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Why does a potential go to zero with infinite distance?

  1. Sep 21, 2013 #1
    in physics II, i learned about how the electric potential goes to zero as r goes to infinity. Okay, well, this was when we were dealing with two positive charges. ie a repulsive force. now i'm learning about gravitational potential now and i see this in my notes: [itex]\Phi[/itex][itex]\rightarrow[/itex]0, r[itex]\rightarrow[/itex]∞.

    what?? why? now we're dealing with an attractive force..

    mathematically, i can see that yes, [itex]\Phi[/itex]=-GM/r and so they are inversely proportional. but conceptually, let's start with the two positive charges from physics II. because the force is repulsive, the closer the test charge is to the source charge, the more potential energy it should have. because, if we consider the points radially extending from the source charge, there is an electric field value/strength associated with each point. since the force is repulsive, the closer the test charge is to the source charge, the more electric field it will be exposed to as it tends toward infinity and therefore the more total force it should experience right?

    equivalently, for the case of an attractive force (gravitational case), if the field is [itex]\vec{g}[/itex], the gravitational acceleration, then the farther the test mass is from the source mass, the more total [itex]\vec{g}[/itex] values the test mass should be exposed to as it tends toward the source body (r=0) and therefore the more acceleration the test mass should have by the time it gets to r=0 right? so why in the world is [itex]\Phi[/itex] inversely proportional to r??

    well.. we did the derivation in class.

    [itex]\vec{g}[/itex]=[itex]\nabla[/itex][itex]\Phi[/itex]

    [itex]\vec{g}[/itex]=-[itex]\frac{GM}{|r|^2}[/itex][itex]\hat{r}[/itex]=-[itex]\frac{dPhi}{d|r|}[/itex][itex]\hat{r}[/itex]

    [itex]\frac{dPhi}{d|r|}[/itex]=[itex]\frac{GM}{|r|^2}[/itex]

    [itex]\Phi[/itex]=-[itex]\frac{GM}{|r|}[/itex]

    i just don't understand why r is inversely proportional to phi conceptually.
     
    Last edited: Sep 21, 2013
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  3. Sep 21, 2013 #2

    Nugatory

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    With any ##\frac{1}{r^n}## central force, the potential at infinity goes to some constant value and the potential at zero goes to plus or minus infinity, depending on whether the force is repulsive or attractive. Because it's hard to relate the potential at any intermediate point to the infinity at ##r=0##, it's convenient to define the potential at infinity to be zero, use that as a reference point.
     
    Last edited: Sep 21, 2013
  4. Sep 21, 2013 #3
    so is that why phi at the end is a negative value? because 0 is the highest number it can reach, ie zero IS the infinity that i suggested it should be?
     
  5. Sep 21, 2013 #4

    Nugatory

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    Yes, zero is indeed infinitely greater than the minus infinity you find at r=0.
     
  6. Sep 21, 2013 #5
    but according to the derivation, we don't actually get to do any defining/setting. ie phi is already cut out to be zero at r[itex]\rightarrow[/itex]∞.

    well, actually that still works out. lol everything is still the same. since [itex]\Phi[/itex] [itex]\rightarrow[/itex] ∞ as r[itex]\rightarrow[/itex] ∞, only thing is that we don't really get a say in it. but that's fine, the negative makes the zero work out.

    but still, one more question: potentials are something we always get to have a say in. we can always define one point to have a particular potential. So in this case, what could we have done and where so that phi does go to infinity as r goes to infinity?
     
  7. Sep 21, 2013 #6

    Nugatory

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    We can't. Our freedom to choose the value of the potential is limited to being able to add or subtract an arbitrary constant everywhere to make the value come out to what we want at a particular point. It's like choosing to measure heights relative to sea level, or relative to the floor of the room you're in, or relative to street level outside the building you're in - they're all right, and different choices are convenient for different problems.

    But there's no arbitrary constant that we can add or subtract to turn a non-infinite value into an infinite one, so our freedom doesn't extend that far.
     
  8. Sep 21, 2013 #7
    would this imply that potential energy isn't relative in this case? that there is an absolute potential value at this point and another absolute potential value at that point? this would seem to be the case since the point that we're starting from is SET in stone.
     
  9. Sep 21, 2013 #8

    WannabeNewton

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    Only the difference in potential is absolute. The potential itself is not absolute because of that freedom in the addition of an arbitrary constant.
     
  10. Sep 21, 2013 #9

    Nugatory

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    Infinity is a very difficult thing to set in stone :smile:.
    It's not really a number at all, and when you hear people (including me, earlier in this thread) saying that the potential is infinite at ##r=0## we're playing a bit fast and loose with the math. It would be more accurate to say that the potential is undefined at ##r=0## but it becomes arbitrarily large as ##r## becomes arbitrarily close to zero.

    As long as you avoid the mathematically problematic and not physically realizable ##r=0## point in a central force problem you'll find that the potential always has some finite value, and that you are always free to add an arbitrary constant to the potential at all points.
     
  11. Sep 22, 2013 #10

    Redbelly98

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    No, I wouldn't say it is set in stone. Setting the potential equal to zero at an infinite distance is a convention. Widely used, yes, but not "set in stone". Nugatory gave an excellent example -- taking height relative to sea level is a similar convention, but not set in stone either.

    Potential is relative. There is no absolute potential. Only the potential difference -- the difference in potentials at different locations -- is physically meaningful.
     
  12. Sep 23, 2013 #11
    Actually, the potential going to zero at infinity is just a convention to make the math easier. If you placed your system in a giant metal box (giant enough to be considered out at infinity) and held the box at a potential V, the problem would go to V at infinity and not zero. When you work with electric potentials, you have to remember that only potential differences have physical meaning. But the potential can be set equal to the potential difference if you define the lower value on the difference operation as 0, call it ground, and measure everything relative to ground.
     
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