the third and the  fourth I find unbearably patronizing, but that is neither here nor there. Christianity *is* weird. And a lot of humans have horribly twisted it and corrupted it and I desperately wish we could make those corruptions a thing of the past, though there is something to be said for the devil you know.

So let's get something out of the way before I get any father into recording my theological thoughts. Just make this the first post.

My faith is not just a comfort in bad time (though it is that), or a I'll-go-someplace-nice-when-I-die wishful thinking, or a philosophy, or a way to connect with a larger community. It is in a very real sense *everything* to me. It defines the universe, my place in the universe, the purpose of the universe and myself; it defines my relationship to God, between myself and my family, between myself and my husband, between myself and every human I will ever encounter; it determines my responsibilities to this world, and everyone and everything in it; it is the entire framework on which my life is built. If you striped everything else away, my faith remains.

"How can you be a scientist and a Christian?" is a question I have heard a (frankly) irritating number of times. From both directions, actually. Scientists who are atheists look askew at my ability to trust science if I also believe in a man-god, and Christians with whom I have strong doctrinal disagreements don't trust my soul to be saved if I think we came from monkeys. The question makes as much sense to me as "how can you be a scientist if you are a woman?". If I really believe that God created the universe, and he created us, how can I *not* believe that this universe would be designed in such a way that we, striving to understand it as we follow our natural, God-given curiosity and using the minds He gave us, could understand? How could I not jump at the opportunity to study a master-craftsman's work? If you think I'm crazy for believing in a Creator, or for believing in a Triune God, or a Savior or whatever particulars of my doctrine baffle you  to the extent you doubt my science, you are welcome to check my math. If you think I'm going to Hell because  when the math and science say the universe is 14 billion give-or-take years old, I trust that it's right,  please point me to the passage in the New Testament where this is named as a salvific issue. I'll wait.

That I am a scientist is not a stumbling block to my faith, and  my faith is not a stumbling block to my science. Though I wont go quite so far as Kepler to say that math is the language of God, or even as far as the Belgic confession in favor of natural theology, I will say with the psalmist that the "heavens declare the glory of the LORD" and with Maltbie D. Babcock that "This is my Father's world".

Basic Physics: Part 0, Section 4: Derivatives

Previously in this series, we covered algebra, trigonometry, vectors and vector multiplication. Now (after more delay than I would have liked) it's time to tackle the elephant in the room--calculus.

No, please don't close this tab! I swear, it's not as scary as you've been told. If you made it through trig and vectors (which, if you are reading this I assume you have) you've really made it through more mind bending stuff than we will need to cover here.

Why cover calculus at all? Aren't there algebra-based physics courses at every university? Yes, yes there are. And anyone who has taught physics with only algebra, trig and vectors will tell you it's actually harder to teach physics without reference to derivatives and integrals. Newton invented/discovered calculus so he could describe his theory of gravity and motion (his notation was abysmal, though). Calculus is the mathematics of change. Algebra is the mathematics of stability. And physics is really boring if nothing ever moves.

Now, depending on when you went to school, learning calculus may have been reserved for the students who were good at math, or who hadn't been told that "math wasn't for them". I am here to tell you this is like telling students who are going to live in another country that they don't need to learn the past or future tense, they can get along just fine with the present tense. Technically, this is true in a lot of cases, but it limits their ability to get everything out of their trip. Try to think of calculus in this way--not as some strange new kind of math, but just a different tense in this language.

We'll begin where most calculus textbooks begin with derivatives. Calculus has a very intuitive explanation of derivatives: they are the slopes of lines. That's it. What makes derivatives interesting is that they give you the slope at any point along a line*. You will generally hear the included caveat that the line must be smooth and continuous, but this isn't a calc class and I'm not going to show you any equations which are not differentiable (capable of having their derivative taken), so we aren't going to worry about that here.

Let's start with the simplest case, a straight line going through the origin of our coordinate system:

In this case, the slope of the line is going to be the same everywhere, and we can find the slope using the tried and true "rise over run" method. In moving 4 units to the right, the line has moved 3 units up, so our slope \(a\) is $$ a = 3/4 = .75 $$ So far so good. Nothing scary or uncomfortable  to date. A little algebra, that's all, and a little reading off a plot. Now, what if we were just given the equation for this line, in the slope-intercept form encountered in algebra class: $$ y = ax$$ $$y = .75 x$$
Still not too bad. And if I had presented this to your first, you probably could have told me the slope of this line just from this--the coefficient of \(x\) gives the slope, so \(.75\). Congratulations, you just took your first derivative without knowing it!

So, if derivatives are that easy, you ask your computer suspiciously, why is there an entire semester of calculus dedicated to it, hmm? Well, two reasons. First of all, because there are way more complicated kinds of lines than straight lines, and second of all, no one dedicates an entire semester to derivatives. They usually also teach limits (proto-derivatives) and numerical integration (proto-integrals) in the same semester. Derivatives are usually 4-5 weeks of the semester, a lot of that learning special cases.

What if I gave you the line with the equation $$ y = x^2 + 3, $$ would you know what it's slope is? It looks similar to the linear equation in slope-intercept form, but you probably have a feeling that the \(x\) being squared complicates things. And it does, since \(x^2\) is a parabola.

Parabola!

Now, you could draw tangent lines at a sampling of points along the parabola, and find the slope of those tangent lines, plot those slope values and approximate the slope of \(y = x^2 + 3\) and you would find that it came close to \(2x\). I can't speak for everyone here, but I find doing that unbelievably boring. Some algebra teacher once made me do that once and  it was tiresome to say the least.

But calculus and the tool of differentiation gives us a much better way.  Remember, mathematicians do not "invent" new kinds of math to torture students and non-mathematicians. They develop new techniques because the old way was inefficient or tedious or just didn't work all that well. Calculus is  a great example of this. Rather than calculating a bunch of individual slope points and extrapolating what we think the slope is, we can find the exact slope with a few simple rules, and a little new notation.

Let's look at our parabola again. So we have the equation $$y = x^2 +3$$ which describes the line itself. If we want to say that we are looking at the equation for the slope of that line we can write it in Leibniz notation as $$ \frac{dy}{dx}$$ which is nice and concise (there is also Lagrange notation and Newton notation). But Leibniz is nice for beginning with because it has a nice math to english translation: the change in \(y\) over the change in \(x\). This is the more formal way to say "rise over run" and is more generally applicable. Also, now that we are finding the slope of the parabola everywhere we call it a "derivative", and we find it by the process of "differentiation".

To find this, we need two rules. The first rule is formally known as the "elementary power rule" but I just learned it as "this is how you do it". For a function \(f(x)\) that has the form (i.e., it looks like or follows the pattern of) $$ f(x) = c x^n $$ where \(c\) is a constant, \(x\) is the variable and \(n\) is a real number (usually integer, but not necessarily) the derivative can always be found in the following manner: $$\frac{df}{dx} = c*n*x^{n-1} $$ If you are wondering what that \(f(x)\) is doing here, since I kinda just started using it, think of it as a way to label a generic equation. You could keep saying \(y=\) such and such, but then it's not always clear which \(y\) you're talking about. If you instead use the notation of Letter(variable) it lets you label both the equation uniquely (function f, function g, function h) and specify which letter is acting as your variable (x, y, z). Neat, huh?

That's it. That is the most basic rule and definition of the derivative. For the special case where there is no variable, just a constant, the derivative of a constant is \(0\). So, to summarize,

1. Given a function \( f(x) = c x^n\), the derivative is \(\frac{df}{dx} = c*n*x^{n-1}\).
2. Given a constant function \(f(x) = c\), the derivative is \( \frac{df}{dx} = 0 \)

So, let's apply these rules to the equation for our parabola.
$$y = x^2 + 3$$
$$\frac{dy}{dx} = (2) x^{(2-1)} + 0$$
$$\frac{dy}{dx} = 2x^1 = 2x$$

And so we find in three lines of calculus the  same answer that a bunch of line drawing and measuring and plotting got you. Let's try another one, that's a little longer.

And really funky looking on a graph.

$$f(x) = 3 x^{5} - 2 x^{2} + x^{-3} $$
$$\frac{df}{dx} = 3*5 x^{(5-1)} - 2*2 x^{(2-1)} + -3 x^{(-3-1)} $$
$$\frac{df}{dx} = 15 x^{4} -4 x - 3 x^{-4} $$

Longer, but still not too bad, right? See, I told you calculus wasn't the terror it was made out to be. One more rule and we've knocked out all the differential calculus we'll need for both physics 1 and physics 2. This rule is called the "chain rule" and it covers almost every other situation we could face outside of a calculus book or more advanced physics. What it is, really, is a short cut when your variable of interest is buried inside a parenthetical expression, instead of having to bother to separate it out by algebra (if it can be separated by algebra at all).

Let's start with something that we could mess around with algebra and get it into a form that our first two rules apply. Let's begin with the equation $$g(x) = (x+2)^2$$ By using the FOIL method, we  find that this could also be stated as $$g(x) = x^2 + 4 x + 4$$ Using the two rules laid out above, we find that it's derivative is $$\frac{dg}{dx}= 2x + 4$$ Now we have something to check the chain rule against.

Displaced parabola!

The chain rule is a way to approach these things methodically, working from the outside in. You start by treating everything inside the parentheses as a block. It does not matter how complicated it is inside the parentheses, or how simple. Treat it all as though it were just the variable. So step one of the chain rule gives us $$\text{ Step 1: } \frac{dg}{dx} = 2 (x+2)^{2-1}$$
Now you take the derivative of what's inside the parentheses, and multiply that result by the result of Step 1. $$\text{Step 2: } \frac{dg}{dx} = 2(x+2)^{1} (1+0) = 2x+4$$
Lo and behold, it's the same result. Now for something this simple is using the chain rule worth it? Maybe, maybe not. But what about something that I don't know how to FOIL, like $$h(x)= (x+2)^{-1/2} =\frac{1}{\sqrt{x+2}} $$

How do you FOIL a square root?! Tell me!

Let's try the chain rule on this and see if it doesn't save us having to look that one up in an obscure algebra text.
Step 1: Ignore what's inside parentheses, take the derivative as if (blah blah) = variable.
$$\frac{dh}{dx} = (-.5)(x+2)^{(-.5 - 1)} $$
Step 2: Take the derivative of what's inside the parentheses, multiply it by Step 1.
$$ \frac{dh}{dx} = -.5(x+2)^{-1.5} (1)$$
Step 3: Simplify if necessary
$$\frac{dh}{dx} = -.5 (x+2) ^(\frac{-3}{2}) = \frac{-1}{2 (x+2)^{\frac{3}{2}}}$$

I can guarantee that that was easier than trying to FOIL a square root. But what about something really nasty, like THIS

Honestly had no idea what this would look like before I graphed it

Behold, the rollercoaster that is $$ k(x) = (x^3 + 2)^{-.5}$$ Surely my nasty, terrifying calculuses gets horrifying and complicated now, heh? Stupid physicistses.

Um, nope. Not really. Let's take a look, shall we?
Step 1: Ignore what's inside parentheses, take the derivative as if (blah blah) = variable.
$$\frac{dk}{dx} = (-\frac{1}{2})(x^3+2)^{(-.5 - 1)} $$
Step 2: Take the derivative of what's inside the parentheses, multiply it by Step 1.
$$ \frac{dk}{dx} = (-\frac{1}{2})(x^3+2)^{-1.5} (3x^2)$$
Step 3: Simplify if necessary
$$\frac{dh}{dx} = (\frac{-3x^2}{2})(x^3+2)^{-1.5}  = \frac{-3x^2}{2 (x^3+2)^{\frac{3}{2}}}$$

Still just 3 bite sized steps.

Ah ha, you say, but what if there are parentheses inside the parentheses? What if I have a russian nesting doll of a problem?

You just repeat step 2 until you run out of parentheses inside parentheses. But I honestly can't say that I've ever seen that happen.

And that's nearly all you really need to know about differential calculus to conquer introductory physics! Hopefully you can see, at least a little bit, why physicists and mathematicians love it. It's like upgrading from a hand drill to a power drill. Or a sheet of sandpaper to a power sander. It might take a little getting used to, but it is a very powerful tool in our toolbox and one that will open up the rules of the physical universe to us in a way that algebra just can't. Because as I said in the beginning, the physical world is dynamic and changing, and algebra is the math of the static and stable.

There are two "special cases" that aren't really special cases that we will need, and they are very easy to use, but a bit lengthy to explain, so I'll cover them in a separate section, partly because they are both really cool, and partly because this post is already pretty long.

If anything is still unclear, or even a little foggy, let me know in the comments and I'll do my best to explain! And I hope to see you next time for integration!

* there are a few significant exceptions to this, which we don't have to be concerned with here. If you are interested in knowing more about these exceptions, brownian motion is a particularly interesting case being continuous everywhere and differentiable nowhere.

Kill the myth of "stupid"

For about a month now, I've been plugging away at a series called "Basic Physics", trying to go through a first year physics curriculum in a way that is understandable to people who aren't in STEM, and may not have even looked at 'math' in years. My mother has kindly been acting as one of the guinea pig for this experiment, reading through posts and giving me feedback on what is or isn't clear, is or isn't helpful. The last post on vector multiplication was particularly difficult for the both of us. It's hard to explain simply, and she really wanted to understand them in the same way she understood the trig section (after some rewrites at her suggestions). Every time we spoke and she said she still didn't get it, she would apologize "for being so stupid".

Now, stupid isn't a word I would use to describe my mother, and I sincerely doubt she has ever honestly been accused of that in her life. I reassured her that these were not easy topics, and pointed out that I had complained to her for at least 2.5 years now that my students, who nominally should walk into my classroom knowing this stuff, don't get it. I added a paragraph of encouragement at the top of the post, which seemed to help because I got this as a response:

ok I realized that I was trying too hard.

I get it now because I accept your math without trying to do it in my head every step.

Bring on the next chapter.

I called her up later in the day to thank her, because I realize that she probably hadn't been looking to learn this stuff before I asked for her help. She is a very gracious woman, and said she was always open to learning, but again apologized for being "stupid".

After we hung up, I realized that this is a refrain I have heard over and over when teaching: "I'm sorry I'm being so stupid". I've heard it from students in class, in office hours, in tutoring sessions back in college, and now from my mother. The general sentiment always seems to be that if they can't get it on the first go round, they are stupid and incapable rather than the reality that the topic is difficult. My students have gone so far as to tell me that I must be far more intelligent than them to understand this stuff.

There is an article in the New York Times today who headline was "Why do Americans Suck at Math?" and I can't help but think that the refrain of "I'm sorry I'm so stupid" and headlines like this are connected. Connected because they reinforce this idea that people "suck" at math in bulk. There is this weird perception that math is something only special people are good at, that you have to have some innate ability to do it and understand it. That people who are good at math look always use the Feynman method of problem solving: write the problem down, think about it, write down the solution. The idea that math people look at a new math topic and go "Oh, of course! Obviously this is true" and run off and use it seems to be weirdly pervasive, both consciously and unconsciously.

Of course, it would be lovely if this were true. I could have whole years of my life back if this were true. And of course it feels nice to be on the math people side of this, because it makes one feel smart and talented when in your work you frequently feel frustrated. It's like payback for the mockery, real or perceived, for being STEM types with all the cultural baggage that goes with it.

But I think it is also incredibly toxic. If math is something only special people can do, then why should ordinary people try? If we ignore or hide away our own struggles with understanding, we encourage this myth and scare people away who, even if they aren't in STEM, might enjoy seeing the beauty of it all. And it is beautiful. Being able to see the world with math and science at your back is awe inspiring, adding a whole new dimension to everything you can look at and experience.

I know very, very few people who haven't struggled to grasp every math and physics concept when they were first introduced. I think I've known two in my entire life. I was on the 'elevated' math track in school, which means I got all the way through AP Calc B in high school. And I still struggled and struggle with math. What my students (and my mother) never saw was me with wikipedia on my laptop and my calc book open as I desperately tried to understand different kinds of integrals, or tests for convergence. They never saw the early mornings, between classes and late nights in the physics lounge with scratch paper everywhere, chalk covered hands, asking anyone who entered the room, "Can you explain this? What is a [cross product, wave equation, probability density, etc]?" The extended arguments that eventually ended up with the stuffed monkey Harold on one of the professor's door in a plea for help. They will never know how much help I got from professors, from other students, from older students as I struggled to learn this stuff that I now seem so natural at. I'm not smarter than them. I was just persistent. When my students see me reduce a fraction on the board, or quickly do a cross product they assume it's just natural to me, like music is natural to my dad. What it really is is 7 years more experience and work.

Now, is there some natural inclination involved? Sure. But not nearly as much as people seem to think. Being good at anything, regardless of natural inclination, requires work above all else. My sister is more naturally inclined than I towards languages; she also studied more and is therefore far more fluent than I am (as in, actually fluent). No matter what your natural talent and inclination, if you never work at it, it will wither and dry up. And while you may never be a prodigy, hard work can get a person far in pretty much anything that's not sports.

People don't seem to believe me when it comes to math and science, so here's an analogy. I enjoy cooking. At this point in my life I am pretty good at it. I can make recipes up on the fly and nine times out of ten they work. I can tell if a cake is done by appearance and a light poke; I know if my steak is done to my liking by touch. Now, is there some natural inclination at work? Maybe. My mother is an excellent cook, and let me mess around in the kitchen at an early age. But mostly it's because I've been cooking for over half my life. Because I read cookbooks and watched masters and purposefully worked on my techniques, my understanding of the underlying food chemistry, the physics of different methods of cooking. Anything I am good at is maybe 5% natural talent, 95% work. Five percent alone gets you absolutely no where. Ninety five percent alone can get you pretty far.

This is something that we need to work on emphasizing more. We need to emphasize fewer Sheldon Coopers and Charlie Epps, boy geniuses grown up and solving MATH. We need to make it clear that what we do is not magic, not the result of some fluke of genetics that gave us special math powers. Something sparked an interest and we pursued it to the best of our abilities. We weren't destined to become mathematicians/physicists/chemists/what-have-you any more than non-STEM people were destined to be librarians/writers/bankers/secretaries/what-have-you. We chose to be what we are, and we worked hard to get here. Of course, this means admitting that we aren't special beings with math vision. But if we want to encourage people to engage with STEM, we need to kill this myth of "stupid".

Basic Physics: Part 0, Section 3: Vector Multiplication

Welcome back to my Basic Physics series! In previous sections, I covered some basic algebra topics, necessary trig functions, coordinate systems and vectors. The last post was getting rather long, and since vector multiplication is a bit tricky, I broke that topic into its own section, which you have before you! And I going to further preface this section with this: if you read through this, and you aren't sure you understand what is going on, don't give up! It's not easy to understand, and it looks down right weird; I think my reaction upon first introduction was something along the lines of 'what new devilry is this?!". I'm pretty sure I was using vector multiplication for at least 4 years before really understanding what it is. It didn't stop me from getting a B.Sci in physics, and it shouldn't discourage you from continuing reading this series, because it is possible to use these things without really understanding. Think of it like a car--you use a car everyday, you can make it do what you want, but almost everything under the hood is a mystery. Doesn't mean you can't use it to get you where you want to go! They also get easier with physical examples, which we will get to in coming posts. 

There are two types of vector multiplication. Each has its own uses and peculiarities. This section is going to cover what they are, their peculiarities, and how to do them. They pop up repeatedly in physics, so we'll discover their myriad uses along the way. Let's bring back our two arbitrary vectors from the last post 

$$\vec{v} = a \hat{x} + b \hat{y} + c \hat{z}$$

$$\vec{w} = d \hat{x} + e \hat{y} + f \hat{z}$$

and see what weird things we can do with them!

The first type of multiplication is called the "dot product" (remember that "product" is whatever results from a multiplication), and we get the new symbol \(\cdot \) to indicate that we are combining two vectors using this method. It is fairly straight forward: you multiply the x-hat components together, you multiply all the y-hat components together, you multiply all the z-hat components together and then you add up the results. 

$$\vec{v} \cdot \vec{w} = (a \hat{x} + b \hat{y} + c \hat{z})\cdot(d \hat{x} + e \hat{y} + f \hat{z})$$

$$\hspace{15 pt} = (a*d) + (b*e) + (c*f) = ad+be+cf$$

Interestingly, a dot product of two vectors does not yield a new vector with magnitude and direction, but rather a "scalar" which just has magnitude. This little quirk becomes very important when we get to electromagnetism. It should also be noted that the dot product is commutative, which is to say that 

$$ \vec{v} \cdot \vec{w} = \vec{w} \cdot \vec{v} $$

The same cannot be said of the other type of vector multiplication, which we will get to shortly. 

So, what does a dot product tell you? Speaking geometrically, it gives you the product of the length of the projection of one vector onto another vector. That's about as clear as mud, so lets look at some pictures. Here we have two vectors, \(\vec{A}\) (red), and \( \vec{B}\) (green), with an angle \( \theta \)  between them. 

If we draw a line between the end of \(\vec{A}\) and the point on \( \vec{B}\) where that line can intersect normally

we now have a right triangle. And we know from two weeks ago in the trigonometric section what to do with right triangles. We are given the angle, so by using the cosine function we can find the length of the projection of \(\vec{A}\) onto \(\vec{B}\), kind of like finding the length of your shadow. Let's label that projection \(A_B\) since it's the shadow of \(\vec{A}\) on \( \vec{B}\). So

$$ A_B = |\vec{A}| \cos{(\theta)} $$

with the vertical lines around indicating that we are using the total length, the magnitude, of A and not the vector form. The direction information is not helpful when dealing with triangles. So now we have the length of the projection of \(\vec{A}\) onto \(\vec{B}\). Assuming we have the length of \(\vec{B}\)  we can now find the geometric value of the dot product

$$ \vec{A} \cdot \vec{B} = |\vec{A}| \cos{(\theta)} |\vec{B}|$$

or, more prettily and more commonly,

$$ \vec{A} \cdot \vec{B} = |\vec{A}|  |\vec{B}|\cos{(\theta)}$$

This is another way to calculate the dot product, and is handy if you have been given magnitudes and angles, and not the component form for your vectors. 

But what does this mean, I can imagine you asking. Recall the idea of orthogonality from the previous section. The dot product allows you to relate two vectors based on the degree to which they are orthogonal to each other. If two vectors are perfectly orthogonal, the angle between them is \(90^{\circ}\), there is no projection, no 'shadow' of one vector onto the other, the cosine is zero, and the dot product is zero. The vectors are completely unrelated to one another, and they have no multiplicative interaction. If, on the other hand, the angle between them is \( 0^{\circ}\), then the cosine between them is 1 and they are parallel. They are both going in the same direction and can have the largest multiplicative effect on one another. (If you happen to dot a vector with itself, the angle will be zero, the magnitudes will be identical and you will get the square of the magnitudes. This is in fact how the magnitude of a vector is defined--the square root of the vector dotted with itself.) For any angle in between the effect is proportionately diminished. This will be easier to see when we can give some physical examples. 

The second type of vector multiplication is the "cross product", and the \( \times \) symbol which you probably learned to use for regular old multiplication back in elementary school gets reserved for this particular operation from here on in. Regular multiplication is usually indicated either by abutting parentheses, by an asterisk, or in the case of variable/coefficient terms just writing them next to each other without a space, as in the dot product example above. However, it is not generally as straightforward a calculation as the dot product, because the result of a cross product is still a vector. 

In order to calculate the cross product from scratch, as it were, we need to borrow a tool from linear algebra, namely the determinant*. The determinant is a way to arrange vectors so that you can easily calculate the cross product, no matter the size of your vectors. It is basically an organizational tool.  Once again, let's use the general vectors \(\vec{v}\), \(\vec{w}\), and calculate the cross product of \( \vec{v} \times \vec{w}\). 

$$ \vec{v} \times \vec{w} =\begin{vmatrix}\hat{x}& \hat{y} & \hat{z} \\ a & b & c \\ d & e & f  \end{vmatrix}$$

First, let's dissect what this thing is, line by line. 

 The first row is a label row of sorts. They label each column and they will help label the results. All the \(\hat{x}\) components go in the column under the \(\hat{x}\); if there is no \(\hat{x}\) component to a particular vector, that column gets a 0 for an entry for that vector's row. The same goes for the rest of the directions, namely \( \hat{y}\) and \(\hat{z}\).

 The second row is where the elements from the first vector, in this case \(\vec{v}\) are placed in their respective columns. Remember, if a vector is lacking an element, it is entered as a zero; the column is not deleted entirely, even if all it contains is the label and zeros.

The third row is treated in the same manner as the second row, except that it contains the second vector, in this case \(\vec{w}\). It's rather like filling out a spreadsheet. 

Now what do we do with this thing? 

You first calculate the \(\hat{x}\) component. To do this, you imagine blocking out everything that shares the \(\hat{x}\) column and row, leaving you with a square of components that are not the \(\hat{x}\) component

With those remaining 4 elements, you multiply the diagonal elements, and subtract the lower left/upper right pairing from the upper left/lower right pairing. So the results of this step are 

$$\hat{x} (bf-ec) $$

Note that the \(\hat{x}\) component of the product contains every element except the \(\hat{x}\) components of the  original vectors. Cool, right?

The second step is a little strange, because by blocking out everything in the \(\hat{y}\) element's column and row, we get a kind of split square, or two rectangles. 

What this means is that you have to subtract this result from the final product. You multiply the elements from the two rectangles as though it were one square, so this component of the product gives us

$$-\hat{y}(af - cd)$$

So far, so good. 

The last component is almost identical to the first. 

And you find the results the same way you did for the \(\hat{x}\) portion of the product. So the final bit is 

$$\hat{z}(ae-db)$$

Weird, but not too horribly complicated.  So the final result is 

$$\vec{v} \times \vec{w} = \hat{x} (bf-ec) -\hat{y}(af - cd) + \hat{z}(ae-db) $$

Some people are able to simply memorize the result given above, and don't bother to use the determinant. I am not one of those people gifted in memorizing formulae.  It's easier for me to remember a compact method or tool than to memorize lines of elements. 

It should be noted that the cross product is not commutative--the order in which things are multiplied matters, unlike the dot product. That is to say, \( \vec{v} \times \vec{w} \neq \vec{w} times \vec{v}\). It is however anticommutative, which means that reversing the order negates the result: \( \vec{v} \times \vec{w} = - \vec{w} \times \vec{v}\). 

So, what does the cross product give you? This is a little easier put than the dot product. The cross product calculates the area of the parallelogram whose sides are defined by the two vectors. 

So what's all that vector information doing? Well, hold on to your socks, it turns out that area is a vector quantity. Yep, and the direction of that area is normal to the surface which the area is a measure of. So if you can imagine standing on that parallelogram, which every direction is straight out of your head is the direction that that area points! This also make clearer an important point about the cross-product, which is that the resulting vector from a cross product is necessarily orthogonal to its two parent vectors. The two vectors must lie in a plane to form a parallelogram, and the normal to that parallelogram must be normal to the two vectors defining that parallelogram. 

The fact that the cross product calculates the area of a parallelogram leads us to our last point. What if you don't need to know which way the area is pointing, for some reason? You just need the magnitude of the result, not the whole thing. Well, if you remember your geometry you might recall that the area of a parallelogram can be found by multiplying the lengths of the sides and the sine of the angle between the sides. The same formula works for the cross product in a way. Assuming you have the magnitudes of each vector and the angle between them, you can find the magnitude of the cross product, at the cost of the direction information. 

$$|\vec{v} \times \vec{w} | = |\vec{v}| |\vec{w}| \sin{(\theta)}$$

That wraps up what you need to know from vectors. I know it may seem like a lot, but remember that it's a tool in our tool kit for physics. We will be using these tools frequently, and like any skill it becomes easier with use, and you get to build more incredible things the better you become!

As always, I hope that I have explained things clearly. If I haven't, please let me know in the comments, and I'll do my best to clarify!

*DH objected to this section, because what physicists call a determinant is technically a 'formal determinant', as it has the right form but does not adhere to the strict definition used by mathematicians. If you should show this to a mathematician, they will twitch, and possibly rant about physicists. This is the normal reaction of mathematicians to physicist notation. 

However, the math editor of this blog, sirluke777, objects to DH's objection, and says it's perfectly fine. His background is math, physics and chemistry, so make of that what you will. If there is an outcome to this math geek debate, I will update here. 

Basic Physics: Part 0, Section 2: Vectors and Coordinate systems

In the previous sections, I cover some basic algebra topics and necessary trig functions.

In this section, I'm going to lay out some basics of coordinate systems and vectors. I'm pairing these concepts because vectors without coordinate systems are a little esoteric for this series, and coordinate systems are necessary, but easily dealt with, at least compared to things like trig functions.

These two concepts are needed because when we talk about a problem and set out to solve it, we need a way to describe where things are, where they are going and how they are getting there. The combination of vectors and coordinate systems allows us to know exactly what we are referring to  and what it's relation is to anything else that might be relevant to the problem. Without this tool, problems in more than one dimension can easily become a hopeless jumble.

Let's start with coordinate systems. There are many possible coordinate systems of which  11 of which are commonly used and  only 1 of which we need for right now. That coordinate system would be the Cartesian coordinate system, apocryphally realized by M. Rene Descartes  (he of "I think therefore I am" fame) as he lay in bed watching a fly buzzing above him. It was also realized by M. Fermat, though he failed to publish it. If you ever had to plot things by hand in a math class, you have used the two-dimensional cartesian coordinate system! The cartesian coordinate system can be thought of as a grid system in 3 dimensions, that lets you specify a  location based 3 numbers, one for each dimension. You can think of it as giving someone a latitude, longitude and altitude. You have given them all the information they need to locate a particular spot on planet earth. (I will officially note that the earth is NOT a cartesian system, since the lines of longitude are not parallel but intersect. But for small distances, say NYC, it is a decent approximation).

Formally, a location in any coordinate system  is the intersection of three orthogonal planes. Orthogonal, for our purposes, means that the lines/planes intersect at \( 90^{\circ}\) to each other. Think of the walls in your house. Your walls (hopefully!) intersect your floors at right angles most of your walls will intersect each other at right angles, unless you have a very interesting house. So your walls are orthogonal to each other and they all orthogonal to the floor.

An example is probably the easiest way to show this. Let's start with a basic three-dimensional (3D) cartesian coordinate system:

In the picture above, I have drawn the same coordinate system from two slightly different perspectives. The top one you are staring down the barrel, so to speak, of the z-axis, looking at the x-y plane straight on. In the bottom one the picture has been rotated \(45^{\circ}\) about the y-axis so you can see along the z-axis as well. This becomes very useful if you are talking about things in 3D, while the top one is fine if you are only worried about two dimensions.

Another word about terminology and notation. What is an axis, and why are those letters wearing hats? As with a lot of math-stuff, it comes down to the dual needs for conciseness and precision. Let's start with the hats. When you draw a coordinate system for a problem and you label the axis, you are defining your directions. It's as if you are creating a mini universe and saying "this is East/West, this is North/South and this is Up/Down". But rather than label things in poetic victorian manner as "easterly direction" mathematicians and their ilk like to label things with letters. So "easterly direction" becomes "x-direction", but that's still too wordy. So 'direction' becomes 'axis', and that can get further shortened with vector notation as \( \hat{x} \) said "x-hat".  So an axis defines the direction of your coordinate system, but it also serves as a point of reference, much like the equator, the Greenwich meridian, and sea level  serve as reference points for finding places on the earth. So if you are on the x-axis, you are not moving in a y-direction or a z-direction. If you want to give a location in the coordinate system, you can notate it either as \( \left\langle a, b, c \right\rangle \) or you can use vector notation, to get a little ahead of ourselves: \( a \hat{x} + b \hat{y} + c \hat{z} \). The latter notation is preferably simply because it is more flexible, as we shall see.

Getting back to our example. Let's say we want to find a point \( \left\langle 3, 2, 2 \right\rangle\) (\( 3 \hat{x} + 2 \hat{y} + 2 \hat{z} \) ). For the moment we don't care what the units are. We start by locating the \( x = 3\) plane, that is, the plane that contains every point of the form \( \left\langle 3, b, c \right\rangle \) where \( b \), \(c\) are every real number. Then our diagram looks like this

with the red dot noting the point where the plane intersect the x-axis  in the bottom view, since it's a little hard to see. Next, we locate the \( y = 2\) plane.

The blue dot notes it's intersection in the y-axis in the bottom diagram because again, it's hard to tell. It's much easier to see in the top image, but there' a reason why the bottom diagram is actual preferable in some ways. This can be most easily seen when we try to add in the last point, the \( z = 2\) plane to give us our three-plane intersection.

Amazing what you can do with a basic drawing program and a little insanity.

The top image doesn't really allow us to visualize that last necessary dimension. You can mentally add it, but you can't draw it into the top one. The bottom one you can see the last plane and pinpoint their intersection (marked with a black dot here). 

That's pretty much all there is to coordinate systems. They let you pick a frame of reference so you can locate things in a mini-universe for the purpose of problem solving. What I find particularly neat is that you can place your coordinate system anywhere you like and the problem will still be solvable. It may be easier to solve from a computational standpoint if you center it nicely, but you don't have to. Why this is the case is something that I'll get into more when we start doing physics properly. 

Now, on to vectors. If a coordinate system gives you a frame of reference, vectors are what let you move around in that frame of reference and deal with more than just static problems. Now, what they are precisely requires linear algebra and is way outside the scope of this series, so we are going to stick to just definitions and not get into the nitty gritty. So, here goes.
 

A vector pairs a quantity with the direction that quantity is in, going or pointing to. A vector has both "magnitude" (quantity) and "direction". So long as you can describe a quantity as having these two qualities, you can express it as a vector*. We've already shown how we can describe position as a vector. You can also describe velocity as as a vector. "He's going 80 miles per hour" gives you a speed (a magnitude). "He's going 80 miles per hour due north" gives you a magnitude and a direction. We'll get more deeply into what physical quantities can be described using vectors in the first section of Part 1. 

For now, let's just work with two arbitrary vectors and see what we can do with them. As discussed in the algebra post, we'll use letters to stand in for numbers that we can plug in later. 

$$\vec{v} = a \hat{x} + b \hat{y} + c \hat{z}$$

$$\vec{w} = d \hat{x} + e \hat{y} + f \hat{z}$$

The little arrow above \(v\) and \(w\) indicates that they are vectors. In math everything has its own shorthand because you never know when you will want to deal with something in its entirety, or just don't want to write out the whole thing for the umpteenth time. 

So, what can we do to these things? Well, we can add them. The trick is that you can only combine things attached to like 'hats'. So you combine all the x-hat components, all the y-hat components and all the z-hat components, but you can't combine x-hat components with non-x-hat components. So 

$$\vec{v} + \vec{w} = a \hat{x} + b \hat{y} + c \hat{z}+d \hat{x} + e \hat{y} + f \hat{z}$$

$$\hspace{10 pt} = (a+d) \hat{x} + (b+e) \hat{y} + (c+f) \hat{z}$$

Subtraction works the same way:

$$\vec{v} - \vec{w} = a \hat{x} + b \hat{y} + c \hat{z}-d \hat{x} - e \hat{y} - f \hat{z}$$

$$\hspace{10 pt} = (a-d) \hat{x} + (b-e) \hat{y} + (c-f) \hat{z}$$

At this point you are probably wondering about multiplication and division, since addition and subtraction have been relatively straight forward. The answer is that there are two types of multiplication for vectors, and no types of valid division. Why this is starts getting into linear algebra and "outside the scope of this course". So I'm going to ask you to trust me on this one, because it's absolutely true even if I can't show you right now why it's true. They are also rather more involved than vector addition/subtraction, so I am going to move them to their own post so we can really take our time with them. 

I hope that this all was clear. If it wasn't, please let me know in the comments!

*There are also a few things that we'll get to over the course of this series that you wouldn't think you could describe as vectors, but they behave identically to the ones we deal with here. 

Basic Physics: Part 0, Section 1: Trigonometry

In Part 0, Section 0 I gave an overview of two basic algebraic skills needed to learn physics effectively. In this post I will explain several trigonometric (hereafter abbreviated to 'trig') ideas that are indispensable when physics problems are in more than one dimension. Since the world we live in is three dimensional, trig is really quite useful. \(\require{cancel}\)

Now, there is a lot more to trig than I am going to discuss here, mainly because we only need this one small portion to be able to tackle physics at this level. But trig is a much deeper subject with really interesting insights, if your fancy is tickled by beautiful interconnected math.

So to begin, we are going to need a triangle. This should be clear just from the name, which comes from the Greek for 'triangle' and 'measure'. More specifically, we need a right triangle, which is a triangle that possesses a \(90\) degree angle (a 'right' angle). You may have run into them before if you ever had to do anything with the Pythagorean Theorem. Right triangles are very nice to work with because you automatically know one angle, so all you need is is either two side lengths or a side and one of the other two angles and you can deduce everything else. So let's begin with a basic right triangle:

You will be subject to the same drawings and handwritten labels as my students are/will be

This is just a run of the mill right triangle. Each of the sides is labeled with a term that both labels it and can be used to indicate or find its length. The little square in the bottom righthand corner of the triangle tells you (as it says in the picture) that that angle is \(90^{\circ}\) (degrees). One of the other angles, in this case the one in the lower left hand corner of the image, is specified using the Greek letter \(\theta \) (pronounced 'theta') to stand in for whatever the actual measurement is.

Before we go any further with triangles, I want to say a few words about the use of Greek letters in math and physics. When I was younger, I thought that weird looking symbols were the sign of TRUE MATH and required REAL SMARTS to use. An equation could look like pure gibberish to me but so long as it included some Greek letter it must contain a great truth about the universe. Then I started taking physics and math classes and realized NOPE. They are, basically, there because we ran out of useful letters in the Carolinian alphabet we use in western European languages. We set aside a bunch of Greek letters to stand for quantities that pop up a lot (e.g. \(\phi, \theta \) are generally reserved for angles) and free up \(a\), \(b\), \(c\), to be used as needed. In other words, Greek letters do not imply fancy-shmancy calculations. They say "we needed more letters and  these were easy to co-opt".

So, getting back to our representative right triangle, which I am going to post again  so you don't have to scroll back up to see what I'm talking about.

 For this triangle, we are given all the information we could possibly want or need. Yes, even that unlabeled corner. First of all, all triangles* have angles which must sum to \(180^{\circ}\). The right angle eats up half of that (\(90 ^{\circ}\) ) automatically. The remaining \(90^{\circ}\) are divided between the other two angles. So if we know \(\theta\) we know the other angle is going to be \(90^{\circ}-\theta\). But secondly,  and most importantly, we don't need that third angle. Trigonometry lets us get away with just one angle, using the power of

Repeated like a prayer at every physics 1 exam

You may have seen or heard this mnemonic before. Maybe carved into a desk in a high school math classroom, maybe muttered by some flustered-looking college student. This acronym is a way to remember the names and definitions of the three most important trig functions: sine, cosine, tangent. Each of these functions describes a ratio, or a particular relationship between two sides of a right triangle and one of its angles.  So in this picture,

our chosen angle is \(\theta\), so all of our ratios, all of our functions will be found with respect to this angle. The side of the triangle that connects our angle and the right angled corner is known as the 'adjacent' side. In our triangle, that would be side '\(a\)'. The other side (or leg) of the triangle that is connected to the right angled corner is called the 'opposite' side, which in our case would be side '\(b\)'. The last side that does not connect to the right angle at all is called the 'hypotenuse', and is the longest side of any right triangle. In this case, side '\(c\)' is our hypotenuse.  The reason that the hypotenuse has to be the longest side is because of the Pythagorean Theorem which states that for a right triangle, the square of the first leg plus the square of the side of the second leg will equal the square of the side of the hypotenuse. Mathematically, $$a^2 + b^2 = c^2,$$ and so the hypotenuse ('\(c\)') will always be bigger than either leg. You can prove this to yourself with a bit of graphing paper. Although you do not have to do so, I can tell you it is easiest to demonstrate this with a 3-4-5 triangle (a triangle whose legs measure 3 units and 4 units and whose hypotenuse is therefore 5 units).

Now, let's get down to the nitty-gritty definitions. This is where most people start to have problems, because while you can express the trig functions as simple ratios, what  they *are* is a little tricky. Lets start by expressing them as simple ratios, and work backwards to what it all means.

The function \( \sin{(\theta)} \) (said 'sine of theta') gives us the ratio between the 'opposite' side ('\(b\)') and the hypotenuse ('\(c\)')**. Explicitly, $$\sin{(\theta)} = \frac{b}{c}, $$ and thus the mnemonic for the sine function being "SOH": Sine [is] Opposite [over] Hypotenuse.

The function \( \cos{(\theta)} \) (said 'cosine of theta') gives us the ratio between the 'adjacent' side    ('\(a\)') and the hypotenuse ('\(c\)'). Explicitly, $$\cos{(\theta)} = \frac{a}{c}, $$ and thus the mnemonic for the sine function being "CAH": Cosine [is] Adjacent [over] Hypotenuse.

And lastly, the function \( \tan{(\theta)} \) (said 'tangent of theta') gives us the ratio between the 'adjacent' side ('\(a\)') and the 'opposite' side ('\(b\)'). Explicitly, $$\tan{(\theta)} = \frac{b}{a}, $$ and thus the mnemonic for the sine function being "TOA": Tangent [is] Opposite [over] Adjacent.

I should note that there are functions for the opposite ratios (hypotenuse over adjacent, etc) but they are rarely, if ever, used in physics at this level and I honestly don't see them pop up in physics at any level.

That's all well and good, you might say to me through your computer screen, but what is 'the sine of theta'? What if I just have an angle? How do I find the sine of it then?

Simple answer? Calculator or an online tool like Wolfram Alpha. But let's face it, that's a cop out, not an explanation.

The way to find the sine (or cosine or tangent) of any angle is using a thing called a unit circle, illustrated so very nicely below.

Via WikiCommons, Public Domain

A unit circle is simple a circle of radius \(1\). Right triangles are created using the radius of the circle as the hypotenuse of the triangle and drawing a vertical line connecting the x-axis and where the radius meets the circle. If you move the location of the radius, 'sweeping' though the full \(360^{\circ}\) of the circle, the triangle changes accordingly.

Now I want you to imagine that at every increase of \(1^{\circ}\), we measure '\(b\)', the vertical leg or the 'opposite' side. We then plot, or map out, that leg height against the number of degrees in our angle. We do the same for  '\(a\)', the horizontal leg or 'adjacent' side. Since our hypotenuse has a value of \(1\), by plotting out the height and length of the two legs we are plotting the sine and cosine functions, as shown in this nifty graphic***.

Via WikiCommons, Public Domain

This goes on forever. One of the cool, and occasionally frustrating to scientists, thing about the trig functions is that they go on forever. They never slow down, approach any sort of limit. If you could keep tracing the unit circle forever, you would keep producing sine and cosine waves.

So, how does this help us with our triangles back there? Remember that at their hearts, trig functions are ratios. And ratios scale, as any cook can tell you (though in cooking you do reach some practical limitations on the ability to scale). If I want to make a double batch of spaghetti sauce, I don't have to reinvent my recipe. If my one-batch recipe calls for one garlic clove for a pint of pureed tomatoes, I know that for my double batch I need two garlic cloves for \(2\) pints of pureed tomatoes, and I will still get the same tomato sauce. Similarly, if I know the sine of an angle, say \( \sin{(30^{\circ})} = 0.5 \), I am saying that on the unit circle, the vertical leg is half a unit long. So for any right triangle that has a \(30^{\circ}\) angle, the side opposite that angle will be half the length of the hypotenuse!

Let's work through an example of this. Here we have right triangle, with one angle given and one side length.

It's considered good practice, and it becomes much easier to do this when pictures become more complex, to label the picture with symbols and give the value to those symbols elsewhere. It makes the image cleaner, and leads to fewer errors.

So, what can we find out about this triangle? We have chosen our angle (or it was chosen for us) and we have the 'adjacent' side length. We could start by finding the 'opposite' side length using the tangent, which gives us the relationship between the angle and the two legs. Let's call the 'opposite' side '\(y\)'. Then we can say $$\tan{(\theta)} =  \frac{y}{l}$$ and we can solve for \(y\) using just a touch of algebra $$l\times \tan{(\theta)} =  \frac{y}{\cancel{l}} \times \cancel{l} $$ $$l \times \tan{(\theta)} =  y$$ and then to keep with convention we flip the sides of the equation to the thing we are solving for is on the left hand side (frequently abbreviated LHS): $$y =l \times \tan{(\theta)}.  $$ Now we can plug in our values if we like to find the value of '\(y\)'. Since \( \theta= 30^{\circ} \) , \( l = 5\, \text{cm}  =  .05\, \text{m}  \) and  $$y=\left(0.05\,\mathrm{cm}\right)\tan\left(30^{\circ}\right)= 0.03\, \mathrm{m} =3 \,\mathrm{cm} $$ You may be wondering why I converted from centimeters (cm) to meters(m) and then back again. This is again just a 'best practice' thing. It does not make a big difference in situations, like this, were there is only one unit (length), but when we start calculating things like forces, we will be combining different units into other units and it becomes very important to make sure all your units are base units (e.g. meters, seconds) or you can easily end up with an answer a thousand times bigger or smaller than it should be.

So now we have the 'opposite' side of the triangle. We now have several options on how to find the hypotenuse. We could use the sine or the cosine to find the hypotenuse (call it '\(h\)') in much the same way that we used the tangent to find '\(y\)'. Or we can use the Pythagorean Theorem to find it, which is my preferred method. Plugging in our particular terms to the formula we get $$l^2 + y^2 = h^2.  $$ To extract the '\(h\)' value we need to take the square root of both sides: $$ \sqrt{l^2 + y^2} = \sqrt{h^2}$$This is the same as raising both sides to the power of 1/2, so \( (h^2)^{1/2} \) becomes just \( (h) \). So  we find that $$ h = \sqrt{l^2 + y^2} = \sqrt{(.05\, \mathrm{m})^2 + (.03\, \mathrm{m})^2 }  = .05773627...  $$ $$ h \approx 6\, \text{cm} $$ And now we know everything about this triangle!

So, why do we need this? What arcane bit of knowledge am I trying to put in your head? What good does it do to know anything about this *one* triangle? How will this help me in physics?

Good questions.  First of all, I want to emphasize that, while math is beautiful and worth knowing for it's own sake, for a student of physics I find it useful to think of math concepts less as "yet another thing I have to learn" and more like a tool in your tool kit. Just like you need multiple, different tools to attack house repair (a hammer to drive nails, a screwdriver for screws) you need different tools to break down a problem and then build up your answer. Having these basic trig functions in your back pocket gives is like having a drill and driver set for a power drill. You can use it to both break down and rebuild the problem.

Secondly, by knowing about this one triangle, you are now able to attack any right triangle, no matter its shape or orientation in space.

All equally susceptible to attack by trig

And by knowing about one triangle, you allow yourself to know about triangles which can be drawn as extensions of that main triangle. Why exactly you would want to do that will be made more clear in a few weeks when we get into physics, so you will have to take me for my word on that for now.

Lastly, by being able to translate between angles and the sides of a right triangle we have set ourselves up to deal aptly with another very important concept for physics--vectors!

As always, please leave any questions in the comments section!

*In Euclidean, or flat, space. If you drew a triangle on a sphere, it would not have angles totaling \(180^{\circ}\)

**You may notice that I kept 'opposite' in quotation marks, but not hypotenuse. This is because while the hypotenuse is fixed, which side is called the 'opposite' depends on which angle you chose for your reference.

***Here's another graphic that shows a little better the cosine function, but it's not as smooth and it's labels are in German

Basic Physics: Editorial Consortium

The next promised post on trigonometry is in the final polishing stages, but in the meantime I would like a post to mention several people who have graciously agreed to help me in this endeavor to bring the first year of a physics majors schooling in physics to a non-math, non-science types audience.

 I know full well that as a grad student in physics I am a very bad judge of what is and isn't understood or common knowledge. Teaching has helped rein me in enormously, but my students are assumed to have at least basic calculus knowledge. So I anticipated myself having a problem recognizing what needed more explanation, what was over-explained or even patronizing. I don't want to be the detective novel criminal who spells all the easy words wrong and all the hard words right (in reverse, kinda). Dear Husband, my usual editor, is too well versed in math to much use in this particular arena, so I reached out to some other family members, specifically my mother, my sister, and my brother, to help make sure I do this right. I asked them to do this because each of them brings something that I felt I really needed on what I am dubbing my Editorial Consortium.

My mother is in the demographic group, you might say, that always gives me deer-in-the-headlights or horrified looks when I say I do physics and protest it was too hard for them. Though very talented, she has not directed her talents in a STEM field direction. She is, however, the only reason that I can do long multiplication or division and light years ahead of me in mental arithmetic (also cooking, social skills, language, and checkbook balancing). She is also a natural copyeditor of high standards who is not shy of letting me know when I have fallen short of the mark.

My sister, hereafter to be referred to as Sylvia, Historian Extraordinaire, just graduated college with an absurd amount of honors with a major in History and a minor in French, her thesis work (yes, thesis for undergrad) being on Dorothy L. Sayers. She has a good math background, but hasn't used it much, having no call to do calculus as a literary historian. Her one and only basic physics class was the same one in high school that inspired me to physics. She is also representing a group that I want to reach--younger adults--and she would know if a reference is too obscure. She is also incredible at calling me out for being obtuse and/or patronising.

Last, but not least, is my brother, who will start high school in the fall. I included him for three reasons. First of all, he has had all of the math that I claim is required to understand the blog, but has never taken a physics class in his life. He's interested in the sciences, but he is yet untainted by misconception and bad teaching (other than my own). Second of all, it turns out he inherited Mother's copy editing skills and is very good at noting my inconsistent use of single and double quotation marks. Thirdly, I'm curious if the explanations are clear enough for younger persons who might be interested, but don't have much of a background. The flip side of my mother, so to speak.

They have all agreed to read, edit and comment every post that I write in this series. Between them all I think there is a fair shot that I will do what I set out to do. But I won't know if I am actually succeeding unless you, the reader, let's me know. You are the other part of this Editorial Consortium. If something is not clear, if I mess something up or forget something or just plain gloss over with the hated "the reader can obviously see", let me know! There is a comments link below each post. I'd love your feedback.

Basic Physics Part 0, Section 0: Algebra

[This post is the first in a series intending to teach basic physics concepts in a blog format.]

As I mentioned in my introductory post, math is the language of physics. Physics cannot realistically be understood or done without math. While advanced physics requires some advanced math, basic first-year type physics requires some relatively basic math and math concepts. The first math topic that I want to cover is one that everyone who graduated high school should have covered at some point: algebra. 

Algebra has a kind of strange reputation. Among STEM people, it's the boring math that you needed to do to do the REAL math, or at least the non-boring stuff. It carries the same emotional connotations as diagramming sentences. Among non-STEM people, it's the boring math that they forced you to do and you never ever used again.

Until I really got into teaching and my research, I was mostly of the opinion that algebra was best left to machines. It was tedious and beneath my dignity to spend hours and pages rearranging symbols. When I started teaching, I began to understand the subtle power of algebra to make or break a solution. When I finally started to understand my research, I saw not only its power, but its beauty. Algebra is a tool that allows order to arise out of chaos.

To do the kind of physics this series is going to look at, you really only need 2 major algebra skills: the FOIL method, and some equation manipulation skills. The quadratic equation can come in handy, but that is one time that I am ok using a math program for because it doesn't pop up as frequently.

But before we get to that, I think some terminology definition is in order. When I speak of a "variable" I am referring to a symbol that can take on any value on the real number line (i.e., any where between negative infinity and infinity) within the confines of the equation and/or is the quantity we are solving for. A coefficient is a symbol that has a fixed value for that particular problem. Most physics texts I've seen and used have the convention that any letter from p-z can be used as a variable, while letters a-m are used as coefficients. The letter 'n' is a special case because it is typically used for integer numbers only. The letter 'd' is sometimes used as a variable because it's just so convenient to use it to stand for 'distance'. The letter 'o' is never used, because in handwritten notes it can all too easily look like a zero. A constant, for our purposes, is a symbol that has a fixed value that does not change from problem to problem. For example, \( \pi = 3.14159...\) no matter what problem we are doing. A 'term' is a catchall, just denoting that a symbol stands for something, without specifying type.

Now, on to algebra!

FOIL Method

The FOIL method (First Outside Inside Last) is one of the first things I was taught in algebra class, way back in 7th grade. It's basically a method for multiplying mathematical expressions together in a way that doesn't let you double multiply or leave something out.  If you are multiplying just two terms together, say \(a\) and \(b\), its easy to know when you got it all.
$$ (a)(b)  = ab$$
But what if you don't have just two items, but two expressions, \( (a+b) \) and \( (c+d) \) ? FOILing the two expressions makes sure you do all available multiplications without double counting.  You multiply the first terms from each expression, here \( a \) and \( c \), then the outside ones, here \( a\) and \( d\). Then you do the inner ones, \( b\) and \( c\), and the last ones from each expression, \( b\) and \( d\). Thus
$$ (a+b)(c+d) = ac + ad + bc + bd $$
This method can be logically extended to cover expressions with more than two terms, with the corresponding result being proportionately longer.

When I first learned this, it seemed incredibly useless. Why on earth would I need such a simple method? The answer is 'everywhere in physics'. From the simplest two-body problems to the most complex problems I've worked on for research, FOILing turns up again and again and again. Becoming not just  proficient, but a master at this technique has been crucial to my work. It is something that my students consistently underestimate, to their detriment, every semester.

Manipulating Equations

This isn't so much a single method as the Rules of Engagement for math. Equations are pretty flexible, but there are some rules. The underlying principle to these rules is that you have to do the same thing to each side of the equation, and you have to do it to everything on each side. For example, lets say we have this equation $$ 5 x + 2 y = 6, $$ and we want to solve for \( y\). We can start by subtracting \( 5x\) from each side like this $$ 5x + 2y - 5x = 6 - 5x$$ where you can see we have explicitly taken \( 5x\) from each side and thus have not changed the equation. By adding the same thing to both sides, we have effectively added zero, just like if you add a one pound weight to either side of a balance scale, it won't change position.  So now we have the equation $$ 2y = 6 - 5x,$$ but we still have not completely isolated \( y \). So now we have to divide both sides by 2, which is the coefficient of the variable \( y\). $$ \frac{2y}{2} = \frac{6 - 5x}{2}$$ Again, it is important to note that we have done exactly the same thing to both sides of the equation and in the case of division or multiplication we have applied that change to every term. $$y = \frac{6}{2} - \frac{5 x}{2}$$ $$ y = 3- \frac{5}{2}x$$ is the correct solution in this case. Do not, I repeat, DO NOT make the mistake I see so often, which is to only apply the division to one (usually convenient) term. The following 'solution' is wrong for this problem: \( y = 3- 5x\)

In certain cases, this also involves remembering the Order of Operations: Parentheses, Exponents, Multiplication, Division, Addition, Subtraction. The Multiplication/Division and Addition/Subtraction orders are less critical, since they are just two sides of the same coin. Division is the same as multiplying by a fraction, subtraction is the same as adding a negative number. But the parentheses-> exponents->multiplication/division->addition/subtraction ordering is inviolate. It is impossible, outside of sheer fluke, to get a correct answer if you do not abide by this rule.

And that's the basics of algebra that you may have forgotten  (accidentally or on purpose) that you need for physics, other than the kind that you, honestly, do intuitively. Next week, we'll cover some basic trig[onometry] that everyone should know.

Basic Physics: Introduction

I have done a few introductory physics posts in the past, but I have never been happy enough with them to continue them as a series. After writing my post yesterday on the problems with science communication, I have thought more deeply about why I wasn't happy with them. I've decided that I didn't like them because they weren't able to adequately explain what I wanted to convey. This is mostly my limitations, but also because I hadn't set in my mind who my audience was, and because I had never done posts that explained what I understood to be background material to the topics.

So, I am going to try again in a more cohesive manner. By which I mean that I am going to do a year-long series of blog posts that roughly coincide with what physics majors (and engineers, and other interested parties) learn in their first year, covering basic classical mechanics, electrostatics, magnetostatics and circuitry.

Why? Why would I do this? First of all, I think it will be good practice for when I (fingers crossed) become a professor. I have mostly been working off of the curricula and methods of other professors--I would like to find my own. Secondly, I want to show that physics isn't "hard". Nearly every time I mention that I do physics for a living, I hear the same story--the person I'm talking to either took it in high school and did miserably, thus putting them off the whole thing, or they never took it because they were never any good at that brainy stuff. I want to write a series that, even if it doesn't make physics converts, gives people the confidence that they understand key physics concepts, and maybe understand why physics geeks geek out over physics.

To that end, I am going to be writing for people who have never taken a physics class, but who have some basic math background. I am going to assume a high school level of education, though even that seems to be a somewhat variable standard anymore. I am not going to hold back on "vocabulary words" as my students put it, because it's a blog and you have instant access to a dictionary, but I will explain any technical terms or words that are used in a manner different from their colloquial usage.

To start out with, I am going to do a series of crash-course algebra, trig, vectors, and calculus, so that we have a common math starting point, and a kind of reference guide. Math is the language of physics and it is very hard to really understand what physics is without being able to speak about it using math. Otherwise you kind of end up with something more like Aristotle's physics than Newton's, simply because it is very difficult to describe it using just words.

Then there will be a longish series on basic classical mechanics, which is the one physics topic most people can grasp with at least a bit of intuition. We have all thrown balls, used a seesaw, and spun in an office chair. It will cover more or less the same material you would see in first semester physics class.

The last part will be on what I have taught for 5 semesters now--introductory electromagnetism, or Intro E&M. This will cover basic electrostatic forces/fields, currents, simple circuitry and basic magnetism.

I'm going to try to stick to a schedule of posting one a week, again, roughly like it would be in a classroom setting. This will also give me enough time (hopefully) to properly proof read them and weed out errors.

So, without further ado, on to Part 0, Section 0: Algebra!