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I am studying the lecture on general relativity for beginners here: https://youtu.be/foRPKAKZWx8?t=1998

I was able to follow the steps up until this point (33:18 in the video). During the deriving of the metric tensor, he is talking about how the vector transforms in a change of basis. He derived a formula using a gradient formula with partial derivatives by simply replacing the gradient by a vector.

enter image description here

The circled formula is what he derived from the boxed formula above. As I understand, X and Y here are frames of reference (rather than coordinates). How is the partial derivative of Y with respect to X obtained?

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  • $\begingroup$ When is gradient ever a scalar? $\endgroup$ – Dvij D.C. Nov 26 '20 at 4:37
  • $\begingroup$ You are right, got confused there, I've removed that part of the question $\endgroup$ – l3utterfly Nov 26 '20 at 4:44
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The coordinates are functions on a manifold. Say you have two sets of coordinates $y^i$, $x^i$ and transformation relation $y^i(x^j).$ Forming a gradient of these functions gives you:

$$dy^i=\sum_j\frac{\partial y^i}{\partial x^j}dx^j$$

Now this gradient is a linear machine (1-form), that takes vector as an input and tells you how quickly does the coordinate changes in the direction of the vector. So let us apply it to vector $\vec{V}=\sum_i V^i_y\vec{e^y_i}=\sum_i V^i_x\vec{e^x_i}$, where $\vec{e^y_i}$ is coordinate basis vector for coordinate $y^i$ (and analogically for $x$ coordinates):

$$dy^i(\vec{V})=\sum_jV^j_ydy^i(\vec{e^y_j})=V^i_y$$ $$\sum_j\frac{\partial y^i}{\partial x^j}dx^j(\vec{V})=\sum_{j,k}\frac{\partial y^i}{\partial x^j}V^k_xdx^j(\vec{e^x_k})=\sum_{j}\frac{\partial y^i}{\partial x^j}V^j_x$$

In the first equality, I have used the fact, that 1-form is a linear machine. In second equality the fact that $dy^i(\vec{e^y_j})=\frac{\partial y^i}{\partial y^j}=\delta^i_j$. The rest is putting these two results into the first formula for gradient and you will get the desired formula.

P.S.

I think calling $df$ a gradient is not correct terminology. $df$ is a 1-form, while gradient is a vector ($\vec{\nabla} f$). Gradient is defined as dual vector through metric.

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  • $\begingroup$ Thank you. Your terminology "transformation relation" from one basis to the other suddenly made it all clear to me! $\endgroup$ – l3utterfly Nov 26 '20 at 6:03
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In physics, a "frame" is nothing more than another word for a coordinate system. There are distinct objects known as frame fields, but by the looks of it, this is not what the lecturer is describing (they are, in my opinion, a more philosophically pleasing description of things, but more mathematically involved.

So, in the end there are some coordinates $x$ and some coordinates $y$. These are related to each other in the sense that one set of coordinates can be written as functions of the other set. For example, we may write $x^\mu=x^\mu(y)$ or $y^\mu=y^\mu(x)$, depending on whether we want to describe things in terms of the $x$ or $y$ coordinates.

As an example, in two dimensions, we may describe things in terms of Cartesian coordinates $(x,y)$ or polar coordinates $(r,\theta)$ and we may write $$ r(x,y)=\sqrt{x^2+y^2},\ \ \ \ \theta(x,y)=\tan(y/x). $$ Similarly, we could invert this coordinate transformation and write $$ x(r,\theta)=r\cos\theta,\ \ \ \ y(r,\theta)=r\sin\theta. $$

So, writing the Jacobian circled in the question out more fully we would have $$ J^\mu_\nu=\frac{\partial y^\mu}{\partial x^\nu}=\frac{\partial y^\mu(x)}{\partial x^\nu}. $$

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  • $\begingroup$ Thanks for the answer. If possible, can you help me by giving me a concrete example please? For example, let's say I have a vector (1,0), how would I apply that formula to get a vector in a different frame of reference, let's say, a frame of reference rotated by 45deg counterclockwise? $\endgroup$ – l3utterfly Nov 26 '20 at 5:53

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