Special Relativity: The laws of physics must be

Special Relativity: The laws of physics must be

Special Relativity: The laws of physics must be the same in all inertial reference frames. Inertial Reference Frame: One in which an object is observed to have zero acceleration when no forces act on it [i.e. a reference frame that moves at constant velocity with respect to the distant stars]. What is so special about the distant stars? General Relativity: The laws of physics must be the same in ALL reference frames. The connection between reference frames that are accelerating with respect to each other involves gravity. Notice that the concept of mass enters Newtonian physics in two different ways, that seem unrelated: Inertial mass: F = ma: The mass of an object is that property which resists changes in its motion. Gravitational mass: The gravitational force between two objects Fgrav = -GMm r /r2

or more generally, the gravitational force on an object in a gravitational field g: Fgrav = mg e.g. outside a spherical mass M, g = - GMr/r2. e.g. near the surface of the earth: g = constant (g = -9.8 m/s2, down) Inertial mass: F = ma Gravitational mass: Fgrav = mg The fact that the inertial mass = gravitational mass is called the equivalence principle and is a coincidence from the point of view of Newtonian physics but the underlying principle of general relativity. It has the important consequence that all objects accelerate together if the only force that is acting is gravity, i.e. a = g for all objects v r

In satellite with speed v distance r from earth, a = v2/r = GM/r2 for all objects a = 9.8 m/s2, down for apple and diver Being in a freely falling reference frame Being in an inertial reference frame weightless In both cases, no mechanics experiment you can do that says that you are moving. Consider an observer at rest in a gravitational field. He observes that objects (that arent held up by an additional force) fall down with acceleration = g.

Consider the following situation in zero gravity: A man standing in a car which is accelerating UP with acceleration --g, lets go of his briefcase. An observer outside the car sees the man accelerating up at g = +gj and the briefcase as stationary. However, the man in the car thinks hes at rest and that the briefcase falls

down with acceleration = g. Consider an observer at rest in a gravitational field. He observes that objects (that arent held up by an additional force) fall down with acceleration = g. The man in the car thinks hes at rest and that the briefcase falls down with acceleration = g. Both men have exactly the same observation: they both think that

they are at rest and that the briefcase is accelerating down with acceleration = g. Because of the equivalence principle, this is true for any (mechanics) experiment the men could do: No mechanical experiment can distinguish between being at rest in an accelerating frame and being at rest in a gravitational field.

Compare the following situations: suppose you (with mass m) are in a closed box with no windows: 1) If the car is accelerating to the right with acceleration a, you will feel the left wall press against you with force F = ma. 2) If the car is stationary (or moving at constant velocity) but there is a gravitational field toward the left with g(now) = a(previous), you will be feel the left wall press against you with force F = mg = ma. If you cannot look outside, there is no way for you to distinguish the two situations (i.e. no (mechanical) experiment which can tell them apart: they are (mechanically) equivalent: being at rest in an accelerating frame is equivalent to being at rest in a gravitational field. Being in an accelerating frame is equivalent to being in a gravitational field (possibly a very complicated one). N Consider a coordinate system that is

rotating with angular velocity . The acceleration of a point at a distance r from the origin a = 2r, toward center. So an observer can stay at rest in this frame by standing on the rim i.e. the rim provides a normal force in just like the floor in a room does to balance gravity. Therefore being at rest in a rotating frame is equivalent to being at rest in a gravitational field which points radially out. N No mechanical experiment can distinguish between being at rest in an accelerating frame and being at rest in a gravitational field. Einstein extended this: No experiment can distinguish between being at rest in an

accelerating frame and being at rest in a gravitational field. Being in a freely falling frame Being at rest in a gravitational field Consider a ray of light: Being in an inertial frame Being at rest in an accelerating frame Although general relativity was formulated as a way to consider accelerating reference frames, it has also changed the way we think about gravity ! Since everything reacts the same way to a gravitational field, Einstein proposed that instead of considering the gravitational field as a force field, consider it as a distortion of space-time that affects everything. For example, the moon revolves around the earth because it is

moving in space-time that is warped by the presence of mass (the earth). Recall that the, the interval ds2 = dx2 + dy2 + dz2 c2dt2 dx12 + dx22 + dx32 + dx42 is invariant under the Lorentz-transformation, i.e. has the same value in all inertial frames. In more general, ds2 = ij dxi dxj , where the metric ij may vary with position (and time) and is a function of the local gravitational field (i.e. including field that is present because of acceleration of reference frame). Note: this means that grid lines may no longer be perpendicular. Being at rest in an accelerating frame is equivalent to being at rest in a gravitational field (possibly a very complicated one). Consider a coordinate system that is rotating with angular velocity . The acceleration at a distance r from the origin a = 2r, toward center. So an observer in this frame, will fall out as if there is a gravitational pull to the outside.

Consider an element of length tangent to the surface. Since it is parallel to the motion , its proper length (i.e.as measured by the rotating observer) will be longer than its value measured by an external (non rotating) observer: the circumference of the paths will be different as measured by the two observers. However, the radius is perpendicular to the motion so does not Lorentz contract. Therefore, if one observer measures the circumference = 2r, the other does not! This is an example of how gravity warps space. Being at rest in an accelerating frame is equivalent to being at rest in a gravitational field (possibly a very complicated one). Consider a coordinate system that is rotating with angular velocity . The acceleration at a distance r from the origin a

= 2r, toward center. So an observer in this frame, will fall out as if there is a gravitational pull to the outside. Consider a clock in the rotating frame, since it is moving with respect to a stationary observer, its clock will be slower. However, since the speed is proportional to the distance from the center, the clocks rate will depend on how far it is from the center. This is an example of how gravity affects time. [These corrections are needed for the accuracy of GPS systems.] Some Consequences of General Relativity: 1) The gravitational force is not exactly proportional to 1/r2 (most noticeable close to massive objects). Einstein correctly calculated the precession of the perihelion of Mercury.

2) Light bends in a gravitational field gravitational lens. 3) Gravity waves (reported for the first time with the LIGO interferometers in 2016) 4) Gravitational red shift: Frequency of light is affected by a gravitational field 5) Black holes Gravitational force not exactly proportional to 1/r2 (most noticeable close to massive objects). Einstein correctly calculated the precession of the perihelion of Mercury. An object moving in an attractive 1/r2 force will move in perfect elliptical orbits (Kepler). However, it was known that the orbit of Mercury precesses 5.74 every year. 5.31 of this precession can be (exactly !) attributed to the other planets tugging on Mercury. To account for the remaining 0.43, scientists thought there was another planet

(called Vulcan) near the sun, but it was never observed. Einstein (1915) calculated the precession of Mercury using general relativity and found that it was (exactly !) 0.43/year. This was the first success of General Relativity. Light bends in a gravitational field: Einstein also predicted this in his 1915 GR relativity paper. It was verified in 1919 by Eddington by observation of a star behind the edge of the sun during a solar eclipse. the stars detected Eddingtons photograph

Gravitational Lensing: For objects directly behind massive objects, one can obtain images in different directions. The gravitational lens can act as a magnifier, and is used to study black holes, dark matter, exoplanets, and distant galaxies (and expansion of universe). One way to observe a black hole is that it may act as lens for object behind it. Gravity Waves: Changes in space-time will travel away from an accelerating mass at the speed of light. If the mass is oscillating (or revolving), the changes will have the form of periodic waves. LIGO Interferometers LIGO signal (9/2015):

2 black holes spiraling into each other Gravitational Red Shift: We saw that a gravitational field affects time therefore it must also affects frequency. As an electromagnetic wave leaves a massive object (and strong g-field), its frequency (F) decreases: F() = F(R) (1- 2GM/c2R)1/2 [F()-F(R)]/F(R) = (1- 2GM/c2R)1/2 -1 Since the energy of a photon E = hF, this can also be viewed as the photon losing (kinetic) energy as it leaves the gravitational field. First observed in 1925 in spectra from massive stars. Example: The mass of the sun 2.0 x 1030 kg and its radius R 7.0 x 108 m, [F()-F(R)]/F(R) - (6.7 x 10-11 Nm2/kg2) (2.0 x 1030 kg)/[(3 x 108 m/s)2(7.0 x 108m)

= - 2.1 x 10-6 Gravitational Red Shift: F() = F(R) (1- 2GM/c2R)1/2 What if E() = hF() = 0? Then light (EM wave) cant escape: Black Hole Black Hole: R < 2GM/c2 Schwarzschild Radius [Interestingly, this can also be predicted from classical physics: Black hole exists if the escape velocity c: For an object of mass m, mvesc2/2 = GmM/R vesc2 = 2GM/R] Examples: 1) What is the minimum mass of a black hole with the radius of the earth (R = 6.4 x 10 6 m)? M > Rc2/2G = (6.4 x 106m) (3 x 108 m/s)2 / [2(6.67 x 10-11 Nm2/kg2)] = 4.3 x 1033 kg i.e. M > 2000 suns White dwarf has radius R R(earth) and M M(sun) 2) An object with the mass of the sun, would be a black hole if its radius R < 2 (6.7 x 10-11 Nm2/kg2) (2 x 1030 kg) / (3 x 108 m/s)2 = 3000 m Neutron star has M M(sun) and radius R 104 m 3) The (supermassive) black hole in the center of the Milky Way has a mass 4 x 106 mass

of sun. Its radius R < 4 x 106 x 3000 m = 1.2 x 1010 m ( 1/10 earth-sun distance).

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