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Subsections

Energy

Why should we study the concept of energy?

Energy is an idea that runs through the entire gamut of science, and defines what is meant by physical reality. Everything that exists has energy, and occupies space and evolves in time. All of physical reality can be said to be the manifestation of the different forms of energy. Energy is also central to the functioning of human civilization; to reshape nature in accordance with human needs involves the expenditure of energy. Hence, engineering and technology are intimately concerned with the properties of energy which make it amenable for human manipulation and utilization. All of things that exist in nature, as encountered in physics, chemical processes, biological entities and so on, regulate and transform energy from one form into another. Consequently, being familiar with the major forms of energy lays the basis for understanding the underlying substratum of apparently diverse phenomena, and prepares one to understand new and unforseen forms of energy. Figure 3.1 shows the organization of this chapter on energy. The various forms of energy are discussed, as these forms are the basis of the physics of a diverse range of phenomena. Every form of energy has its specific features, and although the various may at first seem unrelated, as one progresses deeper into nature's laws, the various forms energy turn out to inter-connected all sorts of strange and unexpected manners. The transformation of one form of energy into another is a major focus in the study of physics, and reveals the inner workings of the various forms of energy.

Figure 3.1: Forms of Energy
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From Force to Energy

The concept of energy is pervasive not only in physics but in the rest of the sciences as well. This was not always the case. Physics in Issac Newton's time (1642-1727) started with the concept of force and it was only after a few centuries that the concept of energy was understood to be more central. One of the most fundamental and useful laws of nature is the conservation of energy. Energy is a concept that took many centuries of science to discover and we briefly recount the reasoning which led to its emergence. The mechanics of Newton emphasized the concept of force, and the century that followed was one of the unchallenged victory of Newtonian, also called classical, physics in all spheres that it was applied to. The motion of planets, the mechanics of solid bodies, and so on were understood in great detail within the framework of classical mechanics. By the end of the eighteenth century, it was becoming clear that classical mechanics was no longer able to address all the questions confronting scientists. All that Newtonian mechanics could tell them was that if they could generate a force, then it would cause acceleration and so on. The pioneers of the industrial revolution, however, needed to know more, in particular, how does one generate a force? Engineers and scientists needed to have a prescription on how to move things, build engines and so on, and Newtonian mechanics is silent on this question for practical applications. Newtonian mechanics was also confronted by the phenomenon of heat, for which it had no explanation. And even more embarrasing was the complete inability of Newtonian physics to explain chemical elements and chemical processes that were being vigorously investigated under the impetus of the industrial revolution. We briefly discuss Newton's laws to understand how the concept of energy goes beyond the idea of force.

Newton's Laws

Newton was the first physicist to try and predict the path that a particle would follow, once its present position and velocity is known. The trajectory of the particle is determined by what is called equations of motion. Newton's law makes a quantitative prediction of the future, something unique in the history of science. The prediction for the particle's trajectory was based on the idea of the particle's momentum, as well as on the idea of the forces acting on the particle. Motion is usually associated with velocity, that is, how fast you are changing your position. In general, velocity has both magnitude (are you exceeding the speed limit?) and direction (are you heading South or North?). However, if a light body like a bullet moves very fast, it has the same amount of motion as a slower moving body which is heavier. Hence, a better measure of motion is not just velocity, but something which also takes into account the mass of the object as well. Momentum is such a measure of motion. Hence we expect that
\begin{displaymath}
\mathrm{Momentum} \propto \mbox{\rm mass} \times \mbox{\rm velocity }
\end{displaymath} (3.1)

Since we have not yet defined the units for mass, we can absorb the proportionality constant into the definition of mass. For a particle with mass $m$ and velocity $v$, momentum $p$ is consequently defined to be the following.
$\displaystyle \mbox{\rm {Momentum}}$ $\textstyle =$ $\displaystyle \mbox{\rm {mass}} \times \mbox{\rm {velocity }}$ (3.2)
$\displaystyle \Rightarrow p$ $\textstyle =$ $\displaystyle mv$ (3.3)

The dimension of momentum is given by $[p]=MLT^{-1}$. If there are many particles, the total momentum of the system is the sum of the individual momenta. Most of us have an intuitive idea of force. We know that for instance, to move a heavy object, we have to exert a lot of ``force''. Newton took this experience of force, and gave it a precise mathematical formulation. Let a force $F$ act on a particle from time $t_1$ to time $t_2$, and during this time interval let the momentum change from $p_1$ to $p_2$. Then the change in momentum, in time interval $\Delta t=t_2-t_1$, is given by $\Delta p=p_2-p_1$. Force is defined by Newton to be that physical quantity which causes the momentum of a particle to change. In other words, the rate of change of momentum, represented by $\displaystyle \frac{\Delta p}{\Delta t}$, is caused by a force $F$ acting on the particle. In other words, Newton's Second Law of motion states
$\displaystyle \mbox{\rm Force}$ $\textstyle =$ $\displaystyle \mbox{\rm rate of change of momentum }$ (3.4)
$\displaystyle \Rightarrow F$ $\textstyle =$ $\displaystyle \frac{\Delta p}{\Delta t}$ (3.5)

Given that in most circumstances mass is a constant, the second law is equivalent to saying that for the object to experience acceleration, which is the rate of change in velocity - be it in its magnitude or direction - there must be a force acting on it. Hence, for acceleration given by $\displaystyle a=\frac{\Delta v}{\Delta
t}$, we have
$\displaystyle F$ $\textstyle =$ $\displaystyle m\frac{\Delta v}{\Delta t}=ma$ (3.6)
$\displaystyle \Rightarrow \mbox{\rm Force}$ $\textstyle =$ $\displaystyle \mbox{\rm mass} \times \mbox{\rm acceleration }$ (3.7)

The force equation looks very reasonable. The more the force, the greater the change in the amount of motion of the particle. The dimension of force is given by $[F]=MLT^{-2}$. Let us look at the units for above equation. Mass is measured in $kg$, acceleration is measured in $ms^{-2}$; hence unit of force, called a Newton, is given by $kgms^{-2}$.


\fbox{\fbox{\parbox{12cm}{In general, Newton's second law is a vector equation,
...
...Newton's
equation of motion and are discussed below as Case I and Case II.
}}}


Figure 3.2: Initial position and velocity specified
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{\bf Case I} Since the d...
...
\mathrm{Given : } x(t_i)&= \frac{dx}{dt}(t_i)=v(t_i)=v_i
\end{eqnarray}
}}}


Figure 3.3: Initial and final position specified
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{\bf Case II} We have an...
...{dt^2}; t_f>t>t_i\\
\mathrm{Given : } x(t_i)&= x(t_f)=x_f
\end{eqnarray}
}}}


We can now state Newton's three laws of motion.

  1. First Law A body continues in uniform motion if no force acts on it.
  2. Second Law A force acting on a body causes it to accelerate (or decelerate).
  3. Third Law The total momentum of an isolated system is constant.
The first law took an enormous leap of imagination, since we know from experience that nothing can continue to be in motion for ever. The reason being that frictional forces slow down any moving body. Newton, however, could imagine the idealized situation and understood the underlying principle of motion. The famous experiment of Galileo in dropping two different masses - and showing that they fell at the same rate - also needed a leap of imagination, since again friction due to air had hindered a clear understanding of the workings of the force of gravity. The second law is the backbone of the predictive power of Newton's laws. One may (correctly) object that this law has no predictive power, since the moment one sees a particle accelerating, one can simply multiply it by the mass of the particle and determine the force that is causing this acceleration. And if one did this, then Newton's second law would indeed have no predictive power. The genius of Newton lies in understanding that nature has a quantity called force, and once we can determine these forces independently from Newton's second law, we can then use this force as input to predict the motion of particles experiencing this force. And sure enough, there are multifarious forces to be found in nature, from gravitational and electrical to subnuclear forces such as strong and weak interactions - which can be then be plugged into Newton's second law to predict the future. The third law is often stated in popular literature as action equals reaction, and its formulation as given in III is more transparent. There is a crucial concept buried in the third law, and that is the concept of an isolated system. We will use this concept time and again, and so it best to say a few words on it. No system in the universe is perfectly isolated, since, if it was, it would not be a part of our universe. What we mean by an isolated system, be it in mechanics or thermodyamics, is that we can isolate and shield the system from all the forces in the environment to any degree of accuracy. This statement is correct except for one exception, and that is gravity. We can never ``shield'' any system from the effects of gravity. However, nature is kind in that the effects of gravity are usually trillions of times weaker than all the other forces of nature, and hence can safely be ignored! A number of subtle assumptions have been made in stating Newton's three laws of motion. For example, what is mass? In which frame of reference are position, velocity and force being measured? If I am stationary and you are moving very fast, and we both observe the same particle, clearly we will observe very different forces and velocities. Newton's response was that all observers who are moving with constant velocity - called inertial observers - will observe the same physics. The question then arises is whether all inertial observers will measure the same flow of time? Although Newton's answer was yes, we know that this is not true. The ideas of time, position, velocity, mass and acceleration were all radically changed by Albert Einstein, in 1905, and form the basis of the Special and General Theory of Relativity.

Back to Energy

We now discuss the limitations of Newton's formulation of mechanics, which hinges on the ideas of force and momentum. Following are some examples to illustrate the limitations :
  1. Consider a bullet fired from a gun. According to Newton's third law, momentum is conserved, so that the recoil of the gun must have the same momentum as that of the bullet, and hence their combined momentum before and after remains zero. But common sense tells us that something very significant and irreversible has happened; something was taken from gunpowder and transformed into motion. Physics must be able to capture the transformation of gunpowder into motion, and Newton's laws are silent on what this connection is.
  2. Car manufacturers need to know how to build cars which can accelerate, coast at a constant speed, and, of course, decelerate as well. Newton's laws are useful in telling us how much force is required if we want to accelerate, and also how much force the brakes must exert to decelerate. But Newton is silent on the great difference between the force that accelerates the car, and the brakes which decelerate.
The forward force requires that we expend fuel. The braking force costs no fuel, and results in the heating of the brake pads. The heat in turn is released into the air flowing over the brake pads. Newton's laws make no distinction between the direction of the force, be it causing acceleration or deceleration. To keep the car moving at constant speed, some force is required to overcome friction and air resistance. Newton's laws have very little to say on this, since net acceleration is zero, although to achieve constant speed the car must constantly burn fuel. And lastly, one can turn a car around bends at a constant speed and with high acceleration, but without hitting the accelerator and hence without the expenditure of any fuel. So clearly we need a new concept which makes up for the inadequacy of the concept of force. To more fully understand the science of motion, we need to have a concept which has the following features.
  1. A non-directional conservation law, that unlike conservation of momentum, does not cancel out for motion in opposite direction.
  2. A quantity that takes into account the direction of force relative to the motion, so that a force along the direction of motion has a positive effect, force against the direction of motion has a negative effect, and force perpendicular to motion has no effect.
  3. And most importantly, the concept must connect apparently unrelated things, such as the explosion of gunpowder with the motion of the bullet, the burning of gasoline with the motion of a car.
The concept of energy fulfills these criteria precisely. We will explain later how the concept of energy clarifies the physics of the examples given above. Energy (in some units) is a pure number and is a measureable quantity that we can assign to every physical system. One may ask, well what is energy? The surprising answer is that we do not know! To quote Richard Feynman [#!feynman!#] "It is important to realize that in physics today, we have no knowledge of what energy is. We do not have a picture that energy comes in little blobs of a definite amount." In other words, energy is not a material thing. Rather, it is a intrinsic property of a material thing. Energy is intrinsic to a body in that it inherent to the very physical existence of the body. All we know at present is that for every physical object there is an abstract quantity called energy which we can compute, and which always remains constant no matter how many changes the object goes through. In other words, the most significant property of energy is the following: in every known experiment performed to date, it is an empirical result that energy is absolutely conserved. We will later discuss the subtleties of energy conservation in quantum theory, and even though there is an uncertainty relation involving energy and time, we will find that energy is absolutely conserved not only in Newtonian physics, but in quantum theory as well. One of the most informative and useful quantity associated with natural phenomenon is the energy involved in the process. In Figure 3.4 we list a number of phenomena to have an idea of the vast range of energy that occur in natural processes.

Figure 3.4: Energy Scales in Nature
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Conservation of Energy

To illustrate the idea of energy conservation, let us start with the example used by Feynman [#!feynman!#] in his famous Lectures on Physics. Consider the case of a 10 year old who has 28 indestructible playing blocks, the weight of each block being 0.1 kg. Everyday, the boy's mother counts the blocks, and it always adds up to 28. One day she finds only 27 blocks, and by looking around she finds one lying under the bed. A few days later she finds only 26 blocks. On searching the room she finds a toy box; when she tries to open the toy box, the boy screams ``Don't open my toy box''. Being a smart mother, she weighs the box when she has 28 blocks and finds that the empty toy box weighs 0.5kg. Henceforth, when she cannot find all the 28 blocks, she weighs the toy box to determine how many blocks it has. The mother now has the following formula for the number of blocks
$\displaystyle \mbox{\rm Total number of blocks}$ $\textstyle =$ $\displaystyle \mbox{\rm number of blocks
seen}$  
  $\textstyle +$ $\displaystyle \frac{{\mbox{\rm (weight of toy box)-0.5kg}}}{{\mbox 0.1kg}}$ (3.8)

After a few days, she finds that the blocks no longer add up to 28, and she finds that the bathtub has water full of soap, submerged in which, she suspects, there are some more blocks. Given that the original height of the bath-water was 0.5m, and that each block raises the height of water by 0.1m, she now has a new formula for computing the number of blocks, namely
$\displaystyle \mbox{\rm Total number of blocks}$ $\textstyle =$ $\displaystyle \mbox{\rm number of blocks
seen}$  
  $\textstyle +$ $\displaystyle \frac{{\mbox{\rm (weight of toy box)-0.5kg}}}{{\mbox {\rm0.1kg}}}$  
  $\textstyle +$ $\displaystyle \frac{{\mbox{\rm (height of water)-0.5m}}}{{\mbox {\rm0.1m}}}$ (3.9)

We see that the calculation for the number of blocks is becoming more and more abstract, and has less and less to do with counting the blocks themselves. Measuring the weight of a toy box or the height of bath water has no direct correlation with the blocks themselves. The application of this example is straightforward to energy. In other words, there are many, many forms of energy, that taken together are conserved, but there is no analogy of what is energy per se, that is, there is no analogy for the ``blocks'' themselves. Just as the total number of blocks was constant, energy is absolutely conserved, that is energy can neither be created nor destroyed. All that a physical process can do is to transform energy from one form into another, analogous to, for example, the weight of the toy box being reduced at the expense of the bath water rising in height, so that the total number of blocks remains constant. Hence, for energy we have the following expression.
$\displaystyle \mbox{\rm Total energy of system}$ $\textstyle =$ $\displaystyle {\mbox{\rm Form A of Energy}}
+ {\mbox{\rm Form B of Energy}}{}$  
  $\textstyle +$ $\displaystyle {\mbox{\rm Form C of Energy}}+ {\mbox{\rm etc.}}$ (3.10)

There are various forms of energy such as solar energy, chemical energy, nuclear energy, wind energy, and so on.


All the forms of energy can be classified under two great headings: kinetic and potential.


Energy due to motion in space is called kinetic energy $T$, and energy due to position or the internal configuration of the material body is called potential energy $U$. For example a stone on a mountain has potential energy due to its position, and a stretched spring has potential energy due to its internal composition. Denoting total energy by $E$ we have the fundamental relation

$\displaystyle \mbox{\rm Total energy of system}$ $\textstyle =$ $\displaystyle {\mbox{\rm Kinetic Energy}}
+ {\mbox{\rm Potential Energy}}$  
$\displaystyle \Rightarrow E$ $\textstyle =$ $\displaystyle T+U$ (3.11)

Suppose the system has energy $E_1$ at time $t_1$ and energy $E_2$ at a later time $t_2$, then, the change in energy is $\Delta E=E_2-E_1$. Conservation of energy implies that
$\displaystyle \Delta E =0$     (3.12)
$\displaystyle \Rightarrow \Delta T + \Delta U = 0$     (3.13)

Note an important fact that since all we know is that $\Delta
E=0$, the absolute value of $E$ has not been fixed. Hence, energy is only defined upto a constant, since $E$ and $E+$constant would both be equally conserved. In both the example of the bullet being fired or the car being accelerated, kinetic energy due to motion was created by expending energy that was present as potential energy in the gunpowder and in the gasoline, respectively.

Kinetic Energy

Kinetic energy is that portion of total energy that is due to movement in space. We all know that a truck moving at high speed is something to be avoided; a head-on collision is violent enough to demolish a wall. The violence that a truck can do is due to the large kinetic energy that it has, and comes both from the fact that it is heavy, and depends on how fast it is moving. No matter in which direction the truck moves in, for a given velocity it has the same amount of kinetic energy. If $m$ is the mass of the truck and $v$ its velocity, its kinetic energy is defined by
\begin{displaymath}
T=\frac{1}{2}mv^2
\end{displaymath} (3.14)

The dimension of energy is $[E]=ML^2T^{-2}$. The units of energy can be deduced from the above equation as $kgm^2s^{-2}$, and is called a Joule, abbreviated as $J$. Example 1 Revisited Recall in Example 1, we had a bullet of mass $m_B$ being fired at some velocity $v_B$ from a gun of mass $m_G$ with a recoil velocity given by $v_G$. Momentum conservation implies that initial and final momentum must be zero. Consequently
\begin{displaymath}
0=m_Bv_B+m_Gv_G
\end{displaymath} (3.15)

This doesn't tell us much, since we are not able to translate this information into how much energy must be expended by the gunpowder. However, the total energy expended in shooting the bullet is given by
$\displaystyle T$ $\textstyle =$ $\displaystyle \mbox{\rm Kinetic Energy of bullet} + {\mbox{\rm Kinetic Energy of
Gun}}$ (3.16)
  $\textstyle =$ $\displaystyle \frac{1}{2}m_Bv_B^2+\frac{1}{2}m_Gv_G^2$ (3.17)

Using eqn.(3.23) we have the result that to design a gun which specifies the velocity of the bullet to be $v_B$, we need to spend an amount of energy given by
\begin{displaymath}
T=\frac{1}{2}m_B(1+\frac{m_B}{m_G})v_B^2
\end{displaymath} (3.18)

For most guns, the mass of the gun is much greater than that of the bullet, that is, $\displaystyle m_G » m_B$, and hence $T$ above reduces to the kinetic energy of the bullet.


Rotational Motion


There are only two kinds of kinetic motion. Namely, rectilinear, or linear, motion, that is, motion in a straight line, and kinetic motion associated with rotation. Rotational motion, in general, is also associated with periodic motion.

Figure 3.5: Solid body spinning. Coordinate $\theta $
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Let a solid object be spinning, around some fixed axis say the direction perpendicular to this piece of paper, with angular velocity $\omega$. Let the angular position of a point on the solid body be given by the angular coordinate $\theta $, with the range $-\pi\leq \theta \leq \pi$. Suppose that at time $t_1$ the angular coordinate is $\theta_1$ and at later time $t_2$ the angular coordinate is $\theta_2$ as shown in Figure 3.5. Then, in analogy with linear motion, we define angular velocity by
\begin{displaymath}
\omega=\frac{\Delta \theta}{\Delta
t}=\frac{\theta_2-\theta_1}{t_2-t_1}
\end{displaymath} (3.19)

The angle $\theta $ is dimensionless. Angular velocity $\omega$ has dimension of $T^{-1}$, and has units of $s^{-1}$.


\fbox{\fbox{\parbox{12cm}{The instantaneous angular velocity is obtained by taki...
... and we obtain
\begin{equation}
\omega=\frac{d\theta}{dt}
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We have , similar to kinetic energy for linear motion, the angular component of kinetic energy given by

$\displaystyle T_{\mbox{\rm {angular}}}=\frac{1}{2}I\omega^2$     (3.20)

$I$ is the moment of inertia of the solid body. From dimensional analysis $I$ has the dimensions of $ML^2$ and has units $kgm^2$. Similar to (linear) momentum, a spinning body has angular momentum given by
\begin{displaymath}
L=I\omega
\end{displaymath} (3.21)

We have $[L]=ML^2t^{-1}$. Angular momentum measures the rotational motion of a rigid body. If a point particle of mass m rotates with constant velocity $v$ at a distance $r$ about, say, the $z$-axis, its angular momentum is given by
\begin{displaymath}
L=mvr
\end{displaymath} (3.22)

Similar to linear momentum, the total angular momentum of a physical system is conserved. In general, for a solid body that is moving with velocity $v$ and spinning with angular velocity $\omega$, the total kinetic energy is given by
\begin{displaymath}
T=\frac{1}{2}mv^2+\frac{1}{2}I\omega^2
\end{displaymath} (3.23)

The moment of inertia of a sphere of radius $R$ and mass $m$ is given by
\begin{displaymath}
I_{\mathrm{sphere}}=\frac{2}{5}mR^2
\end{displaymath} (3.24)

Note to obtain the moment of inertia for a sphere, we did not specify around which axis it is spinning since for a sphere all directions yield the same moment of inertia $I_{\mathrm{sphere}}$. However, for a cylinder of mass $m$, radius $R$ and length $L$, this is not the case, as there are two inequivalent axis around which we can spin the cylinder as can be seen from Figure 3.6. Spinning the cylinder along the axis of the cylinder will have a moment of inertia given by $I_{\mathrm{cylinder}}(\mathrm{parallel})$, and orthogonal to the axis of the cylinder will yield $I_{\mathrm{cylinder}}(\mathrm{orthogonal})$

Figure 3.6: Axis of rotation for the cylinder and for the sphere
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We consequently will have two different moments of inertia, given by
$\displaystyle I_{\mathrm{cylinder}}(\mathrm{parallel})=\frac{1}{2}mR^2$     (3.25)
$\displaystyle I_{\mathrm{cylinder}}(\mathrm{orthogonal})=\frac{1}{4}mR^2+ \frac{1}{12}mL^2$     (3.26)

Work

Work is a crucial concept which links energy to force. How does one increase or decrease the energy of a particle? Intuitively, one would think that by acting with a force on a body, we should be able to change its energy. For example, by acting with a force we can increase the velocity of a particle, and which in turn implies that its (kinetic) energy has increased. So clearly, there is an intimate connection between force and energy. Note an important fact that force acts over a finite period of time, and hence is a dynamic quantity, whereas energy is a property inherent and intrinsic to a body which can be time independent. Hence, the a concept of work distinct from energy is necessary to relate dynamic force to intrinsic energy. Energy is defined in many books as the capacity to do work. Work, denoted by $W$ is in turn defined as
\begin{displaymath}
W=\mbox {\rm {force}} \times
\mbox {\rm {distance moved along direction of force}}
\end{displaymath} (3.27)

The dimension of work $[W]=ML^2T^{-2}$, which is the same as the dimension of energy. Hence we can in principle add work to energy, showing that there is a deep relationship between energy and work. This is an instance where dimensional analysis yields new insights into dimensionally related quantities. Work is a measure of the action of force acting over a distance. Think of pushing a car; the heavier the car, the more the force required to move it. Also, the longer the distance the car needs to be pushed, the larger the amount of work that needs to be done. So the definition seems reasonable. However, many students have an intuitive problem with this definition of work. We all know that if we hold up a big piece of stone, even though the stone is not moving, and consequently no work is being done, we will soon break out into a sweat from the exertion that we are undergoing. So what's going on? How can there be no work done, even though we have had to exert ourselves? The answer to this counter-intuitive result lies in the physiology of human muscle. There are two kinds of muscle cells, one which change over a long period of time and the other which change over a short period.The clam, for example, has a muscle cell which relaxes over a very long interval, and hence a clam can be in an open position and support a large weight without expending any energy. In contrast, human muscle needs constant electrical impulses to hold its position, and consequently to hold a piece of stone requires a large expenditure of biological energy. Hence we are doing internal biological work, and not work on the piece of stone, when we hold it in air. To confirm that this is true, one can just place the stone on a table, and it will sit on the table without any work being done, and with no energy being used up for that purpose. The fact that no work is done on a load put on a table is the same reason why high rise buildings can stand without any work being done. All the high floors are stationary, and hence do not require any expenditure of work (energy) to hold them up. Of course overloading may cause the floor to break, and this then becomes a problem of material science rather than that of mechanics. One cannot store work, since once the body ceases to move, no more work is done. This is a reflection of the dynamic nature of force. However, unlike work, we can store energy. When work is done on, or by, a particle, the result is to increase, or to decrease, its energy. In other words, the deep connection between work and energy is that force results in work , which in turn increases or decreases the energy of a body. Of course, energy conservation tells us that we are simply transforming or transferring energy from one form to another. Consider a moving particle with mass $m$ that at initial time $t_i$ has a position of $x_i$ and speed of $u$. Let a constant force $F$ act on it from time $t_i$ to $t_f$, during which time it travels, along the direction of the force, to the final position of $x_f$. At the end of time $t_f$, it has increased its velocity to $v$. The particle has only kinetic energy, and the conservation of energy then tells us that the increase in the (kinetic) energy of the particle must be due to the work done on it by the action of the constant force $F$. Distance $s$, given by
$\displaystyle \Delta x$ $\textstyle =$ $\displaystyle x_{\mbox{\rm {final}}}-x_{\mbox{\rm {initial}}}$ (3.28)
  $\textstyle =$ $\displaystyle s$ (3.29)

was covered during time given by
\begin{displaymath}
\Delta t=t_f-t_i
\end{displaymath} (3.30)

Hence, we have the conservation of energy
$\displaystyle \mbox{\rm {Change in KE of particle}}$ $\textstyle =$ $\displaystyle \mbox{\rm {Work done on
particle}} {}$  
$\displaystyle \Delta T$ $\textstyle =$ $\displaystyle F \Delta x$ (3.31)
$\displaystyle \Rightarrow \frac{1}{2}mv^2 - \frac{1}{2}mu^2$ $\textstyle =$ $\displaystyle F s$ (3.32)

Power

Power is defined as the rate at which work is done. For the case of the particle moving under constant acceleration, its total energy is equal to its kinetic energy, and work done on the particle is equal to the change in its kinetic energy. Hence, in this case power is equal to the rate of change of energy of the particle. From eq.(3.40) we have that the work done in time $\Delta t$ is given by
\begin{displaymath}
\Delta W=F\Delta x
\end{displaymath} (3.33)

We rewrite (3.40) in the following manner. The change in the particle's kinetic energy is seen to be the work done on it.
$\displaystyle \Delta T$ $\textstyle =$ $\displaystyle \frac{1}{2}mv^2 - \frac{1}{2}mu^2$ (3.34)
  $\textstyle =$ $\displaystyle \Delta W$ (3.35)

The change in $T$ comes about by constant force acting over a distance s. Power, denoted by $P$, is then defined as
$\displaystyle \mbox{\rm Power}$ $\textstyle =$ $\displaystyle \mbox{\rm Rate of doing work}$ (3.36)
$\displaystyle \Rightarrow P$ $\textstyle =$ $\displaystyle \frac{\Delta W}{\Delta t}=\frac{\Delta T}{\Delta t}$ (3.37)

Since the time interval over which the kinetic energy has changed is $\Delta t$, we have $\displaystyle \frac{\Delta x}{\Delta
t}=v$, and hence
$\displaystyle \frac{\Delta W}{\Delta t}$ $\textstyle =$ $\displaystyle F\frac{\Delta x}{\Delta t}$ (3.38)
$\displaystyle \Rightarrow P$ $\textstyle =$ $\displaystyle F v$ (3.39)

This is an important equation, since it tells us that the rate of change of kinetic energy of a particle, that is the power being expended on the particle, is equal to the force times the velocity of the particle. The dimension of power is $ML^2T^3$. One unit of power is defined to be 1 Watt (W) defined to be 1 Joule per second, and W has units of $kgm^2s^{-3}$. Example. Consider a typical CD system that uses 250 W of power per hour. If you play the system for 3 h how much energy do you use? If electricity costs $0.08/kWh, how much is the electric bill? We know that Energy= Power x Time, and hence $E=250W \times 3h=0.75 kWh$. Hence, the electric bill is ${\displaystyle \frac{0.08}{kWh} \times 0.75
kWh=\$0.06}$. Example 2 Revisited We can now answer the questions raised in Example 2 regarding the acceleration of a car. When the car accelerates, $F$ and $v$ are in the same direction, and hence $Power=F v$ required is positive since power has to be supplied to the car , and that is why the car has to burn fuel to provide this power. When we brake, $F$ and $v$ are in opposite directions, with the brake causing acceleration in the direction opposite to velocity $v$. Hence the required $Power=-F v$ is seen to be negative, since the car is providing power to the brake, causing it to heat, and hence no fuel needs to be burnt. For a typical passenger car moving at 25 $ms^{-1}$ on a level surface requires a force of 1000 Newtons to keep it moving. Hence the power required $P=Fv=1000 \times 25=25,000W=33\frac{1}{3}$ horsepower. Since most car engines have much more power, the car can easily coast at constant speed. Since the car is not accelerating, the force of resistance is exactly equal to the power $P$ supplied by the engine, that is $\displaystyle P=vF_{\mbox{\rm {friction}}}$. Consequently, we see that the velocity of the car $\displaystyle
v=\frac{P}{F_{\mbox{\rm {friction}}}}$. The frictional forces $F_{\mbox{\rm {friction}}}$ dissipate fuel energy being expended by the car into heating. We will see later that the conversion of low entropy fuel into high entropy heat follows from the Second Law of Thermodynamics, and causes the process to be irreversible. When the car turns, say in a perfect circle, the force is perpendicular to the displacement of the car, and the distance moved by the car along the direction of the force is hence zero. From eq.(3.36), we then have that no power needs to be expended for making this turn. Question.Why is the force perpendicular to displacement when the car is moving in a circle? We see from above that it is the idea of power rather than force that is required to explain the motion of a car. Notice the questions regarding the expenditure of power does involve the use of force $F$, but we will see later when we discuss potential energy, that the concept of force will be replaced by concept of potential and then eq.(3.48) will be seen to be a re-statement of the principle of conversation of energy.

Constant Acceleration

From Newton's second law, we know that a constant force causes motion with constant acceleration, say a. That is
\begin{displaymath}
F=ma
\end{displaymath} (3.40)

Hence, from (3.41) and above, we have
$\displaystyle \frac{1}{2}mv^2 - \frac{1}{2}mu^2$ $\textstyle =$ $\displaystyle mas$ (3.41)
$\displaystyle \Rightarrow v^2$ $\textstyle =$ $\displaystyle u^2+2as$ (3.42)

From eq.(3.51) we see that we have recovered the result of mechanics of how velocity and acceleration are related to distance covered. Since acceleration $a$ is a constant, we also have
$\displaystyle a$ $\textstyle =$ $\displaystyle \frac{\Delta v}{\Delta t}=\frac{v-u}{t_f-t_i}$ (3.43)
$\displaystyle \Rightarrow v$ $\textstyle =$ $\displaystyle u+a(t_f-t_i)$ (3.44)

Hence, combining eqs.(3.51) and (3.53), for $t=t_f-t_i$, we have
\begin{displaymath}
s=ut+\frac{1}{2}at^2
\end{displaymath} (3.45)

We see, from the equations (3.51),(3.53) and (3.54), that all the equations of motion for constant acceleration can be derived from the principle of energy conservation.


Potential Energy

Kinetic energy is easy to visualize. Potential energy, denoted by $U$ is a more complicated concept, since potential energy means energy which is in a latent (``hidden'') form, and capable of being ``released''. Potential energy is energy stored in the physical shape and configuration of the body, and is different from kinetic energy in that it is present in the body without any motion or movement. The dimension and units of potential energy $U$ is the same as energy (of which it is one form). Replacing the concept of force by that of the potential is the essence of why the idea of energy is so useful. Can we always replace $F$ by $U$? The answer is that if the force is conservative such that energy is conserved, we can always replace $F$ by $U$. For such forces, if one compares the value of the potential $U$ at two neighboring positions $x_1$ and $x_2$, we obtain the force at that point given by
\begin{displaymath}
F\simeq \frac{ U(x_2)-U(x_1)}{x_2-x_1}
\end{displaymath} (3.46)

One may correctly object that since energy is always conserved, it should always be possible to express force in terms of a potential. In principle, this observation is true. However, there are many non-conservative cases, such as in the study of viscous fluids, that one uses an approximate formulation of the problem where for example energy losses due to friction are not included in the problem. For such cases, we have to directly use the concept of force, and it cannot be then replaced by a potential term.

Figure 3.7: Path dependence and path independence
\begin{figure}
\begin{center}
\epsfig{file=core/figure3.eps, height=6cm, angle=-90}
\end{center}
\end{figure}

Conservative and non-conservative forces are characterizing by whether the work done in going between two points in space is path dependent or not. Consider for example traveling between two points, say $E1$ and $E2$, by two different paths, one which is a gravel road and the other which is a paved road. This situation is shown in Figure 3.7 with $1$ being the label for say the gravel path and $2$ the label for the paved path. Clearly, traveling via the gravel road will entail doing more work than the paved road in getting from from $E1$ to $E2$, and hence the work is path dependent in that it depends on which path we take, and we conclude that the force experienced in traveling is non-conservative. On the other hand, if in Figure 3.7 the work done on the particle by a force $F$ which takes it from point $E1$ to point $E2$ does not on the path taken, for example the work along path 1 is equal to that of taking it along path 2, then the force is conservative.


% latex2html id marker 9853
\fbox{\fbox{\parbox{12cm}{In general,a potential $U(...
...at the equation for power is simply a re-statement
of energy conservation.
}}}


A body can come to rest only if there is no force acting on it. Since force is equal to the negative gradient of the potential, that is

\begin{displaymath}
F=-\frac{dU}{dx}
\end{displaymath} (3.47)

we can conclude that no force acts on a particle that is located at the minimum value of the potential. An example of this is shown in Figure 3.9, where at the point $x_E$ is the position of equilibrium. We have the important and general result that


A body reaches equilibrium when its position is at the minimum value of the potential


There are almost unlimited forms of potential energy. We discuss a few of these, in particular, the gravitational and electrical potential energy, mass as potential energy as well as elastic potential energy that is stored in springs and other elastic objects. These examples are chosen to give a flavor of great variety of forms that potential energy can take. Other forms of potential energy such as chemical energy, radiant energy, nuclear energy and so on are more subtle, and require deeper study.

Gravitational Potential Energy

We all know that a stone dropped from a great height picks up a lot of speed before hitting the ground. Since a fast moving piece of stone has a lot of kinetic energy, and it started with zero kinetic energy, clearly its position at a great height must have endowed it with ``potential'' energy, and that becomes ``actual'' (kinetic) energy before it hits the ground. So what is this potential energy? The hint is already given by the connection of potentiality with height. The higher the elevation, the greater must be its potential energy. Also, the heavier the body, the greater its potential energy. Both these intuitive expectations reflect everyday experience when we see objects fall - if they fall from a greater height, they have higher impact, and similarly if they are heavy. If we represent potential energy by $U$, for a body with mass $m$ at a height $h$ we reasonably expect the simplest expression for (gravitational) potential energy to be given by
$\displaystyle U$ $\textstyle \propto$ $\displaystyle m h$ (3.48)
$\displaystyle \Rightarrow U$ $\textstyle =$ $\displaystyle m g h$ (3.49)

We have called the proportionality constant $g$; clearly $g$ is linked to gravity, since if there was no gravitational force, the particle would not fall towards the earth to start with. By dimensional analysis, since $U$ is energy, it has dimensions of $ML^2T^{-2}$, and hence the dimension of $g$ is $LT^{-2}$, which is the dimension for acceleration. From dimensional analysis, $g$ must be proportional to acceleration due to gravity, and in fact $g$ is precisely equal to acceleration arising from earth's gravity, and is equal to $9.8 ms^{-2}$. The gravitational potential given in (3.64) is an approximation, valid only for bodies that are close to the earth's surface. At distances from the earth's surface that are comparable to the earth's radius, the correct expression for $U$ is given by the ``inverse law''. Consider a particle falling under the force of gravity. At a height of $h$, it has velocity $v$, and hence its total energy that is given by
$\displaystyle E$ $\textstyle =$ $\displaystyle T+U$ (3.50)
$\displaystyle \Rightarrow E$ $\textstyle =$ $\displaystyle \frac{1}{2}mv^2+ mgh$ (3.51)

As the particle falls, its velocity increases as its height decreases. Conservation of energy then requires that energy always be a constant, that is, the change in $E$ be zero. Hence
\begin{displaymath}
\Delta E=\Delta (\frac{1}{2}mv^2+m g h)=0
\end{displaymath} (3.52)

Simplifying above equation, we have
\begin{displaymath}
\Delta v^2= -2g \Delta h
\end{displaymath} (3.53)

To understand the content of above equation, at height $h_1$ let the velocity of the particle be $u$ and its energy $E_1$, and at height $h_2$ let it be $v$ and $E_2$ respectively. Energy conservation requires that $E_1=E_2$. We have $\Delta v^2=v^2-u^2$ with $\Delta h=h_2-h_1=s=$height that the particle has fallen through. Hence, from (3.68) we have
\begin{displaymath}
v^2=u^2+2gs
\end{displaymath} (3.54)

We see that we have recovered (3.51), with the additional information that the constant acceleration is due to gravity. Note the significant fact that, unlike the case for (3.51), we did not use the idea of force at any stage in deriving (3.69). This is a reflection of a the general procedure of replacing conservative force $F$ by potential $U$.

Elastic Potential Energy

Consider a spring with a ball, of mass $m,$ attached to one end with the other end of the spring fixed to a wall.

Figure 3.8: Diagram of spring with ball
\begin{figure}
\begin{center}
\epsfig{file=core/springs.eps, height=10cm}
\end{center}
\end{figure}

The system is in equilibrium when the ball is at position $x_E$. What do we expect for the potential energy of the spring? Whether we compress or stretch the spring, in both cases it gains energy. Hence the potential energy should be equal if the new position $x$ of the ball is compressed or stretched by the same distance from its resting position $x_E$. The simplest expression for such a potential is given by
\begin{displaymath}
U=\frac{1}{2}k(x-x_E)^2
\end{displaymath} (3.55)

where $k$ is the spring constant, which is a measure of the stiffness of the spring, and has the dimensions of $MT^{-2}$ and units of $kgs^{-2}$.

Figure 3.9: Potential Energy versus Position
\begin{figure}
\begin{center}
%%
\input{core/potential.eepic}
\end{center}
\end{figure}

Hence the total energy of the ball, moving at velocity $v$, is given by
$\displaystyle E$ $\textstyle =$ $\displaystyle T+U$ (3.56)
$\displaystyle E$ $\textstyle =$ $\displaystyle \frac{1}{2}mv^2+\frac{1}{2}k(x-x_E)^2$ (3.57)

Let the ball be at some position $x \ne x_E$. We can for example ask the following question: What is the force required to displace it by a small distance $d$? Let the ball have zero velocity. To change the energy of the ball by displacing it, an external force $F_{external}$ has to do work on it. Hence we have
$\displaystyle F_{external}\times d$ $\textstyle =$ $\displaystyle \Delta U=\frac{1}{2}k(x+d-x_E)^2-\frac{1}{2}k(x-x_E)^2$ (3.58)
  $\textstyle =$ $\displaystyle kd(x-x_0)+ \frac{1}{2}k d^2$ (3.59)
$\displaystyle \Rightarrow F_{external}$ $\textstyle =$ $\displaystyle k(x-x_0)+\frac{1}{2}kd$ (3.60)
  $\textstyle \simeq$ $\displaystyle k(x-x_E)$ (3.61)

where in obtaining the last equation we have taken $d$ to be so small that it can be completely ignored. For equilibrium, the restoring force due to the spring has to exactly cancel the external force. Hence, if the ball is moved to a position $x \ne x_E$, we have a restoring force acting on the ball that is given by
$\displaystyle F$ $\textstyle =$ $\displaystyle -F_{external}$ (3.62)
$\displaystyle \Rightarrow F$ $\textstyle =$ $\displaystyle -k(x-x_E)$ (3.63)


% latex2html id marker 9873
\fbox{\fbox{\parbox{12cm}{
From eq.(\ref{fu}) we ha...
...
&=&-\frac{d}{dx}[\frac{1}{2}k(x-x_E)^2]\\
&=&-k(x-x_E)
\end{eqnarray}
}}}


In other words, the force required to stretch the spring is proportional to the amount of stretching, namely $-(x-x_E)$. Note that the spring is always a restoring entity since it acts against any change of position of the ball. For $x>x_E$ the force is negative, that is, it acts to pull $x$ back to $x_E$, whereas if $x<x_E$ then the force also acts to push the ball back to $x_E$.


\fbox{\fbox{\parbox{12cm}{{\bf Trigonometric Functions}\\
Trigonometric functi...
...verse of each other), but taken together, as $kx$, they
are dimensionless.
}}}


We solve for the motion of the particle that is moving in the elastic potential. From (3.72) we have for a particle moving with energy $E$ the following.

$\displaystyle v=\frac{dx}{dt}$ $\textstyle =$ $\displaystyle \sqrt\frac{2E-k(x-x_E)^2}{m}$ (3.64)
$\displaystyle \Rightarrow x(t)$ $\textstyle =$ $\displaystyle x_E+A\sin(2\pi f t + \phi)$ (3.65)

One can easily verify, by differentiating (3.87), that to satisfy the energy equation given in (3.72), we have the following
$\displaystyle f$ $\textstyle =$ $\displaystyle \frac{1}{2\pi}\sqrt\frac{k}{m}$ (3.66)
$\displaystyle A$ $\textstyle =$ $\displaystyle \sqrt\frac{2E}{k}$ (3.67)
$\displaystyle \Rightarrow E$ $\textstyle =$ $\displaystyle \frac{1}{2}m(2\pi f)^2A^2$ (3.68)

Note $A$ has dimension of $M$, $f$ has dimension of $T^{-1}$ and the phase $\phi$ is a dimensionless angle measured in radians. Recall $x_E$ is the position of equilibrium of the particle. Suppose the particle starts its oscillations at time $t=0$ from the position $x(0)$. Then $\phi$ is fixed by
\begin{displaymath}
\phi=\sqrt\frac{m}{2E}\sin^{-1}(x(0)-x_E)
\end{displaymath} (3.69)

Figure 3.10: Diagram of x(t)
\begin{figure}
\begin{center}
\epsfig{file = core/figure34.eps, height=5cm}
\end{center}
\end{figure}

One can see from the behaviour of $x(t)$ that if the ball is disturbed from its equilibrium position at $x_E$, it will undergo oscillations about this position; note that if one increases time $t$ by the amount $1/f$, that is, for $t \rightarrow t+1/f$, the particle returns to the position that it occupied at time $t$. In other words, it is undergoing periodic motion about its equilibrium position, with period of oscillation given by $T=1/f$. The maximum departure that it has from $x_E$ is called the amplitude of the oscillation, and is given by $A$. The phase tells us where the oscillations begin at $t=0$, the amplitude of the oscillation $A$ is fixed by the ratio of the total energy and mass of the particle, and the period of oscillation of the particle, namely $T$, is given by the ratio of the spring constant $k$ and the mass of the particle. In other words, the stiffer the spring is, the larger is the value of $k$ and consequently the faster the particle oscillates. The particle will oscillate forever if not for frictional forces, which we have ignored in our discussion. Due to oscillatory motion that the elastic potential energy term produces, it is also called the harmonic oscillator potential. We studied the motion due of the oscillator in some detail since it is of fundamental importance in all kinds of diverse fields. In particular, as we will see later, the underlying explanation of wave motion relies on an understanding of oscillatory motion.

Mass as a Form of Potential Energy

We discuss another form of potential energy. Albert Einstein showed that mass itself is a form of potential energy. For a body with mass $m$, the formula for potential energy of the body at rest is given by
\begin{displaymath}
E=m_0c^2
\end{displaymath} (3.70)

where $c=299,792,458 ms^{-1}$ is the velocity of light. The equivalence of mass and energy stated above is how the mass equivalents to energy were arrived at in Figure 3.4. If the particle is moving at velocity $v$ with respect to an inertial frame, its mass increases. As its velocity approaches $c$ the mass tends to infinity, as shown in Figure 3.11, reflecting that a finite mass particle can never reach the velocity of light. The mass of the particle increases indefinitely with respect to the said inertial frame, and is given by
\begin{displaymath}
m=\frac{m_0}{\sqrt{1-v^2/c^2}}
\end{displaymath} (3.71)

Figure 3.11: Mass as a Function of Velocity
\begin{figure}
\begin{center}
\epsfig{file=core/mass.eps}
\end{center}
\end{figure}

If one thinks of the atomic nature of matter, one might be tempted to think that all forms of energy are ultimately kinetic energy. This example is important to realize that this is not so, and that there are some forms of energy which are irreducibly potential in nature.


Gravitational and Electrostatic Potential Energy

Two of the most important forms of potential energy in physics are the gravitational and electrostatic potential energy.

Figure 3.12: Gravity and electromagnetism
\begin{figure}
\begin{center}
%%
\input{core/figure6.eepic}
\end{center}
\end{figure}

Suppose two particles of masses $m_1,m_2$ and charges $q_1,q_2$ are separated by a distance $r$. The gravitational potential energy and the potential energy between static electrical charges - electrostatic potential energy - also known as the Coulomb potential, are given respectively by
$\displaystyle U_{gravity}(r)$ $\textstyle =$ $\displaystyle -G\frac{m_1m_2}{r} \mbox{\rm { : Gravitational
Potential Energy}}$ (3.72)
$\displaystyle U_{Coulomb}(r)$ $\textstyle =$ $\displaystyle k\frac{q_1q_2}{r} \mbox{\rm { : Coulomb
Potential Energy}}$ (3.73)

where $G$ is Newton's gravitational constant with dimension of $L^3M^{-1}T^{-2}$, and its numerical value is given by
\begin{displaymath}
G=6.67 \times 10^{-11}Nm^2/kg^2
\end{displaymath} (3.74)

Electrical charge is measured in units of coulomb, labeled C, and by an abuse of notation we will denote the dimension and the SI unit for charge by $C$. The Coulomb constant $k$ has a dimension of $ML^3T^{-2}C^{-2}$, with its numerical value being given by
\begin{displaymath}
k=9.00\times 10^9 Nm^2/C^2
\end{displaymath} (3.75)

The Coulomb constant $k$ is also defined in terms of the so called permittivity constant $\epsilon_0$ by the relation
\begin{displaymath}
k=\frac{1}{4\pi \epsilon_0}
\end{displaymath} (3.76)

Question. Show that the gravitational potential energy given in (3.64) can be derived from (3.94). Determine the value of $g$ in terms of the gravitational constant $G$. [Hint: Consider the whole mass of the earth to be as if it were located at the centre of the earth.] An important similarity of the two forms of potential energy is that mass and charge are the source of the gravitational and electrostatic potential energies, respectively. In other words, a mass or a charge at some point in space creates, in the space around it, a gravitational or an electrostatic potential, and other masses or charges gain potential energy as they enter the field created by the mass or charge source. One may wonder as to what is the potential ``made out of'', and by what mechanism does the source exert an influence in space around it. The answer to this question leads one to concept of a field. Masses and charges are the sources of fields, and the gravitational and electrostatic potentials are the physical manifestation of the gravitational and electrostatic forces in nature. We will see later that a field is a physical entity as real as the source, and leads to many new insights into the workings of nature. Although the gravitational and electrostatic potential energy are similar in the sense of having a common mathematical form and are generated by sources, they are also profoundly different as well. The differences are the following.
  1. Mass is always a positive quantity, so gravitational potential energy always gives an attractive force of attraction between any two masses. This is the reason that one can never 'shield' a system from gravity, since there is no way of cancelling it out. So although gravity is an extremely weak force, as we will soon show, its effects keep piling up. Hence for celestial bodies, solar systems, stars, galaxies and so on, the effects of gravity are the most important. Due to the fact that gravity is always attractive, one can imagine that if one gathers together a large enough mass, there could be no force which could be strong enough to stop the inward pull of gravity. This is precisely what happens when we have a star with mass three times bigger than that of our Sun : for such stars, the force of gravitational attraction is so strong, that the star - under the inward pull of gravity - undergoes gravitational collapse, and results in the formation of a black hole.
  2. Another difference between electrical and gravitational potential energy is that every physical entity feels the force of gravity - there are no gravitationally ``neutral'' entities. On the other hand, there are many fundamental entities like the neutron that are electrically neutral, and do not feel the effect of electrical forces.
  3. On the other hand, electrical charge can be positive or negative. The gravitational and Coulomb potential also differ by a minus sign, which implies that although masses attract gravitationally, two particles with like charges repel whereas as opposite charges attract. Unlike gravitational forces, we can completely shield a system from electrical forces by using negative charge to cancel positive charge, since having a net charge of zero means we have effectively cancelled out electrical forces. Recall all elements are made out of electrically neutral atoms. The atoms of each element are made out of a nucleus consisting of an equal number of electrically neutral neutrons and positively charged protons, which is surrounded by a 'cloud' of an equal number of electrons. The Coulomb potential holds the electrons in a bound state with the nucleus. The repulsive Coulomb potential - inside the nucleus - between like charges is overcome by strong nuclear forces which glue together the protons inside the nucleus. The nuclear force acts over very short distances, it can hence cancel Coulomb repulsion only over short distances. However, since the Coulomb potential acts over long distances, it finally wins over strong nuclear forces. Namely, if the number of protons exceeds a certain value close to 140, the Coulomb repulsion becomes so large that the nuclear force can no longer overcome it, and the nucleus falls apart (disintegrates). This repulsion due to the Coulomb potential is the reason that there are only a finite number of stable elements in the universe, numbering about 100 only. Problem Suppose the strong attractive force between protons acts over a distance of $10^{-15}m$, and that the electrostatic potential acts over an infinite distance. Estimate how many protons would make the nucleus unstable.
  4. The greatest difference in the gravitational and electrical potential energy lies in their relative strengths, characterized by the empirical values of the two constants $G$ and $k$. To determine the relative strengths of the two potential energies, we evaluate the ratio of their values for the case of an electron and proton separated by a distance $r$. Consider an electron, having mass $m_e=9\times 10^{-31} kg$ and a charge $e=-1.7\times 10^{-19}C$ , and a proton having mass $m_p=1.7 \times 10^{-27} kg$ and a charge $e=1.7\times 10^{-19}C$. We hence have
    $\displaystyle \frac{U_{Coulomb}(r)}{U_{gravity}(r)}$ $\textstyle =$ $\displaystyle \frac{ke^2}{Gm_em_p}$ (3.77)
      $\textstyle =$ $\displaystyle \frac{9.00\times 10^9Nm^2/C^2\times(1.7\times 10^{-19}C)^2}
{6.67 \times 10^{-11}Nm^2/kg^2\times9\times 10^{-31}
kg\times1.7 \times 10^{-27} kg}$  
      $\textstyle =$ $\displaystyle 2.6 \times 10^{39}$ (3.78)

    We see that, for any distance of separation $r$ of two charged bodies, the gravitational potential energy is almost $10^{39}$ smaller in magnitude than the electrical potential energy. Even if we consider bodies with much bigger masses than an electron or a proton, the effect of gravity is extremely small. So for all cases of interest, we can ignore the effects due to gravity compared to other forces. One may object to the thought of ignoring gravity, considering that we always feel the force of gravity in our daily lives. This objection is correct if the mass involved is immense; for example, what we experience in daily life is the gravitational effect of the earth, which is an immense mass indeed. Compared to the gravitational effect of the earth's mass, the gravitational effect of any other massive body on the earth's surface, on say an object undergoing experiments in the laboratory, are negligible. The weakness of the gravitational force is ultimately the reason why the idea of an isolated system is possible.

Negative Energy:Bound States

An important extension of the idea of energy is the case when a particle has a net negative energy. For all cases where the potential is well defined, the concept of negative energy, that is $E<0$, can be properly defined, and the state with negative energy is said to be a bound state. Bound states are the norm in nature. Neutrons and protons are in a bound state forming the nucleus, the electron is bound to the nucleus forming the atom, atoms bind together to form molecules, all the bodies on the earth are bound to it including ourselves, the earth is in a bound state with the Sun, the Sun itself is bound to the Milky Way galaxy and so on. Since energy is defined only up to a constant, what does it mean to talk of negative net energy? Recall from (3.19) that
$\displaystyle E$ $\textstyle =$ $\displaystyle T+U$ (3.79)
  $\textstyle =$ $\displaystyle \frac{1}{2}mv^2+U$ (3.80)

We have to first remove the arbitrariness in energy, which is only defined upto a constant. Only then can we talk of positive and negative energy. Since there is no arbitrariness in $T>0$, we have to first define what we mean by zero potential energy $U$. Take the case of the gravitational potential. We consider the approximate expression given in eq.(3.64), namely
\begin{displaymath}
U=mgh
\end{displaymath} (3.81)

It is natural to measure the height from the surface of the earth, and that is what has been implicitly assumed in the equation for $U$ above. In other words, we choose the surface of the earth $h=0$ as the height for which $U=0$. Hence, if $h>0$, it has positive potential energy, and if we can drop a particle from this height, it will get transformed into kinetic energy as the particle loses height. All this we have already discussed. We may now ask: what happens if we dig a ditch or a well which has depth $-h_0$, and put a particle inside the well? The height for a particle in a well is negative, that is $-h_0<0$, and hence this particle has negative potential energy, that is $U=-mgh_0<0$. This configuration is shown in Figure 3.13/

Figure 3.13: Particle in a Well
\begin{figure}
\begin{center}
\epsfig{file=core/figure7.eps, height=8cm, width=8cm}
\end{center}
\end{figure}

If the particle has velocity $v$, the total energy is then given, for $h_0>h>0$ (particle inside the well), by
$\displaystyle E$ $\textstyle =$ $\displaystyle \frac{1}{2}mv^2-mgh$ (3.82)

Hence, for a bound state
$\displaystyle E$ $\textstyle <$ $\displaystyle 0$ (3.83)
$\displaystyle \Rightarrow v$ $\textstyle <$ $\displaystyle \sqrt{2gh}$ (3.84)

What is the physics of this result? Negative energy, $E<0$, means that the particle cannot reach the surface, since on the surface its energy is at least zero (and higher if it has non-zero velocity on the surface). Hence, as long as $v<\sqrt{2gh}$, the particle can move inside the well, but cannot escape from it. The particle is consequently bound to the well, and the particle is said to be in a bound state. To free the particle from the well, we would have to give it enough energy so that its net energy would then be at least zero. For example, suppose the particle is stationary at the bottom of the well. Then $U=-mgh_0$ and we would have to supply the particle with a velocity of $\sqrt{2gh_0}$ to make its total energy equal to zero, and hence allow it to escape from the well.

Escape Velocity and Black Holes

Can we escape from a bound state? Since the bound state is a result of the attractive force due to some attractive potential, the only way one can escape from a bound state is by acquiring enough kinetic energy so that the total energy $E=T+U$ becomes at least equal to zero. To understand the mechanism of escape, let us study a particle bound to the earth by gravity. The particle in a ditch is identical to a particle held to the surface by earth's gravitational force. We can find the minimum velocity required for the particle to escape from the earth's gravity. We have to use the correct gravitational potential given by (3.94). Let the mass of the earth be $M_e$ and radius $R_e$. Let the $E$ be the energy of a particle of mass $m$ moving with vertical velocity $v$. Hence, we have
$\displaystyle E=\frac{1}{2}mv^2-G\frac{M_em}{R_e}$     (3.85)

For a bound state, we have
\begin{displaymath}
E<0 \mbox{\rm { : Bound}}
\end{displaymath} (3.86)

The escape velocity $v_{\mathrm{escape}}$ of the particle is achieved when $E=0$. Hence
$\displaystyle E$ $\textstyle =$ $\displaystyle 0 =\frac{1}{2}mv_{\mathrm{escape}}^2-G\frac{M_em}{R_e}$ (3.87)
$\displaystyle \Rightarrow v_{\mathrm{escape}}$ $\textstyle =$ $\displaystyle \sqrt{\frac{2GM_e}{R_e}}$ (3.88)

Example The mass and radius of the earth $M_e=5.89 \times 10^{24} kg$, $R_e=6.37\times 10^6
m$ respectively. The resultant escape velocity is given by
$\displaystyle v_{\mathrm{escape}}$ $\textstyle =$ $\displaystyle 2 \frac{(6.67 \times 10^{-11} Nm^2/kg^2)(5.89 \times 10^{24}
kg)}{6.37\times 10^6 m}$ (3.89)
  $\textstyle =$ $\displaystyle 11.2 \mathrm{km/s} \simeq 25,000 \mathrm{miles/hour}$ (3.90)

From eq.(3.111), we see that the escape velocity depends inversely on the radius of the earth $R_e$. What does this mean? Note we can reduce the radius of the earth by squeezing the mass of the earth into a smaller volume, in other words, by increasing the density of the earth. Suppose we compress the mass $M_e$ of the earth to a new radius $R<R_e$; given that the volume of a sphere is $\frac{4}{3}\pi R^3$ the new density of the earth is given by
$\displaystyle \rho$ $\textstyle =$ $\displaystyle \frac{\mbox{\rm {Mass}}}{\mbox{\rm {Volume}}}$ (3.91)
  $\textstyle =$ $\displaystyle \frac{3M_e}{4\pi R^3}$ (3.92)
  $\textstyle >$ $\displaystyle \rho_{earth}$ (3.93)

The escape velocity from a body with the earth's mass but smaller radius $R$ is given, from (3.111), by
$\displaystyle v_{\mathrm{escape}}$ $\textstyle =$ $\displaystyle \sqrt{\frac{2GM_e}{R}}$ (3.94)
  $\textstyle >$ $\displaystyle \sqrt{\frac{2GM_e}{R_e}} (R<R_e)$ (3.95)

Note that we can keep on increasing the value of $v$ by reducing the value of $R$. What is the maximum value that $v$ can have? We know that the maximum velocity that any object in the universe can have is the velocity of light, namely $c = 3\times 10^8 m/s$. When the escape velocity $v_{escape}$ approaches $=c$, this is an indication that nothing, not even light, can escape from the gravitational pull of a highly dense object. Such an object is called a black hole. The radius into which the mass of the earth has to be squeezed so that it forms a black hole is called the Schwarzchild radius of the earth, called $R_S$, and is given by
$\displaystyle v_{\mathrm{max}}$ $\textstyle =$ $\displaystyle c=\sqrt{\frac{2GM_e}{R_S}}$ (3.96)
$\displaystyle \Rightarrow R_S$ $\textstyle =$ $\displaystyle 2 \frac{GM_e}{c^2}$ (3.97)
  $\textstyle =$ $\displaystyle 2 \frac{(6.67 \times 10^{-11} Nm^2/kg^2)(5.89 \times 10^{24}
kg)}{(3\times 10^8ms^{-1})^2}$ (3.98)
  $\textstyle =$ $\displaystyle 8.7\times 10^{-3}m \simeq 1cm$ (3.99)

Surprisingly, a much more complicated calculation based on General Relativity yields exactly the same result for the Schwarzchild radius as given in eqn.(3.122). So the Schwarzchild radius of the earth is really tiny. For the Sun the Schwarzchild radius is $\simeq 3$km. As mentioned earlier, it is know from astrophysics that stars with masses greater than 3 times the mass of our Sun all end their stellar evolution by undergoing gravitational collapse and forming black holes. Question. You may have noticed that light in a room is spread over all space, whereas a piece of stone is what we usually have in mind when we talk of a particle. Can we extend the notion of energy to the phenomenon of light? Would energy be then distributed all over space instead of being a property of a particle?
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Marakani Srikant 2000-09-11