Physics 53Energy 1What I tell you three times is true.— Lewis CarrollThe interplay of mathematics and physicsThe “mathematization” of physics in ancient times is attributed to the Pythagoreans, who taught that everything true is contained in numbers. But the introduction of algebraic equations as a way of stating laws of nature dates from the time of Galileo. In Isaac Newton science had both a great innovator in mathematics and a great analyst and experimenter on natural phenomena. Because he was (independently of Leibniz) the discoverer of calculus, he was able to think of physical processes in terms of rates, infinitesimal changes, and summations of infinitesimals into finite quantities by means of integrals. Although he published relatively little of his thinking on these matters — his famous book, the Principia, contains no reference to calculus, giving geometrical arguments and proofs instead — it seems clear that he thought analytically, in the modern mathematical sense of that word.In the 18th century great advances were made in mathematical analysis, and many of these were applied to — indeed, arose from — problems in physics. New mathematical formulations were found for the content of Newton’s three laws of motion, making easier the solutions of many physical problems. The work of Euler and Lagrange in particular gave important new insights, and led to the emergence in the 19th century of what is probably the most important single concept in science, that of energy. In this part of the course we deal with energy as it applies to the mechanics of particles and systems of particles. Later we extend the concept to fluids and thermal systems, and in the next course we discuss the energy associated with electromagnetic fields. The scalar productIn our discussion of energy we will need to use the product of two vectors that results in a scalar. Consider two vectors, A = (Ax, Ay, Az) and B = (Bx,By,Bz). Multiplying one component of A by one component of B gives a set of 9 pairs. This set as a whole is not very useful because it does not transform simply when we rotate the coordinate axes. PHY 53! 1! Energy 1Some combinations of the 9 do have simple transformation properties, however. The simplest of these is AxBx+ AyBy+ AzBz. It can be shown to be unchanged by rotation of the axes, so it is a scalar, and is therefore called the scalar product. The standard notation for it uses the “dot” multiplication sign between the two vectors:Scalar product A ⋅ B = AxBx+ AyBy+ AzBzBecause of the notation, it is often called the “dot” product. We can rewrite this formula in terms of the magnitudes and relative direction of the two vectors. Let the vectors be as shown in arrow representation. Then it is easy to derive a very useful formula:Scalar product formula A ⋅ B = ABcosθSome properties of the scalar product A ⋅ B that follow from this:• A ⋅ B is positive if θ<π/ 2, negative if θ>π/ 2, zero if θ=π/ 2.•When θ= 0, A ⋅ B has its maximum value + AB.•When θ=π, A ⋅ B has its minimum value − AB.•The scalar product of a vector with itself is its squared magnitude : A ⋅ A = A2. A useful way to think of the value of the scalar product is this: it is the magnitude of one vector multiplied by the component of the second vector along the line of the first.Power, work and kinetic energyNewtonian analysis of the mechanics of a particle consists of this process: 1. Given the initial conditions of the particle ( r0 and v0); 2. Given the net interaction it has with its environment ( Ftot);3. Determine its position r as a function of time. If we know Ftot as a function of time, Newton's 2nd law gives us a(t) from which (by two integrations and using r0 and v0) we can find r(t). This is what we have done in the simple cases where a is a constant, and in a few other cases. B AθPHY 53! 2! Energy 1Unfortunately, often we do not know the forces as functions of time. More often they are known as functions of the particle’s position, r. From the above argument it seems that to determine r we must know r in advance, which makes the direct approach hopeless. In these cases we use a trick devised by 18th century scientists, notably Euler.We start with the one-dimensional case. Assume we know the total force F(x) as a function of the particle’s position x. The 2nd law then reads F(x) = ma. Now construct the quantity P = Fv, where v is the velocity (not the speed, since v can be negative). Then P = mav. But since! ddt12v2( )= vdvdt= vawe can also write P = mav as!  P =ddt12mv2( ).This equation relates two new quantities that turn out to be useful enough to have names: the power input by a forcePower input by a force (1-D) P = Fvand the kinetic energy of the particle:Kinetic energy of a particle K =12mv2What has been shown (in one dimension) is that P = dK /dt. The power input by the total force is equal to the rate of change of the kinetic energy.All we have really shown is that two quantities we have defined are related. It remains to be shown why this is useful. And it remains to be shown that it is true in 3-D.To find the net change in kinetic energy between two times we integrate over time:! K(2) − K(1) = P dtt1t2∫= Fv dtt1t2∫= F dxx1x2∫.(We have used the fact that the distance dx moved in time dt is given by dx = v dt.) Since we know F as a function of x, in principle we can evaluate the last integral. It also has a name: the work done by the force as the particle moves from x1 to x2:PHY 53! 3! Energy 1Work done by a force (1-D) W(1 → 2) = F dxx1x2∫What we have proved is an important theorem:Work-energy theoremThe work done by the total force is equal to the change in kinetic energy.Through this theorem we can find how the particle's kinetic energy (and from that its speed) depends on its position. This approach tells us the speed when the particle is at a particular place. It does not tell us the time when the particle is at that place. It also tells us only the speed, not the direction of the velocity. We have not found a complete description of the motion, but it is nevertheless very useful.Since the change in K is equal to the total work done, and the rate at which K changes is the total power input, it follows that the relation between power and work is! P = dW / dt.Power input by a force is the rate at which work is done by that force.Work and kinetic energy have the same