Baron Munchausen Meets Rube Goldberg in Einstein's Elevator

PS No matter how hard the Baron fires his rocket motor, his non-geodesic world line must be inside the forward light cone at each point on that world line. The field of light cones is not parallel in curved space-time so that the bundle of non-geodesics available to the Baron is not identical to what it would be in globally flat space-time. Therefore, one must take the connection field of the source in some global coordinates like the asymptotic flat "Book Keeper coordinates" (Wheeler's term) in

ds^2 = (1 - 2M/r)dt^2 - (1 - 2M/r)^-1dr^2 + r^2(dtheta^2 + sin^2thetadphi^2)

to make the problem simpler let theta = pi/2 equatorial orbit

x^1 is radial r, x^3 is azimuthal phi, the only non-zero LC-connection components then are

(^010) = (M/r^2)(1 - 2M/r)^-1

(^133) = -r(1 - 2M/r)

(^100) = (M/r^2)(1 - 2M/r)

(^313) = 1/r

For the fiducial hovering LNIF shell observers at fixed r

dt(shell) = (1 - 2M/r)^1/2dt

dr(shell) = (1 - 2M/r)^-1/2dr

dphi(shell) = dphi

So for an arbitrary motion of the Baron subject to the above causal light cone restriction, one needs to find the GCT from the shell observer to the Baron to do find the Baron's coincident connection field relative to the shell observer.

i.e. symbolically we have the CONTINGENT inhomogeneous non-tensor transformation from non-geodesic LNIF Shell Observer to the non-geodesic LNIF Baron

(Baron) = (GCT)(GCT)(GCT)(Shell Observer) + (GCT)Grad(GCT)

This is not intrinsic. This is not of any fundamental theoretical interest. It is contingent even though one can formally relate it to curvature, it's Fool's Gold. It's a Chimera. It's The Siren beckoning. Also it is a very complex calculation in general not worth the effort since the Baron would do better to directly measure his local g with a scale.


On Sep 3, 2006, at 8:23 PM, Jack Sarfatti wrote:

The word "fictitious" is a bad one like "hidden variables" in Bohm's reinterpretation of quantum theory. You certainly "feel" a "fictitious" g and it will cause a pointer on a suitable detector to move. Therefore, fictitious forces, or inertial forces, are physical in that they are detectable. However, they are not tensors relative to the relevant symmetry groups, therefore, one cannot construct objective frame-invariants from them under those symmetry groups. In this subtle sense I meant "g-forces are not physical" or "g-forces are fictitious" although in a pragmatic experiential sense they are real!



The covariant equation for the NON-geodesic in ALL frames is

D^2x^u(test)/ds^2 = d^2x^u(test)/ds^2 + (Connection)^uvw(dx^v(test)/ds)(dx^w(test)/ds) = F^u(non-gravity)/m(test)

If Alice is LIF her connection vanishes and she sees the special relativity F = ma

D^2x^u(test)/ds^2 = d^2x^u(test)/ds^2 = F^u(non-gravity)/m(test)

What does the Baron see in his own LNIF rest frame when he fires a small rocket motor on his cannon ball?

D^2x^u(Baron)/ds^2 = (Connection Baron)^u00(dx^0(Baron)/ds)(dx^0(Baron)/ds) = F^u(Baron)/m(Baron)

(Connection Baron)^u00(dx^0(test)/ds)(dx^0(test)/ds) - F^u(non-gravity)/m(Baron) = 0

(Connection Baron)^000(dx^0(test)/ds)(dx^0(test)/ds) - F^0(non-gravity)/m(Baron) = 0

(Connection Baron)^i00(dx^0(test)/ds)(dx^0(test)/ds) - F^i(non-gravity)/m(Baron) = 0

i = 1,2,3 spacelike

The experienced "fictitious" inertial nongeodesic g-force in the Baron's frame is simply

g-force = {(Connection Baron)^i00}

= {F^i(non-gravity)/m(Baron)}

Notice that curvature is completely irrelevant i.e. gradients of the connection play no role whatsoever.

Let the Baron in his LNIF rest frame look at a Alice who is on a geodesic. The Baron's version of Alice's geodesic equation is

D^2x^u(Alice)/ds^2

= d^2x^u(Alice)/ds^2 + (Connection Baron)^uvw(dx^v(Alice)/ds)(dx^w(Alice)/ds) = 0

This simplifies to

= d^2x^u(Alice)/ds^2 + (Connection Baron)^u00(dx^0(Alice)/ds)(dx^0(Alice)/ds) = 0

= d^2x^u(Alice relative to Baron)/ds^2

+ F^i(non-gravity on Baron)/m(Baron)(dx^0(Alice relative to Baron)/ds)^2 = 0

Now above assumes Minkowski space-time.

Suppose all of the above happens in the vacuum curvature field of an SSS source

g00 = (1 - 2M(source)/r) = - 1/grr etc.

in the usual asymptotic coordinates.

Let Alice and the Baron be momentarily close to a given r,theta, phi, t.

For example, in the equatorial plane theta = pi/2 "1" = "r"

(SSS connection)^100 = M/r^2(1 - 2M/r) etc.

One would then have to find the GCT connecting the nearly coincident static shell observer to the Baron. If, for example, the Baron adjusted his rocket motor to be a shell observer, then the GCT is the trivial identity transformation and one can then use

(SSS connection)^100 = M/r^2(1 - 2M/r)

Actually doing a detailed calculation is not trivial.


From: Jack Sarfatti
Date: September 3, 2006 4:58:01 PM PDT
To: Sarfatti_Physics_Seminars@yahoogroups.com
Subject: Re: Zielinski fails to grasp the subtle beauties of Einstein's Vision


On Sep 3, 2006, at 1:16 PM, xerberos2 wrote:


Jack wrote:You misread Einstein's text ...

I'm not misreading his text. Einstein's text is very clear. He is
proposing to treat a fictitious inertial field as if it were a real
gravitational field, so that he can pretend that the accelerating
frame K' is not accelerating.

By "real gravitational field" he means in the sense of Newton's theory.

K' that is non-geodesic in curved space-time is locally equivalent to a geodesic inertial frame in flat space-time with a "real" gravity field. This is Einstein's bridge back to Newton's theory. "geodesic" has two different meanings in the same sentence here.

"Real gravitational g-field" is meaningful in Newton's theory in flat Euclidean 3D space with absolute simultaneity (Galilean relativity v/c ---> 0) where the Newtonian geodesics are straight lines in flat Euclidean 3D space with point test particles moving at constant speed along them. That's the Newtonian geodesic. There is zero g-force on Newton's geodesic.

"g-field" in Einstein's theory means exactly the same thing as in Newton's theory except that the notion of geodesic has changed. An Einstein geodesic projected down into 3D space is generally not a straight line nor is the test particle speed constant. For example the Earth's elliptical orbit around the Sun is geodesic relative to the Sun's curvature field - to a good approximation. Curvature is geodesic deviation. g-forces are non-zero only on non-geodesics created by non-gravity (essentially electromagnetic) forces. There is no necessary intrinsic relationship of a g-force event to the local curvature.

Now, what confuses Zielinski is the following: consider a cannon ball in free fall as in


http://www.zonalibre.org/blog/diversovariable/archives/baron-munchausen.jpg

Newton's explanation: the Baron and the cannonball are NOT on a geodesic, therefore, there is a real gravitational force per unit test mass on both the Baron and the cannonball relative to the frame K' (surface of Earth) that is "inertial" to a good approximation. It is the same real gravitational force per unit test mass g for both

g-force(Newton) ~ GM(Earth)/r^2

Therefore, the Baron feels weightless, i.e. no pressure on his behind from the cannonball since each are falling in exactly the same way at every moment. That is, there is zero g-force in the common rest frame of the Baron and the cannonball.

Einstein's explanation: the Baron and the cannon ball are on a timelike geodesic in curved space-time.

The covariant equation for the geodesic in ALL frames is

D^2x^u(test)/ds^2 = d^2x^u(test)/ds^2 + (Connection)^uvw(dx^v(test)/ds)(dx^w(test)/ds) = 0

This is the covariant

F = ma

with

F = 0

This form-invariant (local frame covariant) equation means.

OBJECTIVE TENSOR TEST PARTICLE ACCELERATION = 0

This is the DEFINITION of a GEODESIC!

THIS IS TRUE IN ALL FRAMES FOR ALL POSSIBLE CURVATURES INCLUDING GLOBALLY FLAT ZERO EVERYWHERE-WHEN.

D^2x^u(test)/ds^2 is the GCT tensor acceleration of the test particle. Its local frame-invariant scalar is

g = (D^2x^u(test)/ds^2D^2xu(test)/ds^2)^1/2

g = 0 on a geodesic - universally true!

Look more closely at the meaning of the geodesic equation. Let Baron Munchausen be on the test particle that is the cannonball in the above picture.

D^2x^u(Baron)/ds^2 = d^2x^u(Baron)/ds^2 + (Connection)^uvw(dx^v(Baron)/ds)(dx^w(Baron)/ds) = 0


d^2x^u(Baron)/ds^2 = Newton's flat space + time kinematical acceleration that is not a GCT tensor.

(Connection)^uvw(dx^v(test)/ds)(dx^w(test)/ds) = inertial "force per test mass" that is a contingent artifact of the local frame of reference. This term even exists in globally flat spacetime when K' is accelerating from an electromagnetic force.

Let Alice be a nearly coincident to the Baron geodesic LIF in curved space-time. What Alice sees is

D^2x^u(Baron)/ds^2 = d^2x^u(Baron)/ds^2

= 0

(Connection Alice LIF) = 0

The size of Alice's LIF is such that the gradients in (Connection Alice LIF) are ignorable. You can think of the LIF as a ball at the bottom of a potential well with a very small zero point jiggle - this is only a rough analogy.


http://rsc.anu.edu.au/~sevick/groupwebpages/images/animations/capture_3D0282.jpg

Let Bob be a nearly coincident to the Baron non-geodesic LNIF observer, then in Bob's POV

D^2x^u(Baron)/ds^2

= d^2x^u(Baron)/ds^2 + (Connection Bob)^uvw(dx^v(Baron)/ds)(dx^w(Baron)/ds)

= 0

(Connection Bob) =/= 0

Because an electromagnetic force is acting on Bob.

Finally in the Baron's rest frame, which in this case is also geodesic

D^2x^u(Baron)/ds^2 = d^2x^u(Baron)/ds^2

= 0

dx^i/ds = 0

i = 1,2,3 spacelike

dx^0/ds = 1

This covers all of the cases.

Homework Problem: Put an external force on the Baron. Describe all the cases.

In answer to Z's question about Wheeler. When Wheeler says gravity is curvature he means tensor "geodesic deviation" he does not mean contingent non-tensor non-geodesic "g-force."

"Gravity" and "gravity field" mean different things in different contexts. Usually this is not a problem for physicists to get the nuance intended in each specific. It is a problem for Z.