Soumendra Nath Thakur, Tagore's Electronic Lab, India
February 07, 2025
Abstract
This study explores the extended dynamics of photon motion, challenging the conventional view that light speed (c) is always constant. We propose that at an infinitesimally small scale, a photon transitions from rest to c due to an inherent acceleration phase governed by its negative apparent mass. This acceleration generates a self-exerted force, distinct from external interactions, allowing the photon to maintain equilibrium while escaping gravitational influence. Furthermore, we analyse the continuous nature of photon frequency, distinguishing it from discrete digital signals. Unlike step-like binary transitions, a photon's wave packet exhibits a smooth, incremental frequency pattern, implying alternating cycles of acceleration and deceleration within its propagation. These insights suggest that photon motion and frequency dynamics involve fundamental, phase-dependent changes at quantum and relativistic scales.
Keywords: Photon Dynamics, Negative Apparent Mass, Acceleration Phase, Transmission Delay, Continuous Frequency, Wave-Particle Duality
Force Dynamics on Photons:
• Derive the force equation F = −Mᵃᵖᵖaᵉᶠᶠ for photons using apparent mass and associated acceleration.
• Explore how this equation governs the photon’s motion under varying energy-momentum conditions.
• The derivation of the effective acceleration aᵉᶠᶠ aligns with the methodological exploration of force and acceleration acting on photons. It would complement the discussion of the force equation F = −Mᵃᵖᵖaᵉᶠᶠ and further clarify the dynamics of photons as analysed through the extended classical mechanics framework. The constant effective acceleration: aᵉᶠᶠ = 6 × 10⁸ m/s².
Determination of Constant Effective Acceleration of Photons
The distance travelled by the photon in 1 second is 3 × 10⁸ m, and that the acceleration is constant. The expression for the distance travelled in the case of constant acceleration is given by:
Δd = v₀Δt + (1/2)aᵉᶠᶠ(Δt)²
Where:
• Δd is the distance travelled (3 × 10⁸ m in 1 second),
• v₀ is the initial velocity (0 m/s, at emission),
• Δt is the time (1 second),
• aᵉᶠᶠ is the effective acceleration, which we want to solve for.
Substituting the known values into the equation:
3 × 10⁸ m = 0•1 s + (1/2)aᵉᶠᶠ(1)²aᵉᶠᶠ = 6 × 10⁸ m/s²
Extended Photon Dynamics and Phases of Motion: Transition from Rest to Constant Velocity
• When considering a photon's motion, its apparent mass is negative. As a result, its effective acceleration leads to a force with a negative value. This behaviour is different from that of ordinary matter, which always has a positive mass.
• The commonly referenced distance that light travels in one second does not represent the photon's actual path during that time. Instead, it marks the moment of emission, where the photon, initially at rest in an apparent sense, rapidly attains its full velocity within a brief interval.
• During this transition period, the effective acceleration is determined by the relationship between force and the negative apparent mass. The force involved does not come from an external source but is instead exerted by the photon itself due to its unique mass-energy properties. This results in the photon undergoing a continuous deceleration at twice the speed of light.
• The force generated by the photon serves a dual purpose. It counteracts the gravitational pull of its source while ensuring the photon maintains a constant speed as it escapes. The energy necessary for this process is provided by the photon itself, allowing it to sustain the required acceleration and remain in
Photon Dynamics: Returning to the Force Equation for Photons
• Since the apparent mass is negative (−Mᵃᵖᵖ), the constant effective acceleration aᵉᶠᶠ = 6 × 10⁸ m/s² results in a force term with a negative value. This contrasts with the behaviour of matter mass (Mᴍ), which always remains positive.
• The distance of 3 × 10⁸ m in one second does not represent a photon’s trajectory over that duration. Instead, it corresponds to the initial emission event, where the photon, initially at rest in an apparent sense (t₀, v₀), attains a velocity v₁ at time t₁, with Δt = t₁ − t₀ = 1 second and Δv = v₁ − v₀ = 3 × 10⁸ m/s².
• During this interval (t₁ − t₀), the effective acceleration is given by aᵉᶠᶠ = F/(−Mᵃᵖᵖ). The force F is not an external force but is instead exerted by the photon itself due to its negative apparent mass (−Mᵃᵖᵖ). This implies that the photon undergoes continuous deceleration at twice the speed of light (6 × 10⁸ m/s²).
• The exerted force (F) not only counteracts the gravitational attraction of the source (Fg) but also enables the photon to escape the gravitational well at a constant speed of 3 × 10⁸ m/s². The energy required for this escape is compensated by the photon itself, maintaining the necessary energy balance to sustain its effective acceleration of 6 × 10⁸ m/s².
Explanation of Phases of Motion: Transition from Rest to Constant Velocity
On a number line, there are infinitely many points between any two nearest numbers. When you say "1," you are actually referring to the difference between 0 and 1, with an infinite sequence of points in between.
Similarly, while the speed of light (c) appears constant on large scales, at an infinitesimally small scale, it has a beginning due to transmission delay. This delay occurs because motion progresses incrementally, however small, starting from absolute rest (v=0) before reaching c.
The first phase, where velocity increases from 0 to c, represents acceleration. Motion does not begin with an arbitrary velocity but transitions from rest. The first phase starts at zero (v = 0) and progresses to an initial velocity (v), whereas successive phases continue from an already established velocity (v = v) rather than starting anew from v = 0.
Mathematical Representation
Let v(t) represent the velocity of the object as a function of time. In the first phase of motion:
Initial Phase (Acceleration)
The motion begins from rest, so at t = 0, v(0) = 0. The velocity increases from v=0 to some initial velocity v₁ = c, over some time interval Δt₁. The acceleration a(t) in this phase is given by:
a(t) = dv(t)/dt, where v(t) = ∫a(t)dt
The velocity increases gradually from 0 to c, so during this phase, the object undergoes acceleration.
Subsequent Phases (Constant Velocity)
After reaching an initial velocity v₁ = c, successive phases of motion proceed at this established velocity. In these phases, the velocity remains constant, so for t > Δt₁, we have:
v(t) = v₁ = c, a(t) = 0
In the subsequent phases, the object continues with the velocity v = c, without starting from rest or accelerating further.
Photon Frequency: Continuous Analogous Waves vs. Discrete Digital Signals
Photon frequency is not a discrete, step-like, binary signal. Unlike digital frequencies, which exhibit distinct on-off states, photon frequency is continuous and behaves in an analogy manner. It follows a smooth, incremental, and decimal-like wave pattern within its energy packet.
While digital signals transition between fixed values, a photon's frequency remains constant within its wave-packet, forming an uninterrupted oscillatory motion. This continuous wave behaviour implies that every phase of a photon’s wave structure inherently represents alternating cycles of acceleration and deceleration, rather than discrete jumps between states.
This suggests that the wave characteristics of a photon are not just propagating in a static manner but involve intrinsic dynamical changes at the quantum scale, reinforcing the idea that photon energy and momentum continuously adjust within their wave structure.
