Einstein's General Theory of Relativity | Lecture 12

[willbirful]

love the video really good

[Iberedmas]

the problem is measuring the distance. First solve one dimension x ,y, z alone then you can solve two dimension x and y and then you can solve three dimension x and y and z. so far as I know the problem is you solve the distance of one dimension by two dimension. I have a formula how to solve one dimension which is straight line.

[shoppittsburghnow]

This is a great video

[clairebehun]

some great inforamtion here thanks

[SuperDogbrown]

i enjoyed this vid

[adelle0001]

thanks!

[potpie921]

great vid!

[thebigfootme]

very informative..thanks!

[agapitoflores001]

This can be a good reference for research. very informative.

[zaneacademy]

good stuff

[m4r35n357]

Amazing collection, but very frustrating, after watching the whole series twice in about 6 months, I am really annoyed that there isn't a single worked example for the curvature tensor, or even Christoffel symbols. Yeah I know they are messy, but why not do eg, 2D polar coordinates and a two-sphere just to set the scene. As it is I still don't know how to get started . . . .

[Ikjyot]

@m4r35n357 I would look at Stephani's book on Relativity. It is the only book that actually shows how calculations are done with Christoffel symbols and the Ricci tensor, etc... You could also look at Woodhouse's book on General Relativity. I used both of these in my undergraduate studies.

[nrasandun3]

I really like what I can't understand. Thank you sir! I got a lot of pleasure.

[gamma809]

I've found these lectures helpful. However I'm here because on wikipedia there's a page "Schwarzschild Geodesics" which I think is rarely visited so i thought I'd say this here. There's an error when it tries to show that there exists solutions to the geodesic equation with the schwarzschild metric despite there being the singular cot function in the equations it ends up arriving at a result which still essentially says there's singularities at theta=n*pi its under "Mathematical derivations of"

[Progrose]

Thanks to stanford and prof susskind for a great lecture series. Please post lecture 13

[teamd3ath1]

Why would u watch approximately 18 hours of learning

[scootrider91]

@teamd3ath1 This is not "just" learning:) this is exploring the world:D

[theLaven]

Is Lec 12 the end of the lec of Einstein's General Theory of Relativity or is there further?

[Lunatic4ever]

man u need to have balls to ask this dude some serious binary questions :D btw...why the hell do I watch this ? I feel a lot dumber now

[giftu2b]

So if you fall into a black hole, will you see the end of the universe?

[gruff5]

you generally can't hear the audience's question words clearly, so switching the audio recording between the two microphones is not helpful (and even annoying). Would be helpful if transcript of the questions could be provided.

[ila1964]

one of those 2 questioners is definately kermit

[ScalarPhotonZ]

Presumably in the case where we don't have a real gravitational field but we're just looking at the Rindler accelerated system, nothing physically spectacular (or devastating) occurs at any of the negative R hyperbolas. But when describing the black hole, it's useful to analogously refer to this Rindler system that includes the negative R hyperbolas. But on the other hand, the Rindler horizon is something physically spectacular within the Rindler frame - it's a plane on which time is frozen!

[ScalarPhotonZ]

I'm not quite sure how intelligible my first comment appears. I had to try to cram it into 500 characters. I'm basically speculating that if the variation of gravitational acceleration with position can be related formally to the similar effect in the Rindler accelerating frame by the Equivalence Principle, then even the fact that gravity is attractive rather than repulsive might become explainable - assuming of course that field strength increases the closer one is to the source. Comments?

[ScalarPhotonZ]

Whereas in Rindler space, proper acceleration of a Rindler observer tends to infinity as the distance of the observer from the Rindler horizon tends to zero, in the case of a Schwarzchild black hole, does the "acceleration provided" to keep an object stationary within the gravitational field tend to only a finite value as you tend towards the event horizon? It's obvious that the Newtonian calculation would give a finite value, but is it likewise in the relativistic case?

[ScalarPhotonZ]

In the Rindler frame, both clock rate and proper accel. of a static object are position dependent. If grav. time dilation is meant to follow from the pos. dep. of clock rate in accel. frames, we might similarly consider variation of grav. acc. Now both in the Rindler frame and in the vicinity of a static observer in a static grav. field, prop. acc. decreases the further ahead we look in the direction of prop. acc. But this would reverse in repuls. grav. Valid justification for grav. attractive?

[dannyboy12357]

this man is a brillian human being, he presents such complicated material in a simple way.

[TheHyouth]

and students need to pay money to get this?

[TedDGPoulos]

"I believe that in order to make real progress one must again ferret out
some general principle of nature." - Albert Einstein 1952 . The known universe 14,000,000,000 BCE . Earth 5,000,000,000 BCE . Homo sapiens 300,000 BCE . Fire (applied) 200,000 BCE . The underlying law of nature (discovered & applied) 2003 CE

[JonYodice]

I made a list in which all the videos are in the correct numerical order. screw this list >>>>>>>>>>>>>>>>>>>>>>>>>>>>

[drandersw]

@JonYodice What is the right order?

[estaere]

Perhaps the most interesting from all 12 lectures.

Now. Let's talk about cosmology... (finally:)

[were455]

seriouslyy.....how are you meant to understand all of this stuff.....its sooo harrd!!!!!!

[purpleproductionftw]

However, this does not mean that the graph is not curved, or that the second derivative is zero at that point.Gamma(x) will not be zero but will be proportional to derivatives of Gamma. Does this make sense? In other words, at the infinitesimal level, the curve becomes something that it is not - flat. And the 1st and 2nd derivatives of a flat anything are zero, thus making curvature impossible

[BLANDCorporatio]

In reply to SpiritualAtheist: "at the infinitesimal level, a curve is flat". What you mean is that for a short enough segment, the curve looks "a lot like" a line, right? The shorter the curve segment, the better it is approximated by a line and in simple real calculus that's how you start defining the slope (the 1st derivative). BUT, the 1st derivative is itself a function that varies from point to point: the curve has different slopes at different points. So higher derivatives may not be 0.

[BLANDCorporatio]

In reply to the parallel transport around a small area question: msleifer got it right imo. "Locally flat at a point" means that however small the deviation from flatness you want, there is a region around that point that doesn't deviate more than the amount you tolerate. Usually the smaller the deviation you want, the smaller the region. Also, and this is important, infinitesimal is not zero. That's why equations in infinitesimals make any sense, and can't just be trivially replaced by 0 = 0.

[htytube]

Parallel transporting a vector around a small loop doesn't necessary give you small curvature, e.g. around the tip of a cone. Even if the object you talking about is a the surface of a sphere, small loop still give you small curvature. Until you take limit, the curvature of the point on the sphere goes to zero.

[primeskier]

Have viewed all of Prof. Susskind's lecture. These videos are an incredible gift to all of us from a world class physicist.

[foketesz]

a great value indeed.

[darrenmav]

Hi, a question not pertaining to this particular lecture. Can someone please explain to me why parallelly transporting a vector around an infinitely small closed loop will lead to a rotation of the vector since locally, a manifold looks just like a flat space? If this is the case, then parallel transport around a small closed loop shouldn't be able to detect curvature, can it? Thank you.

[msleifer]

Well, locally you can find a coordinate system such that the metric is flat and the Kristoffel symbols are zero, but this does not mean that their derivatives are zero. Therefore, terms like Gamma(x + dx) - Gamma(x) will not be zero but will be proportional to derivatives of Gamma. Does this make sense?

[msleifer]

Another way to think about it is to draw an analogy to functions in single variable calculus. Roughly speaking, curvature is analogous to the second derivative of a function. Assuming you have a continuous differentiable function, then if you zoom in on small enough segment of the graph of the function then you can make the it look arbitrarily close to a straight line. However, this does not mean that the graph is not curved, or that the second derivative is zero at that point.

[Zantorc]

The degree of rotation is a second order effect related to the area. Take a square loop and divide the rotation by the area and you'll get a finite amount related to the curvature.

[ShootBigBird]

Thats what I thought back in lesson 8 (I think) when he introduced it. But the axis rotation does change although the lengths remain constant. Also, the area within the loop reduces in curved space when compared to flat space, and is directly related to the rotation angle. This is the obstruction.

[SpiritualAtheist]

I know we're not supposed to consider such things, but perhaps there is simply a contradiction at the heart of this whole conceptual enterprise. The contradiction being that a curve, when viewed at an infinitesimal level, is flat. In other words, at the infinitesimal level, the curve becomes something that it is not - flat. And the 1st and 2nd derivatives of a flat anything are zero, thus making curvature impossible. I'm not saying this is correct, but I think we should not rule it out.

[AccountantEli]

i am following the whole lectures, but where is 13? i can´t find it!! thank you very much prof. susskind!!

my respects to the family of poor alice whose atoms disintegrated as she landed on the r=0 hyperbola...

[bbsonjohn]

@AccountantEli I think the series on GR is over

[Photonimus]

TripMaster extraordinaire!

Please sir, I'd like some MORE!

[wrighta04]

2.05, stinky paki

[EDGWEIGHTLIFTING]

I suggest also posting in the future classes of Mechanical Statistics and Electromagnetism. Thank you very much to Leonard Susskind and to Stanford University for this great initiative!!!!!

[EDGWEIGHTLIFTING]

I just finished with the whole series of this wonderful Physics classes posted by Stanford and taught by this extraordinary teacher Leonard Susskind. Leonard Susskind is smart enough to make very heavy stuff look very easy, easy to understand and easy to learn. Thank you Dr. Susskind for this wonderful classes I am very proud to have being one of your students at long distance, hope one day will be in person, and thank you very much also to Stanford University for this great initiative!!!!!

[DoubleDutchBust]

Theory of Relativity X

Einstein's General Theory

Justsaynobscarnineleven

E n d u p i n a t r a i l e r N

T h e o r y R e l a t t i v i t y

[Lambda45]

Thanks Leonard and Stanford for these excellent lectures. We eagerly await for Lecture 13!

[LogicalFlawDetector]

@Lambda45

This is a QUARTER COURSE douchebag! That's it. It's over.

[bscar911]

this is shit i like ta watch stond as fuck

[jdfghjkl]

wow so difficult

[wrighta04]

well, not really

[Wizard4592]

What a brilliant analysis of the Swartzchild metric and blackholes. Susskind also lectures on the event horizon and singularity for more than an hour.

[matu88jiujitsu]

thak you again. Lecture 13 coming up?

[freshramses]

thank you

[videoofmike]

OMG YES