As small as it may be, does every 'thing' have a gravitational pull? That is, something with mass or energy. No matter how obsolete or negligible it may be, is it there? If so, how is it calculated? What does 'it' affect?
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$\begingroup$ Related: physics.stackexchange.com/q/6197/2451, physics.stackexchange.com/q/10612/2451 , physics.stackexchange.com/q/22876/2451 and links therein. $\endgroup$– Qmechanic ♦Commented Apr 4, 2013 at 15:07
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5 Answers
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Yes, everything generates a gravitational field, whether it is massive or massless like a photon.
The source of the gravitational field is an object called the stress-energy tensor. This is normally written as a 4 x 4 symmetric matrix, and the top left entry is the energy density. Note that mass does not appear at all. We convert mass to energy by multiplying it by $c^2$ (as in Einstein's famous equation $E = mc^2$) and then put in the energy. So even a photon generates a gravitational field because although it has no mass it does have energy.
It's surprising what else is in the stress-energy tensor and can therefore generate a gravitational field. For example pressure and shear stress appear. It's even been suggested that a gravitational field could be generated by gravty itself i.e. the energy of the gravitational field generates the curvature that creates the field. The resulting object is called a geon, though I should emphasise that no-one has proved these could exist and most of us think they probably can't.
answered Apr 4, 2013 at 14:47
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$\begingroup$ +1: I think I slept, thereby forgetting the "mass-energy" while writing the answer ;-) $\endgroup$ Commented Apr 4, 2013 at 14:54
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1$\begingroup$ wait - stress-energy can generate a magnetic field? next you tell us anti-gravity is real (SCNR ;)) $\endgroup$ Commented Apr 4, 2013 at 15:02
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As small as it may be, does every 'thing' have a gravitational pull?
Newtonian: Not everything. But, only massive objects. Definitely not a photon...
GR: Every single energetic object curves spacetime the same way as massive ones, as a consequence of mass-energy $mc^2$. So, a photon does curve spacetime, as it contains energy. (as John mentioned...)
No matter how obsolete or negligible it may be, is it there? If so, how is it calculated?
For instance, let's take a proton. As it has charge, it is affected by electric & magnetic fields. So, Mass spectrometers can be used to deflect them. By applying a magnetic field perpendicular to the plane of the particle, it's made to traverse a circular path and in this way, it's mass can be determined.
What does 'it' affect?
This is the most important part. A massive body (though it's negligible) does affect another massive body. The force is much negligible. Even then, there's always the force. If there isn't any force like that, these interstellar clouds wouldn't have had any gravitational pull and no stars, planets, etc...
answered Apr 4, 2013 at 13:48
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2$\begingroup$ Actually photons do have a gravitational field (though you must use GR to calculate it). You can even make a black hole with enough electromagnetic energy together in one place. $\endgroup$– MichaelCommented Apr 4, 2013 at 13:50
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2$\begingroup$ To be fair to you it is an effect that nobody will ever observe. :D $\endgroup$– MichaelCommented Apr 4, 2013 at 13:56
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1$\begingroup$ I'm the kind of person that likes to nitpick about the existence of in-principle-unobservable signals (have you heard of standard model baryon number violation? How about laboratory scale effects of certain dark energy models..?), but I completely understand that there exist people who have better things to do with their time. :) $\endgroup$– MichaelCommented Apr 4, 2013 at 14:08
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1$\begingroup$ @Prathyush: Ahh... This has become a debate. The only reason I didn't revise (now revised) is because I don't wanna bump the question again & again in activity. So, I inserted a quote:
Newtonian;-) $\endgroup$ Commented Apr 4, 2013 at 15:02
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$$F_{gravitational}=G\times \dfrac{m_1 m_2}{r^2}$$ For as much small mass as you get eg. $m_e=9.1\times 10^{-31}kg$.
$m_1,m_2$ are masses of objects between whom force is calculated and $r$ is distance between them.It is always attractive in nature along the line joining two particles.$G$ is a gravitational constant of the order $10^{-11}$.
Though for large distances and small masses this force is negligible.That is why you can't feel a attraction towards your pc :) and for large masses it is too important eg. earth - sun system is bound by the gravitational force only.
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$\begingroup$ I provided the answer to second part
how is it calculated?$\endgroup$– ABCCommented Apr 5, 2013 at 3:50
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Why would mass be negligible or obsolete? You can assume a mass to be negligible to facilitate your calculations, but in reality every mass has its own gravitational field and affects every other mass in its environment. If you want to go more into the the reason why matter has gravity, you'd probably need to research quantum mechanics.
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I think that there is a confusion of time as a metric versus time as a dimension. Metrics are culturally arbitrary but dimensions are real. Cultural metrics for time are so common, they need no proof. Time as a dimension is no more theoretical than spatial dimensions and there are also culturally arbitrary metrics for them as well. Considering that your knowledge of string theory and quantum theory is surely much greater than mine, I think that you will conclude that the time dimension in those theories has been well establish no matter how you measure it.
