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## Main Question or Discussion Point

What is frame dragging?

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What is frame dragging?

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http://simple.wikipedia.org/wiki/Frame-dragging

- #3

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http://cosmology101.wikidot.com/

one particular article will be of use to help your understanding of cosmology. It has zero maths involved and written as more a FAQ style.

http://arxiv.org/abs/1304.4446

I'm posting the article based on some of your other posts, not to say there is particularly anything wrong with your question. Just that the article will help aid your understanding of current main stay cosmology or in other words the concordance model currently represented by LCDM.

Of course other models may or may not be more acurate thats the fun part of science.

- #4

Chalnoth

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I don't think this is strictly correct. Frame dragging is a particular feature of General Relativity. It generally isn't talking about particles, but macroscopic objects. It has been tested for the Earth by Gravity Probe B (http://en.m.wikipedia.org/wiki/Gravity_Probe_B).

http://simple.wikipedia.org/wiki/Frame-dragging

- #5

WannabeNewton

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But how will an observer sitting at infinity see this? Well to him the observers are all hovering in place (constant spatial coordinates in the Kerr chart) so the displacement vector simply points from one static observer to an infinitesimally nearby static observer and doesn't do anything at all as far as he's concerned, meaning the observer at infinity will see the aforementioned static observer's gyroscopes rotate relative to him i.e. he sees the static observer precess in place. This is an example of frame-dragging.

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I don't think this is strictly correct. Frame dragging is a particular feature of General Relativity. It generally isn't talking about particles, but macroscopic objects. It has been tested for the Earth by Gravity Probe B (http://en.m.wikipedia.org/wiki/Gravity_Probe_B).

agreed however the OP has posted numerous threads on the applications of Calabai Yau metric applications in cosmology. My response was more geared towards that application. As that model deals primarily with string theory and particle interactions in a 6d manifold rather than cosmological applications.

I may have been mistaken in that regard however

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- #8

Bobbywhy

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Your description is correct: The Gravity Probe B experiment showed that spacetime near earth is curved by the earth's mass, and it is "dragged" by the earth's rotation.

“According to Einstein’s theory, space and time are not the immutable, rigid structures of Newton’s universe, but are united as spacetime, and together they are malleable, almost rubbery. A massive body warps spacetime, the way a bowling ball warps the surface of a trampoline. A rotating body drags spacetime a tiny bit around with it, the way a mixer blade drags a thick batter around.”

http://einstein.stanford.edu/content/press-media/results_news_2011/C_Will-Physics.4.43-Viewpoint.pdf

- #9

WannabeNewton

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Going back to the example of Kerr space-time, instead of looking at the static observers one can look at the family of observers following orbits of the time-like vector field ##\nabla^{\mu} t## i.e. ##u^{\mu} = \gamma \nabla^{\mu}t## where ##t## is the canonical global time function and ##\gamma## is the normalization factor. These are the observers who are locally non-rotating i.e. (unlike the static observers) these observers have vanishing twist ##\omega^{\mu} = \gamma^{2}\epsilon^{\mu\nu[\alpha\beta]}\nabla_{\nu}t \nabla_{(\alpha}\nabla_{\beta)}t - \gamma\epsilon^{\mu[\nu\beta]\alpha}\nabla_{(\nu}t\nabla_{\beta)}t\nabla_{\alpha}\gamma = 0##; this means that they don't have any precession in the sense described in post #5.

However, notice that ##\frac{\mathrm{d} \phi}{\mathrm{d} t} = \frac{u^{\phi}}{u^{t}} = \frac{g^{\phi \mu}\nabla_{\mu}t}{g^{t\mu}\nabla_{\mu}t} = \frac{g^{\phi t}}{g^{tt}}## so these observers have non-zero angular velocity about the central rotating body. In other words, the rotation of the central body induces an orbital (in this case azimuthal) angular velocity of said observers about this mass. This is

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You're saying:

But how will an observer sitting at infinity see this? Well to him the observers are all hovering in place (constant spatial coordinates in the Kerr chart) so the displacement vector simply points from one static observer to an infinitesimally nearby static observer and doesn't do anything at all as far as he's concerned, meaning the observer at infinity will see the aforementioned static observer's gyroscopes rotate relative to him i.e. he sees the static observer precess in place. This is an example of frame-dragging.

An observer at rest will witness an object having infinitesimal rotation, because of frame dragging (the larger mass near it is rotating), and not because the object is moving deeper into the larger mass's gravity well?

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WannabeNewton

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