# Vector transformation derived from partial derivatives

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.

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?

• When is gradient ever a scalar?
– ACat
Nov 26, 2020 at 4:37
• You are right, got confused there, I've removed that part of the question Nov 26, 2020 at 4:44

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.

• Thank you. Your terminology "transformation relation" from one basis to the other suddenly made it all clear to me! Nov 26, 2020 at 6:03

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 $$J^\mu_\nu=\frac{\partial y^\mu}{\partial x^\nu}=\frac{\partial y^\mu(x)}{\partial x^\nu}.$$

• 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? Nov 26, 2020 at 5:53