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1 Welcome back to PHY 3305 Today s Lecture: Doppler Shift Velocity Transformation Momentum and Energy Christian Doppler

2 Announcements An additional office hour will be held each week on Fridays from 11 am - 12 pm. Reading assignment for this week is Chapter 2, sections 2.3 and 2.5. Homework Assignment 2 is due Tuesday, September 4. Regrade requests for Homework Assignment 1 are due Tuesday, September 4. You staple your original assignment to the back of your newly worked problems. Only rework problems that you want regraded. Problem 1 clarifications: c) According to the pole vaulter, which occurs first... d) What is the time interval between the events in part c?

3 Last Time: The Lorentz Transformations We can use γ to write our transformations. γ ν 1 1 v 2 c 2 Frame S: Frame S : x = γ ν (x vt) t = γ ν ( v c 2 x + t) x = γ ν (x + vt ) t = γ ν ( v c 2 x + t )

4 Last time Proper Time Δt 0 : Occurs in the frame where two events happen at the same location (i.e. x 1 = x 2 or x 1 = x 2 ). t = γ ν t 0 This is the shortest time interval of any measured intervals in any of the frames. Proper Length L 0 : Occurs in the frame where two events occur simultaneously (i.e. t 2 = t 1 or t 2 = t 1 ). L = L 0 γ ν This is the longest length any measured lengths in any of the frames.

5 Review Question 1 a) The event at y A always occurs first. b) The event at y A always occurs last. c) Either event might occur first, depending on spatial separation y B -y A of the events and the observer s speed v. d) The events remain simultaneous for all v. e) None of the above. Quiz # 11.2 Two events A and B occur simultaneously at separated points y A and y B, with y B > y A, on the y-axis of reference frame S. According to an observer S moving with respect to S with a speed v in the positive x-direction:

6 Review Question 1 a) The event at y A always occurs first. b) The event at y A always occurs last. c) Either event might occur first, depending on spatial separation y B -y A of the events and the observer s speed v. d) The events remain simultaneous for all v. e) None of the above. Quiz # 11.2 Two events A and B occur simultaneously at separated points y A and y B, with y B > y A, on the y-axis of reference frame S. According to an observer S moving with respect to S with a speed v in the positive x-direction:

7 Review Question 2 Molly flies her rocket past Nick at a constant velocity v. Molly and Nick both measure the time it takes the rocket from nose to tail, to pass Nick. Which of the following is true? A) Both Molly and Nick measure the same amount of time. B) Molly measures a shorter time interval than Nick. C) Nick measures a shorter time interval than Molly.

8 Review Question 2 Molly flies her rocket past Nick at a constant velocity v. Molly and Nick both measure the time it takes the rocket from nose to tail, to pass Nick. Which of the following is true? A) Both Molly and Nick measure the same amount of time. B) Molly measures a shorter time interval than Nick. C) Nick measures a shorter time interval than Molly. Nick measures proper time because the nose passes Nick event and the tail passes Nick event happen at the same location (x 1 = x 2 ). Proper time is the smallest interval between the two events.

9 What is the Doppler Effect?

10 What is the Doppler Effect? In classical physics sound experiences a shift in pitch (frequency shift) when the sound source is moving relative to the observer.

11 What is the Doppler Effect? In classical physics sound experiences a shift in pitch (frequency shift) when the sound source is moving relative to the observer. Classic example: Police Siren appears to go up in pitch as it approaches you and down in pitch as it goes away.

12 Doppler Shift in frequency The source (S ) is moving relative to the observer (S). t = t + v t c sound additional time to travel to observer.

13 Doppler Shift in frequency The source (S ) is moving relative to the observer (S). t = t + v t c sound f 1 t additional time to travel to observer.

14 Doppler Shift in frequency The source (S ) is moving relative to the observer (S). t = t + v t c sound f 1 t additional time to travel to observer. f = f (1 + v c sound ) -v = source moves towards listener +v = source moves away from listener

15 Doppler Shift in frequency The source (S ) is moving relative to the observer (S). t = t + v t c sound f 1 t additional time to travel to observer. f = f (1 + v c sound ) -v = source moves towards listener +v = source moves away from listener What if the source is moving at an angle to the listener? f = f (1 + v c sound cos θ) o v θ vcosθ

16 Does the same happen for light?

17 Does the same happen for light? The source (S ) is moving relative to the observer (S). Yes! But we must apply special relativity. t γ ν t f γ ν

18 Does the same happen for light? The source (S ) is moving relative to the observer (S). Yes! But we must apply special relativity. t γ ν t f f = f (1 + v c 1 v2 c 2 γ ν cos θ)

19 Does the same happen for light? The source (S ) is moving relative to the observer (S). Yes! But we must apply special relativity. t γ ν t f f = f (1 + v c 1 v2 c 2 γ ν cos θ) Assuming that the source is moving away/towards (not at an angle) this can be simplified (exercise for the student). away: f = f 1 v c 1+ v c towards: f = f 1+ v c 1 v c

20 Implications:

21 Implications: Classical Physics: - Light from a source moving away from an observer is redshifted. - Light from a source moving towards an observer is blueshifted.

22 Implications: Classical Physics: - Light from a source moving away from an observer is redshifted. - Light from a source moving towards an observer is blueshifted. Modern Physics: - All objects are redshifted because time moves more slowly (time dilation) for the moving source. The slowed movement lowers the light frequency. - Small effect which is outweighed by the movement of the source.

23 Problem: Running a Red Light You are driving towards a traffic light at 0.15c. A policeman next to the light sees it as red. What color does the light appear to you?

24 Problem: Running a Red Light You are driving towards a traffic light at 0.15c. A policeman next to the light sees it as red. What color does the light appear to you? Red light has a wavelength of 650nm. The first step is to convert this to frequency.

25 Problem: Running a Red Light You are driving towards a traffic light at 0.15c. A policeman next to the light sees it as red. What color does the light appear to you? Red light has a wavelength of 650nm. The first step is to convert this to frequency. f = c λ

26 Problem: Running a Red Light You are driving towards a traffic light at 0.15c. A policeman next to the light sees it as red. What color does the light appear to you? Red light has a wavelength of 650nm. The first step is to convert this to frequency. f = c λ = m s m

27 Problem: Running a Red Light You are driving towards a traffic light at 0.15c. A policeman next to the light sees it as red. What color does the light appear to you? Red light has a wavelength of 650nm. The first step is to convert this to frequency. f = c λ = m s m f = Hz

28 Now plug this into the appropriate Doppler equation. f = Hz

29 Now plug this into the appropriate Doppler equation. f = Hz f = f 1+ v c 1 v c

30 Now plug this into the appropriate Doppler equation. f = f 1+ v c 1 v c =( Hz) f = Hz c c c c

31 Now plug this into the appropriate Doppler equation. f = f 1+ v c 1 v c =( Hz) f = Hz c c c c f = Hz

32 Now plug this into the appropriate Doppler equation. f = f 1+ v c 1 v c =( Hz) f = Hz c c c c f = Hz Convert back to wavelength. λ = c f

33 Now plug this into the appropriate Doppler equation. f = f 1+ v c 1 v c =( Hz) f = Hz c c c c f = Hz Convert back to wavelength. λ = c f = m s Hz

34 Now plug this into the appropriate Doppler equation. f = f 1+ v c 1 v c =( Hz) f = Hz c c c c f = Hz Convert back to wavelength. λ = c f = m s Hz =535nm

35 Now plug this into the appropriate Doppler equation. f = f 1+ v c 1 v c =( Hz) f = Hz c c c c f = Hz Convert back to wavelength. λ = c f = m s Hz =535nm You will observe the light as green.

36 Notes: A modest speed of 0.15c = 4.5 x 10 7 m/s (~ 100 x 10 6 mph) causes a dramatic shift in the character of light. Although these high speeds are not so common in our daily life here on Earth, they are common in nature. The Doppler shift is an essential part of modern astrophysics.

37 Speed of Quasar Quasar SDSS produces a hydrogen emission line of wavelength λ rest = 121.6nm. On Earth, this emission line is observed to have a wavelength λ obs = nm. Find the speed of the recessing quasar.

38 Speed of Quasar Quasar SDSS produces a hydrogen emission line of wavelength λ rest = 121.6nm. On Earth, this emission line is observed to have a wavelength λ obs = nm. Find the speed of the recessing quasar. First we need to find the appropriate equation for a source which is moving away from the observer.

39 Speed of Quasar Quasar SDSS produces a hydrogen emission line of wavelength λ rest = 121.6nm. On Earth, this emission line is observed to have a wavelength λ obs = nm. Find the speed of the recessing quasar. First we need to find the appropriate equation for a source which is moving away from the observer. f = f 1 v c 1+ v c

40 Rewrite the Doppler equation in terms of wavelength f = f 1 v c 1+ v c f = c λ

41 Rewrite the Doppler equation in terms of wavelength f = f 1 v c 1+ v c f = c λ Thus, c λ = c λ 1 v c 1+ v c

42 Rewrite the Doppler equation in terms of wavelength f = f 1 v c 1+ v c f = c λ Thus, c λ = c λ 1 v c 1+ v c λ λ = 1 v c 1+ v c

43 Rewrite the Doppler equation in terms of wavelength f = f 1 v c 1+ v c f = c λ Thus, c λ = c λ 1 v c 1+ v c λ λ = 1 v c 1+ v c λ 2 λ 2 (1 + v c )=1 v c

44 Rewrite the Doppler equation in terms of wavelength f = f 1 v c 1+ v c f = c λ Thus, c λ = c λ 1 v c 1+ v c λ 2 λ 2 v c + v c =1 λ 2 λ 2 λ λ = 1 v c 1+ v c λ 2 λ 2 (1 + v c )=1 v c

45 Rewrite the Doppler equation in terms of wavelength f = f 1 v c 1+ v c f = c λ Thus, c λ = c λ 1 v c 1+ v c λ 2 λ 2 v c + v c =1 λ 2 λ 2 λ λ = 1 v c 1+ v c v c = 1 λ 2 λ 2 1+ λ 2 λ 2 λ 2 λ 2 (1 + v c )=1 v c

46 Rewrite the Doppler equation in terms of wavelength f = f 1 v c 1+ v c f = c λ Thus, c λ = c λ 1 v c 1+ v c λ 2 λ 2 v c + v c =1 λ 2 λ 2 λ λ = λ 2 1 v c 1+ v c λ 2 (1 + v c )=1 v c v c = 1 λ 2 λ 2 1+ λ 2 λ 2 v = c( 1 λ 2 λ 2 ) 1+ λ 2 λ 2

47 Plug in the numbers: v = c( 1 λ 2 λ 2 ) 1+ λ 2 λ 2 v = c( )

48 Plug in the numbers: v = c( 1 λ 2 λ 2 ) 1+ λ 2 λ 2 v = c( ) v =0.963c or v = m s

49 Redshift Parameter We know there is a relationship between the velocity at which a celestial object moves away from us and the change in the wavelength of the light it emits. Astronomers use the redshift parameter (z) to describe this change in wavelength. z λ obs λ λ If z > 0 the object is receding.

50 2011 Nobel Prize in Physics From the nobel prize website: The Nobel Prize in Physics 2011 was awarded "for the discovery of the accelerating expansion of the Universe through observations of distant supernovae" with one half to Saul Perlmutter and the other half jointly to Brian P. Schmidt and Adam G. Riess. For more information:

51 Saul Perlmutter: SMU Honorary Degree Recipient (2010) Perlmutter was cofounder of the Supernova Cosmology Project which devised methods in 1988 to use distant supernova to measure the expansion of the universe. Schmidt and Reiss where leaders of the High-Z Supernova research team which also used distant supernova to measure the expansion of the universe.

52 Cosmic Distances LIDAR: Objects close to us (moon) can be measured using LIDAR. Basic idea - bounce light off an object using a laser. Measure the round trip time of light to travel and then calculate the distance.

53 Cosmic Distances Geometry: For objects further away (nearby stars), we use trigonometric Parallax (triangulation).

54 Cosmic Distances Standard Candle: An object of known brightness that can be seen from very far away. Analogy: Lightbulb If we have a 40 W lightbulb, we know that there is a relationship between the distance between how bright the lightbulb appears and how far away from the lightbulb we are. Type 1a Supernova: Standard Candles that can be seen from very large distances.

55 Supernova (type 1a) - Binary systems where 1 of the 2 stars is a white dwarf. - The companion accretes material onto the white dwarf until the mass of the white dwarf is 1.4 times that of our sun. - At this mass the white dwarf collapses under its own gravity.

56 Scan the sky over multiple nights, looking for light to appear that wasn t there before.

57 Light curve - Intensity Over a Period of Time

58 Opportunity The universe is constantly expanding. So, as the light from a supernova explosion travels to us, the spacetime underneath it is expanding. As this space-time expands, it changes the wavelength (or color) of the light. redshift z = λ λ

59 Opportunity So, the relationship between measured brightness of the supernova and the redshift of the supernova can tell us about the expansion of the universe.

60 Spectral Lines Atoms give off waves of a certain frequency. Each atom in the universe gives off a unique set of colors. This set of colors is known as spectral lines. Spectroscopy is the science of using spectral lines to figure out what atoms an object contains. This technique is used to determine the composition of distant stars.

61 How to Measure Redshift 1. Supernova are dying stars which are composed of elements. Measure the spectral lines of the elements. Spectral lines are like color signatures of atoms or elements. ( 2. Identify which spectral line was created by which element. 3. Measure the wavelength shift of any one of the lines w.r.t. its expected wavelength as measured in a laboratory on Earth. 4. Use a formula to relate the observed shift to the velocity of the object. z = λ λ = λ 0 λ E λ E

62 Relative brightness fainter magnitude Perlmutter, Physics Today (2003) Supernova Cosmology Project High-Z Supernova Search Calan/Tololo Supernova Survey Accelerating Universe Type Ia Supernovae Blue line is best fit to the data Red lines are theoretical models with vacuum energy without vacuum energy 0 empty mass density 1 21 Decelerating Universe redshift Scale of the Universe [relative to today's scale]

63 Transforming Velocity y y S v Recall, that frame S moves with a velocity v with respect to frame S. A Galilean Transformation gives us r Frame S: u = u + v x Frame S : u = u v S x

64 Transforming Velocity Remember: u = dx dt x = γ ν (x + vt ) t = γ ν ( v c 2 x + t )

65 Transforming Velocity Remember: u = dx dt x = γ ν (x + vt ) t = γ ν ( v c 2 x + t ) From this we can derive the following relations. (Exercise for the student). u = u + v and u = u v 1+ vu c 2 1 vu c 2

66 Quick Check of 2nd Postulate: u = u v 1 vu c 2

67 Quick Check of 2nd Postulate: u = u v 1 vu c 2 Speed of light is the same in all reference frames.

68 Quick Check of 2nd Postulate: u = u v 1 vu c 2 Speed of light is the same in all reference frames. Light is traveling at the speed of light (c) in the u reference frame. Calculate the speed of light in the u reference frame?

69 Quick Check of 2nd Postulate: u = u v 1 vu c 2 Speed of light is the same in all reference frames. Light is traveling at the speed of light (c) in the u reference frame. Calculate the speed of light in the u reference frame? u = u v 1 vu c 2

70 Quick Check of 2nd Postulate: u = u v 1 vu c 2 Speed of light is the same in all reference frames. Light is traveling at the speed of light (c) in the u reference frame. Calculate the speed of light in the u reference frame? u = u v 1 vu c 2 = c v 1 vc c 2

71 Quick Check of 2nd Postulate: u = u v 1 vu c 2 Speed of light is the same in all reference frames. Light is traveling at the speed of light (c) in the u reference frame. Calculate the speed of light in the u reference frame? u = u v 1 vu c 2 = c v 1 vc c 2

72 Quick Check of 2nd Postulate: u = u v 1 vu c 2 Speed of light is the same in all reference frames. Light is traveling at the speed of light (c) in the u reference frame. Calculate the speed of light in the u reference frame? u = u v 1 vu c 2 = c v 1 vc c 2 = c(c v) c v

73 Quick Check of 2nd Postulate: u = u v 1 vu c 2 Speed of light is the same in all reference frames. Light is traveling at the speed of light (c) in the u reference frame. Calculate the speed of light in the u reference frame? u = u v 1 vu c 2 = c v 1 vc c 2 = c(c v) c v u = c Einstein s second postulate holds!

74 Momentum In classical physics, pi = p f Is momentum conserved in classical physics when you apply the Galilean Transformation?

75 Momentum In classical physics, pi = p f Is momentum conserved in classical physics when you apply the Galilean Transformation? YES!

76 Momentum Consider the collision of two objects of mass m 1 and m 2. In frame S: m 1 u 1i + m 2 u 2i = m 1 u 1f + m 2 u 2f In frame S : m 1 u 1i + m 2 u 2i = m 1 u 1f + m 2 u 2f

77 Momentum Consider the collision of two objects of mass m 1 and m 2. In frame S: In frame S : m 1 u 1i + m 2 u 2i = m 1 u 1f + m 2 u 2f m 1 u 1i + m 2 u 2i = m 1 u 1f + m 2 u 2f Use Galilean velocity transformation to rewrite u in terms of u. (u = u - v) m 1 (u 1i v)+m 2 (u 2i v) =m 1 (u 1f v)+m 2 (u 2f v)

78 Momentum Consider the collision of two objects of mass m 1 and m 2. In frame S: In frame S : m 1 u 1i + m 2 u 2i = m 1 u 1f + m 2 u 2f m 1 u 1i + m 2 u 2i = m 1 u 1f + m 2 u 2f Use Galilean velocity transformation to rewrite u in terms of u. (u = u - v) m 1 (u 1i v)+m 2 (u 2i v) =m 1 (u 1f v)+m 2 (u 2f v)

79 Momentum Consider the collision of two objects of mass m 1 and m 2. In frame S: In frame S : m 1 u 1i + m 2 u 2i = m 1 u 1f + m 2 u 2f m 1 u 1i + m 2 u 2i = m 1 u 1f + m 2 u 2f Use Galilean velocity transformation to rewrite u in terms of u. (u = u - v) m 1 (u 1i v)+m 2 (u 2i v) =m 1 (u 1f v)+m 2 (u 2f v)

80 Momentum Consider the collision of two objects of mass m 1 and m 2. In frame S: In frame S : m 1 u 1i + m 2 u 2i = m 1 u 1f + m 2 u 2f m 1 u 1i + m 2 u 2i = m 1 u 1f + m 2 u 2f Use Galilean velocity transformation to rewrite u in terms of u. (u = u - v) m 1 (u 1i v)+m 2 (u 2i v) =m 1 (u 1f v)+m 2 (u 2f v) m 1 u 1i + m 2 u 2i = m 1 u 1f + m 2 u 2f Momentum conservation is invariant under Galilean Transformation

81 What if we use the relativistic Lorentz Transformations instead?

82 What if we use the relativistic Lorentz Transformations instead? In frame S : m 1 u 1i + m 2 u 2i = m 1 u 1f + m 2 u 2f Use our velocity transformation u = u v 1 vu c 2

83 What if we use the relativistic Lorentz Transformations instead? In frame S : m 1 u 1i + m 2 u 2i = m 1 u 1f + m 2 u 2f Use our velocity transformation u = u v 1 vu c 2 m 1 ( u 1i 1 v )+m 2 ( u 2i v? u 1 2i v )=m 1 ( u 1f v u c 2 c 1 1f v )+m 2 ( u 2f v u 1 2f v 2 c 2 u 1i v c 2 )

84 What if we use the relativistic Lorentz Transformations instead? In frame S : m 1 u 1i + m 2 u 2i = m 1 u 1f + m 2 u 2f Use our velocity transformation u = u v 1 vu c 2 m 1 ( u 1i 1 v )+m 2 ( u 2i v? u 1 2i v )=m 1 ( u 1f v u c 2 c 1 1f v )+m 2 ( u 2f v u 1 2f v 2 c 2 u 1i v c 2 ) Momentum is not conserved!

85 If p = mu is wrong, what is the correct form?

86 If p = mu is wrong, what is the correct form? - We need to to reduce to p = mu at low speeds. (experimental evidence tells us so)

87 If p = mu is wrong, what is the correct form? - We need to to reduce to p = mu at low speeds. (experimental evidence tells us so) Do we know of a quantity which behaves this way, which we could multiply by mu?

88 If p = mu is wrong, what is the correct form? - We need to to reduce to p = mu at low speeds. (experimental evidence tells us so) Do we know of a quantity which behaves this way, which we could multiply by mu? γ u = 1 1 u2 c 2

89 If p = mu is wrong, what is the correct form? - We need to to reduce to p = mu at low speeds. (experimental evidence tells us so) Do we know of a quantity which behaves this way, which we could multiply by mu? γ u = 1 1 u2 c 2 So, we can postulate that p = γ u mu

90 Energy Einstein sought an expression for the total energy of a body in motion. Starting with conservation of energy, he made some arguments about what must happen as a consequence of relativity. He also knew that the energy of a body MUST take the form E = constant mu2 What he found was E 2 =(mc 2 ) 2 +(pc) 2

91 Examine Energy Postulates tell us the laws of physics are invariant for observers in relative motion. Classically: E 1b + E 2b = E 1a + E 2a p 1b + p 2b = p 1a + p 2a E = E(0) mu2 p = mu Where energy comes to us from definition of work. W = Fdx = ( dp dt )dx = m( du dt )(udt) = mudu = C mu2

92 Generalize energy and momentum in order to deduce the correct relativistic form: p = M(u) E = E(u)

93 Generalize energy and momentum in order to deduce the correct relativistic form: p = M(u) E = E(u) M & E are unknown functions of the motion of the relative frames. We do know that E & p should approach their classical values as u approaches 0. lim u 0 M(u) =mu lim u 0 E(u) u 2 = 1 2 mu2

94 Momentum: We already saw that this is straight forward. p = M(u) =γ u mu

95 Momentum: We already saw that this is straight forward. Energy: p = M(u) =γ u mu More tricky due to the nonlinear dependence on velocity of the body in the frame (u 2 ).

96 Momentum: We already saw that this is straight forward. Energy: p = M(u) =γ u mu More tricky due to the nonlinear dependence on velocity of the body in the frame (u 2 ). Use Binomial Expansion for γ ν. γ u = u 2 c 2 + O(u4 )

97 Remember - we want something that as u approaches 0 gives γ u = u 2 c 2 + O(u4 ) E = E(u) constant mu2 c 2

98 Remember - we want something that as u approaches 0 gives γ u = u 2 c 2 + O(u4 ) E = E(u) Try γ u. constant mu2 c 2 E = γ u E(0)

99 Remember - we want something that as u approaches 0 gives γ u = u 2 c 2 + O(u4 ) E = E(u) Try γ u. constant mu2 c 2 E = γ u E(0) In the low velocity limit lim u 0 E(u) E(0) E(0)u2 c 2

100 Remember - we want something that as u approaches 0 gives γ u = u 2 c 2 + O(u4 ) E = E(u) Try γ u. constant mu2 c 2 E = γ u E(0) In the low velocity limit lim u 0 E(u) E(0) E(0)u2 c 2 Which leads us to conclude E(0) = mc 2

101 E = mc 2 for an object at rest. This is remarkable. It says that the total energy of a body at rest can be described by its mass. When energy is given off by an object, its mass decreases correspondingly by m = E c 2

102 Example: Lightbulb A lightbulb radiates photons, which themselves have no mass, with a power of 75 Watts. How much mass is lost by the lightbulb in one year? 75W =75 J s

103 Example: Lightbulb A lightbulb radiates photons, which themselves have no mass, with a power of 75 Watts. How much mass is lost by the lightbulb in one year? 75W =75 J s In one year there are 3.2 x 10 7 s. Assuming that the light is left on all the time, the total energy radiated by the lightbulb is 75 J s s = J

104 Example: Lightbulb A lightbulb radiates photons, which themselves have no mass, with a power of 75 Watts. How much mass is lost by the lightbulb in one year? 75W =75 J s In one year there are 3.2 x 10 7 s. Assuming that the light is left on all the time, the total energy radiated by the lightbulb is 75 J s s = J Convert to mass -- m = E c 2 = J m ( s ) 2

105 Example: Lightbulb A lightbulb radiates photons, which themselves have no mass, with a power of 75 Watts. How much mass is lost by the lightbulb in one year? 75W =75 J s In one year there are 3.2 x 10 7 s. Assuming that the light is left on all the time, the total energy radiated by the lightbulb is 75 J s s = J Convert to mass -- m = E c 2 = J m ( s ) 2 m = kg

106 Last Detail E = γ u E(0) E(0) = mc 2 What if the object is moving? Is there a general equation we can use for an object in any inertial reference frame?

107 Last Detail What if the object is moving? Is there a general equation we can use for an object in any inertial reference frame? E 2 = γ 2 ue(0) 2 E = γ u E(0) E(0) = mc 2

108 Last Detail E = γ u E(0) E(0) = mc 2 What if the object is moving? Is there a general equation we can use for an object in any inertial reference frame? E 2 = γ 2 ue(0) 2 = γ 2 u(mc 2 ) 2

109 Last Detail What if the object is moving? Is there a general equation we can use for an object in any inertial reference frame? E 2 = γ 2 ue(0) 2 = γ 2 u(mc 2 ) 2 = 1 1 u2 c 2 (mc 2 ) 2 E = γ u E(0) E(0) = mc 2

110 Last Detail What if the object is moving? Is there a general equation we can use for an object in any inertial reference frame? E 2 = γ 2 ue(0) 2 = γ 2 u(mc 2 ) 2 = 1 1 u2 c 2 (mc 2 ) 2 E = γ u E(0) E(0) = mc 2 Use the Binomial Expansion of 1/(1-x 2 ) to expand (x = u/c). E(u) 2 =(mc 2 ) 2 (1 + u2 c 2 + u4 c )

111 Last Detail What if the object is moving? Is there a general equation we can use for an object in any inertial reference frame? E 2 = γ 2 ue(0) 2 = γ 2 u(mc 2 ) 2 = 1 1 u2 c 2 (mc 2 ) 2 E = γ u E(0) E(0) = mc 2 Use the Binomial Expansion of 1/(1-x 2 ) to expand (x = u/c). E(u) 2 =(mc 2 ) 2 (1 + u2 c 2 + u4 c ) E(u) 2 = m 2 c 4 + m 2 u 2 c 2 + m 2 u 4 + m 2 u 6 ( 1 c 2 )+...

112 Last Detail What if the object is moving? Is there a general equation we can use for an object in any inertial reference frame? E 2 = γ 2 ue(0) 2 = γ 2 u(mc 2 ) 2 = 1 1 u2 c 2 (mc 2 ) 2 E = γ u E(0) E(0) = mc 2 Use the Binomial Expansion of 1/(1-x 2 ) to expand (x = u/c). E(u) 2 =(mc 2 ) 2 (1 + u2 c 2 + u4 c ) E(u) 2 = m 2 c 4 + m 2 u 2 c 2 + m 2 u 4 + m 2 u 6 ( 1 c 2 )+... E(u) 2 =(mc 2 ) 2 + m 2 u 2 c 2 (1 + u2 c 2 + u4 c )

113 Last Detail What if the object is moving? Is there a general equation we can use for an object in any inertial reference frame? E 2 = γ 2 ue(0) 2 = γ 2 u(mc 2 ) 2 = 1 1 u2 c 2 (mc 2 ) 2 E = γ u E(0) E(0) = mc 2 Use the Binomial Expansion of 1/(1-x 2 ) to expand (x = u/c). E(u) 2 =(mc 2 ) 2 (1 + u2 c 2 + u4 c ) E(u) 2 = m 2 c 4 + m 2 u 2 c 2 + m 2 u 4 + m 2 u 6 ( 1 c 2 )+... E(u) 2 =(mc 2 ) 2 + m 2 u 2 c 2 (1 + u2 c 2 + u4 E(u) 2 =(mc 2 ) 2 +(γ u mu) 2 c 2 c )

114 Last Detail What if the object is moving? Is there a general equation we can use for an object in any inertial reference frame? E 2 = γ 2 ue(0) 2 = γ 2 u(mc 2 ) 2 = 1 1 u2 c 2 (mc 2 ) 2 E = γ u E(0) E(0) = mc 2 Use the Binomial Expansion of 1/(1-x 2 ) to expand (x = u/c). E(u) 2 =(mc 2 ) 2 (1 + u2 c 2 + u4 c ) E(u) 2 = m 2 c 4 + m 2 u 2 c 2 + m 2 u 4 + m 2 u 6 ( 1 c 2 )+... E(u) 2 =(mc 2 ) 2 + m 2 u 2 c 2 (1 + u2 c 2 + u4 c ) E(u) 2 =(mc 2 ) 2 +(γ u mu) 2 c 2 =(mc 2 ) 2 + p 2 c 2

115 As a final step, redefine E(u) E and then E 2 = m 2 c 4 + p 2 c 2 This equation describes the total energy of any inertial reference frame.

116 consequences If the object is stationary, this simplifies to E = mc2. - This implies that if you heat an object, (i.e. it gains internal energy through an increase in T, but NOT KE through motion), the mass of the object increases. What if the mass of an object is zero? Is that allowed? - When mass is zero, E = pc = γ umu. The only way the energy of such an object is non-zero is if u = c and γ u = infinity. Light is a massless particle. (Further discussions of momentum and energy of photons later in the course.)

117 The End (for today)