# I want a second opinion

1. Oct 7, 2007

### mlazos

[SOLVED] I want a second opinion

We know the equation

$$F=-\frac{dV}{dr}$$

we want to find the integral from $$r_{0}$$ to $$r$$.

I have seen someone doing this

$$\int^{r}_{r_{0}}Fdr'=-\int^{r}_{r_{0}}\frac{dV}{dr}dr'$$

I am a mathematician and the way I was doing at the university was

$$F=-\frac{dV}{dr}\RightarrowFdr=dV$$

and then I integrate

$$\int^{r}_{r_{0}}Fdr=-\int^{}_{r_{0}}dV$$

Since the potential depends on r we can integrate. So I would like someone who knows the subject to tell me if the first way is correct since I know the second is correct. Its difficult for me to accept the introduction of another variable r' while we have the r itself.

2. Oct 7, 2007

### arildno

Well, the first one is incorrect; it should be:
$$\int_{r_{0}}^{r}Fdr'=-\int_{r_{0}}^{r}\frac{dV}{dr'}dr'$$

Now, the second integral simply equals $$V(r)-V(r_{0})$$, thus, we have identity between the expressions:
$$\int_{r_{0}}^{r}\frac{dV}{dr'}dr'=\int_{V(r_{0})}^{V(r)}dV$$

3. Oct 7, 2007

### mlazos

ok the first way is what a I saw, exactly the way I wrote it. With your correction makes sense. Thank you very much for your answer.