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Tensors Explained Intuitively: Covariant, Contravariant, Rank

5879 ratings | 296391 views
Tensors of rank 1, 2, and 3 visualized with covariant and contravariant components. My Patreon page is at https://www.patreon.com/EugeneK
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Anthony Townsend (3 hours ago)
If the Tensor concept is crucial to Relativity and is a creation of Riemann how is Relativity Einstein's idea?
Areeba Shamshad (13 days ago)
hatts off to ur mindset👏👏👏
umamaheswara reddy (13 days ago)
In this video,you said that a tensor of rank 3 is composed of 3 vectors. Could you pleassse explain what exactly the word "compose " mean? Is it a multiplication or addition or anything else?
Alessandro Rovetta (15 days ago)
Great Music, great Math. The best possible combination.
Abhishek Kumar (23 days ago)
Animation speaks louder than words.
Amruta Poojary (30 days ago)
What is the name of the background music
William_Tell_Overture_by_Rossini from the free YouTube audio library
Yunnik (1 month ago)
So every rank n tensor can be created from some combination of n rank 1 tensors?
Independent Mind (1 month ago)
@Origami and pubs Kk I am also a pianist, Modest player though but crazy enough to venture into the Pathetique, the Moonlight and the Tempest. All them very demanding. Congratulations to you if you can tame LvB’s passion on the piano with any of his sonata’s. However the Pastorale fits better this video.
Neph1l1m999 (1 month ago)
PLease stop this background music :)
Bahar JafariZadeh (1 month ago)
what the hell of the music!
Christian Lingurar (1 month ago)
a tensor of rank one is a vector. got it. just like in thumb down the vector points downward.
Origami and pubg Kk (1 month ago)
Amazing animation. I think you used some open GL or special shaders or something to render it. The animation was a great aid to visualising the vectors and components, which is often hard to do.
c s (1 month ago)
No background music is required. Awesome video. But background music is annoying and distracting. Science itself is musical and rhythmic.
pierluigi desimone (1 month ago)
my bit: better no music, it is distracting. Also no need to introduce the need of a base change to explain contra-variance. If you double the unit length then vector coordinated will half. This hold true also in the Cartesian coordinate system. Thank you
klick2destruct (1 month ago)
Nothing against William Tell but this tutorial combined with the monotonous voice feels like someone's dying. Keep it light on the music, would you?
Áron Ecsenyi (1 month ago)
Very good video, thank you :)
Kevin Wright (1 month ago)
Suppose we just shove some numbers together in some particular order. Not going to say *why*, but hey... at least they're swaying constantly. Suppose we then claim this to be intuitive.
physnoct (1 month ago)
Good explanation of co/contravariant. I like how the music change with rank 3 tensors!
Thanks.
Joseph Noonan (1 month ago)
8:00 I think you mean "every possible permutation".
Maria Callous (1 month ago)
It's official . If I were in charge we wouldn't ever leave the planet.
Mohammad Haithem (1 month ago)
Why does the music have to be so dramatic....
furzkram (1 month ago)
Audio way too low
Hasan Shirazi (2 months ago)
Great explanation of Co-variant and Contra-variant Tensors.
Thanks.
MilesBellas (2 months ago)
Classical music not needed and distracts
KSITIZ KUMAR (2 months ago)
You are a math god !!
Thanks.
Marco Barbieri (2 months ago)
Despite this being a great animation (like the one about Fourier transforms, which is even much better) this video I feel an inconsistency lurking with regard to the statement that the dot product decomposition is covariant. Let's take the most simple example of three orthogonal basis vectors and an arbitrary vector (like the situation around 20 seconds in this video). Now all the components of this vector are the dot product (orthogonal projections) with (on) the basis vectors. So if you make the basis vectors x times longer (or shorter) and giving this new basis vector the value 1 the components of the vector become x times as short (or long). But because the components are the dot product with the basis vectors, also the dot product decomposition becomes x-times as short, and this result is passed on to the case where the basis vectors are not orthogonal. Look for example at the video at around 2:58, where it is said that if you make the basis vector twice as large the dot product becomes twice as large too, but the basis vector you make twice as large gets again the value 1 and the corresponding vector component becomes twice as small (like is explained earlier: if you make the base vectors twice as large, the vector's components get twice as small), so each of dot product of the vector components with the basis vectors becomes x times smaller (larger) if you make the basis vectors x times larger (smaller), hence contravariance. A good example of a covariant vector follows from the (x,y,z) vector. This is a contravariant vector, but the (1/x,1/y,1/z) vector is a covariant one. More concrete, the wavelength vector [which corresponds to (x,y,z)] is a contravariant vector while the wavenumber vector, the number of waves per unit length, is a covariant vector [which corresponds to (1/x,1/y,1/z)]. See Wikipedia's "Contravariant and covariant" article.
Ass Möde (2 months ago)
my brain hurts
stanis083 (2 months ago)
Your video sir/madam is f****g amazing!
Alaa Alajmy (2 months ago)
I am adding a comment here because more people REALLY need to find this in their YouTube search engine :) God bless your soul!
Thanks.
nagesh kothawale (2 months ago)
Animation is the very effective platform to visualise the mathamatics..and this channel is very good to doing that.
Thanks.
Great video but Rossini's music is very nice but does not help to stay concentrated.
SomeRandomPerson (2 months ago)
So then is a rank-n tensor formed by multiplying the appropriate co-variant/contra-variant components of n different vectors?
Webkinz Kz (2 months ago)
great explanation
varun varun (2 months ago)
this is most beautiful video i ever found on youtube.. huge respect for the team who made it
Thanks for the compliment.
Solveig Van (2 months ago)
Oh this was the INTUITIVE explanation. Well, shit. Looks like I need a lot of catching up to do. I'm out of my depth... for now.
柴士童 (3 months ago)
physics is likely to be closely related to math
Three Camels (3 months ago)
No such thing as gravity or "space/time".    Einstein was a hack and a fraud.
Tommy 77 (3 months ago)
I just know that a vector can also be described using its dot product with the bases, but how do I show that such description will give a unique representation of that vector?
Rachel Ginsberg (3 months ago)
Also, I liked the music :) It matched the excitement I felt at finally understanding this!
Rachel Ginsberg (3 months ago)
Thank you so much. I've been trying to get some sort of intuition for what a tensor is, and this is definitely the best video I've found to help me with that.
MrDeep414 (3 months ago)
I dislike this video, due to I can not concentrate on text. Remove music please!
Dan Woodall (3 months ago)
The music is distracting
陈绍伍 (3 months ago)
Watch this video https://www.youtube.com/watch?v=f5liqUk0ZTw together, then you will have a better understanding
Kevin Byrne (3 months ago)
For DECADES I've searched for an explanation of tensors that's as simple as the one that you've presented here in less than 12 minutes. Thank you, thank you, thank you ! I am in your debt.
Glad my video was helpful. Thanks.
Mark Masterburg (3 months ago)
great video but no music needed... its horrible.
Claudio Saspinski (3 months ago)
I have followed the excelents videos of Leonard Susskind, and that concepts were used a lot in general relativity. But today, for the first time, I see a definition and why the names: contravariant and covariant. It is amazing how good is this video. Thanks a lot for that. I think that if that videos were available when prof. Susskind recorded his lectures, he would tell the students to watch that video before the next class!
aka izo (3 months ago)
about 8:00 i am not sure understand it well, we have basis (5 yellow, 2 red, 3 green) so i can use T^11= (4 yellow, 1 red and 2 green), and (5 yellow, 0 red 0 green). what is the exact definition of "every possible combination of these two basis vector?? can you give some clear example of 2 rank tensor has same basis.
Rodrigo E Toobe (3 months ago)
No, we "associate" a number with each basis vector, this number is so called component of the vector/tensor Suppose we have a Force vector F and a displacement vector x, we introduce now a coordiante system, say cartesian or polar=cilindrical , we express these vectors in contravariant form [[In index notation and youtube comments limitations, F_i mean F sub i, Fi mean F super]] That is: Fi and xi (contravariant components, covariant basis asociated) if we make the tensor (Fi xj) with each combination as shown in the video we get a tensor of rank 2 full contravariant, you know what Force dot product distance is, it's work done by the force and its a number (a tensor of rank 1) wich can be calculated from the tensor by the trace of the matrix (sum over i of this multiplication: Fi*xi). But this way of doing it has a problem, in cartesian coordinates is OK, but in any other coordinates the work done by the force is different than cartesian, ¿How to make it in such a way that the work done is the same for every coordinate system? Well the "trick" is to make one of the vector *components* Fi or xi, a covariant component (u need metric tensor to transform from one type to another) so we make a new tensor, Fi*x_i, and this new tensor is called mixed tensor or contra-covariant tensor and this tensor is great because, the trace of this damn tensor is a number wich is the same number for every coordinate system, also de determinant of the mixed tensor is the same The old tensor (the Fi*xi full contravariant components) has another preperty, that is the Determinant of the tensor ||Fi xi|| in one coordinate system (say cartesian) is proportional to a number obtained by the determinant of the Jacobian ||J|| to any other coordinate system (primed system) ||Fi xi|| = ||J|| ||Fi' xi'|| This last one property is more advanced and is incomplete (u need to be carefull wich jacobian to use). The exponent of the jacobian is called weight of the tensor, the metric tensor (in full covariant or full contravariant has a Jacobian of exponent 2 or -2 so the metric (expressed in full covariant or contravariant components) is called a *relative tensor* of rank 2 and weight 2, In a mixed tensor (contra-covariant) the weight is 0, so a mixed tensor is a relative tensor of weight 0 You can play with cartesian and cilindrical coordinates to get used by tensors and stuff the metric (full covariant one ) of cartesian is the identity matrix,,, the metric of polar (the full covariant one) is almost the identity matrix but a factor of r^2 in the middle,(1 0 0, 0 r^2 0, 0 1 0) and the mixed metric of both ones is the identity matrix
M6BrokeMe (3 months ago)
That was some of the most intense mental masturbation I have experienced in a very long time. Thanks!
Cindy Tsai (3 months ago)
Thank you !!
You are welcome and thanks.
Jeff Wiebe (3 months ago)
Appreciate the helpful instruction. Thank you.
Thanks.
Jayarava Attwood (3 months ago)
By the end I am slightly sea sick from the swinging motion of the figures. It adds nothing by distraction. And the William Tell Overture is also just distracting. I'm trying to concentrate, but you keep interrupting me. Any animation effects or music should contribute something to the learning process - not add random bits of non-essential information. This is poor pedagogy.
Jayarava Attwood (3 months ago)
At 5:05 "Suppose we multiply one of the contra-variant of V with one of the contra-variants of P". It's not clear why we would do this or what it achieves.
Iadved Czahnaett (3 months ago)
Explenation of rank 3 tensor *William Tell overture ensues* ayy lmao
The Vegg (3 months ago)
The problem with your nice video is where you increase the dot product to show covariance. Sure, the increase in length of the basis vectors increase the dot product BUT you have then entirely broken the connection whereby the dot product length projected perpendicularly to the tip of the vector actually projects to the tip. Following that logic, I could say the increasing basis vectors INCREASES the contravariant components as long as I'm not constrained to having the projection of the length reach the tip of vector. What am I missing, please.
Jorge Guzmán (4 months ago)
If I describe vector in terms of dot product with each of basis vectors?
Pendragon (4 months ago)
I watched this when it came out but now i actually need it for class, i love this channel!
Glad my video was helpful and I am glad you like my channel. Thanks.
Jacob011 (4 months ago)
This is awesome! I FINALLY understand all that co-variant and contra-variant business. I've never seen it explained so well.
Thanks.
Nathaniel Neubert (4 months ago)
This is making me sick...
Carlos Cardenas (4 months ago)
What a great idea to explain as you do it. Thanks!!!
Thanks.
N VDL (4 months ago)
The music is well matched. Thank you
The Vegg (4 months ago)
Finally, I know of two definitions of covariant that seem to not say the same thing. https://www.google.com/search?q=how+do+you+determine+the+covariant+components+of+a+vector%3F&safe=active&source=lnms&tbm=isch&sa=X&ved=0ahUKEwjW3oKzmr3dAhUJvFkKHXzjA9QQ_AUIDigB&biw=1821&bih=868#imgrc=qUwQ6AUw4nFf5M: 1-The above image states that covariant components are projections perpendicular to the x axis while contravariant is parallel to the y axis. 2-Covariant components are found by taking the dot product of the vector and the basis vector. In 2, I can see how increasing the basis vector increases the scalar result (covariant) In 1, increasing the basis vector has NO effect on the projection since the light source perpendicular to x axis does not change and the vector length does not change. (non-variant).
The Vegg (4 months ago)
Can you please relate the dot product to the explanation in this image? https://www.google.com/search?q=how+do+you+determine+the+covariant+components+of+a+vector%3F&safe=active&source=lnms&tbm=isch&sa=X&ved=0ahUKEwjW3oKzmr3dAhUJvFkKHXzjA9QQ_AUIDigB&biw=1821&bih=868#imgrc=qUwQ6AUw4nFf5M:
The Vegg (4 months ago)
in keeping to my previous question, I notice when the covariant components double because of doubling the basis vectors, you make the (90 degree notation and the line pointing to the tip of the vector) disappear. is that because it is impossible to keep it at 90 degrees if you change the length of the covariant basis vector? Does changing the length of a basis vector from 1 to any other length (in covariant measurement) change the angle from 90 to other degrees? If so, what is the significance of the 90 degrees at the magic (1) for a basis vector? I assumed that the dot product was the projection of the vector on to the basis vector, but if that is true then changing the length of the basis vector would cause no change to the projection and as such the value of the component in the covariant measurement.
The Vegg (4 months ago)
I agree with Jesus. Also, if the dot product of a vector is tantamount to the projection of the vector on the basis, why would increasing or decreasing the basis increase or decrease the number. The projection is the same no matter the length of the basis. Thanks for your great animations.
The dot product is more than just the projection. I explain how to visualize dot products in my video at https://youtu.be/h0NJK4mEIJU Thanks.
Sorry to sound harsh, but I didn't found this very intuitive. Suppose we multiply the co-variant component with the contra-variant components to get this matrix. Sure...you CAN do that, but WHY would you want to that? There is where intuition lies, in my opinion.
Rodrigo E Toobe (3 months ago)
if you do that you get the mixed metric wich is always the Identity matrix. In the video in 8:00 the components of a tensor of rank 2 are a numbers associated to each combination of the basis (idk if you falled in that confusion but i just wrote that just in case) If you multiply the covariant with the contravariant you get a nice matrix, this matrix has a two special properties(and probably more): 1) The trace of that matrix is the same for every coordinate system (is invariant) u can check it by doing cartesian and polar coordinates 2) the determinant is also the same for every coordinate system, determinant is the area enclosed by two vector. If you look at the other tensor variation (full contravariant or full covariant) you cant do that.
I love your vids
Thanks.
m33LLS (4 months ago)
Great video! Could you do a video about stress (or strain) tensors?
Zes (4 months ago)
no such thing as way or usual or not, say anyx
Cat22 (4 months ago)
Eugene Khutoryansky: Can you do 2 videos - 1. on the vacuum energy and virtual particles (quantum foam) what are they, do they have charge? can energy be extracted from the particles? etc 2. on quantum entanglement 3. Bonus: How about one on 'dark flow' where it appears galaxies are moving towards a specific point in space. Your video's are very logical and easy to follow - excellent work!
Swarna Kumar S Naik (4 months ago)
I'm seriously saying this , your all videos are tremendous .
Thanks for the compliment.
ibrahim lakhdar (5 months ago)
contravariant and covariant components are equal in a orthonormal basis,am I wrong?
You are correct.
Philippe durrer (5 months ago)
Absolutely phenomenal video, i really wish we had these to study 8 years ago. I finally understood the difference between co and contravariant .... before i just knew the definition
ckp design solutions (5 months ago)
ma'am may i ask which software do you use to make videos
I make my 3D animations with "Poser."
Lyman Paul (5 months ago)
very cool. great job!
Thanks for the compliment.
Kunal Garg (5 months ago)
Awesome graphics and explanation. Nice work. Keep it up.
Thanks for the compliment.
Steve S (5 months ago)
At last a simple illustration on the difference between co variant and contravariant components along with associated indexing. Brilliant.
Thanks.
Marco Furlan (5 months ago)
Did you notice the cat at 9:10?
Dimi S (5 months ago)
Дикторше очевидно пох на то о чем она ведает...
Adrián Ramírez (5 months ago)
Compairing with the abstract learning in math books and engineering courses, I believe this kind of videos in YTube present a different perspective on math learning, and are very helpful to clarify concepts and equations. Thank you.
Thanks.
Ana Maria Quintal (5 months ago)
Some times the discourse of the music colides with the speech of resoning. Here we have It.
Anyone know how he got the values of 1.826, 5.055 and 3.567 for the dot products at 1:04? Is there enough information on the screen at 1:04 for us viewers to be able to calculate these values? I thought we would at least need to know the angles between each basis vector (red, green, yellow) and the white vector? Also, is the length (magnitude?) of the white vector = square root of (2^2 + 4^2 + 6^2) = square root of 56? Thanks to anyone who can help!
Ken Jenks (6 months ago)
A very clear explanation of tensors. Nice, simple graphics.
Thanks.
rewtnode (6 months ago)
The oral explanations were very good. I realize though that I found the fancy graphics being of barely any use. The explanation of covariant versus contra variant was great. The graphics coming with that explanation however was mostly a distraction. However, when you letter explain 2nd and third order tensor I found the graphics quite useful.
Reliable Beast (6 months ago)
The spinning diagram gave me motion sickness.
Frazer Kirkman (6 months ago)
After 2 minutes, you no longer explain anything, you are just making definitions. I wish you had explained why you would multiply those different basis vectors together.
RAVI SHANKER (6 months ago)
Tensors for Beginners: https://www.youtube.com/playlist?list=PLJHszsWbB6hrkmmq57lX8BV-o-YIOFsiG
Nick (7 months ago)
All very well. But dot product - well of course everyone knows what that means!
I cover dot products in my video at https://youtu.be/h0NJK4mEIJU
umeng2002 (7 months ago)
Having a good instructor makes a night and day difference when learning more advanced subjects. Great video. Making the jump from just dealing with vectors to tensors trips up a good number of people.
-/ChaNNel713/- (7 months ago)
Умоляю, сделайте перевод на Русский Translate in Russian, pleeeease
Akshay Sunil Bhadage (7 months ago)
3:53 Avengers:Infinity War vibes for a moment because of music
Ian Haggerty (7 months ago)
Yessss! Finally an explanation behind the terminology "covariant" and "contravariant". It's alien language like this that can really throw me off learning new topics in physics & math. MAHASIVE Props to you.
Thanks.
Cees Timmerman (7 months ago)
So a tensor (always 3 dimensional?) has n ranks, where n is the number of vectors it combines?
Mj B (7 months ago)
Great video - just for future videos keep in mind that sound//music is contra productive to some types of learners - like me. VERY distracting. And I like the music ;-)
foreverseethe (8 months ago)
I don't find this intuitive in the least. Misleading title. The term intuitive should be reserved only for concepts that laymen can generalize from everyday experience.
Robert Collins (8 months ago)
The video seemed to do a good job, but the background music was so distracting that I gave up about halfway through. (I liked the music, and I liked the video, but I could not understand the narrative because of the music.) Remove the music and you will have a much better video.
carmatic (8 months ago)
can you explain what a dot product is? what does it have to do with 90 degree angles?
carmatic (8 months ago)
super helpful, thank you!
I cover dot products in my video at https://youtu.be/h0NJK4mEIJU
Swapnil Udgirkar (8 months ago)
I found the background music a little distracting. Maybe tone it down a bit? The content was superb though. Great visuals and explanation! Thanks a lot :)
Daniel Ribas Tandeitnik (8 months ago)
Nice video, but the classical music was very odd... thank you very much!!
Thomas Lehner (8 months ago)
Good video but distracting music

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