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Twin Paradox

Consider the following (attributed to Paul Langevin in 1911).

Consider two twins. One twin undertakes a long space journey in a spaceship. She travels at almost the speed of light to a distant star, while the other twin remains on Earth. The traveling twin finally returns to Earth from the distant star, traveling again at almost the speed of light.

The Earth-bound twin says, “I observe that your velocity is very large relative to mine, so I observe your clock running much slower than mine. Therefore, when you return to Earth, you will be younger than me.”

The traveling twin says, “Well, in my frame, your velocity is very large relative to mine. I observe that your clock on Earth is running much slower than mine. Therefore, when I return to Earth, you will be younger than me.”

Who’s correct ?

Tuesday, March 13, 2012

The General Theory of Relativity

Gravitational influence of other planets cause the major axis of Mercury’s elliptical orbit to precess around the Sun relative to

fixed stars at a rate of 574” per century.

But, Newton’s gravitational force law explains only 43” per century. People tried to explain this by hypothesizing there was

another planet within Mercury’s orbit (Vulcan) that was perturbing Mercury

Mercury

Precession is 574” per century


From 1907-1915, Einstein developed a new theory of Gravity. This was not his goal. He was more interested in

“Accelerated Coordinate Frames”.

I.e., Accelerated ≠ Inertial

Consider two objects, separated by a distance r, one of mass m and charge q, and the other of mass M and charge Q.

The force due to gravity is:

mag = G Mm / r2

The force due to electrostatics is:

mae = (1 / 4πε0 ) Qq / r2


The mass m on the left-hand sides is an inertial mass. It measures an object’s resistance to being accelerated. M and m and Q and q on right-hand side are gravitational and

electrostatic “charges”, govern the strength of their respective forces.

Why should an object’s “resistance” to being accelerated be the same as the gravitational “charge” ?


Formally, one can rewrite these formula in terms of their inertial (mi) and gravitational mass (mg).

mi ag = G Mg mg / r2 and mi ae = (1 / 4πε0 ) Qq / r2

The General Theory of RelativityWhy should an object’s “resistance” to being accelerated be

the same as the gravitational “charge” ?

Or, rewriting for the acceleration:

ag = G Mg ( mg / mi ) / r2 and ae = (1 / 4πε0 ) Qq / r2 / mi

Experimentally, physicists find that mg / mi = 1 with an uncertainty of 1 part in 1000 billion (1012).

This is remarkable - proportionality of inertial and gravitational masses means all masses experience the same gravitational acceleration. This is

referred to as the weak equivalence principle.


Einstein (like you) realized that inertial reference frames cannot be defined in the presence of gravity.

In 1907, Einstein realized that gravity and acceleration acted exactly the same and that no experiment by

an observer can differentiate them. He wrote:

... a sudden thought occurred to me: “If a person falls freely he will not feel his own weight.” ... This simple

thought made a deep impression on me. It impelled me toward a theory of gravitation.


Principle of Equivalence: All local, freely falling nonrotating laboratories are fully equivalent for the performance of all physical experiments.

Restriction to nonrotating labs eliminates fictitious forces such as Coriolis and centrifugal. Local, freely falling,nonrotating frames are “local inertial reference frames.

We can eliminate gravity in a laboratory by entering a state of free-fall. In free fall you can do no (local) experiment to determine that you are experiencing acceleration.



Einstein’s new theory is a geometric description of how distances (intervals) are measured in the presence of mass. Near any mass, space

and time must be described in a new way.

Space surrounding a mass becomes curved through a fourth spatial dimension perpendicular to all the three “normal” spatial dimensions (xyz).



Sun

Sun

Actual Light Path

Straight line if no mass

Actual Light Path


The General Theory of RelativityIn 1919, Arthur Stanley Eddington measured the apparent positions of stars close to the Sun during a total eclipse.

Einstein’s Theory predicted that the positions of stars near the Sun should be shifted from their actual positions by 1.75o, in

excellent agreement with Eddington’s measurements !

Actual position of star

Apparent position of star

Earth Sun

1.75o


The General Theory of RelativityEinstein’s Theory predicted that the positions of stars near the Sun should be shifted from their actual positions by 1.75o, in

excellent agreement with Eddington’s measurements !


The General Theory of RelativityAnother aspect of GR. Nothing can move faster between two points than light. Light must always follow the shortest distance between two points. Compare to the solid line, a trip on the dashed line would (even at the speed of light):

1. Cover a longer distance than the trip along the solid line;

2. have slower running time (clocks would run slower nearer the mass).


Curvature of Space



Precession of Mercury:

Einstein applied his theory for the precession of Mercury.

His theory predicted the exact measured precession.

This was a 2nd great triumph for his Theory of General Relativity.

Mercury

Precession is 574” per century


Observer watching from afar sees the box accelerate downward. Photon path

must appear curved.

The Bending of Light

Observer in the box must see photon move in horizontal line across

the box.



must appear curved.

Geometry for the Bending of Light

Angle of deflection of photon, ϕ, is small, exaggerated here :

Arc of deflection subtends an angle ϕ

between the radii OA and OB.

If width of lab is L, then photon crosses lab in

time t ≈ L / c.

In this time, lab falls a distance d = (gt2) / 2.

Triangles ABC and OBD are similar, thus

BC / AC = BD / OD

with BC = gt2/2

AC = L

BD = (1/2) L / cos(ϕ/2)

For small angles, cos(ϕ/2) ≈1 and OD ≈ rC

ϕ

L



must appear curved.

Geometry for the Bending of Light

ϕ

Triangles ABC and OBD are similar, thus

BC / AC = BD / OD

with BC = gt2/2

AC = L

BD = (1/2) L / cos(ϕ/2)

For small angles, cos(ϕ/2) ≈1 and OD ≈ rC

Therefore, BC / AC = BD / OD, becomes,

(gt2/2) / L = L / (2rC)

for t = L / c and g=9.8 m/s, near the surface of the Earth the radius of curvature of a photon’s path is:

rC = c2/g = 9.17 x 1015 m ≈ 1 lyr

Angle of deflection is

ϕ = L / rC = 1.09 x 10-15 rad

= 2.25 x 10-10 arcsecond


Gravitational Redshift and Time Dilation

Photon of frequency ν0

Consider a photon leaving the floor of a lab

at the instant a cable holding the lab is cut.

Observer moving with lab measures light’s frequency at ν0 at top of lab.

From ground, lab falling toward the light, so the light meter has a speed of

v=gt=gh/c when the photon gets there.

Change in frequency is from the doppler shift (v << c, non-relativistic)

Δν/ν = v / c = gh/c2.

But, meter sees no frequency change.


Change in frequency is from the doppler shift (v << c, non-relativistic)

Δν/ν = v / c = gh/c2.

But, meter sees no frequency change.

Only explanation is that a gravitational redshift decreases the light frequency

of the light as it travels up by

Δν/ν = -v / c = -gh/c2.


Photon of frequency ν0

Consider a photon leaving the floor of a lab

at the instant a cable holding the lab is cut.

This has been proven by experiments time and again (see Example 17.1.1 in book).Tuesday, March 13, 2012

= [ 1 - 2GM / (r0c2) ]1/2


Derivation of exact formula (valid for strong and weak gravity) requires more mathematics. Answer is:

When Gravity weak (x = GM/r0 c2 << 1), then (1 - x)1/2 ≈ (1 - x / 2), which gives

ν∞ν0

ν∞ν0≈ 1 - GM / (r0 c2), for (GM / r0 c2) << 1

Gravitational redshift is then, z = (λ∞ - λ0) / λ0 = ν0 / ν∞ - 1

z = (1 - 2GM / r0 c2)-1/2 - 1 ≈ GM / r0 c2.

Where the approximation is for the weak case, G M / r0 c2 << 1.


= [ 1 - 2GM / (r0c2) ]1/2


Δt0

Δt∞

Imagine a clock that “ticks” for each Period of a wave of light. The gravitational redshift says these “ticks” get slower. Δt0 = 1/ν.

Time gets slower as spacetime is more curved!

ν∞ν0

=

Example: Sirius B (white dwarf star talked about before) has R=5.5 x 106 m and M=2.1 x 1030 kg. The radius of curvature at the surface of the star is

rc = c2 / g = c2 / (GM/R2) = c2 R2 / GM = 1.9 x 1010 m = 0.13 AU.

Note that (GM / R c2) << 1, even for white dwarf stars.

Gravitational redshift suffered by a photon emitted from the surface and escaping to infinity is z ≈ (GM / Rc2) = 2.8 x 10-4

Excellent agreement with observations of z = (3.0 ± 0.5) x 10-4 !



Example: Sirius B (white dwarf star talked about before) has R=5.5x106 m and M=2.1 x 1030 kg.

Compared to time measured from a great distance away, time at the surface of the white dwarf star runs slower.

If a distant observer measures a time interval of 1 hr, the recorded at the surface of Sirius B would be less than this:

Δt∞ - Δt0 = Δt∞ ( 1 - Δt0/Δt∞ ) = (3600 s) x ( GM/ Rc2 ) = 1.0 s.

The clock on the surface runs 1 second slower for each hour of time that passes for the distant observer.


Intervals and Geodesics

Spacetime consists of 4 coordinates (x, y, z, t) which specifies an “event”.

Einstein developed the equations for calculating the geometry of spacetime produced by some mass (and energy,

they are the same thing!).

G = -8πG

c4 T

T is the “Stress-Energy” Tensor, evaluates

the effect of a given distribution of mass and energy on the curvature of spacetime.

G is the Einstein Tensor (Gravity)

To learn more about this equation.... take a course on General Relativity

Einstein’s Field Equation


Intervals and GeodesicsSpacetime consists of 4 coordinates (x, y, z, t) which specifies an “event”.

Mathematically, let the distance Δℓbe a straight line between two points A and B,

(Δℓ)2 = (xB - xA)2 + (yB - yA)2 + (zB - zA)2

The spacetime interval between “events” A and B is (Δs)2 = (cΔt)2 - (Δℓ)2,

(Δs)2 = c2(tB - tA)2 - (xB - xA)2 - (yB - yA)2 - (zB - zA)2

Distance between A and BDistance light traveled in time |tB - tA|

(Δs)2 can be positive, zero, or negative. The sign tells us whether the events A and B are causally connected. (Is there enough time for light to get from A to B ?)

(Δs)2 > 0 is a timelike curve - light has enough time to make the trip

(Δs)2 = 0 is a lightlike curve - light has exactly enough time to make the trip

(Δs)2 < 0 is a spacelike curve - light does not have enough time to make the trip


(Δs)2 can be positive, zero, or negative. The sign tells us whether the events A and B are causally connected. (Is there enough time for light to get from A to B ?)

(Δs)2 > 0 is a timelike curve - light has enough time to make the trip

(Δs)2 = 0 is a lightlike curve - light has exactly enough time to make the trip

(Δs)2 < 0 is a spacelike curve - light does not have enough time to make the trip

timelike curves for Event A are within cone. Event A can influence them or be

influenced by them.

spacelike curves are “Elsewhere” of A

lightlike curves are on surface of cone.


Intervals and GeodesicsSpacetime consists of 4 coordinates (x, y, z, t) which specifies an “event”.

Proper Time: time between two events that occur at same location, (Δℓ)2 = 0

τ = Δs / c

Proper distance: distance measured between two events in a reference frame where they occur simultaneously (tB = tA):

ΔL = √- (Δs)2

For a straight rod connecting two locations, the proper distance is the rest length of the rod.

Metrics: metrics are the differential distance formula used to calculate intervals. For flat spacetime, the metric would be:

(ds)2 = (c dt)2 - (dx)2 - (dy)2 - (dz)2

To determine the interval, you take the line integral of this from point A to B.



In flat spacetime, the interval measured along a straight timelike worldline between two events

is a maximum. Other worldlines will have smaller intervals.

For massless particles, like photons, all worldlines are 0.

Example, the straight line connecting A and B is

Δs(A→B) = ∫√(ds)2

A

B

Δs(A→B) = ∫√(cdt)2 - (dx) 2 - (dy) 2 - (dz) 2

= ∫c dt = c(tB - tA)A

B

A

B



Three different World Lines.

(a) object at a constant location

(b) object moving in the y-direction

(c) satellite in circular orbit

around a planet



Straightest Possible Worldlines are Geodesics. Massless particles (photons) always follow null geodesics.

In curved Spacetime, timelike geodesics, (Δs)2 > 0, has either a maximum or a minimum, it is an extremum.

Einstein realized that freely falling objects through spacetime follow geodesics !

Three features of General Relativity:

1. Mass acts on spacetime, telling it how to curve.

II. Spacetime acts on mass, telling it how to move.

III. Any freely falling particle (including photons) follows the straightest possible worldline (geodesic) through spacetime. For massive particles, geodesic has a

maximum or minimum interval. For massless particles, the geodesic = 0.


Metrics: recall that the spacetime metric is (ds)2 = (c dt)2 - (dℓ)2.

In spherical coordinates, this becomes,

(ds)2 = (c dt)2 - (dr)2 - (r dθ)2 - (r sinθ dϕ)2

This is for flat spacetime, that is, spacetime not in the vicinity of a massive object.


Important: variables r, θ,ϕ, t are measured by an observer at rest a great distance from any mass (r≈∞). Now, place a sphere of mass M with radius R at the origin O.

For an observer at r≈∞, spacetime is approx. flat. Light from the sphere will experience time dilation, so (1 - 2GM / rc2 )1/2 factor should be in Metric.

Karl Schwarzschild (1873-1916)

In 1916 (two months after Einstein published his theory) Schwarzschild solved Einstein’s field equations in the

presence of a mass M at the origin.

(ds)2 = (c dt √1 - 2GM/rc2 )2 - ( ) - (r dθ)2 - (r sinθ dϕ)2dr

√1-2GM/rc2

2


If a clock is at rest at a radial coordinate r then the proper time dτ (dr=dθ=dϕ=0) is related to a time dt measured at an infinite distance.

Schwarzschild Metric: .


Schwarzschild Metric contains all the effects of time dilation (in temporal term). The curvature of space is contained in the radial term. The

distance along a radial line (dθ=dϕ=0) is the proper distance (dt=0):

Thus, the radial distance is longer by a factor of 1 / (1 - 2GM/rc2)1/2 compared to flat space.


√1-2GM/rc2

2

dL = √-(ds)2 = dr

√1-2GM/rc2

dτ = ds / c = dt √1 - 2GM / r c2

Note that dτ < dt always.


Black HolesIn 1783 John Michell (1724-1793) considered Newton’s theory of light and gravity. He speculated that if the Sun were 500x more massive, then light

from its surface would not have enough velocity to escape the Sun’s Gravity.

Using Newton’s Force Law (vescape)2 = 2GM/r.

For vescape=c, R = 2GM/c2.

For M=mass of the Sun, R = 2.95 km.

In the 18th century this was too small to be more than a curiosity.

In 1939, J. Robert Oppenheimer and Harland Snyder considered the gravitationally collapse of a star that had exhausted its nuclear fuel.

J. Robert Oppenheimer (1904-1967)


Black Holes

What happens as 2GM/rc2 →1 ? At this point space becomes infinitely curved. This is the Schwarzschild Radius.

Note that at the Schwarzschild Radius, the temporal term in the Metric goes to 0. the Proper time, dτ = 0. Is light frozen in time ? (Does time stop ?)


√1-2GM/rc2

2

RS = 2 GM / c2

Consider the apparent speed of light, the “coordinate speed of light”, the rate at which the spatial coordiantes of a photon change. Start

with lightlike curve: (ds)2 = 0, above. Inserting dθ=dϕ=0.

c dt √1 - 2GM/rc2 = ( ) dr

√1-2GM/rc2 dr/dt = c( 1 - 2GM/rc2 ) = c( 1 - RS / r)

When r>>RS, then dr/dt=c. But at r=RS, dr/dt=0. This is a barrier (the Event Horizon).

At center of a Black Hole is a singularity. The Event Horizon prevents the singularity from being observed.


What is a Black Hole ?For really massive stars (>18 solar masses), nothing stops the core’s gravitational collapse. It forms a Black Hole, an object so powerful that not even light can escape.


Black Holes

Imagine a trip into a Black Hole. Starting at a safe distance (r >> RS) we fire a radio photon at a Black Hole, which is reflected (somehow) from the Event Horizon.

How long does the trip take ? Integrate the formula for the coordinate speed of light.

dr/dt = c( 1 - 2GM/rc2 )

Δt = ∫dr / (dr/dt) = ∫dr / c(1 - RS / r ) = (r2 - r1)/c + (RS / c) ln( )r2 - RS

r1 - RS

valid for r2 > r1. Inserting r1 = RS, Δt=infinity ! Because the trip is symmetric, the same results applies if the the photon starts at infinity.

The photon never makes it to the Event Horizon from the point of view of the observer at infinity.

A photon fired at the Event Horizon takes an infinite time to get there.


Black Holes

Now consider a brave astronaut, who agrees to fall into a black hole. The astronaut starts at r=infinity and falls toward a 10 M⊙ black hole, where RS = 30 km.

What happens from the astronaut’s point of view ?


Black Holes



Progression of radial coordinate from outside observers view

Progression of radial coordinate from astronaut’s view


Black Holes



Physically, tidal forces become extreme and stretch and

squeeze the poor astronaut.

Once inside event horizon, r < RS. For dr = dθ=dϕ=0, the interval is:

(ds)2 = (c dt)2 ( 1 - RS / r) < 0

This is a spacelike curve, which is not permitted for massive particles. At the

singularlity all worldlines converge.


Can Photons “orbit” a Black Hole ?

Classically (Newton):

v2/r = GM/r2

Set v=c, and solve for r, yields:

R = GM / c2

This is smaller than RS = 2GM / c2.

Black Holes


Can Photon’s “orbit” a Black Hole ?

For General Relativity, consider the the coordinate speed of light in the ϕ-direction. Start with the Schwarzchild solution:


√1-2GM/rc2

2

Set dr=dθ=0, and for the speed of light, ds=0:

Which yields:

Black Holes


Black Holes

Lets equate this to the centripetal velocity, v = (GM/r)1/2, for stable circular orbits:

And solve for the radius where this occurs:

This is the “photon sphere” where photons follow circular orbits around the Black Hole. If you looked into the photon sphere, you would literally see the

back of your head....


Black Hole Candidates

Evidence that the center of our Milky Way Galaxy contains a black hole with a mass of several million solar masses (we’ll discuss this more in 2 weeks).

First candidate for a Black Hole was in a binary star system, Cygnus X-1 (X-ray source near bright star η Cygni).

In the best(?) candidate, V 404 Cyg underwent an X-ray outburst in 1989. This object is orbited by a K0 Giant star with a radial velocity of 211 km/s with a

period of P=6.473 days. Kepler’s Laws give a mass for the “unseen” companion of 12 M⊙, which would make it much, much too massive to be a neutron star

(maximum neutron mass is ~3 solar masses).

Other evidence in Active Galactic Nuclei (Black Holes in the centers of other galaxies with masses up to one billion solar masses, much more on this later).

The relativistic effects of the black hole can be seen on the emission lines of gas very near the black hole, which gives the mass.


Most Popular Black Hole candidate is Cygnus X-1.

Do Black Holes Exist ?


Black Hole Candidates


Wormholes

“normal space”

light ray path

light ray path

light ray path

“Wormhole”


How to make a time machine:Postulated by Prof. Kip Thorne (Caltech)

Similar to Twin Paradox of Special Relativity

1. Kip and his wife make a wormhole and hold hands through it.

2. His wife takes the wormhole on a rocket and travels at 0.6 c for 10 hrs,

then stops and returns to Earth taking 10 more hrs.

3. Special Relativity says that in her frame the trip took

20 hrs x (1 - 0.62)1/2 = 16 hrs

So her clock will be 4 hrs behind Kip’s clock.

4. Since they were holding hands through the wormhole the whole time, if anything Kip sends through the wormhole will emerge 4 hrs in the past !


Remember Pulsars ? Pulsar = rapidly rotating Neutron Star. Pulsar at the center of the Crab nebula pulses 30 times per second!

Chandra X-ray Observatory Hubble Space Telescope


Pulsars = Test of General Relativity

In the 1960s, Russell Hulse and Joseph Taylor discovered that the pulsar PSR 1913+16 was a binary system of two neutron stars (or a neutron star and a white dwarf). The orbital separation of the system is a little larger than the Sun’s

diameter.

A 30-year study of this system showed that the orbital period is speeding up as a result of gravitational radiation (gravity waves !). The two stars should merge

in ~300 Myr.

Theoretical : dP/dt = -(2.40242 +/- 0.000002) x 10-12.

Measured: dP/dt = -(2.4056 +/- 0.0051) x 10-12. Excellent !

Hulse and Taylor received the Nobel prize in 1993 for this work.


Orbiting Compact Objects emit Gravitational Waves


LIGO: Laser Interferometer Gravitational Observatory


Hawking Radiation“Semiclassical” result merging Quantum Mechanics and General Relativity

e-

e+

γ

γ

creation

annihilation

annihilation

Virtual Particles are able to “exist” for a time duration given by Heisenberg’s uncertainty principle: ΔE Δt ≈ ħ, or Δt ≈ ħ/E

Stephen Hawking (b. 1942)


Virtual Particles are able to “exist” for a time duration given by Heisenberg’s uncertainty principle: ΔE Δt ≈ ħ,

or Δt ≈ ħ/E

Hawking Radiation

Consider what happens to virtual particles produced near

the Event Horizon.

Hawking worked out that blackholes “radiate” these

particles with spectrum of a black body with a temperature

given by:


Hawking RadiationBlack hole evaporation ! As black holes emit energy, they lose mass (Einstein), so as they accrete virtual particles, black holes lose mass.

Area of a Black hole is:

Luminosity of a perfect black body is:

Schwarzchild radius is:

You can rewrite the Stefan-Boltzman constant as:

rs =2GM

c2

A = 4πr2s = 4π

�2GM

c2

�2

=16πG2M2

c4

Temperature of Black Hole Radiation is:

L = AσT 4

σ =�

π2k4B

60�3c2

�

Which gives a luminosity of:

L =16πG2M2

c4

�π2k4

B

60�3c2

� ��c3

8πGMkB

�4

=�c6

15360πG2M2


Hawking RadiationBlack hole evaporation ! As black holes emit energy, they lose mass (Einstein), so as they accrete virtual particles, black holes lose mass.

L =16πG2M2

c4

�π2k4

B

60�3c2

� ��c3

8πGMkB

�4

=�c6

15360πG2M2

Consider that the luminosity is the change in energy (change in mass) with time: L =

dE

dt=

dm

dtc2

So, Hawking predicts that black holes should evaporate.

For a 1 solar mass Black Hole, the Luminosity minuscule: L = 10-29 W.

You can work out how long it takes a black hole to evaporate:

The evaporation time increases as the cube of the mass ! For a 1 solar mass black hole is tev = 2.1 x 1067 yrs. Much, much longer than the age of the Universe.

For tev = 1 s, the mass is 2.3 x 105 kg (about the size of a large military cargo plane)and the released energy would be 2 x 1022 J = 5 x 106 megatons of TNT

= 200,000 of the most powerful nuclear bombs made.


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