My first chat.

John Rennie

Sep 8, 2016 08:25

@KaumudiHarikumar: hi :-)

user228700

@JohnRennie: Hello :) Firstly, sir, I should tell you that I'm not familiar with the math involved in the probability distribution curve.

John Rennie

Everyone finds probability density functions hard to understand at first.

But they are used in lots of areas of physics so you'll get used to them eventually :-)

user228700

Okay, that's comforting :P My major doubt comes from this whole area being unity thing...

John Rennie

If we go back to the histogram I mentioned in my answer, are you happy with my explanation of what that means?

i.e. each column is the fraction of molecules with velocities in some range.

user228700

I'm so sorry but I still don't get that.

John Rennie

Sep 8, 2016 08:31

Well suppose we have 1000 molecules, all with different velocities.

user228700

Okay, starting from the basics, can you tell what it is that we're plotting on the y-axis?

John Rennie

The $y$ axis is the number of molecules in each velocty range dividied by the total number of molecules.

Let's try and take a concrete example.

Suppose the velocities are all the range 0 to 100 m/s

user228700

So...the probability of finding a given molecule in that range?

John Rennie

Yes

user228700

Okay, go on sir...

John Rennie

Sep 8, 2016 08:33

Suppose my first column is for velocities 0 to 10 m/sec

Suppose I count the number of molecules with velocities in the range, and get the number 15. So 15 out of the thousand molecules have velocities in the range 0 to 10 m/sec.

Then the height of the column, the $y$ value, is 15/1000 = 0.015.

So far so good?

user228700

Yes, OK...

John Rennie

The net column would be velocities in the range 10 to 20 m/sec, the next column would be 20 to 30 m/sec and so on giving me ten columns covering the while velocity range of 0 to 100 m/sec.

And all the 1000 molecules I started with are counted in one of these ten columns.

user228700

Yup, OK...

John Rennie

For each column the $y$ value is the number of molecules in that column divided by the total number of molecules, so if I add together all ten $y$ values I will get the number one.

Still OK?

user228700

Yeah, that's exactly where the confusion comes from!

user228700

Sep 8, 2016 08:40

That's why I kept asking you about just adding $y$ values for all the points on the curve to get unity...

John Rennie

OK. The thing is that those $y$ values depend on the width of the column. In the histogram I've described I have ten columns so I add up ten $y$ values.

user228700

OH...

John Rennie

Suppose I narrow my velocity range to 0-5, 5-10, 10-15, and so one, so each column is now half as wide as they were.

In my example i said that 15 out of the 1000 molecules were in the 0-10 column, so if I split this into two I'm going to get for example 5 in the range 0-5 and 10 in the range 5-10 m/sec.

user228700

OK...

John Rennie

So the $y$ value for my new first column would be 5/1000 = 0.005 and the second column would be 10/1000 = 0.01. The $y$ values for my new columns are smaller than the original ones.

But if I add up the 20 $y$ values in my new histogram they still all add up to one.

OK so far?

user228700

Sep 8, 2016 08:46

Yup :)

John Rennie

If I keep making the columns narrower and narrower then I'll have lots more $y$ values but because each $y$ value is smaller they still all add up to one.

user228700

OK...

John Rennie

To get to the continuous curves like the one you show in your question we, in principle, make the columns infinitely thin so we have an infinite number of them. And adding up an infinite number of infinitely thin columns is exactly what we mean by integration.

user228700

Yes, OK...

John Rennie

So when we integrate the curves we are in effect adding up lots of infinitely thin columns and we are going to get the number one.

That's why Resnick and Halliday say that the integral has the value one.

If you're OK with this so far I need to explain one more thing that I've kind of glossed over so far ...

user228700

Sep 8, 2016 08:52

So, essentially, when we integrate it, we're not multiplying with the $dv$ or anything. We're just integrating the function, correct..?

John Rennie

This is the thing I glossed over ...

user228700

Okay..?

John Rennie

The way I've defined my $y$ values they are the number of molecules in each column. But if we reduce the column width to zero there aren't any molecules in each column because the columns are too thin. So all my $y$ values will be zero - and that isn't much use!

user228700

Yeah! Please go on...

user228700

Wait, you're doing this...why?

John Rennie

Sep 8, 2016 08:56

No hang on, I got that wrong, let me try again.

user228700

OK, sure :)

John Rennie

OK, I'm going to define a new value P that is given by:

P = y/dv

That is, it is my y values divided by my column width dv.

user228700

Riight. Why?

John Rennie

I said above that as I make the columns thinner my values of $y$ are going to reduce. But as I make the columns thinner the column width dv also reduces.

user228700

So this value remains constant..?

John Rennie

Sep 8, 2016 08:59

So in the fraction y/dv both the y and the dx reduce at the same rate and the ratio y/dx remains well defined.

@KaumudiHarikumar yes, the ratio y/dv remains constant.

user228700

Okay...

John Rennie

And that ratio, for any value of v, is how we define P(v)

user228700

OH! P(v) is THAT ratio?!

user228700

Oh God, I didn't know...

John Rennie

Yes. So now you should be able to see why we multiply P(v) by dv to get P(v)dv

Since I defined P as y/dv that means Pdv = y

And when we add up all the y's we get one, so that means when we add up all the P(v)dv's we also get one.

user228700

Sep 8, 2016 09:01

Yes, I see that. Thanks so much! This was just less intuitive than say, integrating the velocity function to obtain the displacement or something like that...

John Rennie

And that's why when we integrate P(v)dv we are going to get the value one.

user228700

Less intuitive...and I didn't know how P(v) was defined.

user228700

Yes sir, I completely understand this now.

John Rennie

It is a bit unintuitive, and that's part of the reason it's so hard to grasp when you first encounter probability density functions.

Have another look at those curves you show in your question, and look closely at the units for the y axis.

user228700

Yeah, s/m. Makes total sense :D

John Rennie

Sep 8, 2016 09:05

The units for the P(v) axis are shown as s/m, but really they are 1/(m/s).

i.e. they are a number (my $y$ number) divided by dv that has the units of speed.

That's why the units are 1/speed.

user228700

I see that now :) Thanks so so much sir. I've been trying to get my head around this for quite some time now and today, I think I finally did.

John Rennie

Like I say, this puzzles everyone when they first meet it :-)

It puzzled me as well about 40 years ago when I was a student :-)

user228700

OK, that's definitely comforting. Thanks, again! You just saved me a lot of hours.

John Rennie

When you learn quantum mechanics you'll find that the wavefunction squared is also a probability density function.

David Z

BTW now that you know, don't forget the difference between "probability" and "probability density"! That's a particularly confusing thing, especially because they're often denoted with the same letter.

user228700

Sep 8, 2016 09:09

Oh God, yes, I was learning a bit of that and it didn't make sense but now it does!

John Rennie

Congratulations, you're now officially 40 years ahead of me when I was your age :-)

And now I'm going off to make a cup of coffee to celebrate.

user228700

All thanks to you! :D

user228700

Okay, enjoy your coffee sir! It's been great.

The h Bar

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