It's probably going to take me a long while to keep all this straight in my head, but I'd just like to keep saying that I really appreciate all of these videos. Thanks so much for putting them all together!
This was very good! This is more abstract than how I learned it (as partial derivatives being used for a change of coordinates). Thanks to you, I’ve learned something new ☺️🙌🏽🎊
@13:50 seems like an error - there is no "B^mu" in the prior equations, it should be B^nu = A^mu (\Lambda^{-1})_mu^nu (and you probably want to explain why indices get switched with the inverse matrix). Agree with the comment below that the occasional swap of mu and nu is jarring, but otherwise this is an excellent series.
thank you for the video. at 18:15, you said we can transform our coordinate basis vectors and components vector with a linear transformation lamda and its inverse lamda^-1. however if we think of the transformation from cartesian to spherical coordinate, the component part transformation is not linear. so does this mean that spherical coordinate transformation is not the case of linear coordinate transformation because its component transformation in non-linear?
Coordinate transformation can be arbitrary and therefore can, an usually are, nonlinear. Cartesian to spherical is a fine example. But you are confusing the coordinate transformation with the *basis* transformation. Basis transformations are always executed linearly. A basis transformation is *derived* from a coordinate transformation, but we have not discussed how that works yet. However, you don’t need to wait for the appropriate lecture here, just go to any book about General Relativity and look for the expression about transforming a tensor component. It is created by taking the coordinate transformation’s various derivatives w.r.t. each coordinate and *evaluating the result at a specific point in space time.* What is left is just a collection of real numbers which can be interpreted as a matrix that operates (linearly) on the components.
@@XylyXylyX thanks. does that mean that at *each point* of space time, basis transformation is linear although coordinate transformation itself can be non-linear? for example, from cartesian to spherical, we can find 3 basis unit vectors (rho_hat, phi_hat, theta_hat) that are from vector space, and coordinate components (rho, phi, theta) *in real number* that can be linearly transformed to a vector in Cartesian coordinate *at a specific point P*(and each specific point in R3).
Burn Koll Yes, that is correct. Every basis vector of the new basis can be expressed as a linear combination of the basis vectors of the old basis. The transformation between the two basis systems is always linear. In fact the very definition of a “linear transformation” is all about changing basis vectors in a vector space.
Thank you for those very helping videos. I'd like some more precision about the manipulation starting at 11:00 with "w". Forgive me I stopped linear algebra a few years ago ; there's something I found non-trivial : to replace f(mu) you write f(nu) and replace Lambda (mu nu) by the transposed. That suppose a constraint on e(mu) to preserve equality between f(nu) and transposed-lambda*e(mu). Is that constraint coherent with the constraint on this equality A(mu)*e(mu)=B(nu)*f(nu). Einstein convention make it actualy look simple, but it's not that intuitive to me. I tried the demonstration with w=A(nu)*e(nu)=B(mu)*f(mu). Transposing lambda is indeed necessary to male it right, but I'm not confortable with it.
the arbitrary swapping of the mu's and nu's is killing me - making my head spin. i like your vids, but you lost me on this one after you went to a clean sheet @ ~8:50. I'll try again tomorrow.
Hi, I have a question about the inverse linear transformation. at 18:43. Lets say B^v is weights in cylindrical coordinates and A^v is a weight in cartesian coordinates. We can find the basis vector conversion matrix from cartesian to cylindrical coordinate easily (ex. rho_hat= cos(phi)x_hat+sin(phi)y_hat and we can find phi_hat, z_hat in the same manner) . Then if we were to express weights in cylindrical coordinates from cartesian coordinates, by using the relation B^v=A^v(T^-1)_v^u, then the inverse of conversion matrix must operate on cartesian weights . However we know that weights in cylindrical coordinate are not linear functions of (ex. rho=sqrt(x^2+y^2), phi=arctan(y/x)). Then how can we say B^v=A^v(T^-1)_v^u is true?
Because everything on both sides of the expression is a *number* not a function. At this point in the lectures we are not treating A and B as a vector field, only as a single member of a single vector space. You have promoted everything to vector fields. Now, if you consider A and B to be vector fields then they are functions of spacetime. But in that case the transformation matrix is a function of spacetime as well. The point is that for any single point in spacetime you can calculate a transformation matrix and also values for A. This will lead to a simply matrix product to give B. The fact that T is a function of spacetime handles all the non-linear nature of the coordinate transformation that concerns you.
I really appreciate you for your quick reply. I will watch your video upto lecture 14, Vector and tensor fields and then I might fully grasp what is actually going on. And your lectures are by far the best ever on youtube. Really, it feels like the materials come out from the lectures and naturally flow into my brain. Thank you!
Quick question, you seem to switch indexes when substituting components for their transformations. For example around 18:52 mark you have w = (B^ nu) (f_ nu) .You then use a substitution going from (A^nu) to (B^mu) for (B^nu) and the transformation from the basis( e_delta to f_mu) for (f_nu.).Is this intentional?
No, it is a *little* sloppy because I am recycling dummy indices and it would be cleaner if I always used different indices, even if they are paired up. In the case you cite I should have used, say, \alpha instead of u on A and \Lambda, then I would have used u instead of \mu on the transformations. Then it would \mu on both sides. It is ok because the indices are dummy indices on both sides of the arrow, but yes, it is a bit confusing as I have written it.
The basis vectors transform covariantly, which is to say that it gets transformed by the matrix not the inverse of the matrix, right? If It's so then can we still call the transfromation of e_{\mu} @13:44 as covariant transformation?
Michael Johnson I will not be providing problems. There are many books on this subject. Try Bowen and Yang “Introduction to Vectors and Tensor” (Dover). Good Luck!
While I understand that the 'f' basis vector transforms covariantly and that f's components 'B' transform contravariantly...based on how you've defined covariant and contravariant, is it then true that the 'e' basis vector is transformed contravariantly and that e's components 'A' transform covariantly? time point 15:43 is a good reference point, as all of the equations relevant to my question are located on the screen ~Thanks
Looking at around 3:55? No! All basis vectors, e or f or any other set of basis vectors will always transform covariantly. Otherwise there would be two classes of basis vectors in V. All basis vectors are basically equivalent and transform the same way. Likewise their components also always transform contravariantly. That is, basis transformations do NOT take covariant things and turn them into contravariant things or the other way around. THey just reorganize covariant things into other covariant things. The index notation reflects this and basis transformations do not change the index positions of the objects they operate on.
I don't know why you are referencing minute 3...but I mostly meant that at around minute 15, you have an equation for the e basis vector that involves the inverse of our transformation matrix (i.e. to get e in terms of f). You also have an equation for A (the components of the e basis vector) that transforms using the non-inverse transformation matrix (in order to get A in terms of B). In accordance with your definitions of contravariant and covariant, where covariant transformations involve the non-inverse and contravariant involves the inverse, I was slightly confused.. But thank you for your quick response. Cheers~
Samuel Cramer Oh, I see. Yes....we always need to track what is the “forward” direction of a transformation. In this case we started with e and transformed to f using \Lambda. Obviously, to go back to e from f requires \Lambda^(-1) but that must be understood as just another basis transformation and you could just rename the transformation \Gamma, say. Notice that the index position does NOT change when going from f -> e so the resulting object is still covariant. The way to understand this is that objects transforming in the same direction but using inverse transformation matrices have a co/contra relationship. So if f->e uses \Lambda^(-1) then its components will use \Lambda^(-1)^(-1)=\Lambda. But f is transforming in the opposite “direction” from e so they cant possibly share a co/contra relationship. Sitting in the f basis the “forward” direction is the matrix \Lambda(-1) which would define the covariant transformation and its contravariant partner is \Lambda. Renaming the transformation to \Gamma might help. I see the possibility of confusion here, you are not seeing things. I have written, correctly, that f-> e uses the inverse of \Lambda, but this is a trivial and obvious use of the inverse transformation. The fact that the components of B=B^ie_i, say, transform using \Lambda^(-1) when e->f via \Lambda is not trivial however. If we started in f and transformed to e we would use the matrix \Gamma. It just happens that \Gamma = \Lambda^(-1)! That is where your confusion arises.
Hey, thanks for your great sophisticated videos! Which literature do you suggest, that handle these topics in a similar way like you do (mathematically accurate, but physics-topic related)? Or do you even orient this video after a specific book?
Sil Sabat Well, the closest book is "Gravitation" by Misner, Thorne, and Wheeler but that isn't entirely the same as these lessons. I am blending that book and "Introduction to Vectors and Tensors" by Bowen and Yang (Dover) and a few other references. The point is that after listening to these lessons you should be able to read any book about GR that isn't too advanced ...
I have a question! Around 9:35, in this video, you wrote a familiar form of linear algebraic formula with vectors and a matrix(NOT tensors). But, I thought.. e_0, e_1, ..., f_0, ..., f_3 things are not "components" of vectors but 8 individual "vectors" themselves(the basis vectors) in the vector space V!! How can you express them like that? I'm very confused right now.
서민식 The matrix elements of Lambda are numbers. The objects e_0 ... f_3 are, as you said, vectors. The expression in interpreted as the individual vectors multiplied by real numbers and then added together. If you try to think of it in terms of linear algebra then you will get confused. You must forget the standard linear algebra and consider e_i and f_j as abstract vectors in V.
in 6 :14-6:21 it stated that "there's no way to screw around and mix 06:12 up these indices with these indices 06:14 because they live in different equations 06:15 or even if they didn't they're in 06:17 different terms of a single equation 06:19 these are called dummy indices and we" could you explain this little bit further to me i get stuck as i am not sure what do you mean by "didnt they are in different terms of single equation" thank
I'm not sure if this is still relevant for you, but here you go: In Einstein sum convention (google it!) you automatically sum over two indices, which appear in the same term. So instead of writing ∑v^nu v_nu you discard the sum-symbol and just write v^nu v_nu. This would be exactly the same as v^mu v_mu, since it doesnt matter what you call the index, you sum over. Also v^nu v_nu + v^mu v_mu in Einstein sum convention is ∑v^nu v_nu + ∑v^mu v_mu. You sum over each term independently and can therefore change the indices in each term. Also in x_nu = Y^mu_nu x_mu you can change the indices to whatever you like, it must not even by letters. E.g.: x_# = Y^®_ # x_®. Just add the sum-symbol, expand it and you'll realise, that those are the exact same equations! Hope this helped!
According to the expression on the right at 13:48 can we say f_nu = lambda_mu^nu e_mu ? If yes then is it not violating einstein convention? I am new to the subject, pardon me if this is a silly question. It would be a great help if you can explain me this. Love your lectures !
A silly review question, but please be patient. At 16:23 can you confirm.... When the components transform contravariantly, it involves TRANSPOSE of the INVERSE of lamda.
OK, then at 13:49, you write e mu =( (lamda)-inverse sub-mu, upper-nu ) * f nu Now I now lamda is not a tensor and the index rules do not apply. However, earlier, you did take note to state that lamda is row, column into the vector on the right. If that is still true, then upper-nu should prceed sub-mu in space. Unless you are not concerned about row/column notatoin and just expect the reader to know it. Which one is it?
Thomas Impelluso I am ambivalent regarding this. F could be a "row vector" I suppose, then it would make sense in the elementary matrix algebra context. Best to think of these transformations as just numbers. No need to cast things into a matrix representation, even if we use those words.
I get that part: consider them just numbers. But then how do you pull off talking about the inverse of lamba? What is the meaning of the inverse if there is no matrix? And I need to reiterate: your presentations are great!
At 14.26, shouldn't the order of lambda inverse and A be reversed according to matrices laws, (even though the Einstein's summation convention works in both representations) keeping the superscripts and the subscripts where they are on lambda and A ? (Equation in the bottom right)
gyan varshnay If you abandon indices and treat these things as abstract matrices, then yes. We use index notation so we don’t have to worry about order.
I have questions at about 10:00 mark... the f0,f1,f2, f3 and e0, e2, e3 and e4, are they individual basis vector( I think this is the likely explanation)? or they are the components of basis vector in term standard cartesian basis?
@@XylyXylyX at the 10:00 mark, it seems that those vectors are organized as row vectors instead, going from top row to bottom row... then according to the matrix multiplication convention, the transformation matrix should be positioned to the right of the matrix (with each column as old basis) . so the overall equation becomes [f0 f1,f2 f3]= [e0 e1 e2 e3]*A
Thanks for these lectures. It is the best I have encountered on this difficult subject. At 18.43 should w->B^mu f mu rather then w->B^ni f ni? If I am not right then I have missed something. Please comment.
very nice videos, but I have a question: I understand that f vectors don't transform, they are always the same, physically they are always the same, I am confused by that, because you sum vectors in another way (using "e" vectors) but then you get the same vector ("f" vectors). Thank you!
I am stuck with the idea of covariant and contravariant transformation. You said the terms with lower indices transform covariantly and terms with upper indices the other way. But in case of transforming from A(mu)e(mu) to B(nu)f(nu) the basis vector e(mu) transforms using an inverse matrix. Isn''t it the contravariant transformation?
Shubham Gothwal Yes! The basis vectors transform in a way that is *opposite* to how the component transforms! In the classical language A^\mu is called a "contravariant vector" but the true vector is the component *and* the basis vector A^\mu e_\mu. The basis vector transforms *co*variantly! This is an important point.
+XylyXylyX So does it mean that using inverse matrix in e(mu) doesn't simply make it contravariant? Like in the case of f(nu), the matrix multiples directly. So what I'm confused with is does the notion of covariant and contravariant depends on whether the inverse matrix is used or not. Or is it a simple convention that we follow and call the basis transforms covariantly and components contravariantly?
Shubham Gothwal Whatever matrix is used to transform the basis vectors of V is called "the covariant transformation matrix". Anything that transforms using V is called "covariant" *by convention*. Then, anything that transforms using the inverse matrix is called *contravariant* by convention. The convention asserts that the "true direction" of the transformation is the direction of the transformation of the basis vectors of V. The "opposite direction" is the inverse of that which also happens to be the direction which transforms the *components* of vectors in V and the *basis vectors* of V*.
This might be an obvious question: in the last step, where you have (\lambda^-1)_ u^\mu(\lambda_\mu^\delta) = \delta_ u^\delta, this is true if all column vectors of the \lambda matrix are orthonormal right? Is this always the case for any linear transformation?
I have read your answers to the same question from other people but I still don't understand why your matrix multiplication is commutative. In one of your answers you say something about multiplicating a vector and a matrix but lambda and A are both matrices. So how come in you lecture A*(lambda^-1) = (lambda ^-1)*A.
Matrix multiplication is not commutative. Real number multiplication is commutative. When we write matrix multiplication in index form, we have created an expression of real numbers. That expression can be shuffled around without any problem.
So, when we move to a new coordinate system (basis), the expression of anything expressed in terms of the old coordinate system transforms opposite to the movement of the system. Let's say I express directions in terms of the way I happen to be facing - using clock face notation or whatever (we got bogies at 8 o'clock!). If I turn *left* by 10 degrees, then the entire world around me spins 10 degrees to the *right*. Of course, the world hasn't changed at all - just the way I am choosing to label where things are. Nevertheless, the positions of things transform contravariantly to how I have transformed my coordinate system - including my original '12 o'clock' vector, my original basis. But we are trying to do this very abstractly and not use those kinds of examples, because they can be misleading and limited. Right brain gets it, left brain is still trying to get on top of the notation. Not sure what an example of a covariant vector might be in my real-world 3d eucledian space. The thing that bugs me is that is that basis vectors are supposed to be elements of the vector space itself, but they have the index at the bottom where all the vectors in the vector space have the index at the top. Does that mean that the basis 'vectors' are actually from the dual space? It kinda makes sense. Things in the dual space are mappings from vectors onto real numbers. And that's what a coordinate system does - it maps a vector down to a set of real number components. If that's right, then the notion that basis vectors are just elements of V is actually not exactly true. They are elements of V*. Basis 'vectors' are actually covectors.
Paul Murray Perhaps this: if you have a rod and you stretch the coordinate system you would have to *shrink* the rod's endpoint's coordinate in order to get its length correct in the new units.
Correct?: if we define (w = mult A^mu e_mu = mult B^nu f_nu) and (f_mu = mult Lambda_mu^nu e_nu) and "(f_mu = mult Lambda_mu^nu e_nu) transforms covariantly with respect to e_nu" since e_nu is a basis for V, then "(B^mu = mult (inv Lambda)_nu^mu A^nu) transforms contravariantly with respect to e_nu". Likewise, since f_mu is a basis for V, "(e_mu = mult (inv Lambda)_mu^nu f_nu) transforms covariantly with respect to f_nu" and "(A^mu = mult B^nu Lambda_nu^mu) transforms contravariantly with respect to f_nu"; it is incidental to the definition of covariance and contravariance that f_nu's transform involves ((inv Lambda)_mu^nu). I.e., doesn't emphasizing by explicitness the basis with respect to which covariance (thence contravariance) is defined eliminate the confusion I and others have had?
It's probably going to take me a long while to keep all this straight in my head, but I'd just like to keep saying that I really appreciate all of these videos. Thanks so much for putting them all together!
This was very good! This is more abstract than how I learned it (as partial derivatives being used for a change of coordinates). Thanks to you, I’ve learned something new ☺️🙌🏽🎊
Glad it was helpful and thank you for your kind comment.
@13:50 seems like an error - there is no "B^mu" in the prior equations, it should be B^nu = A^mu (\Lambda^{-1})_mu^nu (and you probably want to explain why indices get switched with the inverse matrix). Agree with the comment below that the occasional swap of mu and nu is jarring, but otherwise this is an excellent series.
Thanks! keep on doing. It is very good.
Thanks for the great vids man!
Bill Varcho My pleasure! I'm a bit stalled out right now, but should continue the webinar this week.
thank you for the video. at 18:15, you said we can transform our coordinate basis vectors and components vector with a linear transformation lamda and its inverse lamda^-1. however if we think of the transformation from cartesian to spherical coordinate, the component part transformation is not linear. so does this mean that spherical coordinate transformation is not the case of linear coordinate transformation because its component transformation in non-linear?
Coordinate transformation can be arbitrary and therefore can, an usually are, nonlinear. Cartesian to spherical is a fine example. But you are confusing the coordinate transformation with the *basis* transformation. Basis transformations are always executed linearly. A basis transformation is *derived* from a coordinate transformation, but we have not discussed how that works yet.
However, you don’t need to wait for the appropriate lecture here, just go to any book about General Relativity and look for the expression about transforming a tensor component. It is created by taking the coordinate transformation’s various derivatives w.r.t. each coordinate and *evaluating the result at a specific point in space time.* What is left is just a collection of real numbers which can be interpreted as a matrix that operates (linearly) on the components.
@@XylyXylyX thanks. does that mean that at *each point* of space time, basis transformation is linear although coordinate transformation itself can be non-linear? for example, from cartesian to spherical, we can find 3 basis unit vectors (rho_hat, phi_hat, theta_hat) that are from vector space, and coordinate components (rho, phi, theta) *in real number* that can be linearly transformed to a vector in Cartesian coordinate *at a specific point P*(and each specific point in R3).
Burn Koll Yes, that is correct. Every basis vector of the new basis can be expressed as a linear combination of the basis vectors of the old basis. The transformation between the two basis systems is always linear. In fact the very definition of a “linear transformation” is all about changing basis vectors in a vector space.
Thanks! This helps me alot
Thank you for those very helping videos. I'd like some more precision about the manipulation starting at 11:00 with "w". Forgive me I stopped linear algebra a few years ago ; there's something I found non-trivial : to replace f(mu) you write f(nu) and replace Lambda (mu nu) by the transposed. That suppose a constraint on e(mu) to preserve equality between f(nu) and transposed-lambda*e(mu). Is that constraint coherent with the constraint on this equality A(mu)*e(mu)=B(nu)*f(nu).
Einstein convention make it actualy look simple, but it's not that intuitive to me. I tried the demonstration with w=A(nu)*e(nu)=B(mu)*f(mu). Transposing lambda is indeed necessary to male it right, but I'm not confortable with it.
the arbitrary swapping of the mu's and nu's is killing me - making my head spin. i like your vids, but you lost me on this one after you went to a clean sheet @ ~8:50. I'll try again tomorrow.
What kind of problems are you having in terms of understanding?
Hi, I have a question about the inverse linear transformation. at 18:43.
Lets say B^v is weights in cylindrical coordinates and A^v is a weight in cartesian coordinates. We can find the basis vector conversion matrix from cartesian to cylindrical coordinate easily (ex. rho_hat= cos(phi)x_hat+sin(phi)y_hat and we can find phi_hat, z_hat in the same manner) .
Then if we were to express weights in cylindrical coordinates from cartesian coordinates, by using the relation B^v=A^v(T^-1)_v^u, then the inverse of conversion matrix must operate on cartesian weights .
However we know that weights in cylindrical coordinate are not linear functions of (ex. rho=sqrt(x^2+y^2), phi=arctan(y/x)).
Then how can we say B^v=A^v(T^-1)_v^u is true?
Because everything on both sides of the expression is a *number* not a function. At this point in the lectures we are not treating A and B as a vector field, only as a single member of a single vector space. You have promoted everything to vector fields.
Now, if you consider A and B to be vector fields then they are functions of spacetime. But in that case the transformation matrix is a function of spacetime as well. The point is that for any single point in spacetime you can calculate a transformation matrix and also values for A. This will lead to a simply matrix product to give B. The fact that T is a function of spacetime handles all the non-linear nature of the coordinate transformation that concerns you.
I really appreciate you for your quick reply. I will watch your video upto lecture 14, Vector and tensor fields and then I might fully grasp what is actually going on.
And your lectures are by far the best ever on youtube. Really, it feels like the materials come out from the lectures and naturally flow into my brain. Thank you!
Quick question, you seem to switch indexes when substituting components for their transformations. For example around 18:52 mark you have w = (B^ nu) (f_ nu) .You then use a substitution going from (A^nu) to (B^mu) for (B^nu) and the transformation from the basis( e_delta to f_mu) for (f_nu.).Is this intentional?
No, it is a *little* sloppy because I am recycling dummy indices and it would be cleaner if I always used different indices, even if they are paired up. In the case you cite I should have used, say, \alpha instead of
u on A and \Lambda, then I would have used
u instead of \mu on the transformations. Then it would \mu on both sides. It is ok because the indices are dummy indices on both sides of the arrow, but yes, it is a bit confusing as I have written it.
+XylyXylyX Cool that is kinda what I thought was going on. I just wanted to make sure .Thanks again!
The basis vectors transform covariantly, which is to say that it gets transformed by the matrix not the inverse of the matrix, right?
If It's so then can we still call the transfromation of e_{\mu} @13:44 as covariant transformation?
I think these are great tutorials. Have you considered setting problems? Or is there a book at this level that has good problems on this material?
Michael Johnson I will not be providing problems. There are many books on this subject. Try Bowen and Yang “Introduction to Vectors and Tensor” (Dover). Good Luck!
A little confusing about the index. But I get you idea, and that's what matters.
While I understand that the 'f' basis vector transforms covariantly and that f's components 'B' transform contravariantly...based on how you've defined covariant and contravariant, is it then true that the 'e' basis vector is transformed contravariantly and that e's components 'A' transform covariantly?
time point 15:43 is a good reference point, as all of the equations relevant to my question are located on the screen
~Thanks
Looking at around 3:55? No! All basis vectors, e or f or any other set of basis vectors will always transform covariantly. Otherwise there would be two classes of basis vectors in V. All basis vectors are basically equivalent and transform the same way. Likewise their components also always transform contravariantly. That is, basis transformations do NOT take covariant things and turn them into contravariant things or the other way around. THey just reorganize covariant things into other covariant things. The index notation reflects this and basis transformations do not change the index positions of the objects they operate on.
I don't know why you are referencing minute 3...but I mostly meant that at around minute 15, you have an equation for the e basis vector that involves the inverse of our transformation matrix (i.e. to get e in terms of f). You also have an equation for A (the components of the e basis vector) that transforms using the non-inverse transformation matrix (in order to get A in terms of B). In accordance with your definitions of contravariant and covariant, where covariant transformations involve the non-inverse and contravariant involves the inverse, I was slightly confused..
But thank you for your quick response. Cheers~
Samuel Cramer Oh, I see. Yes....we always need to track what is the “forward” direction of a transformation. In this case we started with e and transformed to f using \Lambda. Obviously, to go back to e from f requires \Lambda^(-1) but that must be understood as just another basis transformation and you could just rename the transformation \Gamma, say. Notice that the index position does NOT change when going from f -> e so the resulting object is still covariant. The way to understand this is that objects transforming in the same direction but using inverse transformation matrices have a co/contra relationship. So if f->e uses \Lambda^(-1) then its components will use \Lambda^(-1)^(-1)=\Lambda. But f is transforming in the opposite “direction” from e so they cant possibly share a co/contra relationship. Sitting in the f basis the “forward” direction is the matrix \Lambda(-1) which would define the covariant transformation and its contravariant partner is \Lambda. Renaming the transformation to \Gamma might help.
I see the possibility of confusion here, you are not seeing things. I have written, correctly, that f-> e uses the inverse of \Lambda, but this is a trivial and obvious use of the inverse transformation. The fact that the components of B=B^ie_i, say, transform using \Lambda^(-1) when e->f via \Lambda is not trivial however. If we started in f and transformed to e we would use the matrix \Gamma. It just happens that \Gamma = \Lambda^(-1)! That is where your confusion arises.
much clearer. Thank you, sir.
Hey, thanks for your great sophisticated videos! Which literature do you suggest, that handle these topics in a similar way like you do (mathematically accurate, but physics-topic related)? Or do you even orient this video after a specific book?
Sil Sabat Well, the closest book is "Gravitation" by Misner, Thorne, and Wheeler but that isn't entirely the same as these lessons. I am blending that book and "Introduction to Vectors and Tensors" by Bowen and Yang (Dover) and a few other references. The point is that after listening to these lessons you should be able to read any book about GR that isn't too advanced ...
I have a question! Around 9:35, in this video, you wrote a familiar form of linear algebraic formula with vectors and a matrix(NOT tensors). But, I thought.. e_0, e_1, ..., f_0, ..., f_3 things are not "components" of vectors but 8 individual "vectors" themselves(the basis vectors) in the vector space V!! How can you express them like that? I'm very confused right now.
서민식 The matrix elements of Lambda are numbers. The objects e_0 ... f_3 are, as you said, vectors. The expression in interpreted as the individual vectors multiplied by real numbers and then added together. If you try to think of it in terms of linear algebra then you will get confused. You must forget the standard linear algebra and consider e_i and f_j as abstract vectors in V.
in 6 :14-6:21
it stated that
"there's no way to screw around and mix
06:12
up these indices with these indices
06:14
because they live in different equations
06:15
or even if they didn't they're in
06:17
different terms of a single equation
06:19
these are called dummy indices and we"
could you explain this little bit further to me
i get stuck as i am not sure what do you mean by "didnt they are in different terms of single equation"
thank
I'm not sure if this is still relevant for you, but here you go:
In Einstein sum convention (google it!) you automatically sum over two indices, which appear in the same term. So instead of writing ∑v^nu v_nu you discard the sum-symbol and just write v^nu v_nu. This would be exactly the same as v^mu v_mu, since it doesnt matter what you call the index, you sum over. Also v^nu v_nu + v^mu v_mu in Einstein sum convention is ∑v^nu v_nu + ∑v^mu v_mu. You sum over each term independently and can therefore change the indices in each term.
Also in x_nu = Y^mu_nu x_mu you can change the indices to whatever you like, it must not even by letters. E.g.: x_# = Y^®_ # x_®. Just add the sum-symbol, expand it and you'll realise, that those are the exact same equations!
Hope this helped!
Do you have a book that can supplement your lecture series?
I am not literally following any specific book, but I always recommend “Introduction to vectors and tensors” by Bowen and Yang (Dover).
According to the expression on the right at 13:48 can we say f_nu = lambda_mu^nu e_mu ?
If yes then is it not violating einstein convention? I am new to the subject, pardon me if this is a silly question. It would be a great help if you can explain me this.
Love your lectures !
A silly review question, but please be patient.
At 16:23 can you confirm....
When the components transform contravariantly, it involves TRANSPOSE of the INVERSE of lamda.
Thomas Impelluso No transpose. Just the inverse.
OK, then at 13:49, you write
e mu =( (lamda)-inverse sub-mu, upper-nu ) * f nu
Now I now lamda is not a tensor and the index rules do not apply.
However, earlier, you did take note to state that lamda is row, column into the vector on the right.
If that is still true, then upper-nu should prceed sub-mu in space.
Unless you are not concerned about row/column notatoin and just expect the reader to know it.
Which one is it?
Thomas Impelluso I am ambivalent regarding this. F could be a "row vector" I suppose, then it would make sense in the elementary matrix algebra context. Best to think of these transformations as just numbers. No need to cast things into a matrix representation, even if we use those words.
I get that part: consider them just numbers.
But then how do you pull off talking about the inverse of lamba?
What is the meaning of the inverse if there is no matrix?
And I need to reiterate: your presentations are great!
Thomas Impelluso The inverse will always be the matrix that satisfies A_a^b B_b^a = I.
At 14.26, shouldn't the order of lambda inverse and A be reversed according to matrices laws, (even though the Einstein's summation convention works in both representations) keeping the superscripts and the subscripts where they are on lambda and A ? (Equation in the bottom right)
gyan varshnay If you abandon indices and treat these things as abstract matrices, then yes. We use index notation so we don’t have to worry about order.
Thanks!!
I have questions at about 10:00 mark... the f0,f1,f2, f3 and e0, e2, e3 and e4, are they individual basis vector( I think this is the likely explanation)? or they are the components of basis vector in term standard cartesian basis?
zhang jim They are individual basis vectors *organized* as a column vector! Crazy....I know.
@@XylyXylyX at the 10:00 mark, it seems that those vectors are organized as row vectors instead, going from top row to bottom row... then according to the matrix multiplication convention, the transformation matrix should be positioned to the right of the matrix (with each column as old basis) . so the overall equation becomes
[f0 f1,f2 f3]= [e0 e1 e2 e3]*A
why is the linear transformation matrix of basis not a tensor? Is a linear transformation a 1,1 tensor?
zhang jim It is possible to establish an isomorphism between linear transformations and (1,1) tensors.
Thanks for these lectures. It is the best I have encountered on this difficult subject. At 18.43 should w->B^mu f mu rather then w->B^ni f ni? If I am not right then I have missed something. Please comment.
very nice videos, but I have a question: I understand that f vectors don't transform, they are always the same, physically they are always the same, I am confused by that, because you sum vectors in another way (using "e" vectors) but then you get the same vector ("f" vectors). Thank you!
agusr32 I'm sorry I don't understand your question🤔
I am stuck with the idea of covariant and contravariant transformation. You said the terms with lower indices transform covariantly and terms with upper indices the other way. But in case of transforming from A(mu)e(mu) to B(nu)f(nu) the basis vector e(mu) transforms using an inverse matrix. Isn''t it the contravariant transformation?
Shubham Gothwal Yes! The basis vectors transform in a way that is *opposite* to how the component transforms! In the classical language A^\mu is called a "contravariant vector" but the true vector is the component *and* the basis vector A^\mu e_\mu. The basis vector transforms *co*variantly! This is an important point.
+XylyXylyX So does it mean that using inverse matrix in e(mu) doesn't simply make it contravariant? Like in the case of f(nu), the matrix multiples directly. So what I'm confused with is does the notion of covariant and contravariant depends on whether the inverse matrix is used or not. Or is it a simple convention that we follow and call the basis transforms covariantly and components contravariantly?
Shubham Gothwal Whatever matrix is used to transform the basis vectors of V is called "the covariant transformation matrix". Anything that transforms using V is called "covariant" *by convention*. Then, anything that transforms using the inverse matrix is called *contravariant* by convention. The convention asserts that the "true direction" of the transformation is the direction of the transformation of the basis vectors of V. The "opposite direction" is the inverse of that which also happens to be the direction which transforms the *components* of vectors in V and the *basis vectors* of V*.
+XylyXylyX thank you so much for your help. I'm enjoying following your lecture series.
This might be an obvious question: in the last step, where you have (\lambda^-1)_
u^\mu(\lambda_\mu^\delta) = \delta_
u^\delta, this is true if all column vectors of the \lambda matrix are orthonormal right? Is this always the case for any linear transformation?
It is true for any linear transformation. It is the definition of the inverse transformation.
Hi I enjoy your videos on this topic. Please try to be neat in your writing and layout cause it helps.
I will try harder to think about the clairity as I write. It is a bit late now....sigh...
I have read your answers to the same question from other people but I still don't understand why your matrix multiplication is commutative. In one of your answers you say something about multiplicating a vector and a matrix but lambda and A are both matrices. So how come in you lecture A*(lambda^-1) = (lambda ^-1)*A.
Matrix multiplication is not commutative. Real number multiplication is commutative. When we write matrix multiplication in index form, we have created an expression of real numbers. That expression can be shuffled around without any problem.
So, when we move to a new coordinate system (basis), the expression of anything expressed in terms of the old coordinate system transforms opposite to the movement of the system.
Let's say I express directions in terms of the way I happen to be facing - using clock face notation or whatever (we got bogies at 8 o'clock!). If I turn *left* by 10 degrees, then the entire world around me spins 10 degrees to the *right*. Of course, the world hasn't changed at all - just the way I am choosing to label where things are. Nevertheless, the positions of things transform contravariantly to how I have transformed my coordinate system - including my original '12 o'clock' vector, my original basis.
But we are trying to do this very abstractly and not use those kinds of examples, because they can be misleading and limited. Right brain gets it, left brain is still trying to get on top of the notation. Not sure what an example of a covariant vector might be in my real-world 3d eucledian space.
The thing that bugs me is that is that basis vectors are supposed to be elements of the vector space itself, but they have the index at the bottom where all the vectors in the vector space have the index at the top. Does that mean that the basis 'vectors' are actually from the dual space? It kinda makes sense. Things in the dual space are mappings from vectors onto real numbers. And that's what a coordinate system does - it maps a vector down to a set of real number components. If that's right, then the notion that basis vectors are just elements of V is actually not exactly true. They are elements of V*. Basis 'vectors' are actually covectors.
Paul Murray Perhaps this: if you have a rod and you stretch the coordinate system you would have to *shrink* the rod's endpoint's coordinate in order to get its length correct in the new units.
Correct?: if we define (w = mult A^mu e_mu = mult B^nu f_nu) and (f_mu = mult Lambda_mu^nu e_nu) and "(f_mu = mult Lambda_mu^nu e_nu) transforms covariantly with respect to e_nu" since e_nu is a basis for V, then "(B^mu = mult (inv Lambda)_nu^mu A^nu) transforms contravariantly with respect to e_nu". Likewise, since f_mu is a basis for V, "(e_mu = mult (inv Lambda)_mu^nu f_nu) transforms covariantly with respect to f_nu" and "(A^mu = mult B^nu Lambda_nu^mu) transforms contravariantly with respect to f_nu"; it is incidental to the definition of covariance and contravariance that f_nu's transform involves ((inv Lambda)_mu^nu). I.e., doesn't emphasizing by explicitness the basis with respect to which covariance (thence contravariance) is defined eliminate the confusion I and others have had?