# GR notation, newbie question

1. Sep 6, 2008

### Irid

I'm trying to learn some GR from Carrol's textbook, but I'm a little lost there. For example, this simple problem:

In Euclidean 3-space, let p be the point with coordinates (x,y,z) = (1,0,-1). Consider the curve passing through p:

$$x^i(\lambda) = (\lambda, (\lambda-1)^2, -\lambda)$$

Calculate the components of tangent vectors to these curves at p in the coordinate basis $$\{\partial_x, \partial_y, \partial_z\}$$.

The attempt at a solution

The components of tangent vectors are given by

$$V^i = \frac{dx^i}{d\lambda}$$

It is of course in the basis of x,y,z. But I don't understand what does the basis $$\{\partial_x, \partial_y, \partial_z\}$$ mean. The notation is new to me, but I think that

$$\partial_x \equiv \frac{\partial}{\partial x}$$

so how can this be used as a basis? If you just take these derivatives at each component of the curve, you always get (1,1,1), right?

2. Sep 6, 2008

### Dick

In GR, and differential geometry in general, you often define a vector as a directional derivative. So the vector ai+bj+ck (where i,j,k are the usual unit vectors) is written as a*d/dx+b*d/dy+c*d/dz. Since the derivative along the direction i is d/dx etc.

3. Sep 7, 2008

### Irid

Alright, so this is a trivial problem then, right?

4. Sep 7, 2008

### Dick

Sure. Just like what you did in Calc I but written in GR type notation.