I have read a lot of textbooks.They just told me that df is now a covector,you should get used to it.They didn't tell me why.Finally I got the reason from your course.That's really really awesome!Thank you!
@Chase Roycroft Michael Penn has a wonderful video series on differential forms. He likes the little book A Geometric Approach to Differential Forms by David Bachman. ua-cam.com/play/PL22w63XsKjqzQZtDZO_9s2HEMRJnaOTX7.html
Thanks Chris , you are the maths buddy I sought for the last ten years.. I too started by wanting to understand GR .. having moderate level high school math in 1970..Its so good to be able to follow your explanations .. even tho my memory is a sieve I keep watching Hope we make it to GR tho the workload must be heavy for you
The *directional derivative is the dot product between gradient/vector.* Here, the gradient of f(x,y) is (∂f/∂x , ∂f/∂y), which when dotted with the vector ∂/∂x = (1,0) gives ∂f/∂x Similarly, the gradient (∂f/∂x , ∂f/∂y) when dotted with the vector ∂/∂y = (0,1) gives ∂f/∂y. Given that df(∂/∂x) and df(∂/∂y) are the directional derivatives in th x and y directions respectively, it follows that df(∂/∂x) = ∂f/∂x and df(∂/∂y) = ∂f/∂y.
Excellent work Chris! Thank you for putting in the hours and producing such great lessons! Is there any set of notes or a textbook off of which you based your online lectures?
I'm a self-taught physics student, so today I was wondering how to write the differential form in another coordinate system and if it is invariant under coordinate transformations. Then I remembered that eingenchris must have the answer. Thanks!
All of these tensor videos are excellent and unique! Many thanks for them. One critique tho, as a calc teacher once said ‘math is not a spectator sport!’ Can you point toward a source of ‘homework’ to allow me/us to work thru problems and so gain better familiarity and understanding? Thanks again for all.
Next, I would like to see applications or example problems in tensor calculus so as to build up one's abilities in solving various problems? Very well presented and very much appreciated. You can tell you wanted this to be an excellent introduction to the topic; thanks again, Chris. :) 🤤 🍻
You are a really good teacher. Could you please do some videos on how all this ties into general relativity, curved space, four dimensions, etc.? General relativity has always been a mystery to me and you are the one to explain it clearly, if anyone can...just a beginner's course, please
Thanks. I still don't understand GR myself yet. But I am working on it. Is there anything in particular you want to learn? Maybe I can point you to other online sources.
Can you extend what you are talking about in flat space to curved space? you alluded to this a little in your 5.1 video I believe. You know 1000 times what I know about general relativity--all I know is that mass bends space, I think...don't know how and that there are several tensors in Einstein's field equation, which I have read contains 16 equations, 10 of which are different--but it's a mystery to me. I have only found a very few people trying to explain general relativity and most videos are very old--can barely see them on the screen and the explanations are terrible. You do a fantastic job and use high def video and great graphics. How do Christoffel symbols fit in to what you have presented? or do they? Thanks.
I plan to do videos on all of these topics, but it will take me months to get it all done. The best place I can point you to for immediate answers is here: ua-cam.com/video/foRPKAKZWx8/v-deo.html The Christoffel symbols come up when we talk about "covariant derivatives", which is an upgraded version of the normal derivative that we use when we want to keep track of how tensors change over space. Unlike ordinary derivatives, with covariant derivatives we have to keep track of how tensor components change, and also keep track of how basis vectors change (normally we ignore the basis vectors in cartesian coordinates since they are constant everywhere and don't change). The Christoffel symbols are just a bunch of numbers we use to keep track of how the basis vectors change from point to point, which "fixes" the derivative for us.
"df" is a covector field given by the level sets of f. "partial/partial x" is a vector which points in the x-direction. "df(v)" would be the directional derivative of f in the direction of v, so df(partial/partial x) would be the directional derivative of f in the direction of x. This directional derivative is simply "partial f / partial x".
Yes, this is how I understood it too. I just feel a little uneasy about it :P I feel like I'm making a leap of faith in going from df ( partial / partial x) to partial f / partial x
It's a strange notation, but it's really no different than what you would be doing with tensors in the non-calculus world. If you have a covector alpha and a basis {e1,e2}, then alpha(e1) will give you the 1st component of alpha and alpha(e2) will give you the second component of alpha. In the world of calculus it's the exact same, except the covector is df and the basis is {∂/∂x, ∂/∂y}. df(∂/∂x) is just the first components of df, which is ∂f/∂x, and df(∂/∂y) is just the second component of df, which is ∂f/∂y.
@@eigenchris I understand why this is the case if we look at it purely from like... how it behaves, without considering what it looked like. But it really _feels_ like df(∂/∂x) should equal d^2f/dx (∂df/∂x ?) or something, or maybe it should just stay as the product df(∂/∂x), only becoming reducible when multiplied by a finite quantity (like df(∂/∂x) * u = ∂u/∂x df). It makes sense if we just convert everything to the corresponding tensor algebra notation, but it doesn't make sense to me with my intuition from regular calculus. Or is it like... do ∂ and d behave fundamentally differently, and I shouldn't expect them to multiply together the way I expect? Honestly now that I think of it, I'm not sure I've ever actually gotten a good explanation or intuition for what ∂ and d and ∂x and dx actually are when they aren't inside an integral or derivative
Where did you learn about "piercing the stack" as a geometrical interpretation for covectors? Did you come up with this on your own or did you see it in a textbook?
I think I first came across it in "Gravitation" by Misner, Thorne and Wheeler. I know the "stack" interpretation of covectors is on wikipedia. I'm not sure if they talk about vectors piercing them.
what is confusing is that you are switching between operator definition of dx and traditional multiplication definition of dx... this video is actually very hard, take a lot time to understand
I understand dx, dy, can also be seen as projection maps into the first, second coordinates, i.e., dx(a,b)=a; dy(a,b)= b, as the dual basis to Del/Delx . In this case, dx(\delx)=dx(1,0)=1; dy(1,0)=0 , etc. dx, dy are linearly -independent, so they are a basis for the dual space to vector fields. But I'm having trouble in the opposite direction, using this interpretation , i.e., seeing how Del/Delx(dx)=1 and Del/Del y(dy)=1, etc.
When you apply a covector df to a vector d/dλ, this results in the directional derivative by definition: "df(d/dλ) = df/dλ". The covector "df" basically tracks how fast the function "f" in the direction of a curve given by λ, and this is what the directional derivative is. You can think of dx(∂/∂x) = ∂x/∂x =1 and dx(∂/∂y) = ∂x/∂y = 0 as basically being like the definition of dx.
Another question. I know you said we use derivatives and differential forms to describe the vectors and covector so as not to rely on position vectors as they can be hard to define for something like the flat 2d map of the earth. I was wondering how that fits in with the concept of polar coordinates, namely, how do you go about defining basis vector theta for angular position when you dont have an origin? Or does it actually mean that we are concerned with the magnitude of a vector at any given point in space, and the components of a vector is relative to that point?
As long as you have coordinate curves, you can define tangent vectors (basis vectors) along those curves. So defining basis vectors on polar coordinates isn't much different than any other coordinate system.
@@eigenchris I guess what I don't understand is how you can define tangent vector in polar form without a separate reference point, if the magnitude of the theta basis vector would be 0 at the origin.
Can you extend what you are talking about in flat space to curved space? you alluded to this a little in your 5.1 video I believe. You know 1000 times what I know about general relativity--all I know is that mass bends space, I think...don't know how and that there are several tensors in Einstein's field equation, which I have read contains 16 equations, 10 of which are different--but it's a mystery to me. I have only found a very few people trying to explain general relativity and most videos are very old--can barely see them on the screen and the explanations are terrible. You do a fantastic job and use high def video and great graphics. How do Christoffel symbols fit in to what you have presented? or do they? Thanks.
I think a better word than reinterpreting would be generalizing (perhaps to a different context), but it definitely isn’t a categorical reinterpretation.... maybe a recontextualization of a more general sense of the what a differential multiple is/applies to?
I' honestly not sure. I'd say at least 5. I want to get to the covariant derivative before I move on to something else. But I might choose to add a new video here and there overtime on new topics.
in my textbook of geometry 2 (Edoardo Sernesi) it defines the differential of a differentiable function between manifolds like a linear map between tangent spaces. A linear map is conceivable as a (1,1) type tensor, how could we conciliate this with the statement that the differential is a covector, otherwise known as a (0.1) tensor?
In 1:56, it has been assumed that df operator also acts on vectors from TpM vector space (in the same way as with vectors in real space). Am I missing something? Is this obvious? I think this needs to be established first before proceeding forward. Edit:- I think from this point on, the definition of df operator changes. It is no longer an operator in the real vector space but it is an operator in TpM space. This was not mentioned in the video.
"real space" is in general not a vector space but a manifold. You can't talk about vectors in real space anymore, only in TpM. The only thing real space does is you can have tensor fields defined on every point in real space, and the values of those fields are tensors over TpM.
Does this only apply in an orthogonal system like a sphere parameterized with two angles? Otherwise the tangents of the two parameters, say U and V are not orthogonal and so their gradients or duals aren’t in the same direction as their tangents, making your squared up level set above fail, although the gradients as duals still works as does using the metric to raise the indexes of dR/dU to get gradients(duals)without trying to take derivatives of inverses like du/dx when your R is (u, v, u^2 +2uv)
I believe all of this should apply in all coordinate systems. Generally speaking, basis vectors and their duals do not point in the same direction. If there's a particular point in the video you have questions about, can you give me the timestamp?
Thank you so much for answering. 8:37 you have drawn dx and dy perpendicular to each other, which is fine in the plane, but on a surface, du and dv aren’t perpendicular if g12 isn’t zero so that visual fails. I can’t reconcile grad u, du(v) and the dual to dR/du in the SURFACE. For example, if surface is ( u,v,u^2 -v^2) what is grad u at u=.6,v=.15?( data: G inverse is ( .431,.142,.142,.964), (dR//dU= 1,0,1.2), (dual of dR/dU= .431,.142,.474), dR/dv=(0,1,-.3). If you dot grad u with dR/du do you get one? With dR/dv do you get zero? The angle between u and v is 105 degrees and that makes level set argument for the reimagining of the gradient to du difficult! I can’t figure out how to use gradient or du in place of an actual dual with hard numbers, not hand waving. Thanks again for your amazing work! Thanks again!
I figured it out: 1-you could have skewed your picture of dx dy and skewed the level curves as well to make it more general. 2-Curves of constant u and v are easily confused on a graphic. 3-calculation of partial derivatives inverses is tricky when you can’t separate variables but it can be done implicitly.
@3:55, you refer to the episolons as co-vectors. And this whole section (6, 7, 8) is about co-vectors by title. So why is this using "contravariant" notation? Is this a co-vector, indeed or a contra-variant vector? Please clarify.
I see. So actually there are 3 terms floating around. Covectors, Contravariant vectors, and covariant vectors? Please clarify @5:11, I see you have covariant basis vectors as arguments to alpha and you arrive at alpha1 and alpha2 using e^1 having alpha1. Then alpha2 belongs to e^2. Yet the picture @5:11, only shows e^1. What is the difference between alpha2 * e^2 in the equation and alpha2*e^1 in the picture as well as below the equation alpha = alpha1 e^1 + alpha2 e^2.
Aloha. I don't fully understand why df( e_x) = partial f / partial x. In an earlier video you have said that the directional derivative is v * grad f. If v = ex = part/part_x and grad f(x,y) = (1,2y) then it should be zero. Maybe I am mixing up vectors and vector components here, though. and v = 1*part/part_x + 0*part/part_y = (1,0). In either case I am nit sure how df(part/part_x) = part f/ part_x... What is the rule that makes that work?
Your second guess, where ex -> (1,0), is correct. df(ex) = ∂f/∂x is basically the definition of df. It is the covector field that gives us directional derivatives when we act on a vector with it. I explain this in later videos, but because we have df(v) = ∇f·v, this means that df = ∇f · ___; so df is the covector field associated with the vector field ∇f.
@@eigenchris Hi! Thanks for the reply. If its the definition I will just have to swallow it until I get behind it. The vectors in coordinate form are simple enough to understand but I have some difficulties with the abstract objects. If you go back to that in later videos everything will work itself out, probably. Thank you very much for those videos, btw. I have a similar background and learning everything that GR throws at you at once is pretty difficult :/
@@Simon-ed6zc You can sort of think of the directional derivative as being broken up into 2 pieces: #1 is how fast the function is changing (represented by df) and #2 is how 'fast' we are running by (represented by the vector v). df(v) takes both of these rates of change into account. In expanded form, we could write it as "(∂f/∂x dx + ∂f/∂y dy)(vx ex + vy ey)", which, after distributing, simplifies to "∂f/∂x*vx + ∂f/∂y*vy", which is the directional derivative formula, but with the "length"/"speed" of v taking into account. I'm not sure if that helps more.
@@eigenchris Thats pretty close :D Thanks for the effort you put into helping some random on the internet. My original problem was why dx(∂/∂x) = dx/dx, or df(∂/∂x) = df/dx. But as you said, thats the definition. I thought that was something thats proven and I couldn't wrap my head around how to move prove that. All the best!
At 8:00 this matches directly to multi-variable differentials; which really helped me understand the covector components. See: tutorial.math.lamar.edu/Classes/CalcIII/Differentials.aspx
I still don’t see the parallel between the covector field « df » and the reinterpretation of « g(x)dx » that we see in integrals. Will this come up in the tensor calculus 8?
Tensor Calc 8 will be on covector field transformation rules. Tensor Calc 9 will cover the re-interpretation of integrals. The short explanation is that the integral_(a)^(b) df is just f(b) - f(a). This corresponds to drawing a path from point a to point b in the covector field and counting how many lines we pierce while following the path. This will only depend on the endpoint locations, not the path taken, and hence the answer is f(b) - f(a). When we write g(x)dx, we're really just expanding df = g(x)dx, so g(x) would be equal to df/dx.
@@eigenchris sorry you are right, I mean't d/dλ. I was confused why d/dλ was not ∂/∂λ, i.e why it was a total derivative and not a partial derivative. At the time of asking the question, I had thought the total derivative implied summation of the two basis* vectors i.e. d/dλ=∂/∂x+ ∂/∂y.
I am going to rewatch this while actually focusing on it and then think this over. Because the thumbnail / timestamp 6:11 seems to be in contradiction to the Wikipedia article section on higher differentials: (en.m.wikipedia.org/wiki/Differential_of_a_function). In other words, my misconception appears to be that the derivative of a differential is zero.
Tensor calculus involves re-interpreting df to be a covector (a linear function that takes vector inputs and outputs a number) instead of a differential. At 6:11 the covector df eats the input vector d/d lambda and outputs the number df/d lambda, which is the directional derivative.
@@eigenchris - since lamda is a scalar variable that sort of measures the level of progress or travel along a curve, what direction does δ/δλ point in?? Intuition says the tangent direction, but definition of λ implies no direction that I can see.
mistake in 2:00(?) i think df(p/px) = pf/p(p/px) (p/px=v here) but not pf/px because this is how i use the formula df(v) = pf/pv i think the final result will have a terms with second derivative but i am wrong?? anyone could help me to understand why thank you
It's not a mistake. You need to keep in mind that "df" here is a covector, and the definition of "df" is that "df(v)" is equal to the directional derivative of f in the direction of the v vector. This means that df(e_x) is the directional directional derivative of f in the x direction, which is simply pf/px.
@@eigenchris thank you. i try to think the co-vector in such way. df( ) is something that eat a vector therefore it becomes df(v) and df(v) = pf/px (vx) + pf/py (vy) where vx and vy are the components of v. then df should become pf/px ( ) + pf/py ( ) = df( ) which is a linear map and i can think this is just the "left part of the dot product" since (pf/px.pf/py) dot ( vx,vy) = df(v) therefore i conclude that df itself is actually vectors ( pf/dx , pf.py) which is perpendicular to the lines which is actually the same as the gradient vector but gradient vector is in "normal"vector space(since grad f = df/dx e^x + df/dy e^y) but df is in dual space( df = df/dx * (epsilon^x) + df/dy*(epsilon^y)).wherefore df is not the same as gradient vector f ? is df actually a gradient vector? or co-vector itself is just simply a small change of f. and there is other question i want to ask: i get little confused since there are too many interpretation for the same df. (1)df is just a small change of f in some infinitesimal random direction" (2)df is co-vector which is actually the stack of lines /stack of curves when we zoom in (3)df itself is a linear map which is the left side of the dot product which is perpendicular to the "stack of curves" (but not the stack of lines/curve itself) (1) (2),(3) seems to be related in someway but they are not actually the same... :( which one {(1)(2)(3) }are correct?? by the way thank for your help
i try to link back the "df" into "gradient vector f" but i think they are different gradient vector f is ( pf/px ,pf,py) but df is pf/px dx + pf/py dy what is the relationship between them actually... it seems they are the same thing(they both turn vector into scalar value function such as directional derivative) but obviously they have different structure and meaning.. this paradox comes from the fact you assume dx is a basis of the co-vector.... because both df and grad f "eats" a vector and turn it into the same scalar function... but we know gradient vector is not the same as df which is total differential in cal 3
@@garytzehaylau9432 The following formula "df(v) = pf/px (vx) + pf/py (vy)" is correct in the case of the orthonormal xy coordinate system (it's not true in polar coordinates... there will be extra terms). I have 2 videos on the differences and similarities between "df" and "the gradient of f" (videos 13 and 14 in this series). You should probably watch the 1st one at least. I hope it will clear up the confusion.
@@eigenchris thank you i you are very good teacher one more question: df can be interpreted as a small change in f in some certain direction,right? which is the direction correspond to this "df" thank you
at 8:53 in map, You said length in dx is the same. My brain think it is longer when y closer to zero. So I paused the video and find a ruler. It’s the same. Our visual system is broken and easy to be fooled. 😂
I have read a lot of textbooks.They just told me that df is now a covector,you should get used to it.They didn't tell me why.Finally I got the reason from your course.That's really really awesome!Thank you!
@Chase Roycroft Michael Penn has a wonderful video series on differential forms. He likes the little book A Geometric Approach to Differential Forms by David Bachman. ua-cam.com/play/PL22w63XsKjqzQZtDZO_9s2HEMRJnaOTX7.html
a master lesson as never seen before, exelent edition..as always..
you should really add a donate button somewhere! I'd love to buy you a few coffees for these videos! Honestly I really appreciate them!
Thanks for your support. I think I probably will set something up in the next month or so.
Awesome :) Keep us posted!
Thanks Chris , you are the maths buddy I sought for the last ten years.. I too started by wanting to understand GR .. having moderate level high school math in 1970..Its so good to be able to follow your explanations .. even tho my memory is a sieve I keep watching
Hope we make it to GR tho the workload must be heavy for you
You are gifted dude in simplifying math things, GREAT JOB!
Keep it up!
The *directional derivative is the dot product between gradient/vector.* Here, the gradient of f(x,y) is (∂f/∂x , ∂f/∂y), which when dotted with the vector ∂/∂x = (1,0) gives ∂f/∂x Similarly, the gradient (∂f/∂x , ∂f/∂y) when dotted with the vector ∂/∂y = (0,1) gives ∂f/∂y. Given that df(∂/∂x) and df(∂/∂y) are the directional derivatives in th x and y directions respectively, it follows that df(∂/∂x) = ∂f/∂x and df(∂/∂y) = ∂f/∂y.
Oh yeah. Good explanation. All along there was something wrong with the math, but your explanation clears it up.
Excellent work Chris! Thank you for putting in the hours and producing such great lessons! Is there any set of notes or a textbook off of which you based your online lectures?
We can also imagine those covariant lines as staircase, ascending in positive direction. Closer the steps, more faster we ascend.
I'm a self-taught physics student, so today I was wondering how to write the differential form in another coordinate system and if it is invariant under coordinate transformations. Then I remembered that eingenchris must have the answer. Thanks!
Best explanation I have ever seen.
All of these tensor videos are excellent and unique! Many thanks for them. One critique tho, as a calc teacher once said ‘math is not a spectator sport!’ Can you point toward a source of ‘homework’ to allow me/us to work thru problems and so gain better familiarity and understanding? Thanks again for all.
Finally, some math videos to make our dx hard.
Next, I would like to see applications or example problems in tensor calculus so as to build up one's abilities in solving various problems? Very well presented and very much appreciated. You can tell you wanted this to be an excellent introduction to the topic; thanks again, Chris. :) 🤤 🍻
I don't think I have enough time to do videos with more examples as there are other videos I want to make.
What type of examples were you thinking of?
You are a really good teacher. Could you please do some videos on how all this ties into general relativity, curved space, four dimensions, etc.? General relativity has always been a mystery to me and you are the one to explain it clearly, if anyone can...just a beginner's course, please
Thanks. I still don't understand GR myself yet. But I am working on it. Is there anything in particular you want to learn? Maybe I can point you to other online sources.
Can you extend what you are talking about in flat space to curved space? you alluded to this a little in your 5.1 video I believe. You know 1000 times what I know about general relativity--all I know is that mass bends space, I think...don't know how and that there are several tensors in Einstein's field equation, which I have read contains 16 equations, 10 of which are different--but it's a mystery to me. I have only found a very few people trying to explain general relativity and most videos are very old--can barely see them on the screen and the explanations are terrible. You do a fantastic job and use high def video and great graphics. How do Christoffel symbols fit in to what you have presented? or do they? Thanks.
I plan to do videos on all of these topics, but it will take me months to get it all done. The best place I can point you to for immediate answers is here: ua-cam.com/video/foRPKAKZWx8/v-deo.html
The Christoffel symbols come up when we talk about "covariant derivatives", which is an upgraded version of the normal derivative that we use when we want to keep track of how tensors change over space. Unlike ordinary derivatives, with covariant derivatives we have to keep track of how tensor components change, and also keep track of how basis vectors change (normally we ignore the basis vectors in cartesian coordinates since they are constant everywhere and don't change). The Christoffel symbols are just a bunch of numbers we use to keep track of how the basis vectors change from point to point, which "fixes" the derivative for us.
EigenChris is one of those internet "hidden gems"
I thought I understood this. But could you breifly elaborate on why df(partial / partial X) = partial f / partial X?
"df" is a covector field given by the level sets of f.
"partial/partial x" is a vector which points in the x-direction.
"df(v)" would be the directional derivative of f in the direction of v, so df(partial/partial x) would be the directional derivative of f in the direction of x. This directional derivative is simply "partial f / partial x".
Yes, this is how I understood it too. I just feel a little uneasy about it :P I feel like I'm making a leap of faith in going from df ( partial / partial x) to partial f / partial x
It's a strange notation, but it's really no different than what you would be doing with tensors in the non-calculus world. If you have a covector alpha and a basis {e1,e2}, then alpha(e1) will give you the 1st component of alpha and alpha(e2) will give you the second component of alpha. In the world of calculus it's the exact same, except the covector is df and the basis is {∂/∂x, ∂/∂y}. df(∂/∂x) is just the first components of df, which is ∂f/∂x, and df(∂/∂y) is just the second component of df, which is ∂f/∂y.
Oh you are totally right! Thanks again for all your help :) I really appreciate you doing all of this
@@eigenchris I understand why this is the case if we look at it purely from like... how it behaves, without considering what it looked like. But it really _feels_ like df(∂/∂x) should equal d^2f/dx (∂df/∂x ?) or something, or maybe it should just stay as the product df(∂/∂x), only becoming reducible when multiplied by a finite quantity (like df(∂/∂x) * u = ∂u/∂x df). It makes sense if we just convert everything to the corresponding tensor algebra notation, but it doesn't make sense to me with my intuition from regular calculus.
Or is it like... do ∂ and d behave fundamentally differently, and I shouldn't expect them to multiply together the way I expect? Honestly now that I think of it, I'm not sure I've ever actually gotten a good explanation or intuition for what ∂ and d and ∂x and dx actually are when they aren't inside an integral or derivative
Where did you learn about "piercing the stack" as a geometrical interpretation for covectors? Did you come up with this on your own or did you see it in a textbook?
I think I first came across it in "Gravitation" by Misner, Thorne and Wheeler. I know the "stack" interpretation of covectors is on wikipedia. I'm not sure if they talk about vectors piercing them.
what is confusing is that you are switching between operator definition of dx and traditional multiplication definition of dx... this video is actually very hard, take a lot time to understand
Can you point to a specific time in the video where you are confused?
I understand dx, dy, can also be seen as projection maps into the first, second coordinates, i.e., dx(a,b)=a; dy(a,b)= b, as the dual basis to Del/Delx . In this case, dx(\delx)=dx(1,0)=1; dy(1,0)=0 , etc. dx, dy are linearly -independent, so they are a basis for the dual space to vector fields. But I'm having trouble in the opposite direction, using this interpretation , i.e., seeing how Del/Delx(dx)=1 and Del/Del y(dy)=1, etc.
When you apply a covector df to a vector d/dλ, this results in the directional derivative by definition: "df(d/dλ) = df/dλ". The covector "df" basically tracks how fast the function "f" in the direction of a curve given by λ, and this is what the directional derivative is. You can think of dx(∂/∂x) = ∂x/∂x =1 and dx(∂/∂y) = ∂x/∂y = 0 as basically being like the definition of dx.
Another question. I know you said we use derivatives and differential forms to describe the vectors and covector so as not to rely on position vectors as they can be hard to define for something like the flat 2d map of the earth.
I was wondering how that fits in with the concept of polar coordinates, namely, how do you go about defining basis vector theta for angular position when you dont have an origin?
Or does it actually mean that we are concerned with the magnitude of a vector at any given point in space, and the components of a vector is relative to that point?
As long as you have coordinate curves, you can define tangent vectors (basis vectors) along those curves. So defining basis vectors on polar coordinates isn't much different than any other coordinate system.
@@eigenchris I guess what I don't understand is how you can define tangent vector in polar form without a separate reference point, if the magnitude of the theta basis vector would be 0 at the origin.
Can you extend what you are talking about in flat space to curved space? you alluded to this a little in your 5.1 video I believe. You know 1000 times what I know about general relativity--all I know is that mass bends space, I think...don't know how and that there are several tensors in Einstein's field equation, which I have read contains 16 equations, 10 of which are different--but it's a mystery to me. I have only found a very few people trying to explain general relativity and most videos are very old--can barely see them on the screen and the explanations are terrible. You do a fantastic job and use high def video and great graphics. How do Christoffel symbols fit in to what you have presented? or do they? Thanks.
I think a better word than reinterpreting would be generalizing (perhaps to a different context), but it definitely isn’t a categorical reinterpretation.... maybe a recontextualization of a more general sense of the what a differential multiple is/applies to?
Well, this video does give me a slight headache interpreting all this information, but the info does make sense.
Is d/dλ to be thought as some arbitrary vector or is there something about it being a tangent vector specifically that comes into play here?
It's a tangent vector along a curve parameterized by lambda. So a curve with coordinates ( x(λ), y(λ) ).
@@eigenchris Since the curve is arbitrary, for our purposes here, d/dλ is just some arbitrary vector right?
@@toaj868 Yes. It belongs to a curve, but the curve is arbitrary.
@@eigenchris Thank you. That was the only part that was a bit unclear. Your videos are excellent.
Note to self: 5:21 And differential forms are covector fields.
How many more videos are going to be there in this series?
I' honestly not sure. I'd say at least 5. I want to get to the covariant derivative before I move on to something else. But I might choose to add a new video here and there overtime on new topics.
Waiting eagerly for future videos!
in my textbook of geometry 2 (Edoardo Sernesi) it defines the differential of a differentiable function between manifolds like a linear map between tangent spaces. A linear map is conceivable as a (1,1) type tensor, how could we conciliate this with the statement that the differential is a covector, otherwise known as a (0.1) tensor?
Great!!
In 1:56, it has been assumed that df operator also acts on vectors from TpM vector space (in the same way as with vectors in real space). Am I missing something? Is this obvious? I think this needs to be established first before proceeding forward.
Edit:- I think from this point on, the definition of df operator changes. It is no longer an operator in the real vector space but it is an operator in TpM space.
This was not mentioned in the video.
"real space" is in general not a vector space but a manifold. You can't talk about vectors in real space anymore, only in TpM. The only thing real space does is you can have tensor fields defined on every point in real space, and the values of those fields are tensors over TpM.
I can't imagine how you were able to draw the shape of df curve using that equation per minute 9.41
I think I did all the plotting on the Wolfram Cloud website.
Excellent
Awesome!!!
Any plan for a video on the Hodge operator?
I've been thinking about covering the wedge product, but I want to finish my General Relativity videos first. So not until 2022 at the earliest.
Does this only apply in an orthogonal system like a sphere parameterized with two angles? Otherwise the tangents of the two parameters, say U and V are not orthogonal and so their gradients or duals aren’t in the same direction as their tangents, making your squared up level set above fail, although the gradients as duals still works as does using the metric to raise the indexes of dR/dU to get gradients(duals)without trying to take derivatives of inverses like du/dx when your R is (u, v, u^2 +2uv)
I believe all of this should apply in all coordinate systems. Generally speaking, basis vectors and their duals do not point in the same direction. If there's a particular point in the video you have questions about, can you give me the timestamp?
Thank you so much for answering. 8:37 you have drawn dx and dy perpendicular to each other, which is fine in the plane, but on a surface, du and dv aren’t perpendicular if g12 isn’t zero so that visual fails. I can’t reconcile grad u, du(v) and the dual to dR/du in the SURFACE. For example, if surface is ( u,v,u^2 -v^2) what is grad u at u=.6,v=.15?( data: G inverse is ( .431,.142,.142,.964), (dR//dU= 1,0,1.2), (dual of dR/dU= .431,.142,.474), dR/dv=(0,1,-.3).
If you dot grad u with dR/du do you get one? With dR/dv do you get zero? The angle between u and v is 105 degrees and that makes level set argument for the reimagining of the gradient to du difficult! I can’t figure out how to use gradient or du in place of an actual dual with hard numbers, not hand waving.
Thanks again for your amazing work!
Thanks again!
I figured it out: 1-you could have skewed your picture of dx dy and skewed the level curves as well to make it more general. 2-Curves of constant u and v are easily confused on a graphic. 3-calculation of partial derivatives inverses is tricky when you can’t separate variables but it can be done implicitly.
@3:55, you refer to the episolons as co-vectors. And this whole section (6, 7, 8) is about co-vectors by title. So why is this using "contravariant" notation? Is this a co-vector, indeed or a contra-variant vector? Please clarify.
Basis covectors (epsilons) are contravariant/upper-index. Their components are covariant/lower-index.
I see. So actually there are 3 terms floating around. Covectors, Contravariant vectors, and covariant vectors?
Please clarify @5:11, I see you have covariant basis vectors as arguments to alpha and you arrive at alpha1 and alpha2 using e^1 having alpha1. Then alpha2 belongs to e^2. Yet the picture @5:11, only shows e^1.
What is the difference between alpha2 * e^2 in the equation and alpha2*e^1 in the picture as well as below the equation alpha = alpha1 e^1 + alpha2 e^2.
Aloha. I don't fully understand why df( e_x) = partial f / partial x. In an earlier video you have said that the directional derivative is v * grad f. If v = ex = part/part_x and grad f(x,y) = (1,2y) then it should be zero. Maybe I am mixing up vectors and vector components here, though. and v = 1*part/part_x + 0*part/part_y = (1,0). In either case I am nit sure how df(part/part_x) = part f/ part_x... What is the rule that makes that work?
Your second guess, where ex -> (1,0), is correct. df(ex) = ∂f/∂x is basically the definition of df. It is the covector field that gives us directional derivatives when we act on a vector with it. I explain this in later videos, but because we have df(v) = ∇f·v, this means that df = ∇f · ___; so df is the covector field associated with the vector field ∇f.
@@eigenchris Hi! Thanks for the reply. If its the definition I will just have to swallow it until I get behind it. The vectors in coordinate form are simple enough to understand but I have some difficulties with the abstract objects. If you go back to that in later videos everything will work itself out, probably. Thank you very much for those videos, btw. I have a similar background and learning everything that GR throws at you at once is pretty difficult :/
@@Simon-ed6zc You can sort of think of the directional derivative as being broken up into 2 pieces: #1 is how fast the function is changing (represented by df) and #2 is how 'fast' we are running by (represented by the vector v). df(v) takes both of these rates of change into account. In expanded form, we could write it as "(∂f/∂x dx + ∂f/∂y dy)(vx ex + vy ey)", which, after distributing, simplifies to "∂f/∂x*vx + ∂f/∂y*vy", which is the directional derivative formula, but with the "length"/"speed" of v taking into account. I'm not sure if that helps more.
@@eigenchris Thats pretty close :D Thanks for the effort you put into helping some random on the internet. My original problem was why dx(∂/∂x) = dx/dx, or df(∂/∂x) = df/dx. But as you said, thats the definition. I thought that was something thats proven and I couldn't wrap my head around how to move prove that. All the best!
At 8:00 this matches directly to multi-variable differentials; which really helped me understand the covector components.
See: tutorial.math.lamar.edu/Classes/CalcIII/Differentials.aspx
I still don’t see the parallel between the covector field « df » and the reinterpretation of « g(x)dx » that we see in integrals. Will this come up in the tensor calculus 8?
Tensor Calc 8 will be on covector field transformation rules. Tensor Calc 9 will cover the re-interpretation of integrals.
The short explanation is that the integral_(a)^(b) df is just f(b) - f(a). This corresponds to drawing a path from point a to point b in the covector field and counting how many lines we pierce while following the path. This will only depend on the endpoint locations, not the path taken, and hence the answer is f(b) - f(a).
When we write g(x)dx, we're really just expanding df = g(x)dx, so g(x) would be equal to df/dx.
At 5:50 does d/df = partial/(partial x) +partial/(partial y)? And therefore a unit vector?
The formula for d/dλ is at 6:18. There's no reason to believe it's a unit vector. I'm not sure what d/df there's no curve associated with f.
@@eigenchris sorry you are right, I mean't d/dλ. I was confused why d/dλ was not ∂/∂λ, i.e why it was a total derivative and not a partial derivative. At the time of asking the question, I had thought the total derivative implied summation of the two basis* vectors i.e. d/dλ=∂/∂x+ ∂/∂y.
I am going to rewatch this while actually focusing on it and then think this over. Because the thumbnail / timestamp 6:11 seems to be in contradiction to the Wikipedia article section on higher differentials: (en.m.wikipedia.org/wiki/Differential_of_a_function). In other words, my misconception appears to be that the derivative of a differential is zero.
Tensor calculus involves re-interpreting df to be a covector (a linear function that takes vector inputs and outputs a number) instead of a differential. At 6:11 the covector df eats the input vector d/d lambda and outputs the number df/d lambda, which is the directional derivative.
@@eigenchris - since lamda is a scalar variable that sort of measures the level of progress or travel along a curve, what direction does δ/δλ point in?? Intuition says the tangent direction, but definition of λ implies no direction that I can see.
@@rasraster Yes, d/dλ points in the tangent direction.
@@eigenchris OK. It'll be a fun little project to figure out why a progress variable has a direction 😉
@@eigenchris OK, I see why: when δ/δλ is expanded with the chain rule, it's a linear combo of the basis vectors δ/δx and δ/δy.
shoudnt the cj at 4:38 be subscript?
mistake in 2:00(?)
i think df(p/px) = pf/p(p/px) (p/px=v here) but not pf/px because this is how i use the formula df(v) = pf/pv
i think the final result will have a terms with second derivative
but i am wrong??
anyone could help me to understand why
thank you
It's not a mistake. You need to keep in mind that "df" here is a covector, and the definition of "df" is that "df(v)" is equal to the directional derivative of f in the direction of the v vector. This means that df(e_x) is the directional directional derivative of f in the x direction, which is simply pf/px.
@@eigenchris thank you.
i try to think the co-vector in such way.
df( ) is something that eat a vector therefore it becomes df(v) and df(v) = pf/px (vx) + pf/py (vy)
where vx and vy are the components of v.
then df should become pf/px ( ) + pf/py ( ) = df( ) which is a linear map and i can think this is just the "left part of the dot product" since (pf/px.pf/py) dot ( vx,vy) = df(v)
therefore i conclude that df itself is actually vectors ( pf/dx , pf.py) which is perpendicular to the lines which is actually the same as the gradient vector
but gradient vector is in "normal"vector space(since grad f = df/dx e^x + df/dy e^y) but df is in dual space( df = df/dx * (epsilon^x) + df/dy*(epsilon^y)).wherefore df is not the same as gradient vector f ?
is df actually a gradient vector?
or co-vector itself is just simply a small change of f.
and there is other question i want to ask:
i get little confused since there are too many interpretation for the same df.
(1)df is just a small change of f in some infinitesimal random direction"
(2)df is co-vector which is actually the stack of lines /stack of curves when we zoom in
(3)df itself is a linear map which is the left side of the dot product which is perpendicular to the "stack of curves" (but not the stack of lines/curve itself)
(1) (2),(3) seems to be related in someway but they are not actually the same... :(
which one {(1)(2)(3) }are correct??
by the way thank for your help
i try to link back the "df" into "gradient vector f" but i think they are different
gradient vector f is ( pf/px ,pf,py) but df is pf/px dx + pf/py dy
what is the relationship between them actually...
it seems they are the same thing(they both turn vector into scalar value function such as directional derivative) but obviously they have different structure and meaning..
this paradox comes from the fact you assume dx is a basis of the co-vector....
because both df and grad f "eats" a vector and turn it into the same scalar function...
but we know gradient vector is not the same as df which is total differential in cal 3
@@garytzehaylau9432 The following formula "df(v) = pf/px (vx) + pf/py (vy)" is correct in the case of the orthonormal xy coordinate system (it's not true in polar coordinates... there will be extra terms). I have 2 videos on the differences and similarities between "df" and "the gradient of f" (videos 13 and 14 in this series). You should probably watch the 1st one at least. I hope it will clear up the confusion.
@@eigenchris thank you
i
you are very good teacher
one more question:
df can be interpreted as a small change in f in some certain direction,right?
which is the direction correspond to this "df"
thank you
Your videos are making me sad, because they are just reminding me of the brutally bleak way my teacher in General Relativity is teaching us
May Allah Bless you..
at 8:53 in map, You said length in dx is the same. My brain think it is longer when y closer to zero. So I paused the video and find a ruler. It’s the same. Our visual system is broken and easy to be fooled. 😂