Contains a step by step derivation of the Diffusion Equation following the Einstein approach. Also provides an intuitive explanation of the diffusion equation.
Excellent presentation and I have benifited a lot! A more rigorous statement appears to be that both the notions f and fi by nature represent probability density functions rather than probabilities.
Be confused with the Equation at 6:28, since it implies that the particles around the position x does not participate in any movements outwards. Why is this factor not considered? If considered, next it should done like (f(x, t+\tau) - f(x, t))dx = integral of dx*f(x+\delta, t)*\phi(\delta)d(\delta).
Could you please explain your logic behind the idea that you discussed at 6:24 ? I did not understand why integrating over $\delta$ would give the number of particles at t + $\tau$. Are you just assuming that the particles start at x+$\delta$ at time t and then when they get to x the time s is t+$\tau$? In Einstien's paper, he said that we should easily get it without even explaining it.
Hello! thanks for the comment! Yes that is right. The particles that end up at location x at time t+\tau would have been somewhere, somewhere being any position along the horizontal axis, at the prior epoch t. We are measuring the change in position relatively, i.e., in terms of the distance from their spatial location at time t+\tau. And this distance is \Delta. Each particle could have travelled a different distance over the time interval \tau, and the likelihood of the different distance travelled is captured by the function \phi. So if we integrate f (x+\Delta,t) over all possible distances, which is \Delta from minus \infty to \infty, weight by the density of the distances, we get the number of particles at location x at time t+\tau. Note: f (x+\Delta,t) just represents the number of particles at location \Delta away from x at time t. Hope this helps!
Great video, thanks. One unresolved question, though.... D is defined as a function of Tau at 8:05 ... So, how can we conclude that D is a CONSTANT. What is Tau really, anyway? This video presents it as an "arbitrary" small time... so, that could make D take any arbitrary value we wanted by choosing an appropriate Tau! More realistically, it seems to me that Tau should be chosen so as to have exactly 1 collision during that time span... But wouldn't that be a distribution of times value rather than a single value? The underlying question that I'm trying to resolve is: how can we show from physical principles that the diffusion coefficient D, as defined in your video, or other ways I've seen it defined, is actually a CONSTANT? Every author seems to "hand wave" about it!
Thanks @Julian for the comment! Yes it is very much dependent on \tau, which is the length of small time interval, or one can say it is per unit average(division by \tau can also be interpreted as taking average, height=area/length of time). The key thing here, in this simple case, is homogeneity -homogeneity of particles, medium etc - so one won't expect D to vary. Later on in the series, we will consider cases where D can depend on spatial location, density, and even time. So one gets more complicated equations and processes.
thanks for your geat video I sill have several questions: 1. why does the secod-order of f(x+Δ,t) matters, why not thrid order or higher order? 2. what is the meaing of Φ(Δ)? 3. I am reading the parpers of Einstein relation, could you explain the reason why ”=2Dt", ( denotes the variation of x), are there any good references? thanks, sincerely
I am struggling to understand why the number of particles in a rectangle is f(x,t)dx. If the rectangle is a visualization of a single point (i.e. x=2), then the number of particles at this point is f(2,t). No need for a dx. Any help would be appreciated. Thanks update: It is an integral written in differential notation. Please correct if I am wrong. Thanks
Great question! It is more or less the same thing as the finite approximation of integral. Physically it is a 3D box, but we are looking at one side (rectangle) and measuring location using only the x-axis (we are interested in 1D process). Think of f(x) as the number of particles at location x. We are assuming large number of particles, so f(x) is like a function of x. Number of particles between two vertical slices, one slice at location x, and the other at location x+dx, will be the integral of f over the interval dx. But like Riemann integral (area under the curve), we can partition the interval and then approximate the integral using sum of height times width. For small dx, we can just use one interval, and hence f(x)dx.
@@arkabose1102 hello! Delta is just a variable name, could be z but we follow the original symbols to avoid confusion. x represents an arbitrary location in the space dimension - this represent the location at which we want to determine the number/density of particles at time t. Delta represents the distance of each point in the space dimension from this arbitrary location x, so it also represents space but the origin is in a way x!
At 7:43 I fail to see how that integral goes to zero , you mention delta to be symmetric so that means it’s an even function and then the integral is not zero ?
Many thanks for the question! Here is another way to look at it. The particle is moving randomly and has no preferred direction. It can move \Delta to the left, or \Delta to the right- one would expect the probability of it moving some distance to the left to be equal to the probability of moving the same distance to the right. So the average will be zero: \Delta X Prob-\Delta X Prob=0. The integral is just capturing this over the whole range. Hope this clarifies!
It is a Taylor expansion of 2nd order. If you don't know about it I suggest to you to look for videos which explain what are Taylor expansions because you often use this tool when doing physics (sorry for my english)
Can you go into slightly more detail regarding why all even moments represented by the integrals vanish to zero? Why does the symmetry of the problem dictate the even terms alone vanish?
thanks Robert! Does this relate to the part where we set: \int_{-\infty}^{\infty}{ \Delta \phi \left( \Delta ight) d \Delta}=0, or where we ignored the terms higher than second order in the expansion of f around the x: f \left(x+\Delta,t ight)
@@robertbailey8608 thanks for the clarification! This is because probability is symmetric around zero under the assumed settings. Say a particle is at a given point x. It is equally likely to move to the left or right, and additionally, the probability of the particle experiencing a displacement of minus \Delta is equal to that of the positive displacement of the same magnitude (+\Delta), so the probability weighted average is zero.
@@quantpie why does that effect the second term but not, say the third. I understand why the first is equal to one (normalization) but not in the higher order terms (not shown in the derivation). For clarification, if more terms were shown, the second, fourth, sixth and succeeding even terms would be zero. Why does the symmetry effect those terms specifically but not the odd terms? Sorry, having difficulty understanding that
