# Correct form for d²r?

1. Feb 4, 2014

### Jhenrique

If dr = (dx, dy, dz) so, which is the correct form for d²r? Is (d²x, d²y, d²z) or (dy^dz, dz^dx, dx^dy) ?

2. Feb 4, 2014

what isr ?

3. Feb 4, 2014

### Jhenrique

you must be joking

4. Feb 4, 2014

### Staff: Mentor

Seems like a reasonable question to me.

5. Feb 5, 2014

### MuIotaTau

I imagine he's referring to the "numerator" of a second derivative such as $\frac{d^2 \vec{r}}{dt^2}$ in analogy with his identification of the differential form $d\vec{r}$ with the "numerator" of a first derivative.

6. Feb 5, 2014

### Jhenrique

Is this even. No doubts.

7. Feb 5, 2014

### PeroK

I've seen d²r used as an alternative to dS in surface integrals.

8. Feb 6, 2014

### Mandelbroth

The correct form of $\mathrm{d}^2r=\mathrm{d}(\mathrm{d}r)$ is 0. Your question is incredibly ambiguous, and I wouldn't expect anyone to understand what you're asking without clarification.

9. Feb 6, 2014

### Jhenrique

This is theoretical identity that in the practice is useless.

I just asking which are the components of the vector d²r.

The components of the vector r is (x, y, z);

of dr is (dx, dy, dz)

But, I have doubt if the components of vector d²r is (d²x, d²y, d²z) or (dydz, dzdx, dxdy) or other...