The concept of what a derivative is will probably not come to you easily. There is a real importance and significance behind it which is very subtle (and did not reach me until I was out of high school altogether). I didn't understand it until doing numerous problems involving derivatives and using calculus for proofs. Everyone 'knows' what a derivative is intuitively, it's just difficult to make the connection with the EXPLICIT concept and what you already 'know'.
However, there are simple ways to get an idea of what a derivative is without learning high-level maths. Physically speaking (in a real physics problem, in other words) a derivative is the rate of change of something. This something might be, for example, velocity, or even just plain old position.
As an example:
Consider a ball rolling across the floor. An easy way to get measurements for this ball is to draw an x-y grid on the floor and just use a stop watch to examine it. You can also imagine, for the sake of this problem, that we can pause everything as if it was videotaped - this way we can think better about what is going on.
So let's say you draw an x-y grid with every 'space' being 1 meter.
So at the very beginning of the experiment (at time 'zero' in other words), we have our clock handy and we're ready to go. The ball moves 1 meter and you decide to pause everything. Your clock/stopwatch reads 1 second. So at time '1 second' (in other words t=1), we'll say that x=1 (if it's traveling along your x axis, of course)
But let's continue on. We unpause up until the stopwatch reads 2 seconds (at t=2). At this point, we notice the ball has moved 4 meters altogether (or another way to look at it, it has moved 3 meters in the last second). Interesting, as it seems the ball has somehow sped up. So at t=2, x=4.
Unpausing for another second, we find out that the ball has now moved a total of 9 meters (or 5 meters since the last pause). So at t=3, x=9.
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To summarize, we have:
t=0, x=0
t=1, x=1
t=2, x=4
t=3, x=9
So what is all this analysis telling us? Well we've got these numbers, so we can graph them and find the relationship between them. So let's say we make a graph where x is the x-axis and t is the y-axis.
It's hard to imagine what that would look like without actually graphing it, but I'm going to cheat a little bit and just link you here ( hyperphysics. phy-astr. gsu. edu/hbasees/acons.html sorry about the spaces, it won't let me post a url) to give you an idea. That's skipping ahead, but this would otherwise be a book-length post. Ignore the equations there for the moment and just look at the graphs.
The graph would look like the 'position' graph there at the top. That makes sense, because think about t and x. When t goes up a bit, x tends to go up even more. If you had [t=1,x=1] and [t=2,x=2] and [t=3,x=3] and so on, it would be a straight line (graph those points and see this for yourself if you can't imagine it).
There's a very useful mathematical analysis tool called the slope, analagous to the slope of a mountain or cliff. Normally, you can only take the slope of a straight line (rise over run, or y over x), but in this case, we can just look a bit and I'll cheat at the end to keep this from being too long (as if that isn't already the case ;) ).
You can see, looking at those graphs, the 'velocity' graph has a constant slope. In other words, it's just as steep at the top of the mountain as at the bottom. The angle with the ground doesn't change so it looks like a triangle.
The 'position' graph, on the other hand, gets steeper as you go up the mountain, the angle with the ground gets bigger as you climb up. So this graph doesn't have a single 'slope' value, the slope goes up as you move along the graph.
Very surprisingly and astonishingly, if you work hard (really really hard) and figure out the slope by taking tiny increments in the 'position' graph, you will end up with points to plot the velocity graph! So what does that mean? Does it make sense?
If you think of the velocity graph as the slope of the position graph, it really does make sense. As you move to the right in the position graph, the slope steadily goes up higher and higher. That's exactly what the velocity graph is doing.
So what if we did it again? What if we graphed the slope of the velocity graph? Well the velocity graph has the same slope all the way, through right? So it would be just be some number, and that's what we see on the acceleration graph.
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So what does that all mean? Slopes and graphs, these are just vague concepts, just drawing stuff on paper - what is the meaning behind it?
Remember what I said originally way back up at the top: the derivative is the rate of change of something. Look at the acceleration graph. Is it changing? No, it's the same all the way through. What about the velocity graph? Is it changing? Yeah it's changing at a pretty constant pace. If you go right a bit, it goes up a bit. The position graph? It's changing, but the change isn't constant. It becomes bigger and bigger as you go on. It turns out that the slope IS the derivative!
So now we have a sort of definition to go with, the derivative is the slope/rate of change of something. This can be understood without any physical analysis of any real system, but I find it much easier to think of this stuff in terms of real things going on - baseballs, pots and pans, and boulders instead of just fuzzy concepts.
So let's think back to the ball rolling on the floor. The 'what's the rate of change of position' would translate into 'how fast is it going?'. That makes sense right? I mean if it's position is changing, it must be moving, and therefore it has a speed right? But looking at the velocity graph, it's not constant. If you measure the velocity at one instance, it will be different from another one. At t=0, it turns out, it wasn't moving at all!
'What's the rate of change of velocity' translates to 'how fast is it accelerating?' Well it accelerates at a constant rate, so at ANY TIME you can say the ball has an acceleration of 'blah' - whatever that acceleration is (in this case I chose an acceleration of 2 m/s/s).
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So do physicists sit around all day drawing graphs and taking slopes and whatnot? No way! That takes far too long, as I've demonstrated here. We use equations and their derivatives to help us speed this stuff up.
For example, for a case like this, where we can see that the acceleration is constant, we have a very special beloved equation that goes as follows:
x = x_0 + v_0t + \frac{1}{2}at^2<br />
<br />
<br />
where x_0 is the beginning position, v_0 is the beginning velocity and a is the acceleration - all three of these are just numbers, so you can substitute any numbers you like in there.
In the example above, the beginning position is 0, and the beginning velocity is 0 - that makes it a bit simpler, but you can handle any problem with constant acceleration with this equation.
Taking the derivative directly of this equation (it's a simple mathematical process that you will learn in calculus) gives us an equation for velocity, since velocity is the derivative of position.
<br />
v = v_0 + at<br />
And finally, taking the derivative of that gives us an equation for acceleration since acceleration is the derivative of velocity.
<br />
a = a <br />
(just a number, remember)
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So the idea of a derivative is related to stuff you already understand pretty well. You understand that the rate of change of position is velocity and the rate of change of velocity is acceleration (to an extent, humans aren't so great at second derivatives intuitively). The strange thing is that it works out so well in mathematics in so many relationships, that it's difficult to bridge all the systems and rigor with your every-day experience.
I hope this was more helpful than short ;).
I had P.A.M Dirac as teacher ;)
At least for the massive particles. I haven't tried it for the massless ones, though.
