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
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)
(0.....r2...............0)
(0.....0........r2 sin f)
Further, g ik dqi dqk
=
(1.....0...............0) (dr2 )
(0.....r2...............0) (dq 2)
(0.....0........r2 sin f) (df2)
(0.....r2...............0) (dq 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:
1 = 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 1 = i^ i^ + j^ j^ + k^ k^
1. If 1 = 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:
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.
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