Back in 2011 when I was preparing for my JEE, I followed HCV for physics and tried to derive every equation myself by understanding the concept. My derivation did not give me a 2 in the denominator and I tried discussing it with my lecturer. Sadly, they were all focused only on formulae that mattered in exams and didn't want to discuss it. Now I feel relieved that I was thinking along proper lines and can mark the highlighted doubt as closed (if I ever open my HCV book again)! Thank you
I am preparing for JEE as well and coincidentally even I am experiencing this same thing. Every teachers are purely focused on syllabus and not the reason for studying.
its a shame on the teacher for not doing the correct thing and just do what evere their books says. The eacher must have a reason of everything while deriving any equaton or expression in Physics.
@@mukund5455 if people didn't have imagination, Science wouldn't have been found. And the people who work in the entertainment industry are not very science-friendly.
You need class of HC Verma sir .....first I also used to like that...but more he is master in master.in mathematics....attend his class you will get answer..@@Mahesh_Shenoy
@@alanjamey2777 i never said it isnt, and its not only in science, a mistake is a learning in every aspect of life. but being ignorant about it is not learning anything
This is not religion dude. This is science. People can't make a mistake and create their own stories to hold on to their point. There is nothing like a small mistake here.😊
That was not a small mistake at all either if you see this in an exam point of view, or in real world , in exams , usually option do have double the actuall answer, and half the actuall answer , and if a student followers that formula , he can get -1 and in competitive exam like jee , that means a lot , also the values that would be obtained from this formula would be twice the real world experiments, which may cause a lot of confusion and miss understanding
It enrages me that people in the comments hate mathematical proofs. They think that intuition matters more than the truth. Physics is not always intuition. Sometimes you need to get to the truth the way you are supposed to, the language of physics and probably science- mathematics! Just because people dont want to run their brains to a higher degree they despise mathematics.
back in high school, i used to feel a sigh of relief everytime I understood a mathematical proof, like something clicked. Intuitive proofs never gave me the same satisfaction
While I agree, I only understand something if it's mathematically rigorous. 90% of the viewers here (including me) are here just to casually enjoy physics. And he does a good job creating a narrative story. So all the 90% don't actually care for rigour, but just "intuitive" proofs
There is another mistake realted angular velocity reagrding the radius ...hcv uses r but actually the frame is actually at 2r . It is corrected by my teacher A reputed one senior faculty allen kota iid passout ..teaching since 21 years ❤ He even added that there are many other mistakes he can point out ..man he is a legend ..i mean he is a character of a person
To get that fraction of ts, you need the probability distribution of times between collisions P(t). The numerator would be the integral of P(t)*t^2 dt and the denominator would be the integral of P(t)*t dt. The denominator is exactly tau, so to get Feynmans result the numerator would have to be exactly 2 tau^2 which I'm not sure how he got because I would expect that to depend on the distribution.
I'm very, very confused. I understand why HCV approach is wrong, but the final answer IS correct. The assumption of Tau in HCV's lecture is actually 2 times of the tau in the feynmann lectures, as Tau in HCV is average time between successive collisions while in feynmann its since the last collision. Won't 2Tau(feynmann) = Tau(HCV)? Please help I'm losing my mind over this
True, this was actually stated in many physics forums. Using Newtonian frames itself is not gonna give you the correct answer here, we are applying these equations and results in a frame where they wouldn't hold. They are truly ridiculous approximations, so this will happen.
The factor of 2 comes from rms time over mean time, as we know rms time is sqrt(2) times average time for sinsoidal curves hence the rms^2 is 2 times greater but for that we need to assume that distribution of the collision is a sinusoidal curve which might come from quantum wave function of electron in a substance, hence feynman probably got it right as he worked on field of quantum electrodynamics.
Really got a deep understanding after seeing this, rather listening to this, the intuition. Had to repeat watching this 2-3 times to understand better. Thanks and please keep doing what you're doing 'cause many people aren't doing science the way it was and is meant to be.
Prof. Verma mistakenly assumes = ^2; i.e. average of t-squared = square of average t, which is clearly not true. Assuming time-interval between successive collision is exponentially distributed, we get the first moment = tau and second moment = 2 tau^2.
Also, the diagram in the video is slightly misleading. Drift velocity is the average distance traveled per unit time "in the direction of current"... So more accurately, L1, L2,... must represent the components of the respective distances along the current... Only then, you can use L1 = ½a(t1)² (Coz acceleration is in the direction of electric field which is again in the direction of current) This derivation still works because this fact is already implicitly taken into account in the equations...
my physics teacher (The Physics shelter on YT) also pointed this out in our class and also told that the correct approach is also discussed in E&M of purcell and morin
Hi Mahesh. The part where you have made sigma( u t ) = 0 Can we also not say that each new ut also contains the memory of previous acceleration v = u + at Also the summation of t squares looks like can be done using rms value calculations
the problem is not with the formula but with the notation of TAU but he took t as average time ( TAU ) between collisions which is not a subtle approximation
i cannot speak to the physics, but i can confirm your statistics are perfect: the only way to make Σt²/Σt come out Verma's way is if the variance of t is 0. that's not just sufficient but also necessary
Hi Mahesh! Wonderful video again. I think none of them are 'wrong' per se. Actually, the motion of electrons in a material is a very complex quantum mechanical problem. We have to come up with approximations in order to predict experimentally measurable quantities (like conductivity for example). H.C. Verma chooses a lesser sophisticated model for his calculation, and the one chosen by Feynman (Drude Model) is the just a more sophisticated model. But both of them are still models. They are simplications of the quantum mechanical problem using equations from classical physics. Now, the 'test' for a model is how well it can explain an experiment. As far as I know, people only roughly measure the order of magnitude in case of drift velocities, and for metals both the models work well. Both of them fail in case of other materials. Nevertheless, Feynman's model is indeed more sophisticated and has a subtle catch like you mentioned, which people usually miss. Thanks for making this video!
Thanks for sharing, Sarbajit. Are you the same Sarbajit that I know? :D Surely, we are approximating things. But, H.C. Verma has made a fundamental error in the math. They are both the same models with the same underlying assumption - electrons are tiny balls that bounce of elastically.
Hey again! Yeah, I'm the same Sarbajit xD Well, yeah both do the same assumption in considering electrons as elastic balls. But don't you think there's another level of approximation over it? HCV assumes that each electron is travelling with the same velocity in the same time. On the other hand, Feynman assumes a distribution, i.e, velocities can be random which requires an average that he does later like you said. Now, of course there's no reason to assume what HCV did, so that's why I might call it a less sophisticated approximation.
@@sarbajitdutta5193 good to hear from you, bud! Whether you approximate all electrons to travel with the same velocity or not doesn’t matter :D it should yield the same result. Because, in this model, we assume the electron randomised after each collision. So, a single electron over a billion collisions or a billion electrons over a billion collisions should yield the same result.
@@sarbajitdutta5193accept wrong as wrong.. we human still consider those both model Valid because we are Faliure cannot prove one fully wrong... But Drude model is more correct than HCV sir assumption
@@Mahesh_Shenoy Well yeah. It is still the hard sphere approximation. However, doesn't HCV make an additional call by making all the electrons do exactly the same thing so that the motion of one electron is equivalent to the average motion? Probably, I'm not being able to put it in a nice way. I'm sure there is a better way to put out my point :')
Sir I have a doubt At 9:14 it's an avg time that mean most of electron collide exactly at 6 min not before it is'nt it. Therefore if we apply 2nd eq of motion for all electrons then it shouldn't be wrong, right?
Sir there was a question of kinematics earlier of three particles on the corner of equilateral triangle. Sir you had done that question using relative motion but I want to know other ways to do that question too. Please If you can reupload and show us other ways to do that problem that would be very beneficial for me. Thankyou.
With arbitrary collision statistics, the general drift velocity is proportional to E(T^2)/E(T), where T is the random variable denoting the time between successive collisions. This follows from a simple application of the Strong Law of Large Numbers. This quantity is at least E(T) thanks to Jensen's inequality. Now, assuming T is exponentially distributed with mean t, this works out to be 2t.
t1^2 can be represented as T(tau) +x1(letting the variable be any real number here ) If we do this, then we can clearly see that you cannot just divide the taus becasue there are constants(zero power polynomials) that arise when we calculate (T+x)^2. while t1+t2+t3.... may equal N*T, the above doesnt have to equal tau.
Professor, if you could show some light on independent and identical distribution of these times and then whether we can assume the sum of squares to be actually n times tau^2 according to this rule of statistics.
And since we know the motion is inside a constrained area and very much random and we do have have large no electrons and ions...the deviations of actual time from the mean will be negligible ? Or very small and thus the sum of their squared we can say can we represented as n times their means square
I don't know much but the 1st problem you raised is insignificant as by taking the average time we are also considering the average distance. Therefore, the ratio ultimately remains almost the same with some negligible error (probably). I might be wrong. I am just in class 12. Sorry for the inconvenience caused.
in HC Verma's book note that he is squaring the average time instead of taking the average of time squared. with these simple and similar assumptions, we do not get one result that is half of the other when they are supposed to be equal [v=d/t = (ut+1/2at^2)/t]
Sir why are we applying the equations of motion here? I mean how are we sure of the fact that acceleration of particles is constant, like if we try to delve deeper into this problem can't we say that the electron, as it moves through the conductor would experience a number of different forces like repulsive force due to other electrons and attractive forces due to positive kernels and in the interval between the collision shouldn't these forces vary?
All the forces that you mentioned will be present definitely but the net effect of those forces will be zero on the drift distance similar to the 'ut' part he neglected during the derivation.
Thanks for sharing this valuable information. By thinking so, you actually managed to change the way I look over the derivations in my textbook. Now I can think out of the box.❤
well if you are dealing with this problem in this quantum state, then you can't take the acceleration due to electric field common for all the motions , you will have to take the components of acceleration along the direction of motion to use it in the newton's equation as you did at 11:22 ,so its better to solve it larger dimentions as in hcv
@@Mahesh_Shenoy but the electric field is uniform in a specific direction and is in a single direction through the motion ... but here the same acceleration is used for motion of the electrons irrespective of their direction of motion not being along the net acceleration...... i might be wrong regarding this as i am not as confident as you in this topic ....but according to our class 11 kinematics for calculating the displacement along the motion we consider the component of net acceleration along that motion .....
I derived the equation in slightly different way and I got result without that 2 in the denominator.. That's what I have been teaching to my students.. If you provide your email so that I can send you the derivation to check by yourself, whether it is right or wrong? By the way it was nice explanatory video.
Can U please make a video Schrodinger wave eqn to explain it to 11th standard students...I mean it's there in the syllabus (Idk why they want to add as much syllabus as they can and don't give us time to allow to even think properly)...it's been many days I have been stucked in that topic..I don't understand what does this eqn tell us...does it tell probability?does it tell energy??does it tell momentum??does it tell the shape of orbital??does it tell the size of orbital?does it tell the orientation of orbital??.
@@jeeaspirant9252 Max born did it!!And m not asking what does the wavefunctn mean...I am asking what does it gives us...bcs based upon that only we predicted shape of orbital...energy of e etc.
@@mridulacharya8250 the equation is very complex to understand at a first glance but to put it in easy words it tells us about the energy, like both potential and kinetic energy of an particle. It gives us the idea to where a particle is most likely to be, the probability of the particle location. It is a function like any other and varies over time. So to conclude it'll tell you everything about a particle depending on time. Hope it helps and I would suggest you to read online articles if u want to know more
I'd be quite interested in knowing how the term you obtained (in place of tau) could be proven to be equal to twice the average time. Any hints, please?
I was searching for a video on drift video derivation, I had a test next week and I suddenly remembered this video(I had seen this video for timepass and due to the thumbnail, I had not studied Electric current yet)
I wish I had a Physics teacher like him when I was in my high school. My high school teacher couldn’t explain properly and would scold the class if the students were not able to understand. Teachers should owe up if they don’t know something and try to find solution instead of insulting students. Afterall students want to learn and it is their responsibility to teach them properly and with care.
Why don't we use 1st eq of motion to prove professor's assumption wrong as the equations of motion are only valid for straight line and professor has implemented the equation wrongly, we could use v=u+at where 'u' is the initial velocity which we can assume to be zero just after a collision, and we get v=at where v would be the drift velocity (of that particular time interval ) and further we could find the average d. Velocity
Here's the proof :- From equations of motion v = u + at Therefore, v = aτ ( since, u = 0 ) Since, a = F/m Therefore, v = Fτ/m ---(1) Now, Prof. Verma's method used the equation v = l/t which corresponds to v = S/t in equations of motion. But, here's the catch v ≠ S/t when acceleration is constant, but it only works when an object is moving at constant velocity. The equation which describes an object's velocity when acceleration is constant is as follows: vt + ut = 2S Therefore, in this context v = 2S/t (since, u = 0) Since v is the instantaneous velocity at time t. To calculate the average velocity v_avg = v/2 v_avg = 2S/2t = S/t Thus, on comparing v and v_avg if τ is the actual time period between successive collisions the average time period would be τ/2 (which was used in HC Verma's derivation). If one uses the correct formula v = 2S/t instead of v = S/t one will get the same result of Fτ/m. If you wonder how was vt + ut = 2S derived then here's how :- Since, v - u = at ---(1) And, v^2 - u^2 = 2aS ---(2) On dividing eq(2) by eq(1) we get v + u = 2S/t Thus, vt + ut = 2S.
vt + ut is not equal to 2s on dividing both the equations you mentioned, you'll get (v² - u²)/v - u = 2s/t you can't just cancel the lhs and get v - u you'll have to simplify v² - u² = v² + u² - 2vu this [v² + u² - 2vu] will be the lhs
@@lox1fi Hey there! Lets do the algebra. (v^2 - u^2)/(v - u) = (v - u)*(v + u)/(v - u). Thus, our LHS Simplifies to (v + u) since both (v - u) on numerator and denominator gets cancelled. The key here is to use the algebraic property of (a + b)*(a - b) = (a^2 - b^2).
I too realized the discrepancy while I was preparing for JEE. I followed the physics book by DC Pandey where the correct derivation was there and also a comment on how some books derive the equation with a factor of half. Although theoretically not having the half made more sense to me, I believe there is no real way (not inferred from this formula*) to measure the relaxation time for electrons.
u1t1 u2t2..might not add up to 0. Since the motion is not biased we cant have any biased initial velocity (from the point an elecyron strikes a lattice) so u1+u2+...un should be 0..not their sort of weighted timed mean.
Interesting. My take is that the displacement of an electron, in the absence of any field, over a billion collisions should be zero, because there is no current! What do you think is wrong with that process?
@@Mahesh_Shenoy It tends to zero, you can never say it would be exactly zero, but it would be negligible compared to the velocity when electric field is applied, and assuming that current is zero in absence of external electric field is also not completely accurate, it is zero on average as fields cancel out overall but at some places there are non zero fields which are the reason for van der waal forces but expected value is over infinite trials so even in this case though its over billion collisions there will be slight drift velocity but it will be close to zero.
@@Mahesh_ShenoyAs you are saying that there is no current. Is that current from a single electron or billions of electrons? If it was because of a single electron then the displacement must be zero But if it was because of billions of electrons then you can't simply take the displacement of a single electron zero . You can also think about it in this way like after billions of collisions how much possibility is there for a single electron to come in the same position where it started its motion ?
I didn't find any contradictions here. The invariant between the original model and the simplified model is the total displacement of an electron. Take acceleration due to electric field (E) = a = (eE)/m, time between collosions = t1,t2,.... for total of 'n' collisions of a single electron. Then, L(displacement) = sum of displacements along field = 1/2 * a * Sigma(ti^2) (i = 1,2,3,...,n) So assuming a common time (I think HCV does the mistake of calling it average time, because people usually associate average with 'mean') 'tau' between successive collisions to get the simplified picture, L (displacement) = n * 1/2 * a * tau^2 We should have the same displacement for both scenarios: This gives: tau = sqrt(Sigma(ti^2)/n) (i = 1,2,3,...,n) , i.e. tau = RMS value of actual time intervals (not the mean) The vd (drift velocity) = total displacement/total time = L/(n*tau) = 1/2 * a * tau This also solves the problem of L being small displacement between successive collisions, as now we're talking about total displacement over n collisions. (although individual displacements also work, like given in the book, because all of them are equal in the simplified model) I don't think HCV is assuming t1=t2=t3=...tn, it just fails to derive the value of 'tau' and calls it the average value, which leads to this ambiguity. The only confusion could be calling the RMS value of something as its average, when the common assumption is average implies 'mean'. The true mean that we're calculating is that of the individual displacements. So 1/2 * a * tau^2 is the average (mean) displacement between collisions. Let me know if there's anything wrong with this.
13:27 this inference is wrong, the summation of displacements results in zero when there is no Electric field, but if there is an overall net electric field the summation cant be assumed to be zero, just because 2 expression look same in structure doesn't mean they will evaluate to same value.
The ut in the formula is the component that accounts for when there is no electric field. So it is zero The 1/2 at² accounts for the acceleration due to the electric field
Can you solve it rigorously that how did you get that 2t term as in chat gpt i got the result that if t1 is not equals to t2 and so on then there's no direct proof
@@monke4216 If half can drift the other way, half the collisions should also make the electron displace in the opposite direction giving net displacement zero. Remember, a billion electrons moving randomly within a single collision = the motion of a single electron over a billion collisions.
Some math that made the process slightly more rigorous in my head: What has really stumped us is that sum-of-squares term. But if t1, t2, t3, ... tn are all independent random variables following the same statistical distribution, then there's a nice way to express that sum-of-squares term: (t1²+t2²+...tn²)/n=(variance of t's)+(mean time 'tau')². This comes from the definition of the variance when used to describe random variables. Further, we can assume that the t's follow an EXPONENTIAL distribution with average value tau. This is a decent assumption because in the real world, many 'waiting times' follow this distribution. In this case, t1, t2, ... tn are the times that we have to wait between collisions. In an exponential distribution, the variance is numerically equal to the mean, squared. This means that the sum of squares of t's = n*( tau²+tau²)! That's where the factor of 2 comes from! This isn't anything novel that wasn't covered in the video, but hopefully the math made it slightly more rigorous :)
It should be (Expected length) / time, rather than Expected drift velocity per collision. Meaning you only need to do E(t1^2+....) / E(t1+...) and not E(t1^2+..../t1+...). At least both of these can't give same answer either.
i would love to see the QM derivation of the Sommerfeld model. The problem is really that the classical physics is insufficient and the best we can do classically is to first establish the simple instantaneous velocity function as being qE/me times t . Now we can take the Expectation (averaging) but we have to then employ a "mean free path time". But this is problematic in QM and of course one needs to account for the fact that Expectations under any distribution obey: E[t^2]= E[t]^2 + var[t]. Getting the distribution right is where the real work s and this is why Sommerfeld makes the proper correction over the original Drude model.
The derivation can be done by using the first equation of motion for n collisions of an electron and then averaging it out. It gives what Feynman derived.
I think I am still confused about something, Comparing the derivations in both cases, The feynman lectures assumed the following - "It is just the acceleration F/m (where m is the mass of the S-molecule) times the average time since the last collision. Now the average time since the last collision must be the same as the average time until the next collision, which we have called τ" The fenman lectures said this before that line about τ - "If we start to observe an S-molecule at some instant we may expect that it is somewhere between two collisions. In addition to the velocity it was left with after its last collision it is picking up some velocity component due to the force F. In a short time (on the average, in a time τ) it will experience a collision and start out on a new piece of its trajectory. It will have a new starting velocity, but the same acceleration from F." Meanwhile for Hc verma the τ is defined as - " If τ be the averge time between successive collisions" I feel like in Feynman's definition the τ represents that if you are looking at a particle at a given instance, the avg time until the next collision and the time since the last collision is τ. The highest probility will be that you are looking at the particle when it right between the the past and the next collision. if you look at it like that, then the time between successive collisions will be 2τ, now if you take this value and plug it through hc verma's derivation(which assumed τ to be the time b/w successive collisions), you get L = (1/2)(F/m)(2τ)^2, for the drift velocity, you divide L by 2τ and you get v = (1/2)(F/m)(2τ) = Fτ/m. I think the problem lies in the assumtions sorrounding what τ represents, one assumes that it is the avg time until the next collision and time since the past collision, meanwhile the other assumes it as the time b/w 2 collisions, this why hc verma has a 1/2 in the value, hence his τ(time b/w succesive collisions) becomes τ/2 which then represents the time since the last collision and till the next collision.
But still, clearly the assumption regarding multiple accelerated motions being considered as an averaged accelerated motions displacement is a wrong notion, due to the arrival of the term (t1^2 + t2^2...)/(t1 + t2 + t3...) Let's say for a value of 2 for acceleration, displacement during two times 3 and 4 seconds is given by: What HCV is essentially implying is that: (3^2 + 4^2)/2 = (3.5)^2 (clearly incorrect, gives 12.5 = 12.25)
I think in Richard feynman's book the velocity is of that instant when electron is just collidies with the next electron and in sir hc verma's book he has taken the avg velocity of the electron after colliding.( please correct me if i am wrong )❤
What if we start with v=u+at and for each collision find out v and then average out v. then average v= [(u1+u2+.......) + a(t1+t2+......)]/N (Here, N represents the no of collisions considered) average v = 0 + a tau= a tau = ( e E / m) tau
This will just find the velocity right before the next collision ig, but we wanna find the average drift velocity, I think this is what NCERT did too and I'm not sure if it makes sense
I would debate of not doing Vd = L1 + L2 + L3 +..... Since you cannot add them directly because they are vectors, you need to consider their direction and add for which you need to perform parallelogram addition for every two set (L1 + L2) of distance, get it's resultant and add to L3 and it's resultant to L4 etc. ..
Why so complications? Force= charge × Electric field F= eE Now as per Newton's law of motion Force is change in momentum. mu/t=eE u = eEt/m It is as simple as this. No need for integration and statistics average. Though I didn't study HC Verma during school days.
This explanation is incomplete. You are right that due to the effect of variance, HC Verma's value is an underestimate. But there is no reason why the correction factor is multiplying by 2! The correction factor can be a any arbitrarily large positive number. Proof: Correction factor is [(t1^2 + t2^2 .... t^n))/n divided by [(t1 + t2 + .... tn)/n]^2 suppose t1 = t2 =t3... = tn = 1 (distribution with 0 variance); then correction factor is 1. But suppose t1 = 1 while t2 ... tn is 0. (distribution that maximizes variance). Then correction factor is (1/n) / (1/n^2) = n n can be made arbitrarily large is unbounded. All you've proved is that drift velocity lies somewhere between HC Verma's answer and Infinity. You've not proved that twice of HC Verma's answer is correct.
But t2, t3,.. are non-zero values... So, the second case is not possible.... Even if t2, t3... are infinitesimally small, we are adding infinite number of values, so we can't say t2+t3+... = 0
if t2 ..... tn is 0 then there is no need of correction factor. this also implies the electrons undergo NO collisions. then t = t1 which in this case you took it as 1 there won't even be a concept of AVERAGE time so drift velocity would just be the normal velocity of the electron due to the electric field
If we take infinite electrons at a single time and find drift velocity We get average final velocity of these infinite electrons as eEt/m but this average final velocity. Of we want to find out the drift velocity (which is the average velocity during the whole free path To find that we have to find average of inital average velocity and final average velocity. Means that drift velocity will be equal to (initial average velocity + final average velocity )/2 = (0 + eEt/m)/ 2 So the correct drift velocity will be eEt/2m
I liked your videos on khan academy india but they haven't been arranged in a playlist so I don't know which one to watch after the first 4 videos of this topic(current electricity). Neither on the site nor in youtube playlists. Going back to HCV for now 😅
No Mahesh your argument is totally fine. ProfHC Verma can also make mistakes he is a human I also found some errors in his book The problem is for over many years no one tried to correct it
@aryan4679why are you using bhakt word in insulting sense? You know the literal meaning of word is sacred in Hindu religion? Why people have normalised using the word as insult, would you have used the word namazi, maulvi, rabbi, or bishop as insult?
15:18 the assumption "time gap between collision is equal" is mentioned in his book. And based on the the assumption he is quite correct, but yeah the assumption goes very far way from the actual values.
@Mahesh_Shenoy, @FloatHeadPhysics.. It was a nice video but i feel there is a small mistake in your derivation as well. In this equation of drift velocity, (Vd=L/t), over a long period of time, "L" is the horizontal length covered by the electron (call it drift length). It cannot be equal to (L1 + L2 + L3 + L4.....). It is not even the displacement, displacement would be a straight path connecting the starting and ending points of the electron, in the respective time period being considered. It is just the horizontal length travelled in time "t". I might be wrong, please correct me if i am. Thank you!
@@Mahesh_Shenoy We knew during our preparation that this was wrong but the explanation here is gold so I would really like to watch more of your explanation for different concepts that are there in HCV right or wrong doesn't matter just the thought behind it like you presented here
I'd summarize this as: avg(f(t_i)) is the same as f(avg(t_i)) ONLY if f is linear! Linearity. That's the first thing they tell you in Maths to be able to change the order of the operations! :D The distance calculation (double integral of acceleration) is a square of a, definitely not linear!
I know I'm really late to the video to get a response but for some reason, a program I made to test out this factor of 2 seems to be giving me a different answer, and I'm perplexed as to why. If someone could offer an explanation, that would be great. When I compute the parts of the equation pertaining to tau, I get feynman's value not to be twice verma's, but rather 4/3 times it. I'm using completely randomised intervals and as far as I can tell, the program is logically correct, so why the discrepancy? Below is the program and some sample outputs. It was made in Python and uses the default "random" library : Program : ``` import random collision_count = 10000 # this value can be adjusted for n collisions total_duration = 0 squared_duration = 0 for n in range(collision_count): # generate a random time between collisions t_n = random.random() total_duration += t_n squared_duration += t_n ** 2 verma_tau = total_duration / collision_count feynman_tau = squared_duration / total_duration # feynmann_tau should ideally be twice that of verma_tau factor = feynman_tau/verma_tau print(f" collisions : {collision_count}" + f" duration : {total_duration}" + f" verma : {verma_tau}" + f" feynmann : {feynman_tau}" + f" factor : {factor}") ``` Outputs : #1 collisions : 100000000 duration : 49998360.500493824 verma : 0.49998360500493827 feynmann : 0.6666662707776526 factor : 1.3333762629497983 #2 collisions : 1000000 duration : 500194.6072459398 verma : 0.5001946072459398 feynmann : 0.6668238495570457 factor : 1.3331288260554481 #3 collisions : 10000 duration : 4996.37857230061 verma : 0.49963785723006104 feynmann : 0.6659245185748535 factor : 1.3328143753290993
HC Verma is not wrong both the formulas are correct from their point of view In HC Verma time is taken as a linearly increasing function due to which it's average comes out to be ½(time) So, both the forms are correct
I don't think this is a mistake to be honest, I think it is more like calculating the escape velocity of a rocket without taking air resistance in to account. The assumption that each each collision takes roughly the same time interval is probably an "ideal case scenario". We have to keep in mind that this book is meant for 12th grade students and diving deep into complexities of a topic like drift velocity, might not be necessary at that moment.
but no where is it mentioned that we are assuming the time period between each collision to be the same. in most other cases things like that are mentioned like the frequent use of 'massless' frictionless' in the numerical questions
Drift veocity is average of drift velocity of large number of electrons and for each electron a is qE/m this can be checked in "drift velocity and collision time" by Donal M Title American General of physics. Lastly in my humble opinion in science we are all learning it is about betterment not about being right and wrong.
I observed many did the average of final velocity before next collision to actually derive drift velocity . so how can we relate it llike I didnt get it
Excellent work and please keep up the good work. The problem is that a large cross-section of world population continues to live with tribal mindset and herds mentality even today. And we human beings are not to be blamed for this - indeed we lived the life of herdsmen and tribes million years back. This gives rise to cult following. Once somebody becomes a CULT - he becomes a GOD - other members of the tribe won't allow you to question that CULT - that's why you get to Jaggi Vasudev, Ramdev etc. Two years back I got into an argument with some kids over a RUclips video by H.C. Verma - question was regarding the magnetic effect of a moving charge. I found H. C. Verma's explanation rather naive and incomplete and I have seen many such flaws (of Prof Verma) before. I commented that Prof Verma loves Physics and probably wants to learn Physics, but he is NOT an expert. This irked some kids - some of them appeared and probably about to appear at IIT JEE. They immediately started to throw their weight around by flaunting their IIT JEE ranks and all - which of course I won't cross-verify. I off course won't throw my weight around saying I completed my B.Tech in so and so year - did my Masters in so and so year - happened to work in so and so Semiconductor MNC Majors in so and so places and in so and so positions and took part in so and so R&D projects and that my son is a student at so and so IIT at that point in time. Honestly, when I wrote IIT in 1998 paper pattern was 40 X 2 + 15 X 8 in three hours - my son thinks it was tougher then paper-wise may not be competition wise. And anyway, an IIT rank itself isn't a testimony of one's understanding of Physics. I just tried to be humble. But finally I had to say that Prof Verma is more of a HYPE and a CULT. And it is extremely difficult to convince others that CULT-FOLLOWING isn't a good idea.
Thanks for this very thoughtful comment! But, I think prof. Verma is really good. His books are extremely helpful. It explains things very deeply and in a simple language! It's because he is so good, it's worth pointing out places where I disagree :)
it v = ri while deriving u have to keep it mind it is not a fundamental law it work only in certain range so in approximation both derivation are correct
I think, he uses the equation of motion, s = ut + 1/2 at² Even if we assume that he's correct by saying that sum of all u will become 0, but in such case, we will find S rather V (v is drift velocity while S would be displacement) In order to find the correct formula using the formula HC verma derived, we just keed to differentiate both side with respect to time S = 1/2 a × t² dS/dt = 1/2 a × d(t²)/dt V = 1/2 a × 2t = a × t Then we can replace a by eE/m
@@Mahesh_Shenoy oh true, completely forgot about that Well I haven't ever used HC Verma, the other books I read, have the formula derived using V = U + AT, so the sum of all u becomes 0 and sum of all T becomes Tau
sir the most important part here is that we have mathematical conception of Ohms law and resistance of a conductor, here drift velocity or most probable vel or rms vel may have less importance
The answer is given by HC Verma sir himself. He said that all these derivations are approximeted that Newtons law directly can be applied. But the reality is there quantum physics must be applied. So both the methodes are approximated and hence order is important and not the magnitude.
Bt there is a big mistake in other formula also... He is calculating drift velocity just before the collision... The e- has not travelled full distance between 2 collisions with this... This is the velocity gained till just b4 next collision... And to use this as avg is also wrong v=u+at❌
Shouldn't quantum mechanics instead of newtonian mechanics be used? Moreover the length of path travelled by electron would be uncertain... due to hiesenbergs uncertainity , it would be impossible to accurately determine the position of the electron?
Hey, I think you are also wrong in assuming that you can take out "a" and group (t1² + t2² + t3² + ... tn²) It can be corrected by having a_l = local acceleration vector a_g = global acceleration vector (which you're calling "a") Then, v_d τ = Avg(½ u t + (a_l + a_g)t²) = Avg(½ u t + a_l t²) + a_g Avg(t²) = 0 + a_g Avg(t²) You're assuming that the acceleration between every successive collisions is same which I'm doubtful about. They are colliding with charged bodies which would have a local electric field. This isn't affecting your result because later you assumed Avg(½ u t) = 0, which cancels the effect.
Back in 2011 when I was preparing for my JEE, I followed HCV for physics and tried to derive every equation myself by understanding the concept. My derivation did not give me a 2 in the denominator and I tried discussing it with my lecturer. Sadly, they were all focused only on formulae that mattered in exams and didn't want to discuss it. Now I feel relieved that I was thinking along proper lines and can mark the highlighted doubt as closed (if I ever open my HCV book again)!
Thank you
I am preparing for JEE as well and coincidentally even I am experiencing this same thing. Every teachers are purely focused on syllabus and not the reason for studying.
@@shihab3611Exactly
@@shihab3611these days it's all about solving mcqs
Same here also
its a shame on the teacher for not doing the correct thing and just do what evere their books says. The eacher must have a reason of everything while deriving any equaton or expression in Physics.
I like how you force more on the intuition part rather than pen and paper proof , it at the end just starts to make sense.
Trying to gain and share intuition is what I try to do 90% of the time! It's super fun!
Real physics is pure imagination and Intuition
yes for those who want entertainent
@@mukund5455 if people didn't have imagination, Science wouldn't have been found.
And the people who work in the entertainment industry are not very science-friendly.
You need class of HC Verma sir .....first I also used to like that...but more he is master in master.in mathematics....attend his class you will get answer..@@Mahesh_Shenoy
Be so legendary that when you do even a small mistake people become shocked.
in the world of science there is no small mistakes, a mistake is a mistake, period. stop glazing him
@@ExistingAndLivingin a world of science a mistake is a door for learning.
@@alanjamey2777 i never said it isnt, and its not only in science, a mistake is a learning in every aspect of life. but being ignorant about it is not learning anything
This is not religion dude. This is science. People can't make a mistake and create their own stories to hold on to their point. There is nothing like a small mistake here.😊
That was not a small mistake at all either if you see this in an exam point of view, or in real world , in exams , usually option do have double the actuall answer, and half the actuall answer , and if a student followers that formula , he can get -1 and in competitive exam like jee , that means a lot , also the values that would be obtained from this formula would be twice the real world experiments, which may cause a lot of confusion and miss understanding
It enrages me that people in the comments hate mathematical proofs. They think that intuition matters more than the truth. Physics is not always intuition. Sometimes you need to get to the truth the way you are supposed to, the language of physics and probably science- mathematics! Just because people dont want to run their brains to a higher degree they despise mathematics.
They are going to face a lot of challenges at higher level when things go small (quantum physics). As that is highly counter intuitive prima facie.
back in high school, i used to feel a sigh of relief everytime I understood a mathematical proof, like something clicked. Intuitive proofs never gave me the same satisfaction
While I agree, I only understand something if it's mathematically rigorous.
90% of the viewers here (including me) are here just to casually enjoy physics. And he does a good job creating a narrative story.
So all the 90% don't actually care for rigour, but just "intuitive" proofs
Well said bro ❤
@@Player_is_I thnx bro❤️
There is another mistake realted angular velocity reagrding the radius ...hcv uses r but actually the frame is actually at 2r .
It is corrected by my teacher A reputed one senior faculty allen kota iid passout ..teaching since 21 years ❤
He even added that there are many other mistakes he can point out ..man he is a legend ..i mean he is a character of a person
page number?
Bhai mere teacher to hcv se live padhe the iit k me 😂
Bhai ek example to de
@@arnabmallick0708 amit gupta 😂
M also physics teacher buy recent pass out from Indian education system dear freinds main ko toh mistake hai ye bhi nai pata tha 😂😂
To get that fraction of ts, you need the probability distribution of times between collisions P(t). The numerator would be the integral of P(t)*t^2 dt and the denominator would be the integral of P(t)*t dt. The denominator is exactly tau, so to get Feynmans result the numerator would have to be exactly 2 tau^2 which I'm not sure how he got because I would expect that to depend on the distribution.
I'm very, very confused. I understand why HCV approach is wrong, but the final answer IS correct. The assumption of Tau in HCV's lecture is actually 2 times of the tau in the feynmann lectures, as Tau in HCV is average time between successive collisions while in feynmann its since the last collision. Won't 2Tau(feynmann) = Tau(HCV)?
Please help I'm losing my mind over this
comment for reach
The thing is, they are derived by insane approximation. So derivation by different methods can lead to different results
@@abhimanyulakra3619 But both are making newtonian assumptions. They should be approaching different answers if one of the approaches is wrong.
One is doing it the wrong way that's all stop treating him like a god given figure @@lilac2698
True, this was actually stated in many physics forums. Using Newtonian frames itself is not gonna give you the correct answer here, we are applying these equations and results in a frame where they wouldn't hold. They are truly ridiculous approximations, so this will happen.
The factor of 2 comes from rms time over mean time, as we know rms time is sqrt(2) times average time for sinsoidal curves hence the rms^2 is 2 times greater but for that we need to assume that distribution of the collision is a sinusoidal curve which might come from quantum wave function of electron in a substance, hence feynman probably got it right as he worked on field of quantum electrodynamics.
Came to say the same.. Thanks 👍
Maybe there is a statistical solution to the problem which gets the numerator to be twice that of denominator, leading to the removal of 1/2.
@@bobinkurian3357 there is no maybe, this is a mathematical error
Mahesh sir because of u i got 94 in physics in my board exam tq
Hey, that's so great to hear!
Also, correction! You got 94 because of YOU!
Class 12 boards?
But is every topic covered at Khan academy
Yep it is@@user-pj1wv1ns9x
And where are your rest 6 marks?
I got 85 you gotta pay now Lil bro (joke obviously)@@Mahesh_Shenoy
Really got a deep understanding after seeing this, rather listening to this, the intuition. Had to repeat watching this 2-3 times to understand better.
Thanks and please keep doing what you're doing 'cause many people aren't doing science the way it was and is meant to be.
Prof. Verma mistakenly assumes = ^2; i.e. average of t-squared = square of average t, which is clearly not true. Assuming time-interval between successive collision is exponentially distributed, we get the first moment = tau and second moment = 2 tau^2.
Also, the diagram in the video is slightly misleading.
Drift velocity is the average distance traveled per unit time "in the direction of current"...
So more accurately, L1, L2,... must represent the components of the respective distances along the current...
Only then, you can use L1 = ½a(t1)²
(Coz acceleration is in the direction of electric field which is again in the direction of current)
This derivation still works because this fact is already implicitly taken into account in the equations...
my physics teacher (The Physics shelter on YT) also pointed this out in our class and also told that the correct approach is also discussed in E&M of purcell and morin
I follow him too :)
Did Resnick Halliday made the same mistake? As Prof. Verma often mentions he referred Resnick Halliday before writing Concepts of Physics
No
Hi Mahesh. The part where you have made sigma( u t ) = 0
Can we also not say that each new ut also contains the memory of previous acceleration v = u + at
Also the summation of t squares looks like can be done using rms value calculations
No since the after collision the initial velocity will be zero or at the exact moment it will be in resting position
HC verma use equation of uniform motion L=ut+1/2at^2 and after collision electron has u=0 and it becomes L=1/2at^2.
the problem is not with the formula but with the notation of TAU
but he took t as average time ( TAU ) between collisions which is not a subtle approximation
@@lox1fi
what formula should i use for jee ??
Such a coincidence, I was studying this an hour back!
Same bro :-)
Same
Same
I just revised this chapter few minutes ago
Wow!
Basically we get the more accurate answer if we differentiate wrt T instead of divide by a fixed value T. Is that it?
i cannot speak to the physics, but i can confirm your statistics are perfect: the only way to make Σt²/Σt come out Verma's way is if the variance of t is 0. that's not just sufficient but also necessary
I had the same issues with this derivation back in 2021.
But yeah now I know. Thanks a lot.
Hi Mahesh! Wonderful video again. I think none of them are 'wrong' per se. Actually, the motion of electrons in a material is a very complex quantum mechanical problem. We have to come up with approximations in order to predict experimentally measurable quantities (like conductivity for example).
H.C. Verma chooses a lesser sophisticated model for his calculation, and the one chosen by Feynman (Drude Model) is the just a more sophisticated model. But both of them are still models. They are simplications of the quantum mechanical problem using equations from classical physics. Now, the 'test' for a model is how well it can explain an experiment. As far as I know, people only roughly measure the order of magnitude in case of drift velocities, and for metals both the models work well. Both of them fail in case of other materials. Nevertheless, Feynman's model is indeed more sophisticated and has a subtle catch like you mentioned, which people usually miss. Thanks for making this video!
Thanks for sharing, Sarbajit. Are you the same Sarbajit that I know? :D
Surely, we are approximating things. But, H.C. Verma has made a fundamental error in the math. They are both the same models with the same underlying assumption - electrons are tiny balls that bounce of elastically.
Hey again! Yeah, I'm the same Sarbajit xD
Well, yeah both do the same assumption in considering electrons as elastic balls. But don't you think there's another level of approximation over it?
HCV assumes that each electron is travelling with the same velocity in the same time. On the other hand, Feynman assumes a distribution, i.e, velocities can be random which requires an average that he does later like you said. Now, of course there's no reason to assume what HCV did, so that's why I might call it a less sophisticated approximation.
@@sarbajitdutta5193 good to hear from you, bud!
Whether you approximate all electrons to travel with the same velocity or not doesn’t matter :D it should yield the same result.
Because, in this model, we assume the electron randomised after each collision. So, a single electron over a billion collisions or a billion electrons over a billion collisions should yield the same result.
@@sarbajitdutta5193accept wrong as wrong.. we human still consider those both model Valid because we are Faliure cannot prove one fully wrong... But Drude model is more correct than HCV sir assumption
@@Mahesh_Shenoy Well yeah. It is still the hard sphere approximation. However, doesn't HCV make an additional call by making all the electrons do exactly the same thing so that the motion of one electron is equivalent to the average motion? Probably, I'm not being able to put it in a nice way. I'm sure there is a better way to put out my point :')
Sir I have a doubt
At 9:14 it's an avg time that mean most of electron collide exactly at 6 min not before it is'nt it. Therefore if we apply 2nd eq of motion for all electrons then it shouldn't be wrong, right?
question arises whether it is drift velocity of single or an average d, velocity? isnt it?
Sir there was a question of kinematics earlier of three particles on the corner of equilateral triangle. Sir you had done that question using relative motion but I want to know other ways to do that question too. Please If you can reupload and show us other ways to do that problem that would be very beneficial for me. Thankyou.
Sure, will add it to the list!
@@Mahesh_Shenoy Thankyou sir
With arbitrary collision statistics, the general drift velocity is proportional to E(T^2)/E(T), where T is the random variable denoting the time between successive collisions. This follows from a simple application of the Strong Law of Large Numbers. This quantity is at least E(T) thanks to Jensen's inequality. Now, assuming T is exponentially distributed with mean t, this works out to be 2t.
t1^2 can be represented as T(tau) +x1(letting the variable be any real number here )
If we do this, then we can clearly see that you cannot just divide the taus becasue there are constants(zero power polynomials) that arise when we calculate (T+x)^2. while t1+t2+t3.... may equal N*T, the above doesnt have to equal tau.
a great teacher knows how to teach you well, but a legendary teacher knows where you can mistakes
Professor, if you could show some light on independent and identical distribution of these times and then whether we can assume the sum of squares to be actually n times tau^2 according to this rule of statistics.
And since we know the motion is inside a constrained area and very much random and we do have have large no electrons and ions...the deviations of actual time from the mean will be negligible ? Or very small and thus the sum of their squared we can say can we represented as n times their means square
Use L-Hopital's rule to solve the time equation...you will get 2×tau
I don't know much but the 1st problem you raised is insignificant as by taking the average time we are also considering the average distance. Therefore, the ratio ultimately remains almost the same with some negligible error (probably). I might be wrong. I am just in class 12. Sorry for the inconvenience caused.
Yes you are correct brother
in HC Verma's book note that he is squaring the average time instead of taking the average of time squared.
with these simple and similar assumptions, we do not get one result that is half of the other when they are supposed to be equal [v=d/t = (ut+1/2at^2)/t]
Sir why are we applying the equations of motion here? I mean how are we sure of the fact that acceleration of particles is constant, like if we try to delve deeper into this problem can't we say that the electron, as it moves through the conductor would experience a number of different forces like repulsive force due to other electrons and attractive forces due to positive kernels and in the interval between the collision shouldn't these forces vary?
All the forces that you mentioned will be present definitely but the net effect of those forces will be zero on the drift distance similar to the 'ut' part he neglected during the derivation.
This is an idealized model. If you plan to count every constraints and tiny factors then you won't be getting anything out of science
You need statistical mechanics pov for this. This is a very idealized way.
Thank you for the explanation. Could you provide any reference that proves that the fraction you mention on minute 18:26 is equal to 2.tau?
Thanks for sharing this valuable information. By thinking so, you actually managed to change the way I look over the derivations in my textbook. Now I can think out of the box.❤
well if you are dealing with this problem in this quantum state, then you can't take the acceleration due to electric field common for all the motions , you will have to take the components of acceleration along the direction of motion to use it in the newton's equation as you did at 11:22 ,so its better to solve it larger dimentions as in hcv
Acceleration will be a constant since the electric field is uniform!
@@Mahesh_Shenoy but the electric field is uniform in a specific direction and is in a single direction through the motion ... but here the same acceleration is used for motion of the electrons irrespective of their direction of motion not being along the net acceleration...... i might be wrong regarding this as i am not as confident as you in this topic ....but according to our class 11 kinematics for calculating the displacement along the motion we consider the component of net acceleration along that motion .....
@@rajneeshtetarwal8654 Oh, yes I should have used the vector signs! I always forget to do that!!
The vectors signs take care of them!
@@Mahesh_Shenoy ruclips.net/video/jEKUog6-LiA/видео.html
what hc verma says
I derived the equation in slightly different way and I got result without that 2 in the denominator.. That's what I have been teaching to my students.. If you provide your email so that I can send you the derivation to check by yourself, whether it is right or wrong? By the way it was nice explanatory video.
Can U please make a video Schrodinger wave eqn to explain it to 11th standard students...I mean it's there in the syllabus (Idk why they want to add as much syllabus as they can and don't give us time to allow to even think properly)...it's been many days I have been stucked in that topic..I don't understand what does this eqn tell us...does it tell probability?does it tell energy??does it tell momentum??does it tell the shape of orbital??does it tell the size of orbital?does it tell the orientation of orbital??.
Schrodinger himself couldn't explain it.
@@jeeaspirant9252
Max born did it!!And m not asking what does the wavefunctn mean...I am asking what does it gives us...bcs based upon that only we predicted shape of orbital...energy of e etc.
@@mridulacharya8250 the equation is very complex to understand at a first glance but to put it in easy words it tells us about the energy, like both potential and kinetic energy of an particle. It gives us the idea to where a particle is most likely to be, the probability of the particle location. It is a function like any other and varies over time. So to conclude it'll tell you everything about a particle depending on time. Hope it helps and I would suggest you to read online articles if u want to know more
@@livingbeing8661
Ooo..
Btw I read many
Not an expert on quantun whatever!
Would you folks be okay if I made videos on topics that I do have some perspective to share but am still shaky on?
But there is a mean free distance between collisions that depends on the density of the conductor.
I'd be quite interested in knowing how the term you obtained (in place of tau) could be proven to be equal to twice the average time. Any hints, please?
I am still shaky on that :-/
Same question
@@Mahesh_Shenoy Ah, never mind. I'll think about it, too. Please do make a video if you do figure it out please
@@aadiprasad3167 did you find it out ?
@@physicorum7107 I didn't think a lot about it to be honest 😅
I was searching for a video on drift video derivation, I had a test next week and I suddenly remembered this video(I had seen this video for timepass and due to the thumbnail, I had not studied Electric current yet)
Same Ngl Checking Again After our Sir Told Us To Read The Derivation From HCV
I wish I had a Physics teacher like him when I was in my high school. My high school teacher couldn’t explain properly and would scold the class if the students were not able to understand. Teachers should owe up if they don’t know something and try to find solution instead of insulting students. Afterall students want to learn and it is their responsibility to teach them properly and with care.
Why don't we use 1st eq of motion to prove professor's assumption wrong as the equations of motion are only valid for straight line and professor has implemented the equation wrongly, we could use v=u+at where 'u' is the initial velocity which we can assume to be zero just after a collision, and we get v=at where v would be the drift velocity (of that particular time interval ) and further we could find the average d. Velocity
Thank you for taking me back to high school.
can you link a proof for 18:20 ? I tried to find it online, didn't find it
Lmk if you found one
Here's the proof :-
From equations of motion
v = u + at
Therefore, v = aτ ( since, u = 0 )
Since, a = F/m
Therefore, v = Fτ/m ---(1)
Now, Prof. Verma's method used the equation v = l/t which corresponds to v = S/t in equations of motion. But, here's the catch v ≠ S/t when acceleration is constant, but it only works when an object is moving at constant velocity. The equation which describes an object's velocity when acceleration is constant is as follows:
vt + ut = 2S
Therefore, in this context v = 2S/t (since, u = 0)
Since v is the instantaneous velocity at time t. To calculate the average velocity
v_avg = v/2
v_avg = 2S/2t = S/t
Thus, on comparing v and v_avg if τ is the actual time period between successive collisions the average time period would be τ/2 (which was used in HC Verma's derivation). If one uses the correct formula v = 2S/t instead of v = S/t one will get the same result of Fτ/m.
If you wonder how was vt + ut = 2S derived then here's how :-
Since, v - u = at ---(1)
And, v^2 - u^2 = 2aS ---(2)
On dividing eq(2) by eq(1) we get
v + u = 2S/t
Thus, vt + ut = 2S.
vt + ut is not equal to 2s
on dividing both the equations you mentioned, you'll get
(v² - u²)/v - u = 2s/t
you can't just cancel the lhs and get v - u
you'll have to simplify v² - u² = v² + u² - 2vu
this [v² + u² - 2vu] will be the lhs
@@lox1fi Hey there!
Lets do the algebra.
(v^2 - u^2)/(v - u) = (v - u)*(v + u)/(v - u).
Thus, our LHS Simplifies to (v + u) since both (v - u) on numerator and denominator gets cancelled.
The key here is to use the algebraic property of (a + b)*(a - b) = (a^2 - b^2).
@@lox1fi I hate to mention but you did the algebra incorrectly. We all make mistakes. Happy Learning!
I too realized the discrepancy while I was preparing for JEE. I followed the physics book by DC Pandey where the correct derivation was there and also a comment on how some books derive the equation with a factor of half. Although theoretically not having the half made more sense to me, I believe there is no real way (not inferred from this formula*) to measure the relaxation time for electrons.
u1t1 u2t2..might not add up to 0. Since the motion is not biased we cant have any biased initial velocity (from the point an elecyron strikes a lattice) so u1+u2+...un should be 0..not their sort of weighted timed mean.
Interesting. My take is that the displacement of an electron, in the absence of any field, over a billion collisions should be zero, because there is no current! What do you think is wrong with that process?
Your words are not clear enough. Tell nicely
@@Mahesh_Shenoy It tends to zero, you can never say it would be exactly zero, but it would be negligible compared to the velocity when electric field is applied, and assuming that current is zero in absence of external electric field is also not completely accurate, it is zero on average as fields cancel out overall but at some places there are non zero fields which are the reason for van der waal forces but expected value is over infinite trials so even in this case though its over billion collisions there will be slight drift velocity but it will be close to zero.
@@Mahesh_ShenoyAs you are saying that there is no current.
Is that current from a single electron or billions of electrons?
If it was because of a single electron then the displacement must be zero
But if it was because of billions of electrons then you can't simply take the displacement of a single electron zero .
You can also think about it in this way like after billions of collisions how much possibility is there for a single electron to come in the same position where it started its motion ?
I didn't find any contradictions here.
The invariant between the original model and the simplified model is the total displacement of an electron.
Take acceleration due to electric field (E) = a = (eE)/m, time between collosions = t1,t2,.... for total of 'n' collisions of a single electron.
Then, L(displacement) = sum of displacements along field = 1/2 * a * Sigma(ti^2) (i = 1,2,3,...,n)
So assuming a common time (I think HCV does the mistake of calling it average time, because people usually associate average with 'mean') 'tau' between successive collisions to get the simplified picture,
L (displacement) = n * 1/2 * a * tau^2
We should have the same displacement for both scenarios:
This gives: tau = sqrt(Sigma(ti^2)/n) (i = 1,2,3,...,n) , i.e. tau = RMS value of actual time intervals (not the mean)
The vd (drift velocity) = total displacement/total time = L/(n*tau) = 1/2 * a * tau
This also solves the problem of L being small displacement between successive collisions, as now we're talking about total displacement over n collisions. (although individual displacements also work, like given in the book, because all of them are equal in the simplified model)
I don't think HCV is assuming t1=t2=t3=...tn, it just fails to derive the value of 'tau' and calls it the average value, which leads to this ambiguity.
The only confusion could be calling the RMS value of something as its average, when the common assumption is average implies 'mean'.
The true mean that we're calculating is that of the individual displacements. So 1/2 * a * tau^2 is the average (mean) displacement between collisions.
Let me know if there's anything wrong with this.
13:27 this inference is wrong, the summation of displacements results in zero when there is no Electric field, but if there is an overall net electric field the summation cant be assumed to be zero, just because 2 expression look same in structure doesn't mean they will evaluate to same value.
The ut in the formula is the component that accounts for when there is no electric field.
So it is zero
The 1/2 at² accounts for the acceleration due to the electric field
INITIAL v. BEFORE acceleration came into play
Please make a video to show the full calculation of 9.13.
I will explain why Verma sir has written 1/2....will wait for you reply
Glad this video came in my recommendation
I was so confused between these formula what to use in exam
Can you solve it rigorously that how did you get that 2t term as in chat gpt i got the result that if t1 is not equals to t2 and so on then there's no direct proof
bro is asking chat gpt
12:44 all the electrons can drift automatically . But what if half of them drift the other way ? Then the current would be zero right? 👀
Isn't that why there's no current without potential difference?
he said "all the electrons can drift automatically so we get a current " so why does he assume that all the electrons drift the same way ?
@@monke4216 If half can drift the other way, half the collisions should also make the electron displace in the opposite direction giving net displacement zero.
Remember, a billion electrons moving randomly within a single collision = the motion of a single electron over a billion collisions.
@@Mahesh_Shenoy It still comes down to net displacement = 0 implies current = 0, no?
Some math that made the process slightly more rigorous in my head:
What has really stumped us is that sum-of-squares term. But if t1, t2, t3, ... tn are all independent random variables following the same statistical distribution, then there's a nice way to express that sum-of-squares term:
(t1²+t2²+...tn²)/n=(variance of t's)+(mean time 'tau')². This comes from the definition of the variance when used to describe random variables.
Further, we can assume that the t's follow an EXPONENTIAL distribution with average value tau. This is a decent assumption because in the real world, many 'waiting times' follow this distribution. In this case, t1, t2, ... tn are the times that we have to wait between collisions.
In an exponential distribution, the variance is numerically equal to the mean, squared.
This means that the sum of squares of t's = n*( tau²+tau²)! That's where the factor of 2 comes from!
This isn't anything novel that wasn't covered in the video, but hopefully the math made it slightly more rigorous :)
It should be (Expected length) / time, rather than Expected drift velocity per collision. Meaning you only need to do E(t1^2+....) / E(t1+...) and not E(t1^2+..../t1+...). At least both of these can't give same answer either.
i would love to see the QM derivation of the Sommerfeld model. The problem is really that the classical physics is insufficient and the best we can do classically is to first establish the simple instantaneous velocity function as being qE/me times t . Now we can take the Expectation (averaging) but we have to then employ a "mean free path time". But this is problematic in QM and of course one needs to account for the fact that Expectations under any distribution obey: E[t^2]= E[t]^2 + var[t]. Getting the distribution right is where the real work s and this is why Sommerfeld makes the proper correction over the original Drude model.
The derivation can be done by using the first equation of motion for n collisions of an electron and then averaging it out. It gives what Feynman derived.
I think I am still confused about something, Comparing the derivations in both cases,
The feynman lectures assumed the following - "It is just the acceleration F/m (where m is the mass of the S-molecule) times the average time since the last collision. Now the average time since the last collision must be the same as the average time until the next collision, which we have called τ"
The fenman lectures said this before that line about τ - "If we start to observe an S-molecule at some instant we may expect that it is somewhere between two collisions. In addition to the velocity it was left with after its last collision it is picking up some velocity component due to the force F. In a short time (on the average, in a time τ) it will experience a collision and start out on a new piece of its trajectory. It will have a new starting velocity, but the same acceleration from F."
Meanwhile for Hc verma the τ is defined as - " If τ be the averge time between successive collisions"
I feel like in Feynman's definition the τ represents that if you are looking at a particle at a given instance, the avg time until the next collision and the time since the last collision is τ. The highest probility will be that you are looking at the particle when it right between the the past and the next collision. if you look at it like that, then the time between successive collisions will be 2τ, now if you take this value and plug it through hc verma's derivation(which assumed τ to be the time b/w successive collisions), you get L = (1/2)(F/m)(2τ)^2, for the drift velocity, you divide L by 2τ and you get v = (1/2)(F/m)(2τ) = Fτ/m.
I think the problem lies in the assumtions sorrounding what τ represents, one assumes that it is the avg time until the next collision and time since the past collision, meanwhile the other assumes it as the time b/w 2 collisions, this why hc verma has a 1/2 in the value, hence his τ(time b/w succesive collisions) becomes τ/2 which then represents the time since the last collision and till the next collision.
Yeah you're probably right about this
But still, clearly the assumption regarding multiple accelerated motions being considered as an averaged accelerated motions displacement is a wrong notion, due to the arrival of the term (t1^2 + t2^2...)/(t1 + t2 + t3...)
Let's say for a value of 2 for acceleration, displacement during two times 3 and 4 seconds is given by:
What HCV is essentially implying is that:
(3^2 + 4^2)/2 = (3.5)^2 (clearly incorrect, gives 12.5 = 12.25)
Hi makesh sir plz write physics textbook for class 11 and class 12 Surely ur book will be next halliday and Resnick book
Aww shucks, that's such a compliment!
- or + depends on charge of charge carrier.
For electrons it is -
I think in Richard feynman's book the velocity is of that instant when electron is just collidies with the next electron and in sir hc verma's book he has taken the avg velocity of the electron after colliding.( please correct me if i am wrong )❤
I have a question how do electrons bumb with atoms and go free?, are they not supposed to be attracted to the nucleus of the different atoms
just realized you're the voice behind Indian SyllabusKhan academy videos.
The point of contention is due to the use of different equations of motion to derive the drift velocity.
What if we start with v=u+at
and for each collision find out v and then average out v.
then
average v= [(u1+u2+.......) + a(t1+t2+......)]/N
(Here, N represents the no of collisions considered)
average v = 0 + a tau= a tau = ( e E / m) tau
This will just find the velocity right before the next collision ig, but we wanna find the average drift velocity, I think this is what NCERT did too and I'm not sure if it makes sense
I would debate of not doing Vd = L1 + L2 + L3 +.....
Since you cannot add them directly because they are vectors, you need to consider their direction and add for which you need to perform parallelogram addition for every two set (L1 + L2) of distance, get it's resultant and add to L3 and it's resultant to L4 etc. ..
Why so complications?
Force= charge × Electric field
F= eE
Now as per Newton's law of motion Force is change in momentum.
mu/t=eE
u = eEt/m
It is as simple as this. No need for integration and statistics average. Though I didn't study HC Verma during school days.
This explanation is incomplete. You are right that due to the effect of variance, HC Verma's value is an underestimate. But there is no reason why the correction factor is multiplying by 2! The correction factor can be a any arbitrarily large positive number. Proof:
Correction factor is [(t1^2 + t2^2 .... t^n))/n divided by [(t1 + t2 + .... tn)/n]^2
suppose t1 = t2 =t3... = tn = 1 (distribution with 0 variance); then correction factor is 1.
But suppose t1 = 1 while t2 ... tn is 0. (distribution that maximizes variance).
Then correction factor is (1/n) / (1/n^2) = n
n can be made arbitrarily large is unbounded.
All you've proved is that drift velocity lies somewhere between HC Verma's answer and Infinity. You've not proved that twice of HC Verma's answer is correct.
But t2, t3,.. are non-zero values...
So, the second case is not possible....
Even if t2, t3... are infinitesimally small, we are adding infinite number of values, so we can't say t2+t3+... = 0
if t2 ..... tn is 0 then there is no need of correction factor.
this also implies the electrons undergo NO collisions.
then t = t1 which in this case you took it as 1
there won't even be a concept of AVERAGE time so drift velocity would just be the normal velocity of the electron due to the electric field
If we take infinite electrons at a single time and find drift velocity
We get average final velocity of these infinite electrons as eEt/m but this average final velocity.
Of we want to find out the drift velocity (which is the average velocity during the whole free path
To find that we have to find average of inital average velocity and final average velocity.
Means that drift velocity will be equal to (initial average velocity + final average velocity )/2
= (0 + eEt/m)/ 2
So the correct drift velocity will be
eEt/2m
I liked your videos on khan academy india but they haven't been arranged in a playlist so I don't know which one to watch after the first 4 videos of this topic(current electricity). Neither on the site nor in youtube playlists. Going back to HCV for now 😅
yea thankfully our teacher explained this to us, and he told us that its not 1/2
Sir, which software are you using for writing?
No Mahesh your argument is totally fine. ProfHC Verma can also make mistakes he is a human
I also found some errors in his book
The problem is for over many years no one tried to correct it
@aryan4679why are you using bhakt word in insulting sense? You know the literal meaning of word is sacred in Hindu religion?
Why people have normalised using the word as insult, would you have used the word namazi, maulvi, rabbi, or bishop as insult?
15:18 the assumption "time gap between collision is equal" is mentioned in his book. And based on the the assumption he is quite correct, but yeah the assumption goes very far way from the actual values.
@Mahesh_Shenoy, @FloatHeadPhysics..
It was a nice video but i feel there is a small mistake in your derivation as well.
In this equation of drift velocity, (Vd=L/t), over a long period of time, "L" is the horizontal length covered by the electron (call it drift length). It cannot be equal to (L1 + L2 + L3 + L4.....).
It is not even the displacement, displacement would be a straight path connecting the starting and ending points of the electron, in the respective time period being considered.
It is just the horizontal length travelled in time "t".
I might be wrong, please correct me if i am.
Thank you!
I got 23 in physics...
Out of 25.
Ok we need more of these for all concepts in HCV
I didn't find any major errors besides this! Did you?
@@Mahesh_Shenoy We knew during our preparation that this was wrong but the explanation here is gold so I would really like to watch more of your explanation for different concepts that are there in HCV right or wrong doesn't matter just the thought behind it like you presented here
@@roughero1353 Oh man, that's really encouraging!!
I'd summarize this as: avg(f(t_i)) is the same as f(avg(t_i)) ONLY if f is linear! Linearity. That's the first thing they tell you in Maths to be able to change the order of the operations! :D The distance calculation (double integral of acceleration) is a square of a, definitely not linear!
I know I'm really late to the video to get a response but for some reason, a program I made to test out this factor of 2 seems to be giving me a different answer, and I'm perplexed as to why. If someone could offer an explanation, that would be great.
When I compute the parts of the equation pertaining to tau, I get feynman's value not to be twice verma's, but rather 4/3 times it. I'm using completely randomised intervals and as far as I can tell, the program is logically correct, so why the discrepancy?
Below is the program and some sample outputs. It was made in Python and uses the default "random" library :
Program :
```
import random
collision_count = 10000 # this value can be adjusted for n collisions
total_duration = 0
squared_duration = 0
for n in range(collision_count):
# generate a random time between collisions
t_n = random.random()
total_duration += t_n
squared_duration += t_n ** 2
verma_tau = total_duration / collision_count
feynman_tau = squared_duration / total_duration
# feynmann_tau should ideally be twice that of verma_tau
factor = feynman_tau/verma_tau
print(f" collisions : {collision_count}" +
f"
duration : {total_duration}" +
f"
verma : {verma_tau}" +
f"
feynmann : {feynman_tau}" +
f"
factor : {factor}")
```
Outputs :
#1
collisions : 100000000
duration : 49998360.500493824
verma : 0.49998360500493827
feynmann : 0.6666662707776526
factor : 1.3333762629497983
#2
collisions : 1000000
duration : 500194.6072459398
verma : 0.5001946072459398
feynmann : 0.6668238495570457
factor : 1.3331288260554481
#3
collisions : 10000
duration : 4996.37857230061
verma : 0.49963785723006104
feynmann : 0.6659245185748535
factor : 1.3328143753290993
HC Verma is not wrong both the formulas are correct from their point of view
In HC Verma time is taken as a linearly increasing function due to which it's average comes out to be ½(time)
So, both the forms are correct
I look at this video with so dumb look
How did I pass my exams I wonder😅
I don't think this is a mistake to be honest, I think it is more like calculating the escape velocity of a rocket without taking air resistance in to account. The assumption that each each collision takes roughly the same time interval is probably an "ideal case scenario". We have to keep in mind that this book is meant for 12th grade students and diving deep into complexities of a topic like drift velocity, might not be necessary at that moment.
but no where is it mentioned that we are assuming the time period between each collision to be the same.
in most other cases things like that are mentioned like the frequent use of 'massless' frictionless' in the numerical questions
It’s more like assuming gravity is a constant (say till a particular height and the zero above that) and then deriving the escape speed!
There is a reason why HCV himself considers Resnick to be the best book ever written in physics.
Drift veocity is average of drift velocity of large number of electrons and for each electron a is qE/m
this can be checked in "drift velocity and collision time" by Donal M Title American General of physics.
Lastly in my humble opinion in science we are all learning it is about betterment not about being right and wrong.
wonderful derivation sir.... I had no answer for this dilemma for many years.... NCERT gives right value....
isn't the path between successive collisions parabolic?
Yes it is!
I observed many did the average of final velocity before next collision to actually derive drift velocity . so how can we relate it llike I didnt get it
Excellent work and please keep up the good work. The problem is that a large cross-section of world population continues to live with tribal mindset and herds mentality even today. And we human beings are not to be blamed for this - indeed we lived the life of herdsmen and tribes million years back. This gives rise to cult following. Once somebody becomes a CULT - he becomes a GOD - other members of the tribe won't allow you to question that CULT - that's why you get to Jaggi Vasudev, Ramdev etc.
Two years back I got into an argument with some kids over a RUclips video by H.C. Verma - question was regarding the magnetic effect of a moving charge. I found H. C. Verma's explanation rather naive and incomplete and I have seen many such flaws (of Prof Verma) before. I commented that Prof Verma loves Physics and probably wants to learn Physics, but he is NOT an expert. This irked some kids - some of them appeared and probably about to appear at IIT JEE. They immediately started to throw their weight around by flaunting their IIT JEE ranks and all - which of course I won't cross-verify. I off course won't throw my weight around saying I completed my B.Tech in so and so year - did my Masters in so and so year - happened to work in so and so Semiconductor MNC Majors in so and so places and in so and so positions and took part in so and so R&D projects and that my son is a student at so and so IIT at that point in time. Honestly, when I wrote IIT in 1998 paper pattern was 40 X 2 + 15 X 8 in three hours - my son thinks it was tougher then paper-wise may not be competition wise. And anyway, an IIT rank itself isn't a testimony of one's understanding of Physics.
I just tried to be humble. But finally I had to say that Prof Verma is more of a HYPE and a CULT. And it is extremely difficult to convince others that CULT-FOLLOWING isn't a good idea.
Thanks for this very thoughtful comment!
But, I think prof. Verma is really good. His books are extremely helpful. It explains things very deeply and in a simple language!
It's because he is so good, it's worth pointing out places where I disagree :)
it v = ri while deriving u have to keep it mind it is not a fundamental law it work only in certain range so in approximation both derivation are correct
I think, he uses the equation of motion, s = ut + 1/2 at²
Even if we assume that he's correct by saying that sum of all u will become 0, but in such case, we will find S rather V (v is drift velocity while S would be displacement)
In order to find the correct formula using the formula HC verma derived, we just keed to differentiate both side with respect to time
S = 1/2 a × t²
dS/dt = 1/2 a × d(t²)/dt
V = 1/2 a × 2t = a × t
Then we can replace a by eE/m
That gives you instantaneous velocity after time t since the last collision, and not the average velocity! :)
@@Mahesh_Shenoy oh true, completely forgot about that
Well I haven't ever used HC Verma, the other books I read, have the formula derived using V = U + AT, so the sum of all u becomes 0 and sum of all T becomes Tau
@@sanjeevsinghrajput5593 Yes, but there is a subtle result overlooked!
NCERT, surprisingly, has done a really good job at it.
@@Mahesh_Shenoy surprisingly
sir the most important part here is that we have mathematical conception of Ohms law and resistance of a conductor, here drift velocity or most probable vel or rms vel may have less importance
Get a life
You are comparing a professor with Noble prize winner.
The answer is given by HC Verma sir himself. He said that all these derivations are approximeted that Newtons law directly can be applied. But the reality is there quantum physics must be applied. So both the methodes are approximated and hence order is important and not the magnitude.
ruclips.net/video/jEKUog6-LiA/видео.htmlsi=Y8iR9B1YTSVV4Vjg
Watch it out
@@manojkarmakar4454 The point is even under those assumptions the derivation has an error.
@@Mahesh_Shenoy ruclips.net/video/jEKUog6-LiA/видео.htmlsi=Y8iR9B1YTSVV4Vjg
Here we are averaging distances but we are calling it a velocity... How its possible? Shouldn't we average displacement?
its displacement
@12:33 Aren't you calculating average speed not average velocity? If its average velocity then you cannot add L1+L2+....Ln directly?
Bt there is a big mistake in other formula also... He is calculating drift velocity just before the collision... The e- has not travelled full distance between 2 collisions with this... This is the velocity gained till just b4 next collision... And to use this as avg is also wrong v=u+at❌
I studied the same topic 5 minutes ago and just opened RUclips to see this recommended lol
Haha! YT is scary
Shouldn't quantum mechanics instead of newtonian mechanics be used? Moreover the length of path travelled by electron would be uncertain... due to hiesenbergs uncertainity , it would be impossible to accurately determine the position of the electron?
This is why we are using statistical approach
Just a little difference between instantaneous and avg velocity...
Hey, I think you are also wrong in assuming that you can take out "a" and group (t1² + t2² + t3² + ... tn²)
It can be corrected by having a_l = local acceleration vector
a_g = global acceleration vector (which you're calling "a")
Then,
v_d τ = Avg(½ u t + (a_l + a_g)t²)
= Avg(½ u t + a_l t²) + a_g Avg(t²)
= 0 + a_g Avg(t²)
You're assuming that the acceleration between every successive collisions is same which I'm doubtful about. They are colliding with charged bodies which would have a local electric field.
This isn't affecting your result because later you assumed Avg(½ u t) = 0, which cancels the effect.
Even rajwant sir told about this mistake
what do you prove foriegner is right