Conjecture about the Unification between Electro-Magnetic Lorentz force and Gravity.

# Conjecture about the Unification between Electro-Magnetic Lorentz force and Gravity. 정리 예정. 핵심은 중력만으로 전자기력을 설명하는거. 중력의 다른 형태로 보인게 전자기력일수도 있다는 뜻. 더 확장되면 강력, 약력도 중력으로만 설명이 될수도 있다는거고. 어차피 conjecture 수준이라... Crude 하게 접근. 일반 상대론 쓴다기 보단 그냥 $\vec{E}^2+\vec{B}^2$ 에너지를 질량으로 환산한 뒤 이놈에서 중력이 나올 수 있다는 식으로 접근. 점입자 비스무리 하다면 그 원점에서 \vec{E} field 가 발산하니까, 거리도 짧은데다가 에너지도 무한대로 가서 생각보다 엄청나게 강력한 형태로 힘이 작용 가능. ## TOC ## Can ElectroMagnetic Forces (and possibly Weak & Strong Forces) be explained simply by Gravity? Form of classical EM Force and Gravity. EM force : q (\vec{E} + \vec{v} \times \vec{B}) \frac{k q_1 q_2}{r^2} Gravity : m \vec{g} \frac{G m_1 m_2}{r^2} Just I think that gravity looks more fundamental (since it says about spacetime). So Let’s start from the gravity. Gravity originates from mass. Mass=Energy. E = m c^2 In EM, energy is of $\vec{E}^2+\vec{B}^2$ form. \begin{align*} \frac{\text{Energy}}{\text{Volume}} &=\frac{1}{2} \vec{D} \cdot \vec{E} + \frac{1}{2} \vec{B} \cdot \vec{H} \\ &=\frac{1}{2} \epsilon_0 \vec{E}^2 + \frac{1}{2 \mu_0} \vec{B}^2 \end{align*} First think about positron (not electron, not proton) which is at rest. (Choose an appropriate frame.)

Positron at rest.

\vec{E} = \frac{ke}{r^2} \hat{r} = \frac{e}{4\pi \epsilon_0 r^2} \hat{r} Total Energy diverges if the positron is point particle. E_{\text{T}} = \int_{r=0}^{\infty} \frac{1}{2} \epsilon_0 \vec{E}^2 \cdot 4 \pi r^2 dr \rightarrow \infty Quantum mechanically the Schrödinger wave function of the positron can be thought to be spread over some space. Therefore simply forget about it at this moment. Then apply electric field $\vec{E}_0 = E_0 \hat{x}$ to the positron.

Positron within constant electric field.

Superposition principle says \vec{E} = \frac{e}{4\pi \epsilon_0 r^2} \hat{r} + E_0 \hat{x} Here we are trying to explain Electro-Magnetic Forces by Gravity. So there is no Electro-Magnetic Forces, but only Gravity. Then I just do order estimation. \vec{F}_1 \approx \frac{G m_e m_1}{a^2} \vec{x} \vec{F}_2 \approx \frac{G m_e m_2}{a^2} (-\vec{x}) m_1 c^2 \approx \frac{1}{2} \epsilon_0 \bigg( \frac{e}{4 \pi \epsilon_0 a^2} + E_0 \bigg)^2 a^3 m_2 c^2 \approx \frac{1}{2} \epsilon_0 \bigg( - \frac{e}{4 \pi \epsilon_0 a^2} + E_0 \bigg)^2 a^3 where $a$ is the order of radius of positron. You can simply notice that the direction of the gravity sum agrees with the direction of EM force $e \vec{E}$. \begin{align*} \vec{F}_{\text{Total}} &= \vec{F}_1 + \vec{F}_2 \approx \frac{G m_e}{a^2} (m_1 - m_2) \hat{x} \\ &\approx \frac{G m_e}{a^2} \frac{\epsilon_0}{2 c^2} \frac{e E_0}{\pi \epsilon_0 a^2} a^3 \hat{x} \\ &= \frac{G m_e}{2 \pi c^2 a} e E_0 \hat{x} \end{align*} In order that the above gravity force sum agrees with the EM force $e \vec{E}$, $\frac{G m_e}{2 \pi c^2 a}$ must be order of 1. In other words, if $\frac{G m_e}{2 \pi c^2 a}$ is order of 1, electric force can be explained by gravity!!!??? Then what does that $\frac{G m_e}{2 \pi c^2 a}$ is order of 1 mean? The schwarzschild radius (or the black hole radius) is r_s = \frac{2Gm}{c^2} Particle is just miniscule Black Hole composed of E,B field??? The EM force is simply other form of gravity??? ## RRA

1. My (kipid | 이강수) own unique/special idea.