ERROR at 21:00, the 1st line, there should also be a (partial v1/partial u1)*e_1 term because of the summation over k. I neglected to add this term in. ERROR at 21:11, the 2nd line, 1st component, v2 should be multiplied by 1/2 sin(2u), not a subtraction. Minor error at 19:52; I wrote e_z instead of e_y in the last 2nd order derivative
Hi Chris. You know, I was desperate to understand the whole thing about COVARIANT DERIVATIVE. I had to leave my readings about tensor and started to read only about Differential geometry. I got dizzy with all this mathematical verbiage and had to leave the issue. But now I see your video and believe me, it was like to uncover my eyes, I can see the light again and you did it only in half hour. Why the expositors do not take time to show the origin of their ideas? It is not difficult, and this video is one example of that. Maybe they only repeat like parrots the others dishonesty, maybe there are good books, but is very difficult to find them. The higher you go in math or physics the more complicated they make you the understanding. It is dishonest to do that. Thank you, thank you, very much for your job. Please do not stop, there are still a lot of concepts that need to be clarified and be able to understand Einstein's equation
Gerardo Guevara Pavel Grinfeld’s book on the subject is an excellent introduction that isn’t bogged down by pedantic abstract math formalism. He also has a YT channel MathTheBeautiful with his lectures on the subject (he happens to be a fantastic lecturer as well as a good writer, so an excellent resource). I also recommend Bernard Schutz’s books on GR and mathematical physics as well as Frederic Schuller’s video lectures when you want to start diving into the more abstract side of things (highly recommend some primer on real analysis, topology, and abstract algebra first though).
@@MrPetoria33 I stop watching MathTheBeautiful to watch eigenchris. No one explains Tensor Algebra and carries out the same notation, jargon, concepts, and tools to Tensor Calculus so well on the internet! Thanks eigenChris!!! Fantastic work! you changed my life too!
You don't how much I feel you mate. It looks like the world's been invaded by zombies who repeat a meaningless rigmarole. Those who can explain the meaning and origin of ideas are gems in the desert. I don't even know how some people are convinced to be understanding when to my criteria they are actually not. It is still a truly mysterious subject to me.
Incredible, this is the clearest and most comprensible video I have ever seen on parallel transport. The series is getting really interesting and I am alway waiting for the new video. Thanks for your great effort.
Belíssima aula. Didática, simples e objetiva. Derivada covariante nunca será um conceito difícil, após estudar esta aula. Não vi, até hoje, explicações sequer parecidas com as que você disponibilizou para nós, simples mortais. Parabéns.
Dear Chris I can not express how much grateful I am to you. I am an amateur physicist and electrical engineer. I have been cherishing the dream of learning Non-Euclidean geometry and GTR since I am a kid. But I was unable to grasp the concept of Tensor analysis until the final year of BSc.Since last one and a half year I have taken various initiatives to learn Tensor but failed every time. I found your video series just 2 days ago and watched almost all videos. It has been so useful that I gave me the thrust that no one could not give me ever before. Thank you Sir. I wish you all the best ♥
Thanks. I'm happy this series has made your life easier. I've tried and failed to learn tensor calculus and GTR several times. That is why I started making this series--hopefully it smooths out the journey for others.
Wow! I have been reading multiple resources about parallel transport. This is the most intuitive and the best version, especially at 32:00 when you show the cone. This is crazy!! I love it!! Thank you for making all these great videos.
This is the clearest explanation of the covariant derivative (CVD) I have ever seen. You are especially clear on the connection to parallel transport and the CVD. Most textbooks use parallel transport to vaguely relate it to the CVD. You closed the loop very well.
Outstanding intuition building! You seem to have the only videos on the subject that demonstrate understanding of the source material. ie: you can clearly explain things.
Hi Chris: The Schild's ladder construction (p. 248 of Misner, Thorne, and Wheeler and en.wikipedia.org/wiki/Schild%27s_ladder) so far makes the most sense to me to define parallel transport. It applies to parallel transport along any curve (not necessarily a geodesic). It works for your example of the curve being a circle in flat space (thanks for the example).
one extra question: according to your definition of the covariant derivative (which is the ordinary vector derivative in R^3 space subtract the normal component),it seems the graph in 3:40 fits the definition as well since there is no rate of change of the vector in R^3 space,then if we use x-y-z coordinate to calculate the rate of change of the vector,then it should be zero. The normal component should be zero as well(since the normal component of the rate of change of this vector is zero),therefore the final value will be zero. because you use extrinsic perspective in the graph,therefore you can use normal xyz coordinate to treat the vector as normal,if every point on the sphere assigned with a vector that is pointed into x direction,then the "ordinary part" whould be zero(there is no normal part as well),therefore this fits the definition as well. However this cannot be true since in 3:40 you said that the vector will be pointed into the sky geometrically ,but algebraically we can get zero which fits the definition. there is contradiction between your explanation and the definition?
I should have said that the definition only applies for tangent vector fields. The example at 3:40 involves a vector that is not tangent to the surface.
If more textbooks followed the style presented here (smart use of colored text, good illustrations, lots of examples), math would be way more accessible.
Very nice but I stumbled a little at 14:30 to 15:00 when you say “and since we subtracted the normal component all these just cancel out. “ or something like it. Isn’t it the case that you first computed what would be the normal component, namely vj L_ij times the normalised normal vector - and then declare that it is thrown out because that’s how you defined the covariant derivative, as total directional derivative with the normal component ignored. It seems to me that you compute what amounts to the normal component as the term with the 2nd fundamental form and then say that you want to set that to zero. I just find it weird that you first write minus n all the time and then suddenly have two components that are supposed to cancel out, while not showing why they cancel out (?). I found that spot a bit confusing. Later it seems to make sense again, sort of: basically you have to choose the christoffel symbols such that the whole formula holds. It seems that you just compute the full derivative by using chain rule and what else, and then project the result back in the tangent space.
The "n" with the arrow on top is supposed to represent the generic normal component of the derivative. We have no idea how big it is, we just know that we want to subtract it off. I then do the work of splitting up the derivative into its tangential parts (with the christoffel symbols/gammas) and its normal part (with the second fundamental for/L_ij). This normal part with the second fundamental form is the normal part we wanted to eliminate in the first place. In other words n(arrow) = v^j * L_ij * n(hat). So we just cancel them. The "subracting n(vector)" and "declaring we need to throw the normal part out" are the same. Does that make sense?
eigenchris Thanks a lot for this additional explanation. It became clearer to me now. Actually I began to get it from your next piece where you give the explanation based on intrinsic geometry and then also reproduce the form shown here. I feel like I finally make some progress in understanding differential geometry following your videos.
@@eigenchris Trying to understand better -- This section starts off by finding the covariant derivative of a "tangent vector field". Doesn't that mean that there is no normal component at all? Which would imply that there is no normal component to subtract off? And if this is correct, then IF you happen to have a vector field that is NOT tangent to the surface, THEN you would subtract off any normal part?
@@ronlaspisa450 The tangent vectors themselves don't have a normal component, but the rate of change of the tangent vectors (their derivative) can possibly have a tangential component. This is what we're subtracting off.
@@eigenchris Thanks for making this amazing series of videos! I was also having some difficulty trying to understand this part. But I just figured out the normal vector n(arrow) is the normal part of dv/d(u^j), not v itself. So, the covariant derivative is the projection of the regular derivative onto the tangent plane. (this might seem obvious but I was really stuck on that part, lol)
I think that you should say the covariant derivative is the rate of change of a vector field with respect to the coordinates that describe the surface. In this particular case the two angles of the constant radius sphere.
Covariant derivative notation looks like gradient of v by direction w. But they are different, right? Especially, gradient must include inverse metric tensor, but covariant derivative doesn’t care.
The gradient takes a scalar field and produces a vector field. The directional derivative (dot product of gradient and vector direction) takes a scalar field and a direction, and produces a new scalar field. The covariant derivative is more like a generalization of the directional derivative. The covsriant derivative takes a tensor field and a vector direction amd produces another tensor field.
Hi @eigenchris, really enjoying this series. I know this question is very late now, but is there an explanation for why at 15:04 the normal components v^j L_ij nhat - n go to zero? My suspicion is that the vector produced from v^j L_ij nhat is normalised by the definition of L_ij the 2nd fundamental form? Or is there something about the way the covariant derivative is defined?
Chris, big fan here, but how do the Christoffel symbols tell us how much of the tangent plane vectors we need to produce the vector in the tangent plane corresponding to the three dimensional vector in purple as illustrated here. The Christoffel symbols are partial derivatives of the metric and as such tell us how far the basis vectors deviate from the coordinates. How does this tell us as normal components how much of each basic vector we need. answer this and I'll buy you more coffee. thank you 🤓
I found interesting that unifrom circular movement in physics is actually parallel transporting a vector, in this case the tangent vector, the velocity vector always tangent and in order to get the movement circular there is always an aceleration pointing towards the center. in this case we always get the same vector or the tangent field vector space.
Wouldn't the formula for covariant derivative be d v/ du^i - (n.(d v/ du^i))n=d v/ du^i - v^j L_ij n, in other words you need to scale n according to how much dv/du^i projects on it? See 14:45
Just to be clear, is the superscript on partial of v^j with respect to partial p^i @0:59 seconds supposed to be partial of v^k with respect to p^i (inside of the parentheses) when the covariant derivative in polar coordinates is expanded? It is written this same way in video #17 on covariant derivatives in a flat space. Just wanted to be sure that the index j is swapped for k so that we can factor out e_k from both terms in the sum...
Hello Eigenchris. First of all, I would like to congratulate you for the quality of your explanations and the work you have done on the tensors. I am interested in the geometric interpretation of the parallel transport of a vector along a meridian which is a geodesic. I find the image of the man carrying a javelin excellent. But I have a problem with the explanation you give in step 7m10s (ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-Af9JUiQtV1k.html ) . The difference of the 2 vectors which gives the green vector dv, does not point exactly to the center of the earth from the geometrical point of view (the green vector is neither perpendicular to the first vector, nor to the second). So the dv is not normal to the surface of the earth, and this contradicts a zero covariant derivative. On the other hand it is clear that the norm of the vector (distance of the javelin) is preserved. Could you give me an explanation please?
Great video. I cant really thank you enough for these. But I have a question. This definition really reminds me of the Lie derivative, they say lie derivative is independent of metric,but how? How is the covariant one depends on metric,but the lie derivative doesn't, when both are defined on a "direction" defined by some parameter?
I think video 21 is on the Lie bracket (which is the Lie derivative for vectors). I recommend looking at it, but unfortunately I think I botched the explanation of this exact question somewhat and I will probably need to redo the video eventually. The Lie derivative is defined as L_X(Y) = X(Y) - Y(X). This basically means taking "how Y changes along X" and comparing the result to "how X changes along Y". Because there's a "subtraction" involved, the 2nd derivatives in each term cancel out and you don't actually need to define connection coefficients in order for this quantity to make sense: L_X(Y) = X(Y) - Y(X) = (X^j ∂_j(Y^i) - Y^j ∂_j(X_i))∂_i. In a certain sense, the "comparison of two things" eliminates the need for a connection. The covariant derivative shown in this video is about seeing how a vector changes along a specific path, and this requires a definition of how a vector moves from one place to another. The "connection" defines this for us and so it is needed for the covariant derivative.
Chris, this is a fine series. Someone should give you a medal. However, there's one point which is troubling me in this video - you are relying on the dot product in your calculations, which means you need a metric tensor. AFAICS, to use the dot product with the n normal vector, you need a metric tensor that works in the 3d extrinsic space, rather than just the 2d intrinsic space of the sphere - where are you getting the components of that metric tensor from? Or have I confused myself?
amazing video. I'm having trouble demostrating how the covariant formula you derived simplify when lambda is a geodesic. I read only the first term is still there, and the one with christoff symbols is zero. Why it sort of makes sense i cant understand mathematically why it vanishes. Can anyone help?
If I have a vector field T and a vector field V, where V is projection of T onto the surface tangent plane, is the covariant derivative of T equal to the covariant derivative of V?
@@eigenchris Thanks! I’m interested in computing principal directions / eigenvalue of a rank 2 tensor (the gradient of a tangent vector field). Any chance you can give me some insight into this / point me in the right direction for learning more about this?
@@TheLazyEngineer I'm a bit confused by your question. I'm not totally sure what the "gradient of a vector field" is. I normally think of the gradient as an operation applied to a scalar field. Also, would the procedure for getting eigenvalues be different than the standard method taught in a first year linear algebra class? Maybe I need more context on what you're doing.
@@eigenchris Sorry, I dont have the differential geometry language down yet! Let me try to be more precise. If I take the covariant derivative of a tangent vector field, that gives me a rank 2 tensor right? The basis of this tensor is e_i (X) e^j where (X) denotes the tensor product, e_k is the covariant basis vectors and e^k are the contravariant basis vectors, right? I am interested in the eigenvectors and eigenvalues of this tensor. I think you are correct that it shouldnt differ from the standard method. We can express the tensor as T_ij e_i (X) e^j. And the eigen-guys would satisfy: A_ij e_i (X) e^j * x^k e_k = lambda * x^k e_k. So i'm pretty sure I can just work directly with the components T_ij and apply the standard method. But I want to be sure my understanding is correct hah!
Hi Chris at the end of the video min 31: 48 suppose I want to go through the same curve Lambda and parallel transport a vector in the direction e2, to get a covariant derivative equal to zero the vector field should rotate slightly clockwise correct?
Each term is its own separate j-summations that are independent. We can relabel each pair independently. The fact that both sums had j-indices was just a coincidence.
Dear Chris. Please forgive me with my observations. I just want to confirm one thing. IN minute 21:03 I think you left aside one term in the covariant derivative of (v) respect to e(1) , and it is the partial derivative of v(1) respect to u(1) times e(1). Am I correct? Or that term is also null?
Thankyou for your marvellous lectures. I have just a minor quibble regards your diagram of the tangent vectors diagram (Tensor calculus 18: (23.31min). I thought lambda proceeds 0 -> pi/2 and therefore should the tangent vectors therefore begin at lambda=0 with a vector 1e2 and rise in the northern hemisphere until lambda=pi/2 and a vector -1e1
It's been a while since I looked at this video. Are you confusing the formula for the vector field v with thr formula for the covariant derivative of v? I think the diagram is correct.
One question for the cone illustration at the end: Are you saying that covariant derivative only depends on the first-order (tangent) property, and that's why you can just replace the space by another one that has the same tangent space along the curve and get the same covariant derivative?
We want a derivative that can measure the rate of change of vectors that live on the sphere's surface only. You can imagine trying to measure how a river on the earth changes. Is the river straight? Or does the river zig-zag back and forth? If the river is "straight", there will still be some slight curvature due to the curvature of the earth. We want to ignore this, so we substract the normal component of the derivative. So when the covariant derivative is exactly zero, this tells us that a river on a curved surface is "as straight as possible" on that surface.
You said it’s impossible to define a constant vector field on a curve surface. I’m assuming that you meant by parallel transporting. But what if you parallel transport a normal vector? Would that not be a constant all over the entire surface?
Parallel transport depends on the path you take, so you will end up with multiple possible "constant" vector fields that have the same starting vector. Also, none of these fields will be continuous. They will all involve a constant "jump" at some point on the sphere (usually at the north or south pole).
Hi, Chris, I have returned to read about GR and after seeing this video I wonder in 4-Dim how you can decide which is the normal component? Or in the general case of the covariant derivative of a general tensor what is that part that you must eliminate?. I guess it has relation with the way you parameterize the phat because of many books define the covariant derivative in terms of components of the tensor doing the dot product with the tangent vector. MAybe you will treat these issues in the next videos. I hope not to bother you with my questions
I will cover this in my next video. Spacetime is intrinsically curved, and nor embedded in a higher-dimensional space, so there is no normal component.
I can't understand the transformation (u,v) --> (X=cosvsinu, Y=sinvsinu, Z=cosu) because u and v in X,Y,Z are angles, as you have shown in the example of the sphere in 3D space, instead u and v in 2D space (plane) are the "x" and "y" cartesian axes, not angles....Why?
At 32'24'' wouldn't the same picture with the cone apply if the curve was on the equator like in the previous example? And if that's the case wouldn't the covariant derivative on the equator curve also have nonzero covariant derivative?
A cone that sits tangent on the equator would be "infinitely tall"... in other words it would be a cylinder, and parallel transport on a cylinder is the exact same as parallel transport on a flat plane, so there's no "twisting" of the vectors.
Good Day Chris, in your Video 23:22 stated covarient derivative along equator is non Zero..... I do not really understand it because normally equator is great circle and covariant derivative should be Zero or i miss understand here something?Could you explained it if i Was wrong here.... Thank you..
The covariant derivative requires two things: (1) a path, and (2) a vector field. The equator is just a path. You also need a vector field. Some vector fields along the equator will have non-zero covariant derivative (meaning they are not parallel transported) and other vector fields along the equator will have zero covariant derivative (meaning they ARE parallel transported). The example at 23:47 gives a vector field along the equator with zero covariant derivative.
I don't have a book, but the best lecture notes I found on "classic" differential geometry (before Riemann manifolds were invented) are these notes by Lia Vas: liavas.net/courses/math430/
regarding mistake at 21:00, doesw the e_1 term have any gamma part? Or is it partial v1/partial u1*e_1 + [partial v2/partial u1 + v2*Gamma 221]*e_2 ????
Why use the term parallel? Why not spheres walking? The use of super scriipt 1 and 2 to identify components confuses with the power operation (at first glance)
The concept of parallel transport works in any curved space, not just sphere. The idea is to move a vector while "keeping it as parallel as possible". I agree the upper components do look a bit like exponents, but I usually try to use colour to indicate when a number "belongs" to a symbol as an index, and black when it is an exponent.
Thank you for your very kind reply. I like "Parallel As Possible Transport",(works in any curved space), but that name still does not identify the purpose of COVARIANT DERIVATIVE, would it be Coriolis Effect?
I think the Coriolis Effect is related to rotating reference frames. The covariant derivative measures how much a vector field "moves away" from the vector field we'd get from parallel transporting a vector. As an example, let's say you have a path from the north pole to the equator, and every meter or so along the path you place a spear. The spears can be pointing in various different directions. These spears are a "vector field" along the path. Now, you stand at the north pole, with a special spear in your hand, pointing ahead, and you begin to walking along the path. You will see that the spears on the ground might point in a different direction than the spear in your hand. The "different" between the ground spears and the one in your hand is the result of the covariant derivative... it tells us how much vectors in a vector field deviate away from a perfectly parallel-transported vector. You need to keep in mind derivatives are a the result of a "limit", though, so it only applies in a small region.
Question on 08:00 : The magnitude of the rate of change is 1 -- is it similar to the situation where you demonstrated that the partial derivative of a vector against a coordinate is the unit vector in the direction of the coordinate, i.e. taking the limit as the parameter difference approaches 0 ?
@@eigenchris The derivative is set equal to unit normal vector n^, whitout any scaling factor. So the magnitude of the derivative would be 1 always. Where am I wrong with this ?
Can someone please explain me why, at 20:02 , the fact that so many of the christoffel symbols are 0, leads to a term which is a pure multiple of e_2 on the expansion of grad_e1(v). there is a summation over k as well. aren't we supposed to have (partial v1/partial u1)*e_1 terms in the expansion as well?
Dear Sir. Thank you for your videos from which I benefited. I am little confused about definition of parallel transmitted and its relation with covariant derivative . In the example for a vector directed down word ( e1) along the equator, the rate of change of the vector is zero since it does not change along the curve but the covariant derivative= rate of change - normal = - normal . So how the covariant derivative in this case is not zero contradicting with the definition of parallel transported which requires covariant derivative= rate of change - normal = 0 ?
If you take the vector e1 at the equator, and then move it slightly along the curve, you should see that it has changed. If you align the vector "tails", you should see that there is a gap between the vector "tips". This difference between the tips points in the normal direction, and so this is what gets subtracted off, so that final result should be zero.
Thank you for your reply. I think the vector e1 does not change along the equator. it always points down word so the difference should be always zero. In this case the normal is zero which is subtracted from rate of change of vector e1 which is zero as well. This leads to zero covariant derivative. That makes sense to me now
Sorry... you are correct. I had my basis vectors mixed up... But as you say, since the rate of change is zero amyway, the normal component of the rate of change is zero anyway, so the subtraction of the normal component doesn't change the answer.
@19:08 why don't you apply the d/du^i to the the dR/dX, dR/dY, and dR/dZ? You only apply it to their coefficients, I thought the second order partials had components normal to the tangent space as well.
Yup, you answered it yourself. To be precise, we should use product rule and differentiate the dR/dX and other basis vectors, but they are all constants so their derivatives go to zero.
when people walk around on a curved surface like the earth they always turn and walk in a forward direction no matter which direction they are proceeding.....they do not walk sideways along the equator and they do not walk backwards from the equator to the north pole.....unless they are mathematicians
I arrived at minute 21:20, stop, look better, wait, I think there's an error, mumble, mumble, go back, check all, mumble mumble, after 10 minutes, let's read some comments maybe there's an error. Read the description....
These videos are truly incredible and I'm amazed that I can follow as much as I can, but there's one small problem nagging me and I'd be very grateful if somebody could help. I’m probably missing something obvious but at 21:00 I’m struggling to follow the covariant derivative formula for i=1. I can’t understand why there isn’t an e1 component? If you set k=1 then of course all of the Christoffel symbols equal zero but shouldn’t you still be left with a (∂v^1)/(∂u^1 ) term in the e1 basis?
There are 4 Chrstioffel symbols that have i=1 (the lower-left index). 3 of these are zero, as seen at 20:43. The only non-zero Christoffel symbol with i=1 is (i=1,j=2,k=2). Since this symbol has k=2, there's only a term with the e2 basis vector. There are no terms with the k=1 basis vector.
@@eigenchris Thank you for replying so quickly. I can follow that but the thing that I don’t understand is why isn’t the term (∂v1/∂u1 + 0)e1 included in the expression? I get that all the k=1 Christoffel symbols equal zero but there’s still the partial derivative on the left of the expression in the brackets that doesn’t have to equal zero and this seems to be being multiplied by the basis vector as well.
Oh, sorry. I was reading quickly on my phone and I misunderstood your point... you might be right about that. That's a frustrating mistake for me to have made... I'll have to think about whether I need to re-upload the video or just put a note about the error in the description. It's not really a minor error....
The normal vector for the moving object will be zero: if you only move in a flat 2D plane, it means your velocity vector can be written completely in terms of tangent vectors and you won't need a normal vector.
There's something I don't get with the metric tensor double matrix product. When I try to do the product of the row vector with the metric tensor first, then the product of the result with the column vector I don't get cos²(lambda/sqrt(2)) + sin²(lambda/sqrt(2)). Is this something wrong in my computing ? I ask you that question, because I fear I might have missed something about this metric tensor double product in previous videos.
@@HenriNioto What answer do you get? Could you type out some of your steps? Matrix multiplication is "associative" so (AB)C = A(BC). The order of multiplication shouldn't matter.
@@eigenchris Sorry, I just realized my mistake : I forgot that g22 was sin²(u1) and not sin(u1) (so I computed the product with sqrt(2)/2 instead of 1/2). I think you have forgotten the minus sign in the v expression, though, but in the end, it was probably a deliberate choice on your part because the minus signs cancel out in the product.
