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- Thread starter Za Kh
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In summary: A=-\frac{\partial}{\partial x} \frac 1 2 a (z^2-y^2) \hat x . But this is not a potential function because the gradient does not exist at the origin.In summary, a potential function is a scalar-valued function of whose gradient is equal to the force field.

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As you may know, the work that such a force does on a mass when its moving from point A to point B through a trajectory [itex] \gamma(t) [/itex], is [itex] W=\ \ _\gamma\int_{A}^{B} \vec F \cdot \vec{dr} [/itex] (Where [itex] \vec{dr} [/itex] is an infinitesimal vector which is always parallel to the curve [itex] \gamma(t) [/itex].) Note that I put [itex] \gamma [/itex] under the integral sign because in general the work that the force does on the particle depends not only on the end points of the trajectory but also on the trajectory itself.

Now there happens to be some force fields for which the work is independent of the trajectory taken by the particle and only depends on the end points. This is stated mathematically by [itex] \oint \vec F \cdot \vec {dr}=0 [/itex] for all closed curves. Now from this equation, its possible to prove that [itex] \vec \nabla \times \vec F=0 [/itex].

Then, from Helmholtz decomposition, we know that for any reasonable vector field [itex] \vec F [/itex], there exist a scalar field [itex] \Phi [/itex] and a vector field [itex] \vec A [/itex] such that [itex] \vec F=-\vec \nabla \Phi+\vec \nabla \times \vec A [/itex]. Now if I get the curl of this, I'll have [itex]\vec \nabla \times \vec F=-\vec \nabla \times \vec \nabla \Phi+\vec \nabla \times \vec \nabla \times \vec A [/itex]. But I know that [itex] \vec \nabla \times \vec F=0 [/itex] so I should have [itex] -\vec \nabla \times \vec \nabla \Phi+\vec \nabla \times \vec \nabla \times \vec A=0 [/itex].

From vector calculus, I know that [itex] \vec \nabla \times \vec \nabla \Phi=0 [/itex] identically so only [itex] \vec \nabla \times \vec \nabla \times \vec A=0 [/itex] remains for which [itex] \vec A=0 [/itex] is a reasonable solution. But this means I have [itex] \vec F=-\vec \nabla \Phi[/itex]. Now the scalar field [itex] \Phi [/itex] is called a scalar potential for [itex] \vec F [/itex].

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Sry , but what does it mean that a force has a potential ?

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When we say the force field [itex] \vec F=F_x \hat x+F_y \hat y+F_z\hat z [/itex] has a potential, we mean that we can find a function like [itex] \Phi [/itex] such that [itex] F_x=-\frac{\partial \Phi}{\partial x} [/itex],[itex] F_y=-\frac{\partial \Phi}{\partial y} [/itex] and [itex] F_z=-\frac{\partial \Phi}{\partial z} [/itex]. It sometimes makes things easier because we can work with a scalar instead of a vector. Also it allows us to associate energy with forces so that we can apply conservation of energy.Za Kh said:Sry , but what does it mean that a force has a potential ?

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Thanks :)

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Za Kh said:Sry , but what does it mean that a force has a potential ?

We don't say that a "force" has a potential. We say that a "force field" has a potential. For example, at a point above the surface of the Earth there is a "gravitational field", which is a "force field", but unless there is a mass at that point being pulled by gravity there is no "force of gravity" at that point. The concept of a field is a mathematical abstraction.

A force field is specified by assigning a vector to every point in space. The magnitude of the vector does not represent a force (e.g. Newtons). It has units of force per unit-of-something (e.g. Newtons per something). As Shyan said, it is often simpler to deal with scalar valued functions than vector valued functions. If you are familiar with multivariable calculus, you know that a scalar valued function can have a gradient, which is a vector valued function. A "potential" function for a force field is a scalar valued function of whose gradient (with respect to position variables) is equal to the force field. It's simpler to solve some problems using the potential function than using the force field itself. Some force fields do not have potential functions.

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thank you :)

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In classical electrodynamics, the answer is provided by gauge invariance but what about Newtonian gravitation?

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$$\vec{F}=-\vec{\nabla} \Phi.$$

For Newtonian gravity that's always the case. For a particle of mass ##M## a mass density ##\rho## implies a gravitational force,

$$\vec{F}=M \vec{g}$$

such that

$$\vec{\nabla \times \vec{g}=0, \quad \vec{\nabla} \cdot \vec{g}=-4 \pi \gamma \rho.$$

This implies that there's a gravitational potential $\phi$ such that

$$\vec{g}=-\vec{\nabla} \phi$$

and thus

$$\Delta \phi=+4 \pi \gamma \rho,$$

from which you get

$$\phi(\vec{x})=-\gamma \int_{\mathbb{R}} \mathrm{d}^3 \vec{x}' \frac{\rho(\vec{x}')}{|\vec{x}-\vec{x}'|}.$$

For a spherically symmetric mass distribution around the origin of total mass ##m## this gives for the region outside of the material Newton's famous law

$$\phi(\vec{x})=-\frac{\gamma m}{|\vec{x}|}$$

and

$$\vec{g}=-\vec{\nabla} \phi(\vec{x})=-m \gamma \frac{\vec{x}}{|\vec{x}|^3}.$$

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vanhees71 said:[itex]\vec \nabla \times \vec{g}=0 \quad , \quad \vec{\nabla} \cdot \vec{g}=-4 \pi \gamma \rho[/itex]

This implies that there's a gravitational potential [itex]\phi[/itex] such that

g⃗ =−∇⃗ ϕ

What you said actually only implies [itex] \vec g=-\vec \nabla \phi+\vec B [/itex] for any vector field [itex] \vec B [/itex] that satisfies [itex]\vec \nabla \times \vec{B}=0 \ \ and \ \ \vec{\nabla} \cdot \vec{B}=0[/itex]. Of course [itex] \vec B=0 [/itex] is a solution but its not the only solution. Can it be proved that having the boundary condition [itex] \displaystyle \lim_{r\to\infty} \vec g=0 [/itex], gives us the unique solution [itex] \vec B=0 [/itex]?

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No, its not needed so don't bother. Its just mathematical details. I just wanted to get sure I'm standing on rock here!vanhees71 said:

Thanks

A conservative force is a type of force that, when applied to a system, does not change the total mechanical energy of the system. This means that the work done by the force only depends on the initial and final positions of the system, and not on the path taken. Examples of conservative forces include gravity and elastic forces.

A conservative force is related to potential energy because the work done by a conservative force can be expressed as the change in potential energy. This is known as the work-energy theorem. As a system moves in a conservative force field, its potential energy changes, and this change is equal to the work done by the conservative force.

No, a conservative force cannot do positive work. This is because positive work implies that the force is adding energy to the system, which goes against the definition of a conservative force. A conservative force can only do negative work, which means it is taking energy away from the system and converting it into potential energy.

The main difference between conservative and non-conservative forces is that conservative forces do not change the mechanical energy of a system, while non-conservative forces do. Non-conservative forces, such as friction and air resistance, convert the mechanical energy of a system into other forms of energy, such as heat or sound.

The concept of conservative forces is used in physics to analyze the motion of objects in different systems. By identifying whether a force is conservative or not, we can determine the potential energy associated with the system and use it to calculate the work done by the force. This helps us understand and predict the behavior of objects in various situations.

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