In atoms the electrons occupy orbitals like the 1s, 2s, 2p, etc and the electrons cannot be removed from those orbitals without supplying a lot of energy e.g. in the photoelectric effect. Yes?
When you put the atoms together to form a solid these orbitals interact with each other to form a king of orbital that spreads out across the whole solid. These are called energy bands.
Now, a band can only contain a certain maximum number of electrons, and if the band if full these electrons cannot move. But if the band is only part full the electrons can move.
These exact reason for this is a bit complicated. You would have to read an article about band theory to understand it.
An important thing to remember is that when we charge an object the number of electrons we add is a tiny, tiny fraction of the electrons already there.
If we charge an insulator you might think that because the energy bands are fullthe extra electrons would have to go into the next empty band, then because that band is only part filled the extra electrons will be able to move just like in a conductor.
But the energy of the next empty band is usually considerably higher than the filled band, and instead what happens is the extra electrons get put in various types of states collectively known as defect states. These have a lower energy than the next empty band, and electrons in these states are not free to move.
In a perfect crystal the bands I've talked about are the only states an electron can occupy, but in real crystals there are defects in the crystal structure and the situation is a bit messier.
@JohnRennie Yes but the explanation you told is the correct explanation and I have to go through explanations which confuse me more rather than making me understand points
This means that for a given surface charge density the potential will be different in the middle of a face and at a corner, or conversely for a given constant potential over the surface the charge density will differ at a face and at a corner.
So in any shape that is not symmetrical like a sphere, or a cylinder, or an infinite plane, the charge distribution will be non-uniform and probably quite complicated.
@Lllt btw this is a good read related to the same; unequal charge distribution is taken advantage of by lightning rods as well. There is a result as well which sometimes questions use that $\sigma R$ = constant, where $R$ is the radius of curvature which is an approximation.
Consider the scenario I've drawn below that a fluid in a container is being accelerated horizontally. The angle $\alpha$ made can be easily calculated by equating the pressure due to the horizontal, vertical pressure at a point but I'm interested in calculating it in a different way.
I've commonly seen that the angle is often directly taken as the ratio of the forces/acceleration, but I never really understood this. If you look at the diagram above, from the right triangles you get $\tan(\alpha) = a/g$ as well as $= g/a$. How exactly do you directly calculate angles from forces/accelerations (in general situations, I've taken this as an example)?
A surface has to have a constant potential energy. This is because water is a fluid and can move, so if the potential energy differed across a surface the water would simply move across the surface and would keep moving until the PE settled down to a constant value across the surface. OK so far?
So consider taking some infinitesimal mass of water and moving it along the surface, and calculate the work done on that mass of water. This work has to be zero because we know the potential energy is constant on the surface. Yes?
yes, so you're going to equate m g r sin theta = m a r cos theta. I know this, I was trying to figure out when/when can't we directly take tan theta to be the ratio of the forces/accelerations as I've shown in the above part of the diagram.
i.e, in the biggest right triangle you get tan theta = g/a, and in the smaller one on the bottom left, you get tan theta = a/g. I've never actually understood how you can do this.
@AshishAhuja in the bottom right triangle if you use similarity (with the big right angled triangle), you will conclude that the vertical length is = a and the horizontal length is g so you get tan alpha= g/a there too
The water surface should be perpendicular to mg +ma
in general
and if you work in the fluid frame, (in which the fluid is in in eqb), then ma acts in the opposite direction
in fact the best way to understand this i think is to take a small mass of fluid dm inside water
cross sectional area= A, length=x\
in the ground frame, this dm mass has to be accelerating horizontally with a. so if you look at the horizontal forces, it is due to the pressure at the two ends
@satan29 what you say makes sense, but this makes me think as to shouldn't mg act in the opposite direction as well (even though mg acting in the opposite direction does not make much sense)?
if the surface was horizontal then there would be no P diff. Thats whay the surface kind of becomes diagonal: so that the left end has a higher pressure than the right end
if the vertical difference is y, then the pressure difference rho g y
@AshishAhuja Obviously GR is mega complicated, but the explanation it gives for the directions of the pseudoforces acting on our liquid is actually very simple.
basically for your earlier image(s), I think your diagram simply showed the surface to be along the direction of $\vec{g} + \vec{a}$ which I dont think is the case
the surface in general will be along the direction $X \vec{g} + Y \vec{a} $ where X and Y are scalars
I knew that the surface would be perpendicular to g - a, but never really thought about it here. Since the surface is perpendicular you get tan(90 - alpha) = g/a, so tan(alpha) = a/g
When we talk about giving a positive charge to a conductor what we actually mean is that we remove electrons from it to give it a net positive charge. Apart from some special cases the only charges we move around are electrons so we add electrons to give a negative charge and remove them to give a positive charge.
When we remove an electron we can think about this as creating a positively charged hole, where the hole is the space where the removed electron used to be.
The positive holes do attract electrons, but suppose an electron moves towards the hole and fills it, then that electron left behind a hole when it moved. So all this did is move the hole to where the electron came from.
@JohnRennie Oh so this happens , we bring a spherical shell from infinity and I had been thinking that we bring a point charge dq and placed on shell and then bring another point charge from somewhere else and then place it on shell
Yes it's clear a spherical shell shrinks inwards from infinity , thankyou sir :-D
Now I split the point charge into two halves dq/2 and I move one half to the opposite side of the sphere. So the charge I move starts at a distance r and ends at the same distance r as I move it round the sphere. Yes?
And then we can split the two dq/2 charges into four dq/4 charges and move them round the sphere, and so on until we have formed a spherical shell of charge round the sphere.
Good morning. for a general vector $\vec{r(t)}$ and a general rotation matrix $A$ how do we build $\frac{d}{dt}(A*\vec{r(t)})$? . The result of the dot product is a vector. is it sufficient to say that $\frac{d}{dt}(A*\vec{r(t)})=(A*\dot{\vec{r(t)}} $
I am preparing for an exam and i found out this question, it is asked to check if a certain transformation is symmetrical. and i am having problems with the matrix vector multiplication rules. could someone check if i broke any rules here
$\mathcal{L} = \frac{1}{2}*\vec{r}^2 + 2t*\vec{r} * \dot{\vec{r}}$ With the transformation $\vec{r} \rightarrow \vec{r'}= A*\vec{r}$ It is asked to check if this transformation is "symmetrical".. thus the new lagrangian differs must only differs by a total time derivative. So far i got: $\mathcal{L'} = \frac{1}{2}*A^2\vec{r}^2 + 2t*A\vec{r} *A \dot{\vec{r}}$ $\iff$ $\mathcal{L'} = \frac{1}{2}*A^2\vec{r}^2 + 2t*A^2\vec{r} * \dot{\vec{r}}$ $\iff $ $\mathcal{L'} = \mathcal{L}*A^2 $
Physically speaking, i know that rotations are symmetricla and they conserve the Angular momentum of the system "neothres theorem" however i am failing in this simple algebraic equation to form correctly.
The electric field inside a conductor has to be zero, because otherwise the electrons inside the conductor would move until the field became zero. Yes?
Because the field generated due to the charge movement inside the conductor has to be equal and opposite to the external electric field so the two fields sum to zero. Yes?
@satan29 so if we consider an frame accelerating in the direction of the acceleration of the container with acceleration $a/2$, there would be a pseudo-force of m * a/2 towards the left, and the container relative to this frame is accelerating to the right with acceleration a/2. So in such a frame the liquid would appear to be flat (no tilting)?
@satan29 tbh I'm not sure, but if there is a relative acceleration of a/2 to the right and a pseudo-force causing an acceleration of a/2 to the left, the net horizontal acceleration comes to zero?
@AshishAhuja well if you are using a frame that is accelerating at a rate a/2, then from this frame the acceleration of the container, (and hence the fluid elements) has to be a/2, n0?
pseudo forces are used in order to get the same conclusion as the ground frame
the observer in the acclerated frame and the ground frame will ofcourse observe different dynamics of the system. For example if I'm in a car and I see a pole go past me , then ill conclude that the pole is accelerating and hence is experiencing a force ma, but someone standing besides the pole would disagree and say that the net force is zero. Both conclusions are correct in their own way
pseudo forces (the way I see them) are simply mathematical corrections that allow you to get the same equations as the ground manner, albeit in a different manner
for the person standing beside the pole, in their frame we would apply a pseudo-force which is in the opposite direction of the original force. This allows the net force in their frame to be zero, correct?
a block is accelerating with acceleration $a$ wrt ground. In the block's frame, the pseudo-force allows for the net force in the block's frame to be zero(as it should be), correct?
@caramalizedTomato The mean velocity is zero. You'd have to use the mean of |v| to get a non-zero result and that rarely appears in physics.
Anyhow, the reason we use the RMS velocity is because in statistical mechanics it's the energy that is the important quantity so what we are really doing is finding the average of the energy i.e. the average of ¹⁄₂mv². And since the masses of the molecules are all the same this is ¹⁄₂m times the average of v². So the average of v² appears naturally in statistical mechanics.
@Lllt valence band and conduction band is briefly mentioned in the chemistry textbook with a nice diagram, thou it accompanies with a disappointing explanation
@JohnRennie Sir are all the orbitals forming a single "king of orbitals" or is it like all the valence are in one half-filled orbital and inner electrons are in another filled-orbital?
@AdilMohammed In introductions to band theory we usual say each orbital in the atom forms a separate band in the solid. So all the 1s orbitals merge to form a "1s band", all the 2s orbitals merge to form a "2s band" and so on.
But this is an oversimplification because the bands are usually more complicated than this.
In fact band structure can get very complicated when you start looking closely at it. However in many cases we are only worried about the valence and conduction bands and we don't care too much about other aspects of the band structure.
Good evening ! i have been struggiling for hours with this question, i really hope someone here can help me! given is a hamillton function. $ H (p,q) = p^2/2 - a/q $ and the transformation $ q \rightarrow Q^2 $ Find a function F such that the transformation is canonical.
I have been deriving and integrating for hours with no result... i am lost
@JohnRennie Maybe you can? you can solve everything : )
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