Introduction to Relativistic Mechanics and the Concept of Mass

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1 Introdution to Relatiisti Mehanis and the Conept of Mass Gron Tudor Jones Uniersity of Birmingham CRN HST014

2 Introdution to relatiisti kinematis and the onept of mass Mass is one of the most fundamental onepts in physis. When a new partile is disoered (e.g. the Higgs boson), the first question physiists will ask is, What is its mass? Classial physis ( << ) T = m / m = T/ p = m m = p/ T = p /m m = p /T Knowing any of T, p and, one an alulate m. Same is true in relatiity but we need the generalised formulae. Before that: a brief disussion of = m

3 instein s equation: = m 0 = m = m 0 = m 0 = m 0 where = eloity of light in auo = total energy of free body 0 = rest energy of free body m 0 = rest mass m = mass Q1: Whih equation most rationally follows from speial relatiity and expresses one of its main onsequenes and preditions? Q: Whih of these equations was first written by instein and was onsidered by him a onsequene of speial relatiity?

4 The orret answer to both questions is: 0 = m (Poll arried out by Le Okun among professional physiists in 1980s showed that the majority preferred or 3 as the answer to both questions.) This hoie is aused by the onfusing terminology widely used in the popular siene literature and in many textbooks. Aording to this terminology a body at rest has a proper mass or rest mass m 0, whereas a body moing with speed has a relatiisti mass or mass m, gien by m m 0 1 this terminology had some historial justifiation at the start of our entury, but it has no justifiation today. Today, partile physiists only use the term mass. Aording to this rational terminology the terms rest mass and relatiisti mass are redundant and misleading.

5 There is only one mass in physis, m, whih does not depend on the referene frame. As soon as you rejet the relatiisti mass there is no need to all the other mass rest mass and to mark it with a subsript 0.

6 Letter from Albert instein to Linoln Barnett, 19 June 1948 It is not good to introdue the onept of mass m moing body for whih no lear definition an be gien. m 0 1 of a It is better to introdue no other mass onept than the rest mass m. Instead of introduing m, it is better to mention the expression for the momentum and energy of a body in motion.

7 The two fundamental equations of relatiisti kinematis (Relatiisti generalisations of = p /m and p = m.) Conseration of energy and momentum are lose to the heart of physis. Disuss how they are related to deep symmetries of nature. All this is looked after in speial relatiity if we define energy and momentum as follows: = p + m 4 and p = where = total energy p = momentum = eloity m = ordinary mass as in Newtonian mehanis Next: hope to persuade you to aept these equations.

8 * 0 = m Consider p = m 4. For the situation when the partile is at rest ( = o), the energy is the rest energy 0 and p = 0. So, 0 = m * Show that, for <<, = m + p /m = m + m / Firstly, when <<, p 0 = m. Also, = (p + m 4 ) 1/ = m (1 + p /m 4 ) 1/ = m (1 + p /m 4 + ) For <<, p << m 4. So, = m + p /m = 0 + m / Rest energy Newtonian kineti energy The relatiisti equations for p and redue to the Newtonian ones for <<; so the m in them is the Newtonian mass.

9 * Consider the extreme anti-newtonian limit where m = 0 If m=0, then p p p = Suh bodies hae no rest frame; they always moe with the speed of light. Also m = 0 = p Massless bodies hae no rest energy, just K. (e.g. photon.graiton) Our two expressions for p and desribe the kinematis of a free body for all eloities from 0 to ; and also 0 = m follows from them.

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11 Bak-up slides on relatiisti kinematis

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13 4 4 4 m m p 1 m m k k NWTONIAN 1 1 m m m k RLATIVISTIC Speed and Kineti nergy for Relatiisti letrons by William Bertozzi (Amerian Journal of Physis 3 (1964) )

14 Speed of 7 TeV proton m Substitute p = / into = p + m 4 to get 1 For = 7000 GeV proton (m = GeV) =p (approx); i.e. speed is exeedingly lose to. So (1 / ) = (1 - /)(1 + /) = (1 /) to high auray Hene (1-/) = (m ) / = /( x 7000 ) = 0.9 x And =

15 Inariant ( effetie ) mass of two photons A 1,p 1,p m 1 4 = ( 1 + ) - (p 1 + p ) 4 = p 1 p - p 1 p Now, for photons, = p, so p = 0, and m 1 4 = p 1 p (1 os A) Consider A = 0 and A = 180 o