You answered the question I had about needing a 4-d loop. I had asked @eigenchris the same thing 2 years ago! I'm pleased that I was not alone in my initial confusion. Thanks for a great video.
I struggle with Contraction used to simplify an expression. Yes, I understand what is happening - mathematically. However, is the information 'discarded' able to be physically 'discarded' without a loss of generality/physical meaning? In contracting the Riemann Curvature Tensor to the Ricci Curvature Tensor, you did go on to explain what the Ricci Curvature Tensor can be used to show the change in position of a cloud of dust as it moves on its geodesic path. OK, but what information would the full Riemann Curvature Tensor additionally provide - which is now gone and why is this contracted information appropriate for the Einstein Field Equations. Maybe all that will be revealed but it is not satisfying - to me - to contract without stating that it is a proper maths process. One more like to you, Eddie. Your complete identification of the origin of the Riemann Curvature Tensor and the HUGE number of terms of each component of it (10 Christoffel Symbols which yield even more due to derivatives leading toward the Metric Tensor inclusion) is HUGELY satisfying to me. You have opened the door to such clarity in this whole process - many thanks. FYI, I did find one index error in the graphic shown @49:45, the last row of last column of T(up lambda down mu, nu) should be T(up 3 down 3,3) but is shown as T(up 0, down 3,3). I only state this finding to highlight your superb attention to detail in entire graphic production not this single error; I can only speculate how many hours in this "one man band" approach you have had to expend on our behalf. I am in awe of your energy expended on our behalf.
Great Video. Just wondering if you are familiar with volume form/element and how it is used to prove the Ricci tensor's role in volume changes along geodesics? I've been struggling to get my head around the proof.
Hi Eddie, quick question. As we know a vector has both contra variant and covariant components. When people talk about contravariant they often refer to vectors like velocity. But when they talk about covariant, they talk about gradient. my question is why can’t they talk about velocity as well since velocity is a vector and it would have both contra variant and covariant components. on the flipside why can’t gradient have both contra and covariant relationship relationship?
Hi …. my take on this perennial question is that one should tend only to talk (separately) about covariant and contravariant VECTORS when considering ONE dimension. After all, a ONE dimensional vector only has ONE component. That might seem like a ‘sloppy’ argument, but it does it for me, and I can move on …….. though I understand people wanting to grapple further with the issue. Eddie
@@eddieboyes Thanks for taking the time to respond but your reply was less than satisfying. I seem to be very talented at asking questions that presenters can't or won't answer. This could either be attributed to the questioner being extremely nieve or that the question relates to something responders feel is totally unimportant. Maybe there really are stupid questions. Some of my stupid questions: 1. As we know a vector has both contravariant and covariant components. When people talk about contravariant they often refer to vectors like velocity. But when they talk about co-variant, they talk about gradient. my question is why can’t they talk about velocity as well since velocity is a vector and it would have both contravariant and covariant components. on the flipside why can’t gradient have both contra and covariant relationship? 2. A particular difference vector, when discerning parallel transport has an actual length of 29.394. Would this number EVER show up as a real value in either a metric tensor or a Christoffel symbol? If not, why not. 3. I'm trying to track just one number (for simplicity) say...a covariant component of a single vector, one being analyzed as transforming from one csys to another, and see the pathway to its destination...in a tensor. I want to visually see how it gets there. I may be way off, but this seems like a visual that would answer my questions to those learning about tensors (since it non-stop talks about contravariant and covariant indices. 4. Some authors show covariant component lengths projected on the skewed axis of a skewed csys as ending on the skewed axis. Some show the length continuing onto the dual basis axis. The cause differing lengths. Which one is the covariant component of a vector? The length if you stop at the skewed y-axis or the one continuing on towards the dual basis axis? Thanks
@@thevegg3275 Hi …. I understand that when something ‘bugs’ you, then the kind of answer I gave might not be very helpful. For me, however, the issue you are raising is not a problem ……. and you may have to work at finding a resolution for yourself. Here’s my VERY LAST attempt to help you get through this: START THE WHOLE THING THE OTHER WAY AROUND. What I mean is …… in the usual way: 1) Take (say) a two dimensional vector (ANY type of vector) and work through the two types of components. 2) Give them the names ‘contravariant’ and ‘covariant’ (they’re just names). 3) Find out how they transform (in different ways) from one frame to another. Note that there’s nothing so far about ‘x being on top’ (contra) or on the bottom (co) - anything goes. 3) Now consider a displacement-type vector in ONE dimension. It only has one component (in that one dimension). 4) Find out how that single component transforms from one frame to another and, lo and behold, it transforms like what had been labelled the ‘contravariant’ component of any 2 dimensional vector - so we could name that kind of vector a ‘contravariant vector’ ……. or a vector with a single contravariant component. 5) Similarly, consider what you called a ‘gradient’ type vector in ONE dimension. Again, it only has one component (one dimension). 6) Find out how that single component transforms from one frame to another and this time you find that it transforms like what had been labelled the ‘covariant’ component of any 2 dimensional vector - so give it that name. The important thing is …… not the supposed kind of vector it is, but how the various components of any vector (in 2+ dimensions) transform …. THAT’S the lesson to learn. Maybe teachers do a disservice to students by introducing the idea of ‘contravariant’ and ‘covariant’ vectors (which idea is only really relevant in one dimension), and perhaps (in teaching the subject) it would be better to go straight into the idea of different types of components in 2+ dimensions, and not mention the concept of ‘contravariant’ and ‘covariant’ vectors …… the whole idea of which can be a distraction. Again, it’s down to you. You’re not the first one to battle with this …. and you won’t be the last if we continue to teach things this way. You’re certainly not naïve ….. I think that one of the enjoyments of physics for me is, in good constructivist fashion, grappling with ideas which seem at first to make little sense, but for which the light eventually dawns. As learners, if we’re honest, we are all going through this process the whole of our lives. Eddie
A Great Master! Beautiful lessons, thanks!
You answered the question I had about needing a 4-d loop. I had asked @eigenchris the same thing 2 years ago! I'm pleased that I was not alone in my initial confusion. Thanks for a great video.
Illustrating the derivation by actual example showing the nitty gritty details really helped me understand tensor manipulation!
👍
Thank you for this wonderful introduction to this subject!
Excellent
BUENISIMO TODO. INSUPERABLE ...
Sehr gut
I struggle with Contraction used to simplify an expression. Yes, I understand what is happening - mathematically. However, is the information 'discarded' able to be physically 'discarded' without a loss of generality/physical meaning? In contracting the Riemann Curvature Tensor to the Ricci Curvature Tensor, you did go on to explain what the Ricci Curvature Tensor can be used to show the change in position of a cloud of dust as it moves on its geodesic path. OK, but what information would the full Riemann Curvature Tensor additionally provide - which is now gone and why is this contracted information appropriate for the Einstein Field Equations. Maybe all that will be revealed but it is not satisfying - to me - to contract without stating that it is a proper maths process.
One more like to you, Eddie. Your complete identification of the origin of the Riemann Curvature Tensor and the HUGE number of terms of each component of it (10 Christoffel Symbols which yield even more due to derivatives leading toward the Metric Tensor inclusion) is HUGELY satisfying to me. You have opened the door to such clarity in this whole process - many thanks.
FYI, I did find one index error in the graphic shown @49:45, the last row of last column of T(up lambda down mu, nu) should be T(up 3 down 3,3) but is shown as T(up 0, down 3,3). I only state this finding to highlight your superb attention to detail in entire graphic production not this single error; I can only speculate how many hours in this "one man band" approach you have had to expend on our behalf. I am in awe of your energy expended on our behalf.
Great Video. Just wondering if you are familiar with volume form/element and how it is used to prove the Ricci tensor's role in volume changes along geodesics? I've been struggling to get my head around the proof.
1:56
0:29
Hi Eddie, quick question. As we know a vector has both contra variant and covariant components. When people talk about contravariant they often refer to vectors like velocity. But when they talk about covariant, they talk about gradient. my question is why can’t they talk about velocity as well since velocity is a vector and it would have both contra variant and covariant components. on the flipside why can’t gradient have both contra and covariant relationship relationship?
Hi …. my take on this perennial question is that one should tend only to talk (separately) about covariant and contravariant VECTORS when considering ONE dimension. After all, a ONE dimensional vector only has ONE component. That might seem like a ‘sloppy’ argument, but it does it for me, and I can move on …….. though I understand people wanting to grapple further with the issue. Eddie
@@eddieboyes Thanks for taking the time to respond but your reply was less than satisfying. I seem to be very talented at asking questions that presenters can't or won't answer.
This could either be attributed to the questioner being extremely nieve or that the question relates to something responders feel is totally unimportant. Maybe there really are stupid questions.
Some of my stupid questions:
1. As we know a vector has both contravariant and covariant components. When people talk about contravariant they often refer to vectors like velocity. But when they talk about co-variant, they talk about gradient. my question is why can’t they talk about velocity as well since velocity is a vector and it would have both contravariant and covariant components. on the flipside why can’t gradient have both contra and covariant relationship?
2. A particular difference vector, when discerning parallel transport has an actual length of 29.394. Would this number EVER show up as a real value in either a metric tensor or a Christoffel symbol? If not, why not.
3. I'm trying to track just one number (for simplicity) say...a covariant component of a single vector, one being analyzed as transforming from one csys to another, and see the pathway to its destination...in a tensor. I want to visually see how it gets there. I may be way off, but this seems like a visual that would answer my questions to those learning about tensors (since it non-stop talks about contravariant and covariant indices.
4. Some authors show covariant component lengths projected on the skewed axis of a skewed csys as ending on the skewed axis. Some show the length continuing onto the dual basis axis. The cause differing lengths. Which one is the covariant component of a vector? The length if you stop at the skewed y-axis or the one continuing on towards the dual basis axis?
Thanks
@@thevegg3275 Hi …. I understand that when something ‘bugs’ you, then the kind of answer I gave might not be very helpful. For me, however, the issue you are raising is not a problem ……. and you may have to work at finding a resolution for yourself. Here’s my VERY LAST attempt to help you get through this: START THE WHOLE THING THE OTHER WAY AROUND.
What I mean is …… in the usual way: 1) Take (say) a two dimensional vector (ANY type of vector) and work through the two types of components. 2) Give them the names ‘contravariant’ and ‘covariant’ (they’re just names). 3) Find out how they transform (in different ways) from one frame to another. Note that there’s nothing so far about ‘x being on top’ (contra) or on the bottom (co) - anything goes. 3) Now consider a displacement-type vector in ONE dimension. It only has one component (in that one dimension). 4) Find out how that single component transforms from one frame to another and, lo and behold, it transforms like what had been labelled the ‘contravariant’ component of any 2 dimensional vector - so we could name that kind of vector a ‘contravariant vector’ ……. or a vector with a single contravariant component. 5) Similarly, consider what you called a ‘gradient’ type vector in ONE dimension. Again, it only has one component (one dimension). 6) Find out how that single component transforms from one frame to another and this time you find that it transforms like what had been labelled the ‘covariant’ component of any 2 dimensional vector - so give it that name.
The important thing is …… not the supposed kind of vector it is, but how the various components of any vector (in 2+ dimensions) transform …. THAT’S the lesson to learn. Maybe teachers do a disservice to students by introducing the idea of ‘contravariant’ and ‘covariant’ vectors (which idea is only really relevant in one dimension), and perhaps (in teaching the subject) it would be better to go straight into the idea of different types of components in 2+ dimensions, and not mention the concept of ‘contravariant’ and ‘covariant’ vectors …… the whole idea of which can be a distraction.
Again, it’s down to you. You’re not the first one to battle with this …. and you won’t be the last if we continue to teach things this way. You’re certainly not naïve ….. I think that one of the enjoyments of physics for me is, in good constructivist fashion, grappling with ideas which seem at first to make little sense, but for which the light eventually dawns. As learners, if we’re honest, we are all going through this process the whole of our lives. Eddie
Thank you!
13:01