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.
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?
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basis an angle θ so the vector lies along the x axis, then x becomes the r coordinate and θ is the angle the vector originally made with the x axis.
@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.
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
$$ \hat{x}=\cos{\theta}\,\hat{\theta}-\sin{\theta}\,\hat{r},$$
$$ \...
"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?
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
@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.
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.
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.
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...
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.
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.
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?
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?
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?
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
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?