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Lately I visited a 3T NMR machine (subject to a research project). It got me thinking about maxiums. I know the maximum magnetic fields we have observed are quite a lot stronger than that (field strenghts on some stars), and there seems to be a theoretical maximum value for magnetic fields although the exact value can be disputed.

But, now for the questions: what is the maximum gravity field or value we have observed or possibly calculated? And is there a theoretical maximum?

ghellquist
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In general relativity, we don't measure the strength of the gravitational field as an acceleration $g$, the way we would do in Newtonian gravity. The equivalence principle tells us that $g$ can be anything we like, based on the frame of reference we choose. Instead, we measure gravitational fields using measures of curvature. These measures of curvature naturally have units of inverse length squared (in coordinates that have units of length). The strongest curvature that we expect to be describable by classical general relativity is $1/\ell_{Planck}^2$, where $\ell_{Planck}$ is the Planck length.

  • Gravity has nothing to do with the curvature of space (see Ellis wormhole for example: https://en.wikipedia.org/wiki/Ellis_wormhole), but is caused by a gradient of time dilation, which is infinite at the horizon in any reference frame. – safesphere Nov 28 '19 at 18:03
  • @A.V.S.: Thanks for the correction. Fixed. –  Nov 29 '19 at 19:34
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    @safesphere: Your comment is incorrect. We don't even have a way of defining a gravitational time dilation in a non-static spacetime. –  Nov 29 '19 at 19:35
  • Our inability to define something in certain cases doesn't make my comment incorrect. Gravity is caused by movement in time. When time is non uniform, such as in the gradient of the time dilation, the movement in time results in acceleration. The physics is that things fall to where time moves slower. Our inability to "define" or calculate something doesn't mean this is not happening. – safesphere Nov 30 '19 at 13:30
  • You cannot limit the "measures of curvature" only to the curvature scalars, because they don't uniquely define the spacetime. Other "measures of curvature", such as certain components of the Riemann curvature tensor in the coordinates of physical space and time, diverge at the horizon. See: https://physics.stackexchange.com/questions/295814/ – safesphere Nov 30 '19 at 14:52
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The maximal magnetic (and electric) field is likely set by the Schwinger limit where QED effects become appreciable and pair production starts. The corresponding limit for gravity is when quantum gravity starts mattering. This presumably starts to happen when we approach the Planck scale, for example when the field causes accelerations on the order of the Planck acceleration $5.560\cdot 10^{51}$ m/s$^2$.

These limits do not mean there cannot be stronger fields, just that very different kinds of physics will be going on.

Anders Sandberg
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  • I don't think this is right. The equivalence principle says that just by a change in coordinates, I can make an acceleration be anything I like. I can change to coordinates where the accelerations are $10^{1000}$ m/s2, and classical GR will still work just fine. –  Nov 28 '19 at 16:45
  • And how big will the Unruh effect be then? Note that we are not doing this in just classical mechanics, but in the realm where quantum field effects matter. – Anders Sandberg Nov 28 '19 at 16:48
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    I think you're oversimplifying or misunderstanding the Unruh effect. Its interpretation is controversial, but it clearly can't be used as a criterion for the onset of quantum gravity effects in the way you seem to be imagining. The value of $g$ relative to a static observer approaches infinity as you approach the event horizon of a black hole, but classical GR works just fine at the event horizon and inside it. We don't even have a $g$ inside the horizon, so I don't know how you'd even apply your criterion. These are all symptoms of the fact that $g$ is the wrong way to measure field strength. –  Nov 28 '19 at 17:02
  • @BenCrowell There is acceleration (or more precisely deceleration) inside the horizon. Things moving in space along the Schwarzschild coordinate $t$ decelerate from an infinite speed $dt/dr$ at the horizon to a full stop before hitting the singularity. Assuming the Schwarzschild metric on the inside. – safesphere Nov 28 '19 at 17:51
  • @BenCrowell - My point is, even free-falling observers will note non-inertial effects due to heating of an Unruh vacuum or just using a Forward gravitational sensor to measure the metric tensor. And of course one can watch how nearby particles accelerate away from each other. This has nothing to do with being close to any event horizon. – Anders Sandberg Nov 28 '19 at 23:38
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    @safesphere: What you're defining is a coordinate acceleration, which can be given any value we like by picking whatever coordinates we like. –  Nov 29 '19 at 19:36
  • @AndersSandberg: And of course one can watch how nearby particles accelerate away from each other. Here you're giving an operational definition of tidal effects, which are measured by part of the curvature tensor. This is also true in Newtonian gravity: tidal effects are distinct from gravitational accelerations, and are measured in different units. –  Nov 29 '19 at 19:39
  • @BenCrowell You simply repeat the equivalence principle, but it is local only. You cannot cancel the gravitational acceleration globally by choosing a frame. So your objection does not hold. Your relevant earlier point that "there are no global frames" also is misleading. Global frames may or may not exist depending on the case, but frames certainly exist locally in areas where the gravitational acceleration is non uniform. I can easily define a frame to contain the Solar system, but no such frame would cancel the gravitational acceleration everywhere in this system. – safesphere Nov 30 '19 at 13:51