Are annulled gravitational fields detectable?

[synch]

There is a classic electrostatic example of two slabs of opposite charge in parallel, and the electric field and so on.
I am wondering about the gravitational analogue with two equal slabs of mass. The intuitive view to me, is/was that there will be a gravitational field between the two slabs. However the net effect on a test mass at a point between the two will be zero ...so is the gravitation there detectable ? ( away from edge effects and so on ) .
I assumed the field would be uniform but now I wonder if that is just a combo of two expanding fields (?)

 

[Baluncore]

Maybe one of the Lagrange points, L1 ?
https://en.wikipedia.org/wiki/Lagrange_point#L1_point

There is also the Cavendish Experiment.
https://en.wikipedia.org/wiki/Cavendish_experiment

 

[renormalize]

I am wondering about the gravitational analogue with two equal slabs of mass. The intuitive view to me, is/was that there will be a gravitational field between the two slabs. However the net effect on a test mass at a point between the two will be zero ...so is the gravitation there detectable ?

Are you asking about Newton's gravity or Einstein's (general relativity)?

 

[Nugatory]

The intuitive view to me, is/was that there will be a gravitational field between the two slabs.

Your intuition is mistaken here. Away from the edge effects and if the two slabs are identical the field is zero. That’s just another way of saying

However the net effect on a test mass at a point between the two will be zero

 

[Herman Trivilino]

There is a classic electrostatic example of two slabs of opposite charge in parallel, and the electric field and so on.
I am wondering about the gravitational analogue with two equal slabs of mass.

It doesn't exist because there is no such thing as negative mass. The electric force is can be either attractive or repulsive but gravity is attractive only.

Thus we need two kinds of electric charge (positive and negative) but only one kind of mass (positive).

 

[sbrothy]

Ooooh intuition..... bad gauge.

EDIT: Or perhaps I meant guide. Non-native speaker here.

 

[Nugatory]

Ooooh intuition..... bad gauge.

EDIT: Or perhaps I meant guide. Non-native speaker here.

Native speaker here, agrees with your edit. A gauge produces measurements, a guide points out a path to a conclusion or destination.

 

[Mike_bb]

It doesn't exist because there is no such thing as negative mass. The electric force is can be either attractive or repulsive but gravity is attractive only.

Thus we need two kinds of electric charge (positive and negative) but only one kind of mass (positive).

What do you think about following concepts?
https://en.wikipedia.org/wiki/Negative_mass
Tachyon with imaginary mass: https://en.wikipedia.org/wiki/Tachyonic_field

 

[Matterwave]

It doesn't exist because there is no such thing as negative mass. The electric force is can be either attractive or repulsive but gravity is attractive only.

This is true, and tbh was my first reaction to OPs question. However, I think OP actually means to compare the gravitational situation with the case of two infinite parallel same charged slabs. Between those two slabs, the fields cancel. And in the case of Newtonian gravity, the gravitational fields cancel as well. Post #4 points this out and the OP himself guesses:

However the net effect on a test mass at a point between the two will be zero

In Newtonian gravity, you can still apply Gauss's law and get the same answer as the case in EM (except you only have one charge).

The case using GR is less clear to me since GR is non-linear. I don't even know what the solution for a single infinite slab is in GR. Maybe by pure symmetry you can still work out the same answer, but I'm not sure.

But there is an analogous situation that I do know the analysis of, and that's the case of being inside a hollow spherical shell. By Newton's shell theorem the interior region has zero field and as far as I know this is entirely undetectable (gravitationally) in Newtonian gravity. However in GR, there can be detectable effects if the shell is spinning which gives rise to precession of gyroscopes inside. That's the frame dragging effect which has no analogue in Newtonian gravity.

So perhaps if the infinite slabs are moving in some way, there might be a similar frame dragging effect in GR (if an infinite slab solution is well defined in the first place, which I'm not sure of). It's actually quite an interesting question.

Another point that arises to me, but I haven't thought of in detail at all. In QM we can detect loop integrals of the EM vector potential (via a phase shift) -- the Aharonov Bohm effect. So at least in some sense the gauge potentials are not fully hidden from us. I wonder if there is some analogue in the gravitational case as well.

 

[berkeman]

I think OP actually means to compare the gravitational situation with the case of two infinite parallel same charged slabs. Between those two slabs, the fields cancel.

I may have missed it if the OP has said something different from his first post, but I don't think he was asking about infinite slabs:

The intuitive view to me, is/was that there will be a gravitational field between the two slabs. However the net effect on a test mass at a point between the two will be zero ...so is the gravitation there detectable ? ( away from edge effects and so on ) .

I don't know about infinite slabs of mass, but at least for finite slabs, the gravitational fields would only cancel at the midpoint between the slabs, no?

 

[Matterwave]

I may have missed it if the OP has said something different from his first post, but I don't think he was asking about infinite slabs:

I interpreted "away from edge effects" as meaning "treat the slabs as infinite". I could be wrong of course.

I don't know about infinite slabs of mass, but at least for finite slabs, the gravitational fields would only cancel at the midpoint between the slabs, no?

It is standard in EM to analyze "infinite slabs of uniform charge". This is an idealization of course, but in this limit we can ignore edge effects. Only in that limit can you easily apply Gauss's law with a pillbox geometry plus some symmetry arguments to get the result that the field strength is constant (not dependent on distance to the slab, since the slab is infinite). If you put two (infinite) slabs with same charge density next to each other and you add the fields, they cancel in the region in between the two slabs, not just at the midpoint.

For finite slabs, the analysis is complicated by edge effects, but yes, intuitively at least, one expects only cancelation at the midpoint.

Transferring this analysis to the gravitational field as described by Newton should be fine because there is an analogue of Gauss's law that applies there. We start with the perhaps more familiar Poisson's equation: $\nabla^2 \Phi = 4\pi G\rho$. Now take $\vec{g}=-\nabla \Phi$ and find $\nabla \cdot \vec{g} = -4\pi G\rho$ completely analogous to $\nabla \cdot \vec{E} = \rho/\epsilon_0$.

But as I noted, I don't know how this (infinite slabs) works in the GR. That's why I offered my (rough) analogy with the "inside a spherical shell" case.

 

[renormalize]

But as I noted, I don't know how this works in the GR case.

The physics of planar mass slabs is much more involved (and "weird"!) in general relativity as compared to Newtonian gravity. For example, see:
Infinite slabs and other weird plane symmetric space-times with constant positive density
But the author points out that, under specialized conditions, two planar masses with a vacuum gap between them can indeed have a flat metric metric in that gap (i.e., no gravitational field):

 

[Matterwave]

The physics of planar mass slabs is much more involved (and "weird"!) in general relativity as compared to Newtonian gravity. For example, see:
Infinite slabs and other weird plane symmetric space-times with constant positive density
But the author points out that, under specialized conditions, two planar masses with a vacuum gap between them can indeed have a flat metric metric in that gap (i.e., no gravitational field):
View attachment 374531

Nice! Yeah I was not aware of such papers. I did a mild internet search myself and found this one: https://pubs.aip.org/aapt/ajp/artic...vistic-infinite-plane?redirectedFrom=fulltext (free arxiv: https://arxiv.org/abs/0708.2906)

I wonder if the authors come to the same conclusion or if they cite each other lol. Also interesting that the Arxiv preprints are only 1 month apart in 2007.

 

[Baluncore]

I don't know about infinite slabs of mass, but at least for finite slabs, the gravitational fields would only cancel at the midpoint between the slabs, no?

Yes, if the slabs were parallel and identical.
If different, then the neutral point would not be at the barycenter, it would be at L1.

 

[Markus Hanke]

so is the gravitation there detectable ?

Place a clock anywhere between the two massive slabs, and you will find it being gravitationally time-dilated wrt a reference clock someplace far away from all gravitational sources - even if the metric between the slabs is locally Minkowski.

 

[synch]

Thank you for all your replies, I am reading them with great interest !
De-learning Newton is painful :( ...
:)