Time Energy Potential = 0.5 c²/D² (2024)

c stands for the speed of light, D stands for time dilation rate.

John 9: “Whereas I was blind, now I see.”

In this article I refer to chapters of my free eBook, but you can understand the article without reading this book.

By now you have the impression/intuition that mass does not cause gravity, it alters time dilation around it, and time dilation causes gravity. And on some occasions, we mentioned massless cases of gravity. Thus, there should be a potential formula decoupled from mass.

Let’s agree on some notations first:

• m, when this symbol is bold, denotes some mass;
• m, when this symbol is not bold, denotes meter — unit of length;
• ~ means proportional. For example, F ~ M×m/R² because F = G × M×m/R², where G is a coefficient;
• ~ UOM, where UOM is a unit of measure, notes physical units for a value, for example, G ~ m³/[kg×sec²], M ~ kg, …

We denote force by F, energy by U, potential by V (do not confuse uppercase V with lowercase v that we use for velocity or speed). Quick recap on Newton’s formulas for gravity:

• F = — G×M×m/R²
  — F ~ kg×m/sec² — gravitational force;
  — G = 6.674×10⁻¹¹ m³/[kg×sec²] ~ m³/[kg×sec²] — gravitational constant;
  — M ~ kg — attractor’s mass;
  — m ~ kg — attracted mass;
  — R ~ m — distance between centers of masses;
  — Minus means direction of the force is opposite to the R vector.
• U = — G×M×m/R
  — U ~ kg×m²/sec² — gravitational or potential energy.
• F = — U'(R)
  — Relation between potential energy and force. Apostrophe ' denotes derivative.
• V = U/m = — G×M/R
  — V ~ m²/sec² — potential.

Potential provides the whole picture (that is why the opening quote from John 9): it is easy to derive energy, work, force, and even velocity (using energy conservation principle) from it. Newton’s formula V = U/m = — G×M/R ~ m²/sec² still depends on mass M of the attractor, despite V being measured in massless units. On several occasions, we mentioned massless cases of gravity and gravity as time pressure. It is time to formalize that.

We will start with intuitive derivation of potential formula from another Newton’s formula for speed of sound (here it is denoted by c, but we will replace it by s very soon, not to confuse it with the speed of light) :

• c = sqrt(P/ρ)
  — c ~ m/sec denotes speed of sound wave in a medium,
  — sqrt — square root,
  — P ~ kg/[m×sec²] — pressure in that medium,
  — ρ ~ kg/m³ — density of the medium.

Later, Laplace adjusted this formula to

• c = sqrt(γ×P/ρ),

where γ is a number, heat related. Let’s rewrite Newton-Laplace formula, replacing letter c (which we usually use for the speed of light in space) with letter s:

• s = K1×sqrt(P/ρ) = K2×sqrt(P) ~ sqrt(P),

considering γ and ρ as some constants, K1 is a numeric coefficient, K2~sqrt(m³/ kg) another coefficient (most constants and coefficients in physics are not just numeric, but associated with some units of measurements). And let’s get rid of sqrt:

• s² = K2²×P ~ K3×P, where K3 ~ m³/ kg.

Let’s check what is going on when pressure in medium varies: at a point X and at the point X+Δ, where Δ is a small fixed value (X and Δ are in meters ~ m):

• s²(X+Δ) — s²(X) = K3×[P(X+Δ) — P(X)],
• [s²(X+Δ) — s²(X)] × 1m² = K3×[P(X+Δ) — P(X)] × 1m²,

P(X)×1m² and P(X+Δ)×1m² are forces acting on thin layer of 1m² area in the medium between X and X+Δ:

Thus, [P(X+Δ)–P(X)]×1m²=F(X) ~ kg×m/sec² — is the resulting force acting on this layer. Actually, if account for direction of the force, we should flip its sign:

[P(X+Δ)–P(X)]×1m² = — F(X),

because when P(X+Δ) > P(X), then force F points to the left — opposite to axe X direction, and when P(X+Δ) < P(X), then force F points to the right — the same as axe X direction.

• [s²(X+Δ) — s²(X)] × 1m² = — K3×F(X),
• — F(X) = 1m²/K3 × [s²(X+Δ) — s²(X)] = 1m²×Δ/K3 × [s²(X+Δ) — s²(X)] / Δ ≈
• ≈ 1m²×Δ/K3 × [s²(X)]' = K4×[s²(X)]', where K4 ~ m²×m/(m³/kg) ~ kg.
• F(X) = — K4×[s²(X)]' ~ — [s²(X)]'×1kg,
• F(X)/1kg ~ a(x) ~ — [s²(X)]', where a(x) ~ m/sec² is acceleration.

Apostrophe denotes derivative. Thus, force and acceleration at any point X are proportional to the derivative of s²(X). Besides that, F = — U' in general, where U is potential energy, therefore:

• U' ~ [s²(X)]'×1kg
• U ~ s²(X)×1kg
• V = U/m ~ s²(X), where V is potential,
• V = K5×s²(X), where K5 is a unitless number, since V~m²/sec² and s~m/sec.

And acceleration a(x) ~ — [s²(X)]' depends on the same unitless coefficient K5:

• a(x) = — K5×[s²(X)]'.

Let’s apply this result to the space (vacuum) medium and to light waves instead of sound waves. What is the wave speed in this case? Speed of light? Yes, but because of time variability it is not just the constant c. Let’s denote by D time dilation. D=1 means no time dilation, when speed of light is constant c for us. D=2 stands for time somewhere in space that runs at twice slower pace than our time, and so is light wave there: it goes at speed c/D=c/2 from our perspective (check explanation in chapter 60). Thus, s=c/D is the formula for the wave speed. Therefore, formula for potential V = K5×s² turns into

• V = K5×c²/D²,

and the formula for acceleration a(x) = — K5×[s²(X)]' becomes

• a = — K5×c²×[1/D²]'.

To conclude the derivation, we need to find the exact value of K5. For this we can reuse Dr. Vivian Robinson’s exact formula for gravity, which we already used in chapter 75: a(R)= — G×M/[R²×D²] with time dilation D=exp(G×M/(R×c²)) at the distance R from the mass M. Let’s substitute that in place of a and D in our formula a = — K5×c²×[1/D²]' :

• — G×M/[R²×D²] = — K5×c²×[1/D²]'
• G×M / [R² × exp(2G×M/(R×c²)) ] = K5×c²×[ 1 / exp(2G×M/(R×c²)) ]'
• G×M×exp(–2G×M/(R×c²)) / R² = K5×c²×[ exp(–2G×M/(R×c²)) ]'
• G×M×exp(–2G×M/(R×c²)) / R² = K5×c²×[2G×M/(R²×c²)×exp(–2G×M/(R×c²))]

We can get rid of G, M, exp(…), and R² on both sides, and get rid of c² on the right side:

• 1 = K5×2 => K5 = 0.5

Thus, formula V = K5×c²/D² becomes

V = 0.5×c²/D²,

which we propose for the time potential, which depends on time dilation D in space. Let’s discuss its graph:

Isn’t that strange:

· Difference between potential (and corresponding energy) at D=1 and D –> ∞ (great value D) is finite (though it is still great at 0.5×c²),
· But the difference between potential (and corresponding energy) at D=1 and for D –> 0 (small value D) is not finite?

Why is there such a difference? D=1 corresponds to our time = “now”, and we can think of the past (see chapter 1) as corresponding to D –> ∞, and D –> 0 corresponding to the future. Why is such difference in energy magnitude between the past and the future relative to now (finite gap vs. infinite gap)? All becomes clear, if we recall that all units and constants that are related/expressed in sec, depend on time dilation factor D. For example, the unit of energy Joule J ~ kg×m²/sec², and J(D)=D²×J(1) — Joule was huge in the past, compared to today’s Joule (details are in chapter 20). Simpler example: let’s look at nowadays c and at the graph value V(1)=0.5×c² from big D perspective. If an object now travels almost 300,000 km in our second, then from a 10-times slower time perspective (with D=10), only 1/10 of their second passed, and thus, this object has moved 300,000 km at 0.1 their seconds — at a speed close to 3,000,000 km per their second. Thus, it is relativism that plays games with second-related values, and V(1)=0.5×c² is a huge value from big-D perspective (as nowadays c is). That answers the above two-bullet-point question (about the difference between the left side and the right side of the graph). And be aware: when we move from our current time, which we denoted as D=1 on the graph, to some other place or time in the future with a different D-value, then this new value becomes new D=1 for us, and the whole graph should be rescaled and the graph will turn into the same graph as above, again.

Besides Joule (energy unit) becoming variable in variable time, many physical laws, starting from the first Newton’s laws for motion, are not true anymore, and should be reworked (check, for example, chapter 56 for the third Newton’s law). The same applies to energy conservation formulas — we should be very cautious in trusting math there. You can ask then: “What is the use of having formulas for potential in variable time, if we cannot rely on the energy conservation principle?” Although formula V = 0.5×c²/D² was derived using elementary physics, it holds relativism inside it already, as we explained with the graph. Even better, potential energy

• U = m×V = 0.5×m×c²/D²

already contains restriction on classic/elementary kinetic energy K = m×v²/2:

• K = m×v²/2 ≤ 0.5×m×c² = U(D=1) <=> v≤c.

Now, let’s put time potential to work, in context of chapter 2, where we discussed the Sun moving in and out of Milky Way arms — moving between faster-time and slower-time areas. Potential energy is: U(D) = 0.5×m×c²/D², U(1) = 0.5×m×c². Let’s look at an object moving at a small velocity ≈ 0 (to simplify calculations) from an area with the current time dilation D=1, to an area with slower time, with dilation D: kinetic energy K(1) ≈ 0, K(D) = m×v²/2. What will be velocity v? By energy conservation principle, K(1)+U(1) = K(D)+U(D):

• 0 + 0.5×m×c² = m×v²/2 + 0.5×m×c²/D²
• c² = v² + c²/D² => v² = c²×(1–1/D²) => v = c×sqrt[1–1/D²].

That is the boost value, which the object receives when it comes to slower time (and loses it when it leaves the slower-time area). This value is quantitatively different from the Z×c estimate (where Z is redshift), obtained in chapter 2 using Einstein’s Special Relativity approach. But qualitatively it still correlates the boost value to the redshift Z, since time dilation D and redshift Z are linked by D=Z+1:

c×sqrt[1–1/D²] = c×sqrt[1–1/(1+Z)²] = c×sqrt[1–1/(1+2Z+Z²)] = c×sqrt[1–1+2Z–…] ≈ c×sqrt(2×Z) > c×Z for small values Z≈0 (because 2×Z > Z²).

P.S. Check Simpler Derivation of Gravitational Potential = 0.5×c²/D².