Introduction to Riemannian Geometry - Covariant & Contravariant Vectors

[@hershyfishman2929]

50:14 In case anyone is confused, as I was: if in the case just mentioned where the new "~" scale was increased by a factor of 2 relative to the old one, the factor f was said to be 2 which is 2/1, not 1/2, why then is the factor f = dx1/dx1~ and not its inverse dx1~/dx1? Note that dx1/dx1~ is not the ratio of the old scale to the new scale (which is indeed 1/2), but the rate of change of the vector as given by the old measurement to the vector as given by the new measurement. In the above case it is 1/(1/2) = 2.

[@LibertyAzad]

Though the subject may be hard, I challenge anyone to find a more clear explantion of covariant and contravariant vectors. Brilliant!

[@lukebicknell7702]

I have just started to learn about special relativity in preparation for my relativistic quantum mechanics lectures, and this has helped massively! Thank you

[@faizahmed1601]

these videos are very helpful. it would be nice to upload a course on quantum mechanics and special theory of relativity as well as mathematical physics

[@fabriziorocchi346]

Thank you for the answer. I am going to try to digest it and see if I can grasp this very subtle difference. 😅

[@BuleriaChk]

3D flat cylinddrical space: # = a + b #^2 = [a^2 + b^2] + [2ab] p:= pi p(#^2) = p[a^2 + b^2] + a(2pb) - 2pb is a circumference, a and b are now radii note that a(2pb) is the surface area of a cylinder, but requires the existence of a and b. This equation is complete and consistent and covers all possible curvatures in 3D.

[@johnjobs3027]

I am an Amateur Physicist , on pilgrimage to the Holy Land of General Relativity. Thank you for your time and effort. Also, I was curious to know, if Reimann Geometry has applications outside General Relativity?

[@BuleriaChk]

The affine connection doesn't take into account the area between the geodesics on a sphere. (similar to the fact that the Pythagorean Triple does not take into account the areas of the four angles in a 2D quadrant: # = a+b #^2 = [a^2 + b^2] + [2ab] (Binomial Expansion, Fermat's Last Theorem for the case n=2) Note that 2ab = 4[(1/2)ab] where (1/2)ab is the area of the included right triangles, and is a model of interaction (multiplication, entanglement, entropy; e.g. Beckenstein in Hawking's "A History of Time" So like the "flat" space the "connection" and half of one of the geodesics must be imaginary. But there are no negative numbers: -c = a - b, b > a iff b - c = a, a>0, a+0=a. a-a = 0 If there are no negative numbers, there are no square roots of negative numbers. Note that #^2 = [a^2 + b^2] + [2ab] = [cc*] + [2ab] for c:= a + ib Relativity The correct analysis is that space (x = vt) is not included in the equation to be solved for the "time dilation" equation. (ct')^2 = (ct)^2 + (vt')^2 (solve it for t' for yourselve(s) to understand) and note that this equation cannot be generated from the "space" equation for length in first order (ct') = (ct) + (vt') (draw it on a piece of paper). Hint: If space doesn't exist, the twins don't go anywhere; one of them (the imaginary one) just gets fat in his/her imagination (t'). Which is why Hawking hints that time must be imaginary, but never says why. "Yesterday upon the stair I saw a man who wasn't there He wasn't there again today Oh, how I with he'd go away" - Ogden Nash See my post at "From MM Experiment to STR" at physicsdiscussionforum "dot" org That is, Fermat's Last Theorem is valid for the case n=2 for all positive real numbers c^2 a^2 + b^2 since in second order (I repeat, sigh. ad infinitum, ad nauseam) c= a + b c^2 = [a^2 + b^2] + [2ab] (Binomial Expansion, proved by Newton) [a^2 + b^2] (why) figure it out and you will be enlightened....😎

[@eddieboyes]

Fabrizio I agree that this can seem a bit confusing. The point is, I think, that you have to assume that, although the measuring system (the metre rule/stick) has been stretched to ‘twice its size’, it is STILL a metre rule. One way to look at this is to say that (in a sense) it is the size of the ‘metre’ which has changed - it has doubled in size. It is still called a ‘metre’ and, when using this new ‘metre’, any contravariant vectors will have reduced their MEASURED size’ (to half their original value), whilst any covariant vectors will have increased their MEASURED size (to twice their original value). The actual size of the vector is the same of course. In that sense, there is no “m*” to think about. Best Eddie

[@user-uz7nz2bo2j]

Eddie, I think Fabrizio is right. Bij changing the scale from e.g. meters in millimeters a potential with value 1000 changes to 1. So the numerical value changes (1000 into 1), and the dimension (or unit) changes as well (V/m into V/mm). So the Newtons in g = -5 N/kg in your example, after stretching are not the same "Newtons" in g = -10 N/kg. You cannot say "it is STILL a meter rule", unless you travel to Paris to physically change the meter rule that is kept there. Still, I appreciate the way you explain the subject!

[@user-uz7nz2bo2j]

Eddie, But I have my doubts about my comment... Lately I have googled a lot on "dimension, contravariant and covariant", or versions of that, and I did not get an answer. I think the subject "units (or dimensions) and contravariant/covariant" deserves a special KZbin-video

[@bereketsiz]

Muhteşem. Teşekkürler.

[@undergroundriddimandbassli972]

This is why I love math

[@RoyBurnell-o6n]

Lee Jason Johnson Gary Martinez Daniel

[@DorotheaJacob-c5s]

Gonzalez Brian Walker Linda Taylor Mark

[@joachimachtzehnter4772]

The attempt to develop an intuitive idea of the difference between an intrinsically curved space versus a flat space that is embedded in a higher dimensional space and looks 'bent' from that perspective, is commendable, but the attempt failed. This failure is most obvious when the video claims to show a curved one-dimensional space, when in fact there is no such thing. One-dimensional spaces are always flat. Two-dimensional spaces can be curved, but the example shown was likely flat too. These basic errors at such an early stage don't bode well for what future videos in this series may bring.

[@johngibbons494]

Fantastic explanations. You are a truly gifted teacher and truly a very smart guy. Many can understand the subject but very very few have the extra intelligence to know what the pain points are for people looking at this for the first time . I know the subject but it's a joy to review from your perspective. Thank you for taking the time.. your Liverpool videos are fascinating too. . God bless you.

[@eddieboyes]

Hershey - Good point …. I think that this could have been explained better. I can see how one might be confused. However I don’t intend to add anything to your clarification.

[@Altalex988]

Please Sir, can you explain in more detail this point? If x' are the new basis and x the old ones, why f isn't dx'/dx? So contravariant component follow the rule V'=V/f and covariant the rule V,=V f where f=dx'/dx and 1/f= dx/dx' not viceversa respect of what is show @54.24... thank you for the great content!!!

[@Altalex988]

Ahh ok! dx'/dx are the derivative vector of basis, but dx'/dx are new component of a vector in new basis that transform in the inverse way respect change of basis... Correct?

[@gabriellaprovera8077]

But if we changed metre from x to x tilda, the factor that stretchs vectors component should not be dx tilda / dx for covariant vectors?

[@letslearntogether2847]

@eddieboyes I have one doubt: Suppose if someone living in a 2d flat (cylindrical) space and driving a car in it. Will he/she will feel the centripital force or any kind of accleration?

[@j.k.sharma3669]

Very Intuitive series on general relativity. Your work on explaining basic concepts is fantastic . Please keep it up....

[@thishandledoesntexist]

At 17:38 , I have difficulty imagining how the other two ends were bent to create a curved dome. Could you kindly elaborate that point dear Sir?

[@yancymuu4977]

Have read comments below, and still feel confused. Have also watched other videos on covariant vs contra-variant and can't seem to get it straight. It seems to me that the field and potential should be same (same physics) if we change coordinates. But will continue on. By the way I like your patient way of explains things. This is the forth series of video on the subject that I have attempted. Hope I can get farther this time.

[@fabriziorocchi346]

Hi Eddie. First of all thank you very much for this nice series of videos, very clear and detailed unlike many others on KZbin about same subject. I have a question: at 45:46 you write the two values of g in the two frames respectively as - 5 and - 10 ms2. But since the blue x is not in meter but in a different measurement unit which we can call for example m*, where 1m*=2m, should not be g= - 10m*s2 rather than g=-10ms2? Otherwise if you use the same units you get the strange result that 5=10. Or am I getting this all wrong?