# Tag Info

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

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 ...

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 ...

0

In your last equation $U$ is a function or $n$ variables. Which of these does your $x$, $y$, and $z$ in that equation represent? To find the contribution to the potential energy due to the action of forces on a particular particle, one has to take the partial of the potential with respect to the variables representing the position of that particle. I ...

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.

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Your question doesn't have an answer because it isn't possible to split energy into a potential part and a kinetic part, or at least not in an observer independant way. The Schwarzschild metric is time independant and spherically symmetric, and these symmetries mean there are two conserved quantities that you can think of as total energy, $E$, and angular ...

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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The purpose of all this is to calculate and derive potential equation of the gravitation field. Let us assume that we have symmetrical sphere object that "generates" gravitational field. We want to know if the field depends on the objects "homogeneity" (correct me if this is not quite the right term). So from Gauss theorem (it has different names, but they ...

0

Your presentation is a little confused. I would read your book again very carefully, taking care to understand the meaning of each variable. The integral that you write for potential energy is the definition of potential energy, if $F$ is taken to be an internal force, that is, a force between two objects within the system. For example for free fall at ...

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

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 ...

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 ...

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 ...

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 ...

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: $\frac{dE}{dt}=\frac{d(\frac{1}{2}mv^2+V(x))}{dt}=mv\dot{v}+\frac{dV}{dx}\frac{dx}{dt}$ $=v(ma + \frac{dV}{dx})$

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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. ...

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Whenever there is a battery connected in a circuit we assume that the positive terminal of the battery is at a higher potential then the negative terminal by convention. Also by convention + charge flows from positive to negative terminal .The driving force here is the potential difference. Correct Electrons flow in the opposite direction. ...

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