Math Problem Statement

a) Osoita, että pistemäisen massan aiheuttama voimakenttä F = -(GMm / r^2) e_r on lähteetön origon ulkopuolella laskemalla divergenssi karteesisessa koordinaatistossa. b) Ilmaise voimakenttä ja laske sen divergenssi pallokoordinaatistossa.

Solution

To tackle the problem in the image, here’s the breakdown and explanation:

Part (a): Prove that the force field is source-free in Cartesian coordinates

The force field is given as: F=GMm(x2+y2+z2)3/2(xi^+yj^+zk^),\mathbf{F} = - \frac{GMm}{(x^2 + y^2 + z^2)^{3/2}}(x \hat{i} + y \hat{j} + z \hat{k}), or equivalently: F=GMmr2e^r,\mathbf{F} = - \frac{GMm}{r^2} \hat{e}_r, where r=x2+y2+z2r = \sqrt{x^2 + y^2 + z^2}.

Step 1: Expressing the divergence

The divergence in Cartesian coordinates is: F=Fxx+Fyy+Fzz,\nabla \cdot \mathbf{F} = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z}, where the components of F\mathbf{F} are: Fx=GMmx(x2+y2+z2)3/2,Fy=GMmy(x2+y2+z2)3/2,Fz=GMmz(x2+y2+z2)3/2.F_x = -\frac{GMm x}{(x^2 + y^2 + z^2)^{3/2}}, \quad F_y = -\frac{GMm y}{(x^2 + y^2 + z^2)^{3/2}}, \quad F_z = -\frac{GMm z}{(x^2 + y^2 + z^2)^{3/2}}.

Step 2: Compute Fx/x\partial F_x / \partial x

Using the product rule and chain rule: Fxx=GMmx(x(x2+y2+z2)3/2).\frac{\partial F_x}{\partial x} = -GMm \frac{\partial}{\partial x} \left( \frac{x}{(x^2 + y^2 + z^2)^{3/2}} \right). Let r2=x2+y2+z2r^2 = x^2 + y^2 + z^2, so rx=xr\frac{\partial r}{\partial x} = \frac{x}{r}. Using this: Fxx=GMm[(x2+y2+z2)3/23x2(x2+y2+z2)1/2(x2+y2+z2)3].\frac{\partial F_x}{\partial x} = -GMm \left[ \frac{(x^2 + y^2 + z^2)^{3/2} - 3x^2 (x^2 + y^2 + z^2)^{1/2}}{(x^2 + y^2 + z^2)^3} \right].

A similar computation applies to Fy/y\partial F_y / \partial y and Fz/z\partial F_z / \partial z.

Step 3: Summing the terms

Adding these components, all terms proportional to x2,y2,z2x^2, y^2, z^2 cancel out, leaving: F=0,\nabla \cdot \mathbf{F} = 0, which shows that the field is source-free (F\mathbf{F} has no divergence).


Part (b): Express and calculate divergence in spherical coordinates

In spherical coordinates, the force field is: F=GMmr2e^r.\mathbf{F} = -\frac{GMm}{r^2} \hat{e}_r.

Step 1: Divergence in spherical coordinates

The divergence in spherical coordinates for a radial field is given by: F=1r2r(r2Fr),\nabla \cdot \mathbf{F} = \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 F_r \right), where Fr=GMmr2F_r = -\frac{GMm}{r^2}.

Step 2: Compute divergence

F=1r2r(r2GMmr2).\nabla \cdot \mathbf{F} = \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \cdot -\frac{GMm}{r^2} \right). Simplify: r2GMmr2=GMm,r^2 \cdot -\frac{GMm}{r^2} = -GMm, and: r(GMm)=0.\frac{\partial}{\partial r}(-GMm) = 0. Thus: F=0,\nabla \cdot \mathbf{F} = 0, showing that the field remains source-free in spherical coordinates.


Summary

  1. The force field is source-free in Cartesian coordinates (F=0\nabla \cdot \mathbf{F} = 0).
  2. Expressing the force in spherical coordinates and calculating divergence also gives F=0\nabla \cdot \mathbf{F} = 0, consistent with the source-free nature of the gravitational field.

Would you like additional detailed steps or visualizations for these calculations?


Related Questions:

  1. How does the divergence theorem verify that the gravitational field is source-free?
  2. Can we generalize this proof to other inverse-square law forces (e.g., electric field)?
  3. What role does the delta function play in defining divergence at the origin?
  4. How is the expression for F\nabla \cdot \mathbf{F} different in cylindrical coordinates?
  5. What physical significance does the divergence of zero have in gravitational systems?

Tip: Always cross-check calculations for divergence in Cartesian vs.

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Math Problem Analysis

Mathematical Concepts

Vector Calculus
Divergence of a Vector Field
Spherical Coordinates
Cartesian Coordinates

Formulas

F = -(GMm / (x^2 + y^2 + z^2)^(3/2))(xi + yj + zk)
F = -(GMm / r^2)e_r
Divergence in Cartesian: ∇·F = ∂Fx/∂x + ∂Fy/∂y + ∂Fz/∂z
Divergence in Spherical: ∇·F = (1/r^2)(∂/∂r)(r^2 Fr)

Theorems

Divergence Theorem
Source-Free Field in Gravitational Physics

Suitable Grade Level

Undergraduate Physics or Mathematics