- #1
mlazos
- 49
- 0
[SOLVED] I want a second opinion
We know the equation
[tex] F=-\frac{dV}{dr}[/tex]
we want to find the integral from [tex]r_{0}[/tex] to [tex]r[/tex].
I have seen someone doing this
[tex]\int^{r}_{r_{0}}Fdr'=-\int^{r}_{r_{0}}\frac{dV}{dr}dr'[/tex]
I am a mathematician and the way I was doing at the university was
[tex] F=-\frac{dV}{dr}\RightarrowFdr=dV[/tex]
and then I integrate
[tex]\int^{r}_{r_{0}}Fdr=-\int^{}_{r_{0}}dV[/tex]
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.
We know the equation
[tex] F=-\frac{dV}{dr}[/tex]
we want to find the integral from [tex]r_{0}[/tex] to [tex]r[/tex].
I have seen someone doing this
[tex]\int^{r}_{r_{0}}Fdr'=-\int^{r}_{r_{0}}\frac{dV}{dr}dr'[/tex]
I am a mathematician and the way I was doing at the university was
[tex] F=-\frac{dV}{dr}\RightarrowFdr=dV[/tex]
and then I integrate
[tex]\int^{r}_{r_{0}}Fdr=-\int^{}_{r_{0}}dV[/tex]
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.