0
$\begingroup$

enter image description here

Now, to calculate the gravitational potential due to a ring(or any object for that matter) at a distance $r$ we consider a tiny mass $dm$ on the ring, and calculate the potential $dV$ due to this element giving $dV = Gdm/y$ and then "sum up the $dV$'s" (integrate) $=> V(y) = GM/y$. But really, $dV$ is the infinitesimal change in potential for an infinitesimal change in $y$, and $dV = V(y+dy)-V(y)$ doesn't seem to have any physical significance and doesn't seem to mean the potential due to $dm$. Then, how did we simply treat $dV$ as the potential due the element $dm$ and "sum these up" using integration?

I've always thought of integration as simply a way of solving a differential equation: for an infinitesimal change in the function $df(x)$ in the interval $dx$, $df(x) = f(x+dx)-f(x)$ to which we are supposed to assign a physical meaning and then write the differential equation & solve it. Why can we treat integration as summation of tiny $dF$'s or $dx's$

$\endgroup$
1
1
$\begingroup$

It is not the case that $dV = V(y+dy) - V(y)$.

The problem with your reasoning is that $dV(y)$ isn't the change in potential due to small element of $y$, but rather $dV(y)$ is a small element of potential resulting from the small element of mass $dm$ located at position $y$.

Therefore by summing the $dV$ up, you are adding together the small bit of potential produced by each bit of mass $dm$ that makes up the ring.

$\endgroup$
7
  • $\begingroup$ @Pulzz that's incorrect, by definition dV = Gdm/y. In this context dV is not V(y+dy) - V(y). There is no "dy" in this case at all as y is constant. $\endgroup$ – Kenshin Feb 24 '17 at 9:26
  • $\begingroup$ I've always thought of integration as simply a way of solving a differential equation: for an infinitesimal change in the function $df(x)$ in the interval $dx$, $df(x) = f(x+dx)-f(x)$ to which we are supposed to assign a physical meaning and then write the differential equation & solve it. Why can we treat integration as the summation of tiny $df$'s or $dm's$ or $dV's$ $\endgroup$ – xasthor Feb 24 '17 at 9:54
  • $\begingroup$ @xasthor simply the way you've thought of integration isn't complete. Integration is defined as summing up small elements. In this case it is small elements of potential dV (due to the small elements of mass). When you are doing area under a curve, you are summing small rectangles of width dx and height f(x), so you are summing small areas of area f(x)dx. Integration is defined as summing many small entitites. $\endgroup$ – Kenshin Feb 24 '17 at 10:00
  • $\begingroup$ what kind of a graph would $dV = Gdm/y$ correspond to, and why would the area under that curve give me the total potential due to the ring? $\endgroup$ – xasthor Feb 24 '17 at 10:07
  • $\begingroup$ @xasthor I wouldn't recommend thinking about this situation as a graph. The graph comment was just another example of how integration works as a sum of small elements (rectangles). $\endgroup$ – Kenshin Feb 24 '17 at 10:11
0
$\begingroup$

You have written $ dV = \frac{Gdm}{y}$. So this tells you what dV actually is - It is potential due to small element of mass dm. We write it dV just because, its is not due to whole mass of ring but due to small part of it. We need to integrate this over whole mass to get V. In this case dV is not $$ V(y+dy) - V(y) $$ because y is not the variable with which we are dealing. In fact you must assume y to be constant while doing integration.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.