Relativity 107c: General Relativity Basics - Curvature, Riemann Tensor, Ricci Tensor, Ricci Scalar

Full relativity playlist: • Relativity by eigenchris
Powerpoint slide files: github.com/eigenchris/MathNot...
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You are free to continue watching to the next video, but if you feel you are getting confused, here are some other videos on curvature here:
TC 21 - Lie Bracket, Flow, Torsion: • Tensor Calculus 21: Li...
TC 22 - Riemann Curvature Tensor Meaning: • Tensor Calculus 22: Ri...
TC 23 - Riemann Curvature Tensor Symmetries: • Tensor Calculus 23: Ri...
TC 24 - Ricci Tensor Meaning: • Tensor Calculus 24: Ri...
TC 25 - Ricci Tensor/Scalar Meaning: • Tensor Calculus 25 - G...
TC 26 - Ricci Tensor/Scalar Properties: • Tensor Calculus 26 - R...
0:00 Introduction
1:56 Riemann Curvature Tensor
8:12 Riemann Tensor Components + Symmetries
15:29 Riemann Tensor - Geodesic Deviation
19:04 Ricci Curvature Tensor
23:46 Ricci Curvature Scalar
25:46 Curvature of Rindler Metric
33:19 Summary

[@eigenchris]

I made an unfortunate typo around 32:09. The final result of zero is correct, but the intermediate steps are not. A corrected version is below: R^μ_vαβ = ∂_αΓ^μ_vβ - ∂_βΓ^μ_vα + Γ^σ_vβΓ^μ_σα - Γ^σ_vαΓ^μ_σβ (Set μ=α=t, and v=β=x) R^t_xtx = ∂_tΓ^t_xx - ∂_xΓ^t_xt + Γ^σ_xxΓ^t_σt - Γ^σ_xtΓ^t_σx R^t_xtx = 0 - ∂_xΓ^t_xt + 0 - Γ^t_xtΓ^t_σt R^t_xtx = - ∂_x(1/x) - (1/x)(1/x) R^t_xtx = - (-1/x^2) - 1/x^2 R^t_xtx = 0 I apologize for the mix-up. Staring at indices for hours makes it easy to miss small errors.

[@damienthorne861]

You should be ashamed of yourself! No errors allowed! 😉 Dude I'm joking! 😂 You and this content are amazing and you have advanced my knowledge exponentially. Thank you for all of your videos

[@user-ck1kx5ie6t]

Who would have expected 2 eigenchris videos a day? ;)

[@CrowKunCGS]

This channel is a gold mine 📚💛 There is huge advanced knowledge, teached in a simple and pleasant form, for free. Thank you so much! 😁 I really appreciate your content and I think your work will be really important for many people over the years. Congrats! 👨‍🏫🏅

[@ElliottCC]

Absolutely correct! It's a godsend

[@PM-4564]

I've probably watched your 5 minute topology video about 10 times by now.

[@eigenchris]

You must understand topology really well, then.

[@connorduncan2927]

This is about how dry I would expect eigenchris' sense of humor to be

[@prosimulate]

You are amazing this is superb, I’m a 51yr old Chemical Engineer, but I love this field of maths, but it is brick hard to understand in the books, I have tried, I get some way but your explanation and voice is just perfect! I salute you buddy!

[@eigenchris]

Thanks! I know the experience of not understanding textbooks. It's what prompted me to start making videos in the first place.

[@myaseena]

bruh you are so good at explaining stuff it is amazing and almost relaxing

[@tinkeringengr]

Your instruction style is so powerful. Your impact is great!

[@RichardLibby]

The condensed nature of this presentation is an excellent complement to the various books I've been reading, trying to understand General Relativity. Books of course are invaluable because they are meticulously complete, but the cost of reading them is in the sorting between the main ideas and the necessary but myriad details. Here, we have the main ideas and how they are connected, with occasional brief mentions of where important details are to be found.

[@phillipafrayan8313]

Awesome video. It is One of the best channel on youtube.

[@user-hm5rf8vz6h]

Amazing! Thanks a lot for these all videos about GR. It was really hard to me to understand the meaning of curved geometry and its quantities and couldnt find the appropriate book with simple explanation. By watching your videos finally I have some simple and understandable imagination of what Im calculating and doing now. Good luck for your career)))

[@dorothyyang8590]

When I read the book about GR, I often feel confused. Now you make evething clear for me! Thank you!

[@siddharthpandya7763]

What books are you following? I am reading Landau's second volume and Sean Caroll, but looking for better alternatives.

[@mehdisi9194]

Thank you so much for this great video👌👍

[@jigold22571]

Absolutely Fabulous, Thank You

[@agnaldojunior5538]

these intuitions are amazing!!!

[@signorellil]

Fantastic as usual

[@tomkerruish2982]

Thank you so much! I feel like I'm finally getting a handle on the Ricci curvature and scalar! 24:40 I'm surprised you didn't throw in a TARDIS reference.

[@eigenchris]

I had that in the Tensor Calculus videos I did a couple years ago: kzbin.info/www/bejne/kJmnf5yghKqqZrs

[@tomkerruish2982]

@@eigenchris Thanks, I'll be sure to watch it! Also, any chance for a follow-up video to "Momentum" titled "Spin"?😂

[@eigenchris]

I've been studying spinors for the past few months, and have been considering making a series on them. Maybe I should make a "what is spin?" joke video to go along with it.

[@signorellil]

@@eigenchris please do a series about spinors!

[@Deciptikon]

Incredible!!!

[@justingerber9531]

On the Del_{[V, W]} T term in Riemann curvature: Basically it is exactly as you say about needing to "cover the gap". You can find a clear motivation for defining the Riemann curvature tensor this way in Lee Riemannian Manifolds (I have 1st edition, don't know if there are more) Chapter 7. In flat space we have (for general vector fields X, Y, Z) that: Del_X Del_Y Z - Del_Y Del_X Z = Del_{[X, Y]} Z That is, If you parallel transport Z along X then Y and then again along Y and then X you will see a discrepancy exactly due to the gap [X, Y]. So for the definition to make sense in flat space you need to include that third term. We measure how much the result deviates from the flat space result to describe curved space.

[@HighWycombe]

I love these videos, but confess I'm finding them more and more challenging. There's no shortcut to this. Before trying to follow these GR videos, I need to go back and study ALL the Tensors for Beginners videos, then ALL the Tensor Calculus videos. I've already watched the Special Relativity videos, but for anybody who hasn't I'd recommend watching those too. THEN you'll have a fighting chance to finish these last few GR videos!

[@eigenchris]

Yeah, the mathematics of curvature is kind of terrible. I'm not sure if you need to force yourself to watch every single one of those videos, but for curvature, videos 21-26 of Tensor Calculus might be necessary. I still don't fully understand curvature, but I did the best I could with those videos.

[@user-pe7gf9rv4m]

even though it is challenging ; it is even more rewarding

[@kanikapathak2040]

@eigenchris After seeing your video on "Finite Fields of prime order" , I wonder , will you ever make a series on Abstract algebra ?

[@eigenchris]

I don't plan on teaching abstract algebra beyond what's necessary for error correcting codes.

[@ahmedeloufir]

Gauss passed away in 1855 not 1885 :) Great videos !! Thanks !

[@pong336]

Excellent Video! By the way, what plans do you have for this channel after GR?

[@eigenchris]

I don't want to make any promises, but the thing I have been most interested in lately is quantum spin and spinors. GR will probably take me until the end of the 2021 to finish though, so that's looking far ahead.

[@peter308]

You are the real deal!

[@eric3813]

GREAT VIDEO! thanks! :)

[@tuyenpham1756]

Thank Chris, I found some typo at 10:48 for component parts of vector V and W

[@im54d89]

After finding your channel.. im be like f Netflix . I'm gonna watch eigenchris 💜

[@jurgenanklam3226]

I think I spotted an error at 32:08 where you plug in the wrong gamma \Gamma^{x}_{tt} for \Gamma^{t}_{xx}. The correct Gamma is already 0, the wrong Gamma is non-zero but you derive the term later, where it also becomes zero, so that the error does not impact the final result. Anyway great videos. Thank you for providing them. Hope you will proceed to spinors some time as you indicated in some comment!

[@emiliosani9924]

Excellent classes , marvelous professor. Just a minor biographical typing mistake in the beginning: Gauss died 1855, not 1885.

[@lunam7249]

guass died or dies when eigenchris says he does!!

[@zzzoldik8749]

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

[@tw5718]

I want to first say I really enjoy all of these videos. Secondly I have 2 questions. 1) I apparently misunderstood what the ricci tensor represented. I thought it was more closely related to the change in vectors, with the ricci scalar being similar to change in size, which would make sense to have both of those in the EFE. However they both appear to be a measure of change in size. Why do we need 2 representations of the same quantity in the field equations? 2) I don't see how you got the first step in the derivation of the riemann. Im talking about the del gamma e piece. I tried working it out as the commutator of the covariant derivative and got my double christoffel symbol terms to cancel. Is there any way you could explain why it works out to be del (gamma e) instead of just [d/dx + gamma, d/dy + gamma]?

[@eigenchris]

1) The Ricci Scalar term in the Einstein Field Equations is a mystery to me in terms of the geometry. I just know it needs to be there for algebraic reasons to make sure that both sides of the EFE have zero divergence. For 2), can you give a timestamp for the part of the video you're talking about?

[@tw5718]

Thanks chris. @11:45

[@michaelschnell5633]

Hi EigenChris I studied Math and Physics in University some 40 years ago and never heard about ART nor Quantum Physics at that time. Since some years I watch your (and other's interesting) videos on these topics. So firstly a big THANK YOU for this excellent content ! By this and hours of intensive personal discussions with NarfWhals I got the feeling that I reached a rather decent understanding of the concepts and the Math of the ART. So I started to try to find a definitive algorithm: how to calculate the (observer independent) spacetime interval and the (observer etc depending) spacial distance between two events in spacetime, calculated knowing the the metric tensor field (and using some Coordinate System defined by the observer). This seemed like doable task, even though me wondering that it did not simply is obvious from the content I already digested. When rigorously thinking about it, my spacetime broke down as I realized that I seemingly am lacking understanding of some very basic concepts. E.g. I always thought that the metric tensor "lives" in the spacetime (pseudo-Riemanian) manifold (as seems to be suggested to me by trying to understand the nature and the derivation of the Schwarzschild metric). But on the contrary I found that a manifold is not even a vector space, and as Tensors are defined as multi linear maps on vector spaces, they can't live in Manifolds. Obviously, as a manifold is defined as locally being homomorphous with an Euclidean vector space (the tangent space at that "point" ). So I recon that (regarding the metric) a (pseudo-) Riemanian manifold locally is supposed to be homomorphous to a tangent vector space with the appropriate (flat) metric. Now the metric tensor in the _flat_ tangent space obviously is trivial (constant everywhere, and diagonal in appropriate coordinates), so same can't be an element of the metric tensor field of the manifold. Hence the metric tensor field element associated to a "point" in the manifold is a tensor that lives in the tangent space for that "point" and obviously is not the (unique) metric tensor of that tangent space. So for calculating e.g. a spacetime interval between two events (which is the "length" of a geodesic between the events), we need the metric tensor of all "points" in that curve, and the tangent space at all these "points", as this is where this metric tensor lives (i.e from whose set we need to feed it's inputs with differences of certain vectors). Is there any reference in your videos that in a rigorous way explains, what the metric tensor filed "is" (i.e. as a function of exactly which Domain -> (usually) real numbers) and/or how/why we are allowed to use a (or any) coordinate system defined on the manifold to as well select the metric tensor at a certain "point" (obviously in manifold coordinates), and "feed" this metric tensor with the difference of two vector (from the tangent space, and hence obviously in coordinates defined on the tangent space). Thanks a lot, -Michael

[@eigenchris]

Hi Michael. I'm glad you enjoy my videos! The standard procedure for getting the proper length (or time) of a curve is explained in 105d. It's just the integral over the lengths of the tangent vectors of the curve, where length is calculated using the metric. As you say, the metric changes at every point along the curve, so you need to do an integral to "add up" all the contributions. I do additional proper length calculations in 108b for a black hole and 109e for gravitational waves. As for "what the metric actually is" in mathematical terms, I get into the math of this in my "Tensors for Beginners" video series. I plan to eventually summarize this in a 106b video (I skipped over it for now). The basic idea is that a vector on a manifold at a point lives in a tangent space V. A covector lives in the dual tangent space V*. The metric tensor lives in the tangent space V*⊗V* (two copies of the dual space because the metric is twice-covariant). When we talk about "the metric", we're talking about a metric tensor field, where at every point the metric is a member of the tangent space V*⊗V*. In theory, you could treat each one of the metrics along a curve as a series of flat metrics, but since we're doing an integral over all of them, the result is different than what we'd get in flat space. Does this make sense?

[@michaelschnell5633]

ART (Gemman) means GRT :)

[@michaelschnell5633]

@@eigenchris While all this intuitvely is rather obvious, I still fail to see how to rigorously "map" the two "points" in the said curve (with infenitesimal distance) (which are not vectors in any vector space and hence can't be subtracted from one another) to vectors in the tangent space whose difference is allowed to be used as an input to both of the arguments of a tensor. Thanks again, -Michael

[@eigenchris]

I'm not sure I fully understand your question. Can you point out where you get lost in the below argument? 1. Each point along the curve has a tangent vector that lives in a tangent space attached to that point 2. We calculate the length of a tangent vector by taking the square root of the dot product of the vector with itself (the "dot product" being the same thing as "the metric", so putting the same vector into the metric twice to get its squared length" 3. We integrated all the tangent vector lengths over the curve to accumulate a final result for the curve's length

[@narfwhals7843]

@@eigenchris I think the problem in our understanding is in between step two and three. When we take the magnitude of the tangent vector we do this with the "local" metric, which is only defined on the tangent space. So this is (as far as we've been able to understand) the identity, by definition. But when we integrate we use a metric tensor field, which connects the different tangent spaces who don't know about each other. How do we reconcile using two different metrics for the same calculation? And where does the metric tensor field live? Does it contain dot products of basis vectors of different tangent spaces to connect them? And what Michael(who is my dad, btw :) ) wants to know in detail is what the metric tensor field is a function of (observer coordinates, local coordinates, vectors etc.).

[@jensphiliphohmann1876]

About 22:50ff: This seems to describe tidal forces when applied to spacetime, and PENROSE once wrote that they are connected to another tensor he called WEYL. How do we get that tensor?

[@eigenchris]

I haven't studied the Weyl tensor yet... I've only read about it. Supposedly, since the Riemann Tensor has 20 indepedent components, 10 of these components are in its trace (the Ricci Tensor), which track changes in volume but not shape... and the other 10 components are in a traceless version of it called the Weyl Tesnor, which tracks changes in shape but not volume. I can't say more than that because I haven't studied it. You can look at the formula on wikipedia: en.wikipedia.org/wiki/Weyl_tensor

[@Frostbitecgi]

waiting for energy momentum tensor and einstin field equations.

[@souvikmandal1989]

In Relativity 105f geodesic equation was described as second derivative of postion vector S with respect to path parameter equal to zero. In this video at 16.03 geodesic equation is described as covariant derivative of V equal to zero. Are these two definations same?

[@eigenchris]

Yeah, they're the same. In 105f, the geodesic equation is "d/dλ (dS/dλ) = 0". In curved space, we can't use a position vector S anymore, so we change the formula to just "d/dλ (d/dλ) = 0", where "d/dλ" is taken to be both a derivative operator and a tangent vector. If we rename the tangent vector "d/dλ" as "V", and switch from the "d/d ..." notation to the "∇" notation for the covariant derivative, we get "∇_V (V) = 0". This is the same formula from the start, just written in a different way.

[@rudradityasinghnegi1551]

thanks for giving warning first

[@cinemaclips4497]

At 32:09, the connection coeffecient we have calculated and the one we are subbing in are not the same.

[@eigenchris]

Sorry for the late reply. I included a correct calculating in my pinned comment.

[@lourencoentrudo]

You never miss! Great video as always. I wonder tho. If the accelerated frame yields no space-time curvature, by the principle of equivalence is that also true for a homogenous constant gravitational field? Because I was under the impression that any gravitational field was associated with curvature.

[@eigenchris]

Being in an accelerated frame is equivalent to being pulled against the ground in a uniform gravitational field. But "uniform gravitational fields" don't exist in real life. All real gravitational fields are non-uniform. However, non-uniform gravitational fields can be approximated as uniform in a small region of spacetime. The principle of equivalence tells us that the accelerating Rindler spaceship is equivalent to a gravitational field in a small region of spacetime. This "small region" is a bit like a flat tangent plane on the curved surface of spacetime. On the flat tangent plane, you will never see tidal forces, but a real gravitational field (curved spacetime in a large enough region) you will see tidal forces. The accelerating Rindler observer's spacetime is flat everywhere. It's not just a flat tangent plane approximation of a curved spacetime. It's truly flat everywhere in the universe.

[@lourencoentrudo]

@@eigenchris then would it be correct to say that every observer in a gravitational field locally is a Rindler observer?

[@eigenchris]

@@lourencoentrudo I wouldn't exactly say that. The Rindler observer feels proper acceleration (in other words, they measure proper acceleration on their accelerometer), and not every observer feels this. A "straight line"/"zero acceleration" observer in the standard spacetime diagram feels no proper acceleration. In general relativity, a free-falling observer also measures zero proper acceleration. However, for people standing on a planet, they do feel proper acceleration (a cellphone accelerometer can measure your proper acceleration for you right now). So it is correct to say that everyone standing on a planet is, locally, a Rindler observer.

[@lourencoentrudo]

@@eigenchris I see. So proper acceleration measures the "tangential acceleration"? I'm referring to the definition that in a geodesic the only acceleration is normal to the surface where the geodesic is embedded. So, a free falling observer is tracing a geodesic in space-time, hence he feels no proper (tangential?) acceleration.

[@eigenchris]

@@lourencoentrudo Yes. The "proper acceleration" is the magnitude of the tangential acceleration vector. If the tangential acceleration vector is the zero vector, its length (the proper acceleration) is zero.

[@rudypieplenbosch6752]

How about the Weyl curvature ? I hear Roger Penrose talking about this a lot, it is crucial during a black hole collapse, I'm trying to find mor info about it.

[@eigenchris]

I don't actually know much about Weyl Curvature. My loose understanding is that, since the Riemann Curvature Tensor has 20 independent components, this can by broken up into the Ricci Curvature (10 components that tell you about volume changes, but not shape changes), and the Weyl Curvature (10 components that tell you about shape changes, but not volume changes). That's all I can say.

[@rudypieplenbosch6752]

@eigenchris Yes Roger does mention the Weyl curvature is all about about shape, he also mentions the Ricci curvature works like a magnifying glass. Anyway, your explanations are outstanding, I still need to finish a lot more of them before I understand the Ricci curvature, thanks for your answer, I'm just thinking, if you don't know the Weyl curvature, then it doesn't look so great for me 😉, anyway Roger can't seem to get enough about the Weyl curvature so I have to know what he is talking about.

[@KaliFissure]

It seems to me when I apply any natural curvature (an exponential curve for instance) on a manifold the surface is no longer spherical but a cardioid with a closed limit. Is this the reason there is a visible universe? Since the curve increases with distance from observer and it is the space itself which has the curve wouldn’t that refract and frequency shift light distant to observer exactly mimicking Doppler effect? The more distant the higher the curvature to space and gravity is an acceleration so it would freq shift more at greater distance. What am I getting wrong?

[@eigenchris]

I'm sorry but I don't know what "natural curvature", "cardiod" or "closed limit" mean, so I don't understand your question.

[@KaliFissure]

@@eigenchris apologies for a typo which apparently you were unable to see through. en.m.wikipedia.org/wiki/Cardioid#:~:text=A%20cardioid%20(from%20the%20Greek,epicycloid%20having%20a%20single%20cusp.

[@maujo2009]

Dear eigenchris: I've been struggling for years to find a GR course at this "entrance" level. Which textbook[s] would you recommend that approach GR at the level you use in your videos? Thanks for your great work! I will use your videos in my class!

[@eigenchris]

I haven't come across any textbooks at the level I'm teaching at, which is part of the reason I'm making these videos. You could try looking at Sean Carroll's General Relativity notes. They are more complicated than my videos, but they could be a good start. I've also used "Gravitation" by Misner, Thorne, and Wheeler as a reference, but it is also more complicated and pretty heavy.

[@gman8563]

Do I have this correct? A metric defined over a manifold uniquely determines its curvature… but curvature doesn’t uniquely determine a metric. That is, different metrics can yield the same curvature? Where I’m getting confused is why two different metrics, like the metric belonging to a Rindler observer and the metric belonging to an inertial observer, can yield the same curvature. Does this mean that a metric tensor is coordinate-based? I assumed that, being a tensor, the metric shouldn’t be coordinate dependent. Or am I confusing something here?

[@eigenchris]

For Riemannian manifolds with the Levi-Civita connection: yes, the metric completely determines the curvature (we use the metric to get the connection coefficients, and then we use those to get the Riemann Tensor, which completely describes the curvature). The standard "Minkowski" metric diag(+1, -1, -1, -1) and the Rindler metric are technically the same metric tensor, just described in a different basis (and so they have different component numbers in the matrices). This is sort of like how the same vector will have different component numbers in different bases. In Relativity 103d, I showed how to change basis with the metric... it involves 2 jacobian matrices since the metric tensor is 2-covariant tensor. You can change from the "standard Minkowski" metric components to the Rindler metric components by applying 2 standard-to-rindler-coordinate jacobian matrices. I planned to explain this more in 106b and 106c, but I decided to skip those in favour of going forward with GR. As for the question "can 2 different metrics give you the same curvature?"... I'm honestly not sure. You might be interested in watching "Tensors for beginners 11-15", which is maybe an hour of content. It shows how to write various tensors using basis tensors, and it becomes more obvious how two different matrices can represent the same metric tensor.

[@nightmisterio]

On the next video show us where is the curvature of apples-oranges

[@dottormaelstrom]

10:42 I think there's a problem with your indices, the W^mu should be W^nu

[@eigenchris]

Yeah. Typo. My bad.

[@gmates8537]

why you dont have 200M subs

[@Ravit14]

Plz sir next video fast waiting

[@Salmanul_]

32:19 did you mean to write \gamma^{t}_{xx} instead of \gamma^{x}_{tt}?

[@eigenchris]

Yeah. My mistake.

[@lunam7249]

15:21 "tensors for beginers" just sounds so funny.....like "tumour ablation in lateral hippocampus defects".....

[@korwi7373]

OHHHHH AHHHH EHHHHH THIS SI SO GOOOD. DDDDDD

[@lunam7249]

yes..yes baby..rub it.... caress yourself.....higher order math is very sexy!!

[@phillipafrayan8313]

Sir in your video "Relativity 103e", from the frame of room, electrons are moving(11:40), it means that electrons must be contacted, hence there is a "-ve" charge density in wire, and that's why proton feel a force due to net "-ve" charge. So why peoples does not consider this? And why this net "-ve" charge didn't include in Lorentz force formula?

[@eigenchris]

I explained this badly, but this 2nd example involves electrons that are more "spread out" than the protons in the electron rest frame... and then when they are moving they length-contract to have equal density with the protons. I regret not making this clearer. Does this answer your question?

[@phillipafrayan8313]

@@eigenchrisSir this is not exactly my question's answer. Sir I am not taking about the 1st example but 2nd example. You were said in the 1st situation that there is no net charge density in wire despite having velocity of electron, but how is it possible because of length contraction. And if really there is a net charge density, then why force acting on the proton only due to the magnetic field (and also for velocity) why not both magnetic field and net charge density?

[@phillipafrayan8313]

Please sir i need your opinion.

[@eigenchris]

@@phillipafrayan8313 The 1st and 2nd examples are fundamentally different. In the 2nd example, the moving charges are already length-contracted, so that when they are in the rest frame, the charges are more spread out because they are not length-contracted.

[@khageshtanwar5089]

@@phillipafrayan8313 I posted the same question on the 103e video. I still dont have the answer, if anyone know please answer.

[@phillipafrayan8313]

Sir I think you must add "solution of Einstein Field Equation" into General Relativity videos.

[@eigenchris]

The 108 videos will be on the Schwarzschild solution.

[@phillipafrayan8313]

@@eigenchris ohh.

[@fethilatti5141]

Bien

[@timefactortheoryofgravitya7578]

General Relativity as you never imagined it: kzbin.info/www/bejne/e6WqdHqwZZyGaZo

[@couposanto]

One error at 11:07 W^{ u}

[@DmAlmazov]

Gauss' death year is 1855, not 1885. en.wikipedia.org/wiki/Carl_Friedrich_Gauss =)

[@ShadowZZZ]

Oh god not the Ricci tensors... Up until today I still don't understand it

[@eigenchris]

I don't feel I really get it either. I wish curvature was an easier concept to understand. Most other ideas in relativity I understand, but curvature is more complicated than it has a right to be, in my eyes.

[@jgvermeychuk]

@@eigenchris As Leonard Susskind once said, "General Relativity has the reputation of being very difficult. That's because its very difficult."

[@lunam7249]

😹😹😹🌝🧐👀 hahaha!!! you made me laugh!!❤❤ OMG!!! yes! another RICCI tensor! now pull down your pants!! your gonna get it!!!

[@canyadigit6274]

:)

[@jimtwisted1984]

Gauss lived yo be 108?

[@warrenchu6319]

Carl Friedrich Gauss lived from 1777 to 1855, not 1885.

[@user-te7rf8ik7z]

If your channel name is EigenChris, you should call all your children Chris, so Life(Chris) = n * Chris

[@anthonyC9199]

Spoilers

[@adamsniffen5187]

Gauss was not 108 when he died....1855*

[@homelesshendrix]

you were going good until you mentioned Gravity...