Einstein's General Theory of Relativity | Lecture 8

[msjessypp]

some sweet info here

[thejameskan]

Very enjoyable thank you

[staranjela]

really informative and interesting

[ghostinthemean]

"By now I've lost the thread - I don't know how many times I've changed the sign". HA! The sun is still.

[thegreeensky]

mind bugging... great lecture.. very helpful..

[adelle0001]

nice one! informative video thanks for uploading

[agapitoflores001]

A little more hair then he would pretty much look like Einstein. 

[bobcrunch]

Great! Now the Wikipedia articles on relativity are making sense! The scales have fell from my eyes!

[DragonSlave49]

I'm reasonably certain that these have been posted out of order in the playlist... this is lecture 8/12!

[Cervy18]

"so...it's not so hard, is it?"

lol....

[nilsandrae]

Brilliant! :-)

[bhigr]

This is very technical. I wish he would get back to the physics. O.K. in the end he starts deriving the field equations...

[FryeNL]

@ Charly

You and a friend are standing on the equator, facing north. Your friend walks straight to the north pole. You however DON"T TURN, but walk sideways for a few hundred miles. After which you're still facing north. If you then walk north to meet up with your friend, you'll find him at an angle even though neither of you ever turned.

[Neavris]

@FryeNL Great Explanation, Thank you.

[MrCharlyAndy]

Prof. Susskind, to my mind, never did make clear the concept of parallel transport.  I think that's why questions were always coming up about it. I think it is important to clear this up.

[gruff5]

@MrCharlyAndy yes, agree with you - i was confused, until i looked up a simple diagram explaining it on wikipedia

[fermista]

@MrCharlyAndy Lol, he explained it like, a zillion times in the previous lecture. I think he got a bit annoyed.. Take a curved line through space, put a vector on one end and shift the vector along the line, keeping the line parallel to itself. What that means is keeping the covariant derivative of the vector along that line equal to zero. In a flat space, the final vector is parallel to the initial, but in a curved space the final vector has been rotated relative to the first by a small angle.

[fermista]

@fermista Sorry, *keeping the VECTOR parallel to itself I mean, which means making the covariant derivative of the vector along the line equal to zero.

[ila1964]

very beautiful

(the physics not him)

[TSP105]

So lecture 1 had 241,941 views and lecture 8 has 15,775 views. Glad to have made it this far but I hope the worst of the math is over!

[LogicalFlawDetector]

@TSP105

Do you even understand Riemannian geometry? What mathematics classes and physics classes have you studied before you even started this lecture series?

[TSP105]

So lecture 1 had 241,941 views and lecture 8 has 15,775 views. Glad to have made it this far but I hope the worst of the math is over!

[punshs]

Ugh... this is such lazy math. I hope Susskind is more careful and rigorous in his own research.

[eggistential]

What level of math do I need to understand this? An undergrad in math?

[TheCrocoduck]

these lectures are like holy gospel

only, more awesomer! =D

[atjaybede]

der ist genial!

[LeMegasandwich]

Now we're getting somewhere!

[ogirv101]

This took 12 hours to finish, I'm glad that instead of procrastinating doing stupid things, I actually learned something very important.

[LeMegasandwich]

Nice work

[HaloSakura13]

@ogirv101 But there are twelve lectures and most are almost two hours long...

[asdffdaers]

I didn't understand what justifies using covariant derivatives instead of regular derivatives when analysing how the vector moves from a to a'. Change in V should be a regular derivative; we're not considering space at that moment, just the change in V. Furthermore, since this is parallel transport, doesn't that mean that V is designed, by definition, to have a zero covariant derivative along the whole path?

[kesim]

My answer to asdffdaers, part 1:

You could simply take the ordinary derivate according to your coordinate bases, but then the rate of change might differ in a different coordinate basis: f.e. a constant vectorfield in 2 dim. does not look constant in polar coordinates. Covariant derivative get's rid of rate of change in V due to coordinates.

[kesim]

My answer, part 2:

Now having a vectorfield constant in terms of covariant derivative along a curve - in other words: a parallel transported vector - leaves us with what might actually cause a change/rotation of the vector after beeing parallel transported along a closed path: the curvature of the space.

[barry123www]

1. He didn't take the "regular derivative" because he was calculating the derivative of a vector (not a scalar). To take the derivative of a vector in curved space requires the covariant derivative.

[barry123www]

2. No. Not in curved spaces.

[barry123www]

Parallel Transporting a vector around a closed in circuit in curved space always changes the orientation of the the vector

[xeroiced]

I can travel back in time

[barry123www]

Is the metric of your spacetime Godelian? Measure it for me using only a wristwatch, a ruler, and a flashlight.

[bogeuh]

can't believe some comments, he is not the greatest lecturer but thats not what he's supposed to be, he's great at what he does and the maths is so oabstract/general that its actualluy pretty easy to follow

[foketesz]

true, admirable content, very well explained. Always connecting with the real natural world (when possible)

[miltonbose]

this is for continuing students you smug slauge!

[XionXi]

I think I remember why I don't miss physics...

-_-

[murderface0666]

hahha, i would like to see any of you whiney bitches explain this theory over many hours without getting lost or confused. He is only human

[camino1ca]

My brain exploded

[kap2z]

Instead of gaining brain cells from listenting to this I think they exploded every time I heard dirivideds or whatever the fuck he said