100% agreed There is such a gap in education between linear algebra / mv calc and tensor calculus. This series is the first I’ve seen that addresses tensor calculus as any other mathematical subject - slowly, thoroughly, and with adequate enough examples for you to grasp the fundamental meaning behind each operation. Thanks so much eigenchris! :)
Extremely nice, extremely lucid presentation. As always. (A more precise explanation of the concept of manifold will hopefully turn up in future videos). Thanks!
I honestly wasn't planning to make a video on that. I find the definition of manifolds is very heavy and not required to understand the ideas I'm presenting. Hopefully you can find another video on it.
@@eigenchris I see. Just to explain: I'm watching your vids to refresh my memory on the stuff I learned some years ago, including manifolds. Of course, I agree that explaning the mathematical basics for GR on this level doesn't require plunging into topological spaces etc. I'll be happy to follow the present series wherever you will lead us. The more the better!
At 16:16 you can derive the "intrinsic" formula from the "extrinsic" formula. There is nothing wrong with using the dot product in the "extrinsic" formula. The only thing you can not do is resolve the basis vectors into a Cartesian set of basis vectors. You can use the metric tensor to define the dot product, which is what you did in your derivation.
Thanks Chris, the intrinsic geodesic equation is so elegant. Then by substituting lines of latitude (u1=constant & u2=λ) we get sin(2u1)=0 and we conclude that the only line of latitude that is a geodesic is the equator (u1=𝝅/2) ignoring the north & south poles (u1=0 & 𝝅)
I am preparing an inclusive teaching material with voice, tactile graphs, and Braille. You youtubes are very clear and useful. How can I acknowledge your work?
Thank you for the great explanations. Just wanted to say that Extrinsic, intrinsic formulas are all the same (in terms of the metric tensor) provided that we have metric compatibility and the symmetry of metric tensor and Christoffel symbols in lower indices.
These lectures are exceptionally helpful!! Just a side note, it's indices and not "indexes". Thank you very much for these. I've been watching them on my TV. But I'm going to pull them all up on my phone just so i can select like.
@@trigon7015 -- Nevertheless his lecturers have been exceptionally helpful. I have been studying GR for several years. I've been through the differential geometry part several times but now I understand it better than ever. The graphs are so well done!!
Great series, Just one question. At 14:10 and 15:20 you calculated the christoffel symbols by setting values for i= 2 and i=1 later. Since i does not appear in the christoffel symbol shouldn't you have been summing over i for finding christoffel symbols? in other words shouldn't there have been 2 terms of the entry of metric tensor times the bracket added together? Technically it doesn't make a difference in here due to the non-diagonal terms in the metric tensor being 0 but I was wondering if they were non zero shouldn't we be summing over i for set values of j and k for the christoffel symbol?
Yes, I should have summed over i. Looking back at this video, I'm a bit confused why I wrote things out this way. It would have made more sense if I chose the "m=2" case and then the "m=1" case... I guess I'm relying on the fact that the other terms go to zero as you say. Not sure what I was thinking. I guess when I learn something for the first time, I can sometimes do things in a roundabout away because I don't fully understand it yet. In my more recent relativity videos (where the 4D christoffel symbols are a lot worse), I do the computations the way you'd expect.
Thank you very much! I am watching Andrew Dotson's videos on the same subject as well and I didn't understand where the Christoffel symbol difference came from. Now it is much clearer =)
loved the last five lectures... In math, the approach is to introduce charts, tangent spaces and tangent bundles without motivation... then they go to forms, so I use to feel the whole purpose was to be able to integrate. By teaching the extrinsic model with the specific function R, you really motivated the problem: how can we find 'straight lines'/constant vector fields on curved spaces. Then you also motivate why an instrinsic model is insightful.
Yeah, the mathematical definition of a manifold kind of drives me up the wall. I guess it's nice if you want formal definitions for everything, but in terms of actual understanding, it doesn't help me at all. I wanted to show Riemannian geometry grew out of the differential geometry of 2D surfaces.
I've had to watch these videos alongside XylyXylyXs videos, to get a full picture, and oh man is it nice. They complement each other well. You say things he doesn't and vice-versa.
At 6:25, you just deleted the second fundamental form part of the basis vector derivative because we are now using intrinsic view, why can we just do that? Do we need to prove that the basis vector derivative is a closed operation in the manifold? The vectors and their derivatives should be invariant whether we look at it intrinsically or extrinsically right?
The 2nd fundamental form only exists so that we can remove the vector component that's normal to the surface. Intrinsic geometry has no concept of the "normal" direction, so we can just ignore the 2nd fundamental form.
It's correct because he showed that a vector derivative is a sum between Christophel's symbols + a normal part. Your question is legitimate but he had already shown that you can completely split the vector field into its tangent and normal part. So we just ignore the normal part. The preparatory work allowed us to do this. Had we not shown that we have a natural split of the derivative into a normal vector field and a tangent vector field we would have to do this first before proceeding.
If there is, I don't know it. From this point onward, there are some derivations where I "lose the plot" when it comes to intuition and I just have the algebra, unfortunately.
Coming back after having watched the NEXT video at 25min32sec: If one compares these three equations to components of the covariant derivative of (0,2)- metric tensor in the next video, one sees these are equivalent to saying the covariant derivative of the metric is zero. Therefore one could REVERSE the reasoning and start by demanding that the CD of g is zero. This would make sense since the "backround should not change with respect to itself" (mathematically this means metric compatibatility). This would produce the equations in this video and so on... The choice of reasoning is a matter of taste, but I think the "connection" is interesting in any case.
At 13:03. I don't know if it has already been noticed, but in the formula for the covariant derivative, the derivative of the component vj appears. Shouldn't it be the derivative of vk?
At 12:27 you multiplied by the inverse metric tensor on the right side of the first expression but on the left side of the second one. Is that allowed and why?
When we multiply matrices, the order matters, but g^im isn't a matrix. g^im is just a component number, like 5 or 9, so it doesn't matter which side we multiply on. a*b = b*a. The choice of index that we are summing (either the first or the 2nd) determines the "order of matrix multiplication" in summation notation.
You need it if you want to define the covariant derivative abstractly, as I do in the next video. It's possible to invent connection coefficients which are not metric compatible.
Good Day Chris, in your Video 18:22 parallel Transported of a vector along a circle of Latitude at a higher Latitude... It Look like a coriolis force effect When i See this vector veeing to left... Is it correlated?
They might be related, but I've never investigated it. A quick google for "parallel transport and coriolis effect" gives this article, which says they are related: www.nature.com/articles/s41567-020-1001-y.pdf
Maybe this will become clear when I watch further videos or think about it some more, but how can we get the same answer for e.g. a geodesic in both extrinsic and intrinsic cases, when we remove the subtraction of the normal component from the covariant derivative? It seems to me that, since the formulas for the Christoffel symbols are equivalent, the derivative has changed, and so what constitutes a geodesic must also be different. How is it possible then that they provide the same answer as you say?
Okey so he has made a little some imprezicion(???) what he should say in excestric case normal component is canceling out, with components of \frac{e_j}{u^i} and so in our intresic case we just gonna ignore that.
I explain that in the next video, which should be out this weekend. A connection and a covariant derivative are more or less the same thing. A covariant derivative gives us a way to parallel transport vectors around a space, thereby "connecting" the tangent spaces at each point. There are several different possible covariant derivatives, defined by the choice of Christoffel symbols. The christoffel symbols you have seen in videos 17,18,19 are all just one specific choice of a covariant derivative (called the Levi-Civita connection). There are other possible covariant derivatives as well.
Do the formula for the Christoffel Symbols 14:09 contain sums over i? I understand that the only non-zero term is partial g_22/ partial u^1, but I am not comfortable in directly i = 2, j = 2 and k = 1 into the equation for christoffel symbols directly as i in the equation of the christoffel symbols is just a dummy variable. I went ahead and wrote out the summation over i explicitly and was able to recover the same christoffel symbols as before. Am I correct in doing so?
I'm struggling to wrap my head around torsion right now. I think the problem started with something that can be seen at 9:30. Here it's saying the partial derivative of a basis vector with respect to some component is gives some vector specified by the christoffel symbols. I believe this is a mistake. That's e_i(e_j), or application of e_i to e_j. It should be the covariant derivative of e_j with respect to e_i. When using covariant derivatives, this confusion about "well obviously when you apply two partial derivatives, they commute" doesn't show up because this *isn't* just applying two partial derivatives. So, given that one can then allow the christoffel symbols to be asymmetric again. I guess the next thing to be confused about is how the lie bracket can ever be nonzero since it *is* just applying partial derivatives to each other, but I think I can understand that one. The lie bracket is actually zero whenever the two vector fields are partial derivatives of different coordinates in the same coordinate system. The thing is that the two vector fields might not be partial derivatives with respect to the same coordinate system, and there's no rule in calculus that says a partial derivative in one set of coordinates has to commute with a partial derivative in another set of coordinates.
This video (and the next one) basically skip what torsion is really about. I explain torsion (and the torsion tensor) more properly in video 21. In this video I don't really distinguish much between partial derivatives applied to vectors and covariant derivatives applied to vectors... I treat them as basically the same thing. That might not be mathematically rigorous, but there isn't really any reason to confuse the two. You always need Christoffel symbols to explain what they are. The thing that you need to be careful about confusing is partial derivatives of components and covariant derivatives of components (the latter involves Christoffel symbols but the former does not). Your final paragraph is exactly correct: the Lie bracket of basis vectors is always zero. But for arbitrary vector fields, the Lie bracket can be non-zero. The geometric interpretation of the Lie bracket is checking if two vector fields form "nice, closed 4-sided boxes" with each other. Basis vectors always do this by definition, since coordinate systems are make up of nice, closed 4-sided boxes. This is why their lie bracket is zero.
I am having difficulties in understanding the concept of tangent plane in intrinsic geometry. The whole process of parallel transport and covariant derivative is based on the ability of checking the vector in different tangent planes. But a plane that is tangent to a manifold uses another space coordinate, i.e.: I can describe a sphere surface using the 2 coordinates theta and phi, but as a tangent plane is not curved, it can't be described with the same set of 2 coordinates. Where do I get lost?
When working with a 2D surface in 3D space, the tangent plane is just a literal geometric 2D plane that's tangent to the surface. For intrinsic geometry, the concept of a tangent space is more abstract. The tangent space at a point "p" on a manifold "M" is an abstract vector space called "TpM" that's attached to "p". This vector space is the set of differential operators on functions on the manifold "M". You can pick any basis you like for this space, but the standard choice of basis are the partial derivative operators at the point "p". (So for coordinates "u" and "v", the basis would be ∂/∂u evaluated at "p" and ∂/∂v evaluated at point "p"). So the coordinates on "M" can give us a basis for "TpM" by using the partial derivatives with respect to the coordinates. Does that make sense?
@@eigenchris I would love to say yes, however I think I understand the logic but not the geometric and physical meaning. In your Schwarzshild black hole video you've showed that the space is curved because the Riemann tensor is not 0, and as a consequence the circumference is 2*pi*L0 instead of being 2*pi*r. Of course is not curved in another 4th space dimension, is curved because the ratio between circumference and radius is not 2*pi. Calculating the Riemann tensor in this geometry, this implies that you are parallel transporting a vector field and then evaluating it in different tangent spaces. How do those tangent spaces extend in that geometry? Probably that this question shows that I've not been able to make the jump between the extrinsic and intrinsic geometry yet.
@@lucapellegrini709 The process of "parallel transport" is exactly the process of moving a vector across a series of tangent spaces. The instructions for how to do parallel transport are bundled up in the Christoffel Symbols/Connection coefficients. I'm not sure if you've watch the next video (tensor calc 20) but I use two different connections (the Levi-Civita connection, and another one I call the "boring connection") to parallel transport a vector across the sphere, and the results of the parallel transported vector are different when I used different connection coefficients. If you haven't watched TC 20, give it a look and see if that helps things make more sense.
@@eigenchris unfortunately that doesn't help; I have watched that video several times; I am able to calculate the Christoffel Symbols and the Riemann tensor, but I think that I miss some basic stuff on the real meaning of what I am doing. BTW I am referring to the Levi-Civita connection only. To make a quasi-real case: say that I am near a Schwarzshild blackhole; at that point I can travel along the r coordinate and then get back to the starting point on a closed loop. If while I am travelling I am transporting a vector keeping it straight, when I get back to the initial point I will see that the vector has a different orientation. And the difference can be computed through the Riemann tensor. However the Riemann tensor together with the Christoffel Symbols for the Levi-Civita connection implicitly assume that there are tangent planes at every point in the space; and in order to compare vectors, we have to refer to those tangent planes, connected by Levi-Civita connection. This is done automatically by the Christoffel Symbols and the Riemann tensor, so we don't have to bother about tangent planes. But if I would like to see things more in depth, I could ask myself: what is my tangent plane when I am standing near a black hole on a point A? and then when I move to the point B closer to the blackhole? how those planes extend in space, or what's their equation? I can understand this in extrinsic geometry, where the tangent plane doesn't live in the curved surface (except for one point), but I can't understand in intrinsic geometry
"how those planes extend in space, or what's their equation?" I think this is where you're going wrong. For extrinsic geometry, the tangent planes exist in physical space. For example for a sphere or radius 1, the tangent plane at the north pole is the set of all points (x,y,z=+1). For intrinsic geometry, the tangent space is not a set of points on the manifold. The tangent space at point "p" is an abstract vector space "TpM" that we attach to the point "p". I feel like I'm repeating myself here, so I'm not sure if this is helping. But basically the tangent spaces for intrinsic geometry aren't tangent planes you can see with you're eyes. They're more like abstract mathematical tools for thinking about "velocities" through a given point on the manifold (where the velocities through a point are like the differential operators). You could try reading this page, although in my opinion it doesn't look great for beginners: en.wikipedia.org/wiki/Tangent_space#Definition_via_tangent_curves
so we use the extrinsic view to compute a metric tensor?, just to switch to use the intrinsic view to get the christoffel symbols. are there any circumstances where we have to use the intrinsic formula to acquire the C-symbols?
in 16:05 it said it is intrinsic because it doesn't rely on X-y-Z variables but in my memory when you calculate the value of gij you need to use x- y -z variables as you need dR/du = dR/dx * dx/du +.... to obtain the metric tensor component it still relies on x-y-z variables is that contradicted to what you define "intrinsic" thank you"
The metric is the fundamental starting point for curved geometry. We need it or else we can't do anything. So we either take a metric we already know from extrinsic geometry, or we make one up from our imagination. The important thing is that, after we have a metric, everything else can be computed intrinsically.
@@eigenchris thank i think i understand the "parallel transport on constant latitude" on a sphere when i try to use a ball to roll on the surface if i keep the curve of pi/4 on the sphere "touching" the flat plane continuously and and it will move automatically to left (there is centripetal component) .Finally i can see there is a curve lines on this plane as this plane is tangent to every position on the sphere which has a latitude of pi/4 . In our human frame we cannot feel we exists in the sphere as every point are equally pointed toward the centre,however we feel we exist in a flat plane. Therefore if we roll a ball on a plane (with every points on the great circle touching the flat plane),then we will feel we are moving forward in straight line ,this can be observed that the ball is rolling on a plane and the path is a straight line on the plane. (that is ,we move on equator) if we let every points on the small circle(eg. latitute pi/4 on the sphere) continuously touching the flat plane,the sphere itself will move to the left,if we put some ink overlapped the small circle of the sphere,the plane will left with a curve (bend to the left)with ink which is exactly what the human can feel on earth is that correct thank? this is the other way i can think to describe
@@garytzehaylau9432 This is not an interpretation I had considered before, but it seems to make sense to me. If we roll a ball along a plane so that it is always touching around the pi/4 circle, the ball will indeed roll in a circle.
10:42 you state that the Christoffel symbols are just scalars and therefore can be factored out. However, the Christoffel symbols are functions of (u1,u2), and the unit vectors are derivatives with respect to u1 or u2, so you cannot just take it outside the derivative. @eigenchris
I'm pulling the Christoffel symbols out in front of the dot products, not the derivatives. You're right that you can't pull Christoffel symbols out from derivatives. Also, when I called them "scalars", that's a bit misleading. True scalars are identical in all coordinate systems. The Christoffel symbols are not. It would be better to call them "coefficients".
Defining basis vectors as partial derivative operators immediately brings up three questions for me: Why is this a natural definition of a tangent vector? How doe we define the dot product of two basis vectors ? How can we still intuitively think of these tangent basis vectors as little arrows?
Kind of stupid question Chris, but how do you get the sort of Gothic g symbol for the Inverse Metric Tensor, in Power-Point, Word etc Please? Many thanks again for all these Wonderful Videos. 😀
You type \frakturg and then hit the space bar. You can type \frakturX where "X" is any upper-case or lower-case letter. There's also \scriptX and \doubleX. You can always hover over a symbol in the equation menu to see what the code for it is, or google.
@@JohnJoss1 I think most of them are just stolen from LaTeX. There are many more, like \sqrt \vec \times \dcot, etc. Maybe this article will help. www.pickupbrain.com/ms-word/equation-editor-shortcut-word/
@@eigenchris There's a few Math Symbols you can put in to a Word Document directly, without using the Equation Editor: CTL + SHFT + Q and then any of the letter keys gets you Greek Characters, including the uppercase ones. You can get the Square Root Symbol by typing 221a and then ALT+x, Nabla Symbol 2207 ALT+x, Partial d 2202 ALT+x, Therefore Symbol 2234 ALT+x. There's probably a website somewhere for this stuff. Thanks very much for your link, very useful. 😃
If spacetime is a four dimensions is that mean that the outside world is a five dimension world where we can see the curvature of spacetime and where the normal components would exist And I have a some confusion about the basis vector of manifold been a derivative operator instead of the derivative themselve ,and so since we don't have a position vector is that mean that any thing tangent to the path of the coordinates is the derivative operator ,and thanks
The traditional version of General Relativity only describes spacetime as a 4D space intrinsically. It doesn't take it to be embedded in a 5D space. I'm not sure if this is even mathematically possible. You would have to google the number of dimensions required to embed spacetime in, as thid goes beyond my knowledge. It may be bigger than 5. For the derivative operators being vectors, you can try warching this video to get a better understanding: th-cam.com/video/VHkL5HpL0HY/w-d-xo.html
Sometimes it's just given to you (you invent the metric and see what type of space it is). Sometimes it can be obtained from another equation. In general relativity, the Einstein Field Equations takes the energy-momentum tensor as input, and these solutions are the metric tensor.
hi, eigenchris! I’ve learned tensor from your videos for weeks. Tensor is really a amazing idea to understand our world. But i am confused about how to consider d/d_lamda(d/d_lamda) in vector and covector way since it looks like df(d/d__lamda)=df/d_lamda in #7 video. Oh i get it. d/d_lamda is not a covector when df is. 😅looks like writing the problem down can help figure it out. when seeing d/d_lamda(·), just using the chain rule right? Thanks for the lessons, l really enjoy thinking of it.
at the beginning you left out all parts with a normal component. dont you have to make up for that at some point? or do you just ignore all terms with a normal component when calculating the cov. der. for the intrinsic method? (as i understand you didnt actually calculated the parallel transported vectors, you only checked if the christoffelsymbols are equal)
There's no such thing as the "normal component" in the intrinsic space. This video is somewhat informal... the more "formal" definition of the covariant derivative comes in the next video.
Is the problem with dot product (itself) to derive Christoffel symbol or the procedure of eliminating the perpendicular n vector? (Thanks for great series)
I think the problem is eliminating the perpendicular vector. After all the metric g_ij is by definition e_i dot e_j. We do have "dot product" on the tangent plane.
Hi, I am a bit lost again. Is the covariant derivative along a path the same as the directional derivative? So the gradient of a vector times a vector?
For the covariant derivative of vectors, it's like the directional derivative that only pays attention to the portion tangent to the surface you're on. You'll see in the next video the covariant derivative can also be applied to any tensor, not just vectors.
@@eigenchris Hi, thanks for the answer. I saw it. I got confused with the symbols, but it really is just that, a symbol for a specific type of derivative, isn't it?
@@trigon7015 In extrinsic geometry, you can think of the tangent space as being a 2D plane living in the same 3D space as the surface, with the plane meeting the surface at exactly 1 point. For intrinsic geometry, it's more abstract. For every point p on the manifold, we associate an abstract vector space Tp. It doesn't live in the same "outside space" because there is no "outside space".
I'm not sure what you mean exactly. With most things I present there is an "abstract" version, and then I break things down into coordinates to do concrete calculations. I do talk a bit about thr tangent space in videos 5.1 and 20 in this series, but I don't go into much detail.
@@eigenchris I meant a coordinate free definition of tangent space in this case. I didn't look at the 20th episode, but from what I've seen so far (and that isn't all) you defined elements of a tangent space as the span of d/dx^k at that point. Here you must choose a chart for the definition.
I don't really want to bother with the definition of charts and manifolds. I'm being a bit "loosey-goosey" with the idea of a tangent space, just defining it as the span of derivative operators. I don't plan on getting more complicated than that.
@@eigenchris that's ok. Everybody has different goals. These videos require really little pre-knowledge, but still introduce the new concepts. You make great videos!
I computed all the christoffel symbols by using this method... the difference between your method and mine was that I didn't choose r constant or one... i used 3 by 3 metric of spherical coordinates... all the christoffel symbols came out correctly besides Γ^2 lower23... can you help me out???
Paths likes geodesics are parameterized by a single variable (lambda), and so we use the total (single-variable) derivative instead of a partial derivative.
@@eigenchris Thank you for your reply! And thank you for your videos! They really helped me a lot. There's still a question confusing me: When we take the covariant derivative of a vector (like A^a e_a), should we apply the Leibniz rule (like (nabla A)e+A(nabla e))? If so, there will be two Christoffel symbols, which seems weird to me. If the answer is no, is it safe to say that covariant derivatives are only defined to act on the components?
@@official-zq3bv Yes, we basically use the "Leibniz" rule (product rule). We basically define the covariant derivative of an ordinary function to be the partial derivative (vector components are just considered ordinary functions). The Christoffel symbols only come up when we take the covariant derivative of a vector. So scalars will get 0 Christoffel symbols, vectors will get 1 set of Christoffel symbols, rank-2 tensors will get 2 sets of Christoffel symbols, and so on.
@@eigenchris OK, I think I finally got it. I misunderstood the covariant derivative as a newly defined derivative operator (instead of partial derivative) that acts on the component by adding an extra fix term presented by the Christoffel symbol. Now I understand that the upper index in abla_\mu A^ u does not mean that the operator acts on the A^ u component, but the ^ u component of the _\mu covariant derivative on 'the whole vector A'. I really appreciate your work. Your videos saved me.
I don’t understand why even the bug on the surface would use a partial derivative with no argument. How can you do any calculations with that? If I create a small tangent square of the point P, the bugs can then make u and v their coordinates and calculate basis vectors using a tiny R from P: at least then I can create a little vector space on my computer in that approximate space. Otherwise I don’t see how a partial helps at all without acting on something! Great videos by the way!
The issue is, you can't define a position vector R in a reasonable way on a curved space. The partial derivative operator act exactly like basis vectors, allowing us to expand other partial derivatives using linear combinations, so we use them as basis vectors instead.
Thanks; as soon as you actually put a function in there though, it then won’t give you the metric tensor you need to do problems, plus it gives you a functions “velocity” and I want the velocity of mars moons! It seems like just a placeholder for the real exterior basis, so you do parameter changes, and then you say, sorry, no R, so we’ll just use the metric to get the same result a position vector would have given to begin with. This has been the most confusing part of the series, and the books I’ve looked at gloss over it as well. It’s critical I think though. I’m a medical doctor learning GR for fun. I hope you will teach for a living: you have a gift.
I meant you have the partial in there to make it obvious how to do parameter changes but otherwise it doesn’t help to eliminate R only to bring it back in similar form via metric. I guess metric is easier to guess at than guessing the function R from another dimension lol.
@@robertprince1900 Sorry, I've read your comments but I don't understand exactly what you're getting at. Is your concern where you get the metric from? For intrinsic geometry, it's true that we can't get the metric "automatically" from the outside space like we do with extrinsic geometry. So the metric needs to be given to us. (Or, in the case of general relativity, we get the metric by solving Einstein's Field Equations.)
In curved spaces, you can't really draw a straight vector from an origin point to another point in the space, so the idea of a position vector doesn't make sense. We can only use "tangent vectors" that exist at the tangent space at each point.
@@eigenchris excellent i understand, is there any video where you can find gij components from energy tensor (right side of einsten equation), in order to get rid of the impossible to find ei.ej dot product?, because we dont have extrinsic coordinates out there, we cant really define the o,o,o point, even in that case, which direction would point the x axis? it has no sense there is infinite many possibilities
@@federicopagano6590 I just finished a relativity series (45 videos long). If you search "relativity by eigenchris" you should find it. The 108 videos solve for the metric around a black hole (energy tensor is zero for a vacuum). The 109 videos solve for the metric for gravitational waves (energy tensor is again zero for a vacuum). The 110 videos solve for the metric for an expanding universe (non-zero energy tensor that treats all the galaxies in the universe as a fluid). The 0,0,0 origin point we select is arbitrary, but we often choose a convenient point to make the formulas simple. For example, for a black hole, we choose the origin point to be the center of the black hole.
Conceptually, I always found the notion of ignoring the normal component, on account that the bugs cannot perceive it, tricky. I like to think of the intrinsic geometry like this: a god or gods add a larger space in which to embed the manifold; then they derive all the intrinsic quantities -- the covariant derivative etc -- using the larger space; and finally, they give us all the formulas, which contain only the intrinsic quantities, such as the metric, that we mortals can understand. Of course, since we're better than gods we can just do this our selves.
Yeah, I'm not sure I really have a good understanding of justifying the scribbled-out-normal vector either. Many metrics I have seen "come from" an example involving a higher dimensional space. The problem is that with standard GR, there isn't really an easy-to-imagine outside space. It's hard for me to understand what the covariant derivative really means in GR, other than the fact that it's metric compatible (i.e. keeps vector lengths constant during parallel transport) and torsion-free (i.e. tiny parallelograms "close up" properly).
Hi Chris. Your derivation of the Christoffel symbols appears to use 3 coordinate curves. Why 3? How is it related to the dimension of the space? Thanks
So, for the intrinsic world, we have no way to CALCULATE its metric DIRECTLY. We have to either declare one or solve some other equations, if there is such equation, to get one. It is a really sad story for the people living in the intrinsic world. But the good news is we have some degree of freedom to construct any metric we want as long as the constructed metric can describe the world correctly.
At 10:18 you mention a special product rule for the dot product of vectors. I was just looking at Walter Lewin 18.03 lecture 13 where he shows the curl of a curl of a vector expands out in a way I find very similar except for the negative sign : ∇ x (∇ x A) = ∇ (∇ . A) - (∇ . ∇) A Funny as to how suddenly I can memorize the expansion of the Laplacian.
@@twistedsector (∇ · ∇) A is the Laplacian of a vector field A. So you can do some algebra on the expression given to remember that (∇ · ∇) A = ∇ (∇ · A) - ∇ x (∇ x A), which is the definition of the Laplacian.
I really wish there was a Nobel prize for making the best tensor calculus videos so that Chris could win it.
Accurate
I've been looking for a nice explanation of tensor calculus
couldnt agree more fellow indian
I hope this series is archived in the world's libraries.
100% agreed
There is such a gap in education between linear algebra / mv calc and tensor calculus. This series is the first I’ve seen that addresses tensor calculus as any other mathematical subject - slowly, thoroughly, and with adequate enough examples for you to grasp the fundamental meaning behind each operation. Thanks so much eigenchris! :)
I agree with the other commenters. A masterpiece of exposition so tightly and carefully developed. Many thanks.
Really enjoying this series. It's so hard to come across good resources for more complex topics!
thankyou for all your hard work,please keep going!
Extremely nice, extremely lucid presentation. As always. (A more precise explanation of the concept of manifold will hopefully turn up in future videos). Thanks!
I honestly wasn't planning to make a video on that. I find the definition of manifolds is very heavy and not required to understand the ideas I'm presenting. Hopefully you can find another video on it.
@@eigenchris I see. Just to explain: I'm watching your vids to refresh my memory on the stuff I learned some years ago, including manifolds. Of course, I agree that explaning the mathematical basics for GR on this level doesn't require plunging into topological spaces etc. I'll be happy to follow the present series wherever you will lead us. The more the better!
At 16:16 you can derive the "intrinsic" formula from the "extrinsic" formula. There is nothing wrong with using the dot product in the "extrinsic" formula. The only thing you can not do is resolve the basis vectors into a Cartesian set of basis vectors. You can use the metric tensor to define the dot product, which is what you did in your derivation.
most intuitive video in the series so far!! cheers!!!
18:19 If what you are showing on left side is just bug's view then all the vectors should be pointing in same direction. Right?
Thanks Chris, the intrinsic geodesic equation is so elegant. Then by substituting lines of latitude (u1=constant & u2=λ) we get sin(2u1)=0 and we conclude that the only line of latitude that is a geodesic is the equator (u1=𝝅/2) ignoring the north & south poles (u1=0 & 𝝅)
I am preparing an inclusive teaching material with voice, tactile graphs, and Braille. You youtubes are very clear and useful. How can I acknowledge your work?
I guess you could just include a link to my youtube channel, or just say "eigenchris's youtube channel". Thanks!
Thank you for the great explanations. Just wanted to say that Extrinsic, intrinsic formulas are all the same (in terms of the metric tensor) provided that we have metric compatibility and the symmetry of metric tensor and Christoffel symbols in lower indices.
Gracias desde Lima-Perú! Buena presentación!
These lectures are exceptionally helpful!! Just a side note, it's indices and not "indexes". Thank you very much for these. I've been watching them on my TV. But I'm going to pull them all up on my phone just so i can select like.
Actually, both indices and indexes are valid
@@trigon7015 -- Did not know that.
@@trigon7015 -- Nevertheless his lecturers have been exceptionally helpful. I have been studying GR for several years. I've been through the differential geometry part several times but now I understand it better than ever. The graphs are so well done!!
Great series, Just one question. At 14:10 and 15:20 you calculated the christoffel symbols by setting values for i= 2 and i=1 later. Since i does not appear in the christoffel symbol shouldn't you have been summing over i for finding christoffel symbols? in other words shouldn't there have been 2 terms of the entry of metric tensor times the bracket added together? Technically it doesn't make a difference in here due to the non-diagonal terms in the metric tensor being 0 but I was wondering if they were non zero shouldn't we be summing over i for set values of j and k for the christoffel symbol?
Yes, I should have summed over i. Looking back at this video, I'm a bit confused why I wrote things out this way. It would have made more sense if I chose the "m=2" case and then the "m=1" case... I guess I'm relying on the fact that the other terms go to zero as you say. Not sure what I was thinking. I guess when I learn something for the first time, I can sometimes do things in a roundabout away because I don't fully understand it yet. In my more recent relativity videos (where the 4D christoffel symbols are a lot worse), I do the computations the way you'd expect.
@@eigenchris no worries, its still a great series. I'm using it to review the material and will be shortly using your GR series for review as well.
Came here to ask exactly this, surprised more people haven't pointed this out 14:39
Thank you very much!
I am watching Andrew Dotson's videos on the same subject as well and I didn't understand where the Christoffel symbol difference came from. Now it is much clearer =)
loved the last five lectures... In math, the approach is to introduce charts, tangent spaces and tangent bundles without motivation... then they go to forms, so I use to feel the whole purpose was to be able to integrate. By teaching the extrinsic model with the specific function R, you really motivated the problem: how can we find 'straight lines'/constant vector fields on curved spaces. Then you also motivate why an instrinsic model is insightful.
Yeah, the mathematical definition of a manifold kind of drives me up the wall. I guess it's nice if you want formal definitions for everything, but in terms of actual understanding, it doesn't help me at all. I wanted to show Riemannian geometry grew out of the differential geometry of 2D surfaces.
I've had to watch these videos alongside XylyXylyXs videos, to get a full picture, and oh man is it nice. They complement each other well. You say things he doesn't and vice-versa.
At 6:25, you just deleted the second fundamental form part of the basis vector derivative because we are now using intrinsic view, why can we just do that? Do we need to prove that the basis vector derivative is a closed operation in the manifold? The vectors and their derivatives should be invariant whether we look at it intrinsically or extrinsically right?
The 2nd fundamental form only exists so that we can remove the vector component that's normal to the surface. Intrinsic geometry has no concept of the "normal" direction, so we can just ignore the 2nd fundamental form.
It's correct because he showed that a vector derivative is a sum between Christophel's symbols + a normal part. Your question is legitimate but he had already shown that you can completely split the vector field into its tangent and normal part. So we just ignore the normal part. The preparatory work allowed us to do this. Had we not shown that we have a natural split of the derivative into a normal vector field and a tangent vector field we would have to do this first before proceeding.
12:05 I Wonder if there is some intuitive reason behind this trick?
I would love to know since it is pretty central to the theory.
If there is, I don't know it. From this point onward, there are some derivations where I "lose the plot" when it comes to intuition and I just have the algebra, unfortunately.
Coming back after having watched the NEXT video at 25min32sec:
If one compares these three equations to components of the covariant derivative of (0,2)- metric tensor in the next video, one sees these are equivalent to saying the covariant derivative of the metric is zero.
Therefore one could REVERSE the reasoning and start by demanding that the CD of g is zero. This would make sense since the "backround should not change with respect to itself" (mathematically this means metric compatibatility). This would produce the equations in this video and so on...
The choice of reasoning is a matter of taste, but I think the "connection" is interesting in any case.
At 13:03. I don't know if it has already been noticed, but in the formula for the covariant derivative, the derivative of the component vj appears. Shouldn't it be the derivative of vk?
Yeah I also think so
At 12:27 you multiplied by the inverse metric tensor on the right side of the first expression but on the left side of the second one. Is that allowed and why?
When we multiply matrices, the order matters, but g^im isn't a matrix.
g^im is just a component number, like 5 or 9, so it doesn't matter which side we multiply on. a*b = b*a.
The choice of index that we are summing (either the first or the 2nd) determines the "order of matrix multiplication" in summation notation.
@@eigenchris Ah right. Thank you so much
Why in 10:13 do we need metric compatibility to develop the normal derivative of a product
You need it if you want to define the covariant derivative abstractly, as I do in the next video. It's possible to invent connection coefficients which are not metric compatible.
Good Day Chris, in your Video 18:22 parallel Transported of a vector along a circle of Latitude at a higher Latitude... It Look like a coriolis force effect When i See this vector veeing to left... Is it correlated?
They might be related, but I've never investigated it. A quick google for "parallel transport and coriolis effect" gives this article, which says they are related: www.nature.com/articles/s41567-020-1001-y.pdf
@@eigenchris Thanks a lot Chris.... 🙏
Maybe this will become clear when I watch further videos or think about it some more, but how can we get the same answer for e.g. a geodesic in both extrinsic and intrinsic cases, when we remove the subtraction of the normal component from the covariant derivative? It seems to me that, since the formulas for the Christoffel symbols are equivalent, the derivative has changed, and so what constitutes a geodesic must also be different. How is it possible then that they provide the same answer as you say?
Okey so he has made a little some imprezicion(???) what he should say in excestric case normal component is canceling out, with components of \frac{e_j}{u^i} and so in our intresic case we just gonna ignore that.
How is something they called connection in books is related to this?
I explain that in the next video, which should be out this weekend. A connection and a covariant derivative are more or less the same thing. A covariant derivative gives us a way to parallel transport vectors around a space, thereby "connecting" the tangent spaces at each point. There are several different possible covariant derivatives, defined by the choice of Christoffel symbols. The christoffel symbols you have seen in videos 17,18,19 are all just one specific choice of a covariant derivative (called the Levi-Civita connection). There are other possible covariant derivatives as well.
how can one show that the coordinate lines can be extended to far beyond the local one on The tangent space?
Do the formula for the Christoffel Symbols 14:09 contain sums over i? I understand that the only non-zero term is partial g_22/ partial u^1, but I am not comfortable in directly i = 2, j = 2 and k = 1 into the equation for christoffel symbols directly as i in the equation of the christoffel symbols is just a dummy variable. I went ahead and wrote out the summation over i explicitly and was able to recover the same christoffel symbols as before. Am I correct in doing so?
Yes, you're doing it the correct way. I guess I got too carried away plugging indices in when I was doing this. Sorry about that.
I'm struggling to wrap my head around torsion right now. I think the problem started with something that can be seen at 9:30. Here it's saying the partial derivative of a basis vector with respect to some component is gives some vector specified by the christoffel symbols. I believe this is a mistake. That's e_i(e_j), or application of e_i to e_j. It should be the covariant derivative of e_j with respect to e_i.
When using covariant derivatives, this confusion about "well obviously when you apply two partial derivatives, they commute" doesn't show up because this *isn't* just applying two partial derivatives. So, given that one can then allow the christoffel symbols to be asymmetric again.
I guess the next thing to be confused about is how the lie bracket can ever be nonzero since it *is* just applying partial derivatives to each other, but I think I can understand that one. The lie bracket is actually zero whenever the two vector fields are partial derivatives of different coordinates in the same coordinate system. The thing is that the two vector fields might not be partial derivatives with respect to the same coordinate system, and there's no rule in calculus that says a partial derivative in one set of coordinates has to commute with a partial derivative in another set of coordinates.
This video (and the next one) basically skip what torsion is really about. I explain torsion (and the torsion tensor) more properly in video 21. In this video I don't really distinguish much between partial derivatives applied to vectors and covariant derivatives applied to vectors... I treat them as basically the same thing. That might not be mathematically rigorous, but there isn't really any reason to confuse the two. You always need Christoffel symbols to explain what they are. The thing that you need to be careful about confusing is partial derivatives of components and covariant derivatives of components (the latter involves Christoffel symbols but the former does not). Your final paragraph is exactly correct: the Lie bracket of basis vectors is always zero. But for arbitrary vector fields, the Lie bracket can be non-zero. The geometric interpretation of the Lie bracket is checking if two vector fields form "nice, closed 4-sided boxes" with each other. Basis vectors always do this by definition, since coordinate systems are make up of nice, closed 4-sided boxes. This is why their lie bracket is zero.
I am having difficulties in understanding the concept of tangent plane in intrinsic geometry. The whole process of parallel transport and covariant derivative is based on the ability of checking the vector in different tangent planes. But a plane that is tangent to a manifold uses another space coordinate, i.e.: I can describe a sphere surface using the 2 coordinates theta and phi, but as a tangent plane is not curved, it can't be described with the same set of 2 coordinates. Where do I get lost?
When working with a 2D surface in 3D space, the tangent plane is just a literal geometric 2D plane that's tangent to the surface. For intrinsic geometry, the concept of a tangent space is more abstract. The tangent space at a point "p" on a manifold "M" is an abstract vector space called "TpM" that's attached to "p". This vector space is the set of differential operators on functions on the manifold "M". You can pick any basis you like for this space, but the standard choice of basis are the partial derivative operators at the point "p". (So for coordinates "u" and "v", the basis would be ∂/∂u evaluated at "p" and ∂/∂v evaluated at point "p"). So the coordinates on "M" can give us a basis for "TpM" by using the partial derivatives with respect to the coordinates. Does that make sense?
@@eigenchris I would love to say yes, however I think I understand the logic but not the geometric and physical meaning. In your Schwarzshild black hole video you've showed that the space is curved because the Riemann tensor is not 0, and as a consequence the circumference is 2*pi*L0 instead of being 2*pi*r. Of course is not curved in another 4th space dimension, is curved because the ratio between circumference and radius is not 2*pi. Calculating the Riemann tensor in this geometry, this implies that you are parallel transporting a vector field and then evaluating it in different tangent spaces. How do those tangent spaces extend in that geometry? Probably that this question shows that I've not been able to make the jump between the extrinsic and intrinsic geometry yet.
@@lucapellegrini709 The process of "parallel transport" is exactly the process of moving a vector across a series of tangent spaces. The instructions for how to do parallel transport are bundled up in the Christoffel Symbols/Connection coefficients. I'm not sure if you've watch the next video (tensor calc 20) but I use two different connections (the Levi-Civita connection, and another one I call the "boring connection") to parallel transport a vector across the sphere, and the results of the parallel transported vector are different when I used different connection coefficients. If you haven't watched TC 20, give it a look and see if that helps things make more sense.
@@eigenchris unfortunately that doesn't help; I have watched that video several times; I am able to calculate the Christoffel Symbols and the Riemann tensor, but I think that I miss some basic stuff on the real meaning of what I am doing. BTW I am referring to the Levi-Civita connection only. To make a quasi-real case: say that I am near a Schwarzshild blackhole; at that point I can travel along the r coordinate and then get back to the starting point on a closed loop. If while I am travelling I am transporting a vector keeping it straight, when I get back to the initial point I will see that the vector has a different orientation. And the difference can be computed through the Riemann tensor. However the Riemann tensor together with the Christoffel Symbols for the Levi-Civita connection implicitly assume that there are tangent planes at every point in the space; and in order to compare vectors, we have to refer to those tangent planes, connected by Levi-Civita connection. This is done automatically by the Christoffel Symbols and the Riemann tensor, so we don't have to bother about tangent planes. But if I would like to see things more in depth, I could ask myself: what is my tangent plane when I am standing near a black hole on a point A? and then when I move to the point B closer to the blackhole? how those planes extend in space, or what's their equation? I can understand this in extrinsic geometry, where the tangent plane doesn't live in the curved surface (except for one point), but I can't understand in intrinsic geometry
"how those planes extend in space, or what's their equation?" I think this is where you're going wrong. For extrinsic geometry, the tangent planes exist in physical space. For example for a sphere or radius 1, the tangent plane at the north pole is the set of all points (x,y,z=+1). For intrinsic geometry, the tangent space is not a set of points on the manifold. The tangent space at point "p" is an abstract vector space "TpM" that we attach to the point "p". I feel like I'm repeating myself here, so I'm not sure if this is helping. But basically the tangent spaces for intrinsic geometry aren't tangent planes you can see with you're eyes. They're more like abstract mathematical tools for thinking about "velocities" through a given point on the manifold (where the velocities through a point are like the differential operators). You could try reading this page, although in my opinion it doesn't look great for beginners: en.wikipedia.org/wiki/Tangent_space#Definition_via_tangent_curves
so we use the extrinsic view to compute a metric tensor?, just to switch to use the intrinsic view to get the christoffel symbols. are there any circumstances where we have to use the intrinsic formula to acquire the C-symbols?
General relativity is done intrinsically. The intrinsic metric is determined by the presence of mass/energy/momentum in a given region of spacetime.
in 16:05 it said it is intrinsic because it doesn't rely on X-y-Z variables
but in my memory when you calculate the value of gij
you need to use x- y -z variables as you need dR/du = dR/dx * dx/du +.... to obtain the metric tensor component
it still relies on x-y-z variables
is that contradicted to what you define "intrinsic"
thank you"
The metric is the fundamental starting point for curved geometry. We need it or else we can't do anything. So we either take a metric we already know from extrinsic geometry, or we make one up from our imagination. The important thing is that, after we have a metric, everything else can be computed intrinsically.
@@eigenchris thank
i think i understand the "parallel transport on constant latitude" on a sphere when i try to use a ball to roll on the surface if i keep the curve of pi/4 on the sphere "touching" the flat plane continuously and and it will move automatically to left (there is centripetal component) .Finally i can see there is a curve lines on this plane as this plane is tangent to every position on the sphere which has a latitude of pi/4 .
In our human frame we cannot feel we exists in the sphere as every point are equally pointed toward the centre,however we feel we exist in a flat plane.
Therefore if we roll a ball on a plane (with every points on the great circle touching the flat plane),then we will feel we are moving forward in straight line ,this can be observed that the ball is rolling on a plane and the path is a straight line on the plane.
(that is ,we move on equator)
if we let every points on the small circle(eg. latitute pi/4 on the sphere) continuously touching the flat plane,the sphere itself will move to the left,if we put some ink overlapped the small circle of the sphere,the plane will left with a curve (bend to the left)with ink which is exactly what the human can feel on earth
is that correct
thank?
this is the other way i can think to describe
@@garytzehaylau9432 This is not an interpretation I had considered before, but it seems to make sense to me. If we roll a ball along a plane so that it is always touching around the pi/4 circle, the ball will indeed roll in a circle.
Amazing video! Do you have any book recommendation to use as reference for these series?
Sorry, I don't have a recommendation. Are you looking for exercises?
10:42 you state that the Christoffel symbols are just scalars and therefore can be factored out. However, the Christoffel symbols are functions of (u1,u2), and the unit vectors are derivatives with respect to u1 or u2, so you cannot just take it outside the derivative. @eigenchris
I'm pulling the Christoffel symbols out in front of the dot products, not the derivatives. You're right that you can't pull Christoffel symbols out from derivatives.
Also, when I called them "scalars", that's a bit misleading. True scalars are identical in all coordinate systems. The Christoffel symbols are not. It would be better to call them "coefficients".
Defining basis vectors as partial derivative operators immediately brings up three questions for me: Why is this a natural definition of a tangent vector? How doe we define the dot product of two basis vectors ? How can we still intuitively think of these tangent basis vectors as little arrows?
can you explain the difference between u^i and e_i ? Aren't they both basis vectors?
e_i are thr basis vectors at a given point. u^i are thr coordinates of a given point in the coordinate system (similar to an x,y coordinate system).
Kind of stupid question Chris, but how do you get the sort of Gothic g symbol for the Inverse Metric Tensor, in Power-Point, Word etc Please? Many thanks again for all these Wonderful Videos. 😀
You type \frakturg and then hit the space bar. You can type \frakturX where "X" is any upper-case or lower-case letter. There's also \scriptX and \doubleX. You can always hover over a symbol in the equation menu to see what the code for it is, or google.
@@eigenchris Thanks very much. I didn't know what this font was called. Also \fracturR is useful for this stuff.
@@JohnJoss1 I think most of them are just stolen from LaTeX. There are many more, like \sqrt \vec \times \dcot, etc. Maybe this article will help. www.pickupbrain.com/ms-word/equation-editor-shortcut-word/
@@eigenchris There's a few Math Symbols you can put in to a Word Document directly, without using the Equation Editor: CTL + SHFT + Q and then any of the letter keys gets you Greek Characters, including the uppercase ones. You can get the Square Root Symbol by typing 221a and then ALT+x, Nabla Symbol 2207 ALT+x, Partial d 2202 ALT+x, Therefore Symbol 2234 ALT+x. There's probably a website somewhere for this stuff. Thanks very much for your link, very useful. 😃
If spacetime is a four dimensions is that mean that the outside world is a five dimension world where we can see the curvature of spacetime and where the normal components would exist
And I have a some confusion about the basis vector of manifold been a derivative operator instead of the derivative themselve ,and so since we don't have a position vector is that mean that any thing tangent to the path of the coordinates is the derivative operator ,and thanks
The traditional version of General Relativity only describes spacetime as a 4D space intrinsically. It doesn't take it to be embedded in a 5D space. I'm not sure if this is even mathematically possible. You would have to google the number of dimensions required to embed spacetime in, as thid goes beyond my knowledge. It may be bigger than 5.
For the derivative operators being vectors, you can try warching this video to get a better understanding: th-cam.com/video/VHkL5HpL0HY/w-d-xo.html
How can one calculate the metric tensor intrinsically?
Sometimes it's just given to you (you invent the metric and see what type of space it is). Sometimes it can be obtained from another equation. In general relativity, the Einstein Field Equations takes the energy-momentum tensor as input, and these solutions are the metric tensor.
hi, eigenchris! I’ve learned tensor from your videos for weeks. Tensor is really a amazing idea to understand our world. But i am confused about how to consider d/d_lamda(d/d_lamda) in vector and covector way since it looks like df(d/d__lamda)=df/d_lamda in #7 video. Oh i get it. d/d_lamda is not a covector when df is. 😅looks like writing the problem down can help figure it out. when seeing d/d_lamda(·), just using the chain rule right? Thanks for the lessons, l really enjoy thinking of it.
Yes, I'm glad you figured it out.
at the beginning you left out all parts with a normal component.
dont you have to make up for that at some point? or do you just ignore all terms with a normal component when calculating the cov. der. for the intrinsic method?
(as i understand you didnt actually calculated the parallel transported vectors, you only checked if the christoffelsymbols are equal)
There's no such thing as the "normal component" in the intrinsic space. This video is somewhat informal... the more "formal" definition of the covariant derivative comes in the next video.
Would it be ok to say that geodesics are "locally shortest paths"?
Yes I would agree with that.
Is the problem with dot product (itself) to derive Christoffel symbol or the procedure of eliminating the perpendicular n vector? (Thanks for great series)
I think the problem is eliminating the perpendicular vector. After all the metric g_ij is by definition e_i dot e_j.
We do have "dot product" on the tangent plane.
Hi, I am a bit lost again. Is the covariant derivative along a path the same as the directional derivative? So the gradient of a vector times a vector?
For the covariant derivative of vectors, it's like the directional derivative that only pays attention to the portion tangent to the surface you're on. You'll see in the next video the covariant derivative can also be applied to any tensor, not just vectors.
@@eigenchris Hi, thanks for the answer. I saw it. I got confused with the symbols, but it really is just that, a symbol for a specific type of derivative, isn't it?
This is brilliant !
Surely the tangent spaces live outside the manifold as well?
Yes, the tangent spaces are separate spaces from the manifold.
Ah ok. But doesn't that make them extrinsic?
@@trigon7015 In extrinsic geometry, you can think of the tangent space as being a 2D plane living in the same 3D space as the surface, with the plane meeting the surface at exactly 1 point. For intrinsic geometry, it's more abstract. For every point p on the manifold, we associate an abstract vector space Tp. It doesn't live in the same "outside space" because there is no "outside space".
Ok thanks, that cleared things up
great video, one request, can you make a video only explaining christoffel symbol.
Videos 15 and 16 explain where the Christoffel Symbols come from.
Will you (or did you) talk about defining these notions coordinate free? For example defining the tangent space without the use of coordinates.
I'm not sure what you mean exactly. With most things I present there is an "abstract" version, and then I break things down into coordinates to do concrete calculations.
I do talk a bit about thr tangent space in videos 5.1 and 20 in this series, but I don't go into much detail.
@@eigenchris I meant a coordinate free definition of tangent space in this case. I didn't look at the 20th episode, but from what I've seen so far (and that isn't all) you defined elements of a tangent space as the span of d/dx^k at that point. Here you must choose a chart for the definition.
I don't really want to bother with the definition of charts and manifolds. I'm being a bit "loosey-goosey" with the idea of a tangent space, just defining it as the span of derivative operators. I don't plan on getting more complicated than that.
@@eigenchris that's ok. Everybody has different goals. These videos require really little pre-knowledge, but still introduce the new concepts. You make great videos!
I computed all the christoffel symbols by using this method... the difference between your method and mine was that I didn't choose r constant or one... i used 3 by 3 metric of spherical coordinates... all the christoffel symbols came out correctly besides Γ^2 lower23... can you help me out???
What dod you get as an answer and what did you expect to get?
@@eigenchris i get zero for all components but the answer should be cotangent
@@sufyannaeem2436 Can you write out your calculation? Maybe I can find where it goes wrong.
can you WhatsApp me or something? I will send you notes so you could find where am i messing..
Why when it comes to geodesics, the vector becomes total derivative rather than partial derivatives?
Paths likes geodesics are parameterized by a single variable (lambda), and so we use the total (single-variable) derivative instead of a partial derivative.
@@eigenchris Thank you for your reply! And thank you for your videos! They really helped me a lot. There's still a question confusing me: When we take the covariant derivative of a vector (like A^a e_a), should we apply the Leibniz rule (like (nabla A)e+A(nabla e))? If so, there will be two Christoffel symbols, which seems weird to me. If the answer is no, is it safe to say that covariant derivatives are only defined to act on the components?
@@official-zq3bv Yes, we basically use the "Leibniz" rule (product rule). We basically define the covariant derivative of an ordinary function to be the partial derivative (vector components are just considered ordinary functions). The Christoffel symbols only come up when we take the covariant derivative of a vector. So scalars will get 0 Christoffel symbols, vectors will get 1 set of Christoffel symbols, rank-2 tensors will get 2 sets of Christoffel symbols, and so on.
@@eigenchris OK, I think I finally got it. I misunderstood the covariant derivative as a newly defined derivative operator (instead of partial derivative) that acts on the component by adding an extra fix term presented by the Christoffel symbol. Now I understand that the upper index in
abla_\mu A^
u does not mean that the operator acts on the A^
u component, but the ^
u component of the _\mu covariant derivative on 'the whole vector A'. I really appreciate your work. Your videos saved me.
Alguém poderia fazer legendas em português? As aulas são boas demais para ficarem apenas em inglês!
Não dá pra colocar neste vídeo. Acho que o autor do vídeo tem que liberar.
Another long one. Good!
I don’t understand why even the bug on the surface would use a partial derivative with no argument. How can you do any calculations with that? If I create a small tangent square of the point P, the bugs can then make u and v their coordinates and calculate basis vectors using a tiny R from P: at least then I can create a little vector space on my computer in that approximate space. Otherwise I don’t see how a partial helps at all without acting on something!
Great videos by the way!
The issue is, you can't define a position vector R in a reasonable way on a curved space. The partial derivative operator act exactly like basis vectors, allowing us to expand other partial derivatives using linear combinations, so we use them as basis vectors instead.
Thanks; as soon as you actually put a function in there though, it then won’t give you the metric tensor you need to do problems, plus it gives you a functions “velocity” and I want the velocity of mars moons! It seems like just a placeholder for the real exterior basis, so you do parameter changes, and then you say, sorry, no R, so we’ll just use the metric to get the same result a position vector would have given to begin with. This has been the most confusing part of the series, and the books I’ve looked at gloss over it as well. It’s critical I think though.
I’m a medical doctor learning GR for fun. I hope you will teach for a living: you have a gift.
I meant you have the partial in there to make it obvious how to do parameter changes but otherwise it doesn’t help to eliminate R only to bring it back in similar form via metric. I guess metric is easier to guess at than guessing the function R from another dimension lol.
@@robertprince1900 Sorry, I've read your comments but I don't understand exactly what you're getting at. Is your concern where you get the metric from? For intrinsic geometry, it's true that we can't get the metric "automatically" from the outside space like we do with extrinsic geometry. So the metric needs to be given to us. (Or, in the case of general relativity, we get the metric by solving Einstein's Field Equations.)
What is the need of intrinsic basis on general relativity Why does this exist and for example not go on with life with normal basis I,J,K?
In curved spaces, you can't really draw a straight vector from an origin point to another point in the space, so the idea of a position vector doesn't make sense. We can only use "tangent vectors" that exist at the tangent space at each point.
@@eigenchris excellent i understand, is there any video where you can find gij components from energy tensor (right side of einsten equation), in order to get rid of the impossible to find ei.ej dot product?, because we dont have extrinsic coordinates out there, we cant really define the o,o,o point, even in that case, which direction would point the x axis? it has no sense there is infinite many possibilities
@@federicopagano6590 I just finished a relativity series (45 videos long). If you search "relativity by eigenchris" you should find it. The 108 videos solve for the metric around a black hole (energy tensor is zero for a vacuum). The 109 videos solve for the metric for gravitational waves (energy tensor is again zero for a vacuum). The 110 videos solve for the metric for an expanding universe (non-zero energy tensor that treats all the galaxies in the universe as a fluid). The 0,0,0 origin point we select is arbitrary, but we often choose a convenient point to make the formulas simple. For example, for a black hole, we choose the origin point to be the center of the black hole.
Conceptually, I always found the notion of ignoring the normal component, on account that the bugs cannot perceive it, tricky. I like to think of the intrinsic geometry like this: a god or gods add a larger space in which to embed the manifold; then they derive all the intrinsic quantities -- the covariant derivative etc -- using the larger space; and finally, they give us all the formulas, which contain only the intrinsic quantities, such as the metric, that we mortals can understand. Of course, since we're better than gods we can just do this our selves.
Yeah, I'm not sure I really have a good understanding of justifying the scribbled-out-normal vector either. Many metrics I have seen "come from" an example involving a higher dimensional space. The problem is that with standard GR, there isn't really an easy-to-imagine outside space. It's hard for me to understand what the covariant derivative really means in GR, other than the fact that it's metric compatible (i.e. keeps vector lengths constant during parallel transport) and torsion-free (i.e. tiny parallelograms "close up" properly).
Hi Chris, what do you mean about the tiny parallelograms closing up properly? That sounds like some useful intuition I might've missed
Hi Chris. Your derivation of the Christoffel symbols appears to use 3 coordinate curves. Why 3? How is it related to the dimension of the space? Thanks
@@chrismaudsley127 "Appears to use"??? I would say, be a little more clear and specific.
1番わかりやすい
For those who want the link
math.stackexchange.com/questions/2215084/parallel-transport-equations
Right... I was supposed to do that, wasn't I.
Thank you very much
1:21 Intrinsic vs extrinsic
that formula is terrifying
Thank you a lot!!!
Magic video
So, for the intrinsic world, we have no way to CALCULATE its metric DIRECTLY. We have to either declare one or solve some other equations, if there is such equation, to get one. It is a really sad story for the people living in the intrinsic world. But the good news is we have some degree of freedom to construct any metric we want as long as the constructed metric can describe the world correctly.
At 10:18 you mention a special product rule for the dot product of vectors. I was just looking at Walter Lewin 18.03 lecture 13 where he shows the curl of a curl of a vector expands out in a way I find very similar except for the negative sign :
∇ x (∇ x A) = ∇ (∇ . A) - (∇ . ∇) A
Funny as to how suddenly I can memorize the expansion of the Laplacian.
That's not the Laplacian
@@twistedsector
(∇ · ∇) A is the Laplacian of a vector field A. So you can do some algebra on the expression given to remember that (∇ · ∇) A = ∇ (∇ · A) - ∇ x (∇ x A), which is the definition of the Laplacian.
That’s weird, this didn’t show up in my notifications.
I love you
This is too easy bro, do u have something difficult?