**Preliminary**

What are the equations needed to go from Newtonian Physics to General Relativity? In Is Newtonian Gravitational Fields Valid Within General Relativity? we showed in the case of a weak gravitational field and low velocity how the Einstein Field Equations (EFE)

(A) G

_{μν}= 8πGc

^{-4}T

_{μν}

yield Newton's law of gravity,

(B) ∇

^{2}φ = 4πGρ

However, historically, Einstein guessed equation 16, from the above reference reproduced below,

(C) g

_{00}= 1+ 2φ/c

^{2}

The question is: how do you get from equation C to equation A?

**Thought Experiment**

Einstein was convinced that gravity would affect light. But how? He imagined this thought experiment. Suppose that gravity does NOT affect light. Then one could construct the following scheme:

One could release a particle of mass m at rest at a distance h from the ground (fig. 1a). Its potential energy(PE) would be converted to kinetic energy (KE). Einstein knew that this particle could then be converted to a photon, and then that photon could climb against gravity, reconverted to a particle with KE ≠ 0, and would now fall, picking up more kinetic energy (fig. 1b). Einstein reasoned that the law of conservation of energy demanded that the photon must lose energy when climbing up against gravity. But how, since light always travel at a constant speed c? The only way out was to use what he had already used in his seminal paper on the photoelectric effect,

(D) E = hf, where here h is Planck's constant.

If E has to decrease, then the frequency f must also decrease, or its wavelength increase. This is known as the gravitational redshift.

**Second Thought Experiment**

Einstein arrived as his Equivalence Principle (EP) with the following thought experiment. In a room without windows, one would not be able to distinguish between being pulled up by an inertial force (fig.2a) from being pull down by a gravitational force near a planet (fig. 2b).

**Third Thought Experiment**

In fig.3a,to the observer in free falling spaceship, there is no frequency shift. However, to the observer outside the spaceship (fig.3b), the emitter is moving, and according to the Doppler effect, there should be a blueshift. Einstein reasoned that the null result would come about the cancellation between the Doppler effect and the gravitational redshift.

(1) (Δf/f)

_{doppler}= Δv/c (See equation A10 in the appendix of Relativistic Doppler Effect)

According to the EP,

(2) (Δf/f)

_{gravity}= -(Δf/f)

_{doppler}= -Δv/c

But Δv = gΔt (definition of acceleration)

= g(h/c) (time = distance/velocity)

= Δφ/c (gh = change in gravitational field)

(3) Therefore, (Δf/f)_{gravity}= -Δφ/c

^{2}

We need to convert this last result into a differential equation. We proceed as such,

(4) Δf/f = (f

_{rec}- f

_{em})/f

_{em}

For convenience, the emitted light happens early (em → 1), the received light happens later (rec → 2). Also the frequency is inversely proportional to time ( f ~ 1/τ). Putting this into equation (4), we get,

(5) Δf/f = (f

_{rec}- f

_{em})/f

_{em}

= (1/τ

_{2}- 1/τ_{1})τ_{1}= τ

_{1}/τ_{2}- 1 = (τ_{1}- τ_{2})/τ_{2} = -(τ

(6) Therefore equation (3) becomes,
_{2}- τ_{1})/τ_{2}= -Δτ/τ_{2}= -dτ/τ_{2} (dτ/τ

We also need to relate this result to the proper time τ and the coordinate time t. Consider two points, x at the proper time τ(x), and the second point at infinity, being the reference point where φ(∞)=0, is the coordinate time t(∞).
_{2})_{gravity}= +Δφ/c^{2}Therefore, (7)(dt - dτ)/dt = (φ(∞)- φ(x))/c

^{2}

Or, 1 - dτ/dt = 0 - φ(x)/c

^{2}. Rearranging, we get,

(8) dτ = (1 + φ(x)/c

^{2})dt.

We now need one more step to link this result to the metric tensor, which we have already explored in Relativistic Doppler Effect. The metric tensor can best be expressed as,

(9) ds

^{2}= g

_{μν}dx

^{μ}dx

^{ν}, (μ,ν = 0,1,2,3; or μ, ν = ct,x,y,z)

Separating the time component from the spatial components,

10) ds

^{2}= g

_{00}dx

^{0}dx

^{0}+ g

_{ij}dx

^{i}dx

^{j}, (i,j=1,2,3; or i,j = x,y,z)

= g

_{00}dx^{0}dx^{0}+ g_{ij}dx^{i}dx^{j}, = η

_{00}dx^{0}dx^{0}+ η_{ij}dx^{i}dx^{j}, = -c

Where in the last two lines we used the Minkowski metric for flat space with signature (-1,1,1,1). In the particle coordinate system, it is at rest. Therefore,
^{2}dt^{2}+ dx^{2}+ dy^{2}+ dz^{2},(11) dx

^{2}+ dy

^{2}+ dz

^{2}= 0

and by definition, the time measured in that coordinate is the proper time τ,

(12) Equation 10 reduces to,

ds

^{2}= -c

^{2}dτ

^{2}

= -c

Where we substitute equation (8) in the last line.
^{2}(1 + φ(x)/c^{2})^{2}dt^{2}(13) For low gravity, (1 + φ(x)/c

^{2})

^{2}≈ (1 + 2φ(x)/c

^{2})

Comparing equations (10), (12), and (13), we get

(14) g

_{00}= 1+ 2φ/c

^{2}

Which is (C), what Einstein had identified. Lucky guess or brilliant insight, most likely a combination of both spearheaded Einstein to develop the full theory of General Relativity. But how do we get the EFE,

(15) G

_{μν}= 8πGc

^{-4}T

_{μν}- Equation (A)

Knowing that it must yield in weak gravity field and low velocity, Newton's equation,

(16) ∇

^{2}φ = 4πGρ - Equation (B)

Without delving into differential geometry, we will outline schematically how Einstein arrived at the final destination, equation (15) - which incidentally took nearly 10 years of his life.

**Matter Curves the Geometry of Spacetime**

The left-hand side (LHS) of equation (15) is the geometry part of the Einstein Field Equations (EFE). It is expressed as,

(17)G

_{μν}= R

_{μν}- ½g

_{μν}R

Where g

_{μν}is the familiar metric tensor of equation (14), R

_{μν}is the Ricci tensor, and R is the Ricci scalar. The last two mathematical objects are special objects obtained from the Riemann curvature tensor, which is obtained from the parallel transport of a vector.

We are familiar with the addition of two vectors. Consider two forces acting on a point. We want to know: what is the resultant net force (vector addition).

The assumption that we have made by parallel transporting F

_{2}from the tail of F

_{1}to the head of F

_{1}is that space is flat. But suppose that space isn't flat.

In fig.5, we see that the parallel transport of a vector along a longitude at the equator (1), to the north pole (2), then down a different longitude to the equator (3), back to the initial position (4) brings about a 90

^{0}difference between the vectors at position (4) and (1).

In a space that is curved, the parallel transport of a vector (broken red arrow) at A to B will bring about a deviation (black arrow). Deviations will also occur at C, D and then back to A. The result of the total of these deviations ΔV going around the loop is measured by the Riemann curvature tensor R

_{αμβν}.

The Ricci tensor is a contraction of two indices. First we raise one of the indices,

(18) g

^{ασ}R

_{αμβν}= R

^{σ}

_{μβν}.

Then we contract by making σ = β

(19) R

^{β}

_{μβν}= R

_{μν}

And the Ricci scalar is a further contraction of the Ricci tensor,

(20) R = g

^{ασ}R

_{αμ}= R

^{σ}

_{μ}

= R

^{μ}_{μ}(setting σ = μ)Historically, Einstein had chosen the Ricci tensor for the LHS of equation (15), but soon realized that the gradient of the Ricci tensor was not zero, hence the conservation of energy would not be contained in the EFE. He finally amended by using equation (17).

The right-hand side of equation (15) is a little more of a guessing game. From equation (C), Einstein reasoned that the density of matter ρ must be related to the energy-momentum tensor (See equation 18 in Is Newtonian Gravitational Fields Valid Within General Relativity? ). So Einstein proposed,

(21) G

_{μν}= kT

_{μν}

With the only thing to fix is the constant k, which is done by considering again the case of a weak field and low velocity (the Newtonian limit). See Appendix below for the derivation of k = 8πG/c

^{4}

**Comments**The path from Newton's law of gravity to Einstein's field equations is anything but straight and forward. There were a number of brilliant insights (the three thought experiments) and fortunate guesses ( equations 14 and 17). Nevertheless that path was arduous and far from being obvious. Why is it that the EFE are accepted by the physics community? Empirical evidence gives sufficient support in the anomalous precession of the perihelion of Mercury's orbit, light from a distant star deflected by the sun, and the gravitational redshift - all those confirming that GR is the right stuff. It is also present in every day of our lives with GPS, which without GR the positional satellites would be less than useful. Could GR be supplanted? Of course, if some day a brilliant mind comes along with more brilliant insights, and perhaps a few lucky guesses, and puts forward a new theory of gravity, hopefully one that is compatible with QM. So far, that day doesn't seem to be close at hand.

**Appendix**From now on, we will borrow heavily from Is Newtonian Gravitational Fields Valid Within General Relativity? . In particular, equations 15 and 24 combine to give,

(A1) R

_{00}= -½∇

^{2}g

_{00}

And from the same reference, equation 5, which now reads as,

(A2) R

_{00}= k(T

_{00}- ½Tg

_{00})

Equating A1 and A2,

(A3) -½∇

^{2}g

_{00}= k(T

_{00}- ½Tg

_{00})

LHS of A3:

-½∇

^{2}g

_{00}= -½∇

^{2}(1+ 2φ/c

^{2}) = -∇

^{2}φ

RHS of A3:

k(T

_{00}- ½Tg

_{00}) = ½kT

_{00}- using that T = g

^{00}T

_{00}

Using that T

_{00}= -ρc

^{2}, equation 18 in same reference above.

k(T

_{00}- ½Tg

_{00}) = -½kρc

^{2}

Equating LHS with RHS of A3:

∇

^{2}φ = ½kρc

^{2}

Comparing this with Newton's law of gravity:

∇

^{2}φ = 4πGρ

Then k = 8πG/c

^{4}

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