Quantum Gravity in Temporal Physics.

 

Example Calculation: Single Temporal Quantum in a Harmonic Potential

Step 1: Define the Harmonic Potential

Let's start with a simple harmonic potential defined as:

$$V(\epsilon) = \frac{1}{2} k \epsilon^2$$

where $k$ is the spring constant. For this example, we'll use a common value for $k$.

Assume:

• $k = 1 \, \text{N/m}$ (as an example).

Step 2: Write the Hamiltonian

The Hamiltonian for our single temporal quantum can be expressed as:

$$H = \frac{1}{2} \left(\frac{d \epsilon}{dt}\right)^2 + \frac{1}{2} k \epsilon^2$$

Step 3: Schrödinger Equation

The corresponding time-independent Schrödinger equation is given by:

$$-\frac{\hbar^2}{2m} \frac{d^2 \Psi(\epsilon)}{d \epsilon^2} + \frac{1}{2} k \epsilon^2 \Psi(\epsilon) = E \Psi(\epsilon)$$

where:

• $\hbar$ (reduced Planck constant) $\approx 1.055 \times 10^{-34} \, \text{J s}$
• $m$ (mass of the temporal quantum, assuming $m = 1 \, \text{kg}$ for simplicity)
• $E$ is the energy eigenvalue we want to calculate.

Step 4: Substitute Known Values

For our calculations, we will first consider the ground state of the harmonic oscillator. The energy levels for a harmonic oscillator are given by:

$$E_n = \left(n + \frac{1}{2}\right) \hbar \omega$$

where $\omega = \sqrt{\frac{k}{m}}$.

Calculating $\omega$:

$$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{1 \, \text{N/m}}{1 \, \text{kg}}} = 1 \, \text{s}^{-1}$$

Calculating Ground State Energy $E_0$: For $n = 0$:

$$E_0 = \left(0 + \frac{1}{2}\right) \hbar \omega = \frac{1}{2} (1.055 \times 10^{-34} \, \text{J s}) (1 \, \text{s}^{-1}) = 5.275 \times 10^{-35} \, \text{J}$$

Step 5: Interpret Results

This calculated energy $E_0$ represents the minimum energy of a single temporal quantum in the harmonic potential defined. This value can be interpreted as a measure of the quantum fluctuations inherent in the temporal flow model.

Step 6: Connect with Gravity

To incorporate gravity, we consider the effect of a gravitational field on the harmonic oscillator. In the presence of a gravitational potential $V_g = mgz$ (where $z$ is the height), we can modify our Hamiltonian to include this potential:

$$H' = \frac{1}{2} \left(\frac{d \epsilon}{dt}\right)^2 + \frac{1}{2} k \epsilon^2 + mgz$$

Assuming we have $m = 1 \, \text{kg}$ and the gravitational acceleration $g \approx 9.81 \, \text{m/s}^2$, we would consider a scenario where our quantum is at a height of $z = 1 \, \text{m}$.

Calculate the gravitational potential:

$$V_g = mgz = (1 \, \text{kg})(9.81 \, \text{m/s}^2)(1 \, \text{m}) = 9.81 \, \text{J}$$

Final Result Interpretation

The total energy for the modified system considering both the harmonic oscillator and gravitational potential becomes:

$$E' = E_0 + V_g = 5.275 \times 10^{-35} \, \text{J} + 9.81 \, \text{J} \approx 9.81 \, \text{J}$$