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2:03 AM
yes got it thanks
2:52 AM
@JohnRennie sir did you ever ate an indian dish?
3 hours later…
5:28 AM
@JohnRennie hey
6:02 AM
@psa hi :-)
just got a question about polar coordinates and basis vectors
OK ... ?
Is there an ongoing conversation?
No, I'm free to discuss anything you want :-)
@JohnRennie Can electric field bend pathway of light sir?
6:06 AM
@RishiNandhaVanchi no
Sir can i discuss that beam question?
@PrateekMourya beam question?
Nitrogen beam
Ah, beam of nitrogen atoms.
@JohnRennie above this
6:07 AM
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.
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.
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?
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?
that's from expressing $\hat{x}$ and $\hat{y}$ in terms of $\hat{r}$ and $\hat\phi$
and plugging it into $\mathbf{v} = x\hat{x} + y\hat{y}$
6:13 AM
I know how to calculate change in momentum
But how to calculate time
aren't both in terms of a polar basis?
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.
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
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basis an angle θ so the vector lies along the x axis, then x becomes the r coordinate and θ is the angle the vector originally made with the x axis.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.
there seems to be a subtle difference between coordinate transformations of vectors and bases that I don't understand
6:17 AM
Sir can we continue?
@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.
so this question was asked on math.stackexchange here:
Q: What is the difference between converting vector components from Cartesian to polar, and converting the unit vectors?

aRockStrTo 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:
I think you're correct that the transformation can be more complicated than it seems.
"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?
oh, i remember having the same confusion
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
care should be taken to not confuse the two, youll end up rotating in the opposite direction of what you intended
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.
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?
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.
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
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.
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.
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.
doesn't it depend on if it's a contravariant or covariant vector?
or is that always true?
I'm not sure I have ever really understood the difference between co and contravariant vectors.
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
6:29 AM
This area is called differential geometry and I suspect if you learned it then it would all seem very simple.
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
The apparent difficulty arises from having only a limited grasp of the subject.
My turn
@PrateekMourya yes :-)
Let me repost the diagram:
6:30 AM
Sir how to calculate time
I've drawn the beam hitting a unit area of the surface i.e. 1 m².
@PrateekMourya Is the diagram clear?
Sir why we are taking 1 metre square
Because we are going to calculate the pressure, and pressure is force per unit area.
But it is a beam
So if we start with a unit area and calculate the force then we have automatically calculated the pressure.
6:33 AM
you can take x*x instead of 1*1 if you want, in the end the x's gonna cancel out
@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.
If it's a narrow beam we could take a square micron or a square nanometre then divide by the area at the end.
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.
@RishiNandhaVanchi ya i will do this way also in my notebook
Sir can you skip to calculation of time?
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?
I can calculate the change in momentum
But how to get the time
So o
n dividing we get force
You mean how many atoms hit the surface per second?
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?
So the cross sectional area of the beam is 1m x cos30 m = cos30 m². Yes?
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.
Before you say that number of particle hiting are vcos30
Please clarify this
Q: Why collisions with other molecules not taken in account here?

Prateek Mourya 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...

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
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
Why is the method of calculating number of molecules colliding different in both the scenario
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.
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
Suppose we had just one atom in the box. Obviously this isn't realistic, but I'll use it to explain the calculation.
6:46 AM
The box has a side of length a and the speed of the atom is v.
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?
That what frequency mean right?
In calculation
Yes, if we calculate the time t the the frequency is just f = 1/t.
@JohnRennie ok
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?
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.
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?
6:53 AM
That's where the factor of 2 comes from.
@JohnRennie hi
So in our problem
@ManojGhosh hi :-) I'm in the middle of answering a question for Prateek ...
@PrateekMourya in this problem the gas isn't in a box.
It was not that one particle that was hitting the area
The beam just comes in from far away, bounces off the wall and travels away again.
6:55 AM
We have to calculate how many atoms hit
Not how many tines one atom hit
That one thing
In the beam question each atom only hits the wall once, then bounces off and travels away never to return.
That bothered me for two nights
My teacher roasted me
For asking why the situation were different
@JohnRennie okay. When you are done please inform
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
I'm glad we got this sorted. These simple things can be confusing!
6:58 AM
It's the sort of thing that is obvious, but only when you see it!
Actually i was more bothered aince my teacher declared this to be a stupid question
It isn't a stupid question.
Thanks sir I am going to coaching now bye
All questions look easy if you already know the answer!
@PrateekMourya bye :-)
@ManojGhosh what did you want to ask?
7:00 AM
@JohnRennie are you done ? I have some problem.
@ManojGhosh yes, go ahead.
@JohnRennie I upload wrong problem please wait
@JohnRennie here the problem. I have confussion for option b and c.
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?
7:06 AM
@JohnRennie yes
I can attempt to draw this if you want ...
@JohnRennie no I got it
@ManojGhosh so when you add together the two sets of field lines you're going to get lines that spiral around the axis.
@JohnRennie okay I got it. So option b is correct right ?
If the field lines are spirals then they are not parallel to the axis. Yes?
7:10 AM
@JohnRennie yes yes my mistake I misunderstood it.
OK :-)
@JohnRennie I have another problem
OK ... go ahead, no-one else is waiting.
@napstablook thnx for the link
@ManojGhosh how far did you get with this?
7:12 AM
@JohnRennie I got it. Please delete this problem.
OK :-)
@JohnRennie thank you
@ManojGhosh you're welcome :-)
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
Go for a walk. Walk for about an hour and you'll find you feel better. Watching YouTube will not work!
@user102532 attempt studying something you dont like and let physics distract you from that subject you dont like XD
And you are not able to concentrate for more than 10min
Thank you both of you
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
7:34 AM
if you have like a hobby you can do that
Thanks @JohnRennie and @RishiNandhaVanchi.
I will try these
3 hours later…
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 :-))
Chess is great
When im like "nah man i cant study anymore" I'd go sit with the guitar for sometime then come back
11:01 AM
@napstablook I love Magnus Carlson bro
I just don’t someone to play with but I like to play chess
@user102532 I do some dance,go for playing sports or chat with my friends for some time
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?
@RishiNandhaVanchi yes, though my knowledge of such things is several decades out of date now.
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?
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.
You wouldn't necessarily end up working as a physicist.
11:36 AM
thats very nice to hear
Though it's likely it would be a technical profession because that's where your skills woukd be most useful.
@JohnRennie I'd want to :-) but during intermediate stages, a stable job might be necessary
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.
Thanks sir
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.
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
@RishiNandhaVanchi I don't know I'm afraid.
oh okay, thanks sir. this information actually cleared up lot of misconceptions
1 hour later…
1:04 PM
Is there an explaination why the graph below critical point goes like that?
2 hours later…
2:54 PM
@JohnRennie they seem to make a distinction between BSc and BSc(Research) sir. does the answer change?
7 hours later…
10:02 PM
@RishiNandhaVanchi I think that's because volume cannot increase on increasing pressure.
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?

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