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Thank you so much for your detailed comment and for engaging so thoughtfully with the material! I really appreciate your observations, as well as your suggestions for making the explanation clearer. 1. Clarifying the Transformation Law You’re absolutely correct that the transformation law discussed around 3:40 could be misinterpreted as conflating different points. Thank you for pointing this out-it’s an important nuance that I should have addressed more explicitly. The key assumption is that as 𝜆→0, the two points 𝑃 and 𝑄 (which are initially distinct) merge into the same point in the limit. This assumption underpins the idea that we can compare the transformed quantities at 𝑃 and 𝑄 because, in the infinitesimal limit, the difference between these points vanishes. Without this clarification, the logic could appear inconsistent, as you have rightly noted. I'll aim to be more explicit about this in future discussions. To make it rigorous: - At finite separation, 𝑃 and 𝑄 are distinct points, so their respective transformations need to be treated independently. - In the limit 𝜆→0, 𝑄 approaches 𝑃, and this allows us to meaningfully compare the transformations at these two points under the flow generated by the vector field. I’ll take your suggestion to explicitly mention this limit in future videos to avoid confusion. 2. The Role of the Lie Derivative I completely agree with you that the Lie derivative provides a rigorous definition of this concept. Starting with the Lie derivative allows us to bypass any ambiguity and rigorously define how vectors or tensor fields behave under the flow of a vector field 𝑋. For example: 𝐿_𝑋 𝑇 = lim_(𝜆→0) [Φ_(−𝜆)^∗𝑇 − 𝑇]/𝜆 , where Φ_(−𝜆)^∗ is the pullback associated with the flow of 𝑋. This definition mathematically formalizes the infinitesimal change of the tensor 𝑇 along the flow of 𝑋, resolving any issues with conflating points. 3. Hand-Wave Arguments as Interpretation I also agree with your perspective on hand-wave arguments-they’re best viewed as intuitive interpretations of the Lie derivative rather than rigorous derivations. The transport-based arguments are useful for building intuition about how objects like vectors or tensor fields "change" along a flow, but they shouldn’t replace the formal definition. For those who are more mathematically inclined, starting with the Lie derivative's rigorous definition is the clearest approach. From there, one can return to the approximate, geometric interpretations (like visualizing transporting vectors or forms along flows) to build a more intuitive understanding. 4. Improving the Explanation Thank you for emphasizing the need to fill in the details behind the hand-wave arguments. I’ll aim to strike a better balance between mathematical rigor and intuition in future explanations. For viewers looking for a deeper dive, I’d recommend consulting a textbook or resource that carefully develops the Lie derivative, such as Frankel’s The Geometry of Physics or Schutz’s Geometrical Methods of Mathematical Physics. Thank you again for your thoughtful comment and feedback-it’s incredibly helpful in improving my explanations for this material. I hope this response addresses your concerns and provides a clearer picture of the transformation law and its connection to the Lie derivative!

@@TensorCalculusRobertDavie Thank you so much. I can understand the formal definition at 6:20, but it leaves me uneasy. The numerator of the limit seems like it is the difference between the components of the same vector v, expressed in different coordinate systems, the primed and unprimed. Since these components are real numbers, they can be subtracted, but it still feels a little odd to suppress their associated basis vectors, treat them as raw numbers and subtract them. I understand that in an infinitesimal limit the points coincide and the bases may be the same, so perhaps this isn't so odd.? The definition ‘𝐿_𝑋 𝑇 = lim_(𝜆→0) [Φ_(−𝜆)^∗𝑇 − 𝑇]/𝜆 , where Φ_(−𝜆)^∗ is the pullback associated with the flow of X’ also makes great sense. However, for the 'physical 'argument I am struggling with the intuition at 5:50, starting with the line ‘Now we can compare...’.( Physical intuition is a major weakness for me). It seems obvious that v^i(Q) is the vector v evaluated at Q, but why is v‘ ^i(Q) a valid expression of the vector v at point P? I have a vague mental picture of somehow ‘sliding the the coordinate system underneath the point ’ rather that sliding the point along the coordinate line, and that these motions might be somehow equivalent...but it is still very unclear to me. Many thanks if you can provide the insight. BTW: I think it is rare that you put such care and effort into producing the highest quality lessons possible. It’s made me wonder a bit about who you are and what has inspired you. I’d like to find the spring of your idealism and take a sip of that too.

Thank you and Merry Christmas to you and your family. Thank you for your earlier comments on one of the videos from push forward of vectors on manifolds part five. I replaced that video with two that were vastly better. Your comments made a real difference.

Thank you so much for your thoughtful feedback and for taking the time to engage with the video. I really appreciate your observations, and I’ll do my best to address your points. 1. Clarification of the Steps at 9:30 You are absolutely right that the explanation at 9:30 combines two conceptual steps that deserve to be treated separately for clarity. Let me outline them explicitly: Step 1: Pullback and the equivalence of integrals This step relies on the property of the pullback, which ensures that the integral of a differential form 𝜔 over a manifold 𝑁 is equal to the integral of the pullback 𝑓^∗(𝜔) over the manifold 𝑀: ∫_𝑁 𝜔 = ∫_𝑀 𝑓^∗(𝜔). The pullback 𝑓^∗(𝜔) is expressed in terms of the pullback of the components of 𝜔, and the orientation-preserving map 𝑓 ensures that the geometric meaning of the integral is preserved. Step 2: Expression of 𝑓^∗(𝜔) in local coordinates Once 𝑓^∗(𝜔) is written in terms of local coordinates on 𝑀, it involves a wedge product like 𝑑𝑢∧𝑑𝑣. At this stage, we can invoke the definition of the integral of a differential form over a Euclidean domain to evaluate it as an iterated integral: ∫_𝑀 𝑓^∗(𝜔) = ∫_𝑀 𝑔(𝑢,𝑣) 𝑑𝑢∧𝑑𝑣 = ∫_𝑢1 ^𝑢2 ∫_𝑣1 ^𝑣2 𝑔(𝑢,𝑣) 𝑑𝑢 𝑑𝑣. This connection between the abstract formalism of differential forms and the familiar iterated integral makes it possible to perform explicit computations. I appreciate your suggestion to separate these steps for clarity. I will make an effort to present this process in a more step-by-step fashion in future videos. 2. Typo at 15:45 Thank you for catching this! You’re absolutely correct that ∫_𝑁 𝑧 𝑑𝑥 𝑑𝑦 is not properly defined in the context of integration on a general manifold 𝑁, as 𝑑𝑥 𝑑𝑦 is not a 2-form but rather a product of differentials from traditional calculus notation. The correct notation should indeed be ∫_𝑁 𝑧 𝑑𝑥∧𝑑𝑦, which represents the integral of the 2-form 𝑧 𝑑𝑥∧𝑑𝑦 over the manifold 𝑁. The wedge product 𝑑𝑥∧𝑑𝑦 ensures that the integral is well-defined on a general manifold, where the geometry is more abstract than a Euclidean domain. Without this correction, the expression could be misinterpreted as attempting to apply iterated integrals directly to a manifold 𝑁, which is not always possible. I’ll update my notes and future explanations to clarify this point and ensure that the correct notation is consistently used to avoid any confusion. Apology Not Necessary! There’s absolutely no need to apologize-you’ve raised excellent points, and your observations are spot-on. Discussions like these are incredibly valuable for improving the clarity and rigor of my explanations. Thank you for helping me refine my presentation, and please don’t hesitate to share any further thoughts or questions!

Thanks for that feedback. Cheers.

Hello Daniel and thank you for spotting that. You are quite right. The basis 1-form dz is not not supposed to be there as the question was dealing with the integral of a 2-form over a surface. It should have read: ω=f(x,y,z)dx∧dy. Thanks again for spotting that.

@@TensorCalculusRobertDavie Thank you for this wonderful channel that has taught me so much!

Thank you for that.

You're welcome. I use Paint 2D, PPT and Mathematica.

Hello Daniel and thank you for that! It should be fixed now.

Thank you! If you mean professor as teacher, then yes to that.

Thank you so much for your kind words and detailed feedback! I really appreciate your observations, and I’m glad the explanation was helpful in filling some gaps. Let me address your points one by one: 1. Clarifying the Role of the Vector Field 𝑋 You make an excellent point about potential confusion when introducing 𝑋 as a "vector field" in the general (𝑥,𝑦) coordinates without explicitly tying it to the curve 𝛾(𝑡) until later. This is something I could definitely improve in future videos. To clarify, 𝑋 is indeed just a single vector at the specific point on the curve 𝛾(𝑡) in this example. Although it’s written in terms of a general coordinate basis (∂/∂𝑥, ∂/∂𝑦), it represents the same object that’s derived from the curve via 𝛾′(𝑡). Your suggestion to explicitly connect 𝑋 to 𝛾′(𝑡) earlier in the discussion is very helpful, and I can see how that would prevent the impression of multiple independent vectors or an unrelated field. Thank you for pointing this out! 2. Detailed Connection Between the Two Expressions for 𝑋(𝑓) Great question about the relationship between the two expressions for 𝑋(𝑓)! Here’s the breakdown: From the curve: - We start with a curve 𝛾(𝑡) = (𝑥(𝑡),𝑦(𝑡)) parametrized by 𝑡. - The tangent vector to this curve is 𝛾′(𝑡) = (𝑑𝑥/𝑑𝑡, 𝑑𝑦/𝑑𝑡). - If 𝑓 is a function defined on the manifold, the rate of change of 𝑓 along the curve is given by: 𝑋(𝑓) = 𝑑𝑓/𝑑𝑡 = ∂𝑓/∂𝑥 𝑑𝑥/𝑑𝑡 + ∂𝑓/∂𝑦 𝑑𝑦/𝑑𝑡 . - This 𝑋(𝑓) captures the derivative of 𝑓 in the direction of 𝛾′(𝑡). General formula for 𝑋(𝑓): - A vector field 𝑋 in coordinates is expressed as: 𝑋 = 𝑋^𝑥 ∂/∂𝑥 + 𝑋^𝑦 ∂/∂𝑦. - Acting on a function 𝑓, 𝑋 evaluates as: 𝑋(𝑓) = 𝑋^𝑥 ∂𝑓/∂𝑥 + 𝑋^𝑦 ∂𝑓/∂𝑦. - Comparing this with the previous expression, we identify: 𝑋^𝑥 = 𝑑𝑥/𝑑𝑡 , 𝑋^𝑦 = 𝑑𝑦/𝑑𝑡. Bringing it together: - The vector 𝛾′(𝑡) defines the components 𝑋^𝑥 and 𝑋^𝑦 in the coordinate basis (∂/∂𝑥, ∂/∂𝑦). - Thus, 𝑋 is expressed as: 𝑋 = 𝑑𝑥/𝑑𝑡 ∂/∂𝑥 + 𝑑𝑦/𝑑𝑡 ∂/∂𝑦. The connection lies in the fact that both formulations describe the same vector: 𝛾′(𝑡) gives its components geometrically, while 𝑋(𝑓) expresses its action as a differential operator. 3. Timestamps for Details At 17:30, I describe 𝑋(𝑓) = 𝛾′(𝑡)⋅∇𝑓, which connects the curve's derivative 𝛾′(𝑡) to the function's gradient ∇𝑓. At 21:30, I introduce the explicit coordinate representation 𝑋(𝑓) = 𝑋^𝑥 ∂𝑓/∂𝑥 + 𝑋^𝑦 ∂𝑓/∂𝑦. The key connection between the two is identifying 𝑋^𝑥 and 𝑋^𝑦 as the components of 𝛾′(𝑡) in the coordinate basis. If any of this was unclear in the video, feel free to let me know, and I’d be happy to provide additional explanations. Thank you so much for your kind holiday wishes! I genuinely enjoy sharing these ideas, and your thoughtful engagement is the best reward. I don’t have a Patreon or tip system, but your gratitude and feedback mean the world to me. Happy holidays to you as well, and I look forward to hearing from you again!

You're welcome. Thank you for your comments. They are very helpful for this channel.

Thank you for that feedback. Glad you liked it.

Thank you for saying that!

I know what you mean. Good intentions and all that ......

Thank you for that. Your comment gets to the heart of what I am trying to achieve on this channel.

Thank you for your question because it is a very good one, since it is not obvious why that choice was made! I chose not explain it because I thought it would distract from the rest of the content in the video. At 17:41 in the video, I introduce 𝜙 = Δ𝜙/2 as part of the geometric or analytical setup. Here's the reasoning: 1. Context: In the video, we are likely examining an angular change (or interval) Δ𝜙, which represents the total angular span or difference. To perform certain calculations or analyses, it's often useful to consider the midpoint or symmetry of the angular interval. In this case, 𝜙 is chosen to represent the "half-angle" of the interval, i.e., half of Δ𝜙. 2. Why 𝜙 = Δ𝜙/2? By taking 𝜙 = Δ𝜙/2 , we are essentially centering the angular interval symmetrically around a midpoint. This is especially useful in scenarios where: - The angular interval Δ𝜙 corresponds to a small rotation or change in angle. - We are working with trigonometric functions (like sine or cosine), where halving the angle simplifies certain expressions or calculations (e.g., double-angle formulas). - The geometry or physics of the problem is naturally symmetric, so using the half-angle simplifies visualization and computation. 3. How this is applied: Suppose Δ𝜙 represents the total angular range between two points. Taking 𝜙 = Δ𝜙/2 is a common technique to focus on a representative angle for symmetric contributions or to split the total range into equal halves. For example: - In many integration problems over angular variables, symmetry arguments allow us to calculate over half the interval and then double the result. - In some derivations, the half-angle substitution reduces the complexity of the equations. If you’d like, I can explain this in greater detail or clarify how it fits into the specific context of the video. Let me know!

@TensorCalculusRobertDavie i understand half of what you written here cuz it seems heavy assumptions. So it would be quite helpful if you explain this both geometrically (using this figure and angular representation of the day diagram)and analytically in other video. Thank you for your kind reply

Ahh! I see now! It is correct, because of the symmetry of R under the exchange of the first and the second pairs of indices! I get it..

You're welcome.

Thank you for that!

You're welcome!

Thank you so much for your kind words and thoughtful question! Let’s break this down and clarify the difference between the pushforward operation and a coordinate transformation: 1. Pushforward Operation: The pushforward is a mathematical operation associated with a map between two manifolds, say 𝑓:𝑀→𝑁. If you have a tangent vector 𝑋_𝑝 at a point 𝑝∈𝑀, the pushforward 𝑓_∗ maps 𝑋_𝑝 to a tangent vector 𝑓_∗(𝑋_𝑝) at 𝑓(𝑝)∈𝑁. This is a way to "transport" the action of a vector field from the domain manifold 𝑀 to the codomain manifold 𝑁, based on the behavior of the function 𝑓. To understand it intuitively, if 𝑋_𝑝 represents a direction and rate of change of functions on 𝑀, then 𝑓_∗(𝑋_𝑝) represents how those changes translate to 𝑁 via the map 𝑓. The pushforward depends not on a coordinate system but on the geometry of 𝑓 itself. 2. Coordinate Transformation: A coordinate transformation, on the other hand, is a change of the local coordinate system used to describe the same manifold. For example, you might switch from Cartesian coordinates (𝑥,𝑦) to polar coordinates (𝑟,𝜃) on a 2D plane. Under a coordinate transformation, the components of vectors, tensors, or forms change according to specific transformation rules (e.g., chain rule for vectors). However, the geometric object itself remains the same-only its representation in terms of coordinates changes. 3. Key Difference: - Pushforward: Transfers tangent vectors (or vector fields) from one manifold to another via a map 𝑓:𝑀→𝑁. It involves two different manifolds. - Coordinate Transformation: Changes the representation of vectors or tensors on the same manifold by switching between different coordinate systems. To summarize: - Pushforward deals with mapping vectors between manifolds. - Coordinate transformation deals with describing vectors differently on the same manifold. I hope this clears things up! If you’d like a specific example or further clarification, feel free to ask. Thank you for watching and engaging with these lessons!

Thank you for saying that.

Hello John! Thank you for your thoughtful question! Let me clarify these points for you: What is meant by "physical" components? The term "physical components" refers to the components of a vector (or tensor) that are measured directly in terms of distances, angles, or magnitudes as they would appear in the real, physical space. In contrast to coordinate-dependent components (which can vary depending on the choice of basis or coordinate system), physical components are often associated with orthonormal bases derived from the metric of the space. These components have a direct geometric interpretation in terms of observable quantities like lengths or projections in orthogonal directions. For example, in a curved manifold, generalized coordinates like 𝑥^𝑖 might involve stretching or compressing axes in ways that obscure the underlying physical distances. The physical components correct for this by incorporating the metric tensor, ensuring that they correspond to measurements you could physically observe in the space. Are the generalized coordinates still orthogonal? In the specific example used in the video, the generalized coordinates happen to be orthogonal, meaning that the coordinate basis vectors satisfy 𝑔(∂/∂𝑥^𝑖 , ∂/∂𝑥^𝑗) = 0 for 𝑖 ≠ 𝑗. However, this is not generally true for all coordinate systems on a manifold. In many cases, the coordinate basis vectors are not orthogonal, and the metric tensor 𝑔_𝑖𝑗 captures the angle and length relationships between them. When the coordinates are orthogonal, the metric tensor becomes diagonal (only 𝑔_𝑖𝑖 are non-zero), simplifying many calculations. For more general cases, non-orthogonal coordinates require explicitly accounting for cross terms in the metric tensor, making the relationships between distances and components more complex. I hope this clarifies the distinction between physical components and coordinate components, as well as the role of orthogonality in generalized coordinates.

You're welcome.

Thank you.

You're welcome.

Thank you!

They are sorted in playlists.

@@TensorCalculusRobertDavie From which playlist should I start and please tell the sequence

@@abdurrahmanlabib916 You can start with either the differential geometry or the special relativity playlists depending on your interests.

Thank you!

Thank you for your wonderful comment!

Hi Logan. Nice comment. But what do you mean when you say that a curve is 'graduated'? I've never heard of that (I speak US English).Do you mean 'parameterized'?

It's great! Highly intuitive and approached with appropriate thoroughness! First principles reasoning ftw.

You're welcome.

Thank you for your question. Have a look at the last line on the previous slide and you will see where the negative sign comes from. The negative sign in front of 𝑇_𝑢𝑣 at 18:28 appears because of the Einstein field equations in general relativity, which are typically written as: 𝑅_𝑢𝑣 − 1/2𝑔_𝑢𝑣𝑅 = −8𝜋𝐺/(𝑐^/4)𝑇_𝑢𝑣. The negative sign here arises from the way energy-momentum influences the curvature of spacetime. Specifically: 1. Convention in the Field Equations: The sign is part of the convention used to align the Einstein field equations with physical observations, such as the way gravity behaves (e.g., how masses attract each other). The left-hand side of the equation, which describes spacetime curvature (𝑅_𝑢𝑣 and 𝑔_𝑢𝑣), is equated to the right-hand side, which describes the energy-momentum content (𝑇_𝑢𝑣) that acts as the source of this curvature. 2. Compatibility with General Relativity's Sign Conventions: In most conventions, the stress-energy tensor 𝑇_𝑢𝑣 represents the energy density, momentum flux, and stresses of matter and energy. The negative sign ensures the relationship between curvature and the energy-momentum tensor matches observed physics (e.g., gravitational attraction is caused by positive energy density). 3. Physical Interpretation: The negative sign reflects the fact that the energy density and pressure in 𝑇_𝑢𝑣 contribute to spacetime curvature. This convention ensures that the equations maintain consistency with the equivalence principle and the underlying geometric structure of general relativity. In Newtonian dynamics the gravitational force of attraction is "inwards" so we use a minus sign. Newton's law of universal gravitation states: 𝐹⃗ = −𝐺𝑚_1𝑚_2/(𝑟^2) 𝑟^, where: 𝐹⃗ is the gravitational force, 𝐺 is the gravitational constant, 𝑚_1 and 𝑚_2 are the masses of the objects, 𝑟 is the distance between the two objects, 𝑟^ is the unit vector pointing from one mass to the other. The negative sign indicates that the force is attractive and acts in the direction opposite to 𝑟^ . That is, the force pulls the two masses toward each other, which is interpreted as "inwards" in the context of gravitational attraction. Connection to Potential Energy The gravitational potential energy in Newtonian mechanics is given by: 𝑈 = −𝐺𝑚_1𝑚_2/𝑟. Here, the negative sign reflects that the potential energy decreases (becomes more negative) as the two masses come closer together. The force can then be derived as the gradient of the potential: 𝐹⃗ = −∇𝑈, which reinforces the idea that the force points "inwards" toward decreasing potential energy. Interpretation The negative sign in both the force and the potential energy equations is a way to encode the attractive nature of gravity. "Inwards" means that the force vector is directed toward the center of mass or the object exerting the gravitational pull. If you'd like me to delve deeper into how the sign convention arises mathematically or its connection to physical observations, feel free to ask!

@@TensorCalculusRobertDavie I got it sir . Thank you so much for your brief explanation.

Thank you so much for your thoughtful comment and for taking the time to share your observations! I'm glad you're enjoying the series, and I really appreciate your detailed feedback. Let me address your points one by one: 1. At 10:00 - Definition of ∗𝜔: You're absolutely right; there seems to be a typo in the formula as presented. The definition of ∗𝜔 should involve 𝜔, not 𝜂. I apologize for the oversight, and I'll add a correction or clarification in the video description to ensure viewers aren’t confused. Thanks for catching that! 2. Inner Product Induced by 𝑔: Great question! The inner product induced by the metric tensor 𝑔 on differential forms works by: - Using 𝑔 to lower an index on one of the forms (converting it to a vector field), - Then allowing the other form to act on this vector field, yielding a real number. Essentially, the metric tensor provides the geometric structure needed to define the inner product on forms. I'll consider adding a more detailed explanation in future videos or providing additional examples to make this process more explicit. Your understanding seems spot-on, so I’m glad the examples later on helped clarify this point! 3. At 12:00 - Notation Issue (𝑓,𝑔,ℎ vs. 𝑓,𝑘,ℎ): This indeed looks like another typo-thank you for pointing it out! I'll double-check the notations in that section and include a correction if necessary. The intended focus there was on the relationships and properties of the functions, so the inconsistency in naming wasn’t deliberate and will be clarified. 4. At 14:50 - Summation Convention and Levi-Civita Tensor: You’ve raised a very important point here. In the section with the Levi-Civita tensor and the definition of the form, the summation convention applies to the Levi-Civita tensor's indices, but it wasn’t extended to the form's definition explicitly. This distinction should have been emphasized to avoid confusion. I'll consider including a note about how and where the summation convention applies in this context to ensure clarity for future viewers. Thank you again for your careful observations and for highlighting areas where the explanations could be improved! It's feedback like this that helps refine the content and make it even more accessible. Please let me know if you have additional thoughts or questions-I’m happy to help!

Hello Vicente!

You're welcome.

Thank you for saying that!