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It is said that if the space is homogeneous then momentum is conserved. But I've been thinking in the following situation:

Consider a parallel plates capacitor. In between the plates there is a uniform electric field so that the space is homogeneous. There is no point in between the plates that differs from any other. However an electric charged particle would feel an electric force and its momentum would change.

I know that the potential energy (PE) is not uniform, but at same time I understand PE as just a bookkeeping device (In the sense of section 4-1 of Feynman's lectures of physics, vol 1). Is this true or it has an underlying reality (in the sense PE existence cannot be avoided).

Edit: Another way to put it. Suppose we never used the concept of energy in physics. We only use forces, momentum or anything we can measure directly. How could we say the space in between the plates is not homogeneous so that momentum does not conserve?

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  • $\begingroup$ Can you clarify your question? What do you want to know? $\endgroup$
    – anon01
    Commented Jun 17, 2016 at 13:25
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    $\begingroup$ What do you mean by "just a bookkeeping device"? Is the whole energy a bookkeeping device? What about momentum, position, resistance, or any other concept in physics? Physics doesn't define the notion of a "bookkeeping device" in general and if physicists use it, it's in rather different situations than yours. I understand that the phrase "just a bookkeeping device" is used to dismiss the importance of the concept for reasons that seem utterly incomprehensible to me. It's like you don't want to use the concept at all. That's not a good starting point for understanding its implications. $\endgroup$ Commented Jun 17, 2016 at 13:25
  • $\begingroup$ @LubošMotl By bookkeeping device I mean something that it not strictly necessary but is used to keep things simples. For example: virtual particles. They do not exist but they are usefull. As well as energy. What we measure are forces, but is usefull to think about the abstract concept of energy. Sorry if I am not being clear. $\endgroup$
    – user115652
    Commented Jun 17, 2016 at 13:33
  • $\begingroup$ @user115652 - every concept in physics may be considered a bookkeeping device by your definition because every concept in physics may be omitted and its implications may be formulated using other concepts. In particular, the force is a bookkeeping device, too. One may only talk about the acceleration and positions etc. It's simply not true that the energy or potential energy is in any physical sense "less real" than the force. $\endgroup$ Commented Jun 17, 2016 at 13:36
  • $\begingroup$ Incidentally, just 2 weeks ago, I was answering a question saying that energy is more fundamental than force: physics.stackexchange.com/questions/258757/… - At any rate, for the uniform field, the potential energy is a linear function of the position, and that may also be said to be the reason why the acceleration will be nonzero. It's completely counterproductive to start to convince oneself that there's something wrong with this reasoning - it's exactly the kind of reasoning that shows the power of the concept $\endgroup$ Commented Jun 17, 2016 at 13:38

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I am answering the question formulated after the "edit" in a newer version of the text because that one seems well-defined.

Indeed, a situation with a uniform field $\vec E$ may be said to be "uniform" or translationally invariant in space. Noether's theorem says that this "uniformity" (spatial translational invariance) implies the existence of a conserved quantity, usually named momentum.

How does it agree with the fact that in this uniform field, the momentum of the (charged) particle is not conserved? Well, it's a different momentum.

The conserved momentum resulting from Noether's theorem will not be $m\vec v$ as the normal one but rather $m\vec v-qt \vec E$ where $q$ is the charge and $t$ is time. The time derivative of this quantity is zero which will simply say that $\vec a$ is whatever constant is dictated by the field.

I discussed the electric field but it is a more general result.

In terms of potentials, one may obtain the electric field $\vec E$ from a space-independent potential $\Phi$ (the freedom to make this true may perhaps be described as the potential's being a bookkeeping device – but what we really need is the gauge invariance) as long as the vector potential $\vec A$ is time-dependent (linear in time). The term $qt\vec E$ is nothing else than $q\vec A$ (up to a sign I am not sure about). Quite generally, in electromagnetism, one distinguishes the kinetic and canonical momentum that differ by the $q\vec A$ term. This is also important for the quantum mechanical description of electromagnetic and especially magnetic fields.

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  • $\begingroup$ That was exactly what I was concerned about and now things start to make sense to me. The only thing I did not get is this "new" momentum but I am going to search about it. Thanks! $\endgroup$
    – user115652
    Commented Jun 17, 2016 at 13:58
  • $\begingroup$ I fixed the terminology, they're known as kinetic and canonical momentum. $\endgroup$ Commented Jun 17, 2016 at 13:59
  • $\begingroup$ @LubošMotl I like your answer but the equivalence of $\overrightarrow{E}t$ with $\overrightarrow{A}$ may not apply when the fields are static, I think at that time one needs to use scalar potential. why the vector potential should linearly increase or decrease in time for static field, I can understand the variation in space as charges are present somewhere. $\endgroup$
    – hsinghal
    Commented Jun 17, 2016 at 16:37
  • $\begingroup$ Dear @hsinghal - "the fields" $\vec B,\vec E$ are static for $\vec A = \vec E t$. If the electric field $\vec E(t)$ were time-dependent but uniform in space, one could get the vector potential $\vec A$ by a simple integration of $\vec E(t)$ over time while the scalar potential may still be chosen zero. There would still exist a conserved momentum. $\endgroup$ Commented Jun 17, 2016 at 16:50