Redshift Academy


Gravitational Waves
-------------------

The Einstein Field Equations are:

R_μν - (1/2)g_μνR = 8πGT_μν

R_μν represents the part of curvature that derives from the 
presence of matter (i.e. that part due to the stress energy 
tensor T_μν).  The remaining components of the Riemann tensor 
(the WEYL tensor) represents the part of the gravitational 
field which can propagate as a gravitational wave through a 
region of spacetime containing no matter or nongravitational 
fields.  In other words, these are the parts of the curvature 
derived from the dynamics of the gravitational field itself 
independent of what matter the spacetime contains. An empty 
spacetime containing only gravitational radiation will 
satisfy R_μν = 0 but will also have R_μνσ^λ ≠ 0.  Regions of 
spacetime in which R_μν = 0 are referred to as being RICCI 
FLAT.

Einstein predicted the existence of gravitational waves.  We 
start with the field equations in the absence of any mass.
This is a good approximation if there is a large distance
between the source and the observer.  Therefore,

R_μν - (1/2)g_μνR = 0

Therefore,

g^μνR_μν - (1/2)g^μνg_μνR = 0

Or,

R - (1/2)δ^μ_μR = 0

Now δ^μ_μ = 4.  Therefore,

R - 2R = 0

This implies that R = 0.  Thus, we can write:

R_μν = 0

Note:  R is also the trace of the Ricci tensor.

Now consider empty flat space so that g_μν = η_μν.  Consider a 
small perturbation is spacetime due to some kind of cosmic 
'event'.  The metric tensor can be written as:

g^μν = η^μν - h^μν(x,t)

Where h_μν(x,t) represents the perturbation.

The Ricci tensor is comprised of the first derivatives and 
products of the Christoffel symbols.  Thus, schematicaly and
ignoring all indeces in the interest of brevity:

R ~ ∂Γ + ΓΓ

In turn, the Christoffel symbols consist of the metric tensor
and its derivatives.  Again, schematically:

Γ ~ (1/2)g^-1∂g

Where g^-1 is the inverse metric tensor (g^μν).  Therefore,

Γ ~ (η - h)∂h

If h is very small then the product h∂h can be ignored.

  ~ η∂h

  ~ ∂h

Therefore, the Ricci tensor can be written:

R = ∂²h + ∂h∂h

Again, if is small then ∂h∂h -> 0.  Therefore,

R = ∂²h

This is not just one equation.  There is a separate equation 
for each μ and ν.  Thus,

R_μν = [∂²h]_μν = 0

[∂²h]_μν = 0 has the form of the wave equation:

[∂²h/∂t² - ∂²h/∂x²]_μν = 0   ... 1.

NOTE:  In the case where the source is close by, the above 
equation becomes:

[∂²h]_μν = kT'_μν where k is a constant.

Where T'_μν represents the stress–energy tensor plus quadratic 
terms involving h_μν.  h is involved because the wave field has 
an effect on T_μν.  While the equations are valid everywhere 
they are difficulty to solve analytically because of the 
complicated source term.  Under certain circumstances (flat 
space assumption) the quadratic terms involving h can be 
ignored making the solutions easier.  However, for  most 
purposes, the zero source solutions are perfectly adequate.

Solutions to 1.0 are of the form:

h_μν(t,z) = h⁰_μνsin(k(t - z))

The waves are transverse meaning that waves can only have
components in the plane perpendicular to the propagation of 
the wave. 

           y
           |      |\
           |      | \
           |      |  \
            ---- -----\-------- z
            \     |   |
             \     \  |
               x    \ |
                     \|

Therefore, the time and z components of h are 0.

 -            -
| 0   0 0 0 |
| 0  h_xx h_xy  0 |
| 0  h_yx h_yy  0 |
| 0   0 0 0 |
 -            -

Now, h is a symmetric tensor so h_xy = h_yx.  Furthermore, there 
is another constraint that h_xx = -h_yy (The trace of h_μν = 0).
Therefore, we arrive at 3 equations.

h_xx(t,z) = h⁰_xxsin(k(t - z))

so that,

g_xx(t,z) = 1 + h_xx

This corresponds to coordinates that are stretched and contracted 
in the x direction.  The contraction occurs with the change in sign
of sin(k(t - z).

Similarly, there is stretching and contracting in the y direction:

h_yy(t,z) = h⁰_yysin(k(t - z))

so that,

g_yy(t,z) = 1 + h_yy

So there is stretching and contracting in the y direction:

Thus, in the x-y plane, this looks like.

               y
               ^   
               :    
               ^  
               | 
    <....<---- |----->...> x
               |
               v
               :
               v

Finally, there is a third solution:

h_xy(t,z) = h⁰_xysin(k(t - z))

so that,

g_xy(t,z) = 1 + h_xy

This corresponds to simultaneous stretching and contracting along 
the diagonal direction.

In summary, if a binary-pulsar is being observed, the gravitational
waves produced will be observed in the plane perpendicular to the 
line of sight.

Wave Origins
------------

Gravitational waves are radiated by objects whose motion 
involves asymmetric accelerations.  Two objects spinning about a 
central axis will not produce gravitational waves but if they 
tumble end over end, as in the case of two objects orbiting 
each other, they will radiate gravitational waves.  The heavier 
the masses , and the faster they tumbles, the greater the 
gravitational radiation emitted.

In summary, the the following will/will not emit gravitational 
waves:

- Two objects orbiting each other.
- A supernova if the explosion asymmetric.
- An isolated non-spinning solid object moving at a constant 
  velocity will not radiate (conservation of linear momentum)
  but one that is accelerating will.
- A spinning disk will not radiate (conservation of angular 
  momentum). 
- A spherically pulsating spherical star will not radiate.

Orbiting Bodies
---------------

Gravitational waves carry energy away from their sources and, in 
the case of orbiting bodies, this is associated with a decrease 
in the orbit.  For a simple system of two masses, the radiated 
power is given by:

P = dE/dt = -32G⁴(m₁m₂)²(m₁ + m₂)/5c⁵r⁵

The amount of radiation emitted corresponds to the loss of 
angular momentum of the orbiting body.

In the case of the Earth orbiting the Sun the power emitted is 
about 200 watts!

The emission of radiation first circularizes their orbits and 
then gradually shrinks their radius increasing the orbital 
frequency.  The rate of decrease in the orbital radius is 
given by:

dr/dt = -64G³(m₁m₂)(m₁ + m₂)/5c⁵r³

In the case of the Earth orbiting the Sun, the Earth's orbit 
shrinks by 1.1 x 10^-20 meter per second.   At this rate it 
will take about 10¹³ times more than the current age of the 
Universe to spiral onto the Sun!

Verification
------------

On 2/11/2016 the first experimental evidence of gravitational 
waves was announced to the world, thus proving Einstein's 
original assertion.  The waves were detected by the Laser 
Interferometer Gravitational Wave Observatory (LIGO) that 
consists of widely separated mirrors that are suspended like 
pendulums in a clock.   The apparatus is designed such that 
the tidal forces associated with gravitational waves causes the 
distance between the mirrors to change.  The change in distance 
between the mirrors is detected using the technique of laser 
interferometry.

The discovery of gravitational waves promises to have a
profound impact on our understanding of how the Universe
works.