Discussion in 'Physics & Math' started by curvature, Aug 12, 2018.

  1. curvature Registered Member

    1). It's been suggested the Reyleigh-Plesset equation does not describe appropriate heat

    2). That heat I suggest may come from cyclotron radiation from the spin of the cavity which has been shown to exceed the speed of light inside of the cavity.

    3). I attempt to put corrective factors in including any forces from Van der Waals.

    \(\frac{1}{R^2} (\frac{\Delta P + (\frac{\mathbf{R}T}{(V_m - b )} - \frac{a}{V^2})}{\rho})[1 - (\frac{T_0}{T})^3]\)

    \(= \frac{1}{R^2} (\frac{\Delta P + P_{waals})}{\rho})[1 - (\frac{T_0}{T})^3]\)

    \(= \frac{\ddot{R}}{R} + \frac{3}{2}(\frac{\dot{R}}{R})^2 + \frac{4 \nu \dot{R}}{V} + \frac{2S}{\rho V} + \frac{e^2}{6 \pi mV} \frac{\dddot{R}}{R} + \frac{1}{V}(\frac{e}{m})\frac{\partial \dot{U}}{\partial R}\)

    Distribution of the density and simplifying some terms I get

    \(\frac{\Delta P + \mathbf{P}}{R^2}[1 - (\frac{T_0}{T})^3]\)

    \(= \frac{\ddot{R}}{R}\rho + \frac{3\rho}{2}(\frac{\dot{R}}{R})^2 + \frac{4 m\nu }{V^2}\dot{R} + \frac{2S}{V} + \frac{e^2}{6 \pi V^2} \frac{\dddot{R}}{R} + \frac{\rho}{V}(\frac{e}{m})\frac{\partial \dot{U}}{\partial R}\)

    \(\Delta P = P - (P_0 - P(t))\)

    Integrating the volume element we obtain the simplified version of our equations

    \(\int\ \frac{\Delta P + \mathbf{P}}{R^2}[1 - (\frac{T_0}{T})^3]\ dV\)

    \(= m\frac{\ddot{R}}{R} + \frac{3}{2}m(\frac{\dot{R}}{R})^2 + d\log_V(4 \rho \nu \dot{R} + 2S + \frac{e^2}{6 \pi V}\frac{\dddot{R}}{R} + \rho_q \frac{\partial \dot{U}}{\partial R})\)


    \(\frac{F}{\Delta R} = \frac{E}{\Delta A} \equiv \int\ \frac{\Delta P + (\frac{\mathbf{R}T}{(V_m - b )} - \frac{a}{V^2})}{R^2}[1 - (\frac{T_0}{T})^3]\ dV\)

    \(= \int\ [\Delta P + (\frac{\mathbf{R}T}{(V_m - b )} - \frac{a}{V^2})]\gamma\ dR\)

    \(\gamma = [1 - (\frac{T_0}{T})^3]\)

    [*] SEE NOTES

    With \(\rho_q\) being a charge density and \(c=1\).The dimensions of this equation is force over length or energy over area. It has an ‘’acoustic energy’’ part given by \(m\frac{\ddot{R}}{R}\) and a wall velocity term \(\frac{3}{2}m(\frac{\dot{R}}{R})^2\). This part \(\frac{\Delta P + \mathbf{P}}{R^2}\) can be seen in terms of an ''acoustic intensity'' term. It’s also been known for the surface tension \(S\) to have a coefficient of \((1 - \frac{T}{T_C})\) where \(T_C\) is the critical temperature (known as the Guggenheim–Katayama formula). As temperature increases the surface tension decreases.

    [1] - an alternate simplification from a previous Langrangian of the theory we formalised, requires only the additional binding or repulsive energies from Van der Waals forces

    \(\mathcal{L} = mR \ddot{R} + \frac{3}{2}m\dot{R}^2 + \frac{4 \nu_L m}{R} \dot{R} + \frac{2\gamma m}{\rho_L R} + \frac{e^2}{6 \pi c^3} \dddot{R} + \frac{1}{2}eV + \frac{\Delta P(t)m + \Delta\mathbf{P}}{\rho_L}\)

    \(= mR \ddot{R} + \frac{3}{2}m\dot{R}^2 + \frac{4 \nu_L m}{R} \dot{R} + \frac{2\gamma m}{\rho_L R} + \frac{e^2}{6 \pi c^3} \dddot{R} + \frac{1}{2}eV + \frac{\Delta P(t)m + (\frac{\mathbf{R}T}{(V_m - b )} - \frac{a}{V^2})}{\rho_L}\)

    (which is the Langrangian)

    The repulsive nature of Van der Waals could temporally explain the expanding of the bubble but it seems more likely related to pressures and temperature.

    [2] - Further, there is a part of this equation

    \(\int\ \frac{\Delta P + \mathbf{P}}{R^2}[1 - (\frac{T_0}{T})^3]\ dV = m\frac{\ddot{R}}{R} + \frac{3}{2}m(\frac{\dot{R}}{R})^2 + 4 \rho \nu \dot{R} + 2S + \frac{e^2}{6 \pi V}\frac{\dddot{R}}{R} + \rho_q \frac{\partial \dot{U}}{\partial R}\)

    Namely this expression \(\frac{e^2}{6 \pi V}\) can be fashioned in a different way:

    \(\frac{e^2}{2 \epsilon_0} \frac{e^2}{4\epsilon \hbar_0 c}\frac{1}{ \pi R^3}\)

    This is not too far from the difference of such a potential which actually gives rise to the Lamb shift, a direct consequence itself of the vacuum energy, ie. Casimir effect, par the powers of the fine structure

    \(<\Delta V>\ = \frac{e^2}{4 \pi \epsilon_0} \frac{e^2}{4 \pi \epsilon_0 \hbar c}(\frac{\hbar}{mc})^2\frac{1}{\pi R^3} \ln \frac{4 \epsilon_0 \hbar}{e^2}\)

    Notice, we have encountered this kind of notation before in investigating Anandan's difference of geometries which was part of the topic of my paper to the gravitational research foundation.
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  3. curvature Registered Member

    ''exceed the speed of sound'' that should be. Obviously nothing can exceed the speed of light lol silly mix up here but cannot edit.
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