I am binge watching I'm in the hospital with covid for 2 weeks. I'm going to come out on the other side of this as a relativistic General relativity theorist badass., Boys. 🤘
@@respectpartii6302 she is hot;millions of hot girls though; you're blinded by fame; putting any hot woman in a pedestal; like it's obvious you do will only lead to masturbation; another obvious trait of you
Wow, who would have thought you could use covectors to tease a generalization of the Riemann integral. Your videos are too perfect, +eigenchris! I always post praise in the comment sections because you really deserve it, your teachings are the most crystal-clear I have experienced in a long time. Super excited for what comes next (I'm here for the ride to GR and beyond lol).
Regarding vectors and covectors: Think of a mercury-based glass tube thermometer - the red mercury inside is the vector; the temperature scale lines on the glass tube is the covector. When you read the thermometer, you are functionalizing the mercury on the scale lines and outputing a temperature reading - a scalar. "df" produces the temperature scale from the scalar field; df(v) is the reading.
Due to reasons In university I couldn't fully understand multivariable calculus, Thank You very much Eigenchris, I have a feeling I will be understanding beyond multivariable calculus. I will never forget this
Believe me, it's gonna be highly helpful for me to study further analysis in Differential forms.It's awesome!!!I was just learned differential form in abstract way but now I have a way to visualize this.Plz keep going on...
by far , the most insightfull , exposition of covectors and its relation with the concept of differential and differential forms ever,.......this profesor,.....must came .....from the outer world,...not from dull academy, which specialice in darken any usefull piece of information at hand, gretings great master !!!!!
It's a very good explanation. One thing I recognized is that at 12:10 a 3D visualization could improve the understanding: The vector field represents a mountain with height levels. It is on a plane in x/y direction an a height in z direction. The vector acting with this vector field has a direction [x y] in the plane. Its z direction is dependent on the steepness of the vector field at the given point. The result of the operation of the vector field on the given vector then is the z component of the resulting vector in 3D. Then its easy to visualize the change of output, if the vector direction or length is changed.
I never found a good explanation for the level set interpretation of covectors/differential forms although many of the resources I encountered referenced it and assumed it was understood. Thank you for this excellent explanation!
Just to be clear, differentials such as df, dx, dy, etc do not have to be infinitesimal. dx, dy, etc are arbitrary variables representing the coordinates of the tangent plane to a function at a point and they can be arbitrarily large. The approximation of a function with a tangent plane is only accurate at infinitesimal distances from that point. Unfortunately, educators mixup these two and teach their students that differentials are infinitesimal. Mathematically, they don't have to be; we enforce the limit of differentials going to zero to approximate the function for purposes such as integration. (Reference: See Stuart Calculus 6th edition, Pg 932 for instance). PS - I love your videos. They are very great! Cannot wait for more.
Hi Chris! Im taking a course on tensor analysis and i dont get a thing until i watch your videos, so thanks for making my life easier! Are u going to cover the levi-cevita tensor, and christoffel symbols?
Hi Chis, thank you for putting up these awesome lectures! Question: In your later videos, you seem to have not bothered identifying covectors tangent to curved level sets unlike in your example at 10 min mark in this video to calculate the number of level sets or covector lines pierced by the path. Thank you!
Chris, something is messed up with notation omiting position vector. One side is that derivatives are linear operators so they form itself a vector space, however to represent something meaningful they must have some function to operate on. How do you calculate them without any function substituted isn't it ? So to me as far as I understand we have still position vector but it is now different - it is object defined in intrinsic space or parametric uv-plane. So still there must be some grid defined on map, then partial derivatives give us local tangent space by differentiating grid lines with respect to their own direction. Do you agree with that ? It has its own name - Lagrangian vs Eulerian coordinates.
Videos 15-26 in this series are basically differential geometry videos. I cover geodesics, the covariant derivative, the Lie Bracket, Torsion, and the Riemann and Ricci curvature tensors.
Part of it is having spent a long time looking around on the internet, trying to find explanations I thought made sense (with pictures if possible) instead of just 50 lines of algebra. Otherwise I just explain the topic how I would have wanted it explained to me back when I started learning.
Hi Chris fantastic series, one question thought at 9:40 you're explaining that we should not take the covectors field lines to count the transformation and then on 12:20 with the geo map you seem to count the crossings of the topograoh. levels not straight lines it's confusing me
So the "d sth" is the "dual operator" of the gradient ? Because it ressemble a lot to what you've shown earlier ... The "df" here seems to make the lines and the gradient of f associate a vector to each point in space ...
At 3:26, wherever the outputs of the covectors are negative, the (input) vectors are not oriented exactly opposite to the (orientation of the) covector stacks but are at an angle (pointing backwards). Shouldn’t that be taken into account as well (have the same question pertaining to the corresponding videos in the Tensor Algebra series). Thanks
This is a year later, but I'll answer it for others who may read your comment. The vector can be divided up into two sets of components. The components that are normal (orthogonal) to the covector stacks and the ones that are parallell (tangent) to the covector stacks. Only the normal subcomponents contribute to "piercing" the stacks. After all, the parallell subcomponents all lie parallell to the stacks and neither moves into or out of the stack, thus they do not contribute to the "piercing" value. Just image a pencil held at a slant onto a piece of paper. Depending on the slant, you will get a different amount of piercing pressure on the tip. If you lay the pencil flat, there will be zero piercing pressure. That's the gist of it, and why the "angle" the vector makes with the stack doesn't matter, only the height of its normal component. The visualization with stacks is helpful, but it should be used as a stepping stone to understand the mathematical definition. This argument generalizes to higher-dimensional vectors, covectors and forms.
so for f is a scalar field, df is the corresponding covector field. and v is some arbritrary vector. df(v) is the directional derivative. Does that mean if w is a unit vector along in direction of v, then df(w) gives the scalar field f itself.
df(w) would give the slope of f in the direction of w. If v is longer than a unit vector, there will be a multiplicative factor on top of df(w), because you're "traveling over" the slope more quickly.
Plz tell whether I'm correct or not. .. when we are zooming in at point 'p' the curve of constant value looks like a st. Line and the stacks representing covectors are very very small? And the vector shown is also really small , at 10.17 the stacks and the vector are just shown big as we have zoomed in but in real the stacks are such small that we can approximate curve to tangent at that point ? Is it correct ?? Waiting eagerly for Ur reply. .
So that means, if you want to use an "d" act on a scalar function f to get a differential, the pre-condition is "f" must be a well-behavior continuous function.
My understanding is that the directional derivative is the instantaneous change in the direction of the vector v; the instantaneous change will be the same no matter how long the vector v is. Definitions I've seen always normalize the vector ... as in mathworld.wolfram.com/DirectionalDerivative.html Thanks for the videos ... they are fantastic!
It's true a lot of sources normalize the direction vector. You can think of my definition as a new "expanded" definition which also takes onto account the "speed" of the direction vector. The covariant derivative, which comes a few videos later, usually doesn't nornalize the direction vector. The content in this video can be thought of as a stepping stone to the covariant derivative.
Isn't the following a simpler explanation: the differential form is like the gradient operator, so that grad(f) dot u where u is a unit vector gives the direction derivative of the function f in the u direction. Thus, functionally the gradient is a map of unit vectors to numbers, hence not a vector but a covector. With orthonormal basis vectors they can be conflated, but not otherwise.
I'm having difficulty in understanding how df acts on v around 9.45 .. Please help . As in my other comment previously I have said if we zoom in on the curve , the curve looks like a straight line , so within that approximation at every point , we have very small (in length) covector stacks ... If this correct , I got this. .. the problem is with what about vectors , are they also infinisimally small ,if so how small because how many lines they Pierce will depend on that .. for example, if v= 2i+3j , then at any point to calculate df(v) , we need a very short vector along the dirn of v , correct ? But how short and what vector is that exactly ?
The field is not uniform. Hence covector stacks are oriented differently in different regions. Vector can be of any length. If the vector is long, some portion of it may be in one covector stack & remaining portion will be in another covector stack. My guess is you have to find out the no of lines pierced by the vector in the 2 stacks independently. Add the values & you will get the final value.
Indeed that’s quite a bit of time we are all waiting for the next video; I hope it will come soon too. And still a big thanks to you eigenchris. Looking forward.
This is amazing.....I'm recapping and learning GR (PhD years ago). These videos are hands down the Best explanations I have ever come across... Text book/online or other.....
You can check out this blog post: brickisland.net/cs177/?p=174 Also, this playlist talks about some of the same topics, but goes much more in-depth (with examples): ua-cam.com/play/PLB8F2D70E034E9C29.html
Just an update... Videos 15+16 now cover Christoffel Symbols when used to calculate geodesics. Vdieo 17,18,19, and 20 cover Christoffel Symbols when used for the covariant derivative.
You could argue there are an infinite number of contours for every real number. You could also say a vector field has an infinite number of vectors, one for each coordinate point. But when it comes to visualizing contours and vector fields, you typically use whatever scale is most convenient for visualizing.
@@eigenchris Thanks for the reply. So how many contour line of stacks should we draw? What decides the spacing between two contour lines. (Or what decides the spacing between stacks)
@@ritikpal1491 You could draw 1 contour line for every time the function changes by a value of "1". That way, you can count how many times a vector fields the "sheets" easily. But if a function changes very quickly or very slowly, this might not be the best visualization. If you're dealing with a function that looks like a mountain that's 2500 units tall, maybe you could draw a contour for every 500 units or 250 units. I feel like I'm just repeating myself. Does what I'm saying make sense?
@@eigenchris So, we can make a contour line for every unit change in the function. That makes sense if you are consistently using this measure. If you change this rule for steeper functions, then wont it make the number of stacks passing through a vector inconsistent? If a function changes rapidly, i expect more number of stacks to pass through the vector.
@@ritikpal1491 If you draw a contpur for every 500 units, them each contour is worth 500 when it is pierced by a vector. In theory there's a contour for every possible real number value for the function. When ones you actually draw out are a matter of convenience for visualizarion.
in 9:45 it said that we cannot count the number of line in the space i dont understand why how does "every points in the field have different co-vector" become a factor that you cannot do something like this thank for help
The picture on the screen at 9:45 is just to give you a sense of what the covector field looks like. It can't be used to give accurate numerical answers. If you want to know how the field acts on a vector at some point p, you need the exact covector at point p, which is why you need to "zoom in" on the point p.
@@eigenchris thank for your reply. what is the reason we cannot just count the number of "curves" that the vector cuts through as we did in linear algebra that is,why this is not accurate enough in the graph of level curve but it is OK to do such thing in linear algebra that is what i try to think: because each points on the curves are correspond to different value/direction of co-vector(because it is not lines but curve) therefore it will be inaccurate to do something like this if we dont zoom in ? since the constant vector /linear assumption" fail under these curves because they don't count thing in linear way(we count the line under the assumption they are the lines correspond to the same co-vectors in every points) but the way,what is "co-vector" in geometrical sense in the graph,they are A. some lines in the map (or called level curves?) B.the "little" arrows/vectors that is perpendicular to those lines or curves A or B? which one is correct thank you for your help
I don't think there's a great way to do it. Basically just draw tiny stacks at each point. Similar to how you'd draw a vector field by drawing tiny vectors at each point.
I'm a bit confused on what you call the directional derivative, because in differential geometry I learned to define a tangent vector to a curve y: IR -> M at a point as a linear map: f |-> (f∘y)'(λ). So the curve y: IR -> M (maps to the manifold), f: M -> IR, so (f∘y): IR -> IR. We then call the tangent vector acting on this function f the directional derivative. Are you familiar with this definition?
Sorry for replying to this one late... it slipped by. The more formal definition for a tangent vector is to take a curve through a point on the manifold and look at the differential operators on that curve. Wikipedia talks about it here: en.wikipedia.org/wiki/Differentiable_manifold#Tangent_vector_and_the_differential I find some of the formal definitions in differential geometry to be hard to read and and difficult to think about, so I appeal to intuition instead. Did you have questions about this definition?
@@eigenchris Yea, I think I am a bit confused about the definitions because in your video, we call the directional derivative as a differential (which is a co-vector) acting on a vector, while I have also learned that the directional derivative is the tangent vector (which is a... vector) acting on a function f at a point p along some curve gamma. The confusion lies with the first definition, we use co-vectors while the second definition does not seem to mention that. I think it's basically the definition on the wiki that you linked, but then the one above: en.wikipedia.org/wiki/Differentiable_manifold#Directional_differentiation
@@thedorantor I think the definitions end up being equivalent. The covector I talk about is df, where "f" is the function in your definition. The resulting derivative ends up being the same.
@@eigenchris Ah yes I see now, thanks. I guess I got confused since your definition is a differential acting on a vector, whereas my definition seems to be a tangent vector doing the acting on a function along a curve.
Aren't level sets just another way of representing a function? I don't see how df is meaningfully different from f itself; it seems like they behave in the same way; and by definition, it seems like they have to.
df is a linear function on a vector "v" that produces the scalar "df(v) = v·∇f". The easiest way to visualize df is as the level sets of f, because the density of the stack lines help you figure out the value of df(v).
@@eigenchris I think it clicked now, thank you for replying! I think my misunderstanding was that I thought of the scalar field of being a function of vectors, and the differential as a function of vectors in the same way, as opposed to the differential having a different covector "value" at every point, which then becomes a function of a vector. These videos are really helpful!
f is a scalar field. That is values assigned at each point in xy plane. Example temperature can a scalar field with a unique value at each point in xy plane.
Nah, didn't get it. grad(f) should be the level sets not the differentials ! Let f=f(x,y,z) then df=(∂f/∂x)dx+(∂f/∂y)dy+(∂f/∂z)dz or, df=grad(f)*dr [dr is a vector, * is dot product symbol] And it's grad(f) which takes a function and makes it a level set. See Bernard Schutz for details calculation.
Regarding vectors and covectors: Think of a mercury-based glass tube thermometer - the red mercury inside is the vector; the temperature scale lines on the glass tube is the covector. When you read the thermometer, you are functionalizing the mercury on the scale lines and outputing a temperature reading - a scalar. "df" produces the temperature scale from the scalar field; df(v) is the reading.
Everyone learning differential geometry should watch these videos. They are the best explanations on the internet, and exceed those in textbooks.
absolutely agreed
Yeeeesss!!!
Agreed beyond any doubt
Is it only me who is binge-watching this !!? This should be on Netflix!
yeah? why not? Margot Robbie could point out the Riemann curves, dressed in a one piece bathing suit? Netflix would green light this right away.
@@klam77 Is this a Big Short reference?
@@klam77 You bet, but let's admit... the woman is astonishing.
I am binge watching I'm in the hospital with covid for 2 weeks.
I'm going to come out on the other side of this as a relativistic General relativity theorist badass., Boys. 🤘
@@respectpartii6302 she is hot;millions of hot girls though; you're blinded by fame; putting any hot woman in a pedestal; like it's obvious you do will only lead to masturbation; another obvious trait of you
Wow, who would have thought you could use covectors to tease a generalization of the Riemann integral. Your videos are too perfect, +eigenchris! I always post praise in the comment sections because you really deserve it, your teachings are the most crystal-clear I have experienced in a long time. Super excited for what comes next (I'm here for the ride to GR and beyond lol).
This has to be one of the most lucid, coincise and yet intuitive recap of this entire subject! Kudos!
This lecture provides a perfect start (and complement) to Michael Penn's excellent videos on Differential Forms.
Your video series on tensors and tensor calculus are priceless; you are a uniquely gifted teacher.
Regarding vectors and covectors: Think of a mercury-based glass tube thermometer - the red mercury inside is the vector; the temperature scale lines on the glass tube is the covector. When you read the thermometer, you are functionalizing the mercury on the scale lines and outputing a temperature reading - a scalar.
"df" produces the temperature scale from the scalar field; df(v) is the reading.
Due to reasons In university I couldn't fully understand multivariable calculus, Thank You very much Eigenchris, I have a feeling I will be understanding beyond multivariable calculus. I will never forget this
Can't say thank you enough times to the creator of this series. This series is a god send.
you ve made these concepts crystal clear! Thank you so much!
You move so swiftly, yet you explain the concepts so thoroughly.
like a ninja who stopped throwing his shurikens and started teaching differential geometry instead
The coolest video series on tensors. Thanks a lot!
I can't emphasize enough how good this video is
OMG this is the best course I've ever seen in my love, You are my hero man !
Believe me, it's gonna be highly helpful for me to study further analysis in Differential forms.It's awesome!!!I was just learned differential form in abstract way but now I have a way to visualize this.Plz keep going on...
You are the most intuitive creature in our universe. Thank you for sharing your intuitive wisdom. God bless science.
by far , the most insightfull , exposition of covectors and its relation with the concept of differential and differential forms ever,.......this profesor,.....must came .....from the outer world,...not from dull academy, which specialice in darken any usefull piece of information at hand, gretings great master !!!!!
It's a very good explanation.
One thing I recognized is that at 12:10 a 3D visualization could improve the understanding:
The vector field represents a mountain with height levels. It is on a plane in x/y direction an a height in z direction.
The vector acting with this vector field has a direction [x y] in the plane. Its z direction is dependent on the steepness of the vector field at the given point.
The result of the operation of the vector field on the given vector then is the z component of the resulting vector in 3D.
Then its easy to visualize the change of output, if the vector direction or length is changed.
Thank you for the amazing video.
I never found a good explanation for the level set interpretation of covectors/differential forms although many of the resources I encountered referenced it and assumed it was understood. Thank you for this excellent explanation!
Amazing video. Thank you!
Amazingly helpful
Just to be clear, differentials such as df, dx, dy, etc do not have to be infinitesimal. dx, dy, etc are arbitrary variables representing the coordinates of the tangent plane to a function at a point and they can be arbitrarily large. The approximation of a function with a tangent plane is only accurate at infinitesimal distances from that point. Unfortunately, educators mixup these two and teach their students that differentials are infinitesimal. Mathematically, they don't have to be; we enforce the limit of differentials going to zero to approximate the function for purposes such as integration.
(Reference: See Stuart Calculus 6th edition, Pg 932 for instance).
PS - I love your videos. They are very great! Cannot wait for more.
df,dx,dy don't mean infinitesimal here. They are differential forms. And the author shows that in every example, that they can be "large".
Thankyou very much for these videos
This video is gold
Hi Chris! Im taking a course on tensor analysis and i dont get a thing until i watch your videos, so thanks for making my life easier! Are u going to cover the levi-cevita tensor, and christoffel symbols?
I will cover Christoffel symbols eventually. I don't plan on covering the Levi-Civita tensor.
our boyyyiii thank you so mucchhchchchc i love you oooo
Hi Chis, thank you for putting up these awesome lectures! Question: In your later videos, you seem to have not bothered identifying covectors tangent to curved level sets unlike in your example at 10 min mark in this video to calculate the number of level sets or covector lines pierced by the path.
Thank you!
Chris, something is messed up with notation omiting position vector. One side is that derivatives are linear operators so they form itself a vector space, however to represent something meaningful they must have some function to operate on. How do you calculate them without any function substituted isn't it ? So to me as far as I understand we have still position vector but it is now different - it is object defined in intrinsic space or parametric uv-plane. So still there must be some grid defined on map, then partial derivatives give us local tangent space by differentiating grid lines with respect to their own direction. Do you agree with that ? It has its own name - Lagrangian vs Eulerian coordinates.
Thank you for the effort on making these videos. I like your series very much. But could you please make a series for differential geometry?
Videos 15-26 in this series are basically differential geometry videos. I cover geodesics, the covariant derivative, the Lie Bracket, Torsion, and the Riemann and Ricci curvature tensors.
It's so cool that the covector field of radii looks like the angular dimension and the covector field of angles looks like the radial dimension.
Yes. dr is just the curves of constant radius (so theta-curves) and dtheta is just the curves of constant theta (so the r-curves).
I rarely comment on videos... but dude talk about quality... ur videos are incredibly clear yet concise. How do you do it?
Part of it is having spent a long time looking around on the internet, trying to find explanations I thought made sense (with pictures if possible) instead of just 50 lines of algebra. Otherwise I just explain the topic how I would have wanted it explained to me back when I started learning.
Hi Chris fantastic series, one question thought at 9:40 you're explaining that we should not take the covectors field lines to count the transformation and then on 12:20 with the geo map you seem to count the crossings of the topograoh. levels not straight lines it's confusing me
Yeah, my bad. You'll have to pretend I'm zoomimg in and counting the lines at that point. I skipped that step.
🐐
So the "d sth" is the "dual operator" of the gradient ? Because it ressemble a lot to what you've shown earlier ... The "df" here seems to make the lines and the gradient of f associate a vector to each point in space ...
I'm not familiar with the definition of the dual operator, but df and grad(f) are dual covector/vector fields.
At 3:26, wherever the outputs of the covectors are negative, the (input) vectors are not oriented exactly opposite to the (orientation of the) covector stacks but are at an angle (pointing backwards). Shouldn’t that be taken into account as well (have the same question pertaining to the corresponding videos in the Tensor Algebra series). Thanks
This is a year later, but I'll answer it for others who may read your comment.
The vector can be divided up into two sets of components. The components that are normal (orthogonal) to the covector stacks and the ones that are parallell (tangent) to the covector stacks. Only the normal subcomponents contribute to "piercing" the stacks. After all, the parallell subcomponents all lie parallell to the stacks and neither moves into or out of the stack, thus they do not contribute to the "piercing" value.
Just image a pencil held at a slant onto a piece of paper. Depending on the slant, you will get a different amount of piercing pressure on the tip. If you lay the pencil flat, there will be zero piercing pressure.
That's the gist of it, and why the "angle" the vector makes with the stack doesn't matter, only the height of its normal component. The visualization with stacks is helpful, but it should be used as a stepping stone to understand the mathematical definition. This argument generalizes to higher-dimensional vectors, covectors and forms.
so for f is a scalar field, df is the corresponding covector field. and v is some arbritrary vector. df(v) is the directional derivative.
Does that mean if w is a unit vector along in direction of v, then df(w) gives the scalar field f itself.
df(w) would give the slope of f in the direction of w. If v is longer than a unit vector, there will be a multiplicative factor on top of df(w), because you're "traveling over" the slope more quickly.
cool
Plz tell whether I'm correct or not. .. when we are zooming in at point 'p' the curve of constant value looks like a st. Line and the stacks representing covectors are very very small? And the vector shown is also really small , at 10.17 the stacks and the vector are just shown big as we have zoomed in but in real the stacks are such small that we can approximate curve to tangent at that point ? Is it correct ?? Waiting eagerly for Ur reply. .
That is basically correct.
10:13 The covector at this point is the _stack_ tangent to the curves.
(Just writing little notes to myself for the next time I'm passing through.)
So that means, if you want to use an "d" act on a scalar function f to get a differential, the pre-condition is "f" must be a well-behavior continuous function.
My understanding is that the directional derivative is the instantaneous change in the direction of the vector v; the instantaneous change will be the same no matter how long the vector v is. Definitions I've seen always normalize the vector ... as in mathworld.wolfram.com/DirectionalDerivative.html
Thanks for the videos ... they are fantastic!
It's true a lot of sources normalize the direction vector. You can think of my definition as a new "expanded" definition which also takes onto account the "speed" of the direction vector. The covariant derivative, which comes a few videos later, usually doesn't nornalize the direction vector. The content in this video can be thought of as a stepping stone to the covariant derivative.
This is helpful to know, thanks. I don't think it's a small point, though I guess I'll find out as I go through the rest of the video series.
Isn't the following a simpler explanation: the differential form is like the gradient operator, so that grad(f) dot u where u is a unit vector gives the direction derivative of the function f in the u direction. Thus, functionally the gradient is a map of unit vectors to numbers, hence not a vector but a covector. With orthonormal basis vectors they can be conflated, but not otherwise.
14:17 It tells us how quickly the _height_ changes, ...
I'm having difficulty in understanding how df acts on v around 9.45 .. Please help .
As in my other comment previously I have said if we zoom in on the curve , the curve looks like a straight line , so within that approximation at every point , we have very small (in length) covector stacks ... If this correct , I got this. .. the problem is with what about vectors , are they also infinisimally small ,if so how small because how many lines they Pierce will depend on that .. for example, if v= 2i+3j , then at any point to calculate df(v) , we need a very short vector along the dirn of v , correct ? But how short and what vector is that exactly ?
The field is not uniform. Hence covector stacks are oriented differently in different regions.
Vector can be of any length. If the vector is long, some portion of it may be in one covector stack & remaining portion will be in another covector stack. My guess is you have to find out the no of lines pierced by the vector in the 2 stacks independently. Add the values & you will get the final value.
@@HankGussman Yes, but the guy made a lousy attempt explaining that, because he just started with very uniform fields
Waiting for tensor calculus 7 toooooo long time.
It should be up this weekend.
still waiting......
Indeed that’s quite a bit of time we are all waiting for the next video; I hope it will come soon too. And still a big thanks to you eigenchris. Looking forward.
One of the weird things I noticed on this rewatch is that df(v) and v(f) always give the same thing.
Symmetry of the inner product
This is amazing.....I'm recapping and learning GR (PhD years ago). These videos are hands down the Best explanations I have ever come across...
Text book/online or other.....
9:45 important point
Can you point to a good source of exercises on this?
You can check out this blog post: brickisland.net/cs177/?p=174
Also, this playlist talks about some of the same topics, but goes much more in-depth (with examples): ua-cam.com/play/PLB8F2D70E034E9C29.html
Thanks I knew about the Metzler playlist but not that blog
You are my Hero. Thanks for all videos you made.
Something about christofell symbols???
I will eventually make videos about Christoffel Symbols and the Covariant Derivative.
Just an update...
Videos 15+16 now cover Christoffel Symbols when used to calculate geodesics.
Vdieo 17,18,19, and 20 cover Christoffel Symbols when used for the covariant derivative.
Arent there infinite number of lines in a contour? What decides the density of the stacks, or the spacing between stacks?
You could argue there are an infinite number of contours for every real number. You could also say a vector field has an infinite number of vectors, one for each coordinate point. But when it comes to visualizing contours and vector fields, you typically use whatever scale is most convenient for visualizing.
@@eigenchris Thanks for the reply. So how many contour line of stacks should we draw? What decides the spacing between two contour lines. (Or what decides the spacing between stacks)
@@ritikpal1491 You could draw 1 contour line for every time the function changes by a value of "1". That way, you can count how many times a vector fields the "sheets" easily. But if a function changes very quickly or very slowly, this might not be the best visualization. If you're dealing with a function that looks like a mountain that's 2500 units tall, maybe you could draw a contour for every 500 units or 250 units. I feel like I'm just repeating myself. Does what I'm saying make sense?
@@eigenchris So, we can make a contour line for every unit change in the function. That makes sense if you are consistently using this measure. If you change this rule for steeper functions, then wont it make the number of stacks passing through a vector inconsistent? If a function changes rapidly, i expect more number of stacks to pass through the vector.
@@ritikpal1491 If you draw a contpur for every 500 units, them each contour is worth 500 when it is pierced by a vector. In theory there's a contour for every possible real number value for the function. When ones you actually draw out are a matter of convenience for visualizarion.
in 9:45
it said that we cannot count the number of line in the space
i dont understand why
how does "every points in the field have different co-vector" become a factor that you cannot do something like this
thank for help
The picture on the screen at 9:45 is just to give you a sense of what the covector field looks like. It can't be used to give accurate numerical answers. If you want to know how the field acts on a vector at some point p, you need the exact covector at point p, which is why you need to "zoom in" on the point p.
@@eigenchris thank for your reply.
what is the reason we cannot just count the number of "curves" that the vector cuts through as we did in linear algebra
that is,why this is not accurate enough in the graph of level curve but it is OK to do such thing in linear algebra
that is what i try to think:
because each points on the curves are correspond to different value/direction of co-vector(because it is not lines but curve) therefore it will be inaccurate to do something like this if we dont zoom in ? since the constant vector /linear assumption" fail under these curves because they don't count thing in linear way(we count the line under the assumption they are the lines correspond to the same co-vectors in every points)
but the way,what is "co-vector" in geometrical sense in the graph,they are
A. some lines in the map (or called level curves?)
B.the "little" arrows/vectors that is perpendicular to those lines or curves
A or B? which one is correct
thank you for your help
How can I visualize the 1 form when it is not exact?
I don't think there's a great way to do it. Basically just draw tiny stacks at each point. Similar to how you'd draw a vector field by drawing tiny vectors at each point.
I'm a bit confused on what you call the directional derivative, because in differential geometry I learned to define a tangent vector to a curve y: IR -> M at a point as a linear map: f |-> (f∘y)'(λ). So the curve y: IR -> M (maps to the manifold), f: M -> IR, so (f∘y): IR -> IR. We then call the tangent vector acting on this function f the directional derivative. Are you familiar with this definition?
Sorry for replying to this one late... it slipped by. The more formal definition for a tangent vector is to take a curve through a point on the manifold and look at the differential operators on that curve. Wikipedia talks about it here: en.wikipedia.org/wiki/Differentiable_manifold#Tangent_vector_and_the_differential
I find some of the formal definitions in differential geometry to be hard to read and and difficult to think about, so I appeal to intuition instead. Did you have questions about this definition?
@@eigenchris Yea, I think I am a bit confused about the definitions because in your video, we call the directional derivative as a differential (which is a co-vector) acting on a vector, while I have also learned that the directional derivative is the tangent vector (which is a... vector) acting on a function f at a point p along some curve gamma. The confusion lies with the first definition, we use co-vectors while the second definition does not seem to mention that.
I think it's basically the definition on the wiki that you linked, but then the one above: en.wikipedia.org/wiki/Differentiable_manifold#Directional_differentiation
@@thedorantor I think the definitions end up being equivalent. The covector I talk about is df, where "f" is the function in your definition. The resulting derivative ends up being the same.
@@eigenchris Ah yes I see now, thanks. I guess I got confused since your definition is a differential acting on a vector, whereas my definition seems to be a tangent vector doing the acting on a function along a curve.
df(v) = v(f) is a good definition of the operator d.
Who is the author ?
Where does he lives ?
Why isn't he famous yet ?
This series arouses me intellectually and almost sexually.
this makes me think that your back is actually the farest point from you on the universe unless you turn yourself 180
Aren't level sets just another way of representing a function? I don't see how df is meaningfully different from f itself; it seems like they behave in the same way; and by definition, it seems like they have to.
df is a linear function on a vector "v" that produces the scalar "df(v) = v·∇f". The easiest way to visualize df is as the level sets of f, because the density of the stack lines help you figure out the value of df(v).
@@eigenchris I think it clicked now, thank you for replying! I think my misunderstanding was that I thought of the scalar field of being a function of vectors, and the differential as a function of vectors in the same way, as opposed to the differential having a different covector "value" at every point, which then becomes a function of a vector.
These videos are really helpful!
What is f though, formally?
f is a scalar field. That is values assigned at each point in xy plane. Example temperature can a scalar field with a unique value at each point in xy plane.
Nah, didn't get it. grad(f) should be the level sets not the differentials !
Let f=f(x,y,z) then
df=(∂f/∂x)dx+(∂f/∂y)dy+(∂f/∂z)dz
or, df=grad(f)*dr [dr is a vector, * is dot product symbol]
And it's grad(f) which takes a function and makes it a level set.
See Bernard Schutz for details calculation.
Regarding vectors and covectors: Think of a mercury-based glass tube thermometer - the red mercury inside is the vector; the temperature scale lines on the glass tube is the covector. When you read the thermometer, you are functionalizing the mercury on the scale lines and outputing a temperature reading - a scalar.
"df" produces the temperature scale from the scalar field; df(v) is the reading.