Partial Derivative of a Parametric Equation

[emob2p]

Hi,
I'm getting confused over a few points on the derivative of a parametric equation.
Say we the world line of a particle are represented by coordinates $$x^i$$. We then parametrize this world line by the parameter t. $$x^i = f^i(t)$$.

Now here is where I get confused. The partial derivative $$\frac {\partial x^i}{\partial t}$$ should be zero since x is an independent coordinate and has no explicit time dependence. However, if I take the partial of the RHS above, clearly this is nonzero.

Moreover we define $$\dot x^i$$ as $$\dot x^i = \frac{d x^i}{d t}$$. It seems the RHS will be the same if we take a partial or total derivative, yet the LHS will be zero if we take $$\frac {\partial x^i}{\partial t}$$ but $$\dot x^i$$ if we take $$\dot x^i = \frac{d x^i}{d t}$$. Is my question clear? Thanks.

 

[Perturbation]

Hi,
I'm getting confused over a few points on the derivative of a parametric equation.
Say we the world line of a particle are represented by coordinates $$x^i$$. We then parametrize this world line by the parameter t. $$x^i = f^i(t)$$.

Now here is where I get confused. The partial derivative $$\frac {\partial x^i}{\partial t}$$ should be zero since x is an independent coordinate and has no explicit time dependence. However, if I take the partial of the RHS above, clearly this is nonzero.

Moreover we define $$\dot x^i$$ as $$\dot x^i = \frac{d x^i}{d t}$$. It seems the RHS will be the same if we take a partial or total derivative, yet the LHS will be zero if we take $$\frac {\partial x^i}{\partial t}$$ but $$\dot x^i$$ if we take $$\dot x^i = \frac{d x^i}{d t}$$. Is my question clear? Thanks.

Except t isn't a coordinate, it's a parameter. x does have an explicit time dependance, you've given it one when you equated x to f(t). x is not an independent coordinate in this circumstance, t is. That's like saying y=f(x), so $$\frac{dy}{dx}$$ must be zero because y is an independent coordinate.

The total and partial derivatives with respect to t are the same here because $f:\mathbb{R}\rightarrow\mathbb{R}$ is a one dimensional function.