1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Why cant I get conservation of energy here

  1. Sep 6, 2012 #1
    Okay consider this system one dimensional system:

    We have a positive charge at the origin. Far out at infinity on the x-axis we have a positive charge and at minus infinity on this minus x-axis we have a negative charge. For simplicity we will assume that the plus- and minus charge do not affect each other with their fields - maybe this assumption is what gets it wrong for me. Also we will assume that the positive charge at the origin is stationary throughout the whole time.

    We define the potential energy of either charge relative to infinity. Both sitting at plus and minus infinity the sum of their potential energies is initially zero. We now move the charges simultaneously towards the origin and stop the negative one at a point -a and the positive one at the point a. Let us now assume that the energy gained by the negative charge has been used to do positive work on the positive charge during this transport such that no energy has been put into the system.
    Now the sum of their potential energy must still be zero.
    Now if we let go of holding the charges at point a and -a one will be shot off towards infinity and one drawn towards the origin. How does this match with the conclusion that their total energy is zero - in both cases they are gaining kinetic energy.
    What am I doing wrong?
     
  2. jcsd
  3. Sep 6, 2012 #2

    Doc Al

    User Avatar

    Staff: Mentor

    I would say that was an incorrect assumption. Each pair of charges will have a potential energy.
     
  4. Sep 6, 2012 #3

    jtbell

    User Avatar

    Staff: Mentor

    No. The total potential energy of a three-charge system is the sum of the potential energies of each pair, taken separately:

    $$U = k \frac{Q_1 Q_2}{d_{12}} + k \frac{Q_2 Q_3}{d_{23}} + k \frac{Q_1 Q_3}{d_{13}}$$

    You're not taking into account that the attraction of the "left" and "right" charges to each other contributes to the total potential energy.
     
  5. Sep 6, 2012 #4
    Total kinetic energy goes to infinity, and so does total potential energy. At any finite time total energy is zero.
     
  6. Sep 6, 2012 #5
    Voko: The total kinetic energy is going to infinity. But how is the total potential energy going to minus infinity?

    If it's wrong to neglect the mututal attraction of our two charges then imagine that there is an incredible amount of positive charge at the origin. The surely my assumption should hold to an approximation.
     
  7. Sep 6, 2012 #6

    mfb

    User Avatar
    2017 Award

    Staff: Mentor

    It is negative. The force on the negative charge comes from both positive charges, while the repulsive force on the positive charge is partially cancelled by the force from the negative chage.


    Your potential energy will become extremely negative from the negative and positive charge approaching each other, while the potential energy contribution from the other positive charge gets 0.
    This is independent from the error above.
     
  8. Sep 6, 2012 #7

    Doc Al

    User Avatar

    Staff: Mentor

    The PE goes down and the KE goes up. What's the problem?
     
  9. Sep 6, 2012 #8
    You are right. I don't know what I was thinking. I have one of those moments, where you feel you have thought about something completely the wrong way for a long time without knowing why.
     
  10. Sep 7, 2012 #9
    okay now I know what bothered me and it sort of still does even though the above example really didn't catch the problem.
    Suppose we have a positive and a minus particle lying on top each other so their fields cancel completely (you can view this as effectively nothing though it is important that the particles still exist and no energy is needed to create them).

    Beside them we also have an identical pair of charges lying on top of each other. We can probably all agree that the potential energy of this system is zero and so is the kinetic energy.
    Now suppose we separate one of the pair of charges slowly from each other.
    As this happens the other pair of charges will suddenly have a potential energy associated with the separated pair of charges and will move accordingly. Where did this potential energy come from? Surely we did work in separating the one pair of charges but that is stored in their mutual potential energy.
    I am perhaps thinking of this the completely wrong way. I feel that my question is a bit like asking: "If a rock is lying on top of a cliff still it has no mechanical energy. Yet it can fall down further if it goes off the edge. Where did this energy come from?"
    What goes wrong in my way of thinking about potential energy?
     
  11. Sep 7, 2012 #10

    mfb

    User Avatar
    2017 Award

    Staff: Mentor

    Both have a potential energy, but the sum of those is still 0. As soon as the charges begin to move, they will have different potential energies (with a sum below 0, as both go "down" in potential) and have some kinetic energy (corresponding to the potential energy they lost due to their movement).
     
  12. Sep 7, 2012 #11
    Charges lying on top of each other are a physical impossibility. Their mutual potentials are infinite. No external field, short of being infinite as well, will affect them in any way. To separate such charges, infinite work is required. Infinity minus infinity may result in any finite potential and kinetic energies.
     
  13. Sep 7, 2012 #12

    Dale

    Staff: Mentor

    This is not necessarily true. If the other pair of charges remains exactly on top of each other then they will not have any potential energy. They only have a potential energy if they begin to separate, in which case any work done separating them reduces the overall system's potential energy by partially canceling the dipole field of the other pair.
     
  14. Sep 7, 2012 #13
    hmm yes again you are right. I don't know what bothers me.

    I think it is the fact, that we go from a situation, where the charges cannot move and thus can't gain kinetic energy to a situation where they are able to gain kinetic energy. That just despices my logic. It is like a rock falling down being able to go back up into a high potential and resume its fall.
     
  15. Sep 7, 2012 #14
    This is because you are thinking of an impossible situation. The mathematics of the theory stops making physical sense when charges are very close.
     
  16. Sep 7, 2012 #15
    I did not express myself properly. It is not the mathematics that stops making sense. It is the assumption that one can place charges very closely to one another, and subsequent use of the mathematics under this assumption.

    Mathematics may be compared to a mill of exquisite workmanship, which grinds your stuff to any degree of fineness; but, nevertheless, what you get out depends on what you put in; and as the grandest mill in the world will not extract wheat flour from peascods, so pages of formulae will not get a definite result out of loose data -- Thomas Huxley
     
  17. Sep 7, 2012 #16

    Ken G

    User Avatar
    Gold Member

    Yet I think the question merits more of a physical answer than that, because we do have real situations where, in effect, two charges do completely overlap in some sense. That happens when electrons and positrons annihilate, or when a proton and an electron give rise to a neutron. In such situations, the energy in the fields never goes infinite as the charges get close, and the energy in those fields must be accounted for. It's true that the mathematical simplification that assumes the fields get infinitely large along the way is not physically correct, but the other question remains-- you started out with potential energy in the fields, and you end up with no such energy, so what happened to it demands accounting.

    If we are to give a physical accounting, we must first choose our theory. Hence the OP question can even serve as an entry point into advanced theories like quantum mechanics or relativity or even quantum field theory. If we take the perspective of classical physics, we can still address the issue of what happens to the potential energy of two charges as they approach each other. A standard trick there is to associate with each particle a "classical radius" to get a finite and appropriate field energy. If we do that, we can get a finite potential energy at every stage of the process as the two particles approach each other. The result won't look like the mutual potential energy of two point charges, so it's true that that particular mathematical idealization would break down, but we can endeavor to provide an idealization that would not break down, like the classical radius picture. Then the field energy drops as the charges approach, so the potential energy also drops, but it does not drop to negative infinity, so we would not expect the charges to acquire infinite kinetic energy as they approach.

    Note this brings up a second question, which is that if the charges acquire equal and opposite momentum as they approach, then their total momentum is zero, consistent with a final state where both particles are stationary and overlapping. But that final state does not account for all the kinetic energy they must acquire-- so the OP question also tells us that we have a lot of kinetic energy we must account for. This is an entry point into the issue that two particles must annihilate into two other particles-- it would not conserve energy for them to make a single stationary particle with the rest mass of the two particles. So we get the possibility of neutrino emission when a neutron is made, or photon emission when electrons and positrons annihilate.

    That's all classical thinking, one could also take the OP question to those other realms. So I think it's actually a really nice question.
     
  18. Sep 7, 2012 #17
    These are quantum mechanical effects. In quantum mechanics, you always have some uncertainty in position.
     
  19. Sep 7, 2012 #18

    Dale

    Staff: Mentor

    I don't follow your analogy here at all. If you have a rock which falls, and then you do work to dig a pit next to it, then the rock can certainly fall further. What is illogical about that?

    Here, you have four identical spherical charges A+ A- B+ and B-, all co-located at the origin and motionless. If we assign this initial state 0 energy then to slowly move A+ and A- away from the origin in opposite directions requires work. After they have moved, then we can let B+ move to A- and B- to A+, which will increase the KE of B+ and B-. When they reach the opposite A charge the KE of the B charges will be equal to the work done on the A charges minus any energy radiated off.
     
  20. Sep 7, 2012 #19
    hmm yes everything you say I understand, yet there is just something that bothers me.

    The original question originated from the fact that I have a hard time understanding energy in dielectric systems - although this might seem like far from the topic.

    When we have a dielectric it is just a lump of neutral matter. Then when we put a field over it the charges suddenly have a potential energy making them move accordingly. Where did this potential energy come from if we just imagine the field as having been produced by the separation of two charges. The energy needed to separate those two only depends on their mutual potential energy. That is what I for some reason find illogic. That you, in short, start with something completely neutral. Separate two of the neutral pairs of atoms, and get a whole lot of energy out of it.

    Even though my example of two charges lying on top of each other gave rise to a long discussion it was actually just to simplify this example. I was not going for the mathematical problem there is in having two charges lying on top of each other.
     
  21. Sep 7, 2012 #20

    mfb

    User Avatar
    2017 Award

    Staff: Mentor

    You can always calculate the potential energy of the system by summing over all charge pairs (in the continuous case, integrals are used instead of sums). You will see that some summands can be the same in some cases (maybe with opposite sign), but that does not matter at all.
    A potential does not "appear" somewhere in your setups. The potentials are always there.
     
  22. Sep 7, 2012 #21

    Ken G

    User Avatar
    Gold Member

    You get the energy out that you yourself put in. To understand that physically, you must first choose a theory. Here it suffices to imagine a classical theory where the charges are spheres of finite radius (that avoids any problems with infinities). That is what I was alluding to above-- if you have two opposite charges on spheres that overlap, there is no field anyway, no net field energy there. If the charges came from infinity originally, there was energy released when they came together, which must have been dissipated in some form (like the light an atom emits when it forms, say the recombination epoch in the Big Bang for example). But now that you have the two charges on top of each other, there's no field energy (we often call that a negative potential energy, equal to the energy released if they came from infinity, but it doesn't matter-- the zero of the potential energy scale is of no importance, only changes in potential energy).

    Now you apply the external field, and it turns out that the charges do not want to separate, because they attract each other. But the external field makes them separate, so it does work on them. That work goes into the creation of additional fields, owing to the charges that no longer overlap. So there's no problem with potential energy "coming from nothing", the potential energy is a kind of bookkeeping trick to keep track of the work done by the external field to separate the charges. Another bookkeeping trick is to say that energy "resides in the fields" of the separated charges (though this picture seems to be more than bookkeeping, it has physical implications as well). If you assert all these things, the problems you allude to go away, and a consistent picture emerges. Of course, in a different physical theory, like quantum mechanics, the picture changes, so it's always up to you what level of detail you are shooting for.
     
  23. Sep 7, 2012 #22

    Ken G

    User Avatar
    Gold Member

    It is not necessary to invoke quantum mechanics to account for annihilation of charges, or creation of neutrons, if one simply wishes a basic description of the phenomenon (say, at a level that can conserve energy and momentum). One does have to relax the view that electrons are point charges, for example, so a classical picture can never give a complete description. That's nothing new in physics, we often allow a less than complete description, it simply depends on our choices. This point is something that I think a lot of questioners miss-- they want "the answer" to their question, but physics doesn't work that way. Even so, we can find a mathematical idealization that satisfies the questioner-- we don't need to say that mathematical idealizations cannot answer their question.
     
  24. Sep 7, 2012 #23
    That is my point exactly.The classical theory does not work properly with infinitely close interacting point charges. That has been known for over a century. And quantum mechanics effectively disallows point charges.
     
  25. Sep 7, 2012 #24

    Dale

    Staff: Mentor

    Perhaps the problem is that you don't understand the relationship between work and energy. In order to separate the pairs requires work, and the maximum amount of energy you can get out of it is equal to the work that you put into it. It is only a "whole lot of energy out" if you first put a whole lot of work in.
     
  26. Sep 7, 2012 #25
    well the only work done is that separating our two charges creating the field. That work is only associated with their mutual potential energy.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook