0xVikas

Can someone please explain the figure 6.5 here, it’s supposed to be something related to spherical aberration but i can't figure out what it is exactly, though the text refers it in the context of circle of least confusion

@0xVikas hm I think this is wrong, It should look like the laser at the back was triggered first to the observer on the ground, followed by the laser at the front, so that the observer on the ground will also see both the pulses reach the detector at the same time. Is this correct?

Yet the counter will say 1, making it look like the sensors did not detect the second pulse from the ground frame. How can this be? What am I doing wrong here?

Now the trolley is moving relative to the ground, and both the lasers are triggered simultaneously (relative to the trolley frame), sending a light pulse towards the detector. The detector will sense both the light pulses simultaneously and increments the counter once. Now from the ground frame, it will look like the pulse from the laser at the front of the trolley reached the detector first followed by the other pulse.

Let's say we have a light detector that can detect light falling on it from any direction, and whenever it senses one or more light pulses at a time it will increment some internal counter by 1. This detector is placed at the center of a trolley, with two lasers at the front and back of the trolley facing the detector at equal distances to it.

@0xVikas This kind of implies that the divergence is associated with the velocity of the fluid (or the sawdust here) and that if there's a difference in the velocities of those dust particles, they might diverge or converge. But from what I understand, divergence is about the net amount of fluid flowing in or out of a boundary and it feels like the "sawdust dropped on the fluid" analogy can be very confusing (or even wrong, is it?)

Now you'll get V2L = -U2, as expected

Since you used U2 bar here, you should take U2 bar in the collision equation too.

@123 This is the mistake. Everything works out now.

Outside the earth, Vb = Ue + Ub, don't you think? For example, you're in a bus and you throw a stone towards the driver. From your frame of reference (bus) if the velocity of the stone is u, for someone standing on the road, the stone will have a velocity of Vbus + u.

Thats the problem, you're using same notation for different frames, don't do that.

@123 Is Vb here the velocity from a frame outside the earth?

@123 It doesn't, Vb has Ue + Ub. I suggest drawing a little diagram of this scenario if you're going to assume the sign of velocities.

@123 How did you write these equations?

The final velocity of the ball in earth's frame is obvious, it bounces back with the same speed but in opposite direction. So if initial velocity of the ball in earth's frame was U, final will be -U

In the image you shared, Ub was U2 with a bar, and U = U2. And in some equations you are assuming the sign and in some places you're not.

@naturallyInconsistent Makes sense, thank you so much!

@123 To avoid any confusion, stick to a single notation. Rewrite the equations with proper notation for the velocities in each frame and you'll get this.

@naturallyInconsistent Got it. And in the infinite case, the charge will be uniform. I'm claiming this uniform charge density on the other side will be zero. If this fits in nicely and works out for all the conditions, should this be the solution based on the uniqueness theorem?

@123 Your initial velocity for the ball in a frame outside the earth was $\overline{U_2}

@naturallyInconsistent Why is that the case in finite plane? If it has a significant thickness (like a cuboid) then it won't be uniform right? That should be the case even if we gradually shrink the thickness

@123 that is true. Just check again which frame of reference you're using when you're writing the collision equation. I think it should be V2 = 2U1 - $\overline{U_2}

@naturallyInconsistent uniform in case of a finite plane? or both?

@123 either assume the sign (& direction) everywhere, or do not assume the sign anywhere

Agreed, so the positive charge density will be more at the edge of the plane and less (or 0) at the center. This can explain why there won't be any positive charge distribution in the case of an infinite plane, since the "edge" is at infinity.

@naturallyInconsistent as per my understanding shielding occurs inside the conductor, and in case of an infinite plane, everything on the other side can be considered as shielded (like in the case of a conducting shell with infinite radius, the inside space is shielded) but can that wont apply here right? Since the plane is clearly finite?

I think the equation he wrote for the elastic collision is wrong

I mean you'll get the same equation but the answer would be completely different

@PinkAura would you get the same equation if you were to setup everything and then start rotating the table from rest with some torque?

@0xVikas If we take the expression for r derived here and put it in the angular component of the newton's second law, we can also find the force required to maintain this angular speed, which comes out to be F = 2mrw^2 sinht

In any case it should be the normal force, or more precisely the force that's rotating the table itself, thats causing the particle to accelerate and gain velocity. All we know is that this force is maintaining the constant angular speed of the whole system.

@PinkAura Because if it started from rest, it will take time to gain that angular speed of w. The question is supposed to be solved by just assuming it will start with an angular speed of w right as it is placed in the groove, which makes sense since we don't know anything about the table or the force that's driving it. Check out physics.stackexchange.com/questions/739295/…

@PinkAura I don't think that is what the questions means, if we are to assume it experiences a 'centrifugal' force of mrw^2, then it should already have 'w' angular speed by then. If its starting from rest (from the ground frame) that won't be the case right?

I think it depends on whether it gained velocity after it was placed in the groove, or if it was placed in the groove with a speed of 'a*w' in that point tangentially.

@PinkAura F here is the Normal force which could be anything (in magnitude). There's no constant acceleration applied radially on the particle, even though thats what it appears to be after considering all the constraints on its motion.

Got it, I was really confused and was imagining the lasers firing at the same time in ground frame as well, I guess I didn't fully understand the relativity of simultaneity yet lol.

even though "when" they happened simultaneously could vary.

So at the same point in space, two simultaneous events at that point will remain simultaneous in all frames, right?

@JohnRennie So this is correct? : chat.stackexchange.com/transcript/message/65362030#65362030

@PinkAura The thing is, the acceleration itself is not a constant and is changing direction with time. So its better to stick to the polar coordinates and solve this, as given in the answer you linked here.

Yes I am

@PinkAura But it is the only force being applied on the particle. At a given moment, lets say the normal force is in some direction. In a small interval the particle will gain some velocity in that direction. After sometime the normal force will be acting in a different direction, but because of the the velocity gained in the previous instance in a different direction it will also move radially.

Since the normal force is always perpendicular to the groove at any instance.

@PinkAura The radial acceleration is caused by the normal force between the particle and the groove, I think.

Since the divergence of $\frac{1}{r^2}\hat{r}$ is zero, the flux over a closed surface (without origin) will also be zero. Does this apply to 2d as well? Since the divergence in 2d: $\nabla.\frac{x\hat{i} + y\hat{j}}{(\sqrt{x^2+y^2})^3}$ is also zero, can we say the 2d flux: $\int F.n\,dl$ will also be zero for that field?