## Thursday, December 18, 2014

### A Look at General Relativity and Tensors (Part 1)

1. The Principle of Equivalence.

This could be called the heart of Einstein's General Theory of Relativity.  In its most elementary form, the principle states that gravitational and inertial masses are indistinguishable. Inertial mass refers to that which   initially moves uniformly in a straight line and is then subjected to some acceleration (a) producing a force F (and deviation from uniform motion) such that F = ma (or the mass m = F/a).  In principle, it should be possible to design an experiment to distinguish between inertial masses, say m1 and m2, on the basis of the accelerations imparted to them by the same force F. In general, one can say:

a2/ a1 = m1/ m2

Thus, if one measures the ratio of accelerations: a2/a1 = 2, then the masses are in the exact opposite relation: m2/m1 = 1/2.  In other words, if m2 accelerates at twice the rate of m1, it must have only half the inertial mass of m1.

Gravitational mass, on the other hand, refers to that measured with respect to the force of  attraction (weight or W) of earth's gravity field, thus m = W/g where g is the acceleration of free fall. Thus, the principle of equivalence maintains that the following are interchangeable for all reference frames:

W/g  =  F/a

In other words, the laws of physics are applicable to all  inertial reference frames. More importantly, there is no experiment that can be devised to discriminate between F and W in any of these reference frames. In a broader context, the principle means dispensing with the action of mysterious, long range forces (e.g. force of gravitation) and replacing them with the natural local action due to a deformation in spacetime.[1]  This is a major insight, which hasn’t been fully appreciated.

For example, it implies that optical, electric and magnetic phenomena are subject to the laws of physics. As a case in point, light waves passing near a massive star accelerate by virtue of the fact their energy has mass, thereby disclosing a curved trajectory or deflection. Such ‘bending’ of starlight has been confirmed by measurements from photographic plates made during total solar eclipses, and compared to images of the same star without the Sun in the line of sight.

At a deeper level, the Equivalence Principle motivated the search for a refined mathematical infrastructure, resulting in tensor calculus. This tool enabled easier transformations between differing coordinate systems and reference frames. In the course of tensor analysis of Einstein’s field equations, the Big Bang emerged naturally as a solution with matter present. We will see tensors in the next section and the field equations in Part 3.

It helps to explore further misconceptions to see a more general applicability of the equivalence principle, and this is done using the comparative of two rockets, A and B, shown below.  In rocket (A) one stipulate an “inertial” or non-accelerating reference frame in which there is a gravitational field of intensity g acting. For rocket (B) one has a case of the rocket accelerating uniformly at the rate g i.e. equal to the rate of free fall in Earth’s g-field, or 9.81 ms-2

Einstein’s Principle of Equivalence requires the complete physical equivalence of the two rocket systems.

Fig. 1. Does the Principle of Equivalence apply to two rockets A and B? (The depiction of one critic - Stephen Gift - that it does not- see his explanation below)

According to critic Stephen Gift, referring to the Principle of Equivalence:[2]

In rocket A the observer experiences a force W as a result of the gravitational field. There is also an equal and opposite floor upthrust F that acts at the soles of his feet. These two forces W and F acting on the observer keep him inertial. In rocket B only the upthrust F acts on the observer.  There is no gravitational force as in rocket A. The observer in B has only the force F acting on him thereby rendering the two systems dynamically different. This is completely contrary to Einstein’s Equivalence Principle and it therefore must be wrong.”

But is it really wrong?  Perhaps the most basic error made is in creating an additional force (in rocket A) where only one is needed. Not processed is the fact that it is precisely the upthrust or reaction force which creates the weight W!

For a person standing on a support in such a rocket, we have W = mg. But what if he’s in an elevator instead and the elevator is in free fall?  In this case there is no support so that the acceleration of the elevator a = g  and the relevant force equation is:

F = m(g – a) = m(g – g) = 0

So the observer is in free fall.  If there is an upward acceleration (similar to rocket B) then of course we have, assuming a < g:

F = m(g – a)  = mg – ma

If a is g/ 4,  for example, then:

F = m(g – a)  = mg – m(g/4) = 3mg/4

Again, the existence of an upthrust or reaction force is simply incorporated into the problem with the recognition that the weight is the reaction force from U. This is a basic principle such that forces always occur in pairs! You cannot single out or separate an upthrust from the weight.  In this case, it doesn’t matter if the critic only stipulates a single force F acting on the observer in B, because we know it must be paired with its companion force, W. Hence, the representation is in error.

Thus, the observer in rocket A will be unaware of anything different from the observer in rocket B.  Even though no field intensity downward has been assigned, a uniform acceleration opposite to the direction F (which equates to the weight) has by the very application of Newton’s 3rd law.  The 3rd law again states that “for every action (force) there is an equal but opposite reaction" (counter force). It can be illustrated as shown below for a body resting on a flat surface.

Fig. 2. Illustrating Newton’s Third Law of Motion.

Here: F(AB) = m(A)g (the weight of block A) and F(BA) is the normal force, N acting against it.
Although the critic attempted to portray the conditions of the two rockets differently by resort to Fig. 1, from the point of view of the observers and the forces acting they are equivalent.

2. An Introduction to Tensors.

Before we can proceed much farther it is necessary to introduce some basic features of tensors – which mathematical entities proved to be the keys to Einstein’s creation of general relativity. This introduction is not meant to be in any way comprehensive, only to show the basic properties and then how they are used in his field equations.

We consider, to fix ideas, the motion along some defined curve s as shown below:
Fig. 3. Curve in space associated with particle motion and an element ds.

Along the curve we also find an element ds, we can write for s:

S = s (x, y, z) = s (x1, x2, x3) = s(x i)

Further: ds2 = d x i  d x i = dx2 + dy 2 + dz2

Which can also be expressed:

ds2 x i / q j  x i / q k dqj dqk = g ik  dqi dqk

where the superscripts are used to denote particular contravariant operations. Then  g ik is a matrix which we call the “metric tensor”.  Or:

g ik   =

(1.....0...............0)

(0.....r2...............0)

(0.....0........r2 sin
f)

Further,   g ik  dqi dqk
=

(1.....0...............0)  (dr2 )

(0.....r2...............0) (d
q 2)

(0.....0........r2 sin
f) (df2)

So the operations applied to matrices can be applied to tensors.

In using tensors we take care with the subscripts and superscripts and use the first for  covariant tensors and the second for contra-variant tensors.

Basic terms:

A tensor of rank 2 is a dyad.

A tensor of rank 1 is a vector.

A tensor of rank 0 is a scalar.

The most basic tensor of all is the unit tensor, defined:

=  i^ i^ + j^ j^ + k^ k^  =

(1......0........0)

(0........1.......0)

(0........0.......1)

Also:

1 × C  =  i^ Cx + j^ Cy + k^ Cz

In all the above, of course, we have yet to introduce time, but will in the next section to do with tying geodesics to the Principle of Equivalence.

Further properties:

A tensor is symmetric if:  T i j   = T j  i

A tensor is anti- symmetric if:  T i j   = -  T j  i

The latter will look something like this:

(0 ……a12………a 13)

(-a12…..0……….a23)

(- a13……-a23……0)

Doubled dummy indices, e.g. ii, jj, kk refer to the trace of a matrix, or the sum of the diagonal elements.  For example, if: i^ i^ + j^ j^ + k^ k^  =

(1......0........0)

(0........1.......0)

(0........0.......1)

Then Tr =  (1 + 1 + 1) = 3

Problems:

1. If  =  i^ i^ + j^ j^ + k^ k^

Write out the expression for 1 × D

2. a) Provide a matrix which satisfies:  i^ i^ + j^ j^ + k^ k^  = 7/2

b) Write out the trace for the metric tensor.

c) Give one example of  3 x 3  tensor, then show how it might contain an anti-symmetric and symmetric  tensor (also how to go from one form to the other).

3. Find the trace (Tr) of each of the following matrices:

(2......0........0)

(0......8 p.......0)

(0........0.......p)

----------------------------------------

(2 r......0........0)

(0........2L.......0)

(0........0......r /2 )

-------------------------------------------
(p ..........0  ........0 .........0)

(0 ........ p/ 3....... 0........0)

(0.........0 ......p/ 6........0)

(0.........0........0......p/7)

1. For example, the Earth revolving around the Sun because of a long range gravitational force emanating from the Sun. In the Einstein view, one  visualizes a deep ‘pit’ or ‘well’ surrounding the Sun - as a result of its mass. An analogy would be a lead ball placed in the middle of a suspended rubber sheet. In either case local space-time is deformed by the mass.
[2] Journal of the Barbados Astronomical Society, Vol. XIII, p. 50.