It was literally 40 years ago when I first ran across this material, in the context of General Relativity. I made sporadic attempts to study and understand the formulae, but it was beyond me, the texts were impenetrable, and I didn't have time to take a course. I've pecked at it on and off since then, to no success - until your work came along. You cannot imagine how grateful I am for what you're doing!
Thanks. My study of relativity also involved a lot of banging my head against the material, giving up for a while, and trying again. I'm hoping these videos make it easier for people.
@@eigenchris WHY AND HOW E=MC2 IS NECESSARILY AND CLEARLY F=MA ON BALANCE: Energy has/involves GRAVITY, AND ENERGY has/involves inertia/INERTIAL RESISTANCE. C4 is the proof of the fact that E=mc2 IS F=ma ON BALANCE. This explains the fourth dimension. TIME is NECESSARILY possible/potential AND actual IN BALANCE, AS E=MC2 IS F=MA ON BALANCE; AS ELECTROMAGNETISM/energy is gravity !!! The stars AND PLANETS are POINTS in the night sky. E=MC2 IS F=ma. ("Mass"/ENERGY IS GRAVITY. ELECTROMAGNETISM/energy is gravity.) The EARTH/ground AND what is THE SUN are CLEARLY (on balance) E=MC2 AS F=ma. TIME dilation ULTIMATELY proves ON BALANCE that E=MC2 IS F=ma IN BALANCE, AS ELECTROMAGNETISM/energy is gravity !!! (Gravity IS ELECTROMAGNETISM/energy.) The sky is blue, AND THE EARTH is ALSO BLUE. The stars AND PLANETS are POINTS in the night sky. E=MC2 IS F=ma ON BALANCE. Great !!! This NECESSARILY represents, INVOLVES, AND DESCRIBES what is possible/potential AND actual IN BALANCE, 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 F=ma ON BALANCE; AS ELECTROMAGNETISM/energy is gravity !!! (This ALSO perfectly explains why the rotation of WHAT IS THE MOON matches it's revolution.) It all CLEARLY makes perfect sense. BALANCE AND completeness go hand in hand. Stellar clustering ALSO proves ON BALANCE that ELECTROMAGNETISM/energy is gravity, AS E=MC2 IS F=ma. INDEED, HALF of the galaxies are "dead" or inert; AS ELECTROMAGNETISM/energy is gravity; AS E=MC2 IS F=ma ON BALANCE. Gravity AND ELECTROMAGNETISM/energy are linked AND BALANCED opposites, AS E=MC2 IS F=ma; AS ELECTROMAGNETISM/energy is gravity !! IMPORTANT !! Gravity IS ELECTROMAGNETISM/energy. 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. "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. Accordingly, objects fall at the SAME RATE (neglecting air resistance, of course). Moreover, 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 ELECTROMAGNETISM/energy is gravity; AS gravity/acceleration involves BALANCED inertia/INERTIAL RESISTANCE; AS E=MC2 IS F=ma. Gravity IS ELECTROMAGNETISM/energy. The stars AND PLANETS are POINTS in the night sky. Therefore, it is correct that the planets will (very, very, very slightly) move AWAY (ON BALANCE) from what is THE SUN !! Moreover, I have ALSO CLEARLY explained the cosmological redshift AND the "black holes". GRAVITATIONAL force/energy is proportional to (or BALANCED with/as) inertia/INERTIAL RESISTANCE, AS E=MC2 IS F=ma; AS ELECTROMAGNETISM/energy is gravity. NOW, ON BALANCE, carefully consider what is THE SUN. Very importantly, outer "space" involves full inertia; AND it is fully invisible AND black. SO, again, it ALL does CLEARLY make 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
@@zooj9401 Definitely yes. Teaching stuff forces you go over it very carefully and making sure you understand everything. Also, before starting these videos, I didn't really understand some of the basics of GR. I basically only really learned it through making these videos.
Same story 20 years ago, and I gave up physics. I ended up into ML and make good money tho. This stuff has been discomfort mystery for 2 decades until I find you. You are great of great, top of top. All science and engineering students must watch this channel.
A lot of people are pointing out that we use the extrinsic basis eX eY eZ in our calculation of the intrinsic metric tensor matrix, and feel this is a problem. With intrinsic geometry, we need a metric tensor to measure lengths/angles, so we need to get the metric from somewhere. One option is to invent a metric out of nowhere and use that. Another option is to calculate the metric using the surrounding extrinsic space. Once we have this metric, we can "forget" all about the extrinsic space and do all calculations intrinsically. But we need to start somewhere. For common surfaces like spheres, saddles, and donuts, we usually calculate the intrinsic metric with some help from the outside space, and then forget the outside space exists to do our length calculations.
I did an elective course at ANU, numerous PhD astrophysicists tried to explained this and I think you did a better job in teaching these ideas then all of them :) I don't think anyone in the class had an intuitive understanding of cristoffel symbols or geodesics like you have made so clear in this video.
UNDERSTANDING THE ULTIMATE, BALANCED, TOP DOWN, AND CLEAR MATHEMATICAL UNIFICATION OF ELECTROMAGNETISM/energy AND gravity, AS E=MC2 IS CLEARLY F=ma: The stars AND PLANETS are POINTS in the night sky. E=MC2 IS F=ma, AS this proves the term c4 from Einstein's field equations. SO, ON BALANCE, this proves the fourth dimension. ELECTROMAGNETISM/energy is gravity. Gravity IS ELECTROMAGNETISM/energy !!! TIME is NECESSARILY possible/potential AND actual IN BALANCE, AS E=MC2 IS F=ma; AS ELECTROMAGNETISM/energy is gravity. INDEED, TIME dilation ULTIMATELY proves ON BALANCE that E=MC2 IS F=ma; AS ELECTROMAGNETISM/energy is gravity. Gravity IS ELECTROMAGNETISM/energy. Gravity AND ELECTROMAGNETISM/energy are linked AND BALANCED opposites, AS E=MC2 IS F=ma; AS ELECTROMAGNETISM/energy is gravity; AS gravity/acceleration involves BALANCED inertia/INERTIAL RESISTANCE; AS GRAVITATIONAL force/ENERGY IS proportional to (or BALANCED with/as) inertia/INERTIAL RESISTANCE. Gravity IS ELECTROMAGNETISM/energy. E=mC2 IS CLEARLY F=ma. This NECESSARILY represents, INVOLVES, AND DESCRIBES what is possible/potential AND actual IN BALANCE, AS ELECTROMAGNETISM/energy is gravity. Gravity IS ELECTROMAGNETISM/energy !!! By Frank DiMeglio
@@OllytheOzzy Mr. Boris Stoyanov is a super bright and an HONEST physicist. He has agreed that the following post is "crystal clear": ELECTROMAGNETISM/energy is gravity. This is proven by F=ma AND E=mc2. Accordingly, gravity/acceleration involves balanced inertia/inertial resistance; 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. Gravity IS electromagnetism/energy. That objects fall at the same rate (neglecting air resistance, of course) PROVES that ELECTROMAGNETISM/energy is gravity. Think about it. By Frank DiMeglio
I much prefer this method of derivation using an embedding space, it is much easier to reason about, and then as you say, all of the details about the embedding space can be abstracted away after the fact.
I finally understand the relevance of travelling along a path at constant speed mentioned in previous lectures. Crystal-clear explanation throughout the lecture.
I dont mean to be offtopic but does anybody know a tool to get back into an Instagram account..? I was dumb lost my login password. I would appreciate any help you can offer me
@Alonzo Jonathan Thanks for your reply. I got to the site through google and Im in the hacking process now. Takes quite some time so I will reply here later when my account password hopefully is recovered.
I tend to learn in non-linear fashions, and I randomly stumbled across this as my first encounter with tensor calculus. This was beautifully done. So many new insights! Thank you so much.
ANOTHER SUPERB EXPLANATION OF A COMPLICATED SUBJECT. I FEEL I AM LEARNING THIS SLOWLY BUT SURELY AND I AM DOING IT JUST FOR FUN. I LOOK FORWARD TO YOUR EXAMPLE IN THE NEXT VIDEO. YOUR GRAPHICS AND PRESENTATION ARE FANTASTIC. THANKS FOR TAKING THE TIME TO DO THIS. YOU SHOULD GIVE A COURSE TO ALL MATH PROFESSORS AS TO HOW TO CLEARLY PRESENT COMPLICATED MATERIAL. AND YOU HAVE RIGHT HERE!
G.R. In my opinion, the most difficult part of the theory to understand, that is, to derive it mathematically and understand what it means geometrically, is the derivation of Christoffel Symbols and their geometric meaning. You explained this very well in terms of tangent and normal components. Your proof is very successful...
Great content, brother. Thank you for all the effort put into these videos. I imagine it must take a lot of time out of your day, but just wanted to let you know you have a great knack for sharing this complex material. Keep it up my man!
I'm taking an undergrad course in differential geometry and was close to panicking because I couldn't make sense of everything relating to the Gauss map, covariant derivatives, Christoffel symbols, second fundamental form, etc. This video made me "click" and I've just spent the last hour or so writing on top of my notes everything that's suddenly making sense, thanks to connecting it to what I see in the video. Thanks you a lot!!
No problem. The notes I link in the description are the source I learned from. A lot of relativity courses skip the more common-sense differential geometry of 2D surfaces in 3D space and go straight to the abstract Riemannian stuff, so those notes were very helpful to me.
@@eigenchris Einstein never nearly understood TIME, E=MC2, F=ma, gravity, or ELECTROMAGNETISM/energy. He was, in fact, a total weasel. c2 represents a dimension ON BALANCE, as E=MC2 IS F=ma in accordance with the following: UNDERSTANDING THE ULTIMATE, BALANCED, TOP DOWN, AND CLEAR MATHEMATICAL UNIFICATION OF ELECTROMAGNETISM/energy AND gravity, AS E=MC2 IS CLEARLY F=ma: The stars AND PLANETS are POINTS in the night sky. E=MC2 IS F=ma, AS this proves the term c4 from Einstein's field equations. SO, ON BALANCE, this proves the fourth dimension. ELECTROMAGNETISM/energy is gravity. Gravity IS ELECTROMAGNETISM/energy !!! TIME is NECESSARILY possible/potential AND actual IN BALANCE, AS E=MC2 IS F=ma; AS ELECTROMAGNETISM/energy is gravity. INDEED, TIME dilation ULTIMATELY proves ON BALANCE that E=MC2 IS F=ma; AS ELECTROMAGNETISM/energy is gravity. Gravity IS ELECTROMAGNETISM/energy. Gravity AND ELECTROMAGNETISM/energy are linked AND BALANCED opposites, AS E=MC2 IS F=ma; AS ELECTROMAGNETISM/energy is gravity; AS gravity/acceleration involves BALANCED inertia/INERTIAL RESISTANCE; AS GRAVITATIONAL force/ENERGY IS proportional to (or BALANCED with/as) inertia/INERTIAL RESISTANCE. Gravity IS ELECTROMAGNETISM/energy. E=mC2 IS CLEARLY F=ma. This NECESSARILY represents, INVOLVES, AND DESCRIBES what is possible/potential AND actual IN BALANCE, AS ELECTROMAGNETISM/energy is gravity. Gravity IS ELECTROMAGNETISM/energy !!! By Frank DiMeglio
Okay Eigenchris. Let me thank you with all my heart. I am a PhD student and this is incredible stuff really. To anyone watching, believe me this is the finest exposition on this topic. I have been through 100s of references, books and notes and what have you. Have spent close to 3 months trying to understand covariant derivative, geodesics, parallel transport etc. Now having understood these concepts from these videos suddenly the material in all the crazy abstract books, they have started to make sense. Btw Eigenchris, where did you learn this material from ? Also are you a math/physics professor ? Also coffee seems too small. Please start some crowdfunding thing so that we can help this channel.
Thanks for the high praise. I think my situation was similar to your where I went through many references (not 100, but probably 2-3 dozen) trying to understand these topics. The best one I found was this professor's notes: liavas.net/courses/math430/ These helped me realize there was a version of differential geometry of 2D surfaces by Gauss and others that was more "concrete" and intuitive than the abstract Riemannian Geometry that came afterward. My experience is that it is nearly impossible to understand Riemannian Geometry unless you study classical differential geometry a little bit first. But many books and courses just throw Riemannian Geometry at you, where everything is symbols and no pictures, and expect you to keep up and deal with it. I have a Bachelor's degree in engineering and physics but I'm not a professor. And I have a good full-time job that pays well, so I don't need much money. But I have the tip jar for anyone who wants to leave a small "thank-you".
@@eigenchris Thanks for the reply Eigenchris. So here is what my journey has been so far. As you said, its important to learn classical differential geometry first (from Barret O' Neil). And I did learn a lot of simple differential geometry and differential forms before studying tensors and Riemannian metric and such topics. While doing that, I went totally blank when I first saw the definition of wedge product. Then I came across this book by "Jon Pierre Fortney" called "A visual introduction to differential forms and calculus on manifolds" and it is perhaps the best exposition of differential forms and a beautiful introduction to calculus on manifolds. It is a must for anyone studying these topics. It does have its limitations as in it does not go all the way into Riemannian geometry. I also studied basic differential topology, regular value theorem etc, however Riemannian metric, geodesics and connections were still a challenge. Now that I have seen these videos, I went back to this book on Differential geometry by Boothby and suddenly a lot of explanation in the last chapter on covariant derivative made a lot of sense, in fact a combination of your exposition and Boothby's exposition has been enriching. I am also now going to study the notes that you have provided a link to.
@@tejasnatu90 I started making slides for a video on the wedge product, but I never finished it. I think almost everything involving the wedge product has nice pictures associated with it that make the algebra easier to understand. Perhaps I should try to finish those.
@@eigenchris Consider what is the man (AND THE EYE ON BALANCE) who IS standing on what is THE EARTH/ground. Consider TIME AND time dilation ON BALANCE. What is E=MC2 is taken directly from F=ma, AS the stars AND PLANETS are POINTS in the night sky ON BALANCE. This CLEARLY explains and proves the fourth dimension. c squared CLEARLY represents a dimension of SPACE ON BALANCE !!!! Indeed, E=mc2 is taken directly from F=ma; AS TIME is NECESSARILY possible/potential AND actual ON/IN BALANCE; AS ELECTROMAGNETISM/energy is CLEARLY AND NECESSARILY proven to be gravity (ON/IN BALANCE). Accordingly, ON BALANCE, the rotation of WHAT IS THE MOON matches it's revolution. Notice the TRANSLUCENT blue sky ON BALANCE. Consider what is THE EYE ON BALANCE, AS c squared CLEARLY represents a dimension of SPACE ON BALANCE. Gravity/acceleration involves BALANCED inertia/INERTIAL RESISTANCE, AS TIME is NECESSARILY possible/potential AND actual ON/IN BALANCE; AS ELECTROMAGNETISM/energy is CLEARLY AND NECESSARILY proven to be gravity (ON/IN BALANCE); AS what is E=MC2 is taken directly from F=ma; AS GRAVITATIONAL force/ENERGY is proportional to (or BALANCED with/as) inertia/INERTIAL RESISTANCE. “Mass”/ENERGY involves BALANCED inertia/INERTIAL RESISTANCE consistent with/as what is BALANCED electromagnetic/gravitational force/ENERGY, AS ELECTROMAGNETISM/energy is CLEARLY AND NECESSARILY proven to be gravity (ON/IN BALANCE). Accordingly, ON BALANCE, what are OBJECTS may fall at the SAME RATE. By Frank Martin DiMeglio
GOD BLESS YOU FOR MAKING THIS SERIES. Everything is so clear now! I knew that for geodesics, that was the equation but never thought that's the tangential acceleration that we are equating to zero! Lovely and elegant! Please create a series on general relativity too with Einstein's field equations!!
This is awesome !!! This is what I need to understand the Geodesic equation and apply it to understand the General Relativity. Thank you so so much for this video and maximum respect
You might be interested in the playlist I'm working on now for relativity. I just finished geodesics in special relativity (link below). Will work up to geodesics in general relativity in the next few months. th-cam.com/video/3LBitCErlBE/w-d-xo.html
The explanation of multilinear algebra and matrix operations here are transparent using familiar manifolds or classical differential geometry of surfaces. Can’t believe how deep he goes making things evident.
Excellent video. This series is absolutely fantastic. A small footnote - the RHS of the geodesic equation would only be zero in the case of an affine parameterization of the curve. If a non-affine parameter is used, the RHS will equal some vector that is parallel to the tangent vector being parallel transported. In plain speak, a car can brake and accelerate along a straight road whilst still maintaining a geodesic.
That makes sense. I'm not familiar with the non-affine parameterization along geodesics, so I tried to ensure I insisted on the "constant speed" condition.
I understand the geodesic equation clearly by this video. Thanks a lot for your graphic explanation in this video. I can say this video is really the masterpiece of tensor calculus.
The best lecture on Christoffel symbols I have ever watched or read so far! Question. In the second definition of Geodesic you say . But the vector of velocity is not constant in that straight line on the curved road in the video. The magnitude of velocity could be constant but not the vector. Should we say the speed is constant for the traveler on the curved surface who can not understand the curvature of his space?
I intentionally say "constant speed" and not "constant velocity". I take "velocity" to be a vector/arrow, and "speed" to be the magnitude/length of the velocity vector. So yes, constant speed means the length of the velocity vector stays the same, even though it might change direction. You could think of this as driving around in a car at 50 km/h, where you're allowed to turn the wheels, but you're not allowed to speed up or slow down.
Γειά σου Σωτήρη. Οπου και να ψάξεις θα βρείς Ελληνα. Ασχολούμε με Θεωρηική μηχανική Ηλεκτρομαγνητισμό και Γενική Θεωρία Σχετικότητας. Πουθενά δεν έβρισκα κάποια ολοκληρωμένη ανάλυση σε τανυστές και ο Eigenchris είναι πολύ καλός. Εσύ με τί ασχολείσαι;
- Revisited :) ... - Excellent. Thx. - Well presented. - Amazing how something that seems complicated is actually relatively simple when broken down into parts. Though of course prerequisite knowledge is required (e.g. vectors, calculus, etc.). - So, if one has the necessary background, the matter of the "geodesic", and "shortest distance" is w/i reach of understanding :) - And, the "fun" part is grasping HOW to think, not just what. Then, when considered in combination, the "mystery" is unraveled! :)
21:40 The equation is saying that the "observed" acceleration (the first term) should be opposite to the acceleration resulting from the shifting of the coordinate vectors (the second term). So in a sence it is saying all the observed acceleration (first term) is just due to curved coordinates being opposite to it. Compare to general relativity.
The second term can be understood as follows: When you couple the C. Symbol Gamma with two vectors, it tells you the rate of change of one vector When you move along the other vector (since we assume no-torsion the order of the vector does not matter, check a later video of this series). This Is just the definition of C. Symbol. In our case the velocity vector plays both roles. It is the vector that is changing as we move from point to point, And it also tells us the direction and speed where we move. So the second in the geodesic equations means verbally: "the rate of change of velocity vector as we move along the velocity vector". Nothing very complicated in principle I think. But the New concepts and all the indices may be confusing.
Is there an easier way to find a geodesic on a sphere? Place two points anywhere on a sphere. Using these any-two points you can define a great circle. Useing this great circle, define a cross-section. Viewing this cross-section you will see the two points between which you can trace a curve (following the outer edge of the sphere). And using the length of a curve formula you will determine the shortest distance and straightest path between these two points. No need for a metric tensor or Christoffel symbols. Thoughts?
@eigenchris At 13:23 I'm trying to get a picture in my head of the expression you underline (I'll call it E1): del^2 R/del u^i del u^j ... The first partial del R/del u^j gives the basis vector e sub j so (E1) becomes del e sub j / del u^i ... am I correct in visualizing this as the change in the basis vector e sub j as you move in the u^i direction (i.e. parallel transporting u sub j along u^i)? ... I know that the Christoffel Symbol is symmetric in the lower indices but by convention wouldn't E1 actually have the coordinates gamma k ,ji ? ... so understanding also that the order of differntiation doesn't matter should E1 start out as del^2 R/del u^j del u^i to get gamma k ,ij in the end?
I do that so I can group them together under a single basis vector ∂R/∂u^k. Summation indices are arbitrary. As long as you relabel them consistently, the meaning of the equation doesn't change. (Also note, each term in the sum is its own summation, so I can relabel the indices on the left term without changing the right term.)
1. At 19:03, why did the i's in the left of the R.H.S become k's? that doesn't seem to make sense... Therefore we also can't group the terms with respect to ∂R/∂u^k 2. How does the restriction to moving on the geodesic at a constant speed, come to use in the derivation of the equations? It seems like you stated the tangential component = 0 regardless of ||R'||
1. Sorry I forgot to mention this out loud. The first term of the expression is just a summation, so the summation index doesn't matter. We are free to relable it to be whatever we like. It's really just a sum of the u1 term and the u2 term. 2. If we speed up and slow down, we will get some acceleration in the tangential direction. Forcing the tangential compnent to be zero basically gives us the "constant speed" property for free.
I would like to point out one thing. At 9:19 you have chosen λ to be any parameter describing the curve. But, λ can't be any random parameter. It has to be the infinitesimal path length (or proportional to the infinitesimal path length). If you don't choose λ as proportional to the infinitesimal path length, then the tangent vector would not necessarily be of constant magnitude, and hence you get one more component of acceleration which is parallel to the velocity i.e. which increases the speed. So, we should define the tangent vector of constant magnitude as dR/dl, where dl is infinitesimal path length. For example, on a sphere take this Geodesic parametrized by (r, θ, Ф) = (1, t^2, 0). In this case, the acceleration has a component in the tangent space, more specifically it is along the geodesic itself. So, if you define acceleration vector as d^2R/dt^2 then you will get a non zero tangential component (although we are on a geodesic). But, if you define the tangent velocity vector, in this sphere case, as dR/dl and acceleration as d^2R/dl^2 then you get zero tangential acceleration.
This is implicit in the fact that the parameterisation is stipulated to be a constant speed parameterisation. The actual parameterisation needn't specifically be the integral of the infinitesimal path element lengths, which would result in traversing the path at speed equal to 1 unit; it could be any constant multiple of them, as the resulting parameterisation still traverses the curve at constant speed, just not 1 unit. In particular, scale the infinitesimals by _k,_ and the speed will be _k._
Your explanation on how to detect a geodesic path is excellent. Basically, you said a geodesic path has zero tangential acceleration when we travel along that path at constant speed. What I can't visualize is how tangential acceleration is not equal to zero as a car moves up/down and around a zigzag hill at constant speed. That is, I still can't see how tangential acceleration is not zero as a car moves through a non-geodesic path at constant speed. In flat space, if a car is not accelerating, its tangential acceleration will be zero regardless if the path is curved or not. Can you explain more or draw two vectors on a non-geodesic path like the zigzag hill and get the change in these two vectors to get the resultant acceleration? Maybe I might be able to see it clearer with more pictures.
In 7:25 you mentioned “…curves where the acceleration vector is normal to the surface”.is that just a definition or summarisation?Can you briefly describe the intuition behind this?
It's a definition. Intuitively it means, if you are driving a care, you are driving it as straight ahead as possible, without accelerating to the left or right. So the only accelerations that happen are up and down due to the bumps in the terrain.
@@eigenchris but if the car is ACCELERATING forward in the straightst possible line say from point a to b,then the tangential component(which i pressume is in the direction of travelling)aint zero，since youre acceletaring forwards… however you mentioned tangential acceleration =0 for geodesics.
hi Chris, this is the best course I've ever seen about this topic, thank you so much! Here I got a question about the geodesics: for a space endowed with a metric, does a geodesic connecting two points change if we choose a specific connection other than the Levi-cevita one? Since the geodesic equation contains the connection coefficients, I guess it would chage. For example for the boring connection, it seems the geodesics are just the section circles on the lattitude, but not the great circle. But from another point of view, since the geodesic is the light path for a space with certian metric, it maybe relavant only to the metric, no matter what connection is used. This confuses me. Thanks again for you wonderful job.
So if we use the Levi-Civita connection, I think the "geodesic between two points" naturally ends up being the "minimum distance" between the two points (or "maximum proper time" if we're in spacetime). However if we use another type of non-Levi-Civita connection, I'm pretty sure we lose this "minimum distance" meaning, since the connection no longer comes from the metric. The connection DEFINES what it means to parallel transport a vector. The definition of geodesic according to the connection is "parallel transporting a vector along itself" (i.e. marching forward as straight as possible). So yes, you're right. Geodesics will look different under different connections.
At 8:38, could you explain how you wrote the function of the curve in terms of lambda? I'm confused how you wrote u= lambda and v= lambda and then directly substituted lambda for u,v
Fantastic teaching overall! But I have a question: @14:30 you said something like since we don't know the exact components at all, so we are just going to make them up since they are unknown.😅. I don't understand the reasoning behind such a statement.
I am afraid it is a typo here 18:23, because index i or j appears three times. If it is a typo, how to correct it? Thanks. I am new to tensor calculus, and I have learned a lot from your videoes. Thanks! :)
I wonder, why you choose for the horizontal axis the second variable (later labelled u2). Do you have any reason for it? Many thanks for your great work❤.
Why are all summations between covariant and contravariant components? I would have said that it's to keep lengths and angles invariant. Is that true? Also, are there cases where you have to sum two covariant or two contravariant components?
When you have a sum between a covariant and a convariant index, the result is always invariant. When you change coordinates, the covariant components transform one way (e.g. matrix M), and the contravariant components will transform in the opposite way (e.g. matrix inverse M^-1). The two transformations cancel out and the summation result remains the same. If you are summing two covariant or summing two contravariant components, that's usually a sign you're doing something wrong.
@eigenchris A follow-up question: When you take the trace of the metric tensor, don't you need to sum over two covariant indices? If I had to guess, I would have said that the metric tensor has no trace. Is that true?
The trace is always defined using a covariant and contravariant index. So if you have a tensor with 2 of the same kind of index, you need to raise or lower one of the indices using the metric. In the case of the metric itself, since a lower index is raised using the inverse metric, the (1,1) form of the metric is just the kronecker delta, or identity matrix... and the trace of the identity just equals the dimension of the space. So the trace of the metric is just a way of stating what dimension the space is.
Question, if the Christofell symbols are essentially components with respect to the basis vectors to in the tangent plane and one normal to it, how do I reconcile that with the fact that they are partial derivatives of the metric, and not just like ordinary components which would mean say a certain number of basis vector one plus a certain number of basic vector two. I thought they were used in parallel propagation because the coordinates may be bending and shifting and twisting and turning underneath vectors that are the same, but that would have different components in these different curved coordinate systems. Any help please?
You might be interested in watching videos 17-20, which go through the covariant derivative in a lot of detail and derive the Christoffel symbols in terms of the metric tensor towards the end. The formula in this video has the 2nd derivative of R in it, which means the Christoffel symbols depend not only on the basis vectors (1st derivative), but also on how the basis vectors are changing in the local area (2nd derivative). Christoffel symbols do indeed represent the "bending/shifting/twisting" of basis vectors from point to point. Christoffel symbols can appear in flat space when coordinates are curved, but the space itself has no curvature. However the Christoffel symbols can also tell us if the space is curved (see videos 22-23 on the Riemann curvature tensor for that). Does that answer your question?
@@eigenchris thank you sir. I appreciate your tremendous intellect, your patience, and your obvious and substantial teaching acumen. Thank your quick response to my inquiry. You provide a valuable service whether you realize it or not. intellectual curiosity and scientific inquiry are sadly lacking in a world Replete and content with ignorance. On behalf of the other rabid GR fans, thank you! 🤘
Very interesting way, but I wonder. What are the benefits of doing this way, instead of minimizing the integral that gives distance over a surface with Euler Lagrange ? I bet it would give something very similar.
I think you can derive the geodesic equation from the Euler-Lagrange equations. While E-L is very powerful for getting a lot of results fast, I find it usually lacks intuition. The result kind of just pops out and you have no idea why. I wanted to show you can get the same result using some common-sense rules.
"In flat space a straight path has zero acceleration when we travel along it at constant speed"(3:39) ...shouldn't it be 'zero acceleration at every point along the path'.?....since we can add all the acceleration vectors and get zero ...even if the path is curved..... Also if the speed is not constant then there can be straight paths even for non zero acceleration....then can we say that if all the velocity vectors are parallel to each other then the path is straight?
For your first question: yes, this is basically what I meant. For your second question: there are other ways to define a straight path, but the "zero tangential acceleration" definition is the definition I used.
@@eigenchris Thanks chris for clarifying. I am very grateful to you for working so hard to make this series possible . Also can you suggest any lecture series or online material to learn GTR.
Awesome lecture.. But I have a slight confusion: 1.Is geodesic a length minimising curve or time minimising curve.? 2. Is a geodesic on curved manifold is literally a straight path just like straight path in flat space.?
Geodesics can exist in space and spacetime. Usually in space, geodesics minimize length. And usually in spacetime, geodesics maximize length (or proper time, more technically). I'd say a geodesics is the "straightest possible path" on a curved surface. You can't really draw a straight path on a sphere.
@@eigenchris you said that geodesic in space minimise the length and geodesic in spacetime maximise the length, but i am little bit confused whether it is space or spacetime, geodesic must minimise the length of curve, isn't it so.
Sorry, I don't understand the question. Are you saying you don't know how to tell the difference between space and spacetime? It depends on the signs in the metric. In space, the diagonal components would be (+ + +). In spacetime, there would be a sign change for the time dimension, either (+ - - - ) or ( - + + +).
If n is any tangent vector to a geodesic and s is the arc length, then dn/ds = 0. But what if pi = 3 in this space? Or what if the space is spongy so that ds1 != ds2 at point 1 and point 2?
There is a great idea! For the dark side of the Universe - suppose that it consists of short-term interactions in long-lived fractal networks, the smallest quantum operators in energy, spherical rosebuds, consisting of a large set; 1 - rolled into a sphere, 2 - half collapsed into a sphere and 3 - flat, vibrating quantum membranes relative to their working centers in the sphere
I'm not aware if there's a generalization to N-D space. The 2nd fundamental form goes back to the differential geometry of Gauss, which focused on 2D surfaces embedded in 3D space. The n-dimensional generalization (Riemannian geometry) came later, and I don't know if there's an equivalent of the 2nd fundamental form.
At 12:52, I think you've misidentified the terms; in the second line the 1st and 4th terms are of the single summation form, whereas the 2nd and 3rd aree of the double summation forms. I wouldn't be paying such close attention if I wasn't very interested in your presentation.
It was literally 40 years ago when I first ran across this material, in the context of General Relativity. I made sporadic attempts to study and understand the formulae, but it was beyond me, the texts were impenetrable, and I didn't have time to take a course. I've pecked at it on and off since then, to no success - until your work came along. You cannot imagine how grateful I am for what you're doing!
Thanks. My study of relativity also involved a lot of banging my head against the material, giving up for a while, and trying again. I'm hoping these videos make it easier for people.
@@eigenchris WHY AND HOW E=MC2 IS NECESSARILY AND CLEARLY F=MA ON BALANCE:
Energy has/involves GRAVITY, AND ENERGY has/involves inertia/INERTIAL RESISTANCE. C4 is the proof of the fact that E=mc2 IS F=ma ON BALANCE. This explains the fourth dimension. TIME is NECESSARILY possible/potential AND actual IN BALANCE, AS E=MC2 IS F=MA ON BALANCE; AS ELECTROMAGNETISM/energy is gravity !!! The stars AND PLANETS are POINTS in the night sky. E=MC2 IS F=ma. ("Mass"/ENERGY IS GRAVITY. ELECTROMAGNETISM/energy is gravity.) The EARTH/ground AND what is THE SUN are CLEARLY (on balance) E=MC2 AS F=ma. TIME dilation ULTIMATELY proves ON BALANCE that E=MC2 IS F=ma IN BALANCE, AS ELECTROMAGNETISM/energy is gravity !!! (Gravity IS ELECTROMAGNETISM/energy.) The sky is blue, AND THE EARTH is ALSO BLUE. The stars AND PLANETS are POINTS in the night sky. E=MC2 IS F=ma ON BALANCE. Great !!! This NECESSARILY represents, INVOLVES, AND DESCRIBES what is possible/potential AND actual IN BALANCE, 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 F=ma ON BALANCE; AS ELECTROMAGNETISM/energy is gravity !!! (This ALSO perfectly explains why the rotation of WHAT IS THE MOON matches it's revolution.) It all CLEARLY makes perfect sense. BALANCE AND completeness go hand in hand. Stellar clustering ALSO proves ON BALANCE that ELECTROMAGNETISM/energy is gravity, AS E=MC2 IS F=ma. INDEED, HALF of the galaxies are "dead" or inert; AS ELECTROMAGNETISM/energy is gravity; AS E=MC2 IS F=ma ON BALANCE.
Gravity AND ELECTROMAGNETISM/energy are linked AND BALANCED opposites, AS E=MC2 IS F=ma; AS ELECTROMAGNETISM/energy is gravity !! IMPORTANT !! Gravity IS ELECTROMAGNETISM/energy. 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. "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. Accordingly, objects fall at the SAME RATE (neglecting air resistance, of course). Moreover, 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 ELECTROMAGNETISM/energy is gravity; AS gravity/acceleration involves BALANCED inertia/INERTIAL RESISTANCE; AS E=MC2 IS F=ma. Gravity IS ELECTROMAGNETISM/energy. The stars AND PLANETS are POINTS in the night sky. Therefore, it is correct that the planets will (very, very, very slightly) move AWAY (ON BALANCE) from what is THE SUN !! Moreover, I have ALSO CLEARLY explained the cosmological redshift AND the "black holes". GRAVITATIONAL force/energy is proportional to (or BALANCED with/as) inertia/INERTIAL RESISTANCE, AS E=MC2 IS F=ma; AS ELECTROMAGNETISM/energy is gravity. NOW, ON BALANCE, carefully consider what is THE SUN. Very importantly, outer "space" involves full inertia; AND it is fully invisible AND black. SO, again, it ALL does CLEARLY make 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
@@eigenchris Have you learned relativity better by doing these videos? Understanding concepts even better?
@@zooj9401 Definitely yes. Teaching stuff forces you go over it very carefully and making sure you understand everything. Also, before starting these videos, I didn't really understand some of the basics of GR. I basically only really learned it through making these videos.
Same story 20 years ago, and I gave up physics. I ended up into ML and make good money tho. This stuff has been discomfort mystery for 2 decades until I find you. You are great of great, top of top. All science and engineering students must watch this channel.
A lot of people are pointing out that we use the extrinsic basis eX eY eZ in our calculation of the intrinsic metric tensor matrix, and feel this is a problem. With intrinsic geometry, we need a metric tensor to measure lengths/angles, so we need to get the metric from somewhere. One option is to invent a metric out of nowhere and use that. Another option is to calculate the metric using the surrounding extrinsic space. Once we have this metric, we can "forget" all about the extrinsic space and do all calculations intrinsically. But we need to start somewhere. For common surfaces like spheres, saddles, and donuts, we usually calculate the intrinsic metric with some help from the outside space, and then forget the outside space exists to do our length calculations.
I did an elective course at ANU, numerous PhD astrophysicists tried to explained this and I think you did a better job in teaching these ideas then all of them :) I don't think anyone in the class had an intuitive understanding of cristoffel symbols or geodesics like you have made so clear in this video.
UNDERSTANDING THE ULTIMATE, BALANCED, TOP DOWN, AND CLEAR MATHEMATICAL UNIFICATION OF ELECTROMAGNETISM/energy AND gravity, AS E=MC2 IS CLEARLY F=ma:
The stars AND PLANETS are POINTS in the night sky. E=MC2 IS F=ma, AS this proves the term c4 from Einstein's field equations. SO, ON BALANCE, this proves the fourth dimension. ELECTROMAGNETISM/energy is gravity. Gravity IS ELECTROMAGNETISM/energy !!!
TIME is NECESSARILY possible/potential AND actual IN BALANCE, AS E=MC2 IS F=ma; AS ELECTROMAGNETISM/energy is gravity. INDEED, TIME dilation ULTIMATELY proves ON BALANCE that E=MC2 IS F=ma; AS ELECTROMAGNETISM/energy is gravity. Gravity IS ELECTROMAGNETISM/energy.
Gravity AND ELECTROMAGNETISM/energy are linked AND BALANCED opposites, AS E=MC2 IS F=ma; AS ELECTROMAGNETISM/energy is gravity; AS gravity/acceleration involves BALANCED inertia/INERTIAL RESISTANCE; AS GRAVITATIONAL force/ENERGY IS proportional to (or BALANCED with/as) inertia/INERTIAL RESISTANCE. Gravity IS ELECTROMAGNETISM/energy.
E=mC2 IS CLEARLY F=ma. This NECESSARILY represents, INVOLVES, AND DESCRIBES what is possible/potential AND actual IN BALANCE, AS ELECTROMAGNETISM/energy is gravity. Gravity IS ELECTROMAGNETISM/energy !!!
By Frank DiMeglio
@@OllytheOzzy Mr. Boris Stoyanov is a super bright and an HONEST physicist. He has agreed that the following post is "crystal clear": ELECTROMAGNETISM/energy is gravity. This is proven by F=ma AND E=mc2. Accordingly, gravity/acceleration involves balanced inertia/inertial resistance; 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. Gravity IS electromagnetism/energy. That objects fall at the same rate (neglecting air resistance, of course) PROVES that ELECTROMAGNETISM/energy is gravity. Think about it. By Frank DiMeglio
I much prefer this method of derivation using an embedding space, it is much easier to reason about, and then as you say, all of the details about the embedding space can be abstracted away after the fact.
I finally understand the relevance of travelling along a path at constant speed mentioned in previous lectures.
Crystal-clear explanation throughout the lecture.
I dont mean to be offtopic but does anybody know a tool to get back into an Instagram account..?
I was dumb lost my login password. I would appreciate any help you can offer me
@Keagan Trey Instablaster :)
@Alonzo Jonathan Thanks for your reply. I got to the site through google and Im in the hacking process now.
Takes quite some time so I will reply here later when my account password hopefully is recovered.
@Alonzo Jonathan it did the trick and I now got access to my account again. Im so happy:D
Thanks so much you really help me out :D
@Keagan Trey no problem xD
I tend to learn in non-linear fashions, and I randomly stumbled across this as my first encounter with tensor calculus. This was beautifully done. So many new insights! Thank you so much.
One of the most precise treatment of the subject. Everything flows naturally.
Outstanding treatment, very important for our generation. Please just bookify your videos and many people will buy your book…
ANOTHER SUPERB EXPLANATION OF A COMPLICATED SUBJECT. I FEEL I AM LEARNING THIS SLOWLY BUT SURELY AND I AM DOING IT JUST FOR FUN. I LOOK FORWARD TO YOUR EXAMPLE IN THE NEXT VIDEO. YOUR GRAPHICS AND PRESENTATION ARE FANTASTIC. THANKS FOR TAKING THE TIME TO DO THIS. YOU SHOULD GIVE A COURSE TO ALL MATH PROFESSORS AS TO HOW TO CLEARLY PRESENT COMPLICATED MATERIAL. AND YOU HAVE RIGHT HERE!
G.R. In my opinion, the most difficult part of the theory to understand, that is, to derive it mathematically and understand what it means geometrically, is the derivation of Christoffel Symbols and their geometric meaning. You explained this very well in terms of tangent and normal components. Your proof is very successful...
Finally I understood Christoffel symbols after years. Thank you.
Great content, brother. Thank you for all the effort put into these videos. I imagine it must take a lot of time out of your day, but just wanted to let you know you have a great knack for sharing this complex material.
Keep it up my man!
El nivel de tus explicaciones es admirable! Permite entender conceptos complejos de una forma muy simple. Muchas Gracias !
I'm taking an undergrad course in differential geometry and was close to panicking because I couldn't make sense of everything relating to the Gauss map, covariant derivatives, Christoffel symbols, second fundamental form, etc. This video made me "click" and I've just spent the last hour or so writing on top of my notes everything that's suddenly making sense, thanks to connecting it to what I see in the video. Thanks you a lot!!
No problem. The notes I link in the description are the source I learned from. A lot of relativity courses skip the more common-sense differential geometry of 2D surfaces in 3D space and go straight to the abstract Riemannian stuff, so those notes were very helpful to me.
Absolutely beautiful! It's so important to first understand the math for understanding the actual physics, and u've done such a artistic presentation
One of the greatest educational videos ever made.
In order to commit this content to memory, I will watch this pedagogically excellent video several times.
Oh my God, this is awesome
Thanks.
Hi Reimann!
@@maxwellsequation4887 hey, how you doing
@@eigenchris Einstein never nearly understood TIME, E=MC2, F=ma, gravity, or ELECTROMAGNETISM/energy.
He was, in fact, a total weasel.
c2 represents a dimension ON BALANCE, as E=MC2 IS F=ma in accordance with the following:
UNDERSTANDING THE ULTIMATE, BALANCED, TOP DOWN, AND CLEAR MATHEMATICAL UNIFICATION OF ELECTROMAGNETISM/energy AND gravity, AS E=MC2 IS CLEARLY F=ma:
The stars AND PLANETS are POINTS in the night sky. E=MC2 IS F=ma, AS this proves the term c4 from Einstein's field equations. SO, ON BALANCE, this proves the fourth dimension. ELECTROMAGNETISM/energy is gravity. Gravity IS ELECTROMAGNETISM/energy !!!
TIME is NECESSARILY possible/potential AND actual IN BALANCE, AS E=MC2 IS F=ma; AS ELECTROMAGNETISM/energy is gravity. INDEED, TIME dilation ULTIMATELY proves ON BALANCE that E=MC2 IS F=ma; AS ELECTROMAGNETISM/energy is gravity. Gravity IS ELECTROMAGNETISM/energy.
Gravity AND ELECTROMAGNETISM/energy are linked AND BALANCED opposites, AS E=MC2 IS F=ma; AS ELECTROMAGNETISM/energy is gravity; AS gravity/acceleration involves BALANCED inertia/INERTIAL RESISTANCE; AS GRAVITATIONAL force/ENERGY IS proportional to (or BALANCED with/as) inertia/INERTIAL RESISTANCE. Gravity IS ELECTROMAGNETISM/energy.
E=mC2 IS CLEARLY F=ma. This NECESSARILY represents, INVOLVES, AND DESCRIBES what is possible/potential AND actual IN BALANCE, AS ELECTROMAGNETISM/energy is gravity. Gravity IS ELECTROMAGNETISM/energy !!!
By Frank DiMeglio
2-22-19. This is just so great ; it should be in all the textbooks and GR books. Thank you so much!
Okay Eigenchris. Let me thank you with all my heart. I am a PhD student and this is incredible stuff really.
To anyone watching, believe me this is the finest exposition on this topic. I have been through 100s of references, books and notes and what have you. Have spent close to 3 months trying to understand covariant derivative, geodesics, parallel transport etc. Now having understood these concepts from these videos suddenly the material in all the crazy abstract books, they have started to make sense.
Btw Eigenchris, where did you learn this material from ? Also are you a math/physics professor ?
Also coffee seems too small. Please start some crowdfunding thing so that we can help this channel.
Thanks for the high praise. I think my situation was similar to your where I went through many references (not 100, but probably 2-3 dozen) trying to understand these topics. The best one I found was this professor's notes: liavas.net/courses/math430/
These helped me realize there was a version of differential geometry of 2D surfaces by Gauss and others that was more "concrete" and intuitive than the abstract Riemannian Geometry that came afterward. My experience is that it is nearly impossible to understand Riemannian Geometry unless you study classical differential geometry a little bit first. But many books and courses just throw Riemannian Geometry at you, where everything is symbols and no pictures, and expect you to keep up and deal with it.
I have a Bachelor's degree in engineering and physics but I'm not a professor. And I have a good full-time job that pays well, so I don't need much money. But I have the tip jar for anyone who wants to leave a small "thank-you".
@@eigenchris Thanks for the reply Eigenchris. So here is what my journey has been so far. As you said, its important to learn classical differential geometry first (from Barret O' Neil). And I did learn a lot of simple differential geometry and differential forms before studying tensors and Riemannian metric and such topics. While doing that, I went totally blank when I first saw the definition of wedge product. Then I came across this book by "Jon Pierre Fortney" called "A visual introduction to differential forms and calculus on manifolds" and it is perhaps the best exposition of differential forms and a beautiful introduction to calculus on manifolds. It is a must for anyone studying these topics. It does have its limitations as in it does not go all the way into Riemannian geometry. I also studied basic differential topology, regular value theorem etc, however Riemannian metric, geodesics and connections were still a challenge. Now that I have seen these videos, I went back to this book on Differential geometry by Boothby and suddenly a lot of explanation in the last chapter on covariant derivative made a lot of sense, in fact a combination of your exposition and Boothby's exposition has been enriching. I am also now going to study the notes that you have provided a link to.
@@tejasnatu90 I started making slides for a video on the wedge product, but I never finished it. I think almost everything involving the wedge product has nice pictures associated with it that make the algebra easier to understand. Perhaps I should try to finish those.
@@eigenchris Consider what is the man (AND THE EYE ON BALANCE) who IS standing on what is THE EARTH/ground. Consider TIME AND time dilation ON BALANCE. What is E=MC2 is taken directly from F=ma, AS the stars AND PLANETS are POINTS in the night sky ON BALANCE. This CLEARLY explains and proves the fourth dimension. c squared CLEARLY represents a dimension of SPACE ON BALANCE !!!! Indeed, E=mc2 is taken directly from F=ma; AS TIME is NECESSARILY possible/potential AND actual ON/IN BALANCE; AS ELECTROMAGNETISM/energy is CLEARLY AND NECESSARILY proven to be gravity (ON/IN BALANCE). Accordingly, ON BALANCE, the rotation of WHAT IS THE MOON matches it's revolution. Notice the TRANSLUCENT blue sky ON BALANCE. Consider what is THE EYE ON BALANCE, AS c squared CLEARLY represents a dimension of SPACE ON BALANCE. Gravity/acceleration involves BALANCED inertia/INERTIAL RESISTANCE, AS TIME is NECESSARILY possible/potential AND actual ON/IN BALANCE; AS ELECTROMAGNETISM/energy is CLEARLY AND NECESSARILY proven to be gravity (ON/IN BALANCE); AS what is E=MC2 is taken directly from F=ma; AS GRAVITATIONAL force/ENERGY is proportional to (or BALANCED with/as) inertia/INERTIAL RESISTANCE. “Mass”/ENERGY involves BALANCED inertia/INERTIAL RESISTANCE consistent with/as what is BALANCED electromagnetic/gravitational force/ENERGY, AS ELECTROMAGNETISM/energy is CLEARLY AND NECESSARILY proven to be gravity (ON/IN BALANCE). Accordingly, ON BALANCE, what are OBJECTS may fall at the SAME RATE.
By Frank Martin DiMeglio
Best explaination of Christoffel Symbols applied to GR. Also loved your series on
"Spinors by eigenchris".
Man, you have some talent for explaining things. Thanks for sharing your knowledge!
I asked you for this video and you make it; that's pure gold. thank you.
This is a tour de force. What a lucid explanation of the Christoffel symbols!
Best Video that explains the christoffel symbols. Thank you so much for this great work.
Best tutorial ever on geodesics period.
Clearest derivation of the geodesics eq. I ever have seen. ❤
Thank you very much! I'm a Chinese and not good at English, but I still can understand your video. It's so awesome!
Thanks a lot for the lectures link. This lady prof author is a genius, yet very modest.. she doesn't even put her name on the material.
GOD BLESS YOU FOR MAKING THIS SERIES. Everything is so clear now! I knew that for geodesics, that was the equation but never thought that's the tangential acceleration that we are equating to zero! Lovely and elegant! Please create a series on general relativity too with Einstein's field equations!!
I am working on Relativity videos now. Galilean Relativity is done. Special and General Relativity are coming next.
@@eigenchris that's great, looking forward to watch them too. I can't thank enough for making these videos!!
I bought all your ebooks at Amazon days ago. Thank you for writing them.
I'm not sure what you mean. I've never written a book.
eigenchris I confuse Chris in your username with another TH-camr. My apology.
I just cannot believe how well explained this is!!
Thank u so much, this is awesome!!!
This is awesome !!! This is what I need to understand the Geodesic equation and apply it to understand the General Relativity. Thank you so so much for this video and maximum respect
You might be interested in the playlist I'm working on now for relativity. I just finished geodesics in special relativity (link below). Will work up to geodesics in general relativity in the next few months.
th-cam.com/video/3LBitCErlBE/w-d-xo.html
The explanation of multilinear algebra and matrix operations here are transparent using familiar manifolds or classical differential geometry of surfaces. Can’t believe how deep he goes making things evident.
Excellent video. This series is absolutely fantastic. A small footnote - the RHS of the geodesic equation would only be zero in the case of an affine parameterization of the curve. If a non-affine parameter is used, the RHS will equal some vector that is parallel to the tangent vector being parallel transported. In plain speak, a car can brake and accelerate along a straight road whilst still maintaining a geodesic.
That makes sense. I'm not familiar with the non-affine parameterization along geodesics, so I tried to ensure I insisted on the "constant speed" condition.
Very clear. Better than some books on the topic
That was amazing! Great job, you're a modern day Grossman,
really awesome . really helps. thanks for your fabulous course. look foward to following courses
I'm here from The Portal Book Club - studying Roger Penrose's The Road to Reality. Fantastic and lucid work eigenchris.
Fantastic book. Fantastic lecture series.
I understand the geodesic equation clearly by this video.
Thanks a lot for your graphic explanation in this video.
I can say this video is really the masterpiece of tensor calculus.
Great explanation of a difficult subject. Great series!
I finally understand the mathematics behind geodesics. Thank you.
The best lecture on Christoffel symbols I have ever watched or read so far!
Question. In the second definition of Geodesic you say . But the vector of velocity is not constant in that straight line on the curved road in the video. The magnitude of velocity could be constant but not the vector.
Should we say the speed is constant for the traveler on the curved surface who can not understand the curvature of his space?
I intentionally say "constant speed" and not "constant velocity". I take "velocity" to be a vector/arrow, and "speed" to be the magnitude/length of the velocity vector. So yes, constant speed means the length of the velocity vector stays the same, even though it might change direction. You could think of this as driving around in a car at 50 km/h, where you're allowed to turn the wheels, but you're not allowed to speed up or slow down.
Thanks for your answer and your time.
Γειά σου Σωτήρη. Οπου και να ψάξεις θα βρείς Ελληνα. Ασχολούμε με Θεωρηική μηχανική Ηλεκτρομαγνητισμό και Γενική Θεωρία Σχετικότητας. Πουθενά δεν έβρισκα κάποια ολοκληρωμένη ανάλυση σε τανυστές και ο Eigenchris είναι πολύ καλός. Εσύ με τί ασχολείσαι;
Καθηγητής .Ξεκίνησα π’ερυσι στο ΕΑΠ προχωρημένες σπουδές στη Φυσική.
Τον παρακολουθώ κι εγώ εδώ και καιρό, είναι όντως πολύ καλός.
- Revisited :) ...
- Excellent. Thx.
- Well presented.
- Amazing how something that seems complicated is actually relatively simple when broken down into parts. Though of course prerequisite knowledge is required (e.g. vectors, calculus, etc.).
- So, if one has the necessary background, the matter of the "geodesic", and "shortest distance" is w/i reach of understanding :)
- And, the "fun" part is grasping HOW to think, not just what. Then, when considered in combination, the "mystery" is unraveled! :)
21:40 The equation is saying that the "observed" acceleration (the first term) should be opposite to the acceleration resulting from the shifting of the coordinate vectors (the second term).
So in a sence it is saying all the observed acceleration (first term) is just due to curved coordinates being opposite to it.
Compare to general relativity.
The second term can be understood as follows: When you couple the C. Symbol Gamma with two vectors, it tells you the rate of change of one vector When you move along the other vector (since we assume no-torsion the order of the vector does not matter, check a later video of this series). This Is just the definition of C. Symbol.
In our case the velocity vector plays both roles. It is the vector that is changing as we move from point to point, And it also tells us the direction and speed where we move. So the second in the geodesic equations means verbally: "the rate of change of velocity vector as we move along the velocity vector".
Nothing very complicated in principle I think. But the New concepts and all the indices may be confusing.
Absolutely fantastically presented lecture, thank you very much!
Is there an easier way to find a geodesic on a sphere?
Place two points anywhere on a sphere.
Using these any-two points you can define a great circle.
Useing this great circle, define a cross-section.
Viewing this cross-section you will see the two points between which you can trace a curve (following the outer edge of the sphere).
And using the length of a curve formula you will determine the shortest distance and straightest path between these two points.
No need for a metric tensor or Christoffel symbols.
Thoughts?
These videos are great. I love the visualizations, thanks!
What a fantastic video once again. Thank you so much
11:20 You could’ve clarified that the second derivative of the velocity with respect to lambda would also be tangent.
The colors help a lot with the Christoffel Symbols
This video is absolutely brilliant. Thank you for your service
Intuitive way of explaining it thanks!
Incredible, thank you very much for this so beautiful and didactic video.
Nice video. Looking forward to the example.
Absolutely brilliant, thank you so much for uploading
Very nice and clear explanation, very cool! Thanks.
This is mind blowingly good. Incredible. Thanks for your efforts; you're inspiring a lot of people!
at 19:09, why did the d/dui terms turn into d/duk terms for the first half of the equation?
@eigenchris At 13:23 I'm trying to get a picture in my head of the expression you underline (I'll call it E1): del^2 R/del u^i del u^j ... The first partial del R/del u^j gives the basis vector e sub j so (E1) becomes del e sub j / del u^i ... am I correct in visualizing this as the change in the basis vector e sub j as you move in the u^i direction (i.e. parallel transporting u sub j along u^i)? ... I know that the Christoffel Symbol is symmetric in the lower indices but by convention wouldn't E1 actually have the coordinates gamma k ,ji ? ... so understanding also that the order of differntiation doesn't matter should E1 start out as del^2 R/del u^j del u^i to get gamma k ,ij in the end?
19:05 why did the (i) indices in the first term of the original equation turn into index (k)
I do that so I can group them together under a single basis vector ∂R/∂u^k. Summation indices are arbitrary. As long as you relabel them consistently, the meaning of the equation doesn't change.
(Also note, each term in the sum is its own summation, so I can relabel the indices on the left term without changing the right term.)
Dayum, this video is sooo precious....
1. At 19:03, why did the i's in the left of the R.H.S become k's? that doesn't seem to make sense... Therefore we also can't group the terms with respect to ∂R/∂u^k
2. How does the restriction to moving on the geodesic at a constant speed, come to use in the derivation of the equations? It seems like you stated the tangential component = 0 regardless of ||R'||
1. Sorry I forgot to mention this out loud. The first term of the expression is just a summation, so the summation index doesn't matter. We are free to relable it to be whatever we like. It's really just a sum of the u1 term and the u2 term.
2. If we speed up and slow down, we will get some acceleration in the tangential direction. Forcing the tangential compnent to be zero basically gives us the "constant speed" property for free.
Thank you so much for the derivation. What is wrong with teachers? Why do they pretend that this is not the way people think to understand stuff?
11:35 I think you mean the first derivatives of position vector (rather than velocity vector) are always tangent vectors.
Beautiful explanation!!!! Thank you so much for sharing!!
I would like to point out one thing. At 9:19 you have chosen λ to be any parameter describing the curve. But, λ can't be any random parameter. It has to be the infinitesimal path length (or proportional to the infinitesimal path length). If you don't choose λ as proportional to the infinitesimal path length, then the tangent vector would not necessarily be of constant magnitude, and hence you get one more component of acceleration which is parallel to the velocity i.e. which increases the speed.
So, we should define the tangent vector of constant magnitude as dR/dl, where dl is infinitesimal path length.
For example, on a sphere take this Geodesic parametrized by (r, θ, Ф) = (1, t^2, 0). In this case, the acceleration has a component in the tangent space, more specifically it is along the geodesic itself. So, if you define acceleration vector as d^2R/dt^2 then you will get a non zero tangential component (although we are on a geodesic).
But, if you define the tangent velocity vector, in this sphere case, as dR/dl and acceleration as d^2R/dl^2 then you get zero tangential acceleration.
This is implicit in the fact that the parameterisation is stipulated to be a constant speed parameterisation. The actual parameterisation needn't specifically be the integral of the infinitesimal path element lengths, which would result in traversing the path at speed equal to 1 unit; it could be any constant multiple of them, as the resulting parameterisation still traverses the curve at constant speed, just not 1 unit. In particular, scale the infinitesimals by _k,_ and the speed will be _k._
i love you man. so beautifully explained
Keep up the great work. Very well done
thanks for this wonderful lecture
sorry i want to ask 16:59 metric tenor is lower index but the original k l is upper index ?
is dR/u^k = e underindex k?
Your explanation on how to detect a geodesic path is excellent. Basically, you said a geodesic path has zero tangential acceleration when we travel along that path at constant speed. What I can't visualize is how tangential acceleration is not equal to zero as a car moves up/down and around a zigzag hill at constant speed. That is, I still can't see how tangential acceleration is not zero as a car moves through a non-geodesic path at constant speed. In flat space, if a car is not accelerating, its tangential acceleration will be zero regardless if the path is curved or not. Can you explain more or draw two vectors on a non-geodesic path like the zigzag hill and get the change in these two vectors to get the resultant acceleration? Maybe I might be able to see it clearer with more pictures.
Danke Eigenchis!
Very helpful to get into the matter of GR :)
Excellent explanation. You are the man!
In 7:25 you mentioned “…curves where the acceleration vector is normal to the surface”.is that just a definition or summarisation?Can you briefly describe the intuition behind this?
It's a definition. Intuitively it means, if you are driving a care, you are driving it as straight ahead as possible, without accelerating to the left or right. So the only accelerations that happen are up and down due to the bumps in the terrain.
@@eigenchris but if the car is ACCELERATING forward in the straightst possible line say from point a to b,then the tangential component(which i pressume is in the direction of travelling)aint zero，since youre acceletaring forwards…
however you mentioned tangential acceleration =0 for geodesics.
@@chenlecong9938 it is required the car be moving at a constant speed, so there is no forward/backward acceleration.
hi Chris, this is the best course I've ever seen about this topic, thank you so much!
Here I got a question about the geodesics: for a space endowed with a metric, does a geodesic connecting two points change if we choose a specific connection other than the Levi-cevita one?
Since the geodesic equation contains the connection coefficients, I guess it would chage. For example for the boring connection, it seems the geodesics are just the section circles on the lattitude, but not the great circle.
But from another point of view, since the geodesic is the light path for a space with certian metric, it maybe relavant only to the metric, no matter what connection is used.
This confuses me. Thanks again for you wonderful job.
So if we use the Levi-Civita connection, I think the "geodesic between two points" naturally ends up being the "minimum distance" between the two points (or "maximum proper time" if we're in spacetime). However if we use another type of non-Levi-Civita connection, I'm pretty sure we lose this "minimum distance" meaning, since the connection no longer comes from the metric. The connection DEFINES what it means to parallel transport a vector. The definition of geodesic according to the connection is "parallel transporting a vector along itself" (i.e. marching forward as straight as possible). So yes, you're right. Geodesics will look different under different connections.
@@eigenchrisOhh I got it, thanks ~
Absolute gold. Thank you so much!
Thank you thank you thank you sir I thought I was failing this course
At 8:38, could you explain how you wrote the function of the curve in terms of lambda? I'm confused how you wrote u= lambda and v= lambda and then directly substituted lambda for u,v
I picked a curve parameterized by u(λ) = λ and v(λ) = λ. There are many other curves you could pick, but I chose this one.
Fantastic teaching overall! But I have a question: @14:30 you said something like since we don't know the exact components at all, so we are just going to make them up since they are unknown.😅. I don't understand the reasoning behind such a statement.
It's like labelling the unknowns as variables as "a" and "b", except the symbols we used are the capital Gammas (Christoffel symbols) instead.
I am afraid it is a typo here 18:23, because index i or j appears three times. If it is a typo, how to correct it? Thanks.
I am new to tensor calculus, and I have learned a lot from your videoes. Thanks! :)
I wonder, why you choose for the horizontal axis the second variable (later labelled u2). Do you have any reason for it? Many thanks for your great work❤.
It's necessary that the normal part to be nonzero at the same time that the geodesic equation is satisfied???
Why are all summations between covariant and contravariant components?
I would have said that it's to keep lengths and angles invariant.
Is that true?
Also, are there cases where you have to sum two covariant or two contravariant components?
When you have a sum between a covariant and a convariant index, the result is always invariant. When you change coordinates, the covariant components transform one way (e.g. matrix M), and the contravariant components will transform in the opposite way (e.g. matrix inverse M^-1). The two transformations cancel out and the summation result remains the same.
If you are summing two covariant or summing two contravariant components, that's usually a sign you're doing something wrong.
@eigenchris That's what i thought. Thx for the rapid response.
@eigenchris A follow-up question: When you take the trace of the metric tensor, don't you need to sum over two covariant indices?
If I had to guess, I would have said that the metric tensor has no trace. Is that true?
The trace is always defined using a covariant and contravariant index. So if you have a tensor with 2 of the same kind of index, you need to raise or lower one of the indices using the metric. In the case of the metric itself, since a lower index is raised using the inverse metric, the (1,1) form of the metric is just the kronecker delta, or identity matrix... and the trace of the identity just equals the dimension of the space. So the trace of the metric is just a way of stating what dimension the space is.
Beautiful explanation!
Since vekicuty is an emergent property, it doeasn't exist such thing as a constant speed.
Question, if the Christofell symbols are essentially components with respect to the basis vectors to in the tangent plane and one normal to it,
how do I reconcile that with the fact that they are partial derivatives of the metric, and not just like ordinary components which would mean say a certain number of basis vector one plus a certain number of basic vector two.
I thought they were used in parallel propagation because the coordinates may be bending and shifting and twisting and turning underneath vectors that are the same, but that would have different components in these different curved coordinate systems.
Any help please?
You might be interested in watching videos 17-20, which go through the covariant derivative in a lot of detail and derive the Christoffel symbols in terms of the metric tensor towards the end. The formula in this video has the 2nd derivative of R in it, which means the Christoffel symbols depend not only on the basis vectors (1st derivative), but also on how the basis vectors are changing in the local area (2nd derivative). Christoffel symbols do indeed represent the "bending/shifting/twisting" of basis vectors from point to point. Christoffel symbols can appear in flat space when coordinates are curved, but the space itself has no curvature. However the Christoffel symbols can also tell us if the space is curved (see videos 22-23 on the Riemann curvature tensor for that). Does that answer your question?
@@eigenchris thank you sir. I appreciate your tremendous intellect, your patience, and your obvious and substantial teaching acumen.
Thank your quick response to my inquiry.
You provide a valuable service whether you realize it or not. intellectual curiosity and scientific inquiry are sadly lacking in a world Replete and content with ignorance.
On behalf of the other rabid GR fans, thank you! 🤘
"Who cares about Geodesics?"
Those who think the idea is so freaking *cool!* (Independent of applications)
Very interesting way, but I wonder. What are the benefits of doing this way, instead of minimizing the integral that gives distance over a surface with Euler Lagrange ? I bet it would give something very similar.
I think you can derive the geodesic equation from the Euler-Lagrange equations. While E-L is very powerful for getting a lot of results fast, I find it usually lacks intuition. The result kind of just pops out and you have no idea why. I wanted to show you can get the same result using some common-sense rules.
@@eigenchris I see very interesting I like it a loot, very powerful tool. Great content
"In flat space a straight path has zero acceleration when we travel along it at constant speed"(3:39) ...shouldn't it be 'zero acceleration at every point along the path'.?....since we can add all the acceleration vectors and get zero ...even if the path is curved.....
Also if the speed is not constant then there can be straight paths even for non zero acceleration....then can we say that if all the velocity vectors are parallel to each other then the path is straight?
For your first question: yes, this is basically what I meant. For your second question: there are other ways to define a straight path, but the "zero tangential acceleration" definition is the definition I used.
@@eigenchris Thanks chris for clarifying. I am very grateful to you for working so hard to make this series possible . Also can you suggest any lecture series or online material to learn GTR.
@@Smilenshine25 I've looked at Sean Carroll's lecture note before: www.preposterousuniverse.com/grnotes/
Beautifully done
So didactic ! Thanks for sharing the matter !! regards
U r a great teacher..thank you
Awesome lecture.. But I have a slight confusion:
1.Is geodesic a length minimising curve or time minimising curve.?
2. Is a geodesic on curved manifold is literally a straight path just like straight path in flat space.?
Geodesics can exist in space and spacetime. Usually in space, geodesics minimize length. And usually in spacetime, geodesics maximize length (or proper time, more technically).
I'd say a geodesics is the "straightest possible path" on a curved surface. You can't really draw a straight path on a sphere.
@@eigenchris you said that geodesic in space minimise the length and geodesic in spacetime maximise the length, but i am little bit confused whether it is space or spacetime, geodesic must minimise the length of curve, isn't it so.
Sorry, I don't understand the question. Are you saying you don't know how to tell the difference between space and spacetime? It depends on the signs in the metric. In space, the diagonal components would be (+ + +). In spacetime, there would be a sign change for the time dimension, either (+ - - - ) or ( - + + +).
Tell me please scope of Mathematics in future I want became Mathematics Professor too and working hard.
Great lectures! Thank you!
Excellent presentation
If n is any tangent vector to a geodesic and s is the arc length, then dn/ds = 0. But what if pi = 3 in this space? Or what if the space is spongy so that ds1 != ds2 at point 1 and point 2?
There is a great idea! For the dark side of the Universe - suppose that it consists of short-term interactions in long-lived fractal networks, the smallest quantum operators in energy, spherical rosebuds, consisting of a large set; 1 - rolled into a sphere, 2 - half collapsed into a sphere and 3 - flat, vibrating quantum membranes relative to their working centers in the sphere
Is the second fundamental form being defined in 3D space only? What would be the form in N-D space?
I'm not aware if there's a generalization to N-D space. The 2nd fundamental form goes back to the differential geometry of Gauss, which focused on 2D surfaces embedded in 3D space. The n-dimensional generalization (Riemannian geometry) came later, and I don't know if there's an equivalent of the 2nd fundamental form.
Buen trabajo, muchas gracias...
At 12:52, I think you've misidentified the terms; in the second line the 1st and 4th terms are of the single summation form, whereas the 2nd and 3rd aree of the double summation forms.
I wouldn't be paying such close attention if I wasn't very interested in your presentation.