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Polarisation of a gravitational wave



 
 
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  #1  
Old October 12th 04 posted to sci.physics,sci.physics.relativity
Ziggi
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Posts: 10
Default Polarisation of a gravitational wave

How is this described in GR? In particular, does the polarisation take a
continium of values between 0 and 2pi, or is it somehow constrained? Thanks
in advance.

Ziggi


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  #2  
Old October 12th 04 posted to sci.physics,sci.physics.relativity
Tom Roberts
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Posts: 3,981
Default Polarisation of a gravitational wave

Ziggi wrote:
How is this described in GR? In particular, does the polarisation take a
continium of values between 0 and 2pi, or is it somehow constrained?


The polarization of gravitational waves is unconstrained, but there are
equivalence classes. In GR, gravitational waves are transversely
polarized, and can have nonzero moments from quadrupole on up. A
quadrupole moment with angle \phi relative to some axis is equivalent to
one with angle \phi+pi*N for any integer N. Similarly for higher moments.

Note that the different moments can have different time-dependent
polarizations and amplitudes. Waves with a large number of uncorrelated
moments are essentially unpolarized.


Tom Roberts
  #3  
Old October 14th 04 posted to sci.physics,sci.physics.relativity
John Sefton
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Posts: 230
Default Polarisation of a gravitational wave

Tom Roberts wrote:
Ziggi wrote:

How is this described in GR? In particular, does the polarisation take a
continium of values between 0 and 2pi, or is it somehow constrained?



The polarization of gravitational waves is unconstrained, but there are
equivalence classes. In GR, gravitational waves are transversely
polarized, and can have nonzero moments from quadrupole on up. A
quadrupole moment with angle \phi relative to some axis is equivalent to
one with angle \phi+pi*N for any integer N. Similarly for higher moments.

Note that the different moments can have different time-dependent
polarizations and amplitudes. Waves with a large number of uncorrelated
moments are essentially unpolarized.


Tom Roberts


And when none of that behaves invent dark matter.
What crappolla!!!!!!!!!!!!!!!!!!!!!!!!!!!!

  #4  
Old October 14th 04 posted to sci.physics,sci.physics.relativity
zigoteau
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Posts: 481
Default Polarisation of a gravitational wave

Tom Roberts wrote in message ...

Hi, Tom,


How is this described in GR? In particular, does the polarisation take a
continium of values between 0 and 2pi, or is it somehow constrained?


The polarization of gravitational waves is unconstrained, but there are
equivalence classes. In GR, gravitational waves are transversely
polarized, and can have nonzero moments from quadrupole on up. A
quadrupole moment with angle \phi relative to some axis is equivalent to
one with angle \phi+pi*N for any integer N. Similarly for higher moments.


Could you expand on that, please? I would have said that, since the
field variable of the gravitational field is not a vector (I presume
that people take it to be the Christoffel thingummybob Gamma^i_jk) ,
that it would be meaningless to describe it as transverse.

And what are the 'moments' of a tensor? I wouldn't go as far as
Sefton, but I do find your response verging on taking the ****.

Note that the different moments can have different time-dependent
polarizations and amplitudes. Waves with a large number of uncorrelated
moments are essentially unpolarized.


Cheers,

Zigoteau.
  #5  
Old October 14th 04 posted to sci.physics,sci.physics.relativity
Bilge
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Posts: 13,439
Default Polarisation of a gravitational wave

zigoteau:
Tom Roberts wrote:


Could you expand on that, please? I would have said that, since the
field variable of the gravitational field is not a vector (I presume
that people take it to be the Christoffel thingummybob Gamma^i_jk) ,
that it would be meaningless to describe it as transverse.


Why does it need to be a vector to be transverse? A quadrupole has
components defined by Q_ij = p_i p_j - (1/3)\delta_ij p^2, so
for example,

Q_xx = p_x p_x - (1/3)p^2

or, since there are only 5 independent components, it can be written
in terms of a traceless operator with 5 components,


Q_+/-2 = (1/2)(Q_xx - Q_yy +/- 2i Q_xy)

Q+/-1 = (1/2)(Q_xz - iQ_yz)

Q_0 = sqrt(1/3)[Q_zz - (1/2)(Q_xx + Q_yy)]


Oscillations in the xy-plane only, are the linerar combinations above,
that contain only Q_xx, Qyy and Q_xy, i.e., Q_+/-2.


And what are the 'moments' of a tensor? I wouldn't go as far as
Sefton, but I do find your response verging on taking the ****.


The components of the quadrupole tensor as given above.



  #6  
Old October 15th 04 posted to sci.physics,sci.physics.relativity
Igor
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Posts: 3,903
Default Polarisation of a gravitational wave

(zigoteau) wrote in message . com...
Tom Roberts wrote in message ...

Hi, Tom,


How is this described in GR? In particular, does the polarisation take a
continium of values between 0 and 2pi, or is it somehow constrained?


The polarization of gravitational waves is unconstrained, but there are
equivalence classes. In GR, gravitational waves are transversely
polarized, and can have nonzero moments from quadrupole on up. A
quadrupole moment with angle \phi relative to some axis is equivalent to
one with angle \phi+pi*N for any integer N. Similarly for higher moments.


Could you expand on that, please? I would have said that, since the
field variable of the gravitational field is not a vector (I presume
that people take it to be the Christoffel thingummybob Gamma^i_jk) ,
that it would be meaningless to describe it as transverse.



It's easier to think of the metric tensor components as being
equivalent to the vector potential in this case. For a plane
gravitational wave far removed from any matter and moving in the x
direction, the non-zero metric components are solutions to the
linearized Einstein equations and are given by g_00 = -1, g_xx = 1,
g_yy = 1 + a(x,t), g_zz = 1 - a(x,t), g_yz = b(x,t), where a and b
are oscillating functions of x and t. In this regard, the metric is
Lorentian with a small oscillation occuring in the (y,z) plane. So it
represents a transverse wave.
  #7  
Old October 15th 04 posted to sci.physics,sci.physics.relativity
ande452@attglobal.net
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Posts: 424
Default Polarisation of a gravitational wave

Ziggi wrote:

How is this described in GR? In particular, does the polarisation take a
continium of values between 0 and 2pi, or is it somehow constrained? Thanks
in advance.

Ziggi


Gravitational waves are 2nd rank tensors, not vectors.

Take a look at Misner, Thorne and Wheeler Section 35.6 to see
how polarization of gravitational waves would be measured
experimentally.

John Anderson
  #9  
Old October 15th 04 posted to sci.physics,sci.physics.relativity
Tom Roberts
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Posts: 3,981
Default Polarisation of a gravitational wave

zigoteau wrote:
Tom Roberts wrote in message ...
In GR, gravitational waves are transversely
polarized, and can have nonzero moments from quadrupole on up.


Could you expand on that, please? I would have said that, since the
field variable of the gravitational field is not a vector (I presume
that people take it to be the Christoffel thingummybob Gamma^i_jk) ,
that it would be meaningless to describe it as transverse.


Gravitational waves are self-propagating disturbances in the metric. The
metric determines distance between nearby points. When a gravitational
wave passes a pair of test particles mutually at rest, the distance
between them will vary with the wave if they are oriented transversely
to the wave's direction of propagation, but does not vary if they are
oriented longitudinally.


And what are the 'moments' of a tensor?


This is nto moments of a tensor, but rather moments in the variations of
distnce between points. Look it up in a textbook like MTW.


Tom Roberts
  #10  
Old October 16th 04 posted to sci.physics,sci.physics.relativity
zigoteau
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Posts: 481
Default Polarisation of a gravitational wave

Tom Roberts wrote in message m...

Hi, Tom and Bilge,


And what are the 'moments' of a tensor?


This is not moments of a tensor, but rather moments in the variations of
distance between points. Look it up in a textbook like MTW.


to be taken together with Bilge's response:

Why does it need to be a vector to be transverse? A quadrupole has
components defined by Q_ij = p_i p_j - (1/3)\delta_ij p^2, so
for example,

Q_xx = p_x p_x - (1/3)p^2

or, since there are only 5 independent components, it can be written
in terms of a traceless operator with 5 components,


Q_+/-2 = (1/2)(Q_xx - Q_yy +/- 2i Q_xy)

Q+/-1 = (1/2)(Q_xz - iQ_yz)

Q_0 = sqrt(1/3)[Q_zz - (1/2)(Q_xx + Q_yy)]


Oscillations in the xy-plane only, are the linerar combinations above,
that contain only Q_xx, Qyy and Q_xy, i.e., Q_+/-2.


You don't seem to be saying the same thing. It is perhaps tangential
to sense.

Bilge, it is not clear why Q_0 should not be symmetrical in x, y and
z.
Tom, distances between which points? This is presumably connected to
the metric tensor.

Since you give different answers while both referring to MTW (which I
do not have) I deduce that the subject is not discussed explicitly
anywhere in that textbook, and that this is just a tactic by both of
you to blind me with science and shut me up. Or could either or both
of you give page numbers?

I seem to remember something about elements of tensor space which map
into each other under the operation of the set of rotations. Is there
a connection to spherical harmonics and the total angular momentum
quantum number l? Am I getting warm?


Cheers,

Zigoteau.
 




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