Sunday, September 03, 2006

Newton vs Einstein

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

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


F = 0

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




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.

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


= 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.

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