Problem Solving Strategies

what's wrong with writing a vector $\mathbf{v}$ in polar as $\mathbf{v} = \sqrt{x^2+y^2}\hat{r} + \tan^{-1}\frac{y}{x}\hat{\phi}$, ignoring the problem that $\tan^{-1}$ won't work on all quadrants unless it's redefined.

John Rennie

Yes, I drew a picture to illustrate it. Hang on and I'll post the picture.

@psa that's the standard mapping from Cartesian to polar isn't it?

Why would there be anything wrong with it?

@PrateekMourya there.

psa

well what's the difference between that and transforming the basis vectors to give $\mathbf{v} = (y\cos\theta - x\sin\theta)\hat{r} + (y\sin\theta + x\cos\theta)\hat{\phi}$, then?

John Rennie

That shows the nitrogen beam hitting the surface at 30° to the normal.

@psa Isn't that just a rotation to a diffreent Cartesian basis?

Prateek Mourya

Ok

psa

that's from expressing $\hat{x}$ and $\hat{y}$ in terms of $\hat{r}$ and $\hat\phi$

and plugging it into $\mathbf{v} = x\hat{x} + y\hat{y}$

Prateek Mourya

6:13 AM

I know how to calculate change in momentum

But how to calculate time

psa

aren't both in terms of a polar basis?

John Rennie

Hmm, I can see the confusion. If you take a Cartesian basis and rotate it then it does look like using polar coordinates.

I suppose they are basically the same.

psa

the latter seems preferable to work with since the redefintion of arctan to make it make sense in terms of all 4 quadrants and the origin is rather convoluted

John Rennie

i.e. if you have any arbitrary vector then you can rotate your basis an angle θ so the vector lies along the x axis, then x becomes the r coordinate and θ is the angle the vector originally made with the x axis.

I've never thought about to be honest.

psa

there seems to be a subtle difference between coordinate transformations of vectors and bases that I don't understand

Prateek Mourya

6:17 AM

Sir can we continue?

John Rennie

@PrateekMourya give me a few minutes to finish the discussion with psa.

@psa to be honest I have never worried about it. I have always taken the pragmatic view that in any particular case it was usually obvious how to do the transformation.

psa

right

so this question was asked on math.stackexchange here:

0

Q: What is the difference between converting vector components from Cartesian to polar, and converting the unit vectors?

To transform from a standard (2D) Cartesian coordinate system to polar coordinate system, we have the relations $$ r = \sqrt{x^2+y^2},$$ $$ \theta = \arctan{\frac{y}{x}},$$ for the vector components. We also have the relations $$ \hat{x}=\cos{\theta}\,\hat{\theta}-\sin{\theta}\,\hat{r},$$ $$ \...

however, the answer doesn't really make much sense to me:

John Rennie

I think you're correct that the transformation can be more complicated than it seems.

psa

"The issue has to do with the fact that the first set of transformations are simply transforming the components of the vectors. They are not vectors themselves, but they show how the components (𝑥,𝑦) transform into (𝑟,𝜃). The second set of transformations shows how to transform a basis of the vector space into another basis."

isn't the transformation of the components the same as transforming to a different basis?

RishiNandha Vanchi

oh, i remember having the same confusion

John Rennie

6:21 AM

I think there are two cases:
1. keep the basis constant and rotate the vector
2. keep the vector constant and rotate the basis

RishiNandha Vanchi

care should be taken to not confuse the two, youll end up rotating in the opposite direction of what you intended

John Rennie

And these are different, though I forget the details. In Euclidean space they are effectively the same, but in curved spaces (i.e. general relativity) they are different.

psa

right, so my understanding was that in Cartesian, the vectors are invariant wrt rotations and translations ($\hat{x}$ stays in the same direction, for example), but in polar the basis vectors change as you move

is that right?

John Rennie

Correct.

The change in the vectors is tracked by quantities called the Christoffel symbols.

These are all zero for Cartesian bases but are non-zero for polar coordinates.

psa

so when we transform vectors from one coordinate system to another, do we have to transform only the components, only the basis vectors, or both?

right interesting

John Rennie

6:25 AM

To be honest you've reached the borders of my knowledge now.

Presumably you are tranforming the basis since the vector is just a vector.

psa

no worries. i didn't realize how confusing this would be when i started thinking about it.

i thought i understood coordinate systems and change of basis quite well, i don't think i do though.

John Rennie

i.e. if you have an arrow, i.e. a physical arrow, then choosing to use polar coordinates to describe it can't change the arrow itself.

So you are transforming the basis.

psa

doesn't it depend on if it's a contravariant or covariant vector?

or is that always true?

John Rennie

I'm not sure I have ever really understood the difference between co and contravariant vectors.

psa

me neither, this seemingly simple stuff is terribly confusing

ah well, thank you... I'll keep reading up on it and come back if I have more questions

John Rennie

6:29 AM

This area is called differential geometry and I suspect if you learned it then it would all seem very simple.

psa

right

I'm taking a GR class next year, I think they give an intro to DG in the beginning

so hopefully it's cleared up then

Prateek Mourya

Finally

John Rennie

The apparent difficulty arises from having only a limited grasp of the subject.

Prateek Mourya

My turn

John Rennie

@PrateekMourya yes :-)

Let me repost the diagram:

Prateek Mourya

6:30 AM

Sir how to calculate time

John Rennie

I've drawn the beam hitting a unit area of the surface i.e. 1 m².

@PrateekMourya Is the diagram clear?

Prateek Mourya

Sir why we are taking 1 metre square

John Rennie

Because we are going to calculate the pressure, and pressure is force per unit area.

Prateek Mourya

But it is a beam

John Rennie

So if we start with a unit area and calculate the force then we have automatically calculated the pressure.

Prateek Mourya

6:33 AM

Ok

RishiNandha Vanchi

you can take x*x instead of 1*1 if you want, in the end the x's gonna cancel out

John Rennie

@PrateekMourya we aren't told the width of the beam - it could be 10m wide, or 100m. I'm considering only a small part of the beam from somewhere inside it.

Prateek Mourya

Ok

John Rennie

If it's a narrow beam we could take a square micron or a square nanometre then divide by the area at the end.

Prateek Mourya

Ok

John Rennie

6:35 AM

I'm taking a square metre for simplicity, and we'll assume the beam is wide enough to be uniform over our square metre.

Prateek Mourya

Ok

@RishiNandhaVanchi ya i will do this way also in my notebook

Sir can you skip to calculation of time?

John Rennie

Now, the width of the part of the beam hitting our square metre is actually less than 1m because it strikes the surface at an angle so the beam gets spread out over a larger distance than the beam width.

@PrateekMourya calculation of time? What time?

Prateek Mourya

I can calculate the change in momentum

But how to get the time

So o
n dividing we get force

John Rennie

You mean how many atoms hit the surface per second?

Prateek Mourya

Ya

John Rennie

6:37 AM

That's what I'm doing.

If you look at my diagram then the width of the beam hitting the square metre is cos30. Yes?

Prateek Mourya

Yes

John Rennie

So the cross sectional area of the beam is 1m x cos30 m = cos30 m². Yes?

Prateek Mourya

Yes

John Rennie

And if the beam is travelling at v m/s then in one second the volume of the beam, V, that hits the surface is v times the cross sectional area i.e. V = v cos30.

And now you can see why the question told you the volume per nitrogen atom.

Prateek Mourya

Before you say that number of particle hiting are vcos30

Please clarify this

1

Q: Why collisions with other molecules not taken in account here?

I have few doubts related to this derivation why collisions with other molecules not taken into account since they could also influence the time it takes for one molecule to collide with the wall how can taking average force into account and not the real forces the equation still works very wel...

newtonian-mechanics thermodynamics collision kinetic-theory molecular-dynamics

In the picture

The equation 2

The time calculate is 2× the time to travel to hiting place

But here time taken is just 1× times

John Rennie

6:42 AM

I don't understand what you are asking. This is a beam of particles. They are all moving in the same direction i.e. all parallel to each other. So they don't collide with each other.

@PrateekMourya ah, OK

Prateek Mourya

Why is the method of calculating number of molecules colliding different in both the scenario

John Rennie

That's because in your question the particles are in a box, and the time between collisions is the time to go from the wall to the other side of the box and back again. So if the box size is a that means travelling a distance 2a.

This is a completely different calculation.

Prateek Mourya

But i could just take a tube of certain length and area

Around the hitting area

And calculate time to travel this tube

a V tube

John Rennie

Suppose we had just one atom in the box. Obviously this isn't realistic, but I'll use it to explain the calculation.

Prateek Mourya

Ok

John Rennie

6:46 AM

The box has a side of length a and the speed of the atom is v.

Prateek Mourya

On

Ok

John Rennie

The atom hits the right wall of the box normally, and bounces off. We want to find the time before that atom hits the right wall again so we can calculate the force on the right wall. OK so far?

Prateek Mourya

Ok

That what frequency mean right?

Here

In calculation

John Rennie

Yes, if we calculate the time t the the frequency is just f = 1/t.

Prateek Mourya

Ok

@JohnRennie ok

John Rennie

6:50 AM

The atom bounces off the right wall and starts moving left. It moves a distance a and then hits the left wall. This takes a time t₁ = a/v. OK so far?

Prateek Mourya

Ok

John Rennie

Then it bounces off the left wall and starts heading back towards the right wall again. It moves another distance a across the box then hits the right wall. The time taken to move back from the left wall to the right wall is t₂ = a/v.

Prateek Mourya

Ok

John Rennie

So the time it takes to bounce off the right wall, cross the box, bounce off the left wall then travel back to the right wall is t = t₁ + t₂ = 2a/v. Yes?

Prateek Mourya

Yes

John Rennie

6:53 AM

That's where the factor of 2 comes from.

Manoj Ghosh

@JohnRennie hi

Prateek Mourya

So in our problem

John Rennie

@ManojGhosh hi :-) I'm in the middle of answering a question for Prateek ...

@PrateekMourya in this problem the gas isn't in a box.

Prateek Mourya

It was not that one particle that was hitting the area

Sgain

John Rennie

The beam just comes in from far away, bounces off the wall and travels away again.

Prateek Mourya

6:55 AM

We have to calculate how many atoms hit

Not how many tines one atom hit

John Rennie

Yes.

Prateek Mourya

Ah.

That one thing

John Rennie

In the beam question each atom only hits the wall once, then bounces off and travels away never to return.

Prateek Mourya

That bothered me for two nights

My teacher roasted me

For asking why the situation were different

Manoj Ghosh

@JohnRennie okay. When you are done please inform

Prateek Mourya

6:57 AM

@JohnRennie that's why correct way of approximation was to use how many atoms hit

Not how many time one atom hit

@PrateekMourya yes

Thanks alot sir

OK :-)

God bless you for having such patience

John Rennie

I'm glad we got this sorted. These simple things can be confusing!

Prateek Mourya

6:58 AM

Yes

John Rennie

It's the sort of thing that is obvious, but only when you see it!

Prateek Mourya

Actually i was more bothered aince my teacher declared this to be a stupid question

John Rennie

It isn't a stupid question.

Prateek Mourya

Thanks sir I am going to coaching now bye

John Rennie

All questions look easy if you already know the answer!

@PrateekMourya bye :-)

@ManojGhosh what did you want to ask?

Manoj Ghosh

7:00 AM

@JohnRennie are you done ? I have some problem.

John Rennie

@ManojGhosh yes, go ahead.

Manoj Ghosh

@JohnRennie I upload wrong problem please wait

John Rennie

Deleted

Manoj Ghosh

@JohnRennie here the problem. I have confussion for option b and c.

John Rennie

OK, so you need to add together (vector sum) the two fields.

For R < r < 2R the field lines from the inner cylinder are concentric rings around the common axis, while the field lines from the outer solenoid are straight lines parallel to the axis. Yes?

Manoj Ghosh

7:06 AM

@JohnRennie yes

John Rennie

I can attempt to draw this if you want ...

Manoj Ghosh

@JohnRennie no I got it

John Rennie

@ManojGhosh so when you add together the two sets of field lines you're going to get lines that spiral around the axis.

Manoj Ghosh

@JohnRennie okay I got it. So option b is correct right ?

John Rennie

If the field lines are spirals then they are not parallel to the axis. Yes?

Manoj Ghosh

7:10 AM

@JohnRennie yes yes my mistake I misunderstood it.

John Rennie

OK :-)

Manoj Ghosh

@JohnRennie I have another problem

John Rennie

OK ... go ahead, no-one else is waiting.

Anusha

@napstablook thnx for the link

John Rennie

@ManojGhosh how far did you get with this?

Manoj Ghosh

7:12 AM

@JohnRennie I got it. Please delete this problem.

John Rennie

OK :-)

Deleted

Manoj Ghosh

@JohnRennie thank you

John Rennie

@ManojGhosh you're welcome :-)

user102532

7:30 AM

Hey all

I want to ask how to study when you don’t feel like studying

you know , your head is exhausted

you want to just watch YouTube for a while

Chill

John Rennie

Go for a walk. Walk for about an hour and you'll find you feel better. Watching YouTube will not work!

4

RishiNandha Vanchi

@user102532 attempt studying something you dont like and let physics distract you from that subject you dont like XD

user102532

And you are not able to concentrate for more than 10min

Ok.Ok

Thank you both of you

RishiNandha Vanchi

do something else and get back to studying, but that something else should give you this sense that you took a satisfactory break. Youtube will not give that, youtube will only make you want more break

user102532

Yeah.

RishiNandha Vanchi

7:34 AM

if you have like a hobby you can do that

user102532

Ok.

Thanks @JohnRennie and @RishiNandhaVanchi.

John Rennie

:-)

user102532

I will try these

napstablook

10:41 AM

@user102532 I like to play chess, any chess fans? also JR is on the spot for me, watching youtube doesn't work :-))

RishiNandha Vanchi

Chess is great

When im like "nah man i cant study anymore" I'd go sit with the guitar for sometime then come back

user102532

11:01 AM

Nice

@napstablook I love Magnus Carlson bro

I just don’t someone to play with but I like to play chess

Srijan singh 7

@user102532 I do some dance,go for playing sports or chat with my friends for some time

RishiNandha Vanchi

11:27 AM

chats can actually work, depends though

some people get carried away, thats almost like youtube

some know to handle priorities by themselves

@JohnRennie can I ask you questions about career in Physics?

John Rennie

@RishiNandhaVanchi yes, though my knowledge of such things is several decades out of date now.

RishiNandha Vanchi

Does studying pure physics offer Job options immediately after undergraduate? I heard it's a road start that starts at UG and will go on till Ph.D. I love physics, but beyond ug ill have to become independent, without an occupation that might not work

If not, back when you were involved in research or studying, were discipline changes common?

John Rennie

In the UK physics is an excellent way to get a job.

Physics is a hard subject and people who demonstrate they can do well at it are very much in demand in industry.

RishiNandha Vanchi

ooooooh

John Rennie

You wouldn't necessarily end up working as a physicist.

RishiNandha Vanchi

11:36 AM

thats very nice to hear

John Rennie

Though it's likely it would be a technical profession because that's where your skills woukd be most useful.

RishiNandha Vanchi

@JohnRennie I'd want to :-) but during intermediate stages, a stable job might be necessary

John Rennie

I'm reluctant to make any statements about India as I don't know what the job market is like there.

But if you're good you'll get a good job regardless of what subject you do.

RishiNandha Vanchi

Thanks sir

John Rennie

My instinct is to say that you should do what you really want to do, because you will do best at the things you really enjoy.

5

RishiNandha Vanchi

11:42 AM

@JohnRennie do people get to have discipline changes if they build credits through electives?

like Ive seen lots of mathematicians switch from signals processing, but don't know how much of that happens with physics

John Rennie

@RishiNandhaVanchi I don't know I'm afraid.

RishiNandha Vanchi

oh okay, thanks sir. this information actually cleared up lot of misconceptions

John Rennie

:-)

RishiNandha Vanchi

1:04 PM

Is there an explaination why the graph below critical point goes like that?

RishiNandha Vanchi

2:54 PM

@JohnRennie they seem to make a distinction between BSc and BSc(Research) sir. does the answer change?

Akshat Sharma

10:02 PM

@RishiNandhaVanchi I think that's because volume cannot increase on increasing pressure.

Maj

10:51 PM

Can someone quickly explain for me the use of comoving coordinates in general relativity? When are they used? What are their advantages?

Transcript for

Problem Solving Strategies