Tensor Calculus 24: Ricci Tensor Geometric Meaning (Sectional Curvature)

[@jaeimp]

Nothing, absolutely nothing, makes GR more accessible than your presentations! Thank you for yet another superb effort at making some tough concepts intuitive. These video series is a gift to all curious people out there on par in quality and relevance with the work of Kahn Academy, 3Blue1Brown, Numberphile or Mathloger, and with the important difference that they cover a topic otherwise opaque without solid physics background.

[@jacobvandijk6525]

This isn't bad either: physicsunsimplified.com

[@frankdimeglio8216]

THE MATHEMATICAL PROOF THAT ELECTROMAGNETISM/ENERGY IS GRAVITY (IN WHAT CONSTITUTES A BALANCED FASHION) IS SUCCESSFULLY (AND CLEARLY) DEMONSTRATED, AS E=MC2 IS F=MA: The Maria ("lunar seas") occupy ONE THIRD of the near side of the Moon. The land surface area of the Earth is 29 percent, AND this is EXACTLY between one quarter AND one third. The Moon is about one quarter (27 percent) the size of the Earth in what is a predictable fashion, AS it is FULLY manifest as LAND. The sky is BLUE, AND the Earth is ALSO BLUE. So, consider BALANCED BODILY/VISUAL EXPERIENCE. LOOK at what is the orange Sun, and think LAVA. Gravity/acceleration involves BALANCED inertia/INERTIAL RESISTANCE, AS E=MC2 IS F=ma; AS ELECTROMAGNETISM/energy is gravity. Accordingly, the rotation of WHAT IS THE MOON matches it's revolution. Now, consider the fully illuminated (and setting) Moon in direct comparison with the orange Sun. They are both the size of THE EYE. Notice that the Moon is ALSO BLUE. Consider the man who IS standing on what is THE EARTH/ground. Touch AND feeling BLEND, AS ELECTROMAGNETISM/energy is gravity; AS E=MC2 IS F=ma. The BULK DENSITY of the Moon is comparable to that of (volcanic) basaltic LAVAS on the Earth. The energy density of LAVA IS about three times that of water. Great. The human body is, in fact, about as dense as WATER !!! Think !!! "Mass"/ENERGY IS GRAVITY. ELECTROMAGNETISM/ENERGY IS GRAVITY. THE SUN AND THE EARTH are F=ma AND E=MC2, AS ELECTROMAGNETISM/ENERGY IS GRAVITY. E=MC2 is DIRECTLY AND FUNDAMENTALLY DERIVED FROM F=ma. F=ma AND E=MC2 PROVE that ELECTROMAGNETISM/ENERGY IS GRAVITY, AS ALL of SPACE is NECESSARILY ELECTROMAGNETIC/GRAVITATIONAL (IN BALANCE); AS ELECTROMAGNETISM/ENERGY IS GRAVITY. "Mass"/ENERGY IS GRAVITY. ELECTROMAGNETISM/ENERGY IS GRAVITY. Energy has/involves GRAVITY, AND ENERGY has/involves inertia/INERTIAL RESISTANCE. Gravity/acceleration involves BALANCED inertia/INERTIAL RESISTANCE, AS ELECTROMAGNETISM/ENERGY IS GRAVITY. This explains F=ma AND E=MC2, AS ELECTROMAGNETISM/ENERGY IS GRAVITY. SO, GRAVITATIONAL FORCE/ENERGY IS proportional to (or BALANCED with/as) inertia/INERTIAL RESISTANCE; AS ELECTROMAGNETISM/ENERGY IS GRAVITY. ACCORDINGLY, ALL of SPACE is NECESSARILY ELECTROMAGNETIC/GRAVITATIONAL (IN BALANCE); AS ELECTROMAGNETISM/ENERGY IS GRAVITY. "Mass"/ENERGY involves BALANCED inertia/INERTIAL RESISTANCE consistent WITH/AS what is BALANCED ELECTROMAGNETIC/GRAVITATIONAL FORCE/ENERGY, AS ELECTROMAGNETISM/ENERGY IS GRAVITY. GREAT !!! Gravity IS ELECTROMAGNETISM/ENERGY. ELECTROMAGNETISM/ENERGY IS GRAVITY. INDEED, A PHOTON may be placed at the center of THE SUN (as A POINT, of course); AS the reduction of SPACE is offset by (or BALANCED with) the SPEED OF LIGHT (c); AS E=MC2 IS F=ma; AS ELECTROMAGNETISM/ENERGY IS GRAVITY. GREAT. "Mass"/ENERGY IS GRAVITY. ELECTROMAGNETISM/ENERGY IS GRAVITY. ALL of SPACE is NECESSARILY ELECTROMAGNETIC/GRAVITATIONAL (IN BALANCE), AS ELECTROMAGNETISM/ENERGY IS GRAVITY; AS E=MC2 IS F=ma. Great. The ability of thought to DESCRIBE OR RECONFIGURE sensory experience is ULTIMATELY dependent upon the extent to which THOUGHT IS SIMILAR TO sensory experience. (THOUGHTS ARE INVISIBLE.) Gravity IS ELECTROMAGNETISM/ENERGY. ELECTROMAGNETISM/ENERGY IS GRAVITY. Very importantly, outer "space" involves full inertia; AND it is FULLY INVISIBLE AND black. The perpetual motion of WHAT IS THE EARTH is NOW explained. GREAT !!! The idea that THE PLANETS are "falling" in what is "curved space" in RELATION to what is THE SUN is PROVEN to be NONSENSE. So, the falling objects must be considered in RELATION to WHAT IS THEN THE ORBITING EARTH. GREAT !!! E=MC2 IS F=ma. The stars AND PLANETS are POINTS in the night sky. I have explained why the motion of WHAT IS THE MOON matches it's revolution. E=MC2 IS F=ma. This NECESSARILY represents, INVOLVES, AND DESCRIBES what is possible/potential AND actual IN BALANCE, AS ELECTROMAGNETISM/energy is gravity. Consider the man who is standing on what is THE EARTH/ground. Touch AND feeling BLEND, AS ELECTROMAGNETISM/energy is gravity; AS E=MC2 IS F=ma. Gravity IS ELECTROMAGNETISM/energy. Gravity AND ELECTROMAGNETISM/energy are linked AND BALANCED opposites, AS E=MC2 IS F=ma; AS ELECTROMAGNETISM/energy is gravity. GRAVITATIONAL force/ENERGY IS proportional to (or BALANCED with/as) inertia/INERTIAL RESISTANCE, AS E=MC2 IS F=ma; AS ELECTROMAGNETISM/energy is gravity. Gravity/acceleration involves BALANCED inertia/INERTIAL RESISTANCE, AS E=MC2 IS CLEARLY F=ma; AS ELECTROMAGNETISM/energy is gravity. Accordingly, the rotation of WHAT IS THE MOON matches it's revolution. Great. (Gravity IS ELECTROMAGNETISM/energy.) E=MC2 IS F=ma ON BALANCE. "Mass"/ENERGY involves BALANCED inertia/INERTIAL RESISTANCE consistent WITH/AS what is balanced electromagnetic/gravitational force/ENERGY, AS E=MC2 IS F=ma; AS ELECTROMAGNETISM/energy is gravity. So, objects AND MEN fall at the SAME RATE (neglecting air resistance, of course); AS ELECTROMAGNETISM/energy is gravity; AS E=MC2 IS F=ma. A given PLANET (INCLUDING WHAT IS THE EARTH) sweeps out EQUAL AREAS in equal times consistent WITH/AS E=MC2, F=ma, AND what is PERPETUAL MOTION, AS E=MC2 IS F=ma; AS ELECTROMAGNETISM/energy is gravity. Gravity IS ELECTROMAGNETISM/energy. The stars AND PLANETS are POINTS in the night sky. Excellent !!! E=MC2 IS F=ma ON BALANCE !!! TIME dilation ULTIMATELY proves ON BALANCE that E=MC2 IS F=ma, AS ELECTROMAGNETISM/energy is gravity. INDEED, TIME is NECESSARILY possible/potential AND actual IN BALANCE; AS E=MC2 IS F=ma; AS ELECTROMAGNETISM/energy is gravity. It all CLEARLY makes perfect sense, AS BALANCE AND completeness go hand in hand. Gravity IS ELECTROMAGNETISM/energy. E=MC2 IS CLEARLY F=ma ON BALANCE !!! Great !!! By Frank DiMeglio

[@aliesmaeil1044]

thanks alot, I am a PHD math student ,you are helping me alot with your videos.

[@sweetytarika8068]

hi, whats your area of research? i want to ask some question.

[@aliesmaeil1044]

@@sweetytarika8068 Hi , I am studing the conformal maps between riemannian manifold .

[@swamihuman9395]

- As all commenters have noted (praised), your presentation is superb. - Thx for all you hard effort in creating this clear/concise/understandable content. - I took 'Modern Physics' as part of my engineering degree, so got an introduction to GR, but of course did not go deep into the mathematics. Subsequently, I went on a broad/deep self-study into math (even became a math teacher). So, I have informally studied the Einstein Field Equations (EFE), along w/ tensors, and related matters. As a result, I was able to follow this video. - One important takeaway for me from this vid is the explanation of how gravitation can be described as the consequence of geometry of curved space-time. Thx.

[@pythagorasaurusrex9853]

Cudos! This is the best series about Tensor stuff I have ever seen. You present your videos very visual, clear and focused. Just binge watching your mini series. I wished I had professors at my former university teaching this stuff like you do. You have a subscriber! Best wishes to you for 2021.

[@Cosmalano]

My understanding of the non-vacuum EFE are so close now I can nearly taste them thanks to you. Watching you while I supplemented myself with other resources was the most enjoyable way to learn tensor calculus I think I could’ve had, you boosted my confidence immensely as a mathematician, and you opened up so many doors for me in theoretical physics. Thank you so much for this series.

[@ericbischoff9444]

I did not expect Mary Poppins, but great analogy! Thanks.

[@morganhopkins204]

Amazing content as usual! Keep up the fantastic work. Also RIP to the Ricci scalar term in Einstein's equations at 2:17 😅

[@eigenchris]

whelp.

[@aramrashid5752]

You are the best! I have never seen someone better than you could talk about "Tensor". Your are amazing! and I recommend you to go on and be continue... We all are here so thankful that you do for us these fantastic lectures. We have learnt so so much from your videos. I wish one day after Tensor Calculus you could do some lectures also on Differential Geometry! I'm pretty much sure just you could exactly describe and illustrate how the stuffs works in differential geometry! I wish you the best! We love you man :)

[@lucasmartinsbarretoalves9937]

this series is remarkably good. thank you so much for publishing it!

[@ssrbaqri]

Having watched it just for initial few minutes, I get the intuition that I was so desperate to find but failed despite watching many videos... the moment u say that Ricci tensor gives the change of volume along a geodesic, I get that Eureka feeling.

[@Wooferino]

Thank you. You're the best I've seen at navigating the details.

[@fouziasharmin6218]

im grateful brother. Im just too much grateful

[@redrum41987]

This video is exactly what I've been looking for. Thank you so much

[@silentobserver3433]

It's finally here (again), yay!

[@marks4982]

4:18 - I knew exactly what you were going to say after this came.

[@eigenchris]

It's smaller on the outside.

[@sufyannaeem2436]

Thank god you are back... I thought Ricci Curbastro has put you in dungeon for searching him....

[@izanagi5136]

While watching I had one question, why is it that at 24:37 the v vectors are expanded into linear combinations of the basis vectors? Weren’t they selected specifically as the e_n basis vector, meaning that the vector components v^j = 0 for all j ≠ n and with v^n = 1? In this case then, wouldn’t the curvature K(e_i,v) be 0 except for when j = k = n, in which case you’d continue to sum over i for the Ricci curvature, which would still make sense, but wouldn’t it mean that only one value from the Ricci tensor would impact the Ricci curvature, the R_(nn) entry, with the curvature itself equaling the value of that entry? I assume I made some error, but was just confused by this

[@domenicobianchi8]

amazing content! relatively to the lat part, where after u found Ric(v,v)>0 you said the geodesic always converge, i have a doubt about what it means. Because if you walk from the equator to the poles they do converge, but if go from poles to equator they diverge. What is the right way to interpret this? thanks

[@mad4mathtv699]

If you want to start from the pole to the equator, you should draw parallel lines from the pole. They converge. They will meet at the equator. It means just the same situation as 'from the equator to the pole'.

[@kevinliu3843]

detailed explanations to a terrifying degree.

[@hasanshirazi9535]

Great explanation. Learned a lot from your video. Thanks

[@farzamdadgar-rad8843]

Hi eigenchris. Thanks for your nice videos. Just a comment on the slide proving that “The value of Ricci tensor for a given vector is independent of the orthonormal basis vectors” [time on video: 25:30 to 26:50]. The proof needs to be modified. Basically, the following relation of the video \tilde{R}^b_{abc}=(F^k_a)(F^i_b)(F^j_c)(R^i_{kij})(B^b_i) needs to be replaced by \tilde{R}^b_{abc}=(F^k_a)(F^i_b)(F^j_c)(R^q_{kij})(B^b_q) =(F^k_a)(F^i_b B^b_q)(F^j_c)(R^q_{kij}) =(F^k_a)(delta^i_q)(F^j_c)(R^q_{kij}) =(F^k_a)(F^j_c)(R^i_{kij}) In other words, “i” in R^i_{kij} is a dummy index and must not contribute to the transformation relation (R^i_{kij} are the components of a 2nd-order tensor). Otherwise, we will see four “i”’s in the transformation rule. The rest of the proof is straightforward. Moreover, you have written [delta^b_b=1]. As you know, [delta^b_b=n], where n is the dimension of the space. The term "delta^b_b" does not appear in the proof if the modification is applied.

[@physicsforstudents]

Are you talking of volume expansion in terms of redshift?

[@eric3813]

Thanks so much, i finally Understand it. You Are a Great Teacher 😄👍🙏

[@ceoofracism5713]

and you early mentioned that those are basis vector,nl now you started to do linear combination at 24:40 can you tell what am i missing here?

[@erikstephens6370]

24:35- 24:40: Why expand v into (v^k)ek when we know v is one of the basis vectors?

[@vitrums]

24:46 correct me if I'm wrong. As far as I understood each R falls appart into a summation over the upper index of R-components times basis vectors. But since we are dealing with a special case of an orthonormal basis, then all basis vectors in R-summation but one with index "i" produce 0 in a dot product with e_i from sect.curv. summation. Hence, for each [R(e_i, e_j)e_k] we are left with the sole R-component with upper index "i".

[@raffaeleangrisano5484]

Grazie. Senza questo commento non credo avrei capito perché si arriva alla conclusione che l'indice superiore di R deve essere i. Chiedo scusa se scrivo in italiano.

[@physicsforstudents]

Hello, May I ask you a question. With the geodesics converging and diverging, what do you mean the volume of the ball (sphere) is increasing or decreasing? Thanks.

[@imaginingPhysics]

A ball which does not exert pressure on its boudaries (so unlike a balloon) and does not have rigid structure (so unlike a steel ball) would shrink or change shape when moving. The curvature and converging geodesics would produce a "force" or pressure to it. An actual material ball (or a planet or a star) would not shrink if the internal structural forces can counteract the "geodesic pressure". This pressure will cease as soon as the object comes to rest so it only applies to "moving objects" (remember however that in relativity everything is always moving since passage of time also counts as a movement. ) If the curvature is extreme enough it will break the object.

[@i.m.Q.2]

PS..between this or one of your other vids, Christofel symbols are now of major help to what im prohibited from working on in the privacy of my own "dwelling". 🥴 Seriously, your vids helped me get started with things. 👍

[@rollsreus7727]

I just wanted to say that you have gift in not only being able to understand the topic so well but also you explain it all in such a simple way and I think your videos on tensors are the best explanation that I've found not only on youtube but also in general (textbooks included). Are you still around and will you produce the next series?

[@eigenchris]

I am working on the GR videos now I have rough drafts (with audio problems) of the first videos at the link below. I plan to release proper versions of these within the week. If you want to give me any feedback beforehand, I welcome it: twitter.com/eigenchris/status/1229681192502362114?s=20

[@AlexeyUshakov-y2x]

@eigenchris There is probably an error at 24:21, because v * v should be equal to the square magnitude of the vector. Thank you very much for your videos!

[@waynechau9884]

v is basically a basis vector as defined earlier in the video.

[@A-_--]

24:42 It doesn't make any sense to expand v in terms of the basis if it is orthogonal to the other basis vectors.

[@yeshan6825]

Very intuitive and clear presentation! Could you plz give a simple explanation of "immersion" and "embedding" or "immersible in R3" etc? I see these terms often in books or papers of Differential Geometry but have never ever seen any straightforward explanations like in your videos. Thanks a lot!

[@eigenchris]

I don't really have much real math experience for that, but I think it just means we can find a mapping from the manifold coordinates into some higher-dimensional space, in a way that keeps the metric consistent. So if we took our 2D coordinate manifold description of a sphere, we can map the points of that manifold into R3 in such a way that the distances from the 2D sphere metric "match up" with the standard pythagorean/euclidean distance in R3. So the 2D sphere can be "embedded" in R3. Sorry I can't say more.

[@yeshan6825]

@@eigenchris Thank you very much! Your explanation already helps. Respect!

[@iwonakozlowska6134]

The same relationships ca be obtained using The Gaussian approach to curvature , extended to any number of dimensions. And what's more , the Riemann tensor Rijkl is the minor of the "second fundamental form" matrix.

[@eigenchris]

I never properly taught myself the "classical" differential geometry of Gauss. Thanks for the info.

[@rybaplcaki7267]

Yes, finalllyyyy when I interest in GR after half a year, love

[@kobsterrock]

at 26:39 the Kronecker delta contracted with itself shouldn't be one, because you're summing over each index it should be the trace of the delta=4

[@eigenchris]

You're correct that the kronecker delta contracted with itself should be D, where D is the dimension of the space you're in (D=4 for spacetime). However, in this case, I don't think a contraction is actually happening. It's just a kronecker delta where the top and bottom index are the same, and there it's equal to 1. I realize I may have messed up the notation here--the line above has 4 "i" indexes; there are 2 summations over the "i" but there is no summation over "b"... I probably should have written this out better.

[@kimchi_taco]

24:25 ∂ᵢ·ṽ is not necessary to be 0. As K(∂ᵢ, ṽ) = K(∂ᵢ, ṽ + k∂ᵢ), we can freely remove i-th component in ṽ every summation of i. So Sectional Curvature way can handle arbitrary ṽ. Note: Sectional Curvature way only works on Orthonormal basis, but Ric(ṽ,ṽ) = v¹v²R₁₂, Ricci tensor is tensor, which can be transformed to not-orthonormal basis.

[@waynechau9884]

I am really having trouble following the derivation of the Ricci Tensor Components at 25:26. When I look at other resources, they usually involve contracting the Riemann Tensor components in one of the lower indices. Can you please kindly clarify your method of derivation? I worked on it for a bit by trying to assume v is a general vector and not part of the orthonormal basis (v = e_n). This approach the denominator becomes, ||v||^2- ||e_i||^2 ||v||^2 cos (theta)^2 = ||v||^2 sin(theta)^2 = ||v|| ||v|| sin (theta)^2 = (v x v) sin (theta) = 0 because of the cross product of v with itself. This means that for Ric(v,v) to be defined, v must be an orthonormal basis vector. In this case, I obtained Ric(v,v) = R_nn. What happened to the other elements not on the diagonal?

[@andrewrich9396]

Wayne, I just posted a reply to a comment by Theo Goix that essentially answers your question. In general Ric(X,Y) can be defined as a trace of a linear map using the Riemann Curvature Tensor. How does that square with the definition of Ric(V,V) here? For any vector V, Ric(V,V) is a sum of sectional curvatures multiplied by the square of the component of V that is "not in that section". The formula given in the video at 24:50 (summing over index i with one i raised and one i lowered) is correct for all V, even if the basis is not orthonormal. However, if you want to use a sum with all indices lowered, then you must use an orthonormal basis.

[@aidanmcsharry7419]

Hi eigenchris, hope all is well and thanks a load for these videos. This is probably a bit of a silly question, but I would like to just clarify it: at 24:34, you write out the vector v as a linear combination of the orthonormal basis, despite having said that ej dot v is zero, implying that v is parallel to a basis vector ek where k doesn't equal j (which is what you said a couple slides back). However, how, then, can you write out v as a linear combination of all of the other basis vectors, when it does not depend on them?

[@eigenchris]

Yeah that was an oversight on my part. V would be equal to the final basis vector in this case.

[@theogoix3566]

This series is amazing, thanks eigenchris for making GR so accessible! By the way, at 24:45 I don't understand why the vector v can be expanded as a linear combination of basis vectors, because, if I understood correctly, the basis itself is chosen such that it's orthonormal and v is a basis vector, so it should't have any component except itself. Do I miss something?

[@eigenchris]

That was a "brainfart" on my part. You can't expand it since it's just the nth basis vector.

[@theogoix3566]

Thanks for your so quick reply! Since v is a basis vector (we could denote it e_n), the formula at 24:45 should yield Ric(v, v) =Ric(e_n, e_n)=[R(e_i, e_n)e_n].e_i =R^{i} _{nin} =R_{nn} with the same indices at the bottom of the ricci tensor. Am I right?

[@eigenchris]

That's right. I'm not sure how to give meaning to the off-diagonal elements of the Ricci tensor woth this interpretation. The next video gives another interpretation that helps.

[@theogoix3566]

Thanks!

[@andrewrich9396]

Yes, I noticed the same issue. Is V one of the basis vectors or not? And if it is, there is the issue that one of the terms in the Ricci sum is 0/0. The formula at the bottom at 24:50 is correct for all V, whether it is unit length or not, part of a basis or not. Summing over i, with one upper i and one lower i, indicates that we are taking the trace of a linear map and trace is well-defined, independent of basis (the basis doesn't even have to be orthonormal). What linear map is it? It is the linear map X goes to R(X,V)V. So can we still give a formula for Ric(V,V) in terms of sectional curvatures? Yes. Take an orthonormal basis e_i as in the video. Compute the n sectional curvatures K(e_i, V) as in the video but now multiply each such by the square of the projection of V orthogonal to e_i, that is the square of the length of V when the e_i component has been removed. (Doing so gives R(e_i, V)V dot e_i.) Sum these to get Ric(V,V). So Ric(V,V) is a sum of sectional curvatures multiplied by certain factors. The Ricci tensor is a 2-tensor so Ric(X,Y) should be defined when X is not equal to Y. eigenchris refers to the off-diagonal elements in the comment above. Given a symmetric tensor, there is a standard formula for the off-diagonal elements in terms of the diagonal elements, which in this case yields: Ric(X,Y) = 1/2 times Ric(X+Y,X+Y) - Ric(X,X) - Ric(Y,Y) . Since we have a geometric interpretation of the diagonal elements as sums of sectional curvatures as given in the previous paragraph, this yields a geometric interpretation of Ric(X,Y) altho I admit that it is so convoluted as to not really leave one feeling satisfied.

[@qbtc]

This series along with the Stanford Susskind General Relativity lectures make for a great elementary course on GR!

[@naturematters08]

and trin tragula general relativity

[@AlexanderB41]

At 27:47 you wanted to instead say "Positive Ricci Curvature means that a ball starting between initially parallel geodesics shrinks when travelling along them", right? Since converging geodesics are only derived from positive sectional curvatures. (Although on a sphere, the only sectional curvature is positive, too.)

[@behnamparsaeifard3883]

Amazing video! Thanks. When the geodesics converge or the volume of the sphere shrinks, can we ever see this schrinking of volume in spacetime near earth or a mass? Can we measure it?

[@eigenchris]

Unfortunately, the only spacetime geometry I'm familiar with is the "schwarzschild metric", which is the geometry around non-rotating spherical bodies (moon, earth, sun, black holes, if they weren't rotating). In this case, the Ricci Tensor and Ricci scalar are both zero, so no volume changes happen. When a "test sphere" of particles moves forward in time near another spherical body, it will elongate in one direction and shrink in the other directions in such a way that the volume stays constant. This is the explanation for "tidal forces". An everyday example being the ocean tides on earth. You can also google the "Roche limit" which is when a body is torn apart by tidal forces from another body. I'm not familiar with non-zero Ricci scenarios in GR. You'd have to try and search for that online.

[@asiancollegeofteachers5870]

This volume means the classical volume measurement in Newtonian physics or the volume form on a differentiable manifold?

[@imaginingPhysics]

I copy paste my answer from another topic below since it seems relevant: A ball which does not exert pressure on its boudaries (so unlike a balloon) and does not have rigid structure (so unlike a steel ball) would shrink or change shape when moving. The curvature and converging geodesics would produce a "force" or pressure to it. An actual material ball (or a planet or a star) would not shrink if the internal structural forces can counteract the "geodesic pressure". This pressure will cease as soon as the object comes to rest so it only applies to "moving objects" (remember however that in relativity everything is always moving since passage of time also counts as a movement. ) If the curvature is extreme enough it will break the object.

[@kimchi_taco]

Thank you for awesome lecture. BTW, the (traditional) sign choice of R is unfortunate. R sign comes from lim_rs w - DCBAw in previous video. However, DCBAw - w is more natural because derivative df/dx is lim f(x+dx) - f(x). If R sign is opposite, d_v d_v s = R(s,v)v, which is more straightforward.

[@YuvalDagan38]

Amazing videos! Question, if the Ricci is computed with respect to an orthonormal basis, shouldn’t it be the same across the sphere due to symmetricity? Maybe you should have normalized one of the basis vector elements by sin theta and then the curvature would be the identity?

[@hannahlongsdale1725]

Hi Chris,when calculating the Ricci curvature how can you assume that the manifold will allow an orthonormal basis if it's not flat. Are you assuming just local flat co-ordinates rather than a global orthonormal coordinate system?

[@eigenchris]

Yeah, in differential geometry we always use the flat tangent space to make basis vectors.

[@deepbayes6808]

Yay, it's back. Thanks Chris, once again!

[@KeenestObserver]

Chris you are amazing. Thank you.

[@darkdevil905]

when you do the switching to the cross product couldnt you have easily have used the wedge product instead of cross product? since both cross product and wedge product will give you the expressions for the area of a parallelogram. You would get the same result and it would in fact prove the same statement for an number of dimensions

[@eigenchris]

I could have, but I neglected to introduce the wedge product in this series, so I tried to make do without it.

[@Wooferino]

Chris, the formula for G.R. at 2:30 is missing an R in the term showing -1/2 g_uv.

[@laincoubert9816]

you make tensor physics look so easy

[@hieudang1789]

I have a question. So ricci tensor take two vectors input and output a scalar right? But why in the video, both input are the same vector v?

[@raffaeleangrisano5484]

Instant 14:41. In the expression of R(e1,e2)s, the last term is the covariant derivative of s in the direction of the vector provided by the parenthesis of LIE [e1,e2]. But the LIE bracket of the base vectors e1,e2 is not zero? And is not a derivative along a null vector zero? Why is this fact not mentioned at all? Thank you for your attention and please forgive me if I have written a lot of nonsense.

[@chenlecong9938]

Hi at 15:53,did you just multiply the (R(e1,e2)w)•(u) with the R(u,w) to get (R(u,w)w)•u?? If you did,how was that possible?I see you did not add on any remark to that so I’m assuming you’re doing the abovementioned

[@eigenchris]

I show the trick of getting the 1st factor of "ad-bc" at 14:16. The 2nd factor of "ad-bc" just comes from the standard linearity rules for tensors, and the dot product distributivity rules.

[@Panardo777]

Wonderful content as always !!! Thank you very much Chris for helping me achieve my ultimate goal of understanding GR, after having a good intuition (thanks to your incredible work) of Riemann Tensor I could not imagined that Ricci Tensor would be so hard to grasp. I have a question concerning Ricci components at 25:19 the implicit sum on i excludes n so how can you achieve the equality between the contracted Riemann Tensor components and Ricci Tensor ?

[@doctormeister]

Why do we say at 18:50 that Ric(v,v) = sum_i(k_i,v) if the RHS only depends on one free variable. Couldn't we just write Ric(v)? Whats the second v for?

[@warrenchu6319]

For the Ricci Tensor Ric(V,V): 1. It is the sum of sectional curvatures K, which is the dot product of the Riemann Curvature Tensor for geodesic deviation -R(S,V)V and the separation vector S, correct? 2. At 25:30, why do the Ric components have only 2 indexes, j and k? What does this mean in the sphere example at 27:05: R_ab=R_i_aib?

[@eigenchris]

1. I think you have an extra negative sign in there. I use the dot product of +R(s,v)v and s, as seen on wikipedia: en.wikipedia.org/wiki/Sectional_curvature 2. The Ric components should only have 2 indexes, as seen at 25:04. Do you have problems with this?

[@warrenchu6319]

@@eigenchris Tensor contraction: R is composed of (i x i) sub-matrices, each composed of (k x j) components. We're summing the (k x j) submatrices on the diagonal of R to get *one* (k x j) matrix. It's like taking the trace of a matrix; in this case the diagonal elements are submatrices. R has gone from a rank 1-3 tensor to a rank 0-2 tensor but must be a mixed dyadic tensor for it to be contracted. Correct?

[@eigenchris]

@@warrenchu6319 That's one way to think of it, yes.

[@warrenchu6319]

@@eigenchris OK. Now: How did [R(ei,ej)ek] . ei become R_upper_i_lower_kij, from the 3rd-from-bottom equation to the 2nd-from-bottom equation at 25:02? According to Video 23 at 4:34, Riemann Tensor R(ei,ej)ek is: R_upper_m_lower_kij times em. Then (R_upper_m_lower_kij times em) . ei = ? Is it: Sum of (em . ei) = Sum of (kronecker delta_m_i . ei) = ei? Therefore R[(ei,ej)ek] . ei = R_upper_i_lower_kij, as you have it as?

[@warrenchu6319]

See 10:29 on geodesic deviation

[@kenxchenutube]

@19:00 why Ric(v,v) not Ric(s,v) or Ric(u,w)

[@eigenchris]

I don't know the geometrical meaning of R(u,v), unfortunately.

[@waynechau9884]

In the summation for the Ricci Curvature, you set the range of the summation as i = 1 to D. Shouldn't the range be from i = 1 to n - 1? What is D here? Is D = n -1?

[@allandavis6116]

At 24:43 you 'expand the v vector as a linear combination of the basis vectors' ... but I thought up to this point that v was the nth basis vector en and the summation was was over all the other basis vectors, and D = n-1. So you can't expand v=en as a linear combination of the other n-1 basis vectors ??? Apparently v is now a general vector? Also, at 28:05 you say that on a sphere the geodesics are always converging, but if you'd started at the north pole and gone down a geodesic the geodesics would be diverging .... :) ... but the 2nd derivative of the separation vector would be negative no doubt.

[@eigenchris]

You're right about 24:43... I hadn't really considered that when I wrote out the formula... I don't have an acceptable answer for this right now. As for 28:05, is is true that the geodesics will always converge if the two starting vectors are "parallel transported" versions of each other. In other words, if I start with v, then parallel transport it over to the left a little, and then watch how both geodesics move, they will always converge.

[@allandavis6116]

@@eigenchris I think what it amounts to is that the formula for Ric(v,v) at the top of the screen is the definition for Ricci curvature for any vector v.

[@zzzoldik8749]

When gravity make thing shrink, it happen in one reference frame or for all reference frame? Could you answer this question please

[@riadhalrabeh3783]

It is also possible to think of the Newton system without forces. F=ma. then f=F/m=a. So the force is acceleration if m is kept 1. Then the law F=Gm1m 2/r^2 would be f=GM/r^2. That is the acceleration of a mass is a function of the distance (inverse square) to another mass and the value of the mass. That is to say that space is curved around masses. Or using the famous phrase; masses tell space how to bend and space tell them how to move. regards.

[@ericdantona6591]

Is there a book or another reference on which I can read more about this "Mary Poppins effect"?

[@anantbadal6045]

At 24:50 how the i index goes upstairs in the Riemann tensor?

[@orktv4673]

Can R_mn (in particular the off-diagonal components) be interpreted as the changing of the size of a small area with a normal vector in the m-direction, as it is taken along geodesics pointing in the n-direction? And to put it all together, I envision the Riemann tensor R^d_cab as follows: (c) is the orientation of a surface, (b) is the direction that the surface of a small cube is moving into (i.e. the direction of the geodesics), (a) is the direction in which the geodesics are displaced (their divergence/convergence causes curvature), and d is the component of the tidal force that is felt in this construction. At least, this seems to intuitively lead to the picture of the Ricci tensor measuring size change.

[@imaginingPhysics]

Good idea about the Ricci tensor. I dont know but sounds plausible I guess. On the interpretation of the Riemann tensor I would like to point put that a 2D surface in 4D cannot be determined by a single normal vector. So the Riemann interpretation might not work.

[@person1082]

16:30 this is why i prefer the exterior/wedge product over the cross product for area/volume of parallelograms/parallelepipeds

[@eigenchris]

I agree. But I neglected to introduce the wedge product in this series, and it would seem pretty random to introduce it for just this small proof.

[@haroldhawaiki]

Fascinating topic. You are the best, thank you so much.

[@rodrigomartinorduna7408]

Does anyone know how to derive the Ricci tensor components from what he's got in 24:52? I don't see how [R(ei,ej)ek]•ei = Rkj , since it seems as if we were ignoring the dot product, or am I wrong?

[@muhammedustaomeroglu3451]

you can think it as definition of Ricci tensor components.

[@active285]

Maybe a stupid question: I understand the example given at 1:50. But: What I we would move from the north pole to the equator south but not from the equator to the north pole? Then we would be drawn apart. How does this coincide with curvature? I normally think about positive curvature of having triangles on the circle with sum of the angles > 180° which (for me) coincides independently of the points chosen.

[@eigenchris]

The example requires that the two geodesics be "initially parallel". So at one starting point, have an arrow pointing north. Then take a copy of that arrow and parallel transport it slightly to the right. It will also be pointing north. The geodesics created by these two arrows will eventually converge on a sphere. In your example with the two geodesics moving apart from the north pole, the geodesics are not initially parallel. Your definition of angles in a triangle summing to > 180° also makes sense, but I'm not sure how to express it using tensors.

[@active285]

@@eigenchris Thanks for your swift reply! If I understand correctly, the same holds true, if we would start from a "smaller" great circle (i.e. one closer to the North Pole) and move from there to the equator? Because vectors might be transported along the "smaller" circle from one point x to y as well. If we then take two vectors attached to the points x and y, and follow a geodesic to the equator, it's the same issue.

[@eigenchris]

​@@active285 The key issue is that there are no "smaller" great circles on a sphere. All great circles on a sphere are the same size (and have a center at the sphere's center). If you draw a circle on a sphere that's "smaller" than a great circle, it's not a geodesic. The two "initially parallel" vectors need to be separated by a geodesic.

[@active285]

@@eigenchris Thx, finally got it!

[@andrewrich9396]

In the argument around 26:00 and following, you don't need the basis to be orthonormal. The argument works fine without ever using orthonormality.

[@Onegod40-v4h]

Sorry for a silly question Sir.But I'm curious about it.I've completed this video lectures today,now am I prepared to start general relativity or I need more mathematics to grasp GTR??

[@eigenchris]

I think this is basically all the math you need. I'm doing a relativity series now, but I probably won't reach GR for another few month. I'm currently covering non-inertial frames in special relativity.

[@Onegod40-v4h]

@@eigenchris Thank you Sir.I started STR playlist too & it's an fantastic presentation. During lockdown,I'm really enjoying self -study with this channel 😁😁

[@souvikmandal1989]

in cylindrical coordinate system the geodesics in both parallel (theta) and perpendicular (z) direction will have zero separations, which means the space is flat. But in reality the space in not flat in cylindrical coordinate. How do you interpret this.

[@eigenchris]

In differential geometry, mathematicians do consider a cylinder to be "flat". The reason is that you can unfold a cylinder and lay it flat on a table. With a sphere, there is no way you can unfold the surface without distorting sizes someone (all the flat maps of the world you have seen will distort sizes of countries somewhat). So even if a cylinder looks curved to your eyes, differential geometry considers it to be flat.

[@souvikmandal1989]

@@eigenchris Thank you

[@naturematters08]

@27:11 how did you get thr riemann tensor from the metric thensor?

[@j.k.sharma3669]

Hell Chris ! At 10:53 why did you use square of area of parallelogram for normalisation. I think square should not be used.

[@eigenchris]

The numerator contains 2 factors of "s" and 2 factors of "v". Since the area has one factor of "s" and "v", we need to square the area to make all the factors cancel.

[@official-zq3bv]

I really appreciate your videos. But I dont understand this: If geodesics can spread with a non-zero first derivative as mentioned in 9:16, then volume moving along these geodesics should change I suppose? Like if you look at a cone and take the top as the origin of polar coordinates, and take 'slices' of the cone (which should be a circle). These 'slices' will have different areas as the radius grows, so the Ricci tensor should be negative? But eventually this is in flat space so it should be zero. What is wrong? I think that maybe in a flat space geodesics should just be parallel straight lines. As you mentioned in previous videos the shortest path between two points is given by geodesics, and if we take non-parallel lines, to keep another group of geodesics to be orthogonal to this group of geodesics (otherwise there may be geodesics from these two groups dont intersect), that group of geodesics must be 'curved', like theta coordinates lines. Since the shortest path between two points in a flat space is straight, these 'geodesics' are not actually geodesics. I dont know if Im right.

[@damienthorne861]

Ok stupid question but please indulge me. The second covariant derivative ( the acceleration) of the separation vector in the direction of the tangent vector ( Geodesic deviation) to produce the negative of the Riemann Curvature Tensor why dotted with s? I know I'm going to regret asking this but...

[@eigenchris]

Not sure the exact part of the video you're referring to, but the "dot s" is to make the result independent of direction. s accelerating to the right and decelerating to the left both have identical 2nd derivatives, so that alone is not enough to tell us if the two geodesics are growing apart or not. We need to dot with s (the original direction of s) to see if the s-vector is growing or shrinking. Does that answer your question or did I misunderstand you?

[@damienthorne861]

@@eigenchris you are the f**king man ! Exactly what I needed thank you! 👍

[@JgM-ie5jy]

" ad - bc " : the familiar look of the determinant of a 2 x 2 matrix. Iwould really love to see your insight as to how this part of the normalization relates - if at all - to the determinant of a 2 x 2 matrix. It can't just be a coincidence. Regards

[@eigenchris]

For the denominator part, the explanation is straightforward. We are just using a linear map with coefficients a b c d to move the parallelogram vectors around. The resulting change in area is the determinant ad-bc, and so the square of that is just (ad-bc)^2. I have no idea why the (ad-bc)^2 comes out of the numerator.

[@frankdimeglio8216]

THE ULTIMATE, TOP DOWN, AND CLEAR MATHEMATICAL PROOF REGARDING THE FACT THAT E=MC2 IS F=MA: Time dilation ultimately proves ON BALANCE that E=mc2 IS F=ma, AS ELECTROMAGNETISM/ENERGY IS GRAVITY. Time is NECESSARILY possible/potential AND actual IN BALANCE, AS E=MC2 IS F=ma; AS ELECTROMAGNETISM/energy is gravity ON BALANCE. Gravity is ELECTROMAGNETISM/energy. Great !!!! QUANTUM GRAVITY !!!! E=MC2 IS F=ma. This NECESSARILY represents, INVOLVES, AND DESCRIBES what is possible/potential AND actual IN BALANCE. What are the EARTH/ground AND the SUN are CLEARLY E=MC2 AND F=ma IN BALANCE. Very importantly, outer "space" involves full inertia; AND it is fully invisible AND black. The stars AND PLANETS are POINTS in the night sky. GRAVITATIONAL force/ENERGY IS proportional to (or BALANCED with/as) inertia/INERTIAL RESISTANCE, AS E=MC2 IS F=ma; AS ELECTROMAGNETISM/energy is gravity. Gravity/acceleration involves BALANCED inertia/INERTIAL RESISTANCE, AS E=MC2 IS F=ma; AS ELECTROMAGNETISM/energy is gravity. BALANCE AND completeness go hand in hand. It does ALL CLEARLY make perfect sense. GOT IT !!!! THE SKY is BLUE, AND THE EARTH is ALSO BLUE. Great !!! Now, think about the man who IS standing on what is THE EARTH/ground. Perfect !!!! By Frank DiMeglio

[@00000ghcbs]

@@frankdimeglio8216 r u ok bud

[@Shirsh]

I don't see how you can replace separation vector by basis vector. Separation vector changes length based on how geodesics change (their definition). But basis vectors are not the distance between geodesics and they do not change correspondingly.

[@siddharthjain3078]

the shrinking of ball along geodesics is just some hypothetical situation we use to visualise or does this actually happen? Is this same effect as length contraction of STR?

[@eigenchris]

This is related to General Relativity, not Special Relativity. When you release a ball of dust in a gravitational field, it will experience "tidal forces", where the top and bottom are stretched apart, and the sides are squished together. The overall immediate volume change is zero, so the Ricci Tensor is zero. This is what you see outside Schwarzschild black holes (or any spherical gravitational body). I cover this in my Relativity 107d video.

[@siddharthjain3078]

@@eigenchris thanks for the clarification. So finally, a ball does deform actually when traveling along geodesics?

[@eigenchris]

If the Riemann Curvature Tensor is non-zero, then yes.

[@pomeron490]

At 25.22 you say "when the Ricci tensor acts on the same vector twice the result is ...the Ricci curvature." Should it not be the the Riemann tensor rather than the Ricci tensor?

[@eigenchris]

The "Ricci Curvature" is a number that we get when we put the sams vector twice into the Ricci tensor. You can look up "Ricci Curvature" on wikipedia if you want to learn more.

[@tongzhu6714]

24:43 I don't get it... what is v again? It's one of the ei's, no?

[@eigenchris]

It is. I got a bit lost in the algebra when making the notes for this video and missed that v is just one of the e_i. Although you can technically scale e_i by any number, so you could technically write v = 5e_i or something like that.

[@tongzhu6714]

@@eigenchris Yes thanks for the clarification... I figured that out not very long after I saw that... I don't know maybe because I am from pure math... that's sort of important for me... Great videos (and nice robot voice）

[@y0n1n1x]

Tysm

[@ehsankiani7563]

Thanks for your fantastic videos. Why K(e1,v)=-K(e2,v) at 22:47? We know one of them is negative and the other one is positive, Why their size is equal to each other?

[@eigenchris]

I just forced them to be equal in this example so that we got zero as a result. There's no real reason why it's true--I just wanted to get zero for the purposes of the example.

[@ehsankiani7563]

@@eigenchris Thanks for the quick reply!

[@胡敬超]

i have a question a 2D cone’s Ricci curvature is 0 But the volume of disk will shrink How to explain it?

[@eigenchris]

I am not 100% sure but this is my guess: This is because the geodesics on a cone can be straight lines. On a cone, the volume of a disk will change with the first derivative if the geodesics are straight lines, but this is not because of curvature. The volume will not change with the second derivative (the true indicator of curvature).

[@胡敬超]

eigenchris Thank you for replying Can I understand it as the change of volume is accelerating Actually I think the sign of K also indicate the acceleration of the converging or diverging

[@eigenchris]

Yes, that's correct. +K means converging and -K means diverging.

[@samirbouguerra2626]

Absolutely awesome

[@MohammedAhmed-qv3fj]

Why the magnitude of R(s,v)v dotted with s equal the squared area of the parallelogram

[@eigenchris]

The proof is in the video. Are you asking for an intuitive reason why? I'm not sure about that.

[@sweetytarika8068]

sir, i want to contact you, i have many queries, its request please.

[@eigenchris]

Can you just ask your questions in a comment here? That is my preference.

[@amithkumar3176]

Your videos are simply "fantastic", waiting for "general relativity"......😁😅

[@sanidhyasinha5735]

If the volume is converging if we move along vector v, then if we move in the reverse direction (-v), the volume should diverge, right? I know the equation is suggesting that it will always converge, but intuitively it's suggesting otherwise, what's happening?

[@eigenchris]

There are two things that determine what curvature is: a point p, and a 2D plane (determined by the v and s vectors). Changing +v to -v will not change the curvature since the plane stays the same. The geodesics will converge or diverge in the same way that they did with +v. The key is that, in my definition, the "v" vectors are all near point p and parallel to each other. If we follow the geodescis, we are no longer at point p, and the "v" vectors are not parallel any more, so we aren't measuring curvature at any particular point p. Does that make sense?

[@schokolade1735]

I actually got a question: How would I imagine a difference vector in curved space or better what does such a difference vector in the curved space e.g. between two geodesic points (i.e. for equal lambdas) represent? Is this a vector that has the same length as the arc length of the geodesic connecting these two points? (Is this geodesic unique?) Where does this vector point to? ...Tangential to start of the connecting geodesic? Am I completely off or does this question makes sense?

[@eigenchris]

I think if you took the "separation curves" (which can exist on curved space) and took a tangent vector on one of them, this would be the equivalent of a "separation vector" in curved space. The vector length wouldn't be equal to the geodesic connecting the two points (I'm don't think separation curves need to be geodesics but I am not 100% sure). However, you can integrate the length of the tangent vectors along the curve to get the length of the curve. (integral of ||dR/dλ|| dλ, which I show how to do in video 12)

[@schokolade1735]

@@eigenchris thanks for that quick response

[@teezettsb]

cool, now I know how the Tardis works

[@vaevfunc]

Hi Why is the d_(b)^(b) equals 1 on 26:45? d11+d22=2 for example

[@HankGussman]

Chronecker delta by definition is equal to 1, when upper & lower indices are equal. Here both the indices are "b", hence Chronecker delta is 1.

[@chenlecong9938]

Wait a minute,there’s a major hole in 14:16.If you think about it,if R(u,w)=( ad-bc)R(u,w ),and on the premise that you can’t just cancel out the R(u,w),then theoretically, you can repeat that process infinite number of time,then (ad-bc) becomes infinitesimally large,if not zero.But in both cases,they’re not possible since you’re gonna divide that by the other constant of the Denominator of the Sectional Curvature which can also turn out to be zero or infinity as per above mentioned Can you clarify on that PLEASE???

[@eigenchris]

Sorry, I'm not sure I follow what you're saying. All I'm saying is that the value the formula for sectional curvature doesn't depend on which vectors in the plane you use. So if you switch from (u,w) to (au+bw, cu+dw), which is still in the same plane, the value of the sectional curvature is unchanged. How does infinity come into this?

[@chenlecong9938]

@@eigenchris so the fuss is that,seemingly the conclusion in 14:16 is that R(u,w)=(ad-bc)R(u,w),let’s just call that Eqn ① Then if I just substitute Eqn ① into the Right Hand Side of Eqn ①,I’ll get R(u,w)=(ad-bc)²R(u,w) If I repeat that nasty process unremittingly,doesn’t that imply that (ad-bc) will be raised to the power of infinity??? If that term is -1＜x＜1 thatll be zero,by contrast that’ll ±∞

[@eigenchris]

I wasn't trying to say R(u,w)=(ad-bc)R(u,w). I was trying to show that when we change the input vectors from (u,w) to (au+bw, cu+dw), then R(u,w) changes to R(au+bw, cu+dw)=(ad-bc)R(u,w). The "ad-bc" part will get cancelled out in the denominator in a later step.

[@chenlecong9938]

@@eigenchris yah I can fathom your argument.but my man,I’m wondering if you understood the point(the fuss) I was trying to address over there…. I know it’s trivial indeed,but hey,I still couldn’t quite get over the infinity argument-why isn’t it justifiable?It really seems to me there isn’t anything wrong in thinking that way Is that how Math at this level works-you have two or more arguments that’re all ,to some extend, “YES” ,but you simply choose the one you desire?? As far as I understand,that’s how Theoretical Physics works,as much as Einstein simply postulated speed of light is constant

[@ceoofracism5713]

isn't R2121 reimann components is same as R1212? at 27:19

[@chenlecong9938]

You’re assuming the 12 and 34 symmetries,but one of them has first indice being a subscript(can’t remember which one). Therefore no.

[@silentobserver3433]

So, I think I missed something. At 21:09 size of the ball in v direction is getting smaller, at 21:55 it is getting bigger (in v direction) and at 23:06 it stays about the same. So what determines size change in the direction parallel to the geodesics? Other two dimensions change according to geodesics getting closer or farther apart, that's clear, but what about v direction? As always, thank you very much for making such high-quality videos on this topic!

[@eigenchris]

It's possible I should have used a 2D disk instead if a 3D ball in order to track the changes. I'm honestly not sure. This reflects my own lack of understanding. I'll see if I can find an answer.

[@silentobserver3433]

@@eigenchris Wikipedia says it has something to do with cross section of the cone, emitted from a point and also mentions distortion of circular cross section to elliptical, similar to your animation at 21:55. So yeah, I think it should be a disk in original definition. Maybe distortion in the direction parallel to the geodesics is impossible? Because when we transport this ball, initially of, say, 1 unit "long", we keep it 1 unit "long" by definition of transporting something, 'cause else we could just scale it and get absolutely different Ricci tensor. It seems like the best solution for that

[@eigenchris]

I'll add a correction in my next video. I read an article that described the ricci tensor using a ball of ground coffee floating in space, so I thought it had to do with balls. But I think in 3D it should be a disk in the same plane as the e1 and e2 basis vectors.

[@wolf3213]

In Rici Tensor Definition u sum over 1,2,..n-1 cuz n term which is e_n=v, and is uncessery. (in a another words K(v,v) would be zero) Imma right?

[@eigenchris]

Yeah, you're right. I was trying too hard to make it look like equations I had seen in textbooks, but the e_n = v thing makes the components of v not needed.

[@philamras3732]

​@@eigenchrisHi Chris. This is a great video and I have learned a lot from this series. However, having read these comments, I am confused about this now. Is the formula at 24:54 incorrect? If so, how does one correct it,?

[@acespade5378]

at 25:00 I didn't understand why we do "contraction\summation" over i and remove it from the Ricci Tensor, any help please :l .

[@eigenchris]

The contraction is already in the formula bwcause there arw two e_i basis vectors, so we just replace the Riemann Tensor with the Ricci tensor.

[@acespade5378]

@@eigenchris thank you :)

[@Snake_In_The_Box]

yay new video！

[@a52productions]

Why is the Ricci tensor a (0,2) tensor if it's generally only used with one vector input? It seems like you should be able to express it as a (0,1) tensor.... Wait, no, because if it's a (0,2) tensor, R(kv,kv) = k^2 R(v,v), as opposed to R(kv) = k R(v). Huh.

[@eigenchris]

Yes. Also it transforms with 2 covariant rules. I don't know the meaning of the Ricci temsor acting on 2 different vectors.

[@alaindevos4027]

sign convention. One can not put a minus sign where he wants so something else must also have been defined otherwise.

[@eigenchris]

You're probably right.

[@lucasferreira5170]

In minute 10:00, if I change the sign of V, the sign of the second derivative won't change and so we will have a diverging situation with a positive dot product

[@eigenchris]

I'm not sure I follow. I think the dot product will still have the same sign? If you change the sign of V, the 2 negatives in the v terms for the Riemann tensor will cancel each other out and become positive. If we change the sign of s, the negative s in the Riemann tensor and the negative s on the other side of the dot product will cancel to be positive as well. I don't think the sign of s or v should affect the result here.

[@lucasferreira5170]

@@eigenchris If you change the sign of V but not change the sign of s. For example, suppose V is parameterized by "t" and s is parameterized by "u". And you only change t -> - t . (when we have a family of geodesics defining a surface, we have two parameters so that each point is described by x(t,u). V is the derivative of x wrt to "t" and points along the geodesics and s is the derivative of x wrt "u": it's the separation. )

[@eigenchris]

@@lucasferreira5170 I don't think you can take the family of geodesics and just walk over them backwards. The idea is that you start with a family of "parallel" geodesics in a small neighbourhood and watch where they go. If at a point p and direction v, the geodesics diverge, then geodesics will also diverge for the same point p with reversed direction -v, but diverging in the opposite direction. Does that make sense?