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Time Has Potential: Physicists Are Back to the Drawing Board

Sample sections from Classical Physics Beyond Einstein’s chapter 3

Bertrand Russell: “Physics is mathematical not because we know so much about the physical world, but because we know so little.”

3.1. Potential = 0.5×c²/D²
3.2. Gravity g = – (0.5c²/D²)' ≈ c²×D' ~ ∇D
3.3. Gravitational Time Dilation D = exp[G×M/(R×c²)]
3.4. Escape velocity = c × √ [1–exp(–2G×M/(R×c²))] < c
3.5. Action = – D² × Reaction
3.6. Relativistic Time Dilation D = 1 / √ (1–v²/c²)

3.1. Potential = 0.5×c²/D²

In chapter 1, we got formula (1.3) c/D for apparent speed of light, or light wave (where c = 299,792,458 m/sec and D is time dilation factor). In classical mechanics, there is Newton–Laplace formula for a wave speed in any medium:

c = sqrt(γ×P/ρ).

“c” denotes here speed of wave, P — pressure in the medium, ρ — density of the medium, and γ is a numerical factor, which we will discuss at the end of this section. By replacing the speed of wave “c” with speed of light wave c/D, we have:

c/D = sqrt(γ×P/ρ).

Now, remove sqrt by squaring both sides: c²/D² = γ×P/ρ.

At point X: c²/D(X)² = γ×P(X)/ρ.

At point X+Δ: c²/D(X +Δ)² = γ×P(X +Δ)/ρ.

Subtract those: c²/D(X +Δ)² — c²/D(X)² = γ×(P(X +Δ) — P(X))/ ρ.

Divide both sides by Δ, and multiply and divide the right side by 1m²:

(c²/D(X+Δ)² — c²/D(X)²) / Δ = γ×(P(X+Δ) — P(X)) ×1m² / (ρ×1m²×Δ) (3.1.1)

(P(X+Δ)–P(X))×1m² = –F(X), where F is the resulting force acting on 1m²×Δ layer. Negative sign before F(X) indicates opposite-to-X direction of F, when P(X+Δ) > P(X):

When P(X+Δ) < P(X), P(X+Δ) — P(X) is negative, — F(X) is negative too, therefore, F(X) is positive, thus, resulting force F points in the same direction as X.

ρ×1m²×Δ represents the mass of this layer. Force divided by mass is acceleration or derivative of potential with negative sign, thus:

γ × (P(X+Δ) — P(X)) ×1m² / (ρ×1m²×Δ) = γ × Potential'(X).

Apostrophe ' denotes derivative by location (by X in this case).

And since (c²/D(X+Δ) ² — c²/D(X) ²) / Δ = (c²/D(X)²)', formula (3.1.1) simplifies to:

(c²/D(X)²)' = γ × Potential'(X).

Getting rid of derivatives on both sides: c²/D² = γ × Potential (plus a constant, but in physics, +constant is ignored for a potential).

Now, to the factor γ: it is linked to the degrees of freedom f of the medium “particles”, as γ = 1+2/f (see details in Wikipedia: Relation with degrees of freedom). For a particle moving at the speed of light, there are only two degrees of freedom: it can oscillate only in the plane perpendicular to the velocity of the particle, because oscillation in the direction of motion will change the speed of particle (or plane wave propagation), and this speed (which should be the speed of light) cannot change. More formal explanation check in Quote from a theoretical physicist:

“In summary, massless particles have only two degrees of freedom because their spin states are representations of the two-dimensional rotation group SO(2), which is Abelian…”.

Thus, with f = 2 and γ = 1+2/f = 2, we have:

c²/D² = γ × Potential = 2 × Potential

Potential = 0.5 × c²/D².

3.2. Gravity g = — (0.5c²/D²)' ≈ c²×D' ~ ∇D

Gravity g depends only on D’ or ∇D (derivative by location or gradient of D).

By classical mechanics, derivative by location of –Potential is acceleration. So, derivative by location of the time potential gives acceleration g:

g = –(0.5c² / D²)' = c²×D' / D³ When time dilation D≈1 (for example, near the Sun D=1.000002): g = –(0.5c² / D²)' = c²×D' / D³ ≈ c²×D' ≈ 9×10¹⁶×D' m²/sec² ~ ∇D

“~” denotes proportionality: in the line above g is proportional to ∇D (or D') with 9×10¹⁶ coefficient. We know that g ≈ 9.8 m/sec² on the Earth, thus, near the Earth:

g ≈ 9.8 m/sec² ≈ 9×10¹⁶×D' m²/sec² => D'×1m ≈ 10⁻¹⁶.

We just explained the book cover images.

3.3. Gravitational Time Dilation D = exp[G×M/(R×c²)]

Newton’s law for gravitational acceleration g at distance R from mass M

g = — G×M/R²

does not account for time dilation D. Formula (1.2) from chapter 1 says that apparent acceleration is local acceleration divided by D², so observed acceleration is:

g = — G×M/R²/D² = — G×M/(R²×D²). (3.3.1)

Previous section we started with another formula for g: g = — (0.5×c²/D²)'. Thus,

g = — (0.5×c²/D²)' = — G×M/(R²×D²)

0.5×c²×(D⁻²)' = G×M/(R²×D²)

0.5×c²×(–2D'×D⁻³) = G×M/(R²×D²)

– c²×(D'/D) = G×M/R²

– c²×(ln(D))' = G×M×(–1/R)'

c²×ln(D) = G×M×(1/R)

ln(D) = G×M/(R×c²)

D = exp[G×M/(R×c²)], where exp(x) = eˣ. (3.3.2)

From (3.3.1) and (3.3.2) we get Dr. Robinson’s gravitational formula:

g = — G×M/( R²×exp[ 2GM/(Rc²) ] ). (3.3.3)

Now, finding time dilation D near the Sun becomes an easy exercise, just by replacing letters in formula (3.3.2): G = 6.674×10⁻¹¹, M= 2×10³⁰, R= 7×10⁸, c=3×10⁸:

D ≈ exp[6.674×10⁻¹¹×2×10³⁰/(7×10⁸×3²×10¹⁶)] ≈ exp(2.12×10⁻⁶) ≈ 1.000002.

And you can calculate the time dilation near the Earth for M=5.97×10²⁴ kg and R=6.37×10⁶ m, using Wolfram calculator.

P.S. Formula D=exp[G×M/(R×c²)] debunks “theoretical Black Holes”, because it shows that there is no so-called “Event Horizon” where time stops (where time dilation D is infinite). Time can slow down by a finite factor D, but it does stop: D ≠ ∞ !

P.P.S. Formula D = exp[G×M/(R×c²)] with a constant mass M works outside that mass, when R ≥ R₀, where R₀ is the radius of the attractor. By Newton’s shell theorem, mass m at distance R < R₀ from the center of the body is attracted only by mass of the ball of radius R (not by the whole R₀–ball), which has mass less than M. Mass of R–ball is approximately, proportionally less than the whole mass M of R₀–ball:

M×(4/3×π×R³) / (4/3×π× R₀³) = M×R³ / R₀³, where M and R₀ are constants.

Thus, formula for time dilation D inside the object at distance R< R₀ from its center is:

D ≈ exp[G×(M×R³ / R₀³)/(R×c²)] = exp[G×M×R²/R₀³/c²)] < exp[G×M/R₀/c²)].

Even inside a celestial body, time dilation D is finite (because no division by R near 0 in exp[G×M/R₀/c²)] ).

P.P.P.S. Newton’s gravity g = –G×M/R² keeps planets in elliptical orbits, but with g = –G×M/(R²×D²(R)) and force (by Newton’s 2nd law)

F = –G×m×M/(R²×D²(R)), where D(R) = exp[G×M/(R×c²)]

D(R) increases when the distance R decreases (and vv., D(R) decreases on R increase), elliptical orbits shift/undergo precession:

• On the way to the Sun, with increasing D(R) (in the force denominator), gravity is gained at a lower than by classical formula rate, that is not enough to retain a planet in its classical orbit. That causes the planet to shift outside the anticipated classical orbit.
• Away from the Sun, with increasing R and decreasing D(R) (in the force denominator) gravity weakens slower than classically expected, so the planet is pulled inside the classical ellipse by larger than expected force:

It is good exercise for a CS or Physics freshman/sophomore to write an animation in Python for Mercury’s precession, using g = –G×M/(R²×exp[2GM/(Rc²)]) formula only.

Continue reading sections 3.4–3.6 in the 📖 Book PDF, on Amazon or Google Books.

Time Matters