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6

Potential and potential energy are defined for pairs of objects, not individual objects. It's meaningless to say "the potential energy of A". One must say "the potential energy of the system consisting of A and B". There is only one potential and potential energy in your problem. Perhaps the confusion comes from the way potential is introduced in ...


4

Extremum of potential vs. extremum of action Yes, the action principle is in a special case equivalent to the principle of extremum of potential energy (the maximum of a potential also presents an equilibrium, even though it is an unstable one!). Consider the action principle of a point particle in a potential $V(\vec{x})$, then the Hamilton's action ...


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I think your problem is in how you understand the gradient. Define your z-axis. Lets say it points up. Now $$V(z) = mgz$$. Thus gradient of V(z), that is $$\nabla V(z) = mg \hat z$$. Here, $$\hat z$$ is an unit vector pointing up. So the force will be a vector of length mg, pointing down, because of the minus sign.


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Your question What is the potential energy of a black hole? doesn't make sense because energy is a somewhat tricky concept to deal with in GR. If we treat the black hole as fixed we can study the motion of a test particle falling into it, and we find that there is a quantity analogous to total energy that is constant as the particle falls in. So in ...


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The reason is mainly in order to be able to write the total energy of the system as $$ E=T+V $$ where $T$ is the kinetic energy of the system. It is far more useful to choose the signs of the terms in the total energy to be $+1$ once and for all, rather than setting the force equal to plus the gradient of the potential. The link between the conventions is ...


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Remember that potential energy is defined up to a constant. Here it's defined so that when $r \to \infty$ the potential energy is worth 0, for convenience. So that there's nothing special about a negative potential energy per se. What really matters here is that there's a "well", i.e. a region for $r$ where the potential energy has a minimum and that ...


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There is no conflict here. Let the two charged particles ($M,Q$) be the system with no external forces acting. Momentum is conserved and so for all time $M_BV = M_A V_{Af} + M_BV_{Bf}$ Energy is also conserved and so for all time $\frac 12 M_B v_B^2 + \dfrac{kQ^2}{R_i} = \frac 12 M_B v_{Bf}^2 + \frac 12 M_A v_{Af}^2 + \dfrac{kQ^2}{R_f}$ The ...


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You have got your forces and potentials mixed up. If the force is given by: $$ F(r) = \frac{GMm}{r^n} $$ then the potential is the integral of this: $$ V(r) = -\frac{GMm}{(n-1)r^{n-1}} $$ In the question you link the force has an inverse cubic dependance, $n=3$, while the gravitational force has an inverse square dependance, $n=2$. We get a circular ...


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Black holes are in the realm of General Relativity. In GR even the law of conservation of energy is under question when approaching singularities of the GR solution. Potential energy is a concept that comes with conservation of energy. Where the singularity in the black hole solutions is dominating, one cannot talk in terms of energy conservation and ...


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I think what is missing is the definitions of $F$ and $V$. Consider first the definition of potential energy. The potential energy at a point relative to another point is the work done by a external force (eg the force exerted by you on the mass, $\vec F_{my}$) in taking the mass from the first point to the second point. That force which you exert on the ...


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I know this answer is pretty late, but I'm hoping it'll help at least a little. The idea here is, of course, using Green's Theorem: $$\oint \mathbf F \cdot \mathrm{d}\mathbf r = \int_A (\mathbf \nabla \times \mathbf F ) \; \mathrm{d}\mathbf a $$ Here, you're asking about the line integral around a closed path. When using this equation, you end up making ...



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