E.p. implies no gravitational shielding?; Feynman?

[atyy]

By g(x) do you mean the externally imposed field that we're trying to cancel out? Then the restriction doesn't help, and nulling of the first derivative is still impossible. For example, let the Newtonian gravitational constant be G=1, and let the externally imposed field be created by a non-exotic mass m=+1/2 placed at x=1. Then this externally imposed field has g'(0)=1. Since this g' is positive at x=0, you can't null it at x=0 using further chunks of exotic mass. In one dimension, the gravitational field of any point mass always has g'>0, and therefore you can't use the derivative of one mass's field to null the derivative of another mass's field.

I'm having the naive intuition that if you're allowed to actively place mass, then you can always cancel out the field by adding mass so that the total mass distribution in the universe is a spherical shell. Then you just sit inside it.

 

[bcrowell]

Fair enough, but do you agree that in the limit which defines this form of the equivalence principle, the gravitational effects of anything inside the box must approach having no measurable effect, so any "shielding" relying on the gravitational field of the shield itself won't work in this limit? Of course apart from all discussion of the equivalence principle, it's interesting to think about what form of "shielding" is possible in a larger region of spacetime where the gravitational fields of masses inside the region does have a significant effect, but I wanted to distinguish the two topics of discussion.

Sorry, now I'm losing track of what we were originally debating. As the number of posts in a thread approaches the triple digits, it gets really tough to know who's claiming what. Do you think you could try to make a stand-alone post encapsulating what you're claiming, with careful attention to how the relevant limits are taken? For me, phrases like "no measurable effect" are too loose to be useful in inferring what limiting process you have in mind. We have multiple versions of the e.p. running around here, multiple statements of what is meant by "antigravity" or "gravitational shielding," etc.

 

[JesseM]

Sorry, now I'm losing track of what we were originally debating. As the number of posts in a thread approaches the triple digits, it gets really tough to know who's claiming what. Do you think you could try to make a stand-alone post encapsulating what you're claiming, with careful attention to how the relevant limits are taken?

I'm not sufficiently knowledgeable about GR math to do that, I'm just making an argument from authority based on pages 372-375 of the textbook I cited: there's some common form of the equivalence principle that says measurements in a small room in curved spacetime are indistinguishable in some limit from measurements in a small room in flat spacetime, no? (presumably part of this limit is that the volume of the room and the time period in which the measurements are made are both approaching zero, but there are probably some other conditions as well, like maybe one detailing how \epsilon approaches zero in the quote we can choose U small enough to make all the \Gamma_{jk}^{i} smaller than any given \epsilon > 0 If \epsilon is given by the lower limit of our ability to detect curvature using, say, geodesic deviation, then the result is a region U that we may cautiously consider "flat for practical purposes") And by definition nothing inside a room in flat spacetime can have any gravitational effect whatsoever, so whatever the precise nature of this limit, it must be true that there are no measurable gravitational effects from objects in the room in curved spacetime in this limit.

 

[PAllen]

I understand that we're only zeroing out n derivatives at one point. What you haven't realized yet is that the idea in your #72 doesn't work. To see why it doesn't work, go ahead and try to do the thing I told you was impossible in #74. You will not be able to find a combination of masses that zeroes out the first derivative of g, *even at one point*.

Ok, I finally see your point. But then I kept thinking there must be some escape from this conundrum, because a spherical shell of matter cancels its own gravity (not gravity from outside, as argued to atyy) precisely; and clearly, if the field is zero over a large region, its gradients of any degree must also be zero. How to resolve the paradox? With some real effort (it's been a long time) I see the key is Newtonian gravity is a vector field, and that the problem is actually easy to solve if you go outside the artificial restriction to one line.

In particular, to neutralize g and g' (along a given line) from some source at one point on the given line requires only two masses not the three proposed in my erroneous scheme. The key is that the masses are off the line. Say the source is to the left of the point of interest and we want to neutralize g and the gradient along the source to given point line. Then place two masses on either side of this line, equidistant from it, to the right of the point of interest. Each off axis mass to the right has a gradient component along the line opposite that from the source, and the symmetric placement cancels other gradient components. The angle to the off axis masses allows you to adjust the gradient canceling efficiency (near 45 degrees is the most effective), while adjusting the mass gives you a degree or freedom to cancel g.

I lack the expertise to generalize this to efficient solutions for higher derivatives etc., but observe that the following shows there must be a solution (this was a suggestion of atty in a different context):

build a shell including the source as part of it, and you have canceled the field precisely in a large region by adding mass.

Given we only want point canceling of n derivatives to some precision, I expect there are much simpler solutions.

Thus active gravity shielding is possible (active, in the sense that the masses must be held in place by some external force).

 

[atyy]

But then I kept thinking there must be some escape from this conundrum, because a spherical shell of matter cancels its own gravity (not gravity from outside, as argued to atyy) precisely

My argument is to arrange matter so that there is no matter outside the shell, and no matter inside the shell, except for the laboratory, whose gravitational influence is negligible.

 

[atyy]

In Faraday shielding we wait for t=infinity. In our universe, since it is expanding, won't we automatically get shielded at t=infinity?

 

[bcrowell]

Hi, PAllen,

#94 is interesting. My proof was only for 1 dimension, and was only meant to prove that *some* externally imposed fields were impossible to null out. As you point out, one can construct examples in three dimensions that can be nulled out. For example, the field inside a hemispherical shell can be nulled by adding another hemisphere to complete the sphere.

I'm having trouble visualizing the geometry you're describing in #94. Let's say that I have an externally imposed field \textbf{g}=x\hat{\textbf{x}}, and I want to produce \partial g_x/\partial x=0, while leaving the field itself equal to zero there. Can you tell me what masses you would put at what coordinates in order to accomplish this? (Let's work in units where G=1.)

Nulling in the sense of #66 is definitely impossible in general, in three dimensions without some further restriction on the externally imposed fields, if we say that all partial derivatives up to order n have to vanish. The reason is that I could give you an external field \textbf{g} that had a negative divergence at a point P (as a field created by normal matter typically will), and ask you to null all of the field's first derivatives at P. If you could do this by adding on a nulling field \textbf{h}, then we would have \nabla\cdot\textbf{g}<0, \nabla\cdot(\textbf{g}+\textbf{h})=0, so your field \textbf{h} would have to have a positive divergence, which is impossible without exotic matter.

I've added the one-dimensional thing as an example in my book http://www.lightandmatter.com/html_books/genrel/ch08/ch08.html#Section8.1 (subsection 8.1.3, example 1). There is an acknowledgment to P. Allen at the end of the example. I hope this is OK with you (i.e., you agree that the statements I make in the example are true), and that this is the right form of your name to use.

-Ben

 

[PAllen]

Hi, PAllen,

#94 is interesting. My proof was only for 1 dimension, and was only meant to prove that *some* externally imposed fields were impossible to null out. As you point out, one can construct examples in three dimensions that can be nulled out. For example, the field inside a hemispherical shell can be nulled by adding another hemisphere to complete the sphere.

I'm having trouble visualizing the geometry you're describing in #94. Let's say that I have an externally imposed field \textbf{g}=x\hat{\textbf{x}}, and I want to produce \partial g_x/\partial x=0, while leaving the field itself equal to zero there. Can you tell me what masses you would put at what coordinates in order to accomplish this? (Let's work in units where G=1.)

Nulling in the sense of #66 is definitely impossible in general, in three dimensions without some further restriction on the externally imposed fields, if we say that all partial derivatives up to order n have to vanish. The reason is that I could give you an external field \textbf{g} that had a negative divergence at a point P (as a field created by normal matter typically will), and ask you to null all of the field's first derivatives at P. If you could do this by adding on a nulling field \textbf{h}, then we would have \nabla\cdot\textbf{g}<0, \nabla\cdot(\textbf{g}+\textbf{h})=0, so your field \textbf{h} would have to have a positive divergence, which is impossible without exotic matter.

I've added the one-dimensional thing as an example in my book http://www.lightandmatter.com/html_books/genrel/ch08/ch08.html#Section8.1 (subsection 8.1.3, example 1). There is an acknowledgment to P. Allen at the end of the example. I hope this is OK with you (i.e., you agree that the statements I make in the example are true), and that this is the right form of your name to use.

-Ben

Hi,

Formalities first: fine if you acknowledge me, P. Allen is my actual (partial name). In some more private communication, I could give you more complete attribution, if you want (I have no relation to Microsoft, though we are in the same industry and the same age; bank balance differs).

I don't really want to work out a precise numeric example, but thinking more since post #94, I can make my argument much tighter, and I hope clearer.

Assume a point mass at x=-10, our point of interest (where we want to null g and g' from the source at x=-10) at x=0, and equal (to each other) masses at candidate positions x=1,y=1, and x=1, y=-1. What I am going to argue is by moving the balancing masses further away or closer (along the same line from the origin), and/or changing their angle to the x-axis at the origin (in all cases keeping their distances from the x-axis the equal to each other), we have more than enough degrees of freedom to get any magnitude ratio of g'/g that we want, and that in all cases, the sign of g' for these side masses (taken together) is the opposite of g' from the source at -10 (as is the sign of g). Given that, we find a choice that matches g'/g magnitude for the source we are canceling, then choose masses to match the magnitude for the source values (the ratio g'/g for the balancing masses depends only on distance and angle not on mass).

Ok, put simply (for me), the problem with colinear masses is that g' is positive no matter the sign or value of g. With the off axis masses placed to the 'right' of the origin, we have net g with the correct sign for cancelation, but g' negative, just what we need. This is easily seen by noting that as you move from the left of these masses to the point directly between them, g goes to zero. So we have g decreasing as a funcion of x rather than increasing as for a colinear balancing mass.

To see that we have freedom to match any g'/g magnitude, note that g'/g goes infinity as the balls are moved closer to perpendicular to the x-axis (g goes to zero effectively as cos(angle to perpendicular), g' as sine (same angle)). Note that for some given angle between the balancing balls and the x-axis at the origin, we can make g'/g approach zero by moving the balls further away at the same angle (because g' goes r**-3, while g as r**-2).

Thus, I believe all elements of my argument are established.

My gut feel is that procedures like this will work for higher derivatives, using more balancing masses. The additivity of everything suggest it should work to neutralize any collection of given point source fields. In Newtonian gravity, motion of the source makes no difference (all is instant possition dependent force) except balance balls must move in some complex way.

In GR, my gut is that all this breaks down for moving sources. One line of thought is that I was led to believe this was possible by the spherical shell case. In Newtonian gravity, if this shell rotates, nothing changes. I suspect (you may know for sure) that in GR you get small frame dragging or similar effects. I see no reason to believe these can be canceled. Of course, there are also gravity waves. For E/M waves, the field goes through +/- sign and you can cancel any wave with a phase shifted wave. For gravity waves, there is no negative gravity influence, so I don't see, offhand, how any gravity wave can be cancelled. Thus moving sources in GR seem to produce several types of uncancellable effects. Of course, for weak, slow sources, the Newtonian scheme should work as well in 'practice', given a large supply of self propelled dense matter balls and software to move them as needed.

 

[bcrowell]

PAllen, your argument in #98 makes sense to me, if all you want to do is null out g_x and \partial g_x/\partial x. However, I proved in #97 that you cannot simultaneously null out all nine first derivatives \partial g_{x_i}/\partial x_j in all cases. Because you haven't specified any restriction that would evade that proof, I don't believe that your method works in general. Basically the way I know that your method can't be as general as you think is that if we don't have exotic matter, then you can't independently vary the three i=j derivatives \partial g_{x_i}/\partial x_i, because there are constraints on the divergence of the field.

 

[PAllen]

PAllen, your argument in #98 makes sense to me, if all you want to do is null out g_x and \partial g_x/\partial x. However, I proved in #97 that you cannot simultaneously null out all nine first derivatives \partial g_{x_i}/\partial x_j in all cases. Because you haven't specified any restriction that would evade that proof, I don't believe that your method works in general. Basically the way I know that your method can't be as general as you think is that if we don't have exotic matter, then you can't independently vary the three i=j derivatives \partial g_{x_i}/\partial x_i, because there are constraints on the divergence of the field.

I am not in a position to seriously investigate what can be achieved with balancing masses, so I don't dispute what you say. All I said was 'gut feel' anyway.

Anyway, this thread has clearly established that the relationship between the e.p. and 'no gravity shield' is much more complex than many sources have claimed, starting with pinning down both e.p. and what you mean by a gravity shield; continuing with additional assumptions (but ones the most would accept: no exotic matter, and conservation rules).

 

[bcrowell]

One interesting point to mention about all this is that a bunch of this discussion changes completely in Brans-Dicke gravity. In B-D gravity, you do get gravitational shielding effects, and the pedagogical device described by MTW for defining Lorentz frames fails. It's probably not a coincidence that B-D gravity also lacks the equivalence principle.

 

[PAllen]

One interesting point to mention about all this is that a bunch of this discussion changes completely in Brans-Dicke gravity. In B-D gravity, you do get gravitational shielding effects, and the pedagogical device described by MTW for defining Lorentz frames fails. It's probably not a coincidence that B-D gravity also lacks the equivalence principle.

According to the particular way Clifford Will classifies equivalence principles, he claims Branse-Dicke satisfies WEP (weak equivalence principle), EEP (Einstein equivalence principle), but not SEP (strong equivlence principle. He claims a particular theory of Nordstrom is the only case besides GR that he knows of, that satisfies SEP; but it makes radically false predictions (e.g. light not deflected by gravity).

On my construction in #98: I believe I cancel total g, but only one gradient direction. I don't know offhand how to improve the construction. I thought about a ring of matter, but I think that doesn't really work.