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We introduce a minus sign to equate the mathematical concept of a potential with the physical concept of potential energy. Take the gravitational field, for example, which we approximate as being constant near the surface of Earth. The force field can then be described by $\vec{F}(x,y,z)=-mg\hat{e_z}$, taking the up/down direction to be the $z$ direction. ...


3

If the resultant force acting in a body is give by minus the gradient of potencial you can show that $\frac{dE}{dt} = 0$. Where E is the total energy of particle. So total energy, kinect + potencial is conserved. In 1-dimensional case: $\frac{dE}{dt}=\frac{d(\frac{1}{2}mv^2+V(x))}{dt}=mv\dot{v}+\frac{dV}{dx}\frac{dx}{dt}$ $=v(ma + \frac{dV}{dx})$


3

The equation is just the kinetic and rest energy, it does not include potential energy. But potential energy in relativity is not the proper concept. The linked question has some useful answers, but I think your true question is about how to learn to do things relativistically that you used to do non relativistically. And since the other answer so far takes ...


3

I am not sure what is the path $C$ you are integrating over? In your definition you evaluate $U(C)$ which in the present case of force is independent on the explicit path you choose but still depends on initial and final point, i.e. $U(p_1,p_2)$. In your final result it seems you are actually 'walking' three times the path $p_1=(-\infty,y,z)$ to ...


2

In classical mechanics there is no distinction between free and bound as far as this relation is concerned. In relativistic quantum mechanics (i.e. QFT), a particle that satisfies this relations is said to be "on-shell" or a physical observable asymptotically free particle. It is certainly not satisfied for virtual particles, but they are as their name ...


2

The black machine is a weight lifting machine. It is self contained with no power source. If it can lift an external weight and return to its original state as shown below, it is a perpetual motion machine. Suppose the blue weight is water. We could add a water wheel and generator on the right. You start at the top and work your way to the bottom. Then you ...


2

It's not taking partial derivatives with respect to an observed particle's position, but rather the space of all possible positions of that particle. Think of the potential energy as being defined prior to the particle having an actual path. Really, at heart, these things are defined on a phase space not on ordinary physical space.


1

I agree with everything John Rennie said but let me just take a slightly different direction. Note that Schwarzschild space-time has a time-like Killing field $\xi^{\mu}$. In a stationary space-time such as this, one can define the Newtonian analog of gravitational potential by $\phi \equiv \frac{1}{2}\log(-\xi_{\mu}\xi^{\mu})$. One can then easily show ...


1

In short, while $q_2$ does exert a force against $q_1$, this force does not perform any work because $q_1$ does not move. The work performed is the product of the force exerted times the distance the particle moves against (or with) it; since $q_1$ does not move there is no work performed on it. The potential energy is defined to be the work required to get ...


1

here is what I think he is trying to explain. You have two machines A and B. A is reversible B is not necessarily reversible. Both these machines are placed side by side. let both machines A and B be initially horizontal. (Left , right pans of machine A will hold 3 unit and 1 unit mass respectively. Similarly, the left and right pans of machine B will hold ...


1

The whole energy-concept is a reformulation of Newtons laws. Starting from $\vec{F}=m.\vec{a}$, you could wonder about the effect of a force during a displacement. You call the concept 'work' and do the math $$W=\int_{\vec{x}_1}^{\vec{x}_2} \vec{F} d\vec{x} = \ldots = \frac{1}{2}mv_2^2-\frac{1}{2}mv_1^2$$ To save yourself some work you define $$T = ...


1

You are correct, they are different. $k$ is not a property of the material, its a property of the entire object. Imagine having a small amount of a fairly tight spring. It takes a lot of effort to extend it even a centimeter or two. Now without changing the material, connect a few hundred of the springs together. Extending it a centimeter now will take ...


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It's always possible to expand the potential in Taylor series around any local minima (in this example $U(x) $ has local minima at $x_0$ , thus $U'(x_0)=0 $ ) $$ U(x) \approx U(x_0)+\frac{1}{2}U''(x_0)(x-x_0)^2 $$ Setting $ U(x_0)=0 $ and $ x_0=0$ (for simplicity, the result don't depend on this) and equating to familiar simple harmonic oscillator ...



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