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Chapter 4: Problem 49

Both the Coulomb and gravitational forces lead to potential energies of the form \(U=\gamma / | \mathbf{r}_{1}-\) \(\mathbf{r}_{2} |,\) where \(\gamma\) denotes \(k q_{1} q_{2}\) in the case of the Coulomb force and \(-G m_{1} m_{2}\) for gravity, and \(\mathbf{r}_{1}\) and \(\mathbf{r}_{2}\) are the positions of the two particles. Show in detail that \(-\nabla_{1} U\) is the force on particle 1 and \(-\nabla_{2} U\) that on particle 2.

Short Answer

Expert verified
The negative gradient of potential energy gives the forces on particles: \(-\nabla_1 U\) is force on particle 1 and \(-\nabla_2 U\) is force on particle 2.

Step by step solution

01

Understanding the Gradient

The gradient, denoted as \(abla\), is a vector operator used to find the rate and direction of change in scalar fields. For the potential energy \(U\), \(abla U\) is used to find the force associated with the potential energy for any position \(\mathbf{r}\).
02

Expressing the Potential Energy

Given the potential energy \(U = \frac{\gamma}{| \mathbf{r}_{1} - \mathbf{r}_{2} |}\), where \(\gamma\) could be \(kq_1q_2\) for Coulomb's force or \(-Gm_1m_2\) for gravitational force. Here, \(| \mathbf{r}_{1} - \mathbf{r}_{2} |\) is the distance between the two particles.
03

Finding the Gradients \(-\nabla_1 U\) and \(-\nabla_2 U\)

The gradient with respect to \(\mathbf{r}_1\) is given by\[ -abla_1 U = -abla_1 \left( \frac{\gamma}{| \mathbf{r}_1 - \mathbf{r}_2 |} \right) \]where \(abla_1\) represents the gradient with respect to \(\mathbf{r}_1\). Applying the chain rule and noting \(\mathbf{r}_1 - \mathbf{r}_2\) gives a force magnitude, modifying the direction to point from \(\mathbf{r}_1\) to \(\mathbf{r}_2\). Use similar logic to find \(-abla_2 U\).
04

Calculating \(\nabla_1 U\)

By finding the derivative of \(\frac{\gamma}{| \mathbf{r}_1 - \mathbf{r}_2 |}\) with respect to \(\mathbf{r}_1\), we obtain:\[ abla_1 U = \frac{-\gamma \cdot (\mathbf{r}_1 - \mathbf{r}_2)}{| \mathbf{r}_1 - \mathbf{r}_2 |^3} \]Multiplying by -1 gives the force on particle 1:\[ -abla_1 U = \frac{\gamma \cdot (\mathbf{r}_1 - \mathbf{r}_2)}{| \mathbf{r}_1 - \mathbf{r}_2 |^3} \]
05

Calculating \(\nabla_2 U\)

Similarly, differentiate \(\frac{\gamma}{| \mathbf{r}_1 - \mathbf{r}_2 |}\) with respect to \(\mathbf{r}_2\):\[ abla_2 U = \frac{\gamma \cdot (\mathbf{r}_2 - \mathbf{r}_1)}{| \mathbf{r}_1 - \mathbf{r}_2 |^3} \]Thus, the force on particle 2 is:\[ -abla_2 U = \frac{-\gamma \cdot (\mathbf{r}_2 - \mathbf{r}_1)}{| \mathbf{r}_1 - \mathbf{r}_2 |^3} \]
06

Relating Force to Potential

According to classical mechanics, force \(\mathbf{F}\) is related to potential energy \(U\) by the negative gradient, such that \(-abla U = \mathbf{F}\). This implies that the forces obtained from \(-abla_1 U\) and \(-abla_2 U\) correspond to that on particles 1 and 2, respectively.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Coulomb Force
The Coulomb force is one of the fundamental interactions in nature and governs the electric force between charged particles. It is described by Coulomb's law, which is mathematically expressed as:
  • \( F = k \frac{q_1 q_2}{r^2} \)
Here, \( F \) is the magnitude of the force, \( q_1 \) and \( q_2 \) are the charges of the interacting particles, \( r \) is the distance between the charges, and \( k \) is Coulomb's constant.
The direction of the Coulomb force depends on the nature of the charges. If both charges are similar (either both positive or both negative), the force is repulsive, pushing the particles apart. Conversely, if one charge is positive and the other negative, the force is attractive, pulling them together.
Potential energy associated with Coulomb's force is defined as:
  • \( U = \frac{k q_1 q_2}{|\mathbf{r}_1 - \mathbf{r}_2|} \)
This expression is crucial when calculating the force vector using the gradient, as demonstrated in the original problem. Understanding this helps in predicting how charged particles interact in different conditions.
Gravitational Force
Gravitational force is the attractive force acting between any two masses. According to Newton's law of universal gravitation, the gravitational force \( F \) can be defined as:
  • \( F = G \frac{m_1 m_2}{r^2} \)
Where \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses, and \( r \) is the separation distance between the centers of the two masses.
Despite being one of the weakest forces when intrinsic to a single pair of subatomic particles, gravity becomes significant on a large scale, especially in the context of planetary and cosmic interactions. This force is always attractive, bringing masses closer together.
The gravitational potential energy between two point masses is given by:
  • \( U = -\frac{G m_1 m_2}{|\mathbf{r}_1 - \mathbf{r}_2|} \)
The negative sign indicates that work would need to be done against the gravitational attraction if the masses were to be separated. The calculation of gradient forces connected to this potential energy allows us to relate to the dynamics of orbits and other gravitational phenomena.
Gradient in Vector Calculus
The gradient is a crucial concept in vector calculus, representing the rate and direction of change in a scalar field. When applied to potential energy functions like those from Coulomb or gravitational forces, the gradient becomes a powerful tool understanding field dynamics.
  • The notation for the gradient is \( abla \).
  • It transforms a scalar function into a vector.
  • The vector indicates the direction of the steepest ascent of the function, while the magnitude represents the rate of increase.
In the context of the original exercise, calculating gradients \( -abla_1 U \) and \( -abla_2 U \) points us to the direction and magnitude of the force exerted on particles due to their positions. Calculating these involves understanding how the position vector changes the potential field, allowing forces to be calculated as gradients of potential energy.
Using the gradient, we can identify the field forces acting on particles and systems, essential for problems in classical mechanics, especially those involving fields and potential energies, such as electrostatic and gravitational fields.

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Most popular questions from this chapter

A particle of mass \(m_{1}\) and speed \(v_{1}\) collides with a second particle of mass \(m_{2}\) at rest. If the collision is perfectly inelastic (the two particles lock together and move off as one) what fraction of the kinetic energy is lost in the collision? Comment on your answer for the cases that \(m_{1} \ll m_{2}\) and that \(m_{2} \ll m_{1}\).

Find the partial derivatives with respect to \(x, y,\) and \(z\) of the following functions: (a) \(f(x, y, z)=\) \(a x^{2}+b x y+c y^{2},(\mathbf{b}) g(x, y, z)=\sin \left(a x y z^{2}\right),(\mathbf{c}) h(x, y, z)=a e^{x y / z^{2}},\) where \(a, b,\) and \(c\) are constants. Remember that to evaluate \(\partial f / \partial x\) you differentiate with respect to \(x\) treating \(y\) and \(z\) as constants.

An interesting one-dimensional system is the simple pendulum, consisting of a point mass \(m\), fixed to the end of a massless rod (length \(l\) ), whose other end is pivoted from the ceiling to let it swing freely in a vertical plane, as shown in Figure \(4.26 .\) The pendulum's position can be specified by its angle \(\phi\) from the equilibrium position. (It could equally be specified by its distance \(s\) from equilibrium \(-\) indeed \(s=l \phi-\) but the angle is a little more convenient.) (a) Prove that the pendulum's potential energy (measured from the equilibrium level) is \(U(\phi)=m g l(1-\cos \phi)\). Write down the total energy \(E\) as a function of \(\phi\) and \(\dot{\phi}\). (b) Show that by differentiating your expression for \(E\) with respect to \(t\) you can get the equation of motion for \(\phi\) and that the equation of motion is just the familiar \(\Gamma=I \alpha\) (where \(\Gamma\) is the torque, \(I\) is the moment of inertia, and \(\alpha\) is the angular acceleration \(\ddot{\phi}\) ). (c) Assuming that the angle \(\phi\) remains small throughout the motion, solve for \(\phi(t)\) and show that the motion is periodic with period \(\tau_{\mathrm{o}}=2 \pi \sqrt{l / g}\).

The proof that the condition \(\nabla \times \mathbf{F}=0\) guarantees the path independence of the work \(\int_{1}^{2} \mathbf{F} \cdot d \mathbf{r}\) done by \(\mathbf{F}\) is unfortunately too lengthy to be included here. However, the following three exercises capture the main points: \(^{16}\) (a) Show that the path independence of \(\int_{1}^{2} \mathbf{F} \cdot d \mathbf{r}\) is equivalent to the statement that the integral \(\oint_{\mathrm{T}} \mathbf{F} \cdot d \mathbf{r}\) around any closed path \(\Gamma\) is zero. (By tradition, the symbol \(\oint\) is used for integrals around a closed path \(-\) a path that starts and stops at the same point.) [Hint: For any two points 1 and 2 and any two paths from 1 to 2 , consider the work done by \(\mathbf{F}\) going from 1 to 2 along the first path and then back to 1 along the second in the reverse direction. \((\) b) Stokes's theorem asserts that \(\oint_{\mathrm{T}} \mathbf{F} \cdot d \mathbf{r}=\int(\nabla \times \mathbf{F}) \cdot \hat{\mathbf{n}} d A,\) where the integral on the right is a surface integral over a surface for which the path \(\Gamma\) is the boundary, and \(\hat{\mathbf{n}}\) and \(d A\) are a unit normal to the surface and an element of area. Show that Stokes's theorem implies that if \(\nabla \times \mathbf{F}=0\) everywhere, then \(\oint_{\mathrm{T}} \mathbf{F} \cdot d \mathbf{r}=0 .\) (c) While the general proof of Stokes's theorem is beyond our scope here, the following special case is quite easy to prove (and is an important step toward the general proof): Let \(\Gamma\) denote a rectangular closed path lying in a plane perpendicular to the \(z\) direction and bounded by the lines \(x=B, x=B+b, y=C\) and \(y=C+c .\) For this simple path (traced counterclockwise as seen from above), prove Strokes's theorem that \(\oint_{\Gamma} \mathbf{F} \cdot d \mathbf{r}=\int(\mathbf{\nabla} \times \mathbf{F}) \cdot \hat{\mathbf{n}} d A\) where \(\hat{\mathbf{n}}=\hat{\mathbf{z}}\) and the integral on the right runs over the flat, rectangular area inside \(\Gamma\). [Hint: The integral on the left contains four terms, two of which are integrals over \(x\) and two over \(y\). If you pair them in this way, you can combine each pair into a single integral with an integrand of the form \(F_{x}(x, C+c, z)-F_{x}(x, C, z)\) (or a similar term with the roles of \(x\) and \(y\) exchanged). You can rewrite this integrand as an integral over \(y\) of \(\partial F_{x}(x, y, z) / \partial y\) (and similarly with the other term), and you're home.]

Calculate the gradient \(\nabla f\) of the following functions, \(f(x, y, z):\) (a) \(f=x^{2}+z^{3} .\) (b) \(f=k y\), where \(k\) is a constant. (c) \(f=r \equiv \sqrt{x^{2}+y^{2}+z^{2}} .\) [Hint: Use the chain rule.] (d) \(f=1 / r\).
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