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In the Feynman Lectures, vol. II, chapter 4, Feynman discusses electric potential and says:

If we carry a charge from point $a \to b$, $$W = -\int_{a}^{b} \mathbf{F} \cdot ds.$$ Now, in general, what we get with this kind of integral depends on the path we take. But if the integral were depended on the path , we could get work out of the field by carrying the charge to $b$ along the path for which $W$ is smaller & back along the other, getting out more work than we put in. There is nothing impossible, in principle, about getting energy out of a field. It could be that as you move a charge you produce forces on the other part of the "machinery". If the "machinery" moved against the force, it would lose energy. For electrostatics, however, there is no such "machinery". We know what the forces back on the sources of the field are. If the other charges are fixed in position, the back forces can do no work on them. So, there is no way to get energy from them.

I couldn't understand why, by going to $b$ along the path where $W$ is smaller and back along the other, we could extract energy.

And also, what is meant by back forces? And what is machinery?

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    $\begingroup$ Could you provide context? Is this a quote from Feynman (as the title suggests)? $\endgroup$ – pyramids Apr 12 '15 at 14:46
  • $\begingroup$ Feynman explains the idea of (the impossibility of) extracting free energy by going in a closed path, in more detail in volume I, chapter 4 - when he discusses gravitational potential energy, and again in chapter 13, when he discusses work done by gravity. $\endgroup$ – nir Apr 12 '15 at 21:51
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I suppose you read this passage in the famous Feynman Lectures. I am fairly certain that what Feynman is referring to (and what you are looking for) is a proof that an electrostatic field is conservative. There are a number of equivalent ways of stating that a vector field is conservative, each of which can be taken as a definition. Let $\vec{F}(x)$ be a conservative vector field. Then

  1. $\vec{F}$ is irrotational: $\vec\nabla\times\vec F=0$

  2. $\vec{F}$ is the gradient of a scalar potential $\Phi$: $\vec F = \vec\nabla \Phi$. Here, we technically need some assumptions on differentiability and the decay as $\lvert x\rvert\to \infty$, but in physical situations one typically does not worry about those.

  3. Integrating the vector field along a closed curve yields zero: $$\oint \vec F\cdot d\vec s=0$$

1 is equivalent to 2 by Helmholtz' theorem, while 2 and 3 are can easily seen to be equivalent by the fundamental theorem of calculus for line integrals.

It is a well known fact that the electrostatic field $\vec{E}(x)$ is the gradient of the electric potential $V$ and therefore conservative. Therefore, point 3 on our list tells us that $$ W=\oint \vec F\cdot d\vec s=q\oint \vec E\cdot d\vec s=0$$ Physically, we know that $W$ represents (up to a sign) the work extracted by "going around a closed loop". Thus, we see that one cannot return to the initial state while extracting some surplus energy: Everything is always balanced out and we cannot get "free" energy from an electrostatic field.

Finally, let me make some comments on your confusion regarding the words that Feynman uses in the text. The "machinery" Feynman mentions is just some experimental set-up that he is trying to extract energy with in a thought experiment, and is not important for the argument. By the "back forces", he means the changes in the field as one moves a charge around. By saying that "the other charges are fixed in position, [hence] the back forces can do no work on them" he is emphasizing that he is considering an electrostatic setup.

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  • $\begingroup$ Sorry, sir, I couldn't contact due to poor internet connectivity. Feynman told that the back forces are Coulombic but they can't do anything because of the assumption that the field is due to static distribution. Now, how far is this assumption true? $\endgroup$ – user36790 Apr 13 '15 at 18:13
  • $\begingroup$ It can be true that if the test charge is very minuscule, then it wouldn't effect the field. But when a charge of sufficient magnitude to distort the "machinery" ie. the static field source is brought, still then the source would be static? $\endgroup$ – user36790 Apr 13 '15 at 18:17
  • $\begingroup$ @user36790 It's not really an assumption: You should view it as just a part of how Feynman chooses to set up his thought experiment. He just decides to consider a static set-up, because then we have (by what I explained in my answer) an easy way to see that the field is conservative, and no energy can be extracted. Whether this is a realistic situation or not is not of importance in what he's saying. $\endgroup$ – Danu Apr 13 '15 at 18:36
  • $\begingroup$ So, the bigger charge wouldn't effect the field? Is it sufficient to tell that it is more than an assumption?? BTW, can we extract energy from any field(of course not conservative) as admitted by Feynman? $\endgroup$ – user36790 Apr 13 '15 at 19:10
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    $\begingroup$ @user36790 Yes, we can extract energy from some fields (e.g. gravitational, by using what is known as the Penrose process). Like I said, the questions about backreaction and such are not really relevant here. $\endgroup$ – Danu Apr 13 '15 at 19:39

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