Potential energy always seemed weird to me. Like it's not a real type of energy, just something made up so that the Law of Conservation of Energy stays true. So I want to know if the existence of potential energy can be proven without the use of conservation of energy. And if that is the only way to prove it, then how can conservation of energy be proven without potential energy? Basically I'm looking for how this circular argument, which is the only I've ever gotten, is justified.
[Physics] How to one prove the existence of potential energy
energyenergy-conservationpotential energy
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The answer is ultimately no, but this is a reasonable idea, although old. This idea was floating around in the late 19th century, that the mass of the electron is due to the energy in the field around the electron.
The concept of potential energy is refined in field theories to field energy. The fields have energy, and this energy is identified with the potential energy of a mechanical system, so that if you lift a brick up, the potential energy of the brick is contained in the gravitational field of the brick and the Earth together.
This is important, because unlike kinetic energy, it is difficult to say where the potential energy is. If you lift a brick, is the potential energy in the brick? In the Earth? In Newton's mechanics, the question is meaningless both because things go instantaneously to different places, and also because energy is a global quantity with no way to measure the location. But in relativistic physics, the energy gravitates, and the gravitational field produced by energy requires that you know where this energy is located.
The upshot of all this is that potential energy is field energy, and you are asking if all the mass-energy of a particle is due to the fields around it.
This model has a problem if you think of it purely electromagnetically. Using a model where the electron is a ball of charge, and all the mass is electromagnetic field, you would derive, along with Poincare, Abraham and others that the total mass is equal to 4/3 of the E/c^2. The reason you don't get the right relativistic relation is because of the stresses you need to hold a ball of charge from exploding. The correct relation really needs relativity, and then you can't determine if the mass is all field.
The process of renormalization in quantum field theory tells you that part of the mass of the electron is due to the mass of the field it carries, but there are two regimes now. There is a long-distance regime, much longer than the Compton wavelength of the electron, where you get a contribution to the mass from the electric field which blows up as the reciprocal of the electron radius, and then there is the region inside the Compton wavelength, where you get the QED mass correction from electrons fluctuating into positrons, which softens the blowup to a log. The compton wavelength of the electron is 137 times bigger than the classical electron radius, so even with a Planck scale cutoff, not all the mass of the electron is field, because the blow-up in field energy is so slow at high energy.
So in quantum field theory, the answer is no--- the field energy is not the entire mass of the particle. But in another sense it is yes, because if you include the electron field too, then the total mass of the electron is the mass in the electron field plus the electromagnetic field.
Within string theory, you can formulate the question differently--- is there a measure of a field at infinity which will tell you the mass of the particle? In this case, it is the gravitational field, so that the far-away gravitational field tells you the mass.
But you probably want to know--- is the mass due to the combination of gravitational and electromagnetic field together? In this sense, since this is a classical question, it is best to think in classical GR.
If you have a charged black hole, there is a contribution to the mass of the black hole from the field outside, and a contribution from the black hole itself. As you increase the charge of the black hole, there comes a point where the charge is equal to the mass, where the entire energy of the system is due to the external fields (gravitational and electromagnetic together), and the black hole horizon becomes extremal. The extremal limit of black holes can be thought of as a realization of this idea, that all the mass is due to the fields.
Within string theory, the objects made out of strings and branes are extremal black holes in the classical limit. So within string theory, although it is highly quantum, you can say the idea that all the mass-energy is field energy is realized. This is not very great in giving you what the mass should be, because in the cases of interest, you are finding particles which are massless, so that all their energy is the energy in infinitely boosted fields. But you can take comfort in the fact that this is just a quantum regime of a system where the macroscopic classical limit of the particles are classical gravitational systems where your idea is correct.
In relativity we only use the rest mass, also known as the invariant mass, of an object. In days past the concept of a relativistic mass was used, but this is now strongly deprecated as it has caused endless confusion. For example an obvious question is whether the increase of relativistic mass with speed can cause an object to become a black hole (tl;dr it cannot).
The rest mass does not include contributions from kinetic energy or potential energy, but it does include contributions from changes in internal energy. In the case of your spring suppose it start off with some mass $m_0$ corresponding to an energy $E=m_0c^2$. If we compress the spring we are doing work on it - the work done is $W = \tfrac{1}{2}kx^2$ where $k$ is the spring constant and $x$ is the distance we've compressed the spring. The internal energy of the spring has increased by $\tfrac{1}{2}kx^2$ and that means its mass has increased by $W = \tfrac{1}{2}kx^2/c^2$.
Anything that changes the internal energy of an object, e.g. heating and cooling it as well as doing work on it, will change the rest mass, though usually by too small an amount to be detectable.
The rest mass is useful because it is a Lorentz scalar, that is all observers will agree on its value. However kinetic energy is not an invariant because obviously it depends on the relative speed of the observer and the object. Potential energy is ill defined because it has a global gauge symmetry i.e. we can choose any value we want to use as the zero for our potential energy.
Best Answer
Let $\mathbf{F}(x,y,z)$ be a vector field: if a function $V(x,y,z)$ exists such that the above vector field can be expressed as $\mathbf{F}(x,y,z) = -\textrm{grad} V(x,y,z)$ in any point of its domain, then the function $V(x,y,z)$ is defined as the potential energy associated to the vector field $\mathbf{F}(x,y,z)$, that in turn is said to be conservative.
As you can see from the above definition, conservation of energy is by no means invoked in the definition of potential energy.
Whenever a field is conservative in a simply connected domain, it can be shown that the work done by the field along any path does not depend on the form of the path, rather it only depends on its extrema. As a consequence, the work done by a conservative field on a closed curve vanishes.
Let us now consider a point particle subject to external forces, obeying the Newton's equations; let moreover $T(v_x,v_y,v_z) = \frac{1}{2}m |v|^2$, a function of the velocity of the particle that we refer to as kinetic energy. It can be shown that the work done by a point particle along any path can be written as the difference of the above function in the extrema of the path, namely $$ W_{\gamma}(A\to B) = T(x=B) - T(x =A). $$ However, in general, the work can be always decomposed as the sum of the work performed by the conservative forces plus the work performed by the non conservative forces: $W = W_{\textrm{cons}} + W_{\textrm{non-cons}}$; plugging the above in one obtains: $$ T(B)- T(A) = W_{\gamma}(A\to B) = W_{\textrm{cons}} + W_{\textrm{non-cons}} = V(A)-V(B) + W_{\textrm{non-cons}} $$ where we have used the fact that, if any non-conservative force exists, then it has to be generated, by definition, by its own potential energy. The above becomes: $$ W_{\textrm{non-cons}} = (T(B) + V(B)) - (T(A) + V(A)) = E(B) - E(A) $$ where we have defined the total energy of the particle in any point $(x,y,z)$ as the sum of its kinetic term plus the potential energy of the field calculated in that point. As such, one obtains that the work done by the non-conservative forces equals the difference of the total energy calculated in the extrema of the path. If no non-conservative forces come into play, then the left hand side vanishes and so does the right hand side; we therefore say that in those cases the energy is conserved, as its value in $A$ must equal its value in $B$, $A, B$ being *two any points$.
There is no such thing as "real energy". Energy is defined as a function of the coordinates $(x,y,z)$ as $T(v_x,v_y, v_z) + V(x,y,z)$ and it is, thus, simply a function and no more. It can be related, though, to the work done by the non-conservative forces, as seen above.