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I know what "Potential Energy" is: A function like $U(x)$ whose negative gradient is equal to the force $F(x)$ generating it: $$F(x)=-\nabla U(x).\tag{1}$$

But the definition of the "Potential Function" itself seems to depend totally on the type of the field.

For example:

  1. Electric Potential(Electric Potential Energy per unit charge): $$V_{e}=\frac{U_{e}}{q}.\tag{2}$$

  2. Gravitational Potential(Gravitational Potential Energy per unit mass): $$V_{g}=\frac{U_{g}}{m}.\tag{3}$$

As you see every Potential is defined by the corresponding Potential Energy.

My question is:

1. Is there a way to define Potential of a field independent of the "Nature of the Field"? Whether being gravitational, electric, etc? (Again I'm not talking about "Potential Energy" just "Potential".)

2. Is it possible to derive specific "Potentials" like $V_{e}$ and $V_{g}$ as special cases of the first quantity?

Qmechanic
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Hamed.Begloo
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    What is wrong with $\vec E(x)=-\nabla V(x)$ where $\vec E$ is the field strength? – Farcher Nov 03 '16 at 18:57
  • @Farcher I didn't knew this definition exists. But I want a definition which potential of certain fields(e.g. electric) could be extracted from it as special cases of the primary definition. – Hamed.Begloo Nov 03 '16 at 19:02
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    Isn't this what you want? I could have written is as $\vec g(x)=-\nabla V(x)$ – Farcher Nov 03 '16 at 19:08
  • @Farcher So that means potential is a quantity which its dimension/unit is variable. Since gravitational potential has different dimension than electric potential. Is it right? Now another question which quantity is more fundamental/primitive? "Field strength" or "Potential"? – Hamed.Begloo Nov 03 '16 at 19:16
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    In Physics until relatively recently it was thought to be force (field strength) but now it is thought to be potential. – Farcher Nov 03 '16 at 19:25
  • @Qmechanic Because it defines "Potential energy" not "Potential function". Is there any general definition of "Potential function" whether it caused by any arbitrary field? – Hamed.Begloo Nov 26 '16 at 12:00
  • @Farcher That's a very bold claim. Potentials are not physical. Only gauge-invariant combinations of the potentials ever show up in any measurable quantity--in fact, we demand so. So, I think the popular position is that potentials are useful tools to make our theories manifestly Lorentz invariant and local but they are not physical. Correct me if I am wrong. –  Nov 19 '19 at 21:27
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    @DvijMankad In part my statement was based on this post What is the more fundamental quantity? The electromagnetic field F or the potential A ? and the links therein.but I am on shifting sands here. – Farcher Nov 19 '19 at 23:11
  • @Farcher Thanks for the interesting collection of links. It seems that the heart of the issue is that while only gauge-invariant quantities are observable, it's very difficult (impossible?) to formulate a local theory without relying on potentials--making it hard to say if potentials are really unphysical or not. –  Nov 20 '19 at 00:34

1 Answers1

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Yes. We can define the (scalar) potential of a vector field as the scalar function whose gradient equals that vector field (or its negative, if we so choose). The "nature" of the vector field does not matter, but note that not all vector fields possess a potential. In general we must have a continuosuly differentiable vector field $\vec{V}$ such that $\vec{\nabla}\times \vec{V}=0$. See this Wikipedia article.

Pirx
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  • Thank you for your answer. It seems in physics, that Vector field is called "Field strength ". Then How we define Field strength? And also after finding the definition of "Scalar potential" and "Field strength", what determines its nature? – Hamed.Begloo Nov 26 '16 at 13:18
  • We need to demonstrate that the quantity in question can be described by a physical vector (or "polar vector") at each point in space. If that is the case then the magnitude and direction of such a quantity can be measured and consequently used to determine the field strength in question. The term "nature" as applied to properties or characteristics of physical quantities is undefined. – Pirx Nov 26 '16 at 13:27
  • So does it mean I can study time evolution of a system with having its potential and without knowing nature of the field? Also if I reached a general formulation of the system's evolution and then I wanted to assign this formulation to effects of some special fields and reduce it to a field-theoretic viewpoint what should I do? – Hamed.Begloo Nov 26 '16 at 13:40