Tensor Calculus 22: Riemann Curvature Tensor Geometric Meaning (Holonomy + Geodesic Deviation)
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- Опубликовано: 6 сен 2024
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Video 21 on the Lie Bracket: • Tensor Calculus 21: Li...
Videos on Covariant Derivative:
17 - Flat Space: • Tensor Calculus 17: Th...
18 - Curved Surfaces: • Tensor Calculus 18: Co...
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The day this guy has more subs and views than tiktokers will be the day humankind will make a giant leap forward
Such a great video man. Keep it up!
Thanks!
Andrew Dotson Andrew? You’re here? Amazing!
Papa Dotson
@@canyadigit6274 CanYaDigIt? You're here? Amazing!
@@eigenchris Einstein never nearly understood TIME, E=MC2, F=ma, gravity, or ELECTROMAGNETISM/energy.
He was, in fact, a total weasel.
c2 represents a dimension ON BALANCE, as E=MC2 IS F=ma in accordance with the following:
UNDERSTANDING THE ULTIMATE, BALANCED, TOP DOWN, AND CLEAR MATHEMATICAL UNIFICATION OF ELECTROMAGNETISM/energy AND gravity, AS E=MC2 IS CLEARLY F=ma:
The stars AND PLANETS are POINTS in the night sky. E=MC2 IS F=ma, AS this proves the term c4 from Einstein's field equations. SO, ON BALANCE, this proves the fourth dimension. ELECTROMAGNETISM/energy is gravity. Gravity IS ELECTROMAGNETISM/energy !!!
TIME is NECESSARILY possible/potential AND actual IN BALANCE, AS E=MC2 IS F=ma; AS ELECTROMAGNETISM/energy is gravity. INDEED, TIME dilation ULTIMATELY proves ON BALANCE that E=MC2 IS F=ma; AS ELECTROMAGNETISM/energy is gravity. Gravity IS ELECTROMAGNETISM/energy.
Gravity AND ELECTROMAGNETISM/energy are linked AND BALANCED opposites, AS E=MC2 IS F=ma; AS ELECTROMAGNETISM/energy is gravity; AS gravity/acceleration involves BALANCED inertia/INERTIAL RESISTANCE; AS GRAVITATIONAL force/ENERGY IS proportional to (or BALANCED with/as) inertia/INERTIAL RESISTANCE. Gravity IS ELECTROMAGNETISM/energy.
E=mC2 IS CLEARLY F=ma. This NECESSARILY represents, INVOLVES, AND DESCRIBES what is possible/potential AND actual IN BALANCE, AS ELECTROMAGNETISM/energy is gravity. Gravity IS ELECTROMAGNETISM/energy !!!
By Frank DiMeglio
Before seeing this series, I never imagined I could self-learn tensors intuitively. This is what makes Chris' teaching so amazing! Thank you for all your effort!
Small error around 24:50 where you say that for the sphere the geodesics are originally accelerating away from each other and then towards each other. No. With positive curvature, the geodesics are always accelerating towards each other.
Thanks for such a well-done series.
Hi this is not so clear to me. Apparently Andrew you are right only if you travel to poles from the equator. Is it a matter of "sign" maybe?
@@AndreaPancia1 Remember that velocity is the first derivative and acceleration is the second derivative. They are originally moving apart (positive velocity) but less and less rapidly the further you go (negative acceleration). As you cross the equator the velocity of moving apart is zero and after that both velocity and acceleration are negative.
Pay this man! I just did. I'm very selective about where my money goes. I can't think of a better investment than Chris's hard work and incredible intrinsic understanding. 🤘
Me too!!
i love your videos. You present all the mathematical steps and the intuition. I have 300 math and physics books. Your explanations are much better than the explanations in any of my books. I bought you a cup of coffee and will continue to support you.
Thanks a bunch! That's a lot of books. Are they physical or ebooks?
Physical
Thank you Chris this is just wonderful. I've bought you a coffee, hope this will help you to produce more videos
Thanks for your donation!
A lot of "Heavy Lifting" (hard brain work!) has obviously gone into producing this Superb Video Chris. Thankyou.
Amazing, unique and clear videos, they should be standard viewing for anyone starting in this field, nothing like them exists.
Thank you Chris for this incredible work. Here I try to solve the bracket mystery and I hope it will help. I will use "Why Alex Beats Bobbie at Poker" notation in this commentary (with L the parallel transport along [u,v] ) that i hope will help achieve showing the equality with [u,v] not being the null vector . If you consider the points where the covariant derivatives is taken before going to the limit and using the vector field argument it helps leaning towards the correct formula. Let''s note w(A) for instance the vector w at point at the arrival of A and O the Identity at the origin, u_v the vector u at v and Cov(x,y) the covariant derivative along x of y. Let's consider also C' and D' as invert C and D, and d =||[u,v]||. Basically we want to show R(u,v)w= cov(u,cov(v,w)) - cov(v,cov(u,w)) - cov([u,v],w). So applying your line we have Riemann Tensor R = (w(O)-DCBLAw(O)/rs = DC * (C'D' - LBA)w(O) / rs and we have (C'D' - LBA)w(O) = C'D'w(O)-C'w(C)+C'w(C) -w(L)+w(L)-LBAw(O)-LBw(A)+LBw(A)+Lw(B)-Lw(B)+w(L)-w(L) and then it is possible to cancel extra term w(L) which was not possible before. Then we can form the covariant derivatives (C'D' - LBA)w(O)=C'(D'w(O)-w(C)) + (C'w(C)-w(L)) -LB(Aw(O)-w(A))-L(Bw(A)-w(B))-(Lw(B)-w(L)) and we already see that the last term will lead to the missing term cov([u,v],w). Now we get (C'D' - LBA)w(O)=s*C'cov(v,w(C))+r*cov(u_v,w(L))-r*LBcov(u,w(A))-s*Lcov(v_u,w(B))-d*cov([u,v],w(B)) and by rearranging terms we have (C'D' - LBA)w(O) = s*( C'cov(v,w(C)) - Lcov(v_u,w(B)) - r * ( LBcov(u,w(A)) cov(u_v,w(L))) d*cov([u,v],w(B)). Now we see that to take covariant derivatives you have to be at the same place an for example L in C'cov(v,w(C)) - Lcov(v_u,w(B) helps transport the covariant derivative along v to the correct place by filling the gap so you can take the covariant derivative along u of this covariant derivative along v at the same point (arrival of L), and then we understand easily that C'cov(v,w(C)) - Lcov(v_u,w(B) = r* cov(u,cov(v,w)) and LBcov(u,w(A)) cov(u_v,w(L)))=s * cov(v,cov(u,w)) . So you get cov(u,cov(v,w)) - cov(v,cov(u,w)) - (d/rs) cov([u,v],w) and the only missing point is that d must be equivalent to rs when (r,s)->(0,0) so the equality is true, and I have confidence it is true as the decomposition above allows to evaluate all the covariant derivatives in the correct place (indeed the correct Tangent Spaces ) and this cannot be a simple coïncidence.
An incredible idea
written like this is super hard to follow. u should consider making a video or pdf about it! cause it seems a pretty important equation without a rigorous demonstration even in the black book with thousands pages
This is nothing short of incredible. Thank you eigenchris!
Glad you found it helpful!
Great job, thank you a lot for sharing this! I love that you don't pretend to know everything (the part with the lie braket), it really reinforces that you don't teach things you don't fully understand, and it makes me trust this videos more than other sources I found! I'll continue watching them even after my exam, this is really fun :]
could not contribute because the link failed! Could that be fixed? Best wishes,
Prof. Dasgupta
Columbia University
Which link?
url to contribute
15:20 I think the third term exists so that the curvature tensor doesn't care about the value of u and v anywhere other than the point where it's being evaluated. The lie bracket can be different even if two pairs of vector fields have the same value at some point in the manifold, meaning that if that term was removed here, that would supposedly mean that two pairs of vector fields with the same value at some point could give different linear transformations when put into the curvature tensor. Because of this, applying the riemann curvature tensor to vectors directly wouldn't be as well defined (since each vector can be extended to different vector fields which makes the tensor give different results).
I think this would make it not qualify as a tensor field, yet it would still be a coordinate invariant object. A change in coordinates would not change the linear transformations from this modified curvature tensor when two vector fields are put in, but again, this modified curvature tensor asks for vector fields rather than vectors and can't just be defined as being a tensor at each point.
Thanks so much Chris for this amazing opportunity to learn you are giving us.
Oh damn, I didn't realize this was an ongoing series still. Stoked for more!
Isn't he amazing! I listen to him before I go to sleep. Very relaxing.
Chris, first of all, this is some spectacular work! Precise, intuitive and very well explained!
Also, I would like to offer a hopefully intuitive way to justify the Lie Bracket term of the Riemann curvature Tensor. Basically, redo the proof but this time calculate w - DCLBAw where L is the operator that lets you transport along the "gap" vector [u,v]. All the calculations remain largely the same, except that we should instead write LB - 1 = (LB - B) + (B - 1). If we do everything carefully, there should be an extra term (BAw - LBAw)/rs in the end, which should be equal to exactly the Lie Bracket term (The difference of transporting all the way to the end of the gap, versus the beginning of it should be exactly equal to the covariant derivative over the gap). This makes sense since we need to "pay" the extra price of having to move along the gap over the (now) pentagon instead of the rectangle. I hope this helps and thank you again for the wonderful work!
Thanks for pointing this out. I see w get an extra (BAw - LBAw)/rs term. Although I'm not sure if there's a way to show this equals the derivative in the direction of the Lie Bracket.
@@eigenchris I am not an expert on the subject, but I think the way to do this is to use the DC from outside, which (ignoring the factoring scalars for now) yields DCBAw-DCLBAw which is the same as w - Lw (by preservation of angles and lengths by parallel transport operators). After proper (order 2) scaling this should give our desired Lie Bracket term.
why you can see DCLBA as simply L?@@Why_Alex_Beats_Bobbie
I love this series! It's great to see more content!
your videos are treasure, they really help seeing things and what's hidding behind the definitions. Thank you !
Thanks!
YOUR EXPLANATIONS SHOWS THAT YOU REALY UNDERSTAND DIFFRENTIAL GEOMETRY AND GR
Muchisimas gracias profesor por todos sus videos. Reciba un saludo desde 🇪🇸 España
Thanks for all your videos. I had a thought about the lie bracket term. Let's think about orthonormal polar cords with the basis vectors representing velocities of the tangent to the flow curves. Now, start at (r1, theta1) and execute the following: move for delta t in radial direction, then move for delta t in theta direction, then move for delta t in minus radial direction and finally move for delta t in minus theta direction. The distance traversed in space will be velocity x time so we will move a distance specified by the basis vectors: (r1, theta 1) to (r1+r basis vector, theta 1) to (r1 + r basis vector, theta 1 + theta basis vector) to (r1 + r basis vector - r basis vector, theta 1 + theta basis vector) to (r1, theta 1 + theta basis vector - theta basis vector). Now note that because we are using an orthonormal basis we haven't executed a closed loop. This is because the theta basis vector is normalised to the speed of 1 so we cover the same distance of arc length when we move + theta basis vector and minus theta basis vector, however to close the loop we have to move a smaller distance in the minus theta direction because the radius we are at, r1, a smaller radius than the radius we were at when we executed the move in the + theta direction, r1 + r basis vector. The Lie bracket is required to close the loop for a non co-ordinate basis, like the orthonormal polar example above. If however, you were using the orthogonal polar basis which is not normalised, then the lie bracket term would go to zero. In conclusion, if you have a coordinate basis you don't need the 3rd term and if you have a non coordinate basis you need the 3 rd term to close the loop. The lie bracket is the closer of quadrilaterals - see MTW, page 250 for geometric derivation.
Great videos! Thanks! Regarding "flatness," this would seem to be a property of a region of space, rather than a point in space, and this observation might should impact on the presentation somewhere. Hence we could also consider the metric tensor being an unchanging identity matrix over a region, for example.
The "flatness" here is the result of a limiting process, so it can be applied to single points. If you look at the example of the surface of a donut/torus, the "inside" of the donut has "negative curvature" (like a horse's saddle), and the "outside" of the surface has "positive curvature" (like the surface of a sphere). On the top and bottom of the donut, there is a ring that is one point in thickness where the curvature is zero, as it transitions from - to +.
I am going back and forth between your videos and "Visual Differential Geometry and Forms" by Tristan Needham. Both are excellent! Needham derives the curvature tensor, including the gap, on pages 286-292. You may be interested in how he includes the covariant derivative with respect to the Lie bracket.
From your initial explanation I had the same feeling as stated in the textbook that, to define a curved space, the torsion of the connection must be non-zero.
It seems like saying that in a curved space the the connection is irrotational.
Keep up ur free education service Ur videos are awesome
I had gone through a lot of videos , how to learn mathematics for general theory of relativity i thought tensors are difficult to understand but now i can assure that your videos changes my opinion.thanks
Thank you so much eigenchris for your work. There is a small typo at 21:29: a basis term e_i is missing in the 4th row. Thank you again!
Excellent explanation of the tensor notations and the applications for the Riemann curvature tensor
Thanks. It is very clear for the explanation. Excellent!
These videos are very helpful for learning the math behind GR. Thanks.
Great explanations ! You are really talent in lecture. Keep going and hope that your chanel will be more successful
Did you check R. Wald's book on general relativity? He gives a sort of a proof for why the lie bracket term vanishes. He argues that the vectors u and v are two linearly independent basis vectors of the local inertial frame and by definition their commutator/lie bracket is equal to zero.
I used 'by definition' rather loosely here; in the book he doesn't really give a full fledged proof for why the lie bracket is zero but only provides arguments which are pretty convincing and converting them to a formal proof is more or less trivial.
Ironically though, I came here because I found the section on the curvature tensor in the book incomprehensible lol.
well explained sir,,ur videos making life easy,
Hi Chris! One small formal remark. As written at 13:00, all the numerators inside the four summands (Aw-w), (Bw-w), (C^(-1)w-w) and (D^(-1)w-w) will result into the covariant derivatives of w along u and v with a negative sign in front (-∇w). The same holds also at 14:05 for the repeated covariant derivatives (-∇∇w). However, both steps will produce the same formula for R(u,v)w at the end. :)
I'm looking forward to the rest of the video series on relativity. Excellent work!
Yes I thought also but his convention are opposing
why this channel is so underrated ??
I LIKE YOUR VIDEOS VERY MUCH - YOU ARE GOOD TEACHER.
15:12-15:25. I think I have an explanation. First off, I'm pretty sure there's an error at 12:47-13:04. The Covariant derivative Nabla(u)(w) represents the deviation of the w field (at (r,0)) from its parallel transported w. This means that for a small step r: w(r,0) ≈ rNabla(u)(w) +Aw, meaning Aw-w ≈ -rNabla(u)(w). A similar mistake can be found with the rest of the differences between parallel transported w's and regular w's. At 14:08, the limit shown at the top is the negative of what it should be, but a similar set of mistakes are made when producing the bottom formula, giving the correct limit. This simply means that when I convert a difference between a parallel transported vector and its normal counterpart, I actually get the negative of the covariant derivative (times a small scalar).
We start by adding a fifth linear map which I'll call E, that parallel transports w along [v,u] (the [u,v] vector at 16:02 is pointing the wrong way. You'll see why it starts at u(v) and ends at v(u), making it v(u)-u(v)). Then the Riemann tensor becomes R(u,v)w=lim(r,s->0):(1/rs)(w-DCEBAw). Adding and subtracting (1/rs)(DCBAw) gives lim(r,s->0)(1/rs)(w-DCBAw+DCBAw-DCEBAw). This can be split into lim(r,s->0):(1/rs)(w-DCBAw)+lim(r,s->0):(1/rs)(DCBAw-DCEBAw). The first limit becomes Nabla(u)(Nabla(v)(w))-Nabla(v)(Nabla(u)(w)) as we've already derived, meaning the second limit should give our correction term. We have a limit of lim(r,s->0):(1/rs)(DCBAw-DCEBAw). factor out DC, and substitute BAw for w' (we can first parallel transport w over to BAw, then define a new w' field starting at the location of BAw). This gives lim(r,s->0):(1/rs)DC(w'-Ew').
Now, we know E performs a parallel transport along [v,u]. If you recall, the change of a vector field a, as we travel a tiny amount e along a field b, can be expressed in terms of the covariant derivative. If we move from point p to point q, we get (parallel transported a(p)) + e(Nabla(b)(a))(p) ≈ a(q). This is because the covariant derivative of a vector field along a path measures the field's deviation from parallel transport. We know the direction of the E parallel transport map is along [v,u] (up to a minus sign), but what is the size (the e value) and exact direction?
We can figure this out by using a flat-space approximation and keeping track of all the translations we've done. To start, let's find where the start of the path of the E map is: we start at a given starting point *p*. We move r units along *u* giving *p*+r(*u*). We then move along the *v* vector as it is located at *p*+r*u*, but this isn't the same as the v vector at p. This vector is different by a factor of v's derivative with respect to the u direction, giving *v*+r*u*(*v*) as our new v. we walk along this new vector, by s units giving a new position of *p*+r*u*+s*v*+rs*u*(*v*), and we are at the start of the E path. To get to the end of the E path, we walk along v first, then the changed version of u, giving an end position of *p*+s*v*+r*u*+rs*v*(*u*). subtracting final-initial gives a vector of rs[v,u] telling us the approximate size and direction. This size approximation gets better as r and s go to zero.
Knowing this, we now know the size and direction of the E path. We also know that w' (at the end of the path) ≈ rsNabla([v,u])(w')+Ew', giving w'-Ew' ≈ rsNabla([v,u])w' = -rsNabla([u,v])(w'). This means our limit for the correction term from before is lim(r,s->0): (1/rs)DC(-rsNabla([u,v])(w'))=lim(r,s->0)DC(-Nabla([u,v])(BAw))=-Nabla([u,v])(w), which is what we see in the video, and on the Wiki page.
Fantastic work Chris.
Third term measure that change of W vector if we go through that lie bracket gap, then after measuring that change we are ready to compare those two transports.
Great lecture, thanks for the good explinations
Brilliant exposition as usual by Chris. Just a question to Chris or anyone who's reading this ? Is there a book where your discussion on holonomy can be found ?
Thank you very much, you are an excellent teacher, I am a Spanish retiree and I do not know English, I am helped by the google translator. I would like to thank you in some way, I try to open the link but it is not possible
Depending on the parameter value the vector field is present over the edges of the loop..... It's value and direction changes at every point on the loop.....
Hello, this is an excellent resource for learning differential geometry, thanks a lot for your efforts! I have a small comment on the calculation at 21:00. While the first covariant derivative of a scalar is indeed the ordinary derivative, this is now a vector, so the second covariant derivative does involve the Christoffel symbols. This doesn't change the result, however, because of the torsion free property. All the best to you!
Can you explain what you mean further? I was under the impression that the covariant derivative doesn't change the tensor's type. (i.e. covariant derivative of a scalar gives a scalar, covariant derivative of a vector gives a vector, and so on.)
@@eigenchris Looking at the components, if you write D_i v^j , where D_i is the cov. derivative in the i basis direction, these are (the components of) a (1,1) tensor. In the same way, the derivative of a scalar is a vector (the gradient). The second cov. derivative would then be the cov. derivative of the gradient, which has the Christoffels in it.
@eigenchris Actually, I think in my notation, your Grad_v would be just v^i D_i, and this dot product would give back a scalar again, so I guess there is no problem after all! Sorry about the confusion!
@@leonardodelima3133 I see what you're saying, but D_i (v^j) isn't just one covariant derivative. It's a set of covariant derivatives, one for each "i". In the case of D_v, there's only one direction: v. So it doesn't change the type of the tensor it acts on.
@@eigenchris Yeah, I got it now, it was a notational mishap. Sorry!
14:06, so if the expressions inside the green and purple boxes are the secondary c-derivative in u and v, where does the primary c-derivative inside the bracket on 3rd line come from? these expressions are supposed to be equal to each other? Is there something i'm missing?
Simply wonderful video
Great series of videos. To determine if a space is flat, can't you also measure circumference and radius of a circle, and check if their ratio is 2 pi?
Yes. The Riemann Curvature Tensor is the standard way of determining the curvature at a single point, though.
Hi Chris. Why are we assuming that a parallelogram can be drawn on a curved surface. Normally this should not be possible. Are we assuming that the curved space is locally flat and therefore this should be possible? Are we also assuming that the manifold is torsion free so when you parallel transport anyone of the basis vectors along the direction of the remaining others the sides of the quadrilateral should close and since parallel transport preserves the magnitude of the vector the resulting quadrilateral shape should be a parallelogram?
At right around 6 min re holonomy, you give example pf paralell transport. Is paralell transport only valid with the soldier's arrow is pointing tangent to the curve? If so, then i get why we can't say "What if it points normal to the surface...then it paralell transports perfectly." So in summary, I'm guessing paralell transport is only paralell transport when the spear (tangent vector) is pointing tangent to the surface. This leads me to my next question. If the spear is point normal to the surface and you paralell transport...is that a thing???
Appreciate all your valuable work!
as we are doing parallel transport of w vector throughout the parallelogram, then why not all the 4 covariant derivative of w are zero (0) at 13:50.
As I explain around 11:30, we use a vector field w to calculate the covariant derivatives, which are non-zero. The choice of vector field doesn't matter in the end because thr end result only gives us thr changr in w at the beginning and end of thr loop.
@@eigenchris thanks for clarification.
Thanks . It is very easy expaination and helpful to us.
At 12:20 if follow the rule from Tensor Calculus 21 17:21 the direction of nabla u(w) should be pointed from the end of Aw (parallel transported) toward the end of w actual vector field. But here is vise verse.
Shouldn't it be nabla-u(w) covariant derivative of w in -u direction?
If covariant derivative is evaluated at (r,0) w is brought back from (0,0) by A matrix. It is like we are evaluating the change of w along -u direction. Is that right?
Around 15:15 you mention that you find it hard to justify the 3rd term in your expression for the Riemann curvature tensor. Frederic Schuller (a fantastic lecturer) directly proves precisely why you need a term of that form for the torsion tensor. He doesn't explicitly do it for the Riemann tensor (it is assigned as an exercise), but the procedure follows closely what was done for the torsion case. I highly recommend Schuller's lecture series here, great stuff! (Timestamp where he begins the torsion proof here: ruclips.net/video/2eVWUdcI2ho/видео.html)
Quick question: when you say linear map around 7:34, are you not implicitly making the assume that r and s have been taken to zero already, as these aren't really linear maps as they move the vector to other coordinates? Is there a way to find the same result without the need for such an assumption, or is it a fine one to make? Cheers :))
A linear map associates one vector field with another vector field in a linear way, and the transformation of parallel transporting w from (0, 0) to (0, r) is linear in that the result of scaling is linear and transport of w+x is equal to transport of w plus transport of x.
This doesn't mean parallel transporting "halfway" down one side of the parallelogram will give you half the change, that's a different linear map entirely. Though this is a better approximation as you take the limit to 0 where you effectively just get the derivative, which is a linear operator.
Which book for tensor calculus and GR do you suggest as a support for your lectures?
absolutely great video
9:10 my simpler (but rough) derivation: R(ũ,ṽ)ỹ = ỹ - DCBAỹ = DC(C'D'ỹ - BAỹ) = C'D'ỹ - BAỹ = (1+▽ũ)(1+▽ṽ)ỹ - (1+▽ṽ)(1+▽ũ)ỹ = ▽ũ▽ṽỹ - ▽ṽ▽ũỹ
16:25 my derivation can handle ▽[ũ,ṽ] term. R(ũ,ṽ)ỹ = ỹ - DC(1+▽[ũ,ṽ])BAỹ = C'D'ỹ - (1+▽[ũ,ṽ])BAỹ = C'D'ỹ - BAỹ - ▽[ũ,ṽ]ỹ = ▽ũ▽ṽỹ - ▽ṽ▽ũỹ - ▽[ũ,ṽ]ỹ
13.23 it’s really my first time hearing of Inverse of a Vector.What’s more you alleged its direction is just the opposite of the original….can you explain on that?
Sorry if it’s stupid,I took my Linear Algebra five years ago….can’t remember most
Great to have you back Sir.... Respect from pakistan
The third term in the expression $R(X,Y)Z =
abla_X
abla_Y Z -
abla_Y
abla_X Z -
abla_{[X,Y]} Z$ arises because of the non-commutativity of the Lie bracket of vector fields.
In the case where the Lie bracket $[X,Y]$ of the vector fields $X$ and $Y$ is zero, then the third term $
abla_{[X,Y]} Z$ in the definition of the Riemann curvature tensor disappears. This is because in this case, the covariant derivatives $
abla_X
abla_Y Z$ and $
abla_Y
abla_X Z$ are equal, since the Levi-Civita connection is torsion-free, and the Riemann curvature tensor reduces to:
$R(X,Y)Z =
abla_X
abla_Y Z -
abla_Y
abla_X Z$
However, when the Lie bracket $[X,Y]$ is non-zero, the two covariant derivatives $
abla_X
abla_Y Z$ and $
abla_Y
abla_X Z$ no longer coincide, and the difference between them is given by the third term $
abla_{[X,Y]} Z$ in the definition of the Riemann curvature tensor. This term arises because the Lie bracket of the vector fields $X$ and $Y$ measures the failure of the covariant derivative to commute, and it is responsible for the curvature of the manifold.
Intuitively, the third term can be interpreted as a correction term that accounts for the non-commutativity of the covariant derivatives. It represents the curvature induced by the non-zero Lie bracket of the vector fields $X$ and $Y$, and it quantifies how much the manifold is curved in the direction of the vector field $[X,Y]$.
There is a mathematical proof for the arising of the third term in the definition of the Riemann curvature tensor when the Lie bracket of the vector fields $X$ and $Y$ is non-zero.
The proof is based on the following observation: the covariant derivative of a vector field $Z$ along the Lie bracket $[X,Y]$ is given by the commutator of the covariant derivatives of $Z$ along $X$ and $Y$, i.e.,
$
abla_{[X,Y]} Z = [
abla_X,
abla_Y] Z$
This identity follows from the definition of the covariant derivative and the properties of the Lie bracket of vector fields.
Using this identity, we can write:
$
abla_X
abla_Y Z -
abla_Y
abla_X Z -
abla_{[X,Y]} Z =
abla_X
abla_Y Z -
abla_Y
abla_X Z - [
abla_X,
abla_Y] Z$
Now, recall that the commutator of two covariant derivatives $
abla_X$ and $
abla_Y$ of a vector field is given by the curvature tensor $R(X,Y)$. In other words,
$[
abla_X,
abla_Y] Z =
abla_X (
abla_Y Z) -
abla_Y (
abla_X Z) -
abla_{[X,Y]} Z = R(X,Y) Z$
Substituting this expression into the previous equation, we obtain:
$
abla_X
abla_Y Z -
abla_Y
abla_X Z -
abla_{[X,Y]} Z =
abla_X
abla_Y Z -
abla_Y
abla_X Z - R(X,Y) Z$
This equation shows that the third term $
abla_{[X,Y]} Z$ in the definition of the Riemann curvature tensor arises precisely because of the non-commutativity of the covariant derivatives, and it is equal to the curvature tensor $R(X,Y) Z$. Therefore, the Riemann curvature tensor measures the deviation from the commutativity of the covariant derivatives, and it quantifies the curvature induced by the Lie bracket of vector fields.
14:09 Took a while to realize that you meant it became a covariant derivative with respect to u of the preexisting covariant derivative with respect to v operating on w.
Is our definition of intrinsic geometry intrinsic enough? I especially felt this when you were going through the geodesic deviation concept. I can't get out of my head the idea that true inhabitants of a curved space would perceive geodesic deviation as always zero because their sensory organs would be shrinking/expanding as their elementary particles followed geodesics. I know this idea is wrong but I don't know why! 😂
The third term at (15.20). Is that because it is a general formula including the situation where there may be torsion so the contribution of non-zero torsion has to be subtracted in the formula for R? See Penrose "The Road to Reality" around page 316.
Would it be possible to elaborate on the insight on how Riemann came up with the formula ?
The absolute top value of all your excellent videos is the deep insight you provide. But for this particular video, I find it hard to swallow that Riemann found the formula of the curvature tensor by right off the bat multiplying linear maps D C C -1 D -1 and proceeding until the final formula was discovered. This quadruple multiplication by linear maps is orders of magnitude beyond simply "squaring the square". Maybe the quadruple part of it can be explained by the 4 sides of the parallelogram but calling in linear maps ?
This approach feels more like the standard way math textbooks prove a formula. Sort of like : already knowing the answer, find the most direct way to prove it correct.
I spent around 3 weeks trying to find a good explanation of how the Riemann Tensor formula is derived. The unfortunate truth is that there just aren't that many intuitive explanations of the Riemann tensor online (I'm happy to be proven wrong about this if you find a good explanation).
Here are some alternatives:
Sean Carroll takes the parallelogram approach but focuses on components, and doesn't talk about the Lie Bracket term at all (see page 75): preposterousuniverse.com/wp-content/uploads/grnotes-three.pdf
The essence of Gravitation 11.4 is in this question by me on Physics StackExchange: physics.stackexchange.com/questions/485217/understanding-riemann-curvature-tensor-in-misner-thorne-and-wheelers-gravitati
Section 6.5 of Schultz's "A first course in general relativity" tries to explain the parallelogram approach using integration and I find it quite ugly: www.fis.unam.mx/~max/mecanica/b_Schutz.pdf
The "most convincing" explanation was the one I present in this video, from Theorem 5.11 in this PDF: math.uchicago.edu/~may/REU2016/REUPapers/Wan.pdf
If it makes you feel any better, instead of comparing w-DCBAw, you can imagine compare (C-1)(D-1)w - BAw, which would mean taking 2 vectors at the origin and bringing them to the opposite corner of the parallelogram using different paths, but this is ultimately equivalent to what I did (I just factored out DC, which becomes the identity in the limit).
I'm sorry I can't offer anything better, but I looked pretty hard for something better and found nothing. At this point I think I should just move forward and get to GR.
You might be interested in reading Chapter VII Section 2 of this 1900 textbook "Absolute Differential Calculus" (the old term for tensor calculus) by Levi-Civita and Ricci (page 194 of PDF): download.tuxfamily.org/openmathdep/Calculus_Advanced/Absolute_Differential_Calculus-Levi-Civita.pdf
They use an old notation. The Christoffel Symbols Gamma^k_ij are denoted {ij,k} and the Riemann Tensor components R_abcd are demoted {ab,cd} (I think... could be wrong about the exact indexes).
Thank you for your reply to my harsh criticism of your work. I know you are putting a lot of effort for the benefit of people you don't even know, I will look over the reference you kindly provided. I think I let my pet peeve about standard math textbooks take too much importance.
There is a straight forward approach to verify that the component formula is correct, and it would be simple and intuitive except for the usual plethora of symbols that all the GR formulas generate, making everything look like a bowl of alphabet soup. Note that the formula for parallel transport of a vector, dvx + sum of terms involving Christoffel symbols = 0 gives a formula for dvx = sum of terms involving Christoffel symbols, and you can make a linear approximation of the change in v parallel transported on the paths a->b->c and a->d->c and take the difference which involves the Christoffel symbols evaluated at different points as well as products of the Christoffel symbols, divide by rs, and take the limit and you get the component formula for Rhijk. To verify I tried it for the sphere where most of the Christoffel symbols are 0, so the formulas are manageable, and it wasn't bad. I don't know where to find the proof written out.
In description, you should correct .com instead of /com
Can't wait to have time to start watching all your tensor videos, I already studied them but in the old physics way from old books, but...just wanna really understand them in the deepest way possible.
Thanks.
Thanks. Should be fixed now. I hope you enjoy the series.
@@eigenchris This site can’t be reached www.ko-fi’s server IP address could not be found.
DNS_PROBE_FINISHED_NXDOMAIN
Sorry to bother no rush!
Hi chris, I have a little problem about the geodesic deviation. While deriving the formula we assumed the torsion-free tensor, and my question is that why can we assume it is a torsion-free tensor field beforehand and how to determine a tensor field is torsion-free or not? thanks!
The concept of "torsion free" describes the connection (upper case gammas) that we're using. It's not about any particular tensor field that lives in our curved space. You can see in video 21 that the definition for a connection being "torsion free" is Γ^k_ij = Γ^k_ji.
We're free to use any connection we want on the space we're dealing with. We're choosing to use the "Levi-Civita connection", which is torsion-free by definition. So the Riemann Curvatuve Tensor we talk about here is only defined for the Levi-Civita connection.
@@eigenchris Got it! Thanks for answering.
Dear Chris when you do the formula for R(u,v)w minute 9:58, usually every formula is done taking the final value minus the initial value, but you are doing the opposite. Is it correct?
when i studied your video 21 i developed the idea that if Lie bracket≠0，such that the loop fails to close,then the space is curved.
Not until i saw your video 22,in particular 26:26,when you simply postulated the torsion free.
So,is there a version of the Eqn for Geodesic Deviation when you don't use that postulation?what I'm implying is…is it necessary??if it is,why?
Hi Chris. At 8:38 you say that "We also divide by the product r and s to make the answer independent of the area of the parallelogram". But the product rs is NOT the area of the parallelogram. That has got me confused. Can you explain? Thankyou
It's proportional to the area of the parallelogram. So as the result shrinks, rs will shrink by the same amount.
@@eigenchris Thanks for the quick reply Chris. I can see that the product rs is directly proportional to the area of the parallelogram described by the general formula= The product rs multiplied by the sine of the angle between the vectors u and v. However the sine of the angle is determined by our choice of the vectors u and v that make up the parrallelogram and therefore the Riemann curvature tensor is not entirely independent of our choice of the vectors u and v i.e. it is not "normalised" with respect to u and v. Have i understood you correctly?
In 4:27, why the Chris. coefficients being zero is equivalent to that the metric looks like Euclidean ? I don't get it. May someone explain it in more details?
I think the fact that "all zero coefficients" gives "flat space" is true for the Levi-Civita connection. In this case we are not using the Levi-Civita connection.
How do we know there exists such a parallelogram in a curved space? I mean, how do we know vector u here and there are the same or parallel?
And if the answer is that the space is locally flat when the parallelogram is infinitesimally small, then I'm having a hard time understanding how we find non-zero curvature in this way, as flat means no curvature, unless the curvature goes "slower" to 0 than the vectors become parallel, mind boggling for me.
Thanks to anyone with insights into this, and of course to Eigenchris for the great great videos !
Erez
It's not really a "parallelogram" in the sense that opposite sides are parallel. It's just a 4-sided shape that you get when you travel along u, then v, and then backward along u and backward along v. If the lie bracket is non-zero, the loop is a 5-sided shape because the loop requires the lie bracket result to close properly.
@@eigenchris Thanks for the clarification and (very quick) response !
11:46
Hey I actually devise of another theory in lieu of the vector field stuff since I didn’t find that convincing(perhaps you just glossed through it so it’s not very elaborating).
I think that since we’re saying r and s both approach zero,such that the area of the parallelogram is infinitesimally small,,then any curved surface under the face of infinitesimally small area simply flattens out to effectively become a flat space.
Thus we can just subtract vectors in 11:35
Moreover,if you listen to 8:38,what you alleged had automatically implied that RS gives the area of the parallelogram,but we’re dealing with manifold,so RS really doesn’t give area ain’t it.
So had we used the my self-constructed theory,we won’t have all these problems anymore.
What do you reckon?
this is incredible
Can "as straight as possible" be made more mathematically precise? My thinking about parallel transport is as follows: every curve has normal, binormal and tangent vectors which form the so-called Frenet-Serret frame. Parallel transport means that one keeps the arrow at a fixed orientation relative to this frame. Is this right?
In the Tensor Calculus 18 videos, I define parallel transport in a more "common sense" way using extrinsic geometry, where you can imagine someone walking along a path with a harpoon, trying to hold it as straight as possible. I don't think I can agree with that Frenet-Serret frame definition. As an example, if you have a flat circular path in 3D space, the Frenet-Serret frame will be constantly rotating about it. But parallel transporting a vector around that circle would keep the vector pointing in a constant direction all around the loop. Maybe give the TC #18 video a shot if you haven't watched it.
@@eigenchris Hi Chris: I should have said that I was thinking of a loop whose pieces are all geodesics (as in the Viking walking on the sphere example). So instead of a circle in flat space, I would want a closed polygon. I will watch #18 and have found your videos really helpful---thanks!!
Has the mystery of adding the cov derivative in direction of lie bracket being solved?
Sorry, the hand-wavey explanation in this video is still how I think of it now. I got tired of trying to find a better reason so I haven't thought about it in a while.
@@eigenchris Don’t worry. Besides the quality of these videos, it is crazy the interaction I can have with you even after 4 years. Thanks for your incredible job, I will support it with PayPal
Wonderful video
This is very advanced math men. I cant understand whats going on but im interested in general theory relativity. Looks challenging like quantum mechanics.
You might want to try starting at the beginning od this series if you're confused.
@@eigenchris thanks sir.
this dude is a genius almost einstein level
Just amazing!!
At 7:25 the A 'linear map that parallel transports w from (0,0) to (r,0)' does not seem to be completely defined, that is you've only specified how it transforms 1 vector.
It would be a linear map that parallel transports ALL vectors from (0,0) to (r,0). If you like you could parallel transport a basis and define the linear map from there.
It's an amazing video but I still can not understand why the covariant derivatives of rieman tensor don't go to zero. As I understand W is a vector field and also we want the vectors which are described from this vector field to be parallel transported along the rectangle so W vector field is a field which parallel transports W so it's covariant derivative shouldn't be zero?Also every W field which parallel transports the vectors along the rectangle will have a zero covariant derivative so a zero rieman tensor,but it may be curved because as I understand parallel transport can occur and to curved manifolds like the sphere of video 21.
Thank you very much for your time your videos are very helpful!
The Riemann Tensor is a function that takes 3 input vectors and outputs 1 vector. The output vector can be zero even if the components of R are non-zero. This is similar to how we can have a non-zero covector "alpha" (row vector [1 1]) act on a non-zero vector "v" (column [1, -1]) and still end up with alpha(v) = 0. Does that answer your question?
Sehr gut
I am lost a little. How does the limit as r and s go to zero result in zero, when r,s are in deniminator? Also, making w a vector field arbitrarily? How is that done?
@eigenchris Thanks for this video. However I find your conclusion @ 22:12 a bit misleading. If a space has a metric which in one coordinate system has delta(i,j) coordinates i.e. is the Id., for all points in that space, then the space is flat, correct ?
If the metric is the identity at all points in the space, the I would agree the space is flat. But I tired to say on the slide that when the metric is the identity only at some specifoc point p, this is not enough to tell if the space is flat.
Clear explanation. thank for the great lectures.
i just watch the lecture 22.
There are one question i get stuck .
[update:i am still waiting for the solution in question 2.
i don't understand because the first step is the same for W(V) and Nablaw V(they are both defined to be the change vector V in direction W) and it is hard to see the difference.
in the previous lecture he said the [V,W] is the "ordinary derivative" part but i am not very sure what does he actually mean here.
Since two points have different vector space,i am not sure how does W(V) works in a way that is "ordinary" (we know that Nabla w V is a derivative of vector compared with the parallel one in different points such that they are compared in the same point)
SO MY MAIN QUESTION IS : HOW DOES [V,W] ACTUALLY WORKS in "ORDINARY WAY"]
in the previous lecture,i understand in this way :the main difference is the " connection factor k i j " is not equal to dui duj( ) ,however i dont understand because in the first step they seems to be the same:
Nabla w V = w^i * d/du^i ( V^I e^i) = w^j e^j { V^i (ei )} = W(V)
it seems this logic fails when e^j is not equal to d/du^i
that mean the e^i is not equal to d/du^i under different condition?
may be e^i cannot be written in differential operator form or ...??
since d/dx * d/dy = d/dy * d/dx implied they are commutative..
is that the main criteria here? or [V,W] is something that is independent of the connection factor here?
i am not sure .......
and i also know how they interpret the [V,W] in different ways(2)other question:what is the different between Nabla v W and V(W) in lecture 21
it seems they are the same thing ?(it is obviously different)
what is the "difference" if they are not torsion free
however they are expressed in a form:
v^i ei ( u^j ej ) = V(U)
and
nabla v U = v^i d/du^i ( U^j e_j) = V^j e^i (U^j e_j )
however we know they are not the same but there is a extra factor call T[ V,U]...
(since you assume d/du^i = ei)
in pure algebra,i cannot see the difference between Nabla v W and V(W)
however we know the [V,W] is not equal to Nabla v W - Nabla w V = VW-WV in non-torsion free form
therefore i just wonder why they are different algebraically
thank you
Gary Tze Hay Lau Those problem bother me as well, I would like to know the solution, looking forward to hear that. Thanks! :(
hm so you could write down the equations that would prove after moving on two straight paths at a relatively different direction, that it was mathematically impossible for the earth to be flat (though wed have to account for elevation differences by subtracting the normal velocity components continously etc to make such an experiment practicable).
I am a bit confused by the argument about why Lie bracket is not linear (at 18:25). Isn’t the derivative a linear operator? Why do you need to use a product rule when multiplying by a scalar? Isn’t this like multiplying by a number, that we can just pull in front of the operator?
it is not said that it is a constant, its derivative could be non-zero
24:49 if universe expand with acceleration more than 0 thats mean universe is curve not flat. This General Einstein Theory is like Second Archimedes theorem. This is Archimedes' principle for mass and space. I just wonder if this Differential geometry can bee aplied that same way in Archimedes principle. Question how many stars explode per 1 s in all Universe.
5:19 Is that a LEGO version of Éomer from Lord of the Rings?
The eqn at 15:15 just doesn’t make any sense….
it seems to me the third term can be expressed as the two preceding terms,provided we’re still using the Intrinsic Geometry argument,such that partial is covariant derivative.
If so,the third term evens out with the preceding.Therefore the whole equation goes to zero,which is certainly not what we want.
Hello everyone. PLEASE, TELL ME, ANYONE: If we have infinitesimal loop - TRIANGLE, AND, FURTHER, WE HAVE PARALLEL DISPLACEMENT AROUND THAT LOOP, SO , IN THIS INFINITESIMAL CASE VECTOR DO NOT CHANGE! SO, IN INFINITESIMAL CASE, RIEMANNIAN TENSOR IS EQUAL TO ZERO!!!!!.
SO , THE GENERAL DEFINITION OF CURVATURE IS NOT THERE . AM I RIGHT - ?
16:29 Shouldnt the third term be automaticly zero if we use levi civita connection, since its torsion free? Its general form of Rieaman tensor so it probably has to accout for other connctions that dont absorb torsion would be my guess, but than again what do i know :D It looks to me like the third term accounts for tangetial change of w along lie bracket ''vector'' , thats basicly what you mantioned as one reason.
Best qualitative explenation of relativity a saw , great job.
The 3rd term is zero if you input basis vectors but to can be non-zero if you use vector fields that aren't basis vectors. My next video should hopefully clear this up.
9:09 is it R(u,v)w the Riemann Curvature Tensor or is it just R(u,v)??
The Riemann Curvature Tensor itself is the function R. It takes 3 vector inputs and outputs a vector. So R(u,v)w will output a vector. R(u,v) will output a linear map... something that takes a vector (the missing w) and output the final vector expected from R(u,v)w.
This is tight.
18:35 I don’t see how this makes it non bilinear, the derivative if a constant is 0
At 11:35 the introduction of a vector field to allow subtracting the paralleled transported vectors is mindblowing in that it doesn't seem to have any constraints. Can you elaborate on this, or give some accessible reference?
It's a technique they use in section 11.4 of the Gravitation textbook I mention. However, that reference fails to give a convincing proof of the formula for the Riemann Tensor.
Do you have any specific issues with it?
@@eigenchris I'm learning all this material very passively from resources like your (fantastic) series or XYLYYLYX- I read parts of Gravitation, but I got discouraged reading reviews of it that question its effectiveness in making GR accessible.
@@jaeimp Gravitation has been a good resource for me for some parts of this series. I appreciate that it tries to use pictures a lot. The last part of this video on geodesic deviation is lifted directly from it. However, it is a pretty heavy book, and some parts just aren't explained very well in my opinion (the "derivation" of the Riemann Tensor via the parallel transport around a "small loop" is not very convincing to me). I'm not sure what the most "accessible" source for GR is. Part of the reason I'm making this series is to try and make tensor calculus/differential geometry easier for people to swallow.
Hey,was thinking the same thing.you got any wise explanation yet?
is the connection in the definition of Riemann tensor could be any connections, including those are not compatible with the metric?
I think it is technically possible to use other connections, but some of the properties I derive later in this series only apply if you use the Levi-Civita connection.
Wonderful explanation!
But would you plz tell me that how can we use matrix i.e. rotation matrix to represent this holonomy or rotation of vector after being parallel transported in a close curve?Is there any analogy here between rotation matrix what we learn in linear algebra class and the rotation of vector in curved space?
I'm not toally sure what you're asking, but the R(u,v) is basically a linear map (matrix) that acts on w, and the output is the change w undergoes after parallel transport around the closed curve.
I don't think this change is always a rotation--it can be any generic lineae map/matrix.
@@eigenchris Hello, sorry for late,your all video series are visually Awesome!!!!!Mind blowing!and very simple!
And Actually here is the link of the video of Leonard suskind talking bits about curvature in terms of rotation ,plz see from 8:40
ruclips.net/video/xOzVmC-4gpw/видео.html
I also learned a few months ago that parallel transport or covariant derivative of vector along a close loop on Riemannian manifold just cause "rotation" of vector about some angle,but doesn't change it's magnitude (assuming no torsion).Now I already know a little bit more about lie group SO(2or 3) and their algebras when I deal with rigid body and it always has to do with rotation of vectors.My question is that if parallel transport or covariant derivative just causes "rotation" of vector when it moves,How can I express this Parallel transport or covariant derivative in terms of just SO(2 or 3) acting on vector on 2 or 3D reimanm manifold as covariant deriva. dealing with just "rotation" of vector when it moves on manifold?
I also think but exactly don't know (don't find in book) that connection one form may has something to deal with rotation.
Thanks. Links for previous videos 17,18,19,20, in the description are not working. The link for video 21 is ok.
Thanks for pointing that out. Links should be fixed.
As far as I understand, if we are parallel transporting the vector, shouldn't the covariant terms be all zero? And Riemann tensor turned out to be, as a result, zero too?
The covariant derivative measures how much a vector deviates AWAY from parallel transport. When I take the arbitrary vector field w, I am measuring how much w deviates away from parallel transport, so the covariant derivatives may not be zero.
@@eigenchris Yeah I got it now, after some thinking. Thanks. You're doing amazing work.