# Mach's principle and local fields

1. Sep 18, 2014

### e2m2a

One argument against Mach's principle is the speed of light restriction. How could the distant cosmic mass of the universe instantaneously have a local effect on an accelerating mass? But could we view this from the perspective of a field that is already presently locally at all points in space, a gravitational field that originates from the combined cosmic mass of the universe? Thus, a mass experiences an inertial force (dynamic gravitational force) instantly because a field is already present at the point of acceleration.

This would be similar to an electron that is initially at rest (setting aside Heisenberg's uncertainty for a moment) within a magnetic field, sourcing from a magnet 1 light year away. As soon as the electron moves perpendicular to the field, instantly the electron experiences the Lorentz force.
However, the disturbance of the electric field of the electron would take 1 year to "affect" the magnet.

Any thoughts on this, anyone?

Last edited: Sep 18, 2014
2. Sep 18, 2014

### Simon Bridge

Mach's principle is not usually very well or clearly stated - it expects the changing circumstance to have an immediate effect as in rotation of the stars having a local effect right away you start spinning. But this would require the distant stars to establish a potential about you which has the curious property of changing in your vicinity when you start spinning ...

But what you are approaching is the normal Einstein description where the stress-energy of the whole universe determines the local metric.

3. Sep 18, 2014

### e2m2a

So, could it be stated that inertia is due to an acceleration with respect to the local metric, a metric which is already tempered by the stress-energy of the whole universe? To me this would eliminate the speed of light cause and effect conundrum. Can the metric be viewed as a "real" thing that has "real" properties that affect local acceleration, just as a magnetic field affects the actions of moving particles? Another words, is inertia a gravitational effect due to an asymmetry that arises with respect to the local metric caused by acceleration with respect to the metric?

4. Sep 18, 2014

### Simon Bridge

The metric is determined by the stress energy...

GR in it's current formulation is not compatible with the details of Mach's principle - which I'm having trouble finding well described since it is only of historical interest.
i.e. MP does not allow gravity waves, and local effects depend on the instant-by-instant updates in relative positions with the distant objects and not a local field. If you modify the principle to allow for locality, getting around the light-speed limit, you end up with general relativity.

That's it's job yes. But it does not make Mach's principle work. It is instead of Mach's principle.
Since Mach's principle is so often poorly worded, it is easy to confuse them.

The metric is a bit of math that can be used to derive a lot of the local behaviors.
It's not real in the sense that you can touch it.

No. That's Mach's principle. The stress-energy tensor gets you the gravitational effect.

I'll give you some notes:
http://preposterousuniverse.com/grnotes/grtinypdf.pdf [Broken]

Last edited by a moderator: May 6, 2017
5. Sep 18, 2014

### Staff: Mentor

No, none of that makes any sense.

The metric is not a real thing, it is a mathematical object that describes the geometrical relationship between nearby points in curved spacetime, and it provides the mathematical description of the gravitational forces at a given point. Saying "An acceleration with respect to the local metric" is like saying "an acceleration with respect to the set of all odd numbers" - makes no sense at all.

There is an important relationship between inertia and gravity, but inertia is most certainly not a gravitational effect. At the sound-bite level, we could say that mass curves spacetime, and gravitational effects appear when objects try to move inertially in a curved spacetime... but that's a soundbite, not a serious mathematical explanation.

You might want to search through this forum for a very short animation that member A.T. has created explaining Newton's apple from the perspective of general relativity... It's a better starting point than most of what you'll find on teh web.

6. Sep 18, 2014

### Jonathan Scott

I totally agree with this. I don't know why some people associate Mach's Principle with instantaneous action at a distance, although Woodward (of the reactionless drive idea) may have had something to do with it.

GR predicts linear and rotational frame-dragging effects which are strongly reminiscent of Mach's Principle, and the rotational effects have been confirmed experimentally.

However, in GR, the gravitational constant G is a universal constant whereas in most forms of Machian theory it is a property of the distribution of masses in the universe and cannot be constant. The experimental evidence says that G does not seem to be varying detectably with time at all, at least not enough to be consistent with most forms of Machian theory.

Personally I find it very odd that GR is apparently compatible with Machian results yet apparently denies the possibility of this being more than some form of temporary coincidence.

7. Sep 18, 2014

### PAllen

There are arguments that GR is as Machian as it could be and still be consistent with observation (see Julian Barbour, among others). There is also a more general idea that in a closed universe the distribution of matter completely determines what is inertial versus non-inertial motion. In other cases, it is boundary conditions plus overall mass distribution that determine what is inertial motion.

Where Machian analogies break down is that there is no limit (in GR) of zero inertia for an isolated body in an empty universe. Boundary conditions completely determine what is inertial, which is not Machian at all.

However, if you go beyond vague 'Machian character', you need to formally specify what a Machian theory is. One of the few plausible, fully worked out, examples is Brans-Dicke theory. So far, all experiments favor GR over Brans-Dicke theory.

Last edited: Sep 18, 2014
8. Sep 18, 2014

### stevendaryl

Staff Emeritus
Hmm. I think that's an interpretation of the physics, rather than being an inherent part of the physics. You can view the metric as a tensor-valued field, like the electromagnetic field. It's special in that it couples to absolutely everything, but it still can be viewed as a field.

9. Sep 18, 2014

### stevendaryl

Staff Emeritus
Feynmann worked out a version of electrodynamics (his absorber theory) in which, in a sense, only electrons (and other charged particles) were real. The electromagnetic field was completely determined by the history of the locations and velocities of charged particles. So there was no independent degrees of freedom associated with the electromagnetic field. It was an interesting exercise, although it was rejected as not a very good way of doing electrodynamics. It required a somewhat odd additional assumption that goes beyond our observations (although it's consistent with our observations, I think): He had to assume that every photon is eventually absorbed by a charged particle. If there are "free" photons that never get absorbed, then they constitute degrees of freedom unaccounted for by the motions of charges.

Anyway, it seems to me that it should be possible in theory to do the same thing with General Relativity. In GR, the motions of matter/energy do not uniquely determine the spacetime metric. There can also be source-free fluctuations in the metric (gravity waves). But maybe if it is assumed that all gravity waves are produced by matter and are eventually absorbed by matter, then in a similar way as with Feynmann's theory, the gravity wave degrees of freedom (captured by the Weyl tensor, which is the part of the Riemann curvature tensor that is not determined by the stress-energy tensor) can be similarly removed. What would be left over would be a pure matter theory, with no independent gravitational "fields".

10. Sep 18, 2014

### PAllen

Interesting, but wouldn't you need additional assumptions to get rid of boundary conditions? For example, Kruskal manifold is source (matter) free, no GW, yet has a non-trivial pattern of inertia. It seems if you try to go the direction of banning this, you just end up with the well known view that a closed universe can be considered Machian.

11. Sep 18, 2014

### stevendaryl

Staff Emeritus
That might be a requirement. In an open universe, it's probably not a reasonable assumption that all gravity waves are eventually absorbed--there could be an outward-propagating shell of gravity waves that goes out to infinity without every disappearing entirely.