PHY2061 Enriched Physics 2 Lecture Notes Relativity 4. Relativity 4

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1 PHY6 Enriched Physics Lectre Notes Relativity 4 Relativity 4 Disclaimer: These lectre notes are not meant to replace the corse textbook. The content may be incomplete. Some topics may be nclear. These notes are only meant to be a stdy aid and a spplement to yor own notes. Please report any inaccracies to the professor. Relativistic Momentm Newton s nd Law can be written in the form p F = d dt where the non-relativistic momentm of a body is p x = m where = d dt. However, becase of the Lorent transformation eqations, d x is measred differently in different dt inertial frames. Ths, Newton s nd Law wold not have the same form in different frames. We need a new definition of momentm to retain the definition of force as a change in momentm. x Sppose p = m d, where τ is the proper time in the object s rest frame. Every observer will agree on which frame is the rest frame. Also, since y = yand =, the transverse momentm (p y and p ) will be invariant for a Lorent transformation along the x axis. (This wold not be the case if we did not se the proper time in the definition). We can rewrite this momentm definition as follows: x x p = m d = m d dt dt dt t = γτ = γ From time dilation p= γ m γ = / c Recall that momentm is a vector qantity. Conservation of momentm, which still applies in Special Relativity, implies that each component of momentm is conserved. Note that is the velocity of the object in a reference frame, not the velocity of a reference frame relative to another. In this definition of momentm, the mass m=m is the rest mass. That is, it is the mass of an object in its rest frame. Sometimes γ m is referred to as the relativistic mass, sch that we can retain the Newtonian definition of momentm as p= m. In this sense, the mass of an object grows as its velocity increases. Bt this convenient trick can be problematic. As we shall see, the kinetic energy, for example, is not ½ mv. D. Acosta Page //5

2 PHY6 Enriched Physics Lectre Notes Relativity 4 Relativistic Force With the previos relativistic definition for momentm, we can retain the sal definition for force: dp d dx d dx F= = m = ( γ m) where = and γ = dt dt dt dt / c It is sefl to consider how force transforms nder a Lorent Transformation: y S y' v S' x ' x' According to the addition of velocity formlae, the transformation of the velocity perpendiclar to the direction of the Lorent Transformation is: = where γ = γ v c v x v ( + v / c ) / So for the perpendiclar force, which can be written as: F d dx d = m = m dt it transforms as: F F d d = m = m γ = γ v ( + v / c ) γ ( + v / c ) v x v x F ( + v x / c ) where we assme no acceleration in the direction parallel to the transformation D. Acosta Page //5

3 PHY6 Enriched Physics Lectre Notes Relativity 4 Relativistic Energy Now work is defined as force applied over a distance. It corresponds to the expended energy to accelerate a body. If the force and path are constant, W = F d More generally, if the force and path vary, then a line integral mst be performed from initial position to final position. W = F ds The work applied to a body translates to a change in the kinetic energy since energy mst be conserved. If we assme that the body is initially at rest, then the final kinetic energy is eqal to the work expended: d W = K = γ m dt where we have sed ds= dt dt K = m dt d γ dt γu K = m d γ Integrate by parts: K = γmu m γ d = γ mu m a af a f U U d = γ mu + mc / c / c = γ mu + mc U / c mc c = γ mu + mc U / c mc f U Ths, we get for the relativistic kinetic energy: K = γ mc mc = γ mc h a f Yo can check this integral by differentiation d s This final expression for the kinetic energy looks like nothing like the non-relativistic eqation K =. However, if we consider velocities mch less than the speed of m light, we can see the correspondence: D. Acosta Page 3 //5

4 PHY6 Enriched Physics Lectre Notes Relativity 4 c h f / γ = + / c / c + sing the binomial expansion a K = γ mc mc = m for << c c So at low velocities there is no difference between the definition of kinetic energy in Special Relativity from that in Newtonian Mechanics. Now let s consider the opposite limit when the velocity approaches the speed of light. In that case, the kinetic energy becomes infinite as the relativistic factor γ goes to infinity. This is another way of saying that objects cannot exceed the speed of light, becase it wold take an infinite amont of energy. Now let s rewrite the eqation involving the kinetic energy: E γmc = K+ mc This eqation has the form of kinetic energy pls potential energy eqals total energy. What is the potential energy? It is the term: E = mc which we refer to as the rest energy. As yo know, this is Einstein s famos eqation that tells s that mass is another form of energy. Mass can be converted into energy and vice versa. How mch energy? Let s see: Example: Sppose that a kg mass moves at a velocity = m/s. The kinetic energy is ½ m = ½ J. (We can se the non-relativistic eqation becase the velocity is mch mch smaller than the speed of light.) The rest mass energy is mc = 9. 6 J. Clearly there is a tremendos amont of energy in kg of mass. That is why nclear weapons have the power that they do, becase they convert a significant amont of mass into energy. Conservation of Energy: We have learned in earlier physics corses that kinetic energy does not have to be conserved in an inelastic collision. Likewise, mass does not have to be conserved since it can be converted into energy. However, the total energy (kinetic, rest mass, and all other potential energy forms) is always conserved in Special Relativity. Momentm and energy are conserved for both elastic and inelastic collisions when the relativistic definitions are sed. D. Acosta Page 4 //5

5 PHY6 Enriched Physics Lectre Notes Relativity 4 Relationship between Energy and Momentm Using the Newtonian definitions of energy and momentm, E = m and p= m, we can write: Now consider the relativistic definitions: E = γmc p= γm p = γ m pc mc mc 4 4 = γ = γ = γ mc c γ pc = γ mc mc 4 4 Bt So p E = m E = γmc pc = E mc 4 F HG Ths the eqivalent relationship between energy and momentm in Relativity is: E = p c + m c 4 or eqivalently m c 4 = E p c This is another example of Lorent Invariance. No matter what inertial frame is sed to compte the energy and momentm, E p c always given the rest energy of the object. Energy and momentm take the role of time and space in the other Lorent invariant qantity s. In fact, we refer to ( t, x, y, ) and ( E, px, py, p) as for-vectors, and the lengths of these vectors are these Lorent-invariant expressions we derived. Particles withot mass are a special case E = pc E and pc can also be written: E = γmc and pc = γ mc. The only way we can reconcile these last two definitions with E = pc is to set the velocity to c. Massless particles mst travel at the speed of light. As we will learn, light itself is composed of particles (photons). To travel at the speed of light, these particles mst be massless. I KJ D. Acosta Page 5 //5

6 PHY6 Enriched Physics Lectre Notes Relativity 4 The Electron-Volt Energy Unit qa The Lorent force law is F= E+ v Bf, where E is the electric field and B is the magnetic field. The work done to move a charged particle in an electric field only is: W = F ds= q E ds ( V) = qv The electric potential is φ (sch that the electric field work done by; W = q V V = potential difference E = V ). We can smmarie the Consider the work done to move an electron across a potential difference of Volt? 9 c ha f 9 J W =. 6 C V =. 6 This is a very small nit! We define it as a new nit of energy, the electron-volt: V e -- q=-.6 x -9 C ev =. 6 9 J Example: Express the electron rest mass energy in this new nit: E E c hc h 3 8 ev = mec = 9. kg 3. m / s -9.6 J = 5, ev (or 5 kev,.5 MeV,.5 GeV) We also can define new nits for mass and momentm. For example, the mass of the electron can be expressed m = 5 e. MeV / c. In other words, if yo mltiply the mass by c, yo get the rest energy in electron-volts. Similarly, we know that pc has nits of energy, so momentm can be expressed in nits like MeV / c. In other words, if yo mltiply by c, yo get an energy in electron-volts. D. Acosta Page 6 //5

7 PHY6 Enriched Physics Lectre Notes Relativity 4 Invariant Mass We can now apply the relativistic definitions of energy and momentm to calclations of particle collisions. In particlar, we can compte the rest mass of a particle formed when two particles annihilate into pre energy and then form a new particle. Example: An electron and a positron (an anti-electron) annihilate with eqal and opposite momentm: p = 55. GeV/c. (Note the new momentm nit). The collision + prodces a new particle called the J/ψ in the following reaction: e + e J /ψ. What is the mass of this new particle? We need to compte the invariant mass of the electron-positron initial state to determine the rest mass of the new particle: Mc = E p c where E and p are the total energy and momentm p = p + p = 55. GeV / c 55. GeV / c= by conservation of momentm E = E + E E = p c + m c = 55. GeV +. 5 GeV 55. GeV E tot tot = E tot tot tot tot 4 Mc = E = GeV = 3. GeV tot by conservation of energy b g b g becase the magnitde of the momentm (and mass) is the same The J/ψ particle has a mass of 3. GeV/c. Note that we have made extensive se of the new Lorent-invariant qantity involving energy and momentm D. Acosta Page 7 //5

8 PHY6 Enriched Physics Lectre Notes Relativity 4 Binding Energy As we have learned, mass is a form of potential energy. It can be converted into energy, or energy can be converted into mass. Becase of this, mass does not have to be conserved in reactions. If yo throw two balls at each other and they stick together (an inelastic collision), the reslting mass is not necessarily the sm of the individal masses of the two balls. This srprising reslt makes sense when we consider that mass is jst another form of potential energy. When two balls stick together, there mst be some attractive force holding the composite system together. In the case of the hydrogen atom, an electron and proton are bond by an attractive electromagnetic force. To separate the electron and proton (i.e. ionie hydrogen), one mst overcome the attractive force, and that takes energy. In other words, the particles have larger electromagnetic potential energy when separated than together. This potential energy is: e V = 4πε r which increases as the separation distance r increases. Where does this increase in energy go, since we know the total energy mst be conserved? It goes into the rest mass energies of the electron and proton in the case of hydrogen. Another way of ptting it is that the hydrogen atom has a mass that is less than the sm of the separate masses of the electron and proton. The difference in the rest mass energies of the separate objects from the combined system is called the binding energy: m b g b gr BE = M separate M bond c In the case of hydrogen, the binding energy is 3.6 ev; that is, hydrogen has a mass that is 3.6 ev less than the sm of the masses of the electron and proton. Let s consider another example. The deteron is a bond system of a netron and a proton. The binding energy is given by: o m BE = M n + M p M H c BE = MeV / c MeV / c MeV / c BE =. MeV bg bg c ht Clearly nclear binding energies are mch larger than atomic binding energies! We will explore this more when we stdy nclear physics toward the end of this corse. r D. Acosta Page 8 //5

9 PHY6 Enriched Physics Lectre Notes Relativity 4 Reaction Energy Closely related to binding energy is the concept of reaction energy. Not all composite systems have a mass less than the sm of its constitent masses, and some fndamental particles spontaneosly decay into particles whose combined mass is less than that of the parent. In these cases, energy is released in the decay or reaction becase of the difference in rest mass energies. We define this reaction energy as: m b g b gr Q= M initial prodcts M final prodcts c As yo can see, it is jst the negative of the binding energy. If Q is positive, we say that the reaction or decay is exothermic; that is, it releases energy. If Q is negative, the reaction or decay is endothermic; it takes energy to make it happen. Example: Consider the spontaneos decay of a netron: n p+ e + ν e. We can calclate the energy released in this decay by taking the difference in mass of the lefthand side from the right-hand side. The netrino ( ν e ) will be discssed later in the nclear and particle physics sections; what is relevant here is that its mass is essentially ero. o m Q= M n M p M e c Q= MeV / c MeV / c 5. MeV / c Q = 78. MeV b g b g c ht r c D. Acosta Page 9 //5