Tensor Calculus 17: The Covariant Derivative (flat space)

[@eigenchris]

Typo at 9:30. the e_x basis vector should be equal to dr/dx e_r + d(theta)/dx e_theta.

[@filter80808]

It's like you've opened the cryptic vault of differential geometry! Thank you, thank you!

[@ΔΗΜΗΤΡΗΣΖΑΡΜΠΟΥΤΗΣ]

There are no words to describe how much valuable are these videos!!! Thank you so much Mr Chris!!!!

[@ozzyfromspace]

Just adding this for anyone that doesn’t have a sense of what “flat” means in this mathematical context. We all know intuitively that if you pick any local basis on your manifold, any latter basis can be expressed entirely in terms of the former one, except at finitely many points (eg. “center” of polar coordinates, where basis vectors are undefined). A curved manifold would have derivatives of basis functions that cannot be expressed in any old random basis at a point on the surface. Such a basis derivative is “sloping down the hill of the manifold” so it would have components that are normal to the reference basis. This is what distinguishes a flat space from a curved space, mathematically, and such information is encoded in the Christoffel symbols. Further, the christoffel symbols can be expressed entirely in terms of the metric tensor field of the space under study, so curvature is a property of the space, and not an artifact of observation. Hope this was useful for someone.

[@canyadigit6274]

Float Circuit This is a very interesting explanation. The definition of “flat space” that I was taught was that if a vector does not change orientation as you transport it around the space then the space is flat, otherwise it is curved. Seeing how this ties in to the Christoffel symbols is very eye opening. Thanks for the explanation.

[@justincampbell3261]

Thank you so much for this! I'm on a journey to understand relativistic equations to use as a stepping stone on to quantum field theory. Trying to learn on my own as much as I can, hoping to not hit a wall where I'd need formal college classes to understand the subject matter. You're videos have been so helpful to that end and wonderfully organized, developed, and easy to follow, I eagerly await your continuation of this subject.

[@eigenchris]

Thanks. Can't say I know much about QFT. If you find any accessible videos on it, I'd be interested to see them.

[@justincampbell3261]

I've been reading books on the subject, especially Brian Greene's works, he explains QFT very well conceptually and has references to the mathematics (which were/are still beyond my grasp). Apparently, QFT (and subsequently, String and M-theory) evolved from Einsteins equations, general relativity led Schroedinger and Dirac and others to develop QFT mathematically. I'm hoping to trace their steps up to modern day, or at least as far as i can make it on my own.

[@doctor-mad]

@@eigenchris Maybe: theoreticalminimum.com/courses/advanced-quantum-mechanics/2013/fall/lecture-6

[@dannytman1]

@@eigenchris I really recommend Alex Flournoy kzbin.info/www/bejne/oGWtpoabgZasiaM

[@jonnymahony9402]

no nonsense quantumfieldtheory is the best

[@dXoverdteqprogress]

I like the clarity and patience of your style. Looking forward to watching your videos on the other definitions.

[@eigenchris]

Thanks. Your videos have been very helpful to me in understanding all this. I'm still working my way through this.

[@taibilimunduan]

These videos are great. I am getting to understand many things which I could never grasp before. Your use of colors is very intelligent, as well as the handwriting on the slides. The graphics are simple but illustrative, andyour narrative lively. All this makes the explanations easy to follow. But the greatest thing is how you combine mathematical proof with graphical intution and examples. One can really get it, while not giving up on the mathematical rigour. I thought it was not possible to give any intuition of the Kristoffel symbol. You managed to do it. Thanks and congrats.

[@eigenchris]

Thanks. I'm glad to hear it. The covariant derivative confused me for years. I'm happy if these videos are helping people get un-confused.

[@eamon_concannon]

This playlist has been very useful review and consolidation, having studied most of the material here from other online videos. Hope to see more from you soon. Thanks a lot

[@HilbertXVI]

Oh my God this is amazing

[@bejoysen4468]

17:57 On the right, the index j in the numerator of the partial derivative on the last line should be k.

[@jasonbroadway8027]

I officially have the concepts of this video in my long-term memory. I had to use a considerable amount of paper and ink to accomplish this feat. Great video!

[@anonymousgeek9072]

Absolute gem of a playlist

[@rsalehin]

Obviously, lots of preparation and effort went into these lecture series. I really appreciate your effort. Looking forward to the "Grand Finale": General Relativity.

[@ThomasKundera]

Thank you for this wonderful, clar and precise video. That's exactly what I was looking for: explaining covariant derivative in a way both very accurate but also intellectually meaningful. I'll spend some time more on that first step, and move to your next video toward the generalized approach. Videos like yours are the reason YT is still a great place to visit :-)

[@valeriobertoncello1809]

Just discovered your channel, and it's amazing. Binge watching this differential geometry series of yours rn

[@rahulsadhukhan4591]

A clear visualization was needed for christoffel symbol, and I have got this. Thank you very much.

[@ResetToZero3210]

Such a nice, clear explanation. Man, write a book!

[@ASCENSiON1989]

Einstein Field Equations here we come!

[@paulhowrang]

Nerd gang :B

[@donitarose4225]

Nothing short of amazing sir! My dad is back to studying gen relativity again!

[@gguevaramu]

Hi Chris This way you have to go into detail about the concepts is very helpful. Few books do this. No one has said before in books I have read about different forms of the covariant derivative. When I saw this concept for the 1st time I thought was only a complete derivative of a vector considering the change in components and bases, So what is the big deal?. Following your videos, I could clarify my own ideas, hope you get enough impulse to post more explanations like this. Especially hope you reach the point to explain the way to relate intrinsic measures with the curvature of spacetime in regard with mass and energy. The Riemann, Ricci and Einstein tensors. Maybe is too much, but your point of view can be very helpful to say that all these concepts can be intuitively acceptable. Thanks again for your job.

[@JohnJoss1]

At 16:52, Third Line Down, the first v^j should be v^k now. Same at Bottom Right, at 17:57. Drives me insane trying to figure out all these summations, even for 2D, and Chris' Videos are the clearest exposition anywhere! 😀

[@thoughtexperimentsbybk1830]

thanks for this explanations and for the time spent for coordinate transformations.👍

[@rajanalexander4949]

Such a beautiful, clear, succinct video!

[@mad_vegan]

15:58 The Christoffel symbols were defined on video 15 using extrinsic geometry and the coordinates u. It's not clear to me that the two definitions are equivalent.

[@rariyaboorg3a142]

You are the best one explain the tensor

[@vkoptchev]

A great series. Just want to point out an inaccuracy in one of the pictures. At 13:33, the direction of δeθ/δθ vector should be horizontal to the left, which is the opposite to the r basis vector at the point where the derivative is taken.

[@eigenchris]

Yeah, that's a good point.

[@miguelaphan58]

..a master piece...wanderfull edition..as always...!!!!!

[@HighWycombe]

Wonderful "Columbo" moment at 16:00 when you say "Just one more thing to mention.." then you introduce us to Christoffel Symbols as if they were an after-thought. Christoffel Symbols are are a really important notation aren't they, as we head towards General Relativity!

[@viliml2763]

Christoffel Symbols were already introduced in an earlier video, I suggest you go in order

[@federicopagano6590]

9:34 u can find the jacobian matrix in terms of r theta, its easy and then invert the matrix to find those coeffitients instead of inverting the functions

[@camb5274]

The covariant derivative (in flat space at least) looks awfully similar to the Euler-Lagrange equation from the Calculus of Variations. This is probably not a coincidence, since both are used to construct minimum or maximum paths through a manifold.

[@AMADEOSAM]

Very good work! Many thanks.

[@phb1955]

I love your lectures. But I am confused somewhat by this lecture because.... Given that you introduced the tensor product idea for tensor representations (an idea that really helped me understand tensor manipulations) and given that a derivative of a scalar gives vector (1st order tensor), I would have expected the derivative of a vector to give a 2nd order tensor (0,2). Did you avoid this in your example(s) by choosing constant components and using cartesian basis when taking the derivatives the polar vectors? I just watched the next video and this answered my question. The use of tensor products (or one example of tensor products) here would really have helped (me). Also, I realized that the derivative of a vector by a contravariant variable leads to a covector.... kind-of makes sense.

[@armannikraftar1977]

Good work Eigen. just wanna point out that at 16:52 when you put the component and basis vector derivative parts in the same parenthesis, you should have also changed the index we're summing over on the components. i think it should be "k" instead of "j" . same thing at 17:55 for polar coordinates.

[@thevegg3275]

Is there any connection at all between covariant components of a vector defined by perpendicular project on to dual basis vectors AND covariant indices of tensors AND OR covariant derivatives of vectors and functions?

[@JgM-ie5jy]

Question on the direction of the change in the eθ basis vector at 13:24 : The difference from two vectors is taken from the "second" vector tip towards the "first" vector tip position. If the blue vector is the first vector and the red vector is the second vector, then the red minus blue vector difference does point towards the origin. But if the "first"vector is the red vector and the blue vector is the "second" vector, the direction of the difference from blue to red points in the opposite direction. What am I missing ?

[@eigenchris]

The initial vector is the red one, and the final vector (after sliding it along the theta curves) is the blue one. The change is "final - initial", or in other words, "blue - red".

[@dl569]

you saved my life, thanks!

[@jaeimp]

These lectures are a work of art! From the color-coded LaTeX that makes the stultifying chain rules palatable, to the many points of entry afforded to the novice, every detail is taken care of so that a listener endowed with a fair amount of persistence and some background in basic calculus can follow. Thank you! Also, a quick question... Could we say that the covariant derivative is a gradient vector field with pointwise changing coordinates? en.wikipedia.org/wiki/Gradient#Definition

[@eigenchris]

Thanks. I'm not totally sure what you mean by "gradient vector field with pointwise changing coordinates". It's true that the basis vectors can be changing from point to point when we apply the covariant derivative... however I normally think of the gradient as an operation that turns a scalar field (an ordinary function) into a vector field. The covariant derivative turns a vector field into another vector field. I would say the covariant derivative is better thought of as a generalization of the directional derivative that works on vector fields.

[@jaeimp]

@@eigenchris I am watching and re-watching your lectures because they are excellent, so this is not meant as a challenge; however, for completeness-sake here is a post on the topic of a gradient of a vector field: math.stackexchange.com/a/509853/152225. As for "pointwise changing coordinates" I simply meant "changing basis vectors."

[@eigenchris]

@@jaeimp This is the first time I've heard the expression "gradient of a vector field". I guess there are approaches to calculus I simply haven't seen before. I read parts of this page to try and learn what the gradient of a vector field means: en.wikipedia.org/wiki/Tensor_derivative_%28continuum_mechanics%29#Curvilinear_coordinates From what I can tell, the "Covariant derivative" is the nabla operator, with two inputs: the vector field itself, as well as the "direction vector", that we take the derivative along. On the other hand, the "gradient of a vector field" is the nabla operator with just one input: the vector field itself... no direction vector is supplied. So the gradient vector of a vector field is like a covariant derivative that still "waiting" to eat a direction vector. Other than this difference, the formulas seem very similar and use the Christoffel Symbols in the same way. The covariant derivative outputs a vector (rank-1) field. But since the gradient is still "waiting" to eat a direction vector, it outputs a rank-2 tensor field (this will become a vector/rank-1 tensor field after it eats the direction vector... and of course this vector/rank-1 field is the covariant derivative).

[@ilredeldeserto]

in 9:34 the second derivative in e_x should be d\theta/dx instead of dr/dx, I think...

[@LucienOmalley]

indeed

[@naturematters08]

#ilredeldeserto Yeah, you're right i had the same doubt to.

[@Simon-ed6zc]

Just a quick and probably fairly stupid question: At 16:45 when expanding the derivative of the basisvector in terms of christoffel symbols, there is no nromal component because its flat space, correct?

[@eigenchris]

Yes, that's right.

[@ernestkirstein6233]

Thank you for pointing this out. I was wondering what happened to the second fundamental form component in that equation.

[@davidprice1875]

Excellent description

[@xingangli9791]

This guy is amazing!!!

[@keyyyla]

You are awesome.

[@philwatson698]

Metoo! Wow. Brilliant. You've done it again :)

[@chymoney1]

I love you

[@ajaypotdar7161]

In the first definition of the Christoffel symbols in a previous video, you related it to the second order derivative. That was in the curved space. In this video, towards the end (16:20) you relate the Christoffel symbols to a first order derivative. Is this because it's a flat space? Want to say Thanks too, these videos are amazing!

[@eigenchris]

The basis vectors themselves are derivative operators, so in a sense there are 2 derivatives here. The function of Christoffel symbols is the same in flat space and curved space: it keeps track of the changes in basis vectors over space.

[@ajaypotdar7161]

@@eigenchris Okay, yes, it makes sense now. Thanks!

[@j.k.sharma3669]

We got different vector fields in cartesian and polar coordinates , so does it mean that vector field depends on co-ordinate system ?

[@abnereliberganzahernandez6337]

I ended you course on tensor calculus, and it was great the only thing that was not clear for me is how exactly the riemna curvature tensor is formed by one contravariant and 3 covariant vectors, so I am reviewing here but cant find the solution clearly

[@eigenchris]

The Riemann Curvature Tensor is a multilinear map that eats 3 vectors as input and outputs a vector. So, when written in a basis, it is a linear combination of covector-covector-covector-vector tensor products. The 3 covectors will each eat the 3 input vectors, leaving only the vector part, which is the output.

[@abnereliberganzahernandez6337]

@@eigenchris thanks a Lot for answering you are a legend favorito professor. Ok I see now this More clearly. As a linear maps Is formed of a vector covector pair, then the multilinear maps must be formed Also of vector covector Pairs. I am still a Little bit lost on the details

[@charliedalca4116]

You are godlike

[@siddharthabandyopadhyay5964]

At 16:10 can we write the partial derivative of the basis vector using christoffel symbols and also second fundamental form or is the latter a concept meant for only 3D vectors?

[@chenlecong9938]

Hi i have two questions: 1)at 16:12 bottom equation,would that eqn be derived and explained in one of the following videos?if not could you please provide a link to where i can refer to for further explanation on that since im a bit befuddled by the fact you simply dump that into the video(no hard feeling though)…i dont even know what to google for😂 2)16:22 when you first introduce the dummy variable K(bottom eqn),i take it as K≥J (correct me if im wrong).Then when at 16:33 when you replace the two J by K(you use a red pen),i take that as summing the largest possible value.Then there comes the question(still,at 16:33):why isnt the other J of V^J(that term on the right of the red pen marked)replaced by K?

[@eigenchris]

1) The equation at 16:12 is just a definition. The Γ symbols are just coefficients used to write the derivative of a basis vector as a linear combination of other basis vectors. 2) I'm not sure what you mean by K≥J. They are both summed from 1 to 2. So you can get 4 combinations between them: (j=1,k=1) (j=1,k=2) (j=2,k=1) (j=2,k=2). Each of the 2 terms in that expression has a different summation over J, so they are different summations. I only changed the first summation from j->k. If you like, you can try writing out all the terms explicitly. The first summation gives 2 terms (j=1,2) and the second summation gives 4 terms (j=1,2 and k=1,2). Maybe that will help you understand which terms each summation index represents.

[@chenlecong9938]

2)so essentially what youre impying is that J and K are all the same dummy variables,but the reason (correct me if im wrong)you even define another K，follow by J is to distinguish which term youre summing over in 16:33 third sentence,second term.am i right?if not,i really dont know whats the reason for TWO dummy variables which are essentially the same thing as ive understood(correct me if im wrong)

[@martinnjoroge6006]

Great playlists here mate. Why does the e(theta) basis vector grow with the constants shown in timestamps 12:00 to 12:20

[@eigenchris]

e_θ is defined as the tangent vector along the θ curves. The farther you go from the origin, the more distance one unit of θ is worth, and so the tangent vector gets longer.

[@martinnjoroge6006]

@@eigenchris Yes, I got that, it's just that I am finding it hard to conceptualize/visualize those exact values of 2, 1.5 and 1.33

[@eigenchris]

To get from the 1st e_θ to the 2nd e_θ, you double the length (factor of 2/1 = 2). To get from the 2nd e_θ to the 3rd e_θ, you multiply the length by 3/2 = 1.5. To get from the 3rd e_θ to the 4th e_θ, you multiply the length by 4/3 = 1.33

[@jigold22571]

ThankU for sharing and posting.

[@姚命宏]

great videos

[@N-Tuple]

what is the difference between ordinary derivative and co-variant derivative.????

[@pacchutubu]

I have a question about basis in polar coordinates. when we represent a vector, in polar coordinates, shouldn't we also say to 'which' point the basis corresponds to?

[@eigenchris]

Generally, yes. For curvilinear coordinates, the "basepoint" of the vector matters.

[@clopensets6104]

I absolutely love your lectures! However, I have a question about the equations for the x and y basis vectors at 10:10, the way I have calculated them seemed to yield a different result! What I did was dot product each of the radial basis vectors with the cartesian basis vectors and then multiplied the result by that radial basis vector, for each radial basis vector to construct the cartesian basis vectors in terms of the radial basis vectors. What I got was: e_x = cos(theta) e_r - r sin(theta) e_theta and e_y = e_r sin(theta) + r cos(theta) e_theta. This is very confusing since you got an extra 'r' in the denominator for the second terms, even though (from my knowledge), my method is equivalent! Please tell me where I went wrong!!!

[@eigenchris]

From what you wrote, I assume you did the below calculations? e_x = cos(θ) e_r + (-sin(θ)/r)e_θ e_x·e_r = cos(θ) (e_r·e_r) + (-sin(θ)/r) (e_θ·e_r) e_x·e_r = cos(θ) e_x = cos(θ) e_r + (-sin(θ)/r)e_θ e_x·e_θ = cos(θ) (e_r·e_θ) + (-sin(θ)/r) (e_θ·e_θ) e_x·e_θ = -sin(θ)/r (r^2) e_x·e_θ = -r*sin(θ) The key mistake here is that e_x·e_r and e_x·e_θ are NOT the components of the vector. These are the PROJECTIONS of e_r and e_θ onto ex. The components of a vector are not projections; vector components are just the amount of each basis vector you need to build the vector. In cartesian coordinates, "vector components", and "projections projections onto basis vectors" end up being the same, but this is only true when the metric tensor is the identity matrix. In this video, the metric for polar coordinates is not the identity... because as you probably know, (e_θ·e_θ) = r^2. Let me know if this answers your question.

[@Mysoi123]

Is there a case where the derivative of vector components results in non-zero?

[@eigenchris]

Sure. You could just set the components to something like "x" or "y^2".

[@Mysoi123]

@@eigenchris Thanks! 😊

[@HenriNioto]

There's something I don't understand in polar basis : in video 11, we've shown that the dot product of e_theta with itself was equal to r², which means that the square of its magnitude is r², which means that its magnitude should be equal to r. So I don't understand why e_theta should vary less and less as r grows, because it should always be equal to r, no ?

[@eigenchris]

The short answer is: if we move from r=1 to r=2, e_theta changes from length 1 to length 2, a 100% increase. If we move from r=10 to r=11, e_theta changes from length 10 to length 11, a 10% increase. If we move from r=100 to r=101, e_theta changes from length 100 to length 101, a 1% increase. The rate of increase gets smaller and smaller. This happens because we're measuring the rate of change of e_theta using e_theta itself. It we were measuring the rate of change of e_theta using a constant basis vectors like ex and ey, then I agree the rate of change would be constant. In fact, it is constant, as shown at 9:20, if we pick a constant theta angle and look at the rate of change as we move away from the origin. However when measuring the rate of changing using e_theta itself (which grows as we move away from the origin) we are measuring the rate of change using a ruler that gets bigger and bigger as we move away from the origin. And so the rate of growth, as measured by e_theta itself, decreases.

[@HenriNioto]

@@eigenchris Oh yes of course ! I had a vague intuition of it, but now it makes total sense ! Thank you. :)

[@bankaikun94]

At 16:29, how are you allowed to change the summation index to k for the vector components? Wouldn't that have any side effects? If anything I thought you would also have to change the index for the vector component on the basis term as well.

[@viliml2763]

What's in a name? That which we call a j, by any other letter would could cover the same indices

[@SillySussySally]

Made my day

[@TheCoyoteWayOverland]

If the basis vectors for polar are derivatives, why do we write p in terms of r, theta? Wouldn't writing p in terms of r and theta mean we are using r and theta as the bases? 1:55 I have this question left over from previous videos. I get we could use either but don't you have to choose one and use it throughout?

[@eigenchris]

Sorry, I don't understand the question. "p1" is just another name for "r" and "p2" is just another name for "theta". What's the question?

[@TheCoyoteWayOverland]

yes but you also say the the basis vector e-tilda_theta "is equivalent to" partial of R wrt theta, which is not theta. Isn't the basis vector in theta = [r=0, theta=1]. I get that you COULD use the def you have but then all vectors would need to be written as values of partial of R wrt theta, not values of theta. I'm sure this is as you wanted (not a mistake) but I missed the reasoning in this and video #2. @@eigenchris

[@eigenchris]

@@TheCoyoteWayOverlandA pair like (r=0, theta=1) is more like a position coordinate in the plane. I've been trying to emphasize that components alone are not the same thing as vectors. So something like [0,1] is no a vector, it's just vector components. You always need to include both basis vectors and components together when writing a vector as a linear combination. In this case, the basis vectors are obtained using the partial derivative method I've shown. I'm not sure if this answers your question.

[@TheCoyoteWayOverland]

Yes, understood--I gave the unit vectors. No matter how you call the basis vector, in one place it is derived from the partials in the other you use raw (r, theta). Is there no mismatch in these? Does not using (r, theta) mean you are using r and theta as bases? @@eigenchris

[@TheCoyoteWayOverland]

Ok, try this for an answer to my question: you define the e and e-tilda basis vectors as the "tangent basis vectors" in tensor calculus # 18. I take this to mean that every time I see e and e-tilda as the basis it is the "tangent basis" or perhaps better called the "basis defined by the partials". There are an infinite number of valid basis vectors that work to define a space but the one we choose to use is this special one. (r, theta) defines the same space as (e-tilda_r, e-tilda_theta) so using either is interchangeable in the cases where Ive seen them both used.

[@N-Tuple]

What is the difference between basis vectors and co-ordinate basis?

[@eigenchris]

In my videos they are the same thing. But I think it is possible to choose basis vectors which don't come from a coordinate system. I don't talk about tjat in my videos though.

[@Palisade5810]

Yo Chris! Why are derivatives covariant only? Is there an analogous contravariant derivative? If I raised the indices of the covariant derivative operator with a metric tensor what geometric properties would it have?

[@eigenchris]

The components of the covariant derivative are covariant because differentiation by the c^i coordinate adds an extra lower index (lower indexes transform covariantly). I'm not aware of any sort of contravariant derivative.

[@ASCENSiON1989]

math.stackexchange.com/questions/1908008/why-isnt-there-a-contravariant-derivative-or-why-are-all-derivatives-covarian

[@micheledonvito6223]

Thank you

[@gabrielalmeida2234]

I can't see how the first definition of the Christoffel symbol and the one in the end of video match up. What am I missing?

[@eigenchris]

This is explained in video 17.

[@gabrielalmeida2234]

@@eigenchris Thank you. Keep up the good work, I'm loving these videos

[@jawadahmad2185]

Why contravariant derivatives do not exist?

[@MasterHigure]

8:00 Do you really have to express everything in Cartesian coordinates? Wouldn't it be easier to just say e_r=dR/dr, and then differentiate that with respect to r and theta (and same for e_theta)? I.e. formalise your "intuitive sense" section at 11:20.

[@canyadigit6274]

MasterHigure yesssss this is exactly what I was thinking

[@phb1955]

I think that would require considering a higher order tensor....what is dtheta/dr.... not obviously zero?

[@7DYNAMIN]

jesus man, great stuff

[@that_guy4690]

I don't get it. We have a plane which is flat, use polar coordinates on it and get non-zero basis vector derivatives which means that the space is not flat. What?

[@eigenchris]

Non-zero basis vector derivatives don't imply the space is curved. You need to calculate the Riemann Curvature Tensor to determine if a space is curved.

[@that_guy4690]

@@eigenchris thanks!

[@raresneagu6928]

You are the best!!!

[@gargirath6428]

Thanku sir

[@mdtoha5455]

please suggest a book that you have read

[@eigenchris]

Unfortunately I can't recommend any. I learned from google mostly.

[@JgM-ie5jy]

Is there really a "zero vector" ? The magnitude is zero but the direction is totally undefined. Unless it is implied that the direction is zero degrees ?

[@eigenchris]

If you define a vector as "something with length and direction", the idea of a zero vector might sound weird since it has no direction, as you say. If you instead define vectors as "things that can be added together and scaled", the zero vector makes sense. It's just a vector that leaves other vectors unchanged when you add it. Usually university-level math classes use the second definition (you can google "Vector Space" for more info).

[@JgM-ie5jy]

@@eigenchris Yes, you're right. I was still locked in the "pointy stick" view of vectors.

[@naturematters08]

@9:29 you wrote esubx basis vector is equal to (doe(r)/doe(x)) * esubr + (doe(r)/doe(x))) * esub(theta) but shouldn't it be esubx = (doe(r)/doe(x)) * esubr+ (doe(theta)/doe(x))) * esub(theta)

[@eigenchris]

Yes, that's an error.

[@folepi7995]

i found it very hard to follow the red line in this video.

[@eigenchris]

Red line? Which part confused you?

[@folepi7995]

@@eigenchris i understood the steps you were doing but at some points i couldnt see the motivation or the bigger picture (could just be me). I dont have a specific question, just general feedback. I really appreciate it, it helped nontheless and thanks for the answer.

[@jimbussey1888]

This video is based on a true story.

[@Gordy-io8sb]

That thumbnail.

[@j.k.sharma3669]

Why christoffel symbol gamma ^ theta _theta theta = 0

[@eigenchris]

The rate of change of the theta basis vector in the theta direction only has changes in the r-component, not the theta-component.

[@MsKhch]

Ок

[@adeyeyeisaac7113]

Please what is your email i would like you to explain a question that i have

[@eigenchris]

Can you just ask your question in a comment? I prefer not giving out my e-mail.

[@adeyeyeisaac7113]

Sorry for the inconvenience Its actually a picture I took from a textbook I'm using. The way they transformed covarient and contravarient vectors looks different and I would like you to explain

[@herwig160265]

Can you please release the Mathematica statement that generates the chart at around 4:58? Thanks a lot!

[@eigenchris]

It's been a long time since I did wrote this example, and I'm wondering if this command is 100% accurate to the example I used in the video or not... either way this is the command: VectorPlot[{x/Sqrt[x^2+y^2] - y, y/Sqrt[x^2+y^2] + x}, {x, -2, 2}, {y, -2, 2}]

[@herwig160265]

@@eigenchris Just tried and it works perfect!!! I got stuck in trying because I wanted to do it the "smart" way, using CoordinateTransform["Cartesian" -> "Polar"]. Happy to see that you did it straightforward 😎

[@ВиталийГром-л6л]

8

[@Sharikkhursheed]

hello. bro when will you upload new video.. i want you to move to General relativity..

[@walidkhier]

Beautiful, but too fast.