Professor Ramamurti
Shankar: This is a first part of the year-long course
introducing you to all the major ideas in physics,
starting from Galileo and Newton right up to the big
revolutions of the last century, which was on relativity and
quantum mechanics. The target audience for this
course is really very broad. In fact, I've always been
surprised at how broad the representation is.
I don't know what your major is; I don't know what you are going
to do later so I picked the topics that all of us in physics
find fascinating. Some may or may not be useful,
but you just don't know. Some of you are probably going
to be doctors and you don't know why I'm going to do special
relativity or quantum mechanics, but you don't know when it will
come in handy. If you're a doctor and you've
got a patient who's running away from you at the speed of light,
you'll know what to do. Or, if you're a pediatrician
with a really small patient who will not sit still,
it's because the laws of quantum mechanics don't allow an
object to have a definite position and momentum.
So these are all things you just don't know when they will
come in handy, and I teach them because these
are the things that turn me on and got me going into physics
and whether or not you go into physics,
you should certainly learn about the biggest and most
interesting revolutions right up to present day physics.
All right. So that's what the subject
matter's going to be, and I'm going to tell you a
little bit about how the course is organized.
First thing is, this year it's going to be
taped. You can see some people in the
back with cameras as part of an experimental pilot program
funded by the Hewlett Foundation and at some point they will
decide what they will do with these lectures.
Most probably they'll post them somewhere so people elsewhere
can have the benefit of what you have sitting in the classroom.
So I've been told that from now on we just ignore the camera and
do business as usual. Nothing's going to be changed.
I tried to negotiate a laugh track so that if the jokes don't
work we can superimpose some laughter.
I was told "no." I just got to deal with it as
it happens. So it's going to be--it's like
one of the reality shows where things are going to be as they
are and hopefully after a while we'll learn to act and behave
normally and not worry about its presence.
Then, coming to the rest of the details of the course.
By the way, there are more details on the website that I
posted, that was given to me by the university,
if you want to know more about what all this is about.
The course organization is fairly simple.
We're going to meet Monday and Wednesday in this room,
11:30-12:45. I will give you some problems
to do on Wednesday and I'll post them on the website.
You guys should get used to going to the class' website.
I'm really, really dependent on that now.
I finally learned how to use it. I will use that to post
information, maybe once in a while send e-mail to the whole
class. If you want to get those
e-mails, you got to sign up for the course because I push a
button and it goes to anybody who's signed up there.
The homework will be given on Wednesday and it's due before
class the following Wednesday. Let me introduce you to our
head TA, Mara Daniel, who's recently Mara Baraban. So Mara's going to be the
person who will see you after class and she will take the
problem sets that you have submitted before class and
she'll give you the graded ones after class.
Okay? That will be sorted up,
it'll be up there. So you should drop the homework
before you come into class, rather than furiously work on
it during class, and the solutions will be
posted the same afternoon. So there is not much point in
giving homework that's late. But once in a while,
you know, you will come up with a reason that I just cannot
argue with. You got married,
you're getting a transplant, whatever it is.
That's fine. You got a transplant,
I want to see the old body part.
You got married, I want to see your spouse.
If something happened to a grandparent, I'm counting.
Up to four I don't get suspicious.
Go five, six, seven, eight,
I will have to look into the family tree.
But, you know, any reasonable excuse will be
entertained. Relative importance given to
these different things, there's 20% for your homework,
30% for the Midterm, which will be sometime in
October, and 50% for the Final. That'll be the weighted average.
But I have another plan called the "Amnesty Plan" in which I
also compare just your final grade,
what you did on the Final exam, and whichever is higher of the
two is what I will take to determine your overall course
grade. This is something I used to
announce near the end but then some people felt that it's not
fair not to know this from the beginning.
So, I'm telling you from the beginning, but don't dream and
think that somehow the Final's going to be so much different
from your regular day-to-day performance,
but to give you some reason to live after the Midterm.
So, you feel there is hope. I can change everything
overnight; it does happen.
I put that in for a reason because sometimes some of you
have not taken a physics course and you don't know how to do
well in physics and slowly you catch on and by the time it's
Final exam you crack the code; you know how to do well.
As far as I'm concerned, that's just fine.
If at the end of the semester you take a three-hour exam in a
closed environment and you answer everything,
I don't care what you did in your homework or your Midterm.
That's not relevant. So that's how the grading will
be done. We have Mara's group of TAs.
She is the head TA and she's the one you should write to
whenever you have a problem. Then we also have two faculty
members. One is a Postdoctoral Fellow,
Mark Caprio. So he will have a discussion
section on Tuesdays between 1:00-2:00 in Sloane Lab.
And Steve Furlanetto--I don't know if Steve is here or not.
There's Steve, our new Assistant Professor.
He will have his section on Tuesday night in Dunham Lab,
in Room 220. Tuesday night is the night when
you people realize homework is due on Wednesday.
So we know that, so he will be there to comfort
you and give you whatever help you need.
All right. My own office hours I've not
determined yet. I will have to find out when it
is good for you. You know, I live and work out
of Sloane Lab up on the hill and it was easy to have office hours
before or after class but now you have to make a special trip.
So, just give me a little bit of time to find out maybe by
soliciting e-mail responses from you what would be a good time
for my office hours. But for any procedural things,
like, you know, this problem set was not graded
properly, and so on, there's no point
e-mailing me because I'm going to send it to Mara anyway.
So directly deal with the powers that be.
Okay, finally I want to give you some tips on how to do well
in this course and what attitude you should have.
First, I advise that you should come to the lectures.
It's not self-serving; it's not so much for my benefit.
I think there is something useful about hearing the subject
presented once orally. Secondly, the book,
you can see, one of you had a book here,
it's about 1,100 pages and when I learned physics it was,
like 300 pages. Now, I look around this room,
I don't see anybody whose head is three times bigger than mine,
so I know that you cannot digest everything the books
have. So I have to take out what I
think is the really essential part and cover them in the
lecture. So, you come to class to find
out what's in and what's not in. If you don't do that,
there's a danger you will learn something you don't have to,
and we don't want that. Okay, so that's why you come to
class. Second thing,
most important thing for doing well in physics,
is to do the homework. The 20% given to the homework
is not a real measure of how important it is.
Homework is when you really figure out how much you know and
don't know. If you watch me do the thing on
the blackboard, it looks very reasonable.
It looks like you can do it but the only way you're going to
find out is when you actually deal with the problem.
That's the only time you're going to find out.
So, I ask you to do the problems as and when they're
posted. So if I post it on Wednesday to
cover the material for that week, then you should attempt it
as quickly as possible because I'm going to assume you have
done the problems when you come for the next few lectures.
And in doing the homework, it is perfectly okay to work in
groups. You don't have to do it by
yourself. That's not how physics is done.
I am now writing a paper with two other people.
They are my experimental colleagues who write papers with
400 other people, maybe even 1,000 other people.
When they do the big collider experiments in Geneva or
Fermilab, collaborations can run into hundreds.
So, it's perfectly okay to be part of a collaboration,
but you've got to make sure that you're pulling your weight.
You've got to make sure that if you explain to others how to do
this problem, then somebody else contributes
to something else, but you know what everybody
contributed in the end. So the game is not just to
somehow or other get the solution to the problem set but
to fully understand how it's done,
and the TAs will be there to help you.
Every day there's going to be a TA in the undergraduate lounge.
I would urge you to use that. That's a beautiful new lounge
that the Provost's Office allowed us to build for
physicists and chemists, or whoever happens to be in the
building. If you go there on the third
floor of Sloane, you may run into other people
like you who are trying to work on problems.
You may run into upper-class students, students who are more
advanced, you will run into your TA.
So that's a good climate. There are coffee machines and
there are lounge sofas and everything else.
There are computers, there are printers,
so it's a good lounge, and I think if you go there one
day a week to do your problem sets,
more often that's a good meeting place,
I recommend that. The final piece of advice,
this is very important so please pay attention to this,
which is, I ask you not to talk to your neighbors during
lecture. Now, this looks like a very
innocuous thing, but you will find out,
it is the only thing that really gets my back up.
Most of the time I don't really care.
I'm really liberal, but this disturbs me because I
am looking at you, I'm trying to see from your
reaction how much of my lecture you are following,
and then it's very distracting when people are talking.
So please don't do that. If you talk,
I am going to assume you are talking about me.
If you laugh, I'm going to assume you are
laughing at me. That's not really what I think,
but that's how disturbing it is when people talk,
and very nice students who do not realize this often disrupt
my line of thinking. So I ask you to keep that to a
minimum. Once in a while you'll have to
talk to your neighbor and say, "Can you please pass me my
pacemaker that fell down?" That's fine.
Then you go back to your business.
But don't do too much of that. Finally, there is this ancient
issue about sleeping in class. Now, my view is,
it's just fine, okay.
I know you guys need the rest and interestingly,
the best sleepers are in the first couple of rows.
I haven't met you guys. It's not personal.
I have found some people really have to come to the first and
second row because they claim that if they don't hear me they
cannot really go to sleep. Now, that was true in Sloane
but I think Luce has got very good acoustics so you can
stretch out in the back. But my only criterion is if you
talk in your sleep, now that's not allowed because
talking is not allowed. Next, if you're going to sleep,
I ask you to sit between two non-sleepers because sometimes
what happens, the whole row will topple over.
We don't want the domino effect. Now, it's going to be captured
on tape and that's going to be really bad for my reputation,
so spread yourself around other people.
All right. So that's it in terms of class,
you know, logistics and everything.
I'm going to start going into the physics proper.
I will try to finish every lecture on time,
but sometimes if I'm in the middle of a sentence or the
middle of a derivation, I may have to go over by a
couple of minutes; there's no need to shuffle your
feet and move stuff around. I know what time it is.
I also want to get out like you guys, but let me finish
something. Other days I may finish a few
minutes before time. That's because the ideas of
physics don't fall into 75-minute segments and sometimes
they spill over a little bit. Also, I'm used to teaching this
course three times a week and now it's suddenly twice a week,
and so things that fell into nice 50-minute units are now
being snipped up different ways so it's pretty difficult.
So, even for me, some of it will be new and the
timing may not be just right. I should tell you first of all
that in this class, the taping is not going to
affect you because the camera is going to be behind your head.
I mentioned to you in the website that this is not the big
opportunity you've been looking for to be a star.
Only the back of your head will be seen.
In some cases, the back of the head could be
more expressive than the front, in which case this is your
opportunity and I wish you luck. But otherwise,
just don't worry about it because you will be only heard.
You may not even be heard. So, I've been asked that if a
question is not very clear, I should repeat it so that
people listening to it later will know what the question was. Let me make one thing very
clear. That is, I'm not in favor of
your talking to each other because you're distracting.
Your stopping me at any time is just fine.
I welcome that because I've seen this subject for God knows
how many years. The only thing that makes it
different for me is the questions that you people have.
You can stop me any time and you should not feel somehow you
are stopping the progress of the class.
There is no fixed syllabus. We can move things around and
it's far more exciting for me to answer your questions than to
have a monologue. So, don't worry about that.
So stop me anytime you don't follow something,
and don't assume that you're not following something because
there's something wrong with your level of comprehension.
Quite often, you guys come up with questions
that never cross my mind, so it's very interesting.
And things we've been repeating year after year after year,
because they sound so reasonable,
suddenly sound unreasonable when some of you point out some
aspect of it that you didn't follow.
So, it could be very interesting for all of us to
have issues to discuss in class, and quite often some questions
are very common and your classmates will be grateful to
you that you brought it up. Otherwise, you know,
TAs get ten e-mails, all with the same question.
Okay. So I'm going to start now.
Anybody have any questions about class?
The format? The Midterm?
The exams? All right.
Yes? Student:
You said there's going to be two hours to be announced.
How do we wait for [inaudible] Professor Ramamurti
Shankar: Oh, you mean my office hours?
Student: No. I thought there was an
[inaudible] Professor Ramamurti
Shankar: No, the discussion sections are
Tuesday afternoon from 1:00-2:00,
and Tuesday night from 8:00-10:00, and the website has
got all the details on when and where.
Yes? Student:
So the lab times will still be [inaudible]
Professor Ramamurti Shankar: Yeah.
There are many, many lab times and you have to
go to the website for the lab. And, by the way,
that reminds me. I've got here lots of flyers
given to me by the director of the laboratories which will tell
you which lab is the right lab for you,
and they're offered many times a week.
Yes? Student:
As far as knowing the material, just from your class,
how important is taking a lab concurrent with this class?
Professor Ramamurti Shankar: I think it's a good
idea to take the lab, particularly in this particular
class because I don't have any demonstrations.
They're all in the other building.
So, this will remind you that physics is, after all,
an experimental science and you will be able to see where all
the laws of physics come from. So, if you're going to take it,
you should take it at the same time.
Yes? Student:
Could you please talk about when you expect [inaudible]
Professor Ramamurti Shankar: Ah,
very good. This is a calculus-based class
and I expect everyone to know at least the rudiments of
differential calculus. What's a function,
what's a derivative, what's a second derivative,
how to take derivatives of elementary functions,
how to do elementary integrals. Sometime later,
I will deal with functions of more than one variable,
which I will briefly introduce to you,
because that may not be a prerequisite but certainly
something you will learn and you may use on and off.
But there are different ways of doing physics.
Mine is to demonstrate over and over how little mathematics you
need to get the job done. There are others who like to
show you how much mathematics you could somehow insinuate into
the process, okay. There are different ways of
playing the game, and some of us find great pride
in finding the most simple way to understand something.
That's certainly my trademark; that's how I do my research
also. So, if you feel there's not
enough math used, I guarantee you that I
certainly know enough eventually to snow the whole class,
but that's not the point. I will use it in moderation and
use it to the best effect possible rather than use it
because it is there. Okay.
So I don't know your mathematical background,
but the textbook has an appendix, which is a reasonable
measure of how much math you should know.
You've got to know your trigonometry,
you've got to know what's a sine and what's a cosine.
You cannot say, "I will look it up."
Your birthday and social security number is what you look
up. Trigonometry functions you know
all the time. Okay.
I will ask you, and you do. All right.
And of course, there's trigonometric
identities you know from high school.
Pages and pages of them, so no one expects you to know
all those identities, but there are a few popular
ones we will use. All right.
Anything else? Yes?
Student: This may be a bit early,
but when will we be having our Midterm?
Professor Ramamurti Shankar: Yeah.
Midterm will be sometime around 20th of October.
I have to find out exactly the right time.
We have 24 lectures for this class and the first 12 roughly
will be part of the Midterm, but after the 12th lecture I
may wait a week so that you have time to do the problems and get
the solutions. Then I will give you the
Midterm. Yes?
Student: If wanting one of the two lab
courses, which one do you recommend?
Professor Ramamurti Shankar: Yeah,
this tells you in detail. This flyer answers exactly that.
Okay, there was one more question from somebody?
Yes? Student:
A few people I've talked to have recommended that we start
taking the lab second semester instead of first.
Would that be advisable or should we take both
concurrently? Professor Ramamurti
Shankar: I don't have a strong view.
I think you should take the lab sometime but I don't know how
many semesters that you have to take.
But I would say the advice of your predecessors is very
important. If they tell you this is what
works, that's better than what somebody like me can tell you.
Also, you should talk to Stephen Irons,
who is the director of the labs.
He has seen every possible situation.
He will give you good advice. Let's start now.
Okay. So we are going to be studying
in the beginning what's called Newtonian mechanics.
It's pretty remarkable that the whole edifice is set up by just
one person – Newton -- and he sent us on the road to
understanding all the natural phenomena until the year
18-hundred-and-something when Maxwell invented the laws of
electromagnetism and wrote down the famous Maxwell equations.
Except for electromagnetism, the basics of mechanics,
which is the motion of billiard balls and trucks and marbles and
whatnot, was set up by Newton. So that's what we are going to
focus on, and you will find out that the laws of physics for
this entire semester certainly can be written on one of those
blackboards or even half of those blackboards.
And the purpose of this course is to show you over and over and
over again that starting with those one or two laws,
you can deduce everything, and I would encourage you to
think the same way. In fact, I would encourage you
to think the way physicists do, even if you don't plan to be a
physicist, because that's the easiest way
to do this subject, and that is to follow the
reasoning behind everything I give you.
And my purpose will be not to say something as a postulate,
but to show you where everything comes from,
and it's best for you if you try to follow the logic.
That way, you don't have to store too many things in your
head. In the early days when there
are four or five formulas, you could memorize all of them
and you can try each one of them until something works,
but after a couple of weeks you will have a hundred formulas and
you cannot memorize all of them. You cannot resort to trial and
error. So you have to know the logic.
So the logical way is not just the way the physicists do it,
it's the easier way to do it. If there is another way that it
will work for non-physicists, I won't hesitate to teach it to
you that way if that turns out to be the best way.
So try to follow the logic of everything.
Okay. So, Newtonian mechanics is our
first topic. So, Newtonian mechanics has two
parts. All of physics is a two-part
program. The plan, every time,
is to predict the future given the present.
That's what we always do. When we do that right,
we are satisfied. So the question is,
"What do you mean by ‘predict the future?'"
What do you mean by the future? What do you mean by the present?
By "present," we mean--we will pick some part of the universe
we want to study and we will ask,
"What information do I need to know for that system at the
initial time, like,
right now, in order to be able to predict the future?"
So, for example, if you were trying to study the
motion of some object, here is one example.
[throws a piece of candy for someone to catch]
Professor Ramamurti Shankar: See,
that's an example of Newtonian mechanics.
I'll give you one more demonstration.
Let's see who can catch this one.
[throws another piece] Professor Ramamurti
Shankar: That's a good example.
So, that was Newtonian mechanics at work,
because what did I do? I released a piece of candy,
threw it from my hand, and the initial conditions have
to do with where did I release it and with what velocity.
That's what he sees with his eyes.
Then that's all you really need to know.
Then he knows it's going to go up, it's going to curve,
follow some kind of parabola, then his hands go there to
receive it. That is verification of a
prediction. His prediction was,
the candy's going to land here, then he put his hand there.
He also knew where the candy was going to land,
but he couldn't get his hand there in time.
But we can always make predictions.
But this is a good example of what you need to know.
What is it you have to know about this object that was
thrown, I claim, is the initial location of the
object and the initial velocity. The fact that it was blue or
red is not relevant, and if I threw a gorilla at him
it doesn't matter what the color of the gorilla is,
what mood it is in. These are things we don't deal
with in physics. There is a tall building,
a standard physics problem. An object falls off a tall
building. Object could be a person.
So we don't ask why is this guy ending it all today?
We don't know, and we cannot deal with that.
So we don't answer everything. We just want to know when he's
going to hit the pavement, and with what speed.
So we ask very limited questions, which is why we brag
about how accurately we can predict the future.
So, we only ask limited goals and we are really successful in
satisfying them. So, we are basically dealing
with inanimate objects. So the product of Newtonian
mechanics of predicting the future given the present,
has got two parts, and one is called kinematics
and the other is called dynamics.
So, kinematics is a complete description of the present.
It's a list of what you have to know about a system right now.
For example, if you're talking about the
chalk--if I throw the chalk, you will have to know where it
is and how fast it's moving. Dynamics then tells you why the
object goes up, why the object goes down and
why is it pulled down and so on. That's dynamics.
The reason it comes down is gravity is pulling it.
In kinematics, you don't ask the reason behind
anything. You simply want to describe
things the way they are and then dynamics tells you how they
changed and why they changed. So, I'm going to illustrate the
idea of kinematics by taking the simplest possible example.
That's going to be the way I'm going to do everything in this
course. I'm going to start with the
simplest example and slowly add on bells and whistles and make
it more and more complicated. So, some of you might say,
"Well, I've seen this before," so maybe there is nothing new
here. That may well be.
I don't know how much you've seen, but quite often the way
you learned physics earlier on in high school is probably
different from the way professional physicists think
about it. The sense of values we have,
the things that we get excited about are different,
and the problems may be more difficult.
But I want to start in every example, in every situation that
I explain to you, with the simplest example,
and slowly add on things. So, what we are going to study
now is a non-living object and we're going to pick it to be a
mathematical point. So the object is a mathematical
point. It has no size.
If you rotate it, you won't know.
It's not like a potato. You take a potato,
you turn it around, it looks different.
So, it's not enough to say the potato is here.
You've got to say which way the nose is pointing and so on.
So, we don't want to deal with that now.
That comes later when we study what we call "rigid bodies".
Right now, we want to study an entity which has no spatial
extent. So just a dot,
and the dot can move around all over space.
So we're going to simplify that too.
We're going to take an entity that lives along the x
axis. [draws a line with integrals]
It moves along a line. So you can imagine a bead with
a wire going through it and the bead can only slide back and
forth. So, this is about the simplest
thing. I cannot reduce the number of
dimensions. One is the lowest dimension.
I cannot make the object simpler than being just a
mathematical point. Then, you've got to say,
"What do I have to know about this object at the initial time?
What constitutes the present, or what constitutes maximal
information about the present?" So what we do is we pick an
origin, call it zero, we put some markers there to
measure distance, and we say this guy is sitting
at 1,2, 3,4, 5. He is sitting at x = 5.
Now, of course, we've got to have units and the
units for lengths are going to be meters.
The unit for time will be a second, and time will be
measured in seconds. Then we'll come to other units.
Right now, in kinematics, this is all you need.
Now, there are some tricky problems in the book.
Sometimes they give you the speed in miles per hour,
kilometers per year, pounds per square foot,
whatever it is. You've got to learn to
transform them, but I won't do them.
I think that's pretty elementary stuff.
But sometimes I might not write the units but I've earned the
right to do that and you guys haven't so you'll have to keep
track of your units. Everything's got to be in the
right units. If you don't have the units,
then if you say the answer is 19, then we don't know what it
means. Okay.
So here's an object. At a given instant,
it's got a location. So what we would like to do is
to describe what the object does by drawing a graph of time
versus space and the graph would be something like this.
You've got to learn how to read this graph.
I'm assuming everyone knows how to read it.
[draws a graph of x versus t]
This doesn't mean the object is bobbing up and down.
I hope you realize that. Even though the graph is going
up and down, the object is moving from left to right.
So, for example, when it does this,
it's crossed the origin and is going to the left of the origin.
Now, at the left of the origin, it turns around and starts
coming to the origin and going to the right.
That is x versus t.
So, in the language of calculus, x is a function
of time and this is a particular function.
This function doesn't have a name.
There are other functions which have a name.
For example, this is x = t,
x = t^(2), you're going to have x = sin
t and cos t and log t.
So some functions have a name, some functions don't have a
name. What a particle tries to do
generally is some crazy thing which doesn't have a name,
but it's a function x (t).
So you should know when you look at a graph like this what
it's doing. So, the two most elementary
ideas you learn are what is the average velocity of an object,
as then ordered by the symbol v-bar.
So, the average is found by taking two instants in time,
say t_1 and later t_2,
and you find out where it was at t_2 minus
where it was at t_1 and divide
by the time. So, the average velocity may
not tell you the whole story. For example,
if you started here and you did all this and you came back here,
the average velocity would be zero, because you start and end
at the same value of x, you get something;
0 over time will still be 0. So you cannot tell from the
average everything that happened because another way to get the
same 0 is to just not move at all.
So the average is what it is. It's an average,
it doesn't give you enough detail.
So it's useful to have the average velocity.
It's useful to have the average acceleration,
which you can find by taking similar differences of
velocities. But before you even do that,
I want to define for you an important concept,
which is the velocity at a given time, v (t).
So this is the central idea of calculus, right?
I am hoping that if you learned your calculus,
you learned about derivatives and so on by looking at x
versus t. So, I will remind you,
again, this is not a course in calculus.
I don't have to do it in any detail.
I will draw the famous picture of some particle moving and it's
here at t of some value of x.
A little later, which is t + Δt.
So Δt is going to stand always for a small finite
integral of time; infinitesimal interval of time
not yet 0. So, during that time,
the particle has gone from here to there, that is x +
Δx, and the average velocity in that interval is
Δ x/ Δt. Graphically,
this guy is Δ x and this guy is Δt,
and Δx over Δt is a ratio. So in calculus,
what you want to do is to get the notion of the velocity right
now. We all have an intuitive notion
of velocity right now. When you're driving in your
car, there's a needle and the needle says 60;
that's your velocity at this instant.
It's very interesting because velocity seems to require two
different times to define it -- the initial time and the final
time. And yet, you want to talk about
the velocity right now. That is the whole triumph of
calculus is to know that by looking at the position now,
the position slightly later and taking the ratio and bringing
later as close as possible to right now,
we define a quantity that we can say is the velocity at this
instant. So v of t,
v(t) is the limit, Δt goes to 0 of
Δx over Δt and we use the symbol dx/dt
for velocity. So technically,
if you ask what does the velocity stand for--Let me draw
a general situation. If a particle goes from here to
here, Δx over Δt, I don't know how
well you can see it in this figure here,
is the slope of a straight line connecting these two points,
and as the points come closer and closer,
the straight line would become tangent to the curve.
So the velocity at any part of the curve is tangent to the
curve at that point. The tangent of,
this angle, this θ, is then Δx over Δt.
Okay, once you can take one derivative,
you can take any number of derivatives and the derivative
of the velocity is called the acceleration,
and we write it as the second derivative of position.
So I'm hoping you guys are comfortable with the notion of
taking one or two or any number of derivatives.
Interestingly, the first two derivatives have
a name. The first one is velocity,
the second one is acceleration. The third derivative,
unfortunately, was never given a name,
and I don't know why. I think the main reason is that
there are no equations that involve the third derivative
explicitly. F = ma.
The a is this fellow here, and nothing else is given
an independent name. Of course, you can take a
function and take derivatives any number of times.
So you are supposed to know, for example,
if x(t) is t^(n), you're supposed to know
dx/dt is nt^(n-1). Then you're supposed to know
derivatives of simple functions like sines and cosines.
So if you don't know that then, of course, you have to work
harder than other people. If you know that,
that may be enough for quite some time. Okay, so what I've said so far
is, a particle moving in time from point to point can be
represented by a graph, x versus t.
At any point on the graph you can take the derivative,
which will be tangent to the curve at each point,
and its numerical value will be what you can call the
instantaneous velocity of that point and you can take the
derivative over the derivative and call it the acceleration.
So, we are going to specialize to a very limited class of
problems in the rest of this class.
A limited class of problems is one in which the acceleration is
just a constant. Now, that is not the most
general thing, but I'm sure you guys have some
idea of why we are interested in that.
Does anybody know why so much time is spent on that?
Yes? Student:
[inaudible] Professor Ramamurti
Shankar: Pardon me? Student:
[inaudible] Professor Ramamurti
Shankar: Right. The most famous example is that
when things fall near the surface of the Earth,
they all have the same acceleration,
and the acceleration that's constant is called g,
and that's 9.8 meters/second^(2).
So that's a very typical problem.
When you're falling to the surface of the Earth,
you are describing a problem of constant acceleration.
That's why there's a lot of emphasis on sharpening your
teeth by doing this class of problems.
So, the question we are going to ask is the following,
"If I tell you that a particle has a constant acceleration
a, can you tell me what the
position x is?" Normally, I will give you a
function and tell you to take any number of derivatives.
That's very easy. This is the backwards problem.
You're only given the particle has acceleration a,
and you are asked to find out what is x?
In other words, your job is to guess a function
whose second derivative is a,
and this is called integration, which is the opposite of
differentiation, and integration is just
guessing. Integration is not an
algorithmic process like differentiation.
If I give you a function, you know how to take the
derivative. Change the independent
variable, find the change in the function, take the ratio and
that's the derivative. The opposite is being asked
here. I tell you something about the
second derivative of a function and ask you what is the
function. The way we do that is we guess,
and the guessing has been going on for 300 years,
so we sort of know how to guess.
So, let me think aloud and ask how I will guess in this
problem. I would say,
okay, this guy wants me to find a function which reduces to the
number a when I take two derivatives,
and I know somewhere here, this result,
which says that when I take a derivative,
I lose a power of t. In the end, I don't want any
powers of t. It's very clear I've got to
start with a function that looks like t^(2).
This way when I take two derivatives, there will be no
t left. Well, unfortunately,
we know this is not the right answer, because if you take the
first derivative, I get 2t.
If I take the second derivative I get 2, but I want to get
a and not 2. Then it's very clear the way
you patch it up is you multiply it by this constant and now
we're all set. This function will have the
right second derivative. So, this certainly describes a
particle whose acceleration is a.
The a is not dependent on time.
But the question is, is this the most general
answer, or is it just one answer, and I think you all know
that this is not the most general answer.
It is one answer. But I can add to this some
number, like 96, that'll still have the property
that if you take two derivatives,
you're going to get the same acceleration.
So 96 now is a typical constant, so I'm going to give
the name c to that constant.
Everyone knows from calculus that if you're trying to find a
function about which you know only the derivative,
you can always add a constant to one person's answer without
changing anything. But I think here,
you know you can do more, right?
You can add something else to the answer without invalidating
it, and that is anything with one power of t in it, because if you take one
derivative it'll survive, but if you take two
derivatives, it'll get wiped out.
Now, it's not obvious but it is true that you cannot add to this
anymore. The basic idea in solving these
equations and integrating is you find one answer,
so then when you take enough derivatives, the function does
what it's supposed to do. But then having found one
answer, you can add to it anything that gets killed by the
act of taking derivatives. If you're taking only one
derivative you can add a constant.
If you're taking two derivatives you can add a
constant and something linear in t..
If you knew only the third derivative of the function,
you can have something quadratic in t without
changing the outcome. So, this is the most general
position for a particle of constant acceleration,
a. Now, you must remember that this
describes a particle going side to side.
I can also describe a particle going up and down.
If I do that, I would like to call the
coordinate y, then I will write the same
thing. You've got to realize that in
calculus, the symbols that you call x and y are
completely arbitrary. If you know the second
derivative of y to be a, then the answer looks
like this. If you knew the second
derivative of x, the answer looks like that.
Now, we have to ask what are these numbers,
b and c. So let me go back now to this
expression, x(t) = at^(2)/ 2 + c +
bt. It is true mathematically,
you can add two numbers, but you've got to ask yourself,
"What am I doing as a physicist when I add these two numbers?"
What am I supposed to do with a and b?
I mean, with the b and c?
What value should I pick? The answer is that simply
knowing the particle has an acceleration is not enough to
tell you where the particle will be.
For example, let's take the case where the
particle is falling under gravity.
Then you guys know, you just told me,
acceleration is -9.8, my g is -9.8.
We call it "minus" because it's accelerating down and up was
taken to be the positive direction.
In that case, y(t) will be
-1/2gt^(2) + c + bt. So, the point is,
every object falling under gravity is given by the same
formula, but there are many, many objects that can have many
histories, all falling under gravity, and what's different
from one object and the other object is,
when was it dropped, from what height,
and with what initial speed. That's what these numbers are
going to tell us and we can verify that as follows.
If you want to know what the number c is,
you say, let's put time t = 0. In fact,
let me go back to this equation here.
You'll put time t = 0, x(0) doesn't
have this term, doesn't have this term,
and it is c. So I realize that the constant,
c, is the initial location of the object,
and it's very common to denote that by x_0. So
the meaning of the constant c is where was the object at the
initial time? It could've been anywhere.
Simply knowing the acceleration is not enough to tell you where
it was at the initial time. You get to pick where it was at
the initial time. Then, to find the meaning of
b, we take one derivative of this, dx/dt,
that's velocity as a function of time, and if you took the
derivative of this guy, you will find as at + b.
That's the velocity of the object.
Then, you can then understand that v(0) is what
b is, which we write as v_0. Okay,
so the final answer is that x(t) looks like
x_0 + v_0 t + 1/2 at^(2). Okay.
So what I'm saying here is we are specializing to a limited
class of motion where the particle has a definite
acceleration, a.
Then, in every situation where the body has an acceleration
a, the location has to have this form,
where this number (x_0) is where
it was initially, this (v_0 )
was the initial velocity of the object.
So, when I threw that thing up and you caught it,
what you are doing mentally was immediately figuring out where
it started and at what speed. That was your initial data.
Then in your mind, without realizing it,
you found the trajectory at all future times.
Now, there is one other celebrated formula that goes
with this. I'm going to find that,
then I'll give you an example. Now, I'm fully aware that this
is not the flashiest example in physics, but I'm not worried
about that right now. You'll see enough things that
will confound you, but right now I want to
demonstrate a simple paradigm of what it means to know the
present and what it means to say this is what the future behavior
will be. We want to do that in the
simplest context, then we can make the example
more and more complicated, but the phenomenon will be the
same. So, what we have found out so
far, I'm purposely going from x to y because I
want you to know that the unknown variable can be called
an x or can be called a y.
It doesn't matter, as long as the second
derivative is a; that's the answer.
Now there's a second formula one derives from this.
You guys probably know that too from your days at the daycare,
but I want to derive the formula and put it up,
then we'll see how to use it. Second formula tries to relate
the final velocity of some time, t, to the initial
velocity and the distance traveled with no reference to
time. So the trick is to eliminate
time from this equation. So let's see how we can
eliminate time. You know that if you took a
derivative of this, you will find v(t) is
v_0 + at. What that means is,
if you know the velocity of the given time and you know the
initial velocity, you know what time it is.
The time, in fact, is v - v_0
over a. If I don't show you any
argument for v, it means v at time
t and the subscript of 0 means t is zero.
So what this says is, you can measure time by having
your own clock. A clock tells you what time it
is, but you can also say what time it is by seeing how fast
the particle is moving because you know it started with some
speed. It's gaining speed at some rate
a. So, if the speed was so and so
now, then the time had to be this.
So time can be indirectly inferred from these quantities.
Then you take that formula here (t) and you put it here,
(y(t)) to see a times t,
you put this expression. So what will you get?
We'll get an expression in which there is no t;
t has been banished in favor of v.
So, I'm not going to waste your time by asking what happens if
you put it in. I will just tell you want
happens. What happens is,
you will find that v^(2) = v_o^(2) + 2a times
(y- y_0). [Note: The Professor said x
when he meant y] How many people have seen this
thing before? Okay.
That's a lot. Look, I know you've seen this.
At the moment, I have to go through some of
the more standard material before we go to the more
non-standard material. If this part's very easy for
you, there's not much I can do right now.
So let me draw a box. Drawing a box to you guys means
important. These are the two important
things. Remember, I want you to
understand one thing. How much of this should you
memorize? Suppose you've never seen this
in high school. How much are you supposed to
memorize? I would say,
keep that to a minimum, because what the first formula
tells you should be so intuitive that you don't have to cram
this. We are talking about particles
of constant acceleration. That means, when I take two
derivatives, I want to get a, then you should know
enough calculus to know it has to be something like
at^(2), and half comes from taking two
derivatives. The other two you know are
stuff you can add, and you know where you're
adding those things, because the particle has a head
start. It's got an initial position.
Even at = 0, and it has an initial velocity,
so even without any acceleration,
it will be moving from y^(0) to y^(0) + vt.
The acceleration gives you an extra stuff, quadratic in time.
Once you've got that, one derivative will give you
the velocity, then in a crunch you can
eliminate t and put it into this formula.
But most people end up memorizing these two because you
use it so many times. It eventually sticks in you but
you shouldn't try to memorize everything.
So, we are now going to do one standard problem where we will
convince ourselves we can apply this formulae and predict the
future given the present. So the problem I want to
do--there are many things you could do but I just picked one,
and this is the one with round numbers so I can do it without a
calculator. Here's the problem.
There is this building and it's going to be 15 meters high,
and I'm going to throw something and it's going to go
up and come down. It's something I throw up has
an initial speed of 10 meters per second. So we have to ask now,
now that my claim is, you can ask me any question you
want about this particle and I can answer you.
You can ask me where it will be nine seconds from now,
eight seconds from now, how fast will it be moving.
I can answer anything at all. But what I needed to do this
problem was to find these two unknowns.
So, you've got to get used to the notion of what will be given
in general and what is tailor-made to the occasion.
So, we know in this example the initial height should be 15
meters and the initial velocity should be 10,
and for acceleration, I'm going to use -g and
to keep life simple, I'm going to call it -10.
As you know, the correct answer is 9.8,
but we don't want to use the calculator now so we'll call it
-10. Consequently,
for this object the position y, at any time t
is known to be 15 + 10t - 5t^(2). That is the full
story of this object. Of course, you've got to be a
little careful when you use it. For example,
let's put t equal to 10,000 years.
What are you going to get? When t is equal to
10,000 years or 10000 seconds, you're going to find y
is some huge negative number. You know, that's not right,
what's wrong with that reasoning?
Student: [inaudible]
Professor Ramamurti Shankar: So you cannot use
the formula once it hits the ground because once it hits the
ground, the fundamental premise that
a was a constant of -9.8 or -10 is wrong.
So that's another thing to remember.
Once you get a formula, you've got to always remember
the terms under which the formula was derived.
If you blindly use it beyond its validity,
you will get results which don't make any sense.
Conversely, if you get an answer and it doesn't seem to
make sense, then you've got to go back and ask,
am I violating some of the assumptions, and here you will
find the assumption that the particle had that acceleration
a is true as long it's freely falling under gravity but
not when you hit the ground. Now, if you dug a hole here
until there, and of course it may work until that happens,
okay. But you've got them every time.
This is so obvious in this problem, but when you see more
complicated formula, you may not know all the
assumptions that went into the derivation and quite often you
will be using it when you shouldn't.
All right. See, this you agree,
is a complete solution to this miniature, tiny,
Mickey-Mouse problem. You give me the time and I'll
tell you where it is. If you want to know how fast
it's moving at a given time, if you want to know the
velocity, I just take the derivative of
this answer, which is 10 - 10t.
So let me pick a couple of trivial questions one can ask.
One can ask the following question.
How high does it go? How high will it rise?
To what height will it rise? So, we know it's going to go up
and turn around and come down. We're trying to see how high
that is. So, that is a tricky problem to
begin with because if you take this formula here,
it tells you y if you know t,
but no, we're not saying that. We don't know the time and we
don't know how high it's rising so you can ask,
"How am I supposed to deal with this problem?"
Then you put something else that you know in your mind,
which is that the highest point is the point when it's neither
going up nor coming down. If it's going up,
that's not the highest point. If it's coming down,
that's not the highest point. So at the highest point it
cannot go up and it cannot go down.
That's the point where velocity is 0.
If you do that, let's call the particular time
t*, then 10t* - 10 = 0, or t*
is 1 second. So we know that it'll go up for
one second then it will turn around and come back.
Now, we are done because now we can ask how high does it go,
and you go back to your, and y (1) is 15 + 10 -
5, which is what? Twenty meters.
By the way, you will find that I make quite a lot of mistakes
on the blackboard. You're going to find out,
you know, one of these years when you start teaching that
when you get really close to a blackboard, you just cannot
think. There's definitely some inverse
correlation between your level of thinking and the proximity to
the blackboard. So if you find me making a
mistake, you've got to stop me. Why do you stop me?
For two reasons. First of all,
I'm very pleased when this happens, because I'm pretty
confident that I can do this under duress,
but I may not do it right every time.
But if my students can catch me making a mistake,
it means they are following it and they are not hesitating to
tell me. Secondly, as we go to the more
advanced part of the course, we'll take a result from this
part of the blackboard, stick it into the second part
and keep manipulating, so if I screwed up in the
beginning and you guys keep quiet,
we'll have to do the whole thing again.
I would ask you when you follow this thing to do it actively.
Try to be one step ahead of me. For example,
if I'm struck by lightning, can you do anything?
Can you guess what I'm going to say next?
Do you have any idea where this is going?
You should have a clue. If I die and you stop,
that's not a good sign, okay.
You've got to keep going a little further because you
should follow the logic. So, for example,
you know, I'm going to calculate next when it hits the
ground. You should have some idea of
how I'll do it. But this is not a spectator
sport. If you just watch me,
you're going to learn nothing. It's like watching the U.S.
Open and thinking you're some kind of a player.
You will have to shed the tears and you've got to bang your head
on the wall and go through your own private struggle.
I cannot do that for you. I cannot even make it look hard
because I have memorized this problem from childhood,
so there is no way I can make this look difficult.
That's your job. All right.
So, we know this point at one second is 20 meters,
so let's just ask one other question and we'll stop.
One other question may be, "When does it hit the ground
and at what speed?" -- a typical physics question.
So when does it hit the ground? Well, I think you must know now
how to formulate that question. "When does it hit the ground"
is "When is y = 0"?
By the way, I didn't tell you this but I think you know that I
picked my origin to be here and measured y positively to
be upwards and I called that 15 meters.
You can call that your origin. If you call that your origin,
your y_0 will be 0, but ground will be called
-15. So, in the end,
the physics is the same but the numbers describing it can be
different. We have to interpret the data
differently. But the standard origin for
everybody is the foot of the building.
You can pick your origin here, some crazy spot.
It doesn't matter. But some origins are more equal
than others because there is some natural landmark there.
Here, the foot of the building is what I call the origin.
So, in that notation, I want to ask,
when is y = 0? I ask when y = 0,
then I say 0 = 15 + 10t - 5t^(2).
Or I'm canceling the 5 everywhere and changing the sign
here I get t^(2) - 2t - 3 = 0.
That's when it hits the ground. So let's find out what the time
is. So t is then 2 + or -
or + 12 over 2, which is 2 + or - 4 over
2, which is -1 or 3. Okay, so you get two answers
when it hits the ground. So it's clear that we should
pick 3. But you can ask,
"Why is it giving me a second solution?"
Anybody have an idea why? Student:
Because there was an entire parabola [inaudible]
Professor Ramamurti Shankar: That's correct.
So her answer was, if it was a full parabola,
then we know it would've been at the ground before I set my
clock to 0. First of all,
negative time should not bother anybody;
t = 0 is when I set the clock, I measured time forward,
but yesterday would be t = -1 day,
right? So we don't have any trouble
with negative times. So the point is,
this equation, it does not know about the
building. Doesn't know the whole song and
dance that you went to a building and you threw up a rock
or anything. What does the mathematics know?
It knows that this particle happened to have a height of 15,
a time 0, and a velocity of 10, a time 0, and it is falling
under gravity with an acceleration of -10.
That's all it knows. If that's all it knows,
then in that scenario there is no building or anything else;
it continues a trajectory both forward in time and backward in
time, and it says that whatever seconds,
one second before you set your clock to 0, it would've been on
the ground. What it means is if you'd
release a rock at that location one second before with a certain
speed that we can calculate, it would've ended up here with
precisely the position and velocity it had at the beginning
of our experiment. So sometimes the extra solution
is very interesting and you should always listen to the
mathematics when you get extra solutions.
In fact, when a very famous physicist, Paul Dirac,
was looking for the energy of a particle in relativistic quantum
mechanics, he found the energy of a
particle is connected to its momentum, this p is what
we call momentum, and its mass by this relation.
It's a particle of mass m and momentum p
has this energy so you solve for the energy, you get two answers. Now, your temptation is to keep
the first answer because you know energy is not going to be
negative. Particle's moving,
it's got some energy and that's it.
But the mathematicians told Dirac, "You cannot ignore the
negative energy solution because it tells you there's a second
solution and you cannot throw them out,"
and it turns out the second solution, with negative energy,
was when the theory is telling you,
hey, there are particles and there are anti-particles,
and the negative energy when properly interpreted will
describe anti-particles. So the equations are very smart.
The way the physics works is you will find some laws of
motion in mathematical form, you put in the initial
conditions of whatever, you solve the equations,
and the answer that comes, you have no choice.
You have to accept the answer, but there are new answers
besides the one you were looking for.
You've got to think about what they mean, and that's one of the
best things about physics because here's a person who is
not looking for anti-particles. He was trying to describe
electrons, but the theory said there are two roots in the
quadratic equation and the second root is mathematically as
interesting as the first one. It has to be part of a theory,
and then trying to adjust it so it can be incorporated,
you discover anti-particles. So always amazing to us how we
go into the problem, our eye or mind can see one
class of solutions, but the math will tell you
sometimes there are new solutions and you've got to
respect it and understand and interpret the unwanted
solutions, and this is a simple example
where you can follow what the meaning of the second solution
is. It means that to the problem
you pose, there's more than the answers that you could imagine.
Here it meant particle that was released from the ground
earlier. There it meant something much
more interesting, mainly anti-particles
accompanying particles. They are going to accompany
particles surely as every quadratic equation has two
solutions. All right, so now in this
problem, we can do something slightly different,
and let's use this expression here,
and I will do that, then I'll stop for today.
If you were asking questions, like, how high does it go,
but you don't ask when does it go to the highest point,
then you don't have to go through the whole process of
finding the time at which it turned around.
I don't know where that is, that disappeared on the
blackboard, then putting the time equal to 1 second into this
formula. If the question of time is not
explicitly brought up, then you should know that you
have to use this formula. So how do we get it here?
Well, we say at the top of the loop, when it goes up and comes
down the velocity is 0. Therefore, you say 0^(2)
= initial velocity^(2) + 2 times -g,
that's my acceleration, times y - y_0.
If you solve for that, you find y - y_0 =
v_0^(2) over 2g,
and if you put in the v_0 I gave you,
which was what, 10?
That's 100 over 20, which is 5 meters.
So y = y_0 + 5 meters, and that was the height
to which it rises. I think we got it somewhere
else. We found the maximum height to
be 20 meters. Another thing you can do is you
can find the speed here. If you want to find the speed
there, you put the equation v^(2) = v_0^(2)
+ 2 times -g (y - y_0).
What is y - y_0?
The final y is 0, the initial y is
15. You solve for that equation and you will find the
final velocity. So, if time is not involved,
you can do it that way. I want to derive the last
result in another way, then I will stop,
and that's pretty interesting because it tells you the use and
abuse of calculus. So I'm going to find for you
this result using calculus in a different way.
So, from the calculus we know dv/dt = a.
Now, multiply both sides by v. Now you have to know from
elementary calculus that v times dv/dt is
really d by dt of v^(2) over 2.
Now, I hope you guys know that much calculus,
that when you take a derivative of a function of a function,
namely v^(2) over 2 is a function of v,
and v itself is a function of t,
then the rule for taking the derivative is first take the
v derivative of this object,
then take the d by dt of t,
which is this one. On the right-hand side,
I'm going to write as a dx/dt.
This much is standard. I'm going to do something which
somehow we are told never, ever to do, which is to just
cancel the dts. You all know that when you do
dy/dx, you're not supposed to cancel
that d. That's actually correct.
You don't want to cancel the d in the derivative.
But this happens to be completely legitimate,
so I'm going to assume it's true and I'll maybe take a
second and explain why it's legitimate.
What this really means is in a given time, Δt,
the change in this quantity is a times the change in
this quantity. Therefore, you can multiply
both sides by the Δt, but the only thing you should
understand is Δt, as long as it's small and
finite, will lead to some small infinite errors in the formula,
because the formula is really the limit in which Δx
and Δt both go to 0. So what you have to do is
multiply both sides by Δt, but remember it's
got to be in the end made to be vanishingly small.
As long as we understand that, we can do this cancellation and
this says on the left-hand side the change in the quantity
v^(2) over 2 is a times the change in the quantity
x. So add up all the changes or what you mean by
integral. Same thing.
Add up all the changes. The change in v^(2) over
2 will be the final v^(2) over 2 - the initial
v^(2) over 2 and the other side will be a
times change in x; x - x^(0) and that's the
formula I wrote for you: v^(2) is
v_0^(2) + 2a (x - x^(0)). So,
the point is whenever you have derivatives with something over
dt, do not hesitate to cancel the dts and think
of them as Δv^(2) over 2 is equal to a times
Δ of x. This will be actually true as long as both
quantities are vanishingly small.
They will become more and more true as Δx and
Δv^(2) become vanishingly small,
in the limit in which they are approaching 0,
the two will be, in fact, equal.
If Δx is a finite amount, like 1 second,
this will not be true because in the starting equation,
Δx and Δt and Δv^(2) were all assumed
to be infinitesimal. So don't hesitate to do
manipulations of this type, and I will do them quite often.
So you've got to understand when it's okay and when it's not
okay. What this means is,
in a time Δt, this quantity changes by some
amount, and in the same time,
Δt, that quantity changes by some amount,
then keeping the Δt equal to some number we may
equate the changes of the two quantities,
provided it is understood that Δv^(2) over 2 is a
change in v^(2) over 2 in the same time in which the
particle moved a distance, Δx.
Adding the differences, we eliminate time and we get
this final result. All right.
So if you go to your website today, you will find I've
assigned some problems and you should try to do them.
They apply to this chapter. Then next week we'll do more
complicated problems that involve motion in higher
dimensions, how to go to two dimensions or three dimensions.
