## Philosophy 167: Class 6 - Part 10 - Uniformly Accelerated Motion- the Primacy of Time, and the Mean Speed Theorem.

Smith, George E. (George Edwin), 1938-

2014-10-7

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Now he has a problem. Why equal increments of speed and equal time can equal times instead of equal increments of speed over equal increments of distance? And that problem is posed by Sigretto. First, the way I've done it now. I'm now starting and I'll do this constantly to you.

I will provide you the formulas that you all learned in high school, et cetera. And every time I do it, the number one thing to recognize is I'm being anachronistic, okay? Not wildly anachronistic, because I put a proportion sign up there where it really mattered. And what does that mean when I put a proportion sign?

It doesn't literally mean v is proportional to t. What it really means is if I compare two vs they're proportional to the two times in the same distance. That is, everywhere you see a proportion sign, you're really comparing two ratios of like quantities. Now, gradually that will disappear, as this century goes on, and you'll just start talking about one quantity varying as another.

But what's always underlying it is comparisons of two examples and like ratios. So, every time I put a proportion sign up there and write it this way in our modern form, there were no equal signs at this time yet. They hadn't been invented. They come just a little later.

But the texts will always refer to vary as, proportional to, equal in ratio to, same ratio as, et cetera. When I write it that way, I'm doing a shorthand for something that is being misrepresented, but not necessarily so badly. So, he offers an argument that the second of these two is incoherent.

And the argument doesn't work and there's a footnote in Drake explaining why it doesn't work. I'm not gonna dwell on that for the moment. There is a problem. If speed acquired is proportional to distance traversed, then indeed, the solution that turns into a differential equation, needless to say, dvds is proportional to s.

It's an exponential and of course exponentials didn't exist for Galileo yet. They had actually been devised, but he didn't seem to, if he knew of them, something important I didn't tell you. All the work that's been appearing in this book was done before he left Padua almost certainly, and is being written up 30 years later.

How much of the math that got developed between 1610 and 1638 he knew is a totally open question. He may of known next to none of it. But a lot happened in symbolic math during that period. But he just, he shows no signs of using it. At any rate, there is a shortcoming to the second.

Speed can't start from zero. If it starts from zero, it's always zero. If it's going with distance. That's what the exponential gives you. The interesting question here, and Drake has a wonderful story that I'm about to tell you. How Galileo decided In practice, in actual fact that he wanted equal increments of speed and equal increments of time rather than equal increments of distance.

There is a sheet in Galileo's notebooks where there's a series of numbers. And Stillman Drake looked at the numbers and came up with the following claim. He wanted to do inclined planes to distinguish between these two possibilities. And how do you get equal increments in time? Well, he had a background in music and his father wrote books in music.

So he knew musicians could keep very, very close time. So he got some musicians and he put comparable to the cross bars on a guitar, strands along the incline plane and kept moving them until the object was hitting those strands exactly in the beats of the musicians. And that turned out to be the first unit.

The next unit is three times longer than the first, the next unit is five times longer than the first, the next unit is seven times longer than the first, which is exactly what you end up getting with the thing at the top. And that is what caused him to discover it's equal increments in time, not equal increments in distance.

Then he came up with a kind of argument into new sciences arguing if the second one is incoherent. When in fact it's not incoherent though it is problematic. And his argument as Drake says, doesn't really work because it involves a one to one mapping of distances. If you're concerned with that, I think I have an account of why it's a plausible argument in the notes.

But I don't want to spend a lot of time on it now. The point I'm making though is he's using experiment to, on Drake's account, to figure out equal increments in time versus equal increments in speed on an inclined plane. So you've heard the story. All right. And now he’s got the mean speed theorem.

I’ll read it, I’m not gonna go through. Well, I’ll sketch the proof. The time in which a certain space is traversed by a movable and uniformly accelerated movement from rest is equal to the time in which the same space would be traversed by the same movable carried in uniform motion whose degree of speed is one half the maximum and final degree of speed of the previous uniformly accelerated motion.

That's the celebrated mean speed theorem. Oresme had proved it. The Mertonians and Birodan had worked on it. The proof here in effect says, notice what it says. Let line ab represent the time. So, that's the vertical line next to cd. In which the space cd is traversed by a movable and uniformly accelerated motion from c.

Let eb represent the maximum final degree of speed. That's the bottom piece down here. Okay, now the idea is you're compounding increments of time and speed to get distance. Well that's like multiplying along the abscissa and the ordinate are what we would call integrating. Right? So, the argument becomes one of draw this piece through a midsection.

See this triangle and this triangle? They match one another. And his argument, therefore, each increment of speed added on down here, could, instead, have been added on up here. And you get exactly the same distance. And the problem is, he's assuming the capacity to do a one to one mapping from lines in one triangle to lines in another triangle.

And that's not really rigorous. And it was recognized at the time not to be fully rigorous. Though it's a nice, clean geometric argument, as you can see, because the two balance out. Huygens ends up giving you a rigorous argument for this in the 1673 book. It's a double reductio, as a reductio ad absurdum of a reductio ad absurdum, inside of one.

And that's the first rigorous argument. With Calculus of course, it's child's play as you can well realize. That's why there's the one half and one half at squared. It's where the one half is coming from. Now, the important thing is why is this important at all? Why do you need a mean speed thereom?

And the answer, you may have already figured it out. But the important thing is you realize how this kind of math works. We've got a successful account of uniform motion, right? That's the first part of the third day. The math is all there. Thanks to mean speed theorem, we need no new math to be accelerated motion.

Every problem in the uniformly accelerated motion is immediately reducible to an equivalent problem in uniform motion through the mean speed theorem. It's a classic device in mathematics. You have a new realm. If you can do math in the new realm by simply reducing it to the math in a realm you already have, you don't have to worry about rigor, et cetera.

You can just go marching right ahead. Okay, so it's not just a trivial little thing. It's a very deep point to be able to reduce all problems of uniformly accelerated motion to problems of uniform motion through the main speed theorem. Does everybody see that? It's a very, very nice device.

So therefore, if I know the speed in uniform motion, if I know how long it would have to get a certain length, then I know the mean speed of that time, but then I know the final speed if it started from rest, thanks to the mean speed theorem.

And I don't have to worry about instantaneous speeds or all sorts of things, I can bypass. And I hope you all noticed his efforts to avoid Zeno paradoxes with remarks like, well the object doesn't stop at any one spot. It passes through it, et cetera. He's worried about all the classical problems and the conceptual hurdles that they pose, including one of them that I've largely skipped over but I do in the notes.

Why isn't all of the speed gathered instantaneously? You let the object go, it's already going at its full speed. And for that, he drops the object from different heights, showing it does different indentations. And says, therefore you see from the different heights, it's not getting all of its speed at once.

It's building the speed as you go.