Physics 2203, Fall 2012 Modern Physics

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Physics 2203, Fall 2012 Modern Physics. Monday, Aug.

2: Rela@vis@c Momentum, Energy, Conversion of mass

Do classical laws of momentum and energy conserva@on

Announcements: I posted the in class exercise: my

1 Physics 2203, Fall 2012 Modern Physics. Monday, Aug. 27 th, 2011: Start Ch. 2: Momentum, Energy, Conversion of mass and Energy. Do classical laws of momentum and energy remain valid in Einstein s rela@vity? Announcements: I posted the in class exercise: my calcula@on of γ was wrong! Wednesday or Friday we will have a Quiz. Tutorial session 4:30 pm Tuesday in Nicholson 102 Issac may hit us on Wednesday!

2 Physics 2203 Modern Physics Class Ac5vity, Aug. 24 th, 2012 Two powerless rockets are on a collision course. The rockets are moving with speeds of 0.80c and 0.60c and are ini@ally 2.52x10 12 m apart (t=0) as measured by Liz, an Earth observer. Both rockets are 50m in length as measured by Liz. a) What are their respec@ve proper Lengths. b) What is the length of each rocket measured by an observer in the other rocket? c) According to Liz, how long before the rockets collide? d) According to rocket #1 how long before they collide? u ' x = u ' y = u x v 1 vu x γ u y 1 vu x u x = u y = u' x + v 1+ vu ' x γ u ' y 1 + vu ' x x' = γ ( x vt) y' = y z' = z x = γ ( x'+ vt ') y = y' z = z' u ' z γ u z 1 vu x u z γ u ' z 1+ vu ' x t ' = γ t vx t = γ t '+ vx'

3 Consider the collision of two equal mass objects shown in the figure. We will assume that p = mu p(before)=mv+m(-v)=0 p(after)=0 Momentum is conserved Now consider the S system moving with object #1 at a velocity of v. we need to find v 2 'and V' using the transform v 2 ' = v 2 v 1 v 2 v = V ' = V v 1 Vv = v v = 2v 1 v2 1 v2 0 v 1 ( 0)v = v u' x = u x v 1 vu x Look at momentum p', before and after p'(before) = mv 1 '+ mv 2 ' = 0 2mv p'(after) = 2mV ' = 2mv : Momentum not conserved 1 v2 What did we do wrong? How can we fit this?

What did we do wrong? How can we fit this? Lets back up and look at the defini@on of momentum!

4 What did we do wrong? How can we fit this? Lets back up and look at the of momentum! Classical definition p = m dr dt : This did not work for relativistic transformations u' x = u x v 1 vu x Lets be very careful about a proper measurement of time, t o p = m dr dt 0 p = m dr dt 0 : but dt 0 = dt γ p = m γ dr dt = mu 1 u2 Rela@vis@c momentum Classical momentum

5 Prove that using the new defini5on of momentum that momentum is conserved. Consider the collision of two equal mass objects shown in the figure. Now consider the S system moving with object #1 at a velocity of v. p = mu

If S is an iner@al frame and if a second frame S moves with constant

OR the laws of physics are invariant as we change from one reference frame

In all iner@al frames, light travels through the vacuum with the same speed,

6 If S is an iner@al frame and if a second frame S moves with constant velocity rela@ve to S, then S is also an iner@al frame. OR the laws of physics are invariant as we change from one reference frame to another iner@al frame. In all iner@al frames, light travels through the vacuum with the same speed, c=299,792,458 m/s in any direc@on. How does momentum transform? How can we conserve momentum? mu p = What will the relativistic Energy look like? 1 u2 m E =

7 If S is an iner@al frame and if a second frame S moves with constant velocity rela@ve to S, then S is also an iner@al frame. We must get classical limit! Momentum u<<<c? p x = mu x 1 u 2 x = mu x 1 + u 2 x 2 mu x Relativistic Energy for u<<<c? E = m =m 1+ u2 2 = mc2 + mu2 2 E = m

8 4 meter tall sculpture of Einstein s 1905 E=m Walk of Ideas, Berlin, Germany

9 p = mu 1 u2 = γ mu How should we think about this equa@on?? You will see this done many places. We will do this in subsequent problems. But this is not really consistent. One ahrac@ve way is to consider γm as the mass. This has the advantage that the equa@on for momentum looks like the classical equa@on. m 0 u p = Lets write the kinetic energy using this form of the mass. ( ) 2 E K = mv2 2 = p2 2m = γ m v 0 2γ m 0 = γ m 0 u = mu = γ m 0 v2 2 We will show that E K = m ( γ 1)

$Ques5on: For what value of u/c will the measured mass of an object γm exceed the rest mass by a given frac@on f?$

10 Ques5on: For what value of u/c will the measured mass of an object γm exceed the rest mass by a given frac@on f? Solving for u/c: f = γ m m 1 = γ 1 = 1 m 1 u2 c = 1 2 ( f + 1) : u2 2 c = ( f + 1) 2 u c = f ( f + 2) ) f = 1: γ m=2m f = 3: γ m=4m f + 1 u c = u c = We can generate the table at the lej. How do we find u/c?? Can u/c>1???? u c = 1+ 2 f 1+ 1 f 1+ 1 f 1 1 f = 1 2 f

11 Today we will show how to calculate v/c, but a 1.3 GeV electron accelerator produces electrons with v/c= : γ=2545 f = γm m m = γ 1 = = 2544!

12 f = γm m m = γ 1 = = 2544! Find γ for the 8 Gev Spring 8 in Japan.

13 Ques5on: A high speed interplanetary probe with a mass m=50,000 kg has been sent toward Pluto at a speed of u/c=0.8. (a) What is it momentum as measured by mission Control on Earth? (b) If, preparatory to landing on Pluto, the probe s speed is reduced to 0.4c, by how much does its momentum change? (c) How does this compare the classical answer? mu p = p x = mu x ( a) : p x = mu x ( b) : p x ( 0.4c) = ( )( 0.8c) = 2.0x10 13 kg m / s 1 ( 0.8) 2 = 50,000kg mu x 1 u2 ( b) : p x 0.4c 0.8c) p x ( ) ( ) = 0.33 ( c) : Classically ( ) ( 0.8c) ) = 0.5 p x 0.4c p x ( )( 0.4c) = 6.67x10 12 kg m / s 1 ( 0.4) 2 = 50,000kg

14 We will have two requirements for the form of the Energy 1) The total energy E of any isolated system is conserved. 2) E will approach the classical value when u/c approaches zero. Let us go back to our classical defini@on of F, p, and E. d p F = = d ( γ mu ) : This is the relativistic contribution dt dt ( ) u u Work = E K = F dx = d γ mu u dx = ud γ mu dt You show that d( γ mu ) = 0 mdu 0 0 ( ) u 3/2 3/2 m 1 u2 1 u du = m 0 1 u2 1 E K = m ( γ 1) = γ m m

15 E K = m ( γ 1) = γ m m m Rest Energy Mass and Energy interchangeable Make sure this works in classical limit E K = γ m m = 1 + u2 2 mc2 m E K mu2 2 OK We can now define the total energy E E = E K + m = γm = E = m 1 u2 mc2

16 p E We can derive another very useful equa@on = u β pc E p = mu 1 u2 E = m Square both equa@ons in yellow (p and E) ( pc) 2 = m2 u 2 E 2 = m2 c 4 1 u2 E 2 pc ( ) 2 1 = m 2 c 4 E 2 ( pc) 2 = ( m ) 2 E 2 = ( pc) 2 + m u2 / ( ) 2 If pc is much smaller than m, the Looks like Pythagorean rela@onship. energy is mostly rest energy

17 p E We can derive another very useful equa@on = u β pc E p = mu 1 u2 E = m Square both equa@ons in yellow (p and E) ( pc) 2 = m2 u 2 E 2 = m2 c 4 1 u2 E 2 ( pc) 2 = m 2 c 4 E 2 ( pc) 2 = ( m ) 2 E 2 = ( pc) 2 + ( m ) 2 If pc is much larger than m, the energy is mostly Kine@c.

18 E K = m ( γ 1) = γ m m Ques@on: What is the rest energy of an electron. We can think of m as the proper energy, measured at rest. E = γ m = m ( )( 3x10 8 m / s) 2 = 8.19x10 14 J m = 9.11x10 31 kg Terrible Units ( ) 1.0V 1.0eV = e( 1.0V ) = 1.602x10 19 C ( m (electron) = 8.19x10 15 J)eV 1.602x10 19 J m (electron) = 0.511MeV ( ) = 1.602x10 19 J = 5.11x10 5 ev E e = γm = 0.511MeV

19 E K = m ( γ 1) = γ m m Ques@on: What is the rest energy of a proton. We can think of m as the proper energy, measured at rest. E = γ m = m ( )( 3x10 8 m / s) 2 = 15.03x10 11 J m = 1.67x10 27 kg Terrible Units ( ) 1.0V 1.0eV = e( 1.0V ) = 1.602x10 19 C ( m (proton) = 15.03x10 11 J)eV 1.602x10 19 J m (proton) = 938 MeV ( ) = 1.602x10 19 J = 938 MeV E e = γm = 938 MeV

20 p = p mu = mu 1 u2 1 E 2 = ( pc) 2 + m ( ) 2 ( ) 2 = K + ( m ) K 2 + 2Km + ( m ) 2 = p 2 + ( m ) 2 p 2 2K = m + K 2 ( E = K K ) p/mv

21 E K = 1 1 u2 1 ( 0.511MeV ) K = p 2 + ( m ) 2 m

22 E K = ( γ 1) ( 0.511MeV ) Ques@on: What is the velocity, momentum, and total energy of an electron in the storage ring at CAMD? The machine energy is 1.3 GeV. 1.3 Gev is the Kinetic Energy, so 1.3Gev = γ 1 ( )( 0.511MeV ) γ 1 = 2545 : rest mass irrelevant γ = 2545 = 1 1 β 2 1 β 2 = 1.54x10 7 v = c Momentum E = 1.3Gev MeV pc = E 2 ( m ) 2 = GeV E(electron) = 0.511MeV Almost like a massless particle pc = E 2 ( m ) 2 = GeV pc E Mass mc = γm 0 = 2545m 0

23 Ques5on: Remember that we used muons produced by cosmic rays to illustrate both length They enter our atmosphere with v=0.998c. The rest energy of the muon is ~207 5mes that of an electron. (a) What will an observer on Earth measure as the total energy? (b) What will an observer measure as the mass of the muon? The rest Energy m is: ( ) = 105.7MeV 207x 0.511MeV β = 0.998c γ = 15.8 (a) E = γ m = 15.8( 105.7MeV ) = 1670MeV (b) measured m = γ m = E MeV = c Be very careful with this!

24 Conserva5on of mass energy Classically momentum is conserved in the total collision shown above, but the energy is not conserved. Before it is 2mv but ajer it is 0. Conservation of total enery mass-energy E i = m i c 2 conserved i 1 u i / c ( ) 2 Lets not assume that mass is conserved, only that E before = E after m i 2 ( ) + m i c 2 1 ( u i / c) = 2 Mc2 1 u i / c The lost Kine@c Energy is converted into mass!! Define ΔM = M 2m ΔM = 2m ( u i / c) 2 Lets not assume that mass is conserved only that E before = E after M = 2m 1 ( u i / c) : M>2m 2 ΔM = 2mc2 ΔM = 2K i ( 1 γ)

25 E = γ m = m = E 0 = E K + E 0 A simple experiment: two blocks of wood with equal mass m and kine@c energy K, are moving toward each other with velocity v. A spring placed between them is compressed and locks in place as they collide. Lets look at the conserva@on of mass energy Mass Energy before: E=2m + 2K Mass Energy after: E=M HW: What is the mass of a compressed spring? Since Energy is conserved we have E=2m + 2K=M M is greater than 2m because K went into mass f r = M 2m 2m = K m for real systems ~10 16 ΔM = M 2m = 2K

26 Examples of Energy to Mass Exchange that follow Chemical Binding Energy Fusion Fission

27 Binding Energy of the Hydrogen Atom: The binding energies of electrons to the nuclei of atoms are much smaller than nuclear binding energies. The binding energy for an electron to a proton (Bohr model) is 13.6 ev. How much mass is lost when an electron and proton from a hydrogen atom? This is again a before and ajer problem: Before you have an isolated electron and proton. ATer you have a H atom whose binding energy is 13.6 ev. Δm = E(binding) = 13.6eV m H m p = 938.3MeV Δm m H 1.4x10 8

28 (a) How much lighter is a molecule of water than two hydrogen atoms and an oxygen atom? The binding energy of water is ~3eV. (b) Find the frac@onal loss of mass per gram of water formed. (c) Find the total energy released (mainly as heat and light) when 1 gram of water is formed? (a) ΔM= ( m H + m H + m O ) M H2 O = E(binding) ( ) 1.6x10 19 J / ev ΔM= 3.0eV (b) M H2 O ΔM M H2 O ( ) = 5.3x10 ( 3.0x10 8 m / s) 36 kg Really small! 2 = E(binding) M H2 O c2 = 18u : u is 1/12 of the mass of Carbon (6 protron, 6 neutrons): Wednesday! ( ) M H2 O = x10 27 kg ΔM M H2 O 5.3x10 36 kg = x10 27 kg ( ) 2 = 1.8x10 10 S@ll small! (c) E=Δm = ( 1.8x10 10 )( 10 3 kg) ( 3x10 8 m / s) 2 = 16kJ Big!

29 Fusion: Energy is gained by taking two light atoms and combining them into an atom with a heavier nuclei later this semester. For example: 2 1H H = 4 2 He + Energy E c = mass 2 2 1H H E = ( ) mass( 4 2 He) ( ) MeV E = 23.9MeV Do you understand this nota@on? For He, there are 4 nucleons, and two are protons.

30 Fission Reac5ons: The decay of a heavy radioac@ve nucleus at rest into several lighter par@cles emihed with large kine@c energies is a great example of mass energy conversion. A nucleus of mass M undergoes fission into par@cles with masses M 1, M 2, and M 3, with speeds of u 1, u 2, and u 3. The conservation or Relativistic Energy Requires that M = M 1 1 u + M u + M u 2 3 This is a very important equation to remember Example in our text (pg 61) 232 Th 228 Ra + 4 He The offspring have 4 Mev Kinetic Energy The Equation above is true if ( ) M > M 1 + M 2 + M 3 Disintegration Energy Q defined ( ) Q= M M 1 + M 2 + M 3 Appendix D: 1 u=1.66x10 27 kg: Wednesday ( ) ( ) 1.7x10 27 kg ΔM = 0.004u u ΔM = 7x10 30 kg ΔM = 6.3x10 13 J = 4MeV

Chapter 2. Relativity 2

Chapter 2. Relativity 2 Chapter 2 Relativity 2 Acceleration transformation x = γ x vt t = γ t v x u x = u x v 1 vu x a x = u y = u y γ 1 vu x γ 3 a x 1 vu x 3 u z = u z γ 1 vu x F = m a?? Conservation of momentum p is conserved