this is absolutely fantastic btw. I am a first year in a physics phd program, and i never really covered tensors in undergrad. The graduate program i am in didn't really explain how they worked well either. the students in my cohort also have had serious trouble explaining them nicely. but this, this is a good explanation of where all the ideas come from. this is well grounded and easy to digest. thank you very much
The whole point of inventing and using tensors is the last theorem: To arrive at invariants irrespective of the coordinate system used and that's why tensor calculus used to be called the absolute differential calculus.
I used a different trick to remember where the indicies go. Think of the "cov-" in covariant as "covert" or hidden, thus down below. Think of "contra-" in contravariant like you would in a crowd of protesters (contra = against) waving flags or shaking their fist up above their heads.
I am doing my master's in advanced physics and I'm writing a test 2 days from now in classical electrodynamics with special relativity. The proffessor didn't explain almost anything. This is genius explaination. Thank you so much you are saving me
Can you explain why the contravariant and covariant transformation laws are defined that way. For example in the context of how contravariant and covariant vectors were described in the previous videos.
Surprised this video is getting good comments. Must be alot of smart people out there. Compared to the first video, you totally lost me after the first minute.
I mean, if something can’t be explained to everyone, that doesn’t make a reasonably understandable explanation bad. So why would you complain if something is pretty much as good as it can get
Great video! About the example for covariant vector definition at 8:58: It might be worth to mention than F(x) is a scalar and therefore its value at a certain position in space does not change in the bar - system just by switching to the bar-coordinates - and that this is the reason why in the chain rule dF_bar/dx_n can be replaced by the original dF/dx_n in order to fulfill the covariant transformation condition. Or do I have a misconception?
@ Faculty of Khan thank you for this great video! yet, can you please explain why we can write the yellow formula in 9:00? -> why can we substitute partial_derivitive_of_F_bar_with_respect_to_x_i_bar with u_sub_r ? (u_sub_r = partial_derivitive_of_F_(NO bar)_with_respect_to_x_i_bar )
i was wondering, too, but i think, in the expression of del(F_bar)/del(x_bar^i) with the chain-rule, the factors of the del(x^r)/del(x_bar^i) - terms shouldn´t be denoted as del(F_bar)/del(x^r), but rather as del(F)/del(x^r), as F_bar is the composition of F and x(x_bar). Then, the yellow formula follows immediately.
Can you please explain how did you get that transformation relation for contravariant and covariant vectors at 1:55 ? Everything was fine so far this is where I got stuck and I can't figure it out
I'm confused. Why were you using x super i for the independent variables of the sensor while for the components of the covariant tensor you used U sub i? The basis has subscript, the x has super, and the tensor's components has both?
At around 8:50 how is it that parital F bar by parital x super r is equal to partial F (no bar) by parital x super r as you stated earlier with the gradient definition? Does it not make a difference that F bar and F are not the same?
The covariant proof appears invalid. U_r =! The partial derivative of F bar with respect to x^r. Rather: U_r = partial derivative of F with respect to x^r. Unless, U_r = U_r bar
@8:00 an error in application of chain rule. He wants to calculate the gradient of Fbar(xbar) wrt the gradient of F. The rhs, should be F and not Fbar. This is necessary to relate transform of F to Fbar and show grad F transforms like a covector. In previous video, he said the dot between vector and unit vector, results in size of perpendicular component. That's another mistake. A much more serious one.
@ 8:25 When using the chain rule the partial derivative of F bar with respect to x bar i is equal to the partial derivative of F with respect to x 1 multiplied by the partial derivative of x 1 with respect to x bar i ... ... + the partial derivative of F with respect to x n multiplied by the partial derivative of x n with respect to x bar i? Just wanted to clarify... it looks like that what you wrote but that’s not what you said as you were writing it. Oh and excellent videos! I like watching the Khan Academy videos before I start watching online lectures because you guys do a great job of getting to the point. Thanks!
I gather that F being a scalar and theta being a vector is crucial in these derivations. Is that correct. In other words F equals F bar but theta does not equal theta bar when we change coordinates. Or maybe theta is actually equal to theta bar- not sure about this.
@8:00 you have an error in application of chain rule. You want to calculate the gradient of Fbar(xbar) wrt the gradient of F. The rhs, should be F and not Fbar. This is necessary to relate transform of F to Fbar and shoe grad F transforms like a covector.
i thought that too, but it s kinda trivial tbh. just substitute the formulae for the contra/covariant elements into v^j or u_j respectively and out you get that the result is independent of the transformation function.
As Jendrik mentioned, the proof is quite easy if you use the definitions of contravariant and covariant elements and make the relevant substitutions. But just in case you're really curious, here's the proof (literally a back of the envelope calculation haha): imgur.com/Ztdro1N
@@FacultyofKhan aha! Yeah I see this was quite trivial. We proved this for the center of mass energy in our course this semester, oopsie! You got to continue your dynamical system series again! We studied the ising model of ferromagnetism and I see so many similarities to it (Minimizing free energy in Landau (Ginzburg) theory---> Minimizing potential energy, normal form of the phase transition and so on) I will definitely take a non lin dynamics course, stability analysis is really exciting
I have a question: from the previous episode, I learned that a vector could be represented in a contravariant or covariant form, but here 9:49 said certain physical parameters (velocity, acceleration..) are contravariant while few (gradient) are covariant. Are there any contradictions?
Thanks for pointing that out! What I meant was that mathematically, you can have contravariant and covariant vectors (e.g. a vector [1 2] could be written in the contravariant form or covariant form). However, specific quantities like velocity and acceleration and gradient have to obey specific transformation rules by their physical nature, so velocity/acceleration must be contravariant and gradient must be covariant by virtue of its physical nature. It's a difference between mathematical quantities and physical quantities (which are more restricted). Hope that helps!
@@FacultyofKhan Oh right! Are there any books or videos that detail how the specific transformation rules enforcing these physical vectors to be contravariant or covariant? Thank you for your nice videos. You are awesome!
may be you can slow down a bit while explaining the concepts so that we can catch up with the concepts while you are explaining. Not like pause the video watch it all over again.
I find adjusting the play speed works well. You can do this buy clicking the little 'cog' icon in the bottom right of the screen. Even better for is to pause every now and again to stop and think about what is being covered. This is the luxury of having youtube videos right? You have a top quality lecture that you can follow at your own pace. I think if this video was covered in a class room it would usually take 30 minutes or more for the lecturer to cover, not including time to answer student questions. Make life easier on yourself! you don't have to try and take everything in at the speed it's delivered.
but, i think he means a vector v belongs to the vector field V (which defined above). en.wikipedia.org/wiki/Covariance_and_contravariance_of_vectors#Definition
this is absolutely fantastic btw. I am a first year in a physics phd program, and i never really covered tensors in undergrad. The graduate program i am in didn't really explain how they worked well either. the students in my cohort also have had serious trouble explaining them nicely. but this, this is a good explanation of where all the ideas come from. this is well grounded and easy to digest. thank you very much
The whole point of inventing and using tensors is the last theorem: To arrive at invariants irrespective of the coordinate system used and that's why tensor calculus used to be called the absolute differential calculus.
I used a different trick to remember where the indicies go. Think of the "cov-" in covariant as "covert" or hidden, thus down below. Think of "contra-" in contravariant like you would in a crowd of protesters (contra = against) waving flags or shaking their fist up above their heads.
This is a very good lecture, you are dong a great service to physics and mathematics…
I am doing my master's in advanced physics and I'm writing a test 2 days from now in classical electrodynamics with special relativity. The proffessor didn't explain almost anything. This is genius explaination. Thank you so much you are saving me
9:00 how can you do that substution? is F bar the same as F?
I think this is one of most fantastic i have ever watched
for 8:48, remember that F is a scalar filed, so it does not change in coordinate transformation.
the memorizing tool is interesting. I love these clever tricks.
Can you explain why the contravariant and covariant transformation laws are defined that way. For example in the context of how contravariant and covariant vectors were described in the previous videos.
Surprised this video is getting good comments. Must be alot of smart people out there. Compared to the first video, you totally lost me after the first minute.
This video is just a more rigorous definition then in the first video:) Don't let it scare you by all the strange symbols
skill issue
I mean, if something can’t be explained to everyone, that doesn’t make a reasonably understandable explanation bad. So why would you complain if something is pretty much as good as it can get
Great video! About the example for covariant vector definition at 8:58: It might be worth to mention than F(x) is a scalar and therefore its value at a certain position in space does not change in the bar - system just by switching to the bar-coordinates - and that this is the reason why in the chain rule dF_bar/dx_n can be replaced by the original dF/dx_n in order to fulfill the covariant transformation condition. Or do I have a misconception?
Thank you for your interpretation!
Thank you for the explanation. I also had a question about it.
@
Faculty of Khan thank you for this great video!
yet, can you please explain why we can write the yellow formula in 9:00? -> why can we substitute partial_derivitive_of_F_bar_with_respect_to_x_i_bar with u_sub_r ? (u_sub_r = partial_derivitive_of_F_(NO bar)_with_respect_to_x_i_bar )
I had the same question. One year after =)
i was wondering, too, but i think, in the expression of del(F_bar)/del(x_bar^i) with the chain-rule, the factors of the del(x^r)/del(x_bar^i) - terms shouldn´t be denoted as del(F_bar)/del(x^r), but rather as del(F)/del(x^r), as F_bar is the composition of F and x(x_bar). Then, the yellow formula follows immediately.
Extremely well explained. I was so confused about contra / co - variance of vectors / tensors.
Can you please explain how did you get that transformation relation for contravariant and covariant vectors at 1:55 ? Everything was fine so far this is where I got stuck and I can't figure it out
Watch the tensor videos by andrew Dotson
@@athul_c1375 Thank you. I will
@@athul_c1375 can you share a link please. Im also lost at 01:55
@@sphakamisozondi
ruclips.net/p/PLSuQRd4LfSUTmb_7IK7kAzxJtU2tpmEd3
Somewhere in this playlist
Same here
I'm confused.
Why were you using x super i for the independent variables of the sensor while for the components of the covariant tensor you used U sub i?
The basis has subscript, the x has super, and the tensor's components has both?
At around 8:50 how is it that parital F bar by parital x super r is equal to partial F (no bar) by parital x super r as you stated earlier with the gradient definition? Does it not make a difference that F bar and F are not the same?
Got any explanation for that please? i am stuck there also. thanks
Faculty of Khan , can you please tell me what you actually use for making this kind of videos
And a bit about process
The covariant proof appears invalid. U_r =! The partial derivative of F bar with respect to x^r. Rather: U_r = partial derivative of F with respect to x^r.
Unless, U_r = U_r bar
@8:00 an error in application of chain rule. He wants to calculate the gradient of Fbar(xbar) wrt the gradient of F. The rhs, should be F and not Fbar. This is necessary to relate transform of F to Fbar and show grad F transforms like a covector.
In previous video, he said the dot between vector and unit vector, results in size of perpendicular component. That's another mistake. A much more serious one.
Simply beautiful. Thank you.
@ 8:25
When using the chain rule the partial derivative of F bar with respect to x bar i is equal to the partial derivative of F with respect to x 1 multiplied by the partial derivative of x 1 with respect to x bar i ... ... + the partial derivative of F with respect to x n multiplied by the partial derivative of x n with respect to x bar i?
Just wanted to clarify... it looks like that what you wrote but that’s not what you said as you were writing it.
Oh and excellent videos! I like watching the Khan Academy videos before I start watching online lectures because you guys do a great job of getting to the point.
Thanks!
Why are the co-ordinate systems represented with superscripts? Can't we use "x subscript i" for the co-ordinate system?
Amazing explanation, thank you
I gather that F being a scalar and theta being a vector is crucial in these derivations. Is that correct. In other words F equals F bar but theta does not equal theta bar when we change coordinates. Or maybe theta is actually equal to theta bar- not sure about this.
where did you get that formula at 1:55 ?
I never found this out
@@monojitchatterjee3185 I think it will help you ruclips.net/video/bohL918kXQk/видео.html
@8:00 you have an error in application of chain rule. You want to calculate the gradient of Fbar(xbar) wrt the gradient of F. The rhs, should be F and not Fbar. This is necessary to relate transform of F to Fbar and shoe grad F transforms like a covector.
Pleasee upload videos on tensor calculus related orthogonal axes. Please its a request.
What is example of covariant tensor of second order
1:55 if xbar =2*x then vbar=2*v right! Then it should be covariant right? I dont understant where I am going wrong!!!
what is the book of reference?
Does the theorem at the end have a special name? I want to read the proof.
Thank you!
i thought that too, but it s kinda trivial tbh. just substitute the formulae for the contra/covariant elements into v^j or u_j respectively and out you get that the result is independent of the transformation function.
As Jendrik mentioned, the proof is quite easy if you use the definitions of contravariant and covariant elements and make the relevant substitutions. But just in case you're really curious, here's the proof (literally a back of the envelope calculation haha): imgur.com/Ztdro1N
@@FacultyofKhan aha! Yeah I see this was quite trivial. We proved this for the center of mass energy in our course this semester, oopsie!
You got to continue your dynamical system series again! We studied the ising model of ferromagnetism and I see so many similarities to it (Minimizing free energy in Landau (Ginzburg) theory---> Minimizing potential energy, normal form of the phase transition and so on)
I will definitely take a non lin dynamics course, stability analysis is really exciting
I love you!
I have a question: from the previous episode, I learned that a vector could be represented in a contravariant or covariant form, but here 9:49 said certain physical parameters (velocity, acceleration..) are contravariant while few (gradient) are covariant. Are there any contradictions?
I don't understand why these physical values cannot be expressed in both forms but choose specific one?
Thanks for pointing that out! What I meant was that mathematically, you can have contravariant and covariant vectors (e.g. a vector [1 2] could be written in the contravariant form or covariant form). However, specific quantities like velocity and acceleration and gradient have to obey specific transformation rules by their physical nature, so velocity/acceleration must be contravariant and gradient must be covariant by virtue of its physical nature. It's a difference between mathematical quantities and physical quantities (which are more restricted).
Hope that helps!
@@FacultyofKhan Oh right! Are there any books or videos that detail how the specific transformation rules enforcing these physical vectors to be contravariant or covariant? Thank you for your nice videos. You are awesome!
@@FacultyofKhan I watched your second video again and understand it. It is clearly explained. Thanks!
@@FacultyofKhan Velocity is a contravariant vector because it's a tangent field. But what can be said about position?
Is this from Schaum's outlines? Nice work!
It is!
which Schaum?
@@pierluigidesimone6971 Schaum's outline of theory and problems of tensor calculus by David Kay
THANK YOU!
may be you can slow down a bit while explaining the concepts so that we can catch up with the concepts while you are explaining. Not like pause the video watch it all over again.
I find adjusting the play speed works well. You can do this buy clicking the little 'cog' icon in the bottom right of the screen. Even better for is to pause every now and again to stop and think about what is being covered. This is the luxury of having youtube videos right? You have a top quality lecture that you can follow at your own pace.
I think if this video was covered in a class room it would usually take 30 minutes or more for the lecturer to cover, not including time to answer student questions. Make life easier on yourself! you don't have to try and take everything in at the speed it's delivered.
1:31 How can a vector field be a vector? A vector field is a set of vectors.
I also have the same question...
but, i think he means a vector v belongs to the vector field V (which defined above). en.wikipedia.org/wiki/Covariance_and_contravariance_of_vectors#Definition
YOU GET A COMPLETE CHAOS SO NOLIKE FIRST LEARN THEN TEACH?!!!
U SUB r. U SUB i. NOOOOOOO!!!!!!!!
too many adds you must be a billionaire by now
You teach too mechanically, bro. But thanks all the same.