Vectors as directional derivatives

This video deals with curves and scalar fields defined on manifolds and how an associated directional derivative can produce partial derivative operators that can be used as coordinate basis vectors. This result is of some importance to General Relativity.

[@fuzdaduck1]

I am working on a hypothesis that a concept called homogenous infinitesimals can replace Calculus, Euclidean and non-Euclidean geometry. A homogenous infinitesimal basically means that infinitesimal elements are required in order to sum them up, elements of length for length, elements of area for area, and voluminal elements for volume. Your explanation here gives me an opportunity for a counterargument, so I hope you don't mind. In your image of the manifold, this would be three directional elements of volume where we can only see the one side of the volumes. Basis vectors are redefined as dx^1, dx^2, and dx^3 is not visible. The length of a line is defined as n*dx where n is the relative cardinality of the line, more simply put as the number of infinitesimal dx in the line. Your vector r(x1,x2) would be replaced by r=(n^1*dx^1, n^2*dx^2).It gets a bit more complicated, but essentially I think your equations for basis vectors at 4:30 could be flawed notation as they should simplify down to just the infinitesimal dx^1 and dx^2 for the basis vectors. It would be the equivalent of asking what is the change in the number n of dx^1 for each single dx^1. The answer should be for each dx^1 there is a change of 1 dx^1 (and add in dx^2 if r isn't along dx^1 direction). If you are interested in the background of my logic I am posting on reddit at www.reddit.com/r/numbertheory/comments/1iclwxy/theory_calculuseuclideannoneuclidean_geometry_all/?

[@Mikey-mike]

Excellent I like your graphics and presentation. I like too how you introduced the intrinsic form for the basis vector.

[@TensorCalculusRobertDavie]

Thank you Mikey. It is helpful to me to receive the feedback you have given. Nice to know I am on the right track.

[@patrikengas6479]

i'm having a hard time understanding how partial derivatives are equal to basis vectors, so i'm wondering is there an easy proof for this? my initial attempt was to follow the premise to get a proof for a spesific case: if f(x,y)=x^2 + y^2 and ∂/∂x = e_x => ∂/∂x (f(x,y)) = e_x (f(x,y)) => 2x = [1,0,0]*[x, y, f(x,y)] = x , Now this doesn't seem to make sense. The assumption i made here was that e_x acted on the scalar field as a dot product seeing as this would preserve the category of the output object, based on the partial derivative outputting a scalar when acting on a function. So my main question is if there is a simple general proof to this and if there isn't, do you know what i did wrong/why these were the results? and a minor question regarding the curve on 1:30, is this curve defined by a position-vector from an origin outside the surface? and if not where would the origin be on the surface itself, for the curve to be able to be defined using the basis of the surface?

[@TensorCalculusRobertDavie]

Thank you for your detailed and insightful questions! Let’s address them step by step: 1. Why are partial derivatives equal to basis vectors? This is a foundational concept in differential geometry, so let’s break it down. Partial derivatives ∂/∂𝑥^𝑖 are interpreted as basis vectors in the tangent space because of how they act on scalar functions. Specifically: - The tangent space at a point 𝑝 on a manifold is the space of all possible "directions" in which one can move away from 𝑝. - A vector in this tangent space is defined operationally as a differential operator that acts on smooth functions 𝑓 defined near 𝑝. In local coordinates (𝑥^1, 𝑥^2 , … , 𝑥^𝑛), the operators ∂/∂𝑥^𝑖 naturally arise from infinitesimal variations in each coordinate. To "prove" this equivalence, consider: - The definition of ∂/∂𝑥^𝑖 as a directional derivative: it measures the rate of change of a function 𝑓 along the 𝑥^𝑖-axis. - In the coordinate basis {∂/∂𝑥^1 , ∂/∂𝑥^2 , … , ∂/∂𝑥^𝑛} , any tangent vector 𝑣 can be written as a linear combination 𝑣 = 𝑣^𝑖 ∂/∂𝑥^𝑖 . The coefficients 𝑣^𝑖 correspond to components of the vector, and the basis vectors are precisely the partial derivative operators. Where your example went astray: The mistake lies in interpreting the action of 𝑒_𝑥 (a basis vector) as a dot product. Instead: - 𝑒_𝑥 , as ∂/∂𝑥 , acts on 𝑓(𝑥,𝑦) by taking its derivative with respect to 𝑥, yielding 2𝑥 for your example 𝑓(𝑥,𝑦) = 𝑥^2 + 𝑦^2 . - The dot product interpretation doesn’t apply here because 𝑒_𝑥 (as ∂/∂𝑥) is not a vector in the traditional Euclidean sense; it’s a differential operator. 2. Is there a simple general proof? Here’s a conceptual outline: i. Define a vector in the tangent space 𝑇_𝑝(𝑀) at 𝑝 as a derivation: a linear map 𝑣 : 𝐶^∞(𝑀) → 𝑅 satisfying the Leibniz rule 𝑣(𝑓𝑔) = 𝑣(𝑓)𝑔 + 𝑓𝑣(𝑔). ii. In local coordinates, show that the partial derivatives ∂/∂𝑥^𝑖 satisfy these properties and form a basis for all derivations at 𝑝. iii. Conclude that the partial derivatives ∂/∂𝑥^𝑖 serve as the coordinate basis for the tangent space. 3. About the curve at 1:30 You’re asking whether the curve is defined relative to an origin off the surface or on it. Great question! - Generally, a curve on a manifold is defined intrinsically, meaning it doesn’t depend on an "external" origin. It’s described as a mapping 𝛾 : 𝐼 → 𝑀 , where 𝐼 ⊂ 𝑅 is an interval and 𝑀 is the manifold. - If you’re working in an embedded setting (e.g., the surface exists in 𝑅^3), you can think of the curve as being described by a position vector relative to some origin in 𝑅^3. However, the intrinsic description relies only on the coordinates within the surface itself, so the "origin" would be irrelevant or purely local. Summary 1. Partial derivatives are basis vectors in the tangent space because they naturally represent infinitesimal motions along coordinate directions. 2. Your example misinterpreted ∂/∂𝑥 as a dot product; it’s instead an operator that acts on functions. 3. The curve can be understood intrinsically without needing an origin outside the surface. Thanks again for your questions, and I hope this clarifies things! Feel free to reach out if you’d like further discussion.

[@patrikengas6479]

I appreaciate your response I feel like i sort of am beginning to gain an intuition about it, the partial derivative operator when acting on a function tells the change in the function value with respect to a change / step in a particular direction (𝑥^𝑖), and what else takes a step in a particular direction? well a basis vector.. I am following all up until 7:40, where we get d/dλ = t^μ ∂/∂x^μ, (i agree with this), but then you're mentioning the earlier definition of a tangent vector t = t^μ e_μ, and since the components match and the derivative operators measures in same direction that d/dλ = t, but can we really say that when we have 1 variable being the same and having 3 total variables in the equation?, to illustrate what i mean here: 6 = 3 * 2 and 9 = 3 * 3 where 3 is the common factor like t^μ. Now i'm not a pure math or physics student, so i don't quite understand the formal proof. I'm also quite new to (differential geometry, so i don't have the best knowledge about all the concepts).

[@AlulaZagway]

just subscribed....keep it up...thank you

[@TensorCalculusRobertDavie]

Thank you for that!

[@rorysmith4843]

This is a beautifully clear explanation! Thank you!!

[@TensorCalculusRobertDavie]

Hello Rory and thank you for that.

[@exxzxxe]

Very well done- clear and concise. Thank you!

[@TensorCalculusRobertDavie]

You're welcome.

[@TensorCalculusRobertDavie]

Coordinate lines mark out the space or manifold we are dealing with and the coordinate curves are the lines marked out by each coordinate. In Euclidean space the x and y axes are the coordinate lines of this space in 2D and, for example, the x = 5 is an example of a coordinate curve in this space. Having said all of that, both words can be used interchangeably since a line or curve mean the same thing.

[@joeydifranco0422]

I'm a bit confused. There are two sets of x¹ and x² here right? the x¹ and x² coordinate curves and then the x¹(λ) and x²(λ) magnitudes of the specific curve x_tilde which is parameterized by λ. Those are indeed two different things, right? Because I am having trouble telling which is being used at certain times. Are the x's in the scalar f(x¹, x²) the ones parameterized by λ? I thought not, but the fact that df/dλ=(∂f/∂x)(dx/dλ) seems like the chain rule where x¹ and x² do depend on λ...

[@TensorCalculusRobertDavie]

Hello Joey. Where did you get "curve x_tilde" from? The point of the video is that you can think of the partial derivative operators as forming a set of basis vectors at each point of the manifold. These basis vectors belong to the tangent space to the manifold at each point p (T_pM).

[@joeydifranco0422]

@@TensorCalculusRobertDavie at 1:32, the position vector x with the wavy line beneath it, which has components whose magnitude is x¹(λ) and x²(λ). I couldn’t find the character for an x with a tilde beneath it (tilde is the name of the wavy line) so I just typed out "x_tilde" meaning "x with a subscript tilde".

[@TensorCalculusRobertDavie]

@@joeydifranco0422 My apologies, you are quite right. In the first part of the video up to 1:29 looks at how the coordinate curves are obtained from some position vector (no λ's at this point). Next, we consider some curve on the manifold using some parameter that could, for example, be time as measured by the object's own clock or its path length etc. This parameter is λ. So at λ = 1 second, the object's coordinates might be (x¹, x²) = (1, 1.25) and two seconds later it might be (x¹, x²) = (5, 9). Next, I pick an arbitrary point P on the manifold and consider the tangent vector to the curve at this point. But, this tangent vector lies in the tangent space at that point P, which, for our 2-D example is a plane (hence tangent plane). This takes us up to 5:29. Then at 5:30 a totally arbitrary scalar function f is introduced. This function is a function of the coordinates of the manifold f(x¹, x²) and at this stage it has nothing to do with λ as it depends on the coordinates x¹and x² only. In the next part starting at 6:28 we want to know how f changes in the direction of the coordinate curves x¹and x². We note that these directions coincide with those of the basis vectors for this 2-D space. Still no λ at this stage. Then, at 7:40 we ask ourselves how does f change as we move along the curve (x¹(λ), x²(λ))? From here on I establish the argument for why we can view the partial derivative operators as a set of basis vectors for this tangent space instead of the usual basis vectors we label with the e's. Then remembering that each point P will have its own unique tangent space we can define a basis using these partial derivative operators. Finally, on a manifold we cannot add or subtract vectors at different points in the usual way that we do in Euclidean space because these vectors exist in their own unique tangent space at each point with its own set of basis vectors. Vectors that belong to different vector spaces cannot be added or subtracted because they have different basis vectors. But we can compare the directions these vectors point in using various techniques such as the covariant derivative. If direction is the important quantity then these partial deriviative operators are of more relevance to us than an object with a fixed "length". Hope that helps?

[@joeydifranco0422]

@@TensorCalculusRobertDavie It does a lot. Thanks for your detailed response.

[@robertprince1900]

A basis vector in flat space gives me a direction arrow, and even if I use curvilinear coordinates which change everywhere I still have an arrow. A partial derivative is a scalar; I can’t see how it can be a basis or how you could solve real problem but putting some random function into the partial. In fact what I see done is the partial is replaced by an actual basis vector along v or u when problem solving, like parallel transport calculations, begin. Please help! Thanks!

[@robertdavie1221]

A partial derivative of some scalar function gives the rate of change in the direction of the given coordinate involved. It gives us the rate at which that function is changing per unit of the positive direction of the coordinate involved. Now remove the scalar function and look at the object acting on it. This object points in the positive direction of the coordinate involved and is of unit length since it tells us about the rate of change of the scalar function per unit of the coordinate involved. I hope that helps?

[@shantambajpai4944]

Great explanation robert ! Just one question, is the tangent vector intrinsic to the manifold ?

[@TensorCalculusRobertDavie]

Hello Shantam. Yes. Can I refer you to the following discussion: math.stackexchange.com/questions/3600335/tangent-vector-to-a-point-on-the-surface

[@shantambajpai4944]

Thank you robert ! Will look at it !

[@TensorCalculusRobertDavie]

The tangent vector is indeed a concept intrinsic to the manifold itself. In differential geometry, we define tangent vectors at each point of the manifold, and these vectors are independent of any particular coordinate system or embedding into a higher-dimensional space. Here's why the tangent vector is intrinsic to the manifold: Invariance Under Coordinate Transformations: Tangent vectors are defined solely in terms of how curves on the manifold pass through each point. They are invariant under changes of coordinates. This means that the tangent vector's components may change when we switch to a different coordinate system, but its geometric meaning, representing the direction of a curve at a point, remains the same. No Reference to External Space: Tangent vectors do not rely on any reference to an external space. Unlike vectors in Euclidean space, which we often visualize as arrows pointing in a particular direction, tangent vectors are purely geometric objects defined within the manifold itself. Geometry of the Manifold: Tangent vectors are crucial for understanding the geometry of the manifold, such as curvature and geodesics. They play a central role in defining concepts like the metric tensor, connection, and curvature tensor, which describe how the manifold curves and bends locally. In summary, the tangent vector is an intrinsic property of the manifold, representing the direction of curves at each point. It is independent of any particular coordinate system or embedding and is essential for understanding the geometry and dynamics of the manifold. I hope this helps clarify the concept of tangent vectors and their intrinsic nature within the manifold!

[@abcdef-ys1sb]

at 0:34 what does co-ordinate lines mean ? I can understand co-ordinate curve. x1, x2 is co-ordinate curve. Because in a curved manifold our axis will be also a curve, not straight-line like Euclidean space. But, what does co-ordinate line mean ?

[@TensorCalculusRobertDavie]

Thank you for your question! Let's clarify the difference between coordinate curves and coordinate lines: Coordinate Curves: In the context of a curved manifold, such as spacetime in general relativity, coordinate curves refer to curves traced out by keeping all coordinates except one fixed. For example, in two-dimensional spacetime described by coordinates x and t, a coordinate curve along the x axis would keep the t coordinate fixed while varying the x coordinate. similarly, a coordinate curve along the t axis would keep the x coordinate fixed while varying the t coordinate. These curves help us visualize how the coordinates change as we move along the manifold. Coordinate Lines: Coordinate lines, on the other hand, refer to straight lines or curves traced out by keeping one coordinate fixed while allowing all others to vary. In Euclidean space, coordinate lines are straight lines along each axis (e.g., the x-axis, y-axis, and z-axis). However, in a curved manifold, such as spacetime in general relativity, coordinate lines may not necessarily be straight. Instead, they follow the curvature of the manifold and may appear curved or distorted. So, while coordinate curves trace out curves along individual coordinates, coordinate lines trace out lines or curves by fixing one coordinate and allowing the others to vary. In a curved manifold, both coordinate curves and coordinate lines help us understand the geometry and coordinate system used to describe the space. I hope this clarifies the distinction between coordinate curves and coordinate lines!

[@abcdef-ys1sb]

at 10:00, you have told v^mu is Lorentz invariant. How we can check it ? Is there any trick to check it quickly ?

[@TensorCalculusRobertDavie]

Thank you for your question! Let's explore how we can check if v^μ is Lorentz invariant and if there are any tricks to do so quickly: Lorentz Invariance of v^μ: v^μ represents the components of a vector in spacetime, and for it to be Lorentz invariant means that its components remain unchanged under Lorentz transformations. In other words, if we transform to a different inertial reference frame using a Lorentz transformation, the components of v^μ should remain the same. Checking Lorentz Invariance: One way to check if v^μ is Lorentz invariant is to apply the Lorentz transformation directly to its components and see if they remain unchanged. The Lorentz transformation equations for the components of a vector v^μ are well-defined and by plugging in the components of v^μ and verifying if they transform correctly, we can confirm its Lorentz invariance. Tricks for Quick Checks: While there may not be a single trick to quickly check Lorentz invariance, there are some strategies that can make the process more efficient. These include: Using symmetry properties of the components of v^μ to simplify calculations. Checking special cases or limits where Lorentz transformations are known to simplify, such as when v ^μ represents the velocity of an object relative to the speed of light. Mathematical Consistency: Ultimately, the key to confirming Lorentz invariance is ensuring mathematical consistency between the original components of v^μ and their transformed counterparts. If the components satisfy the Lorentz transformation equations, then v^μ is indeed Lorentz invariant. In summary, to check if v^μ is Lorentz invariant, we apply Lorentz transformations to its components and verify if they remain unchanged. While there may not be a single trick for quick checks, employing symmetry properties and exploiting known limits can streamline the process. I hope this helps clarify how we can check the Lorentz invariance of v^μ!

[@igorbecker2323]

Very instructive!

[@TensorCalculusRobertDavie]

Thank you Igor.

[@simewn]

Very usefull!

[@TensorCalculusRobertDavie]

Thank you Simos.

[@bananichattoraj8441]

What does mean by basis vector?

[@TensorCalculusRobertDavie]

Hello Banani and thank you for your question. A basis vector is a vector that is one of a set of vectors that span some space and are linearly independent and from which any other vector in that space can be formed.

[@dXoverdteqprogress]

Does a position vector make sense on a curved manifold? It certainly doesn't transform like one: x'^i e'_i not= x^ie_i

[@TensorCalculusRobertDavie]

A position vector in the form of a directed line segment exists on Euclidean manifolds but not on curved manifolds where the basis vectors vary from point to point.

[@RTD553]

00:40 when he says lines, he means surfaces. This is the fundamental difference: one vector system follows lines (is tangent to them), the other (the 'dual') forms vectors perpendicular to the coordinate surfaces.

[@TensorCalculusRobertDavie]

Hello Nick and thank you for your comment. I am referring to lines and not surfaces because when one of the coordinates is fixed and the other free to vary then lines are formed. However, when both coordinates are free to vary then a surface is formed. Also, this video is dealing with vectors and no mention is made of one-forms.