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

This isn't bad either: physicsunsimplified.com

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

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

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

whelp.

It's smaller on the outside.

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

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

and trin tragula general relativity

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

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?

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.

Thanks!

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.

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

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.

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.

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

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.

@@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.

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

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.

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.

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

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.

@@eigenchris Thank you

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.

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

@@frankdimeglio8216 r u ok bud

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.

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

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.

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

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

you can think it as definition of Ricci tensor components.

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).

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

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

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.

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.

@@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）

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?

@@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?

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

@@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?

See 10:29 on geodesic deviation

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.

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.

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.

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?

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

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

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.

@@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.

​@@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.

@@eigenchris Thx, finally got it!

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.

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

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

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?

@@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 ±∞

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.

@@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

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.

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.

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?

You're probably right.

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.

@@eigenchris Thanks for the quick reply!

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.

@@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

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.

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

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)

@@eigenchris thanks for that quick response

I think it's because a change of basis will change each expression by a factor of the area increase, and the determinant gives the formula for area increase for a linear transformation.

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.

@@eigenchris thank you :)

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.

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.

@@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. )

@@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?

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.

​@@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,?

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.

@@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 😁😁

This is a valid problem others have pointed out, but basically this entire slide "breaks" Einstein notation somewhat. You can see in the first line that I'm doing a summation over "i", but the "i" index in the expression appears twice as a lower index (all summations in Einstein notation normally have 1 upper index and 1 lower index). So already at the start of this expression I'm breaking Einstein notation. It's just that on the last line I magically "fix" the notation to give us the answer we should get, with 1 upper "i" and 1 lower "i" so that the summation works properly.

It’s honestly so awesome to see you respond so soon. Thanks for being so helpful! I guess my main problem is that we said that R(e_i, e_j)e_k = R^m_ijk e_m. So if you then take the dot product, you’d get R^m_ijk(e_m . E_i) sorry the notation sucks here :P I’m not sure if I’m misunderstanding what you’re saying, but I can’t see how to resolve this problem (unless I define R_mimk = R_ik = which seems to disagree with every other resource)

Perhaps the question I should ask is. Why can you say that [R(ei, ej)ek].ei = R^i_kij? Did we define this somewhere?

Re-reading what you wrote, I'm seeing your point. I might have to read up on this again to make sure I haven't made a mistake.

eigenchris Awesome, thanks! It has been tormenting me for the last week XD