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- Thu, 21 Mar 2019 02:08:39 EST aGo2dCNY No.57582
File: 1553148519011.png -(16747B / 16.35KB, 710x420) Thumbnail displayed, click image for full size. zOMG it spins!
lets say you were somewhat nearby a rapidly rotating neutron star such that the star's diameter was a significant portion of the distance from the star. Would the star's effective center of gravity be offset towards the approaching limb because of the relative velocities and redshifts of the approaching side versus the retreating one?
If its real, how significant would the effect be? Does the effect imply that the gravity well isn't symmetrical?
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Friedrich Bessel - Thu, 21 Mar 2019 13:54:15 EST 457vC2+I No.57591 Reply
>>57582
>> Would the star's effective center of gravity be offset towards the approaching limb
Yup.
>>Does the effect imply that the gravity well isn't symmetrical?
Yes, but of course it is. We model gravity based around a sphere with a radius, assuming that the center point of the sphere is the point of gravitation, but this is a simplification (that works most of the time when dealing with large relative masses.) In reality every single molecule that composes the mass is part of its gravity well, with the strength varying with density and molecular weight.

Think about a dyson sphere with no central star for an extreme example. From the outside of the sphere, the spherical-well model works well for modeling how objects would interact with its gravity well, but from within the sphere it would be exactly wrong; objects in free fall within the sphere would move at the angle the spherical-well model predicts, but with the exact opposite direction (i.e. they would move along the line of the normal connecting the object and the center point of the sphere, but move to the circumference instead of away from it.)
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Urbain Le Verrier - Thu, 21 Mar 2019 19:22:53 EST L/MOWw3u No.57592 Reply
>>57591
I guess it must be the case that the shape of the gravity well changes depending on the observer's reference frame. All of that convenient math having to do with spherical symmetry in gravitation disappears and the closer you get, the more dense the gravity field lines in direction of the approaching limb gets and the gravity field lines rotate through to a 90º offset at the surface. The matter in the star presumably doesn't feel this effect because it rotates as a rigid body
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Otto Struve - Fri, 22 Mar 2019 20:16:34 EST 5eVdRvlO No.57595 Reply
How does the enormous horizontal velocity component of the neuton star's surface where the star is closest to the observer accept the observability of the star? Light from that part of the star wouldn't travel straight down the observer's telescope, it would enter the aperture and smash into the interior walls if the telescope tube was long enough or otherwise it would reflect off the primary mirror and miss the secondary mirror due to it's peculiar velocity.
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George Airy - Sun, 24 Mar 2019 18:29:50 EST 457vC2+I No.57601 Reply
>>57595
I'm trying to visualize what you're suggesting. So the neutron star would appear to have a dark center and bright edges? Or would only the approaching edge be bright, with the receding edge being dark too?
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Stephen Hawking - Sun, 24 Mar 2019 21:07:58 EST cMYUYpep No.57602 Reply
>>57601
No the post you're replying to was someone (possibly me) who got excited and forgot everything they were ever taught about vector addition and relativistic beaming. I still didn't quite grok how a photon leaves a sphere perpendicular to the surface and ends up traveling in an otherwise direction until I remembered to think hard about the various relevant rest frames
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Giuseppe Piazzi - Mon, 25 Mar 2019 19:12:40 EST 457vC2+I No.57610 Reply
1553555560599.jpg -(98161B / 95.86KB, 1280x720) Thumbnail displayed, click image for full size.
>>57602
It's cool bro, our monkey meatware never evolved to process this relativity shit, it's why I needed to make a picture in my brain to even try to think about it
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William Hartmann - Tue, 14 May 2019 17:38:07 EST 457vC2+I No.57702 Reply
>>57701
Mhm. That's why I said
>>We model gravity based around a sphere with a radius...
>> ...but this is a simplification
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Urbain Le Verrier - Thu, 22 Aug 2019 01:39:29 EST aGo2dCNY No.57781 Reply
With respect to the receding limb, as the mass is accelerated away from the observer at relativistic speeds, the mass exerts less gravitational influence because of the redshift. The receding limb appears to effectively lose mass as well as getting dimmer and cooler. Could that effective loss of gravitational pull due to the acceleration be considered "anti-gravity" and if so, what potential uses could be made of a propellant based propulsion system in which the propellant reaches relativistic speeds?
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