So what is the minimum amount of mass that is required for the gravitational lensing effect to be visible? For instance, we can not observe the suns light during a solar eclipse because the moon does not have enough mass to curve space sufficiently for the suns light to be visible here on earth. So what is the minimum amount of mass required for the effect to be visible. does a single star have enough mass, or does a small solar system have enough mass, or does it take the mass of a entire galaxy or or galaxy cluster to make the effect visible. Is there a known amount or is there an equation that tells us the minimum amount of mass required for the effects of gravitational lensing to be visible?
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1$\begingroup$ It depends on what you mean by visible. It makes a difference if you mean visible to the human eye verses visible to a telescope on earth verses visible to the Hubble telescope. It also depends on the distance the source is away. $\endgroup$– J. ShupperdCommented Sep 22, 2016 at 20:01
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$\begingroup$ I believe the MACHO surveys of the 90's and forward found objects of mass similar to that of our gas giant planets, but I don't have references or any notion of what the lower limit might be. This is of course for the micro-lensing of star light from nearby dwarf galaxies and globular clusters by objects inside the Milkyway. $\endgroup$– dmckee --- ex-moderator kittenCommented Sep 22, 2016 at 20:48
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Your question is a bit vague: a massive body deviates a beam of light by a certain angle, and therefore whether the deviated beam will end up on the observer or not depends on the position (distance) of the observer from the massive body. Anyway the equation that you are looking for comes directly from general Relativity and is the following:
$$\delta\phi=\frac{2r_g}{\rho}=\frac{4GM}{c^2\rho}$$ This gives the deviation from the unperturbed trajectory as a function of the mass of the body $M$ and the minimum distance from the body $\rho$. For example for a beam passing close to the sun we have $\delta\phi=1.75''$ (result found on Landau-Lifshitz) which is not so small for modern accuracies. But whether this is observabe (and by who) depends on the position of the observer and the experimental apparatus that he has.
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$\begingroup$ I am not enough of a mathematician to figure this out but say our moon were made out of neutron star material and it was not emitting photons, would the moon then have enough mass to curve space enough so that we could see the sun with our eyes on earth during a solar eclipse? $\endgroup$– BrandoCommented Sep 22, 2016 at 20:49
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1$\begingroup$ The mass of a neutron star is of the same order of magnitude of the mass of the sun. Thus the above estimate applies approximately. Although I haven't done the precise calculation, this means that the shadowed area would be reduced by a very tiny amount, with no visible effect. $\endgroup$– DelCrosBCommented Sep 22, 2016 at 21:02