Physics

In this page, general topics in physics are discussed.

Work and Energy

Energy

Energy is an abstract concept that helps scientists keep track of dynamics of physical systems. In everyday language, we use the idea of energy to determine how power we gain or lose in different processes such as a windmill farm that produces wind power.

But, we need to better define this idea of energy because there are different kinds of energy. The main two categories are:



Potential energy is a type of energy that an object may have where the amount of energy directly correlates to the POTENTIAL of that object TO DO SOMETHING.

Kinetic energy is a type of energy that an object may have where the amount of energy directly correlates to how much the object is MOVING.

Those descriptions might sound too nebulous, so some examples are definitely in order. For Kinetic energy, KE, the amount an object has tells us, more specifically, how fast it moves. For example, a baseball with 40 J (the Joule, J, is a unit of energy) of KE is moving faster than an identical baseball with 20 J of KE. If the only object in the universe is the baseball, then we speak of the baseball as the system and its KE is 40 J. However, we know that we are surrounded by atoms and molecules everywhere that make up matter. This means we need to definition of a system to to keep track of the energy, the forms it takes on and how it can be transferred between objects.

Systems

A system is a collection of one or more particles. Defining a system helps determine the scope of a physical system. For example, a ball and the Earth together can be considered a system. The ball alone and Earth alone can also be systems. A single electron of the baseball can be the system.

Defining systems are useful because they help tell us whether any work done is considered to be a change in potential energy or just work itself.

Systems are either open or closed. An open system means that energy can go into or out of the system. A closed system is when energy cannot enter or leave the system. Realistically, all systems are open. But, if you consider the entire universe as the system, then it is likely closed. However, it is absolutely certain that the universe is closed because we don't know if there is anything beyond it yet.

In any case, the system and the collection therein can be chosen to be whatever may make the analysis meaningful and useful (and easier). We first consider closed systems.

Closed Systems

A closed system is one where no energy can enter or leave the system. The total energy of any system is a sum of Kinetic and potential energy. \[ E_{total} = T + V \] Because the system is closed, this total sum of energy is constant even as the system evolves with time.

A system may also have thermal or internal energy, for which our sum may be written as \[ U + E_{total} = T + V \] \( E_{total} \) is then the total mechanical energy of the system. \( V \) is the potential energy which is due to forces such as electric, gravitational, elastic, etc. We will not worry about thermal energy for the moment. The only comment I will make about thermal energy is that it is directly related to the average kinetic energy, i.e. their average motion, on a microscopic scale.

So, our total energy takes the form \[ E_{total} = T + V \] Because this total is constant, taking the difference in total energy from one time to any other time of the evolution of the system gives \[ \Delta E_{total} = E_{f} - E_{i} = 0 \] or \[ E_{i} = E_{f} \] This makes the analysis of a closed system easy to analyze.

For example, if a rock has is dropped from some height, it will have \( V^{grav} \) energy. As the rock falls, it will speed up. Then \(T\) must increase. However, \( E_{total}\) must stay the same amount. This means that \( V^{grav} \) must decrease by the same amount that \( T \) increases.

At the top, we can write \[ E_{total} = V^{grav}_{top} \] while during the fall, \[ E_{total} = T_{mid} + V^{grav}_{mid} \] and finally, at the bottom just before hitting the ground and there is no more height to fall, \[ E_{total} = T_{bot} \] Note that it is not necessary to write \( E_{total; top, mid, bot, etc.} \) because the total never changes. However, it may be helpful to write out to keep track of the state of the system at different times.

Say we know the initial height and mass. Then we would use the expression for the total energy at the top to estimate the total energy. Or, we know the speed at the bottom, etc. Then if we know the total energy, we can know other properties of the system at any other time. That is, if we know some height, we can determine the potential energy \( V^{grav} = mgh \) and then the kinetic energy by \[ T = E_{total} - V^{grav} \] from which we use \[ T = \frac{1}{2}mv^2 \] to solve for the speed \( v \) at that height.

Or, we know the speed and estimate the kinetic energy. And, in the reverse way, determine the potential energy and with that, the height.

Work & Conservation of Energy

When a system can interact with its surroundings (external system), the system can gain or lose energy. This makes the system an open one. From the interaction, the external system will gain or lose energy while the system of interest will lose or gain that energy exactly. The transfer of this energy is called work. The formula for work is \[ W_{sys}=-\int \vec{F}_{sys}(\vec{r}) \cdot d\vec{r} = -\int F_{sys}(r)\cos (\theta) dr\] where the force is exerted by the system on its surroundings and can be a function of position. \(\theta \) is the angle between the direction of the applied force and the displacement, \( d\vec{r} \). The integral is over the path of the trajectory, usual some initial and final position.

The minus sign is due to the fact the work is done by the system. Some references describe the work more generally as work done by some external force on the system. \[ W=\int \vec{F}_{applied}(\vec{r}) \cdot d\vec{r}\] This is the general definition of work. One can estimate the work done by any force of interest.

If we take a single particle as the system and suppose that everything else is external to it, then, there is no potential energy. The work can only increase or decrease the amount of kinetic energy of the particle. This is called the

Work-Kinetic Energy Theorem.

For a free particle with some velocity, the total mechanical energy is just the kinetic energy. \[ E_{total} = T = \frac{1}{2}mv^2 \] If there is no work, \[ \Delta T = T_{f} - T_{i} = 0 \] But, if there is, \[ \Delta T = T_{f} - T_{i} = W \] Suppose we take the average acceleration of the particle, then with \[ \vec{F}_{net} = m\vec{a} \] the work becomes \[ W = \int \vec{F}_{net} \cdot d\vec{r} = \int m\vec{a} \cdot d\vec{r} \] which is just \[ W=m \int a dr \cos(\theta) = ma \Delta x \] where we assume that the force and displacement are colinear, and in one dimension, \( r=x \). Recall that \[ a = \frac{v_f^2 -v_i^2}{2\Delta x} \] Immediately, we see that we arrive at the expression of kinetic energy as a difference. \[ W = m \left( \frac{v_f^2 -v_i^2}{2\Delta x} \right) \Delta x = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = \Delta T \]

We now address the negative sign when applying the work done by the system. We begin with the work-kinetic energy theorem, \[ \Delta T = W \] If we now allow the external entity doing work on the free particle to be included in the the system, the work that was done is now included in any changes. \[ 0 = \Delta T - W \] Because the work was \[ W=\int \vec{F}_{applied}(\vec{r}) \cdot d\vec{r}\] our running sum of changes in energy becomes \[ 0 = \Delta T - \int \vec{F}_{applied}(\vec{r}) \cdot d\vec{r}\] Now, since the force being applied is now by an interaction in the system, \[ \vec{F}_{applied} = -\vec{F}_{sys} \] we have then \[ 0 = \Delta T + \int \vec{F}_{sys}(\vec{r}) \cdot d\vec{r}\] This says that the negative work done by the system is an increase in potential energy. For example, when a ball is tossed up in the air, the Earth would do negative work on the ball. Or, equivalently, the ball gains potential energy.

The previous example can be thought of this way. While its true the ball has negative work done on it by the Earth, when the Earth is included in the system, it is really increasing the amount of energy that could be used to do positive work later. That is, the potential energy is defined by the work the system can do by \[ \Delta V = - \int \vec{F}_{sys}(\vec{r}) \cdot d\vec{r} =-W_{sys}\] where the force is conservative. This means that the work done by the system is path independent.

Since we are now including the external entity in the system, and the work is defined as the negative change in potential, we might as write the total change in energy of the system as \[ 0 = \Delta T + \Delta V \] which is just \[ 0 = \Delta E_{total} \] We can have other external forces acting on the system at the same time we have potential energy. The most general case of total energy changing in a system is then \[ \Delta E_{total} = W \]

A few other notes are necessary here. We could include thermal energy and the transfer of it. This generalizes the total energy expression. There is also a distinction between forces that depend on position versus forces that depend on velocity where the former are called conservative forces while the latter are non-conservative forces.

Conservative forces obey the conservation of energy principle and the work they do is directly related to potential energy. Particularly, these forces are defined by the gradient of the potential energy they describe. \[ \vec{F}_{cons} = -\nabla V(\vec{r}) \] The negative sign is due to fact that the positive work by the system is the negative change in potential energy. Examples of conservative forces are gravitational, electric, and spring forces. Also, the work these forces do is path independent. This is obvious if you consider a vector force field that is strictly dependent on position alone. Let's suppose we use our definition of a conservative force in terms of the potential energy and integrate the work done. \[ W_{cons} = -\int \vec{F}_{cons}(\vec{r}) \cdot d\vec{r} = \int \nabla V(\vec{r}) \cdot d\vec{r} = V(\vec{r_i}) - V(\vec{r_f}) \] The last equality shows directly that path does not matter. What's more is that if the starting position is the same as the final position, the value of the potential energy is the same at the same location and the work done is zero.

Many references will say then there are only two general statements on conservative forces. \[ W_{loop} = 0 \] and \[ W_{ \vec{r}_f \rightarrow \vec{r}_i } = V(\vec{r}_f) -V(\vec{r}_i ) \]

If we have some conservative forces in the system and some outside, we write the law of conservation of energy as \[ \Delta E_{total} = W_{cons} \]

Now, non-conservative forces are not defined by a potential energy. They are dependent on velocity. An example is friction and because it always opposes motion, it always does negative work to decrease the total energy of the system. Some references may call this a dissipative force. Or, that the work done by friction is dissipated as thermal energy. This type of force does not obey the conservation of energy principle. However, as far as we know, energy is always conserved in the universe. The dissipated energy is not easy to track, but can be accounted for as heat transfer. A more accurate definition is to say that friction is not defined by a negative gradient of a scalar function, i.e. potential energy. As a vector field, it is not conservative. The curl of the vector field is not zero.

In any case, we can write our statement as \[ \Delta E_{total} = W_{cons} + W_{non-cons} \] or \[ \Delta E_{total} = W_{cons} - E_{diss} \] where it is understood that \( E_{diss} \) is always positive or zero and accounts for energy lost due to friction or other non-conservative forces.

If we include the thermal or internal energy, we have \[ \Delta U + \Delta E_{total} = W_{cons} + W_{non-cons} \] This is the most general statement of conservation of energy for a system.

Conservative Forces

A questions then becomes, what type of forces do we that are conservative and what form of potential energy do they have. We start with gravity.

The law of gravity is \[ \vec{F}_{gravity} = -G\frac{m_1 m_2}{r^2}\hat{r} \] where \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses of two objects. Near the surface of the Earth, we approximate the acceleration due to gravity as \[ g = \frac{GM_{Earth}}{R_{Earth}^2} \approx 9.81 \frac{m}{s^2}\] We get the familiar weight due to gravity, \( \vec{F}_{grav} = m\vec{g} \). If we apply our gradient definition, we can determine the potential energy. \[ \vec{F}_{grav} = -\nabla V_{grav}\] We integrate over the vertical direction, for small heights we are okay since mere kilometers are small compared to the over 6 million meters of the Earth radius. \[ - \int \vec{F}_{grav} \cdot d\vec{r} = - \int \nabla V_{grav} \cdot dz\hat{z} \] then \[ - mg \int dz = -\int \frac{dV_{grav}}{dz} dz \] we can take the limits to be some lower height, \( z_i \), and some higher height, \( z_f \). The potential energy is now \[ V_{grav}(z) = mg \Delta z \] This is the expression we had earlier.

We can do the the same thing for the general law of gravity. \[ - \int \vec{F}_{gravity} \cdot d\vec{r} = - \int \nabla V_{gravity} \cdot d\vec{r} \] then \[ G m_1 m_2 \int \frac{1}{r^2} dr = \int \frac{dV_{gravity}}{dr} dr \] Integration then gives \[ V_{gravity}(r) = -G\frac{m_1 m_2}{r} \] Be careful. The limits of the integration must begin from infinity because the potential and force are undefined at \( r=0 \). Also, we expect the potential energy to go to zero for a position infinitely far away such that there is no interaction gravitationally. The negative sign tells us that the force was attractive and the potential energy can be thought of as a binding energy. An object with a kinetic energy equal to the magnitude of this gravitational binding energy will be able to escape the gravitational pull and influence on an object's motion. Note that the total energy would then be zero.

The spring force for small displacements follows Hooke's law, in one dimension, \[ \vec{F}_{spring} = -k x \hat{x} \] where \( k \) is the spring constant and \( x \) is the displacement from equilibrium. We can find the potential energy by \[ \int \vec{F}_{spring} \cdot d\vec{r} = - \int \nabla V_{spring} \cdot dx\hat{x} \] then \[ \int ( -k x \hat{x}) \cdot dx \hat{x} = - \int \nabla V_{spring} \cdot dx\hat{x} \] we then have \[ -k \int x dx = -\int \frac{dV_{spring}}{dx} dx \] which gives \[ V_{spring}(x) = \frac{1}{2} k \Delta x^2 \] Here, \( \Delta x = x_f - x_i \) is the displacement from equilibrium.

In electrostatics, the force between two charges is given by Coulomb's law, \[ \vec{F}_{elec} = k_e \frac{q_1 q_2}{r^2} \hat{r} \] where \( k_e \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the two charges. We can find the potential energy by \[ - \int \vec{F}_{elec} \cdot d\vec{r} = - \int \nabla V_{elec} \cdot d\vec{r} \] the result is similar to that of the gravitational force, as expected. \[ V_{elec}(r) = k_e \frac{q_1 q_2}{r} \] The sign of the potential energy depends on the signs of the charges. Like charges repel and unlike charges attract. The potential energy is positive for like charges and negative for unlike charges.

There are other conservative forces such as magnetic forces, but they are velocity dependent. However, as magnetostatic forces, they can be derived from a vector potential and not as a scalar potential function.