Introduction to 3-Forms
22:03
16 сағат бұрын
Wedge Product
17:15
16 сағат бұрын
Introduction to 2-Forms
20:25
28 күн бұрын
Gradient is a 1-Form
15:42
Ай бұрын
Introduction to 1-Forms
12:07
Ай бұрын
The Levi Civita Connection
26:04
The flow of a vector field
12:38
2 ай бұрын
The Functional Derivative - 2
16:04
The Functional Derivative - 1
10:42
Kerr geodesics - 27
27:54
2 ай бұрын
Kerr geodesics - 26
27:12
2 ай бұрын
Schwarzschild geodesics - 14
16:14
Kerr geodesics - 25
17:06
6 ай бұрын
Kerr geodesics - 24
33:58
6 ай бұрын
Kerr geodesics - 23
18:14
6 ай бұрын
Kerr geodesics - 22
28:13
6 ай бұрын
Kerr geodesics - 19
14:32
6 ай бұрын
Kerr geodesics - 21
27:21
6 ай бұрын
Kerr geodesics - 20
14:26
6 ай бұрын
Kerr metric - 12
9:49
6 ай бұрын
Kerr geodesics - 18
28:15
Жыл бұрын
Kerr geodesics - 17
17:10
Жыл бұрын
Kerr geodesics - 16
20:07
Жыл бұрын
Kerr geodesics - 15
9:44
Жыл бұрын
Kerr geodesics - 14
21:57
Жыл бұрын
Kerr geodesics - 13
26:03
Жыл бұрын
Пікірлер
@thevegg3275
@thevegg3275 4 күн бұрын
Thank you for your answer. I should say I’m no Einstein, lol, so I still have another question. I’m assuming that the value of a particular Christoffel symbol may be the result of a measurement of the change of a basis vector dot producted with a vector that describes a change in angle after parallel transport. Are you saying that if one were to try to represent this graphically, she would define the change directly on the curvy linear coordinate system. The type I’m thinking of can be very messy and irregular so how could it ever be possible to make a measurement of the change in a vector with respect to such cord system axis? I’m creating a graphic so the best I can do would be to somehow break the weird looking curvy linear square if you will with tick marks equally spaced. I did this and for one of the Christoffel symbols I got .3 in the X and .8 in the Y. These are the components that describe the difference vector in terms of the curvy linear coordinate system. I guess my main question is does it sound reasonable to say that lambda bc,a is .3 in the x direction .8 in the y direction? I have created a graphic that will clearly fine the question. If you would be so kind to email me, I will send you a JPEG of the image. I’m trying to define one image that explains Christoffel symbols, visually and clearly hopefully for the benefit of others as well as myself
@antoniopena1183
@antoniopena1183 5 күн бұрын
Why does the transformation with iscos\theta represent rotation around the z axis? What term would we use if we wanted to model rotation around the y axis, for example? Something like ibsin\theta\sin\phi?
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie 4 күн бұрын
Thank you for your question! Let's explore why the transformation using 𝑖cos𝜃 represents a rotation around the z-axis and how you can model rotations around other axes, like the y-axis. Understanding Rotations in 3D Space In 3D space, rotations are typically represented using rotation matrices, quaternions, or complex numbers (in certain contexts). The transformation you're asking about is most likely referring to a complex rotation matrix or a quaternion, which can be used to model 3D rotations. Rotation Around the z-axis When we talk about rotating around the z-axis, we're describing a transformation that keeps the z-coordinate constant while changing the x and y coordinates. The general rotation matrix around the z-axis is: 𝑅_𝑧(𝜃) = ( cos𝜃 −sin𝜃 0 sin𝜃 cos𝜃 0 0 0 1 ) Why 𝑖cos𝜃? The term 𝑖cos𝜃 is often used in complex numbers to simplify the expression of rotations. In the complex plane, a point (𝑥 , 𝑦) can be represented as 𝑧 = 𝑥 + 𝑦𝑖. A rotation around the z-axis by angle 𝜃 can be represented in the complex plane as multiplying by the complex number 𝑒^𝑖𝜃 = cos𝜃 + 𝑖sin𝜃. This effectively rotates the point around the z-axis, keeping the z-coordinate unchanged. Rotation Around the y-axis If you want to model a rotation around the y-axis, you need to use the corresponding rotation matrix: 𝑅_𝑦(𝜙) = ( cos𝜙 0 sin𝜙 0 1 0 −sin𝜙 0 cos𝜙 ) Here, 𝜙 is the angle of rotation around the y-axis. Notice how the y-coordinate remains unchanged, while the x and z coordinates are transformed. Using Complex Numbers for y-axis Rotation If you're interested in using complex numbers similar to 𝑖cos𝜃, you might consider how complex exponentials can be employed for 3D rotations. However, complex numbers are inherently two-dimensional. For a rotation around the y-axis, we rely on trigonometric transformations instead. Quaternion Representation For rotations in 3D, quaternions provide a convenient and efficient way to represent rotations. A quaternion for rotation around the y-axis by angle 𝜙 is given by: 𝑞 = cos(𝜙/2) + 𝑗 sin(𝜙/2) Here, 𝑗 represents the imaginary unit for the y-axis component in quaternion algebra. Quaternions avoid the issue of gimbal lock and provide a smooth interpolation of rotations. Your Proposed Term You mentioned the term 𝑖𝑏 sin𝜃 sin𝜙. Let’s break it down: The term 𝑖𝑏 sin𝜃 sin𝜙 suggests a more complex transformation possibly involving multiple axes. To specifically address the y-axis, you might need to adjust this to fit the rotation matrix around the y-axis as mentioned above. Conclusion To rotate around the z-axis, you use the transformation 𝑖cos𝜃 because it directly modifies the complex representation of the x and y coordinates, keeping the z-coordinate constant. For rotation around the y-axis, you should consider using the 𝑅_𝑦(𝜙) matrix or quaternion representations for more robust calculations. If you have further questions or need more details, feel free to ask.
@parthvarasani495
@parthvarasani495 6 күн бұрын
12:24 , u • v = g(ij) ui vj , not square root of it. I think.(In Numerator)
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie 6 күн бұрын
You are right. Thank you for spotting that.
@parthvarasani495
@parthvarasani495 5 күн бұрын
@@TensorCalculusRobertDavie Thank you for your all efforts, highly appreciated 👍👏👏
@sobrikey
@sobrikey 7 күн бұрын
Thanks Robert , fundamental subject
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie 7 күн бұрын
You're welcome.
@sobrikey
@sobrikey 7 күн бұрын
Thanks a lot from France , Robert
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie 7 күн бұрын
You're welcome.
@edd.
@edd. 7 күн бұрын
I just found your channel. This is great! I love it! I’m a retired electrical engineer and have been teaching myself GR and particle physics for about 8 years.
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie 7 күн бұрын
Thank you for saying that! I'm really glad you like the content. Cheers!
@alandegois1249
@alandegois1249 9 күн бұрын
Very good video. You posted a year ago, do you now have a better explanation why setting 1 when 0 is a problem? I am having an issue, because in the intervalo [0, 0.4] my function is complex, so I was numerically integrating from 0.4 to a superior limit using your method. Should I use 1 even in this case?
@thevegg3275
@thevegg3275 10 күн бұрын
Re the Christoffel symbols related to a curvilinear csys, the superscript on Gamma stands for the components of the displacement vector, whose displacement is a comparison of the original vector (prior to transport) to the vector's new position (after transport). Are these components wrt the A) curvilinear csys or B) wrt to an orthographic csys? I ask because its an easy task to find the components of a vector on an orthographic csys since there are many tick marks. But on most graphic representations of curvilinear csyss, it's just four curves (not broken down into smaller units such that one could measure the change in the displacement vect wrt to said curvilinear csys. To me, that means that the breakdown of a displacement vector must be graphically calculated using an orthograhic cys superimposede under the curvilinear one. Are these components wrt the A) curvilinear csys or B) wrt to an orthographic csys?
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie 4 күн бұрын
Thank you for your question! Let's delve into the Christoffel symbols and how they relate to coordinate systems. Christoffel Symbols and Their Components The Christoffel symbols, denoted as Γ_𝑖𝑗^𝑘 , are mathematical objects used in differential geometry to express how vector fields change as they move along a manifold. They play a crucial role in describing connections and curvature in a space, especially in General Relativity. Reference Coordinate System The superscript and subscripts of the Christoffel symbols relate to the coordinate system in use. In your question: Γ_𝑖𝑗^𝑘 components are with respect to the curvilinear coordinate system (A) rather than an orthographic one. Explanation: 1. Curvilinear Coordinate System: Curvilinear coordinate systems are used to describe spaces where the basis vectors can change from point to point. These systems include spherical, cylindrical, and more general forms seen in manifolds. The components of the Christoffel symbols, Γ_𝑖𝑗^𝑘 , indicate how the basis vectors change as you move along the coordinates of the manifold. The symbols tell us how the coordinate basis vectors ∂/∂𝑥^𝑖 change in the direction of ∂/∂𝑥^𝑗 , and the result is projected onto ∂/∂𝑥^𝑘 . 2. Orthographic Coordinate System: An orthographic (or orthogonal) coordinate system, like Cartesian coordinates, maintains consistent unit vectors throughout the space. While visualizations of curvilinear systems often overlay an orthographic grid for clarity, the calculation of Christoffel symbols does not inherently rely on this orthogonal reference. They are computed solely with reference to the curvilinear system itself. Visualizing and Calculating Components 1. Curvilinear Systems: Often depicted with smooth, continuous curves rather than discrete grids or ticks, making direct measurement less straightforward. However, the Christoffel symbols are computed analytically using derivatives of the metric tensor of the curvilinear coordinates. 2. Orthographic Systems: While it's more intuitive to measure vector components in orthographic systems due to the presence of straight and equidistant lines, superimposing an orthogonal grid on a curvilinear system is more for illustrative convenience rather than for calculating Christoffel symbols. Calculating Christoffel Symbols The Christoffel symbols are derived from the metric tensor 𝑔_𝑖𝑗 of the curvilinear system: Γ_𝑖𝑗^𝑘 = 1/2 𝑔^𝑘𝑙 (∂𝑔_𝑖𝑙/∂𝑥^𝑗 + ∂𝑔_𝑗𝑙/∂𝑥^𝑖 − ∂𝑔_ 𝑖𝑗/∂𝑥^𝑙) Here, 𝑔^𝑘𝑙 is the inverse of the metric tensor, and the partial derivatives are with respect to the curvilinear coordinates. Conclusion The components of the Christoffel symbols, Γ_𝑖𝑗^𝑘 , are with respect to the curvilinear coordinate system (A). While an orthographic system can provide an intuitive framework for visualizing these concepts, the mathematical computation and interpretation of the Christoffel symbols are based on the curvilinear framework. I hope this helps clarify your question! If you have any more queries or need further explanation, feel free to ask.
@DanielFRPater
@DanielFRPater 13 күн бұрын
At 2:59 you transpose the two last index, then the final result is minus the result that you have.
@DanielFRPater
@DanielFRPater 12 күн бұрын
I refer about the two last index of the Riemann tensor
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie 4 күн бұрын
@@DanielFRPater Why Transposing the Last Two Indices Results in a Negative Sign At the point in the video (2:59), you observed the operation where the last two indices of the Riemann tensor are transposed. This specifically refers to the property of anti-symmetry in the last two indices, which means: 𝑅^𝜌_𝜎𝜇𝜈 = −𝑅^𝜌_𝜎𝜈𝜇​ Explanation: 1. Anti-symmetry: This property is a direct consequence of how curvature is defined. The Riemann tensor measures the failure of parallel transport around an infinitesimal parallelogram, and its anti-symmetry reflects the inherent geometric properties of the manifold. When we swap the positions of 𝜇 and 𝜈, it effectively reverses the direction of the path around the parallelogram, hence resulting in a sign change. 2. Geometric Intuition: Imagine the manifold as a curved surface. Moving along one path and then transposing the directions (like swapping two edges of the parallelogram) results in a "mirrored" movement, explaining why the curvature's measure has an opposite orientation, thus yielding a negative sign. Example Calculation Consider the Riemann tensor component 𝑅^𝜌_𝜎12 . If you swap the indices 1 and 2, you get: 𝑅^𝜌_𝜎21 = −𝑅^𝜌_𝜎12​ This is an illustration of the anti-symmetry property in action, showing the resulting negative sign when indices are transposed. Conclusion In this video, the transposition of the last two indices of the Riemann tensor indeed results in a negative sign due to the tensor's inherent anti-symmetry property. This property is essential in describing the manifold's curvature and is a fundamental aspect of differential geometry.
@JanPBtest
@JanPBtest 25 күн бұрын
Wouldn't it be simpler to use motion constants (a technique used for the Schwarzschild case as well)? In the Kerr case there are four of them so the equations describing the geodesics reduce to first order?
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie 20 күн бұрын
The videos that follow on this playlist cover the first two constants while the final videos in this series cover all four constants.
@KippGenerator
@KippGenerator 26 күн бұрын
Vectors point in directions. One-forms measure rates of change etc. What if we define the gradient as the direction in which the function is changing fastest? (Inspired by Spivak prob 4-18, great book). I'm trying to get this right. Now I'm confused.
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie 25 күн бұрын
The question you're asking touches on a fundamental concept in differential geometry and calculus. Vectors and One-Forms: Vectors: In the context of differential geometry, vectors are entities that have both magnitude and direction. They represent directions in space. For example, a gradient vector points in the direction of the steepest ascent of a function. One-Forms: One-forms, on the other hand, are linear functionals that act on vectors. They can be thought of as measuring rates of change of functions in given directions. Mathematically, if 𝜔 is a one-form and 𝑣 is a vector, then 𝜔(𝑣) is a real number representing the rate of change of the function in the direction of 𝑣. Gradient as the Direction of Fastest Change: The gradient of a function 𝑓, denoted by ∇𝑓, is the vector that points in the direction of the steepest increase of the function. Here's why: The gradient vector ∇𝑓 is defined in such a way that for any vector 𝑣, the directional derivative of 𝑓 in the direction of 𝑣 is given by the dot product ∇𝑓⋅𝑣. This means that ∇𝑓⋅𝑣 measures how fast 𝑓 is changing in the direction of 𝑣. The direction in which 𝑓 increases the fastest is precisely the direction where this rate of change is maximized. Mathematically, this happens when 𝑣 is in the same direction as ∇𝑓. Relation to One-Forms: In the language of differential forms, the gradient can also be understood through its relationship to the differential of a function, which is a one-form. If 𝑓 is a smooth function, its differential 𝑑𝑓 is a one-form such that for any vector 𝑣, 𝑑𝑓(𝑣) = ∇𝑓⋅𝑣 Here, 𝑑𝑓 is measuring the rate of change of 𝑓 in the direction of 𝑣. The gradient vector ∇𝑓 is the vector associated with this one-form via the Riesz representation theorem in the context of Euclidean spaces. To summarize: 1. Vectors indicate directions. 2. One-forms measure rates of change in those directions. 3. The gradient vector ∇𝑓 points in the direction of the fastest increase of the function 𝑓. 4. The differential of 𝑓, 𝑑𝑓, is a one-form that relates to the gradient through the action on vectors. Hope that helps?
@Dzjur.
@Dzjur. 28 күн бұрын
So the part I'm really missing is how the pullback operator transpires from the diffeomorphism. I completely understand why we want to drag the vector back along the flow lime, but *why* does that lead to the lie derivative as the difference of two partials or equivalently the difference of two covariant derivatives.
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie 26 күн бұрын
Thank you for your thoughtful question and for watching the video! The pullback operator is a fundamental concept in differential geometry, particularly in the context of mappings between manifolds. When you have a smooth map (diffeomorphism) 𝜑 : 𝑀 → 𝑁 between two manifolds 𝑀 and 𝑁, the pullback operator 𝜑^∗ is a way to "pull back" differential forms (or other tensor fields) from 𝑁 to 𝑀. For a 1-form 𝜔 on 𝑁, the pullback 𝜑^∗𝜔 is a 1-form on 𝑀 defined by: (𝜑^∗𝜔)_𝑝(𝑣) = 𝜔_𝜑(𝑝)(𝑑𝜑_𝑝(𝑣)) for 𝑝 ∈ 𝑀 and 𝑣 ∈ 𝑇_𝑝𝑀. 1-Forms and Pullbacks - Let's look at how this works! A 1-form is a type of differential form that maps a vector to a real number. If 𝜔 is a 1-form on a manifold 𝑁, then 𝜔 can be thought of as a function that takes a tangent vector at any point in 𝑁 and returns a real number. The pullback operator 𝜑^∗ allows us to transfer (or pull back) a differential form from one manifold 𝑁 to another manifold 𝑀 using a smooth map 𝜑 : 𝑀 → 𝑁. This is useful because it lets us relate forms on different spaces. The Definition of the Pullback of a 1-Form Given a smooth map 𝜑 : 𝑀 → 𝑁 and a 1-form 𝜔 on 𝑁, the pullback 𝜑^∗𝜔 is a 1-form on 𝑀. The pullback is defined in such a way that it retains the essential properties of 𝜔 but is now defined on 𝑀 instead of 𝑁. The Formula For a point 𝑝 ∈ 𝑀 and a tangent vector 𝑣 ∈ 𝑇_𝑝𝑀, the pullback (𝜑^∗𝜔)_𝑝 applied to 𝑣 is given by: (𝜑^∗𝜔)_𝑝(𝑣) = 𝜔_𝜑(𝑝)(𝑑𝜑_𝑝(𝑣)) So what is this??? 1. 𝜔_𝜑(𝑝) : This is the 1-form 𝜔 evaluated at the point 𝜑(𝑝) in 𝑁. Here, 𝜑(𝑝) is the image of the point 𝑝 under the map 𝜑. 2. d𝜑_𝑝(v): This is the differential of 𝜑 at the point 𝑝, applied to the vector 𝑣. The differential 𝑑𝜑_𝑝 is a linear map that takes a tangent vector 𝑣 ∈ 𝑇_𝑝𝑀 and maps it to a tangent vector in 𝑇_𝜑(𝑝)𝑁. More formally, if 𝜑 is a smooth map between manifolds, then 𝑑𝜑_𝑝 is the map induced on the tangent spaces, taking 𝑇_𝑝𝑀 to 𝑇_𝜑(𝑝)𝑁. 3. Putting it together: The pullback 𝜑^∗𝜔 is defined such that when it acts on a vector 𝑣 at point 𝑝 ∈ 𝑀, it gives the same result as 𝜔 would give if we first map 𝑣 to 𝑇_𝜑(𝑝)𝑁 using 𝑑𝜑_𝑝 and then apply 𝜔 at the point 𝜑(𝑝). To visualize this: 1. Imagine you have a 1-form 𝜔 on 𝑁. Think of 𝜔 as a field that assigns a real number to each vector in the tangent space at every point in 𝑁. 2. The map 𝜑 takes points from 𝑀 and maps them to points in 𝑁. When you pull back 𝜔 via 𝜑, you are essentially translating the influence of 𝜔 back onto 𝑀. 3. The differential 𝑑𝜑 translates vectors from 𝑀 to 𝑁 in a way that respects the structure of the manifolds. 4. The pullback 𝜑^∗𝜔 then acts on vectors in 𝑀 in a manner consistent with how 𝜔 acts on their images in 𝑁. Summary The pullback of a 1-form allows us to transfer the action of the form from one manifold to another using a smooth map. It ensures that the way the form evaluates vectors is consistent with the map’s induced transformations between the manifolds. This concept is foundational in differential geometry and helps in understanding how different geometric objects transform under smooth maps. Hope that helps?
@Mikey-mike
@Mikey-mike 28 күн бұрын
Excellent. Well done.
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie 28 күн бұрын
Thank you! Cheers!
@saurabhyadav-t6x
@saurabhyadav-t6x Ай бұрын
sir, your videos are very useful. keep it up, thank you
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie Ай бұрын
Thank you for saying that!
@douglasswesty1540
@douglasswesty1540 Ай бұрын
At 3:14 your slide states that "In flat space the basis vectors are constant". This is not true. The fact that the basis vectors are constant in this case is not due to the fact that the space is flat. It is due to your choice of coordinate system, i.e. your choice of Cartesian coordinates. If you had chosen 2-D polar coordinates for the same flat space the basis vectors would not be constant and the components of the connection would not be zero. In general a flat manifold does not imply that the basis vectors are constant.
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie Ай бұрын
Thank you for your insightful comment and for watching the video! You are absolutely correct. My statement on the slide at 3:14 was an oversimplification. The constancy of the basis vectors is indeed a consequence of the choice of coordinate system, specifically Cartesian coordinates, rather than the flatness of the space itself. In flat space, or Euclidean space, it's possible to choose a coordinate system where the basis vectors are constant, such as Cartesian coordinates. However, if we choose a different coordinate system, like polar coordinates in 2-D flat space, the basis vectors will vary with position. Consequently, the Christoffel symbols (components of the connection) may not be zero, even though the space itself is flat. Flatness of a manifold implies that the Riemann curvature tensor is zero, but it does not necessarily imply that the basis vectors in every coordinate system are constant. The properties of the basis vectors depend on the choice of coordinates. Thank you for pointing this out, and I appreciate your attention to detail. I will make a note to clarify this distinction in future content.
@nathsujitkumar
@nathsujitkumar Ай бұрын
Very helpful video. Could you please suggest any reference for this? A paper by Daniel Frederick (PHYSICAL INTERPRETATION OF PHYSICAL COMPONENTS OF STRESS AND STRAIN) and another by C. Truesdell (The physical components of vectors and tensors) seem to give a little different expressions for physical components of second order tensors than what you derive at 11:14 in the video. However, both your derivation and the papers' derivation match for the orthogonal coordinates. But I wanted know the correct expressions in the case of general coordinates. Thanks!
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie Ай бұрын
Thank you for your thoughtful question and for watching the video! I appreciate your interest in understanding the physical components of second-order tensors in different coordinate systems. The expressions for physical components of second-order tensors can indeed vary depending on the coordinate system used. In the case of orthogonal coordinates, the derivations tend to align more closely because the basis vectors are mutually perpendicular, simplifying the mathematical expressions. For general (non-orthogonal) coordinates, the expressions can be more complex due to the interaction between the coordinate system's basis vectors. The key difference arises from the metric tensor, which accounts for the non-orthogonality of the coordinate system. Here are a few references that might help clarify these concepts further: Daniel Frederick's paper: "PHYSICAL INTERPRETATION OF PHYSICAL COMPONENTS OF STRESS AND STRAIN". This paper explores the physical components in various coordinate systems, including non-orthogonal ones. The differences you noticed likely stem from how the metric tensor is incorporated. C. Truesdell's paper: "The physical components of vectors and tensors". Truesdell's work is foundational in the field of continuum mechanics and provides a comprehensive treatment of tensor components in different coordinate systems. For the expressions in general coordinates, the correct approach involves using the covariant and contravariant components of tensors, along with the metric tensor. The metric tensor 𝑔_𝑖𝑗 relates the physical components of the tensor in general coordinates to those in orthogonal coordinates. In summary, while the derivations in my video match those in the papers for orthogonal coordinates, in general coordinates, it is essential to account for the metric tensor and the potential non-orthogonality of the basis vectors. I hope this clarifies the differences you observed. For a more detailed mathematical treatment, I recommend reviewing the aforementioned papers and the sections on tensor calculus in general coordinates. Feel free to ask any more questions or request further clarifications. Thanks again for engaging with the content!
@nathsujitkumar
@nathsujitkumar Ай бұрын
@@TensorCalculusRobertDavie Thank you so much for your clear and elaborate explanation. I will have a deeper look into this and try to clear my ideas using the references. Thanks for making such nice informative videos! Very much appreciated. Keep on doing this. I feel bad for myself that I discovered your channel so late. However, better late than never. I am now a subscriber and a regular visitor to your channel! 🙂
@curtischee2532
@curtischee2532 Ай бұрын
Another wonderful easy to understand video. I am doing self study on differential geometry and manifolds. Please do more videos on these topics.
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie Ай бұрын
Thank you for saying that and more videos will be coming. Good luck with the studies.
@ARBB1
@ARBB1 Ай бұрын
Great video as always mate. What do you think on also speaking about geometric calculus?
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie Ай бұрын
Thank you for that. As to geometric calculus, which is a beautifully compact and powerful way to work with geometric concepts, my focus on this channel is mostly directed at introducing concepts and procedures that many will encounter when first learning GR. That is not to say that I won't make videos with a focus on geometric calculus.
@bobbybannerjee5156
@bobbybannerjee5156 Ай бұрын
What's the difference between tangent space and cotangent space?
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie Ай бұрын
Great question! The concepts of tangent space and cotangent space are fundamental in differential geometry and general relativity. Here's a detailed explanation: Tangent Space The tangent space at a point on a manifold is the set of all possible tangent vectors at that point. Intuitively, you can think of the tangent space as a plane that "touches" the manifold at a given point, containing all the directions in which one can tangentially pass through that point. For a manifold 𝑀 at a point 𝑝, the tangent space is denoted by 𝑇_𝑝𝑀. Example: Consider a 2D surface (like a sphere) in 3D space. The tangent space at a point on the surface is the plane that just touches the sphere at that point, containing all possible directions you can move while staying on the surface. Cotangent Space The cotangent space at a point on a manifold is the dual space to the tangent space. While the tangent space consists of tangent vectors, the cotangent space consists of covectors (also known as one-forms). These are linear functionals that take a tangent vector and return a real number. For a manifold 𝑀 at a point 𝑝, the cotangent space is denoted by 𝑇_𝑝^∗𝑀. Example: If you have a function 𝑓 defined on the manifold, the gradient of 𝑓 at a point 𝑝 is an element of the cotangent space at 𝑝. This gradient can take any tangent vector at 𝑝 and return the rate of change of the function 𝑓 in the direction of that vector. Key Differences Nature: Tangent vectors are elements of the tangent space and represent directions in which one can move on the manifold. Covectors (one-forms) are elements of the cotangent space and are linear functionals acting on tangent vectors. Duality: The cotangent space is the dual space to the tangent space. This means each element of the cotangent space is a linear map that takes an element of the tangent space and returns a scalar. Dimensions: At each point 𝑝 on an 𝑛-dimensional manifold, both the tangent space 𝑇_𝑝𝑀 and the cotangent space 𝑇_𝑝^∗𝑀 are 𝑛-dimensional vector spaces. Application: Tangent vectors are used to describe velocities and directions of curves on the manifold. Covectors are used to describe gradients, differentials, and other quantities that can act on tangent vectors. Visualization Imagine standing on a hill: The tangent space at your feet represents all possible directions you can walk. The cotangent space can be thought of as representing measurements of how steep the hill is in various directions. Each co-vector in the cotangent space can tell you the slope (rate of change) of the hill in a specific direction. I hope this helps clarify the distinction between tangent and cotangent spaces! If you have any further questions, feel free to ask.
@Mikey-mike
@Mikey-mike Ай бұрын
Excellent. Well done. I enjoy your video lectures. Thanks.
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie Ай бұрын
Thank you for saying that!
@punditgi
@punditgi Ай бұрын
Many thanks for your videos! 🎉😊
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie Ай бұрын
Thank you for saying that!
@Person-ef4xj
@Person-ef4xj Ай бұрын
In the more general case when all the coordinates are say u and v and the values in the inverse of the metric tensor are all non 0 if mu is u can lambda be either u or v or is it’s value still fixed by the value of mu?
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie 26 күн бұрын
This video shows the case for a diagonal metric whereas you are asking about the non-diagonal case. In the non-diagonal case mu and lambda can be different.
@edbertkwesi4931
@edbertkwesi4931 Ай бұрын
men you have followers
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie Ай бұрын
Thank you!
@forheuristiclifeksh7836
@forheuristiclifeksh7836 Ай бұрын
1:00
@robertdavie1221
@robertdavie1221 Ай бұрын
?
@forheuristiclifeksh7836
@forheuristiclifeksh7836 Ай бұрын
@@robertdavie1221 Thanks for video. It is my own records that I need to remember. can you recommend me books or problem sheet that maybe I can challenge with together?
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie Ай бұрын
?
@curtischee2532
@curtischee2532 Ай бұрын
In your matrix for h alpha beta why does h oo have a minus sign?
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie Ай бұрын
Thank you for your question and for watching the video! In the context of general relativity, the metric tensor 𝑔_𝛼𝛽 encodes the geometry of spacetime. When we consider a small perturbation around a background metric (typically the flat Minkowski metric), we write the perturbed metric as: 𝑔_𝛼𝛽 = 𝜂_𝛼𝛽 + ℎ_𝛼𝛽, where 𝜂_𝛼𝛽 is the Minkowski metric and ℎ_𝛼𝛽 represents the perturbation. Sign Convention in the Metric Minkowski Metric: The Minkowski metric 𝜂_𝛼𝛽 in standard coordinates (t, x, y, z) is typically written as: 𝜂_𝛼𝛽 = (−1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1). This metric has a minus sign in the time-time component (𝜂_00) and positive signs in the spatial components (𝜂_11, 𝜂_22, 𝜂_33). Perturbation Matrix ℎ_𝛼𝛽: When we perturb the metric, the perturbation ℎ_𝛼𝛽 is added to the Minkowski metric. The component ℎ_00 corresponds to the perturbation in the time-time component of the metric. To maintain consistency with the signature of the Minkowski metric, ℎ_00 typically appears with a minus sign. This is because perturbations in the time component often reflect changes in the temporal part of the metric, which has a negative sign in the Minkowski metric. Why the Minus Sign? Gravitational Potential: In weak-field approximations (such as in the Newtonian limit), ℎ_00 is related to the gravitational potential 𝜙. Specifically, ℎ_00 ≈ −2𝜙 where 𝜙 is the Newtonian gravitational potential. The negative sign reflects the fact that the presence of a gravitational potential reduces the time component of the metric. Consistency with Physical Interpretation: The negative sign ensures that the perturbation correctly modifies the time-time component of the metric in a way that is consistent with our physical understanding of how gravity affects spacetime. For example, near a massive object, the time component of the metric is reduced, indicating time dilation. The perturbation ℎ_00 with a negative sign reflects this effect. Conclusion The minus sign in ℎ_00 is there to maintain consistency with the Minkowski metric's signature and to correctly represent the physical effects of the perturbation, such as gravitational time dilation. This ensures that the perturbed metric accurately reflects the influence of gravitational fields in accordance with general relativity. I hope this helps clarify the reason behind the minus sign in ℎ_00.
@elia610206
@elia610206 Ай бұрын
French mathematician???????
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie Ай бұрын
Sorry about that! I published a correction in the description below the video.
@narfwhals7843
@narfwhals7843 Ай бұрын
Thank you, it was interesting to see the formula for the levi- civita in terms of the metric connection pop out of the conditions. It is important to note that this means that the levi-civita connection is _uniquely_ defined by the metric, which is why it is used in GR. But throughout the video you say thing like "torsion free manifold". I find this confusing. Isn't torsion freeness a property of the connection, not the manifold? We could use a different connection, and even get a different Curvature tensor. Carroll, in spacetime and geometry, states that only in the case of using the levi-civita connection does it make sense to talk about the "curvature of the metric", but using another one is not forbidden. The question is how useful it is. Also a note on the video. It was pretty distracting to watch you look for places to put the camera. I didn't count the time, but I wouldn't be surprised if it was over two minutes over the course of the video. I'd suggest either finding a fixed spot for it in the slides or leaving it out after the introduction. It's nice to have a face to put to the voice, but I don't think it adds much to watch you read to us afterward.
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie Ай бұрын
Thank you for your insightful question and for watching the video! You are absolutely right that the Levi-Civita connection is uniquely defined by the metric and is torsion-free, which makes it particularly useful in general relativity (GR). Let me address your points in more detail: Torsion-Free Property Property of the Connection: Torsion-freeness is indeed a property of the connection, not the manifold itself. A manifold can have various connections defined on it, and each connection can have different properties. The Levi-Civita connection is defined as the unique connection that is both metric-compatible and torsion-free. This means it preserves the inner product defined by the metric and has no torsion (the antisymmetric part of the connection vanishes). Torsion-Free Manifold vs. Torsion-Free Connection: My apologies if the phrasing "torsion-free manifold" caused confusion. It would be more accurate to say "manifold equipped with a torsion-free connection" when referring to the Levi-Civita connection. Other connections on the same manifold might have torsion, leading to different geometric and physical interpretations. Curvature and Connections Curvature Tensor: The curvature tensor depends on the connection used. Different connections on the same manifold can yield different curvature tensors. When we refer to the curvature associated with the Levi-Civita connection, we often say "curvature of the metric" because the connection is uniquely determined by the metric. Other Connections: It is indeed possible to use other connections with torsion or non-metric compatibility. These can lead to different curvature tensors and have various applications in physics, such as in theories of gravity that extend beyond GR (e.g., Einstein-Cartan theory, which includes torsion). However, in standard GR, the Levi-Civita connection is used because it directly relates the geometry of spacetime to the distribution of matter and energy via the Einstein field equations. Practical Considerations Utility of Different Connections: Using different connections can provide insights into alternative geometrical structures and physical theories. However, the Levi-Civita connection's properties make it particularly suitable for describing the curvature of spacetime in GR. As Sean Carroll notes, the concept of the "curvature of the metric" is most meaningful when using the Levi-Civita connection due to its compatibility with the metric and torsion-free nature. Conclusion So, to clarify, the torsion-free property is indeed a feature of the connection, not the manifold. The Levi-Civita connection is special because it is the unique connection that is both metric-compatible and torsion-free, making it central to GR. Using other connections is possible and can lead to different curvature tensors, but the Levi-Civita connection is particularly useful for describing the geometry of spacetime as we understand it in standard GR. Also, I agree with your final point concerning the placement of the video image on the screen and so I will turn it off in future.
@RominPlusFans
@RominPlusFans Ай бұрын
When we vary the metric, does it tell us anything about the underlying manifold? I've always had a hard time separating the concept of the metric from the underlying manifold.
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie Ай бұрын
Understanding the relationship between the metric and the underlying manifold is a crucial aspect of differential geometry and general relativity. Here’s an explanation to help clarify these concepts: The Underlying Manifold vs. The Metric The Underlying Manifold: Think of the underlying manifold as a stage or a canvas. It is a collection of points that can be organized in a smooth and continuous way, forming a shape or space without any additional structure. Mathematically, a manifold is a topological space that locally resembles Euclidean space. For example, a 2-dimensional surface like a sphere or a torus can be considered a 2-dimensional manifold. The Metric: The metric is like the ruler and protractor that we use on the stage. It provides a way to measure distances and angles between points on the manifold. Formally, the metric tensor 𝑔 assigns a scalar product to each pair of tangent vectors at each point on the manifold, allowing us to compute lengths, angles, and volumes. Varying the Metric What Happens When We Vary the Metric: When we vary the metric, we are essentially changing how we measure distances and angles on the manifold, not the manifold itself. Imagine changing the scale or stretching/compressing the ruler we use. The stage remains the same, but our perception of distances and shapes changes. Implications of Varying the Metric: Varying the metric can reveal different geometric properties and physical phenomena without altering the underlying manifold. For instance, in general relativity, the variation of the metric tensor in the Einstein-Hilbert action leads to the Einstein field equations, describing how matter and energy influence the curvature of spacetime. Separating the Metric from the Manifold Conceptual Separation: The manifold provides the framework or the "where" (the set of points and their topological arrangement). The metric provides the "how" (the rules for measuring distances and angles between those points). Understanding Through Examples: Flat Manifold with Different Metrics: Consider a flat 2-dimensional plane. If we change the metric, we can make it look like a cylinder or a cone geometrically, but the underlying set of points (the manifold) remains the same. Curved Manifolds: On a curved manifold like a sphere, changing the metric can alter the perceived curvature but not the fact that it’s still a sphere. Conclusion So, when we vary the metric, it doesn't change the underlying manifold itself. Instead, it changes how we measure and understand the geometric and physical properties on that manifold. This distinction is fundamental in many areas of theoretical physics and mathematics, helping us explore different geometries and their implications while keeping the foundational structure of the manifold intact.
@RuanD
@RuanD Ай бұрын
Thank you, it was very useful!
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie Ай бұрын
You're welcome.
@robertprince1900
@robertprince1900 Ай бұрын
at 1213, commutator formula at top does not look correct. As written two dot products are always zero.I think you should remove the dots.
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie Ай бұрын
The dots here represent multiplication as can be seen on the next slide.
@robertprince1900
@robertprince1900 Ай бұрын
You didn't explain how metric could be known intrinsically, after removing the X. You don't get a number dotting vector operators with each other, and no number no distance calculation for example. Perhaps the surface denizens could measure in different directions and guess G though?
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie Ай бұрын
Please see slides 2 and 3 where I cover the necessary material needed for this video. The concept of an intrinsic metric is indeed subtle but fundamental in differential geometry. Here’s an explanation: Understanding the Intrinsic Metric Intrinsic vs. Extrinsic Perspective: Extrinsic Metric: When we define a surface in an embedding space 𝑋 (like 𝑅^3), we can measure distances using the dot product of vectors in 𝑋. This gives us an "extrinsic" view of the surface. Intrinsic Metric: Even if we remove the embedding space 𝑋, the surface itself retains a notion of distance and angle purely based on its own properties. This is the "intrinsic" view. Metric Tensor: The metric tensor 𝑔 on a surface allows us to compute lengths and angles directly on the surface without referencing the embedding space. For a surface parameterized by coordinates (𝑢,𝑣), the metric tensor 𝑔_𝑖𝑗 (where 𝑖,𝑗 can be 𝑢 or 𝑣) provides the inner product of the tangent vectors to the surface: 𝑑𝑠^2 = 𝑔_𝑢𝑢𝑑𝑢^2 + 2𝑔_𝑢𝑣𝑑𝑢𝑑𝑣 + 𝑔_𝑣𝑣𝑑𝑣^2. This formula enables the surface's inhabitants to measure distances and angles intrinsically. Intrinsic Measurement: Surface denizens can measure distances and angles by considering the deformation of curves and the angles between them. For example, by measuring the circumference of small circles and their radii, they can deduce the curvature and metric properties. Practical Steps to Determine the Metric Intrinsically Local Measurements: Distances: Measure the distance between nearby points along different directions. Angles: Measure the angles between intersecting curves. Constructing the Metric Tensor: By accumulating data on distances and angles, one can construct the components of the metric tensor 𝑔_𝑖𝑗. Suppose denizens measure distances along coordinate directions 𝑢 and 𝑣. They can establish the relationships: 𝑔_𝑢𝑢 = (∂𝑥/∂𝑢)^2 , 𝑔_𝑣𝑣 = (∂𝑥/∂𝑣)^2 , 𝑔_𝑢𝑣 = (∂𝑥/∂𝑢)(∂𝑥/∂𝑣). Verification: Once the metric tensor is deduced, it can be verified by comparing the predicted distances and angles with further measurements. Conclusion So, even without referring to the embedding space 𝑋, the metric tensor 𝑔 allows intrinsic measurements on the surface. It encodes all necessary information to compute distances, angles, and areas directly from within the surface itself. The intrinsic perspective is powerful because it tells us that the geometry of a surface is an inherent property, independent of how it might be embedded in higher-dimensional space. It is a good point and really worth a video in its own right.
@RuanD
@RuanD Ай бұрын
Amazing explanation, thank you for this lecture 🙌🏻
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie Ай бұрын
Thank you for saying so.
@tehreemzahra6406
@tehreemzahra6406 Ай бұрын
1- At 21:15 , it has been written that we are interested in the region that lies inside ring singularity. Why is it so? I couldn't understand the reason behind it. Is it because the trajectories in region r>0 are timelike? That's why we are interested in region inside ring singularity. 2- Also, we have fixed our theta; equatorial plane. What if we only take r<0 with some arbitrary theta (theta not fixed)? Will we still get CTCs? 3- My last confusion is what happens at g_{phiphi}=0?
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie Ай бұрын
1 - The interest in the region inside the ring singularity (specifically 𝑟<0 in the extended Kerr metric) stems from the unique and exotic properties of spacetime in this region. In the context of the Kerr black hole, the region inside the ring singularity can host Closed Timelike Curves (CTCs). CTCs are paths in spacetime that loop back on themselves, theoretically allowing for time travel to the past. These curves do not exist in regions 𝑟>0 outside the event horizon and ergosphere but become possible in 𝑟<0. The presence of CTCs is a direct consequence of the frame-dragging effect caused by the rotating black hole, which becomes more extreme inside the ring singularity. 2 - Yes, CTCs can still exist for 𝑟<0 with arbitrary 𝜃. The presence of CTCs in the Kerr metric is not restricted to a specific plane or angle. While the analysis in the equatorial plane (𝜃=𝜋/2) often simplifies calculations and visualizations, the conditions leading to CTCs are generally valid throughout the region 𝑟<0. The crucial aspect is the metric component 𝑔_𝜙𝜙, which becomes negative in this region, allowing for the existence of time-like loops. Therefore, fixing 𝜃 is not a necessary condition for CTCs to occur; they are a feature of the inner region of the Kerr black hole due to the nature of the spacetime geometry there. 3 - The condition 𝑔_𝜙𝜙=0 is significant because it marks the boundary where the nature of the coordinate 𝜙 (representing the azimuthal angle) changes. In the Kerr metric, 𝑔_𝜙𝜙 becomes zero at the boundary of the ergosphere, known as the static limit. Ergosphere Boundary: Outside this surface, 𝑔_𝜙𝜙>0, meaning the azimuthal coordinate behaves normally, and observers can remain stationary relative to distant stars. However, as you cross into the ergosphere, 𝑔_𝜙𝜙 becomes negative, indicating that all observers are forced to co-rotate with the black hole due to the intense frame-dragging effect. Inside the Ergoregion: Within the ergosphere but outside the event horizon, the negative 𝑔_𝜙𝜙 implies that stationary observers cannot exist. Instead, every observer must rotate in the same direction as the black hole. This is a consequence of the fact that spacetime itself is dragged in the direction of the black hole's rotation. Thus, 𝑔_𝜙𝜙=0 is a critical surface that demarcates these different behaviors in the Kerr spacetime. Hope that helps?
@tehreemzahra6406
@tehreemzahra6406 Ай бұрын
@@TensorCalculusRobertDavie thank you for your response . Your videos have helped me a lot. Just a comment that I think the g-{tt}=0 marks the boundary for the ergo sphere as well as the surface of the infinite red shift not g_{phiphi}. Source: General relativity by M.P Hobson
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie Ай бұрын
@@tehreemzahra6406 I'm glad you found them useful. The reference to 𝑔_𝜙𝜙 refers to its sign in the regions mentioned and how that changes in the different regions and not as a definition of a surface. As you point out g_tt = 0 determines the location of the ergo-sphere. General relativity by M.P Hobson et al. is a great book!
@fdc4810
@fdc4810 Ай бұрын
Hi professor Davie, many thanks for the video. I just have a naive question and hope you won't mind me asking. Could you elaborate on why, for the first case (vector field Y), we generate the vector at T_{p}M by the means of pullback, rather than doing a parallel transport ? And why a pullback would be a meaningful mapping in this case which can preserve Y as it is when moved along X ?
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie Ай бұрын
Great question! The pullback by a diffeomorphism 𝜙_𝑡 generated by a vector field 𝑋 effectively "transports" objects along the flow of 𝑋. For a vector field 𝑌, the pullback 𝜙_𝑡∗𝑌 is defined such that: (𝜙_𝑡∗𝑌)(𝑝) = (𝑑𝜙_(-t))_𝜙_t(p)𝑌(𝜙_𝑡(𝑝)) Here, 𝑑𝜙_(-t) is the differential (Jacobian) of the map 𝜙_(-𝑡), which effectively "pulls back" the vector field 𝑌 from 𝜙_𝑡(𝑝) to 𝑝. This operation makes sense in the context of the Lie derivative because it directly measures how the vector field 𝑌 changes as we move along the flow generated by 𝑋. The pullback preserves the vector field structure in the sense that it maintains the consistency of the flow dynamics dictated by 𝑋. Parallel Transport Parallel transport, on the other hand, is a process defined by a connection on the manifold (typically a Levi-Civita connection in the context of Riemannian geometry). It transports vectors along a curve such that they remain "parallel" with respect to the connection. This means that the covariant derivative of the transported vector along the curve is zero. While parallel transport is crucial in defining geodesics and curvature, it is not as directly related to the flow of vector fields and their infinitesimal changes, which is what the Lie derivative captures. Why Use Pullback for Lie Derivative? Preservation of Flow Dynamics: The Lie derivative 𝐿_𝑋𝑌 measures how the vector field 𝑌 changes as we "flow" along 𝑋. The pullback by 𝜙_(-𝑡) is a natural way to describe this because it directly incorporates the dynamics of the flow generated by 𝑋. Pullback 𝜙_(-𝑡)∗ preserves the structure of the vector field 𝑌 under the flow, meaning that it maintains the relationships between vectors at different points along the flow. Meaningfulness of the Pullback: The pullback 𝜙_𝑡∗𝑌 is meaningful because it respects the underlying manifold’s structure and the flow generated by 𝑋. It ensures that we are looking at the same vector field 𝑌 but at a different point along the flow. This operation is essential in defining the Lie derivative because it provides a way to compare the vector field 𝑌 at 𝑝 and at 𝜙_𝑡(𝑝) in a manner consistent with the flow generated by 𝑋. In summary, we use the pullback in the context of the Lie derivative to capture the infinitesimal change of a vector field 𝑌 along the flow generated by another vector field 𝑋. This approach preserves the dynamics of 𝑌 under the flow of 𝑋 and provides a natural and meaningful way to measure the change, which is the essence of the Lie derivative. Also, I am not a professor.
@fdc4810
@fdc4810 28 күн бұрын
​@@TensorCalculusRobertDavie That's a more than detailed explaination, thank you so much for your work, your series has been more than enjoyable !
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie 26 күн бұрын
@@fdc4810 Thank you for saying that!
@andresvasquez5411
@andresvasquez5411 2 ай бұрын
Really love your work and your videos. They've been really helpful :)
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie 2 ай бұрын
Thank you for saying that. I'm glad you have found them useful.
@sapamniranjansingh8721
@sapamniranjansingh8721 2 ай бұрын
Thanks sir
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie 2 ай бұрын
You're welcome.
@scottdickinson5310
@scottdickinson5310 2 ай бұрын
One thing I don’t quite follow is that the functional derivative is defined as an operator, but here you find it as a quotient of two infinitesimal variations, what’s the link?
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie Ай бұрын
In essence, the functional derivative can be understood both as an operator that acts on variations of 𝑓 and as the limit of the ratio of infinitesimal changes in the functional to those in the function. Both perspectives are equivalent and provide valuable insights into how functionals depend on their underlying functions.
@peterhall6656
@peterhall6656 2 ай бұрын
A compact and clear exposition. Well done.
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie 2 ай бұрын
Thank you for that!
@Mikey-mike
@Mikey-mike 2 ай бұрын
Excellent. Thank you.
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie 2 ай бұрын
Thanks for that!
@AntonioDiMuro-hr1ir
@AntonioDiMuro-hr1ir 2 ай бұрын
Thank you for the explanation! But why n21=1?
@AntonioDiMuro-hr1ir
@AntonioDiMuro-hr1ir 2 ай бұрын
Thank you for the explanation! But why n21=1?
@robertdavie1221
@robertdavie1221 2 ай бұрын
You're welcome. Please look at 1:55 and you will see that n_ji = n_ij so this matrix is anti-symmetric which is the answer you are after.
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie 2 ай бұрын
You're welcome. Notice at 1:55 that n_ji = -n_ij because the matrix is anti-symmetric since n_ji + n_ij = 0. That means if n_21 = 1 then n_12 = -1.
@AntonioDiMuro-hr1ir
@AntonioDiMuro-hr1ir 2 ай бұрын
​@@TensorCalculusRobertDavie thanks for the answer, but why does n21 have that value and not for example n12=2? furthermore the killing vector A=d_t=(1,0,0,0) is covariant, but in the Schwarzschild geodesic N° 1 lesson at 11.44 the same vector is in the contravariant form. It is different? A0 = g00 A^0 and g00=-(1-2GM/(c^2r) but g^00=1/g00 I don't understand Thank you
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie Ай бұрын
@@AntonioDiMuro-hr1ir For the Killing vector 𝐴 = ∂_t , the contravariant components are 𝐴^𝜇 = (1, 0, 0, 0) where 𝜇 runs over the spacetime coordinates. In the Schwarzschild metric, the time component of the metric tensor is 𝑔_00 = −(1−2𝐺𝑀/𝑐^2𝑟). Therefore, the covariant time component of the Killing vector 𝐴_𝜇 is: 𝐴_0 = 𝑔_00𝐴^0 = −(1−2𝐺𝑀/𝑐^2𝑟) ⋅ 1 = −(1−2𝐺𝑀/𝑐^2𝑟).
@AntonioDiMuro-hr1ir
@AntonioDiMuro-hr1ir Ай бұрын
@@TensorCalculusRobertDavie Thanks for the reply, now I understand, sorry if I still abuse your patience and time, but n12?
@smftrsddvjiou6443
@smftrsddvjiou6443 2 ай бұрын
Thanks. I wounder, for calculating the metric tensor always the euclidean basis of the embedding space is used . However idealy we don‘t want to use this, just the affine transformation of the Manifold. So, is it possible to calculate metric tensors just based on the generalized coordinates ?
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie 2 ай бұрын
Yes, it is. Thank you for your question! In differential geometry, the metric tensor on a manifold can indeed be calculated using only the generalized coordinates, without referring to the Euclidean basis of an embedding space. The key idea is that the metric tensor g_{ij} encodes the intrinsic geometry of the manifold and can be expressed directly in terms of the coordinate system used to describe the manifold. It is entirely possible to calculate the metric tensor using only the generalized coordinates of the manifold. This approach is actually the standard practice in differential geometry, where the focus is on the intrinsic properties of the manifold, independent of any embedding space. I hope this clarifies your question! If you have any further inquiries, feel free to ask.
@smftrsddvjiou6443
@smftrsddvjiou6443 2 ай бұрын
@@TensorCalculusRobertDavie Thanks a lot.
@sobrikey
@sobrikey 2 ай бұрын
Avec mes remerciements !
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie 2 ай бұрын
Thank you!
@forheuristiclifeksh7836
@forheuristiclifeksh7836 2 ай бұрын
0:30
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie 2 ай бұрын
???
@forheuristiclifeksh7836
@forheuristiclifeksh7836 2 ай бұрын
@@TensorCalculusRobertDavie Always thanks for video. It is my bookmark. Can you also recommend me books or problem set that I can follow together?
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie 2 ай бұрын
@@forheuristiclifeksh7836 "Introduction to Smooth Manifolds" by John M. Lee: This textbook is widely regarded as an excellent introduction to the topic, offering clear explanations and thorough coverage of the subject. "Differential Topology" by Victor Guillemin and Alan Pollack: While primarily focused on differential topology, this book provides a solid foundation in the theory of manifolds and includes intuitive explanations of key concepts. "An Introduction to Manifolds" by Loring W. Tu: This introductory textbook covers the basics of manifolds in a clear and accessible manner, making it suitable for students with varying levels of mathematical background. "Topology from the Differentiable Viewpoint" by John W. Milnor: In this classic text, Milnor provides an elegant introduction to differential topology and manifold theory, presenting the material in a way that is both rigorous and intuitive. Online Lectures and Tutorials: Websites such as Khan Academy, MIT OpenCourseWare, and KZbin offer a wealth of free resources, including video lectures and tutorials, that cover the basics of manifold theory. Mathematics Stack Exchange: This online community allows users to ask questions and receive answers from experts in the field. Browsing through past questions and answers on the topic of manifolds can provide valuable insights and clarification. Scholarly Articles and Research Papers: While more advanced, scholarly articles and research papers in mathematics journals can offer in-depth discussions and insights into specific aspects of manifold theory.
@forheuristiclifeksh7836
@forheuristiclifeksh7836 2 ай бұрын
@@TensorCalculusRobertDavie Thanks so much! I'll dive into field with the videos!!!!
@charlesfountain3946
@charlesfountain3946 2 ай бұрын
Thee videos are outstanding and are worth their weight in gold for sure... Much (severe under statement ) gratitude by those of us who labor to have the enjoyment of learning physics , mathematics and such enticements with out aid of an instructor... Now our prayers are answered... Thank you for all the hard work involved in getting this material presented in such an enjoyable format... MIT and Princeton eat your heart out: Saved all that tuition and taking endless piles of notes... The author is intellectually blessed indeed... As evidenced by his skill as an orator and communicator...
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie 2 ай бұрын
Thank you for such encouraging feedback. It is greatly appreciated. Most of all I am really glad to hear people tell me that they find these videos helpful. Thanks again!
@aneikei
@aneikei 2 ай бұрын
Hi, nice video. I have a question. At slide 19:33 should that bottom term be (1 + g00)? Since g00 = -(1 + 2phi/c²). As such wouldn't the bottom term would be (1 + (- (1 + 2phi/c²)) no?
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie 2 ай бұрын
The last term is derived by transposing the equation given at the top of the slide.
@sobrikey
@sobrikey 2 ай бұрын
Merci ! from France
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie 2 ай бұрын
Thank you.
@bernardauvity9669
@bernardauvity9669 2 ай бұрын
Thanks a lot for this !
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie 2 ай бұрын
You're welcome.
@subhadipsau363
@subhadipsau363 2 ай бұрын
What should be $\delta g^{\alpha\beta}/\delta g^{\mu u}?
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie 2 ай бұрын
Certainly! Let's go through the derivation in more detail: 1. Starting Point: The inverse metric tensor \( g^{\alpha\beta} \) satisfies the following relationship with the metric tensor \( g_{\alpha\beta} \): \[ g^{\alpha\mu} g_{\mu\beta} = \delta^\alpha_\beta. \] 2. Taking the Variation: To find the functional derivative \( \frac{\delta g^{\alpha\beta}}{\delta g^{\mu u}} \), we consider the variation of both sides of the equation: \[ \delta(g^{\alpha\mu} g_{\mu\beta}) = \delta(\delta^\alpha_\beta). \] The right-hand side is zero because the Kronecker delta \( \delta^\alpha_\beta \) is a constant: \[ \delta(\delta^\alpha_\beta) = 0. \] 3. Applying the Product Rule: On the left-hand side, we apply the product rule for variations: \[ \delta(g^{\alpha\mu} g_{\mu\beta}) = (\delta g^{\alpha\mu}) g_{\mu\beta} + g^{\alpha\mu} (\delta g_{\mu\beta}). \] 4. Setting the Variation to Zero: Since the variation of the Kronecker delta is zero, we get: \[ (\delta g^{\alpha\mu}) g_{\mu\beta} + g^{\alpha\mu} (\delta g_{\mu\beta}) = 0. \] 5. Isolating \( \delta g^{\alpha\mu} \): We want to isolate \( \delta g^{\alpha\mu} \). First, move \( g^{\alpha\mu} (\delta g_{\mu\beta}) \) to the other side: \[ (\delta g^{\alpha\mu}) g_{\mu\beta} = -g^{\alpha\mu} (\delta g_{\mu\beta}). \] 6. Multiplying by \( g^{\beta u} \): To isolate \( \delta g^{\alpha\mu} \), multiply both sides by the inverse metric \( g^{\beta u} \): \[ (\delta g^{\alpha\mu}) g_{\mu\beta} g^{\beta u} = -g^{\alpha\mu} (\delta g_{\mu\beta}) g^{\beta u}. \] Note that \( g_{\mu\beta} g^{\beta u} = \delta_\mu^ u \), the Kronecker delta. 7. Simplifying: This simplifies to: \[ \delta g^{\alpha u} = -g^{\alpha\mu} \delta g_{\mu\beta} g^{\beta u}. \] 8. Functional Derivative: The functional derivative \( \frac{\delta g^{\alpha\beta}}{\delta g^{\mu u}} \) is found by recognizing that the variation \( \delta g_{\mu u} \) can be expressed as a functional derivative: \[ \frac{\delta g^{\alpha\beta}}{\delta g_{\mu u}} = -g^{\alpha\mu} g^{\beta u}. \] Therefore, the functional derivative of the inverse metric tensor with respect to the metric tensor is: \[ \frac{\delta g^{\alpha\beta}}{\delta g^{\mu u}} = -g^{\alpha\mu} g^{\beta u}. \] This result shows how the inverse metric tensor changes with respect to the metric tensor. I hope this step-by-step explanation helps!
@user-ps7gb2ej6i
@user-ps7gb2ej6i 3 ай бұрын
you are amazing thanks
@TensorCalculusRobertDavie
@TensorCalculusRobertDavie 2 ай бұрын
Thanks for that.