Option H: Relativity H7 General relativity

1. Option H: RelativityH7 General relativity The equivalence principle H.7.1 Explain the difference between the terms gravitational mass and inertial mass. H.7.2 Describe and discuss Einstein’s principle of equivalence. H.7.3 Deduce that the principle of equivalence predicts the bending of light rays in a gravitational field. H.7.4 Deduce that the principle of equivalence predicts that time slows down near a massive body.

2. 9.8 ms-2 Option H: RelativityH7 General relativity The equivalence principle Explain the difference between the terms gravitational mass and inertial mass. ●Consider an experiment which uses two identical closed rooms: one on Earth and the other in a spaceship accelerating at 9.8 m/s2 in interstellar space (no gravitation…).

3. 9.8 ms-2 Option H: RelativityH7 General relativity The equivalence principle Explain the difference between the terms gravitational mass and inertial mass. ●Dobson is in the room and cannot see outside. ●He throws up the ball and analyzes its trajectory. ●Whether in the spaceship, or on Earth, the trajectories are identical. y = yo + vot - 0.5gt2

4. 9.8 ms-2 Option H: RelativityH7 General relativity The equivalence principle Explain the difference between the terms gravitational mass and inertial mass. ●On Earth, the net force is given by F = GMm/r2. ●In the spacecraft the apparent force is given by F = ma, where a = g. ●The m in F = mais called the inertial mass. In this formula, it represents the object’s “resistance” to accelera- tion (AKA inertia). ●The m in F = GMm/r2is called the gravitational mass. In this formula it represents the mutual attraction between the mass of Earth M and the object m. Gravitational mass Inertial mass

5. Option H: RelativityH7 General relativity The equivalence principle Describe and discuss Einstein’s principle of equivalence. ●We have often assumed that the two masses were the same. EXAMPLE: Show that the acceleration of a freely-falling body a distance r from a planet of mass M is given by a = GM/r2. SOLUTION: We use Newton’s laws for this: ●From Newton’s law of gravity: F = GMm/r2. ●From Newton’s law of motion: F = ma. ●Equating the two: ma = GMm/r2 a = GM/r2. gravitational mass inertial mass FYI Note that in the last step we just assumed that the gravitational and inertial masses were the same, and we cancelled them out.

6. Option H: RelativityH7 General relativity The equivalence principle Describe and discuss Einstein’s principle of equivalence. ●When formulating his theory of general relativity, Einstein elevated these observations to a principle of physics: ●The equivalence principle states that “No experiment can be conducted which will determine whether you are in a gravitational field or an accelerating reference frame.” ●The equivalence principle can be stated in other ways: “A frame of reference moving at constant velocity (a = 0) far from all masses is indistinguishable from a freely falling frame of reference in a uniform gravitational field.” ●Yet another statement is that “Gravitational and inertial effects are indistinguishable.”

7. Option H: RelativityH7 General relativity The equivalence principle Deduce that the principle of equivalence predicts the bending of light rays in a gravitational field. ●Consider a single photon of light passing from left to right as Dobson accelerates in the direction shown: ●Dobson observed a BENDING light beam… ●Because of the principle of equivalence, he can’t tell the difference between an accelerating reference frame and a gravitational field. ●Thus he may infer that a gravitational field bends light beams! Path of photon

8. Option H: RelativityH7 General relativity The equivalence principle Deduce that the principle of equivalence predicts that time slows down near a massive body. ●Recall the Twin Paradox where we discovered that time dilated only for the twin that accelerated. ●The equivalence principle then predicts that time dilation will occur in a strong gravitational field, which cannot be distinguished from an acceleration. Aging is slower near a large mass like a black hole!

9. Option H: RelativityH7 General relativity Spacetime H.7.5 Describe the concept of spacetime. H.7.6 State that moving objects follow the shortest path between two points in spacetime. H.7.7 Explain gravitational attraction in terms of warping of spacetime by matter.

10. distance in space d2 = x2 + y2 + z2 “I really wouldn't have thought that lazy dog Einstein capable of relativity.” Option H: RelativityH7 General relativity Spacetime Describe the concept of spacetime. ●In 1908, after Einstein’s special theory was published, Hermann Minkowski proposed the concept of spacetime, wherein a fourth dimension of proper time ct was added to the usual three spatial dimensions x, y, and z. ●Recall from geometry that the distance d from the origin to a point P having coordinates (x, y, z) is given by FYI It turns out that the distance to a point varies depending on the frames of reference, because of length contraction and time dilation.

11. interval in spacetime I2 = x2 + y2 + z2 – (ct)2 Option H: RelativityH7 General relativity Spacetime Describe the concept of spacetime. ●Minkowski defined an intervalI having the form ●It turns out that the interval I is invariant, meaning the I is the same regardless of the frame of reference even with relativistic effects. ●Einstein at first was not impressed with Minkowski's idea, calling it "banal" and "a superfluous learnedness." ●Eventually, Einstein used Minkowski’s spacetime view of the geometry of the universe in his general theory of relativity (the relativity of non-inertial reference frames). FYI Note that the time dimension is “ct” instead of c so its dimension is in meters (speed  time).

12. (x,y,z,ct1) (x,y,z,ct2) Option H: RelativityH7 General relativity Spacetime Describe the concept of spacetime. (x,y,z) EXAMPLE: As an example of the difference between a spacetime diagram and a traditional space diagram, contrast the plots of a particle in uniform circular motion in the x-y plane. SOLUTION: ●In the traditional diagram note that the particle keeps repeating its coordinates. ●In the spacetime diagram the particle never repeats its coordinates. It will repeat spatial coordinates as regularly as the in the traditional diagram, but it will never repeat its time coordinate. ●This is more in keeping with how things really are. Think of crossing a busy street- same spatial coordinates, different times! space diagram spacetime diagram

13. Option H: RelativityH7 General relativity Spacetime State that moving objects follow the shortest path between two points in spacetime. ●It turns out that in general relativity not only is spacetime invariant, but objects always follow the shortest paths between two points. ●In other words, objects have a propensity to follow paths through the world in such a way that the I in I2 =x2+y2+z2 –(ct)2 is always minimized. FYI The object simply “follows” the spacetime curvature without “thought”.

14. Option H: RelativityH7 General relativity Spacetime Explain gravitational attraction in terms of warping of spacetime by matter. ●Einstein termed the natural paths followed due to the curvature of spacetime world lines. ●As we learned when we discussed fields, a force like gravity can be either caused by action at a distance (F = GMm/r2) or by the gravitational field (g = GM/r2). ●The spacetime curvature (field) eliminates the need for a planet to “know” where the sun is and to constantly “exchange” coordinate information with it, in order to know which way to orbit. ●Rather, the planet orbits simply by reacting to the field that is the curvature of spacetime. It follow its local world line without the need to know where the source of curvature comes from.

15. Option H: RelativityH7 General relativity Spacetime Explain gravitational attraction in terms of warping of spacetime by matter. ●According to general relativity, the sun curves the spacetime surrounding it, as shown: ●Each planet then moves in response to its local closed world line. The result is orbital motion without action at a distance.

16. Option H: RelativityH7 General relativity Spacetime Describe the concept of spacetime. ct x ●One axis is ct and one is x: ●Far from mass the particle will move in a straight line.

17. Option H: RelativityH7 General relativity Spacetime Describe the concept of spacetime. ct x ●Close to a large mass the particle will move in a curved line, responding to the local curvature of spacetime.

18. Option H: RelativityH7 General relativity Spacetime Describe the concept of spacetime. ●Close to a large mass like Earth the spacetime in its vicinity will become curved. ●All particles follow the shortest path in spacetime. ●The satellite therefore follows the local curvature of spacetime surrounding Earth. ●Orbits are just closed curves in spacetime.

19. Option H: RelativityH7 General relativity Black holes H.7.8 Describe black holes. H.7.9 Define the term Schwarzschild radius. H.7.10 Calculate the Schwarzschild radius. H.7.11 Solve problems involving time dilation close to a black hole.

20. Option H: RelativityH7 General relativity Black holes Describe black holes. ●A black hole is any mass for which the escape velocity exceeds the speed of light c. ●Since nothing can exceed c, not even light can escape from a black hole – hence the name. ●It is thought that in the center of our Milky Way galaxy is a black hole.

22. Option H: RelativityH7 General relativity Black holes Define the term Schwarzschild radius. ●The Schwarzschild radius is that radius which a mass would have to be compressed in order for it to become a black hole. EXAMPLE: For example, if Earth were compressed such that its radius became about 9 mm, light could no longer escape from its surface and it would become a black hole. ●Any mass can theoretically become a black hole if it is compressed enough. ●There is even talk of creating microscopic black holes in the CERN collider!

23. 0 0 Option H: RelativityH7 General relativity Black holes Calculate the Schwarzschild radius. ●The method we use here was used by Laplace in the 18th century, and strictly speaking, has flaws. The final formula is still correct. ●Consider a mass m which is located on a planet of mass M and radius R. We wish to give m an initial kinetic energy sufficient to allow it to just reach infinity and stop. ●From conservation of energy we have EK0 + EP0 = EK + EP mvesc2/2 – GMm/R = m(0)2/2 – GMm/ vesc2 = 2GM/R. ●Note that there are two ways to make vesc big: We can increase M or decrease R.

24. Schwarzschild radius RS = 2GM/c2 Option H: RelativityH7 General relativity Black holes Calculate the Schwarzschild radius. ●We call the radius RS at which a mass M becomes a black hole (i.e.: vesc = c) the Schwarzchild radius. ●Thus vesc2 = 2GM/R becomes c2 = 2GM/RS and we get the following relationship: PRACTICE: Show that the Schwarzschild radius of Earth, whose mass is 5.981024 kg, is 8.88 mm. SOLUTION:RS = 2GM/c2 = 2(6.6710-11)(5.981024)/(3.00108)2 = 0.00888 m = 8.88 mm. FYI RS is also called the event horizon. It is not really a physical radius of a solid object.

25. Option H: RelativityH7 General relativity Black holes Calculate the Schwarzschild radius. PRACTICE: Observe the moon in orbit about Earth. What do you predict will happen to the moon if Earth is shrunk into a black hole? SOLUTION: ●Just because Earth shrinks doesn’t mean its mass changes (its density does). ●Thus the moon maintains its orbit as usual. Its world line does not change! FYI Indeed, this is how a black hole could be detected: A visible object orbiting an invisible one!

26. time dilation near a black hole t0 t = RS r 1 - Option H: RelativityH7 General relativity Black holes Solve problems involving time dilation close to a black hole. ●As stated previously, due to the equivalence principle, time is dilated in the presence of massive objects, such as black holes. ●Without proof, here is the time dilation formula for a distance r from the center of a black hole: FYI Note that the formula has no meaning if r  RS. As r  RS, the event horizon, we see that t  , or time comes to a halt! Time as we know it ceases to exist within the event horizon!

27. time dilation near a black hole t0 t = RS r 1 - Option H: RelativityH7 General relativity Black holes Solve problems involving time dilation close to a black hole. EXAMPLE: Suppose Earth is shrunk to 0.00888 m to become a black hole. An observer from far away observes a miniature astronaut, in orbit about Earth at r = 0.0100 m, cook a 3.00-minute egg. How much time elapses for the observer? SOLUTION: t= t0(1 – RS/r)-1/2 = 3(1 – 0.00888/0.0100)-1/2 = 8.96 min.

28. Option H: RelativityH7 General relativity Gravitational red shift H.7.12 Describe the concept of gravitational red shift. H.7.13 Solve problems involving frequency shifts between different points in a uniform gravitational field.

29. a = 0 Option H: RelativityH7 General relativity Gravitational red shift Describe the concept of gravitational red shift. ●Consider Dobson, at A, and his brother Nosbod, at B, in a rocket traveling at a constant velocity. ●The brothers each have a radar gun which they use to measure the frequency of the light as it reaches them. ●Recalling that light travels at precisely c regardless of the speed of the source, both brothers move at the same speed away from the light, and thus both measure exactly the same red shift. ●Thus at constant velocity fA= fB < f. (a = 0).

30. a = g Option H: RelativityH7 General relativity Gravitational red shift Describe the concept of gravitational red shift. ●Now suppose that the rocket accelerates at a = g. ●Because the brothers are a distance h apart, by the time the light reaches Nosbod at B he is going faster than Dobson was when the light reached him at A. ●In fact, vB2 = vA2 + 2gh shows that vB > vA. ●Thus the red shift at B is more pronounced than the red shift at A. ●Finally, we can deduce that if the rocket is accelerating then fB< fA < f. (a = g). vA vB

31. gravitational red shift f/f = gh/c2 Option H: RelativityH7 General relativity Gravitational red shift Describe the concept of gravitational red shift. ●The previous explanation was for an accelerating reference frame. ●The principle of equivalence of course tells us that the same frequency shift between the brothers will occur if the rocket is stationary and in a gravitational field having a strength g. ●The formula relating the frequency shift f to the distance between the observers h follows: FYI This formula gives the frequency shift as measured by two separated observers in a gravitational field. Don’t mix it up with the Doppler effect from Topic 11.2 (f/f = v/c).

32. Option H: RelativityH7 General relativity Gravitational red shift Solve problems involving frequency shifts between different points in a uniform gravitational field. EXAMPLE: The photons used in the Pound-Rebka experiment at Harvard had 14.4 keV energy, and the building was 22.6 m high. Calculate the change in frequency between the top and bottom. SOLUTION: ●E = (14.4103 eV)(1.6010-19 J/eV) = 2.3010-15 J. ●From E = hf we get f = E/h = 2.310-15/6.6310-34 = 3.481018 Hz. ●Then from f/f = gh/c2we get f = fgh/c2 = (3.481018)(9.81)(22.6)/(3.00108)2 = 8560 Hz.

33. Option H: RelativityH7 General relativity Solve problems involving frequency shifts between different points in a gravitational field. ●A red shift is a lower frequency. ●Since the ship is accelerating, by the time the light reaches Elspeth she will be going faster than Alex was when light reached him. ●Elspeth’s red shift is bigger than Alex’s.

34. Option H: RelativityH7 General relativity Solve problems involving frequency shifts between different points in a gravitational field. ●Gravitational red shift will thus occur. ●From the equivalence principle an accelerating reference frame (the rocket) is indistinguishable from a reference frame at rest in a gravitational field.