0:08 "General relativity has a reputation for being very difficult..." 05:27 On current investigation concerning GR: the detection of gravitational waves 07:34 Importance of GR in other areas of physics 11:17 The problem set out to resolve 12:08 Search for a quantity that indicates us if the space is flat or not (the curvature tensor) 17:59 Quick recap on the metric tensor 23:23 Raising and lowering the indices 25:44 Introduction to tangen spaces (making the metric equal to the kronecker delta locally) 26:33 Theorem on Gaussian coordinates 31:30 Demonstration of the theorem 39:25 The problem with the partial derivative of a tensor field 43:17 The partial derivative doesn't yield a tensor 45:29 Constructing the notion of a tensor derivative 47:33 Definition of the covariant derivative 53:03 The Christoffel symbols 54:14 Symmetry of the Christoffel symbols (not demonstrated) 59:20 Covariant derivative of higher rank tensors 1:05:08 Covariant derivative of the metric tensor (vanishes) 1:06:40 Finding the form of the Christoffel symbols with covariant derivatives of the metric tensor (here we use the symmetry of the Christoffel symbols) 1:17:36 "No no no no no no no no" 1:18:37 "No!" 1:21:44 Parallel transport and how to tell when there's curvature 1:23:54 Scissors 1:29:13 Applying the parallel transport to the covariant derivative 1:38:01 The Riemann curvature tensor

@miriamarajaoalisoa7294

4 жыл бұрын

You cracked me up with the "No no no no" sections!!!

@lucasbuvinic240

4 жыл бұрын

@@miriamarajaoalisoa7294 the sheer desperation this man's voice had to be noted

@hugo-mp1yp

4 жыл бұрын

Man, if you could do this for every GR/SR courses, I would be grateful for eternity and even more

@lucasbuvinic240

4 жыл бұрын

@@hugo-mp1yp i will! But im currently studying for uni so i haven't had time to continue learning GR

@viniciusgoncalves2642

3 жыл бұрын

i just got to the "nononononononono no" part

The way he calmly yet sternly says “no, no, no, no, no…..no” when his students say something incorrect is great lol.

"General Relativity has a reputation for being very difficult. I think the reason is because it's very difficult."

@fluffyfloyd6391

Жыл бұрын

😮😮😢😢😮😢😢😢😮😮😢😢😢😮😢😮😢😢😮😢

@fluffyfloyd6391

Жыл бұрын

yeah but u l

@fluffyfloyd6391

Жыл бұрын

i 😢

@fluffyfloyd6391

Жыл бұрын

😮

@fluffyfloyd6391

Жыл бұрын

😅😮

I am not sure how useful this would be as a first encounter with GR, but as someone who has gone through the rigorous algebraic derivations elsewhere, I really appreciate Susskind's concentration on the physics here. GR must be assimilated from different angles, and these lectures provide!

@zemm9003

7 ай бұрын

In a proper Physics course you will have already learned Riemannian Geometry by the time you get here so it's basically applying the Math but with different notations and perspective.

It's interesting to see the declining number of views as the lecture series progresses, lol.

@marcop3049

3 жыл бұрын

Here I am. Stoicism

@johnhelm6231

9 ай бұрын

Good job five stars 😅😮🎉

lockdown journal day 35 - just been hypnotised by Mr.Susskind, everything is fine

@marcop3049

3 жыл бұрын

Me too! (Old 52 physician from Rome) day 3rd/?? 😁

@crehenge2386

2 жыл бұрын

spam

So cool when he mentions gravitational radiation, and that some day we will detect it. (around 6:43!) Since we now have, of course.

@pedroakjr2371

2 жыл бұрын

I giggled at this point

@noufbouf

2 жыл бұрын

hahahaha me too I smiled up to my ears

This guy is brilliant, he has the time to lecture, do hits for Gus Fring, interact with Walter White, where does he find the time??

@davidr2421

4 жыл бұрын

All while maintaining a side gig as a foul-mouthed, anti-religious standup comedian

As a Mechanical Engineering student trying out a uni course in GR I want to thank you for saving my semester :). Great explanations!

YT algo often leads me here when I leave it playing and go to sleep and honestly, if I had this professor (maybe) I'd like physics a bit more. Seems like a very nice and intelligent person :)

there really are some brilliant students who ask perfect doubts which sir misses to tell just for us.

I think the covariant derivative part of this lecture is the most difficult. It is at the heart of general relativity and you really need to know it well. I recommend the youtube videos "tensor calculus 15", "tensor calculus 16" and "tensor calculus 19" by eigenchris to get a better understanding. Actually, it would be a good idea to watch the whole tensor calculus series.

@DennisBetts1

4 жыл бұрын

Or to get a simple book like the Schaum's Outline of Tensor Calculus. You can learn the same theory in a quarter of the time..

@damienthorne861

Жыл бұрын

Yes! The covariant derivative measures the deviation away from a vector being parallel propagated along a curve. As well, the covariant derivative takes into account the rate of change of the basis vectors

When he says he's defining a metric "at a point", he really means that he's defining a metric on the tangent space at that point. And therefore this is a statement about how the metric changes as it moves from one tangent space to other neighboring ones.

@matthickman2583

3 жыл бұрын

I think he just means there exists a coordinate transformation for the metric at the point. Nothing about tangent spaces is assumed. You are of course free to create a tangent space from those coordinates to 2nd order, but he doesn’t assume that in the math.

Susskind did this whole course of lectures in Jan 2009, and were a lot easier to understand than this 2012 series. He used polar co-ordinates to get the quadratic feel for the displacement derivatives, and he never once mentioned Gaussian normal co-ordinates. This Guassian stuff is not intuitive at all, I recommend watching lecture 3 and 4 of his 2009 lectures. In the 2009 Lecture 5 , he used + signs, not - signs in the Christoffel symbols equation. Another side note, toward the end of both series of these lectures where he explains the integral of the proper distance from a point in space to the Schwartzchild radius notably in the lecture 7 of this series is much more difficult than in 2009. He didn't bother doing the integral in this series, he said do it yourself. I was running into all sorts of weird values using u-substitutions and trig identities. In the 2009 solution to the integral you didn't need all that as I think he used different boundary and terms in the integrand.

@Eldooodarino

2 жыл бұрын

thanks for posting that. I worked through a bunch of stuff transforming vectors in Cartesian space to polar coordinates last night. I'm supposed to already know this stuff but I've forgotten it. Once I'm done with my polar coordinate stuff I might go back and look at the 2009 lectures.

Now the number of vectors that this function (tensor) takes in as input- that is the rank. The more vectors it acts on the more indicies. One important thing I did not mention was that tensors are linear functions since our tangent spaces are linear spaces. The metric tensor for example acts on two small displacement vectors and is thus a tensor of rank 2.

I fell asleep watching car related videos and woke up to this. Now I’m intrigued. Thanks KZread algorithm

Congrats everyone! We made it to lecture 3!

Tensors do depend upon the coordinate system. However one important feature is that tensor equations hold true in every coordinate system. That's the key. Tensors enable one to write down physics equations that are independent of coordinates. Physics should not depend upon who's making the measurements. There are many ways one can define tensors. Because for one there are many different kinds of tensors.

@cwldoc4958

2 жыл бұрын

Actually a tensor is defined on the tangent space of a point in a manifold. A tensor field is defined on the whole manifold and at each point of the manifold evaluates to a tensor on the tangent space at that point. Often when people speak of "tensors" what they are really referring to are tensor fields. No, a tensor or tensor field does not depend on the coordinate system for its definition. For example, a function, f, from R to a Riemannian manifold, M, can be used to define a velocity vector in the tangent space of a point in the image of f. Now, what is true is that the COORDINATES of a tensor depend on the coordinate system.

Could any share the derivative of DsDrVn that was left in the lecture? I don’t know how to do the Ds(Gamma)rn^t Vt.

The tangent space also enables us to define a tensor field. Which is the same as a vector field except at every point in space time instead of assigning a vector you assign a tensor and that is what this metric tensor business is all about. The geometry of the manifold varies from place to place as well as teh function you use to compute "distances" this is where the metric tensor comes in.

A lecture on Riemannian geometry by Leonard Susskind? Yes. Yes please.

Subtitles have not been included in the first five lessons of this video course. Why? Can you insert them? Thank you

Wow, what a transformation of hope into reality, he says gravitational waves may be detected and lo and behold in 2016 if he was giving this lecture, it has transformed into, Gravitational waves detection is a new field leading us to better understanding of astro physcial phenomenon. We'll look into that later. :D

This series of lectures was the best scientific stuff I found in the internet ever ;-) Cannot be topped - great! Unfortunately I cannot see Susskind in real. He explained things in a couple of minutes where it took weeks for me by reading a book.

The tangent space is the set of all tangent vectors to the manifold at that point and it forms a vector space. Why is this important? Well it would be nice to define vector fields over space-time i.e. velocity fields, magnetic, electric... etc If you recall a vector field is a function that takes a point in the space of interest and assigns a vector and these can change as you move throughout the space. At each point the vector in the vector field resides in the tangent space.

Now where do tensors come in? Well like I said there are many different kinds of tensors. And they are build out of the tangent spaces. Each class of tensor is classified by what is called its rank. There are two notions of rank we will discuss one. In this context a tensor is a function on the tangent space (vector space) remember vectors live in the tangent space. So tensors take vectors in and return numbers (in this context real numbers).

In GR you deal with surfaces (manifolds) that have a strange structure to them. They are not your run of the mill. They are what is called pseudo Riemannian manifolds meaning the metric that you use to define the geometry can return distances that are zero and less than zero. This is because we define distance from one event to the other in terms of proper time not your normal notion of distance. Anyway at every point of the space-time pseudo-manifold sets up what's called the tangent space.

Lecture 3 and the math strikes back, realy hard

I think the intuitive disconnect occurs b/c you differentiate the laws that apply to a massive object vs a smaller mass. To a student, this doesn't make sense. They are used to rules. What needs to be explained is the reality of approximation and it's utilitarian function.

i see what you're saying. i think the issue here is that once you start thinking about derivatives you're extending away from the point in question, so it's not self evident that the first derivatives should be zero. that's what he proves with the 40 equations/40 unknowns.this doesn't apply to second derivatives except if the space is flat. so it's the second derivative that really holds information about the curvature of the space. does that make sense?

I could go on however I believe I have said enough. LOng story short tensors enable us to characterize the geometry of complex high diumensionsl objects that are curved. Just like a number cannot define velocity or force. A vector or scalar cannot define the notion of distance or curvature on a manifold. You need more information tensors enable it. The final product is the Einstein Field Equations which relate the energy tensor to the geometry of space-time.

so Einstein did all this and still didn't win a Nobel Prize for it?

I didn't understand why the commuting of covariant derivatives tell us whether the space is flat. Would be great if someone can explain this.

@sardanapale2302

5 жыл бұрын

An easier way (maybe) to see this is to interpret the covariant derivative in terms of parallel transport. Failure of flatness can be defined by the failure of vector to be parallel transported to itself along a closed loop. The covariant derivative of a vector is the difference between the vector at point x+dx and the parallel transport at x+dx of the vector point at x. You can easily see that failure on a sphere. This is related to the Schwartz lemma: order of derivation does not matter in mixed derivatives (i.e. they commute); which works in R^n because R^n is flat (euclidian) and does not work if the metric chosen is not.

Why derivative of the the metric tensor is zero dg_{mn}/dx^{r} = 0 in cartesian space??? Is it because in that case g_{mn}=delta^{m}_{n} (identity matrix I mean) ??? If it is the derivative that gives zero doesn't need to be the derivative of identity matrix: it can also be the derivative of any matrix with some constants in her entries. Am I wrong, why?

@carlosromerofilho6202

10 ай бұрын

You're quite right.

Very Very Very Simple Explanation.

1:34:00 DrDsVm != DsDrVm , but I thought the whole point of finding Cristoffel symbols and finding the covariant derivative was local flatness to the first order, so why does the order mater around the little parallel transport loop..... aahhhhhh ok... comparing the _second covariant derivative.... okokok

At 1:09:52 (3) Is supposed to be Dn gsm = @gsm/@x^n - Tsn^t gtm - Tnm gst = 0 He has it written as (3) Dn gsm = @gsm/@x^n - Tsn^t gtm - Tmn^t gst = 0. Here @ = partial, T = Cristoffel

@jeffreylanz719

3 жыл бұрын

Tnm^t = Tmn^t by an index symmetry. He covered this briefly when he first introduced the T symbol.

What does does he mean when he says calculate the derivative in gaussian normal coordinates and assume it transforms like a tensor to get the derivative in general coordinates? dV_m/dx^r is the derivative in the gaussian normal coordinates defined at that point. Now we know dV_m/dx^r is not a tensor. He had already stated that dV_m/dx^r is not a tensor, then why should we assume it transforms like a tensor?

@HH-hk7nt

4 жыл бұрын

In the normal ones it is A tensor cause the christoffel symbols disappear as the coordinate curves are Autoparalelly transported, so the axis themselves don't change from point to point. And since the vector stays the same, the derivative of it's component is 0 there(dtv of full vector=dtv of comp+ dtv of the bases)

@cwldoc4958

2 жыл бұрын

He is looking for a way that makes sense to quantify the change of a tensor (actually tensor field) as you move in some direction. That is, he wants to define a sort of derivative, which will be called the covariant derivative. In curvilinear coordinates, in general there is no special way to compare tensors at two distinct points. So he considers normal coordinates at the point, P, in question, because at the origin of those coordinates (corresponding to P in our coordinates) the metric tensor is the identity matrix and its first derivative vanishes. In those coordinates it makes sense to say that the rate of change of the vector V_i along (normal) coordinate x_j is dV_i/dx^j, and then he defines the covariant derivative by treating this as a tensor. Yes, there is a bit of arm-waiving involved! In fact, there is a more intuitive explanation of the covariant derivative, at least for a contravariant vector, such as a velocity vector. This explanation requires borrowing a fact from math: Any given Riemannian manifold, M, may be embedded in a euclidean space, R^n for sufficiently large n. In this case embedding means not only that we can view M as a subset of R^n, but that that at any point of M, the metric coincides with the metric of R^n restricted to M. So given a Riemannian manifold, M, embed it in a euclidean space, R^n (picture for example a 2-dimensional surface such as a sphere in 3-space). Suppose there is a vector field, F, defined on M. Then at a given point, P, of M, the derivative in R^n of F in the direction of a vector, v, at P, is dF/dv, which is clearly defined. The only problem is that the vector, dF/dv, may not lie in the tangent space of P. So we project dF/dv onto the tangent space at P, and that is the covariant derivative of F with respect to v! For a simple example, say M is a 2-dimensional sphere of radius 1, embedded in R^3, and it is given that the value of F at any point on the equator is a unit vector pointing east. Let v be a unit vector at some particular point, P, on the equator pointing east. Then dF/dv is calculated in R^3 to be a unit vector at P pointing toward the center of the sphere. This vector does not lie in the tangent plane at P. To get the covariant derivative of F with respect to v in M, we project onto the tangent plane at P, getting the zero vector! In fact this makes sense, since on the surface of the sphere it appears F is not changing as you move along the equator at P.

I don't understand why (around 35:00) he shows that the first derivatives of the metric wrt the coordinates can be chosen to be 0 this way. Isn't it just the statement that at a constant location in space (x=x0) the metric is also necessarily a constant ? And therefore isn't it trivial that its derivative wrt anything is 0 ? I must be missing something..

@klgamit

Жыл бұрын

Answering my own question ten years later... it's clear to me now that a lot of confusion in this lecture arises because when he first stated that coordinates can be chosen such that the first derivative of the metric vanishes, *he didn't emphasize that this is the Covariant derivative that vanishes* !!! In fact, he writes it as the regular derivative that vanishes. If he wrote originally that the Covariant derivative of the metric is the one that always vanishes, it becomes clear why later when we go to general coordinates the regular derivatives of the metric don't necessarily vanish (due to the christoffel symbols being non zero in general coordinates).

It is always great to learn from you sir, and thank you Stanford.

I really hate the whiteboards for some reason, I prefer the chalkboards. Great lecture Professor Leonard!

Susskind is like Brad Pitt in Ocean's 11-13: always eating something. Can't blame him, though, GR makes me hungry too.

So is a cone flat or curved?? On the one hand, if you take a vector around it the direction changes (so the commutator of the covariant derivatives is non-zero, right?), but on the other hand you can cut it up and lay it out flat...I'm confused

@jonawhite17

9 жыл бұрын

A cone is not a differential manifold, as it has a singularity at its apex.

@jonawhite17

9 жыл бұрын

***** The cone is not a differentiable manifold because of the apex. If you remove this singular point, then it does become a differentiable manifold.

@cwldoc4958

2 жыл бұрын

Remove the apex, and the rest of the cone is flat. Alternatively construct an area that is curved as he does in the picture, and then the part of the cone that is not in that curved area is flat. In the second case the parallel transport of a vector around the circle changes direction because the interior the circle is not all flat. (In the first case, there are points "missing" from the interior of the path, i.e. the path with its interior is not compact, so all bets are off.)

did you understand what he was talking about in lectures 1 and 2?

"This called a covariant derivative. Not because its index is downstairs. It's called a covariant derivative ... for ... uhm...Just because it's called a covariant derivative."

@clopensets6104

3 жыл бұрын

It's actually called the covariant derivative, because the derivative changes between different coordinates, so as to allow for the covariant derivative of tensor-fields in one coordinate to be equal to the tensor-field evaluated in Gaussian Normal coordinates.

@bogdansikach7709

3 жыл бұрын

@@clopensets6104 I think the word "covariant" means it is something that transforms "rightly" when written with respect to another reference frame.

can you please insert subtitles?

What are these "Gaussian normal coordinates"? There is nothing about them in Wikipedia

@seanpaul2562

4 жыл бұрын

its just a name for coordinates in vicinity of a point where we can draw almost straight lines some people call it locally affine coordinates or locally euclidean coordinates but in the end all of them mean that we can construct a almost flat surface in small vicinity of a point

"General relativity has a reputation for being very difficult" - So many symbols and equations

The whole stretching bit made me think. If light is headed towards Earth, would the wavelength be stretched to make it seem redshifted in a way? And if a beam is caught by a satellite in a direction tangential to it's orbit around Earth, would it be compressed or blueshifted?

@factshurtfeelings6149

4 жыл бұрын

depends on the relative motion of the two frames, source and observer

@musicalfringe

3 жыл бұрын

Light emitted into space from the Earth's surface would appear redshifted to a relatively stationary observer far away. Light from that observer would hit the Earth blueshifted. I guess you're thinking of the wave as being literally stretched by the tidal forces as it falls in, but remembering that time runs more slowly lower in the gravity well is the key. As for the orbital case, that's purely special-relativistic if you're talking about the source hovering stationary above the Earth (without orbiting). In that case a satellite would see red/blue shift depending on whether its orbit takes it away from or towards the source respectively.

So, from much of the questions of inquisitive students: we are asking them to accept a theory that is fundamentally at odds with the local phenomenon. Such that calculating and understanding in our local frame, is an approximation.

I love this lecture

Isn't the fact that in the rotating frame of reference there are tidal forces due to radial direction of centrifugal force?

@cwldoc4958

2 жыл бұрын

No, if you look at someone who is experiencing no stress from tidal forces, and then you spin around while looking at that same person, you won't suddenly see them being ripped apart by tidal forces.

@bogdansikach7709

2 жыл бұрын

@@cwldoc4958 No, I think I'm right. Consider a body consisting of two masses *m* at a distsnce *D* from each other. in the rotating RF with angular velocity w. Suppose also the initial orientation of the body is so that both masses lie on the same line along radius (on the one side of the center) at a distance R from the center of RF. (So te 1st mass at a distance R-D/2 and 2nd at a distance R+D/2. Let's calculate the accelerations. The total force is 2mw^2R. So the average acceleration is wR. The acceleration of 1st mass is w^2(R-D/2), and of 2nd mass is w^2(R+D/2). So in the RF falling with the averaged acceleration with the body masses are tearing apart. Where am I wrong?

@cwldoc4958

2 жыл бұрын

Just to clarify further, if you are in a rotating frame of reference observing a flexible object in free-fall, you will see different points of that object accelerating in different directions (attributed by you to fictitious forces, centrifugal and coriolis), but the net result will be that the points of the object appear to be rotated and translated in such a manner that the distance between any two points in the object will remain fixed, that is, the object will not experience any tidal forces.

@bogdansikach7709

2 жыл бұрын

@@cwldoc4958 So you tell me that the Coriolis force cancels the deformation I described in my previous comment? Hm... isn't obvious at all, but not impossible :) I'll check.

@cwldoc4958

2 жыл бұрын

​@@bogdansikach7709 My second comment was made before I saw your example, even though it appears after it, so it was only referring to your original comment. Now, let me try to understand your example. There are two objects (object 1 and object 2)of mass, m, separated by a distance D. Am I correct in assuming that they are at rest in some inertial RF? Denote that RF by the letter A. Then consider the point of view from a rotating RF, denoted by the letter B, with angular velocity w. Then at some point in time, we are given that the distance from the axis of rotation of B to object 1 is R-D/2, and the distance from the axis of rotation of B to object 2 is R+D/2. Of course we assume R>D/2. So at that time, the objects lie in a plane perpendicular to the axis of rotation of B, and will remain in that plane indefinitely. Let's use coordinates (x1,y1) and (x2,y2) to describe the position in the B RF of the two objects as a function of time, where we assign t=0 for the point in time mentioned above, and set (x1(0),y1(0))=(R-D/2,0) and (x2(0),y2(0))=(R+D/2,0). Then it is clear that (x1(t),y1(t))=(R-D/2)(cos(-wt),sin(-wt)) and (x2(t),y2(t))=(R+D/2)(cos(-wt),sin(-wt)). The force on object 1 according to B is F1 = m(x1(t),y1(t))" = -(R-D/2) w^2 (cos(-wt),sin(-wt)) and the force on object 2 is F2 = m(x2(t),y2(t))" = -(R+D/2) w^2 (cos(-wt),sin(-wt)). For convenience considering only the forces at t=0, F1(0) = -(R-D/2) w^2 (1,0) and F2(0) = -(R+D/2) w^2 (1,0). The (fictitious) force on object 1 is directed towards the axis of rotation of B and has magnitude (R-D/2) w^2, while the (fictitious) force on object 2 is directed towards the axis of rotation of B and has magnitude (R+D/2) w^2. The forces are different, but so are the paths of the two objects according to B. According to B, the objects are traveling around in circles and the above (fictitious) forces are precisely what is needed to explain why they keep orbiting B instead of continuing on straight paths. The forces observed by B keep them the same distance apart and no tidal forces are observed!

HELP why covarient derivative of metric tensor is zero?

@tracyh5751

4 жыл бұрын

Let's try to calculate the covariant derivative of the metric tensor at a specific point. By the definition of the covariant derivative, we move to Gaussian Normal Coordinates for the point and then calculate the derivative of the metric tensor at that point. But Gaussian Normal Coordinates are defined so that the derivatives of the metric tensor are all zero at the origin (our point in question) !

@carlosromerofilho6202

10 ай бұрын

The fact that the covariant derivative of the metric tensor vanishes everywhere (not only at a single point) cannot be deduced without the assumption that the parallel transport preserves the scalar product of vectors. This assumption leads to the celebrated Levi-Civita theorem of Riemannian geometry, which states that the affine connection components must be the Christoffel symbols. See, for instance, "Riemannian geometry" by M. P. do Carmo (Chapter 2).

Great lecture but few minor things to point out. 31:30 The linear term isn’t y^m in the expansion of x^m in terms of y^m. There should be a coefficient other than 1. Second thing, when you plug that expansion into (del gmn/ del x r) = 0 using chain rules, (actually you gotta go the other way but anyway) those quadratic terms doesn’t survive as the derivatives are evaluated at y=0. So the variable Cmnr’s doesn’t do anything with zeroing the first derivatives.

@thjo1964

2 жыл бұрын

And in my intuition, coordinate transformations like that ( x^m ~= y^m in linear approx.) basically says x’s and y’s are almost same coordinate in a local sense. So upto first derivatives nothing changes when frame changes x to y and vice versa

@cwldoc4958

2 жыл бұрын

He says the C's must be such that the equation del gmn /del xr = 0 must be satisfied at the origin. I assume by gmn he means the coordinates of the metric tensor in the normal coordinates (x). But doesn't the equation gmn = identity matrix also have to be satisfied at the origin? If you use the equation he wrote relating (x) and (y) coordinates to calculate del ym / del xr, you find that at the origin this is the identity matrix, implying that the coordinates of the metric tensor in the normal coordinates (x) are the same as the coordinates of the metric tensor in the arbitrary coordinates (y). So it would seem that in the general case, no choice of the C's could make the coordinates of the metric tensor be the identity matrix in the (x) coordinates! Any thoughts?

lots of things that I don't understand in the first 30mins of this video. Felt like I might have missed GR Lecture 2.5 or something.

@AkamiChannel

4 жыл бұрын

吴邛 Maybe you need to watch the special relativity lectures?

@Iamfafafel

3 жыл бұрын

No this is purely mathematics, no physics

I'm not anywhere near as smart as anyone in that room, let alone the people watching this video. But I love to visualize stuff. And in a field of view where you are free falling, doing the experiment where you measure the drop of light from one wall to the other wall to determine if you are falling. What if you are spinning? Wouldn't that cause the same effect? If so, could you rather use two different beams of light, measuring the distance one both walls rather than just one? And you could then determine if you are falling or spinning? Hopefully one day I can go to college instead of just watching videos during covid-19. 😬

I'm still confused about the Gaussian normal coordinates. Let's say I'm looking at the simplest example of curvilinear coordinates: polar coordinates (r,phi) in the plane with the metric tensor g=diag(1,r^2). Now let's say I want to write down a transformation x^i = y^i + c^i_j,k y^j y^k that takes the metric tensor to the Kronecker delta, where the y^i are (r - r_0) and (phi - phi_0) respectively. The Jacobian of this tranformation is *always* the unit matrix, leaving the metric tensor invariant. What I *can* do is Taylor expand u=r*cos(phi) and v=r*sin(phi) around a point, and then use the expanded u and v as coordinates. This makes the first derivatives of g vanish, but is in general not of the desired form above. - I'm guessing that the least I need to do is include the scale factors in the definition of the y^i, i.e. consider polynomial functions of (r - r0) and r_0*(phi - phi_0).

@randymartin5500

3 жыл бұрын

Susskind did this whole course of lectures in Jan 2009, and were a lot easier to understand than this 2012 series. He used exactly what you described with polar co-ordinates and he never once mentioned Gaussian normal co-ordinates. I'm lost with this Gaussian business. Just goes to show, there are many ways to analyze GR. Another side note, toward the end of both series of these lectures where he explains the integral of the proper distance from a point in space to the Schwartzchild radius in the lecture 7 in this series was much more difficult and he didnt bother doing it, he said do it yourself. The 2009 lecture , the integral was much easier to do because he used different boundary and terms notation in the integral.

@dsmith3112

3 жыл бұрын

@@randymartin5500 Hi. Thanks for the message. It has been a while since I watched this lecture. I never could figure out what was going on with the transformation. I switched to reading the book by Sean Carrol. I think Susskind made a mistake on the board and that such transformations must necessarily involve scale factor. Or he is implicitly using some non-standard notation.

@dsmith3112

3 жыл бұрын

In the book of Carrol things are explained pretty clearly.

@randymartin5500

3 жыл бұрын

@@dsmith3112 Yea I was curious what Susskinds second course would be like in 2012 with the better HD video quality....thanks for the Sean Carrol suggestion I will definitely take a look now that Susskind confused me here lol. I enjoy Sean he does speak very well.

Dr.Susskind,. Boring us a thing you need not fear. You have an undaunted precision so bright and possessed of, or combined with a natural intelligence, that it's rather exciting. My question now is ,"" do we always need spherical geometry in GR.?"" Let's say we are considering electricity travelling in some metamaterial layers that are drastically slowing it down to a few meters per second ?

@redpillmath

Жыл бұрын

The Absurdity of Special Relativity Theory. kzread.info/dash/bejne/fKyuytGMh6Xal9o.html The falsity of Einstein's thought experiments, the thought Light-Clock, Lorent's equations; and the interest of the Elite, a selected coterie, to impose their Relativistic-Chaos stuff in our educational system to manipulate, control, and exploit vulnerable people.

maybe Gmn is a constant to Xr~~my friend,I am not sure

Love it

Love the casual Obama name drop lol

@humanonearth1

5 жыл бұрын

Where was that?

@godnoomgodnoom

5 жыл бұрын

"I tell my friend Barack Obama, never apologize...." at 13:45. At least it is now confirmed that our nation is indeed governed in a region with a curved space-time. Might explain those budgets.... Certainly is a black hole for $$$!

You can compare it to the first and second derivatives of x^2 at the origin - the first derivative is zero but the second derivative isn't. Same with gmn, Just because they're stationary (first derivatives zero) at some point, doesn't mean the first derivatives at that point aren't changing.

Susskind has a IIIZI1CSIIZIECS (If it is zero in 1 coordinate System it is Zero In Every Coordinate System) Criteria before he will call a structure a tensor. He says that the Cristoffel symbol is NOT a tensor because it does not meet IIIZI1CSIIZIECS criteria. Is there a name for such structures that consist of 3 dimensional array of numbers or formulas, that doesn't have further criteria about meeting IIIZI1CSIIZIECS?

@cwldoc4958

2 жыл бұрын

I suspect he would call them 3DAONOFTDHFCAMIIIZI1CSIIZIECS.

i am kinda newbie to tensor, so this basic doubt he often tells that tensors remain same in all coordinates and thats the property of all tensors, if so why isnt the metric tensor same in all the coordinates (if they aint same why are the called 'tensors' in the first place ?)

will be kinda cool coming out of the 4th year hopefully filling in all the holes in my currently insanely fragmented knowledge of this area of physics xD so fascinating tho imo

I dont see how dGmn / dXr = 0 and dGmn / dXr dXs not = 0

@TheTntbomber

5 жыл бұрын

Ist at a given point x_0 which means it ist not zero in all points so that the second derivative does not have to be 0

@Iamfafafel

3 жыл бұрын

y=x^2

"Let's find a thing called Riemann"

@Iamfafafel

3 жыл бұрын

My goal in life is also to be named after a tensor lmao

X & Y axis time and space = faster you go smaller space, slower you go larger space

Why Greenland is Denmark? In Google Earth it is not Denmark there. Year 1977 book is a report for year 1976.

@nurlatifahmohdnor8939

2 жыл бұрын

Bit by bit = se-di-kit de-mi se-di-kit si-kit = little = se-di-kit ba-nyak = many currency = ma-ta wang (often people write ma-ta-wang)

@nurlatifahmohdnor8939

2 жыл бұрын

Page 232 Endnotes of chapter two Book : BLOWN BY BITS Author : PHILIP eVaNS tHOmaS S. WURSteR Copyright 2000 What is BCG in page xiii.

The argument given between 34:00 and 35:00 is incomplete. In the case of 4 dimensions, there are indeed 40 unknowns, but there are in total 50 equations, 40 of which are from the partial derivative condition as he explained, and 10 of which are from the constrain that g_mn = δ_mn. To make the logic work, one must first assume that it is always possible to find a y coordinate system in which g_mn = δ_mn. And then by the way x's are constructed out of y's, this g_mn = δ_mn condition is also automatically satisfied in the x coordinate system. Therefore the 40 unknowns only need to satisfy the 40 partial derivative equations.

@firdausjanoos1367

8 жыл бұрын

+Petra Axolotl I actually didn't understand the argument at all. Firstly, as you point out - for the quadratic expansion case, there are 50 constraints and 40 unknowns. However - if you admit a second order expansion of the coordinate mapping, then there are enough unknowns to satisfy the second order constraints too (and so and so forth) ! So why arbitrarily stop at a quadratic expansion ? That hardly seems like a proof of anything ?

@PetraAxolotl

8 жыл бұрын

+Firdaus Janoos It might be a good idea to read Schutz's A First Course in General Relativity. It gives a pretty good explanation (taking at least a full page and maybe a bit more) on this point. Just beware that there are quite some errors in the book, most of which are non-essential and not included in the official errata. I plan to publish an additional errata (and solutions) at www.petraaxolotl.com in the near future. But I need to first finish the errata/solutions for another textbook (Calculus on Manifolds by Spivak.)

@pouncebaratheon4178

8 жыл бұрын

I skimmed your website. It's obvious that much labor went into it. I'd group your pages into smaller sections though; they're just so big my laptop can barely load them. Are there really so many errors in Spivak? He struck me as so precise... then again I haven't spent much time with his text, and don't own a copy. It's mostly above me at this point, but I plan to return to it when I have the sophistication.

@PetraAxolotl

8 жыл бұрын

Thanks for the suggestion. I was honestly also surprised by the number of errors in Spivak, a book published in the 1970s and used by generations of mathematicians. I know personally the editor of another very popular series of textbooks The Feynman Lectures in Physics and he was not surprised at all. There were even more mistakes in that one when he started to correct the mistakes systematically about 15 years ago. The difference since then is that The Feynman Lectures in Physics was and still is being corrected based on readers' feedback while Spivak was not.

@PetraAxolotl

8 жыл бұрын

Finally I have done what you suggested for A First Course in General Relativity :) I'll do the same for Calculus on Manifolds when I've got time.

Isn't the basic definition of a Riemannian Manifold is that it's everywhere locally isomorphic to R^n? Then by definition you can put a Cartesian metric at any point.

So the connection coefficient relates information of the curvilinear coordinate choice, (nothing about the surface?). It's your 'connection' back to normal Gaussian coordinates? This explains why the Christoffel symbols are non-zero for polar coordinates in a flat-space. I feel like this video answered many of my questions -- thank you!

@daccoa

9 жыл бұрын

+eugeniomyles Pretty much. It's too complex doing a derivative when changing from a flat to a curved space. Instead, we pretend the curved space is flat using Gaussian Normal Points (which as he says, works up to second order), then do the normal derivative, and add a second term which represents the change in each coordinate basis compared to flat coordinates (how do the actual coordinates of a curved coordinate system differ in relation to simple xyz coordinates)

I bet this class has the gnarliest homework

May God bless you

difficult subject, very good lecture

Hang on. Susskind says he is not an expert on GR? I beg to differ.

I just spent a day thinking about physical essence of contra and covariant vectors and now he talks about covariant derivatives of tensors. Can someone please spoonfeed me what that means? If it is just the ordinary derivative of gausian normal metric tensor at a point plus the gamas? then what is essence of "co" in that? and then what is contravariant derivative?

@jimdogma1537

10 жыл бұрын

IMO, This is what is frustrating about these "theoretical minimum" lecture series. There's rarely any concrete examples ever worked out. It's just theory build on abstract theory, etc. Have you noticed that we were given a peek only once at the guts of the metric tensor with its 10 components? But we never used it EVER to calculate some specific example. I "think" I know what it does but I'm not sure because I haven't seen it used in a practical example. Now, for some reason we are on to taking derivatives of this thing of which I'm not even sure what it is. What does it mean to take a derivative of a matrix? And we're not even talking about the Chrisoffel symbols yet. Examples would have helped alot.

@dreamscometrue30989

10 жыл бұрын

Hey there. I don't know if you're still interested, but I have found that there is a text book by Sean Carroll on it to help you. Alternatively, here are some free notes arxiv.org/pdf/gr-qc/9712019v1.pdf . Covariant derivatives are basically normal partial derivatives, but then we add extra terms on the end to MAKE SURE that it transforms like a tensor (and so is coordinate independent). This is because ordinary partial derivatives are highly coordinate dependent! It is called covariant because it has "downstairs" indices. The contravariant derivative would just be equal to the covariant derivative, contracted over the inverse metric (I think, it's not a thing I use a lot). Hope this helped!

Compete Compute Crusoe's diary is a good read. Entertaining. True. I think. Specific time and place stated. Lolland is in Denmark.

@nurlatifahmohdnor8939

2 жыл бұрын

6 g g looks like 9

@nurlatifahmohdnor8939

2 жыл бұрын

7 kids all natural milk. I know from this mag. generous at promoting dates to remember. Issue September 2013. Page 335 R. A. J. said these people is disinterested in knowledge. No. 2 is favourable. disinterested adj 1 free from bias or partiality; objective. 2 not interested. USAGE NOTE Many people consider that the use of disinterested to mean not interested is incorrect and uninterested should be used. Maybe he was in a very angry state. He did not show. I remember disinterested than uninterested.

@nurlatifahmohdnor8939

2 жыл бұрын

Judge #ChocolateFudge

@nurlatifahmohdnor8939

2 жыл бұрын

The #Danone biscuits are all delicious.

@nurlatifahmohdnor8939

2 жыл бұрын

First published 1996. 1997 I was in Science stream class. 4Sc1. 2nd batch. Monograph Series No.1/1996 Project Director Molly N. N. Lee

Where is the 2nd lecture in this series? I have only found 1 and 3

999k subscribers!

3πγ portal to pi nuclear black hole

1:13:15 It probably should be gts not gst

@seanpaul2562

4 жыл бұрын

doesn't matter because metric tensor is symmetric on its 2 indices

@user-gs6lg4gd3b

4 жыл бұрын

@@seanpaul2562 I was think about that. I guess it is not necesary. I agree, not symmetric metric tensor doesn't makes any sense.

(if so why isnt the metric tensor same in all the coordinates) They are the same in all the coordinate system ... they only look different.... They are expressed in different forms exactly as tree or cat expressed differently in different languages although they mean the same.

12:53 13:36 Restart 25:43 1:21:32

at this point the indices look like a soup of letters

kzread.info/dash/bejne/la2Lkq6QdaSplqw.html: Perhaps better than the adjective "stressed", it's "dilated" or "contracted" is better.

ty old man

He kept touching the food for comfort I guess....

He hasn't eaten much during this particular lecture,except a little in the end...and he wasted the coffee too...

My only request is that the prerequisites be listed in the description.

I wish I went to Stanford

@6:10

you should start at lecture 1

I get what you're saying, seemed like a backwards way of deriving the Christoffel II symbol... but it's definitely beneficial to follow it through this way because it can be cryptic from a purely mathematical pov

i wish he can start cooking on the show

Not the easiest nor the most intuitive way of defining the covariant derivative… But I guess Susskind has something else in mind to chose this way :P

Locally flat only means that the metric becomes the kronecker delta at the point right?? Wat to the derivatives have to do with it?? Wat im asking essentially is that how a space will be if it has the property that the metric can take the kronecker delta form locally but its derivatives cannot be transformed away??

I can be Sheldon Cooper if I cram it.

I got this wrong - sorry. There are 8 not 12 equations; the two involving Div are single scalar not vector equations. Only the two with Curl are vector each with 3 components.

@Iamfafafel

3 жыл бұрын

Imagine turning memorizing maxwells eqns into a dick measuring contest

@adrianwright8685

3 жыл бұрын

@@Iamfafafel they have a wonderful symmetry which makes them really rather easy to memorise; a matter of minutes!

@gamestarz2001

2 жыл бұрын

I'm pretty sure there are 8 since 2 of the vector equations involve dot products, so they're really scalar equations already.

I wish I was smart so I could understand any of this..

@hank1519

5 жыл бұрын

Me too!

@koerel

3 жыл бұрын

It's more about being properly educated than being smart imo. You need to have studied quite a lot of different kinds of math to be able understand everything in these lectures. If you haven't, just being smart won't get you anywhere.