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 independant coordinate in this circumstance, t is. That's like saying y=f(x), so $$\frac{dy}{dx}$$ must be zero because y is an independant 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.

 

[tacman]

Hello,

I can understand your confusion regarding the partial derivative of a parametric equation. Let me try to clarify it for you.

First, it is important to understand that the partial derivative of a function with respect to a variable only takes into account the changes in that variable while keeping all other variables constant. In the case of a parametric equation, the independent variable is the parameter t, and the dependent variable is x^i. Therefore, when we take the partial derivative of x^i with respect to t, we are only considering the changes in x^i due to changes in t, while keeping all other coordinates (x^j where j≠i) constant.

Now, in your example, x^i is an independent coordinate and does not have an explicit time dependence. This means that changes in t will not affect x^i, and hence the partial derivative of x^i with respect to t will be zero. However, when we take the partial derivative of the RHS, we are considering the changes in f^i(t) with respect to t, which will not necessarily be zero. This is because f^i(t) can also depend on other variables besides t.

Moving on to your second question, \dot x^i is defined as the total derivative of x^i with respect to t, which takes into account all the changes in x^i due to changes in t, regardless of whether x^i has an explicit time dependence or not. Therefore, in this case, the LHS will not be zero, as it considers all changes in x^i.

In summary, the partial derivative of a parametric equation only considers changes in the dependent variable with respect to the independent variable, while the total derivative takes into account all changes in the dependent variable. I hope this helps clarify your confusion. Let me know if you have any further questions.