Week 03 관련

• REALTIVISTICI MASS
  • in NON-RELATIVISTIC frame (PHYS031)
  • in RELATIVISTIC frame
• DIREC'S DISCOVERY
• PARTICLE COLLISION
  • RELATIONSHIP BETWEEN ENERGY & MASS
  • SLIDES:
  • PROTON & NEUTRON
• QUANTUM PHYSICS: DISCOVERY
  • 1792
• INTERFERENCE
  • standing waves (1-D)
  • vibration of standing wave
• BLAKC BODY

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REALTIVISTICI MASS

$$\begin{align*} m&=\gamma\:m_0,&\left<m_0:\:\text{mass at rest}\right> \end{align*}$$

in NON-RELATIVISTIC frame (PHYS031)

$$\begin{align*} E&=W=\int{\vec{F}\cdot{d}\vec{x}}=\int{\left(m\vec{a}\right)\cdot{d}\vec{x}}\\ &=\int{m\left(\frac{d\vec{v}}{dt}\right)\cdot\vec{x}}=\int{m\left(\frac{d\vec{x}}{dt}\right)d\vec{v}}\\ &=\int{m(v)(dv)}=m\left(\frac{v^2}{2}\right)=\frac{1}{2}mv^2; \end{align*}$$

in RELATIVISTIC frame

$$\begin{align*} E&=W=\int{\vec{F}\cdot{d}\vec{x}}=\int{\left(\frac{dp}{dt}\right)dx},&&\left<F=\frac{dp}{dt}\right>\\ &=\int{\frac{d}{dt}(p)dx}=\int{\frac{d}{dt}\left(\gamma{m_0}v\right)}\\ &=\int{m_0\left(\gamma\frac{dv}{dt}+v\frac{d\gamma}{dt}\right)}\\ &=\cdots \end{align*}$$

$$\left\{ \begin{align*} \gamma&=\left(1-\frac{v^2}{c^2}\right)^{-\frac{1}{2}};\\ \frac{d\gamma}{dt}&=\frac{d}{dt}\left(\gamma\right)=\frac{d}{dt}\left(1-\frac{v^2}{c^2}\right)^{-\frac{1}{2}}\\ &=\left(-\frac{1}{2}\right)\left(1-\frac{v^2}{c^2}\right)^{-\frac{3}{2}}\left[\left(-\frac{2v}{c^2}\right)\left(\frac{dv}{dt}\right)\right]\\ &=\frac{v}{c^2}\gamma^3\frac{dv}{dt} \end{align*} \right\}$$

$$\begin{align*} E&=\int{m_0\left(\gamma\frac{dv}{dt}+v\left(\frac{v}{c^2}\gamma^3\frac{dv}{dt}\right)\right)dx}\\&=\int{m_0\left(\gamma{dv}+\frac{v^2}{c^2}\gamma^3dv\right)\frac{dx}{dt}}=\int{m_0\gamma{dv}\left(1+\frac{v^2}{c^2}\gamma^2\right)\frac{dx}{dt}} \end{align*}$$

$$\left\{ \begin{align*} 1+\frac{v^2}{c^2}\gamma^2&=1+\left(\frac{v^2}{c^2}\right)\left(\frac{c^2}{c^2-v^2}\right)=1+\frac{v^2}{c^2-v^2}\\ &=\left(\frac{v^2}{c^2-v^2}\right)+\left(\frac{c^2-v^2}{c^2-v^2}\right)=\frac{c^2}{c^2-v^2}=\gamma^2 \end{align*} \right\}$$

$$\begin{align*} E&=\int{m_0\gamma(\gamma^2)dv(v)}=\int{m_0\gamma^3v\:dv},\\ &\left<\frac{d\gamma}{dt}=\frac{v}{c^2}\gamma^3\frac{dv}{dt};\:\:\:\:d\gamma=\frac{v}{c^2}\gamma^3dv\right>\\ &=\int{m_0(c^2)d\gamma}=m_0c^2\int{d\gamma}=\gamma{m_0}c^2; \end{align*}$$

$$\begin{align*} \therefore{E}&=\gamma{m_0}c^2 \end{align*}$$

Give the lower and upper bounds for this integral

$$\begin{align*} E&=m_0c^2\left.\left(1-\frac{v^2}{c^2}\right)^{-\frac{1}{2}}\right|_{0}^v\\ &=m_0c^2\left(1-\frac{(v)^2}{c^2}\right)^{-\frac{1}{2}}-m_0c^2\left(1-\frac{(0)^2}{c^2}\right)^{-\frac{1}{2}}\\ &=m_0c^2\gamma-m_0c^2=(\gamma-1)m_0c^2; \end{align*}$$

given velocity "close" to speed of light, the kinetic energy is denoted,

$$\begin{align*} K&=\gamma{m_0}c^2-m_0c^2=E_{\text{total}}-E_{\text{rest}}\\ &=(\gamma-1)m_0c^2 \end{align*}$$

to see if this equation holds true in classical physics.

$$\left\{ \begin{align*} v\ll{c},&\underset{\text{Binomial Expansion}}{\gamma\approx1+\frac{1}{2}\frac{v^2}{c^2}} \end{align*}\right\}$$

$$\begin{align*} K\approx\left[\left(1+\frac{1}{2}\frac{v^2}{c^2}\right)-1\right]m_0c^2=\frac{1}{2}m_0v^2 \end{align*}$$

ENERGY RELATION ($$E_{\text{total}} vs. E_{\text{rest}}$$)

$$\begin{align*} m&=\gamma{m_0};\\ m^2&=\gamma^2m_0^2=\left(\frac{c^2}{c^2-v^2}\right)m_0^2;\\ m_0^2c^2&=m^2(c^2-v^2)=m^2c^2-m^2v^2;\\ (m_0^2c^2)(c^2)&=(c^2)(m^2c^2-m^2v^2);\\ (m_0c^2)^2&=(mc^2)^2-m^2c^2v^2;\\ (mc^2)^2&=m^2c^2v^2+(m_0c^2)^2;\\ &\left<E=m_0c^2\right>\:\:\:\:\left<p=mv\right>\\ (E_{\text{total}})^2&=(pc)^2+(E_{\text{rest}})^2 \end{align*}$$

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DIREC'S DISCOVERY

Basically he proposed that in special relativity, electrons can have both a positive charge and negative energy.

Q: how is it possible to have an electron with energy that is positively charged?

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PARTICLE COLLISION

RELATIONSHIP BETWEEN ENERGY & MASS

$$E=mc^3$$

SLIDES:

$$\bar{p}+p\to{\text{many}}$$

PROTON & NEUTRON

proton:

$$m_p=938.3\:\text{MeV}/c^2$$

neutron:

$$m_n=939.6\:\text{MeV}/c^2$$

deutreon:

$$\begin{align*} m_d&=(m_p+m_n)\\ &=(939.6+939.6)\\ &=1877.05\:\text{MeV}/c^2 \end{align*}$$

Q: But from the investigation, this is simply shown not true. Will it be larger or smaller?

A: it is LESS .

$$m_d=1875.6\:\text{MeV}/c^2$$

What happened? See here.

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QUANTUM PHYSICS: DISCOVERY

1792

Thomas Wedgwood

• pioneer of photography.
• We used to work for pottery shop owned by his dad, Josiah Wedgwood.
• Discovery: Regardless of shape, thing glowed in the same colord and at the same temperature (generally high).

Kirchoff

• spectroradiancy

$$\frac{E}{A}\propto{R_T}(f);\\ \begin{align*} \left<\underset{\text{power flux}}{R_T(f)}=1\left[\frac{\text{W}}{\text{m}^2}\cdot\frac{1}{f}\right]=\left[\right]\frac{1}{f}\right> \end{align*}$$

• black body
  • black body tends to absorb more heat.

black body radiation graph

$$\begin{align*} \left<\underset{\text{power flux}}{R_T(f)}=1\left[\frac{\text{W}}{\text{m}^2}\cdot\frac{1}{f}\right]\right> \end{align*}$$

$$\begin{align*} R_T(f)&=\frac{E}{t\cdot{A}}\cdot\frac{1}{f}\\ &=\frac{E}{t}\cdot\left(\frac{L}{\text{vol}}\right)\cdot\frac{1}{f}\\ &=\frac{E}{\text{vol}}\cdot\frac{L}{\text{t}}\cdot\frac{1}{f}\\ &=\underset{\text{energy density}}{\left(\rho_T(t)\right)}\left(v\right)\left(\frac{1}{f}\right)\\ \therefore{R}_T(f)&\propto\left(\rho_T(t)\right)\cdot\frac{1}{f} \end{align*}$$

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INTERFERENCE

standing waves (1-D)

$$\begin{align*} y_1&=A\:\sin{(kx+\omega{t})}\\ y_2&=A\:\sin{(kx-\omega{t})} \end{align*}$$

$$\left\{ \begin{align*} y&=A\sin{\underline{\phi}}\\ \phi&=\underline{kx}+\omega\:t,&&\left<k:\text{wave }\#\right>\\ kx&=d\pm\omega\:t\\ x&=\frac{d}{k}\pm\frac{\omega}{k}t,&&\left<v=\frac{\omega}{k}\right>\\ &=C\pm{v}\:t; \end{align*} \right\}$$

$$\begin{align*} y_1+y_2&=A\sin{(kx+\omega{t})}+A\sin{(kx-\omega{t})}\\ &\left<\sin{(a)}+\sin{(b)}=2\left[\sin{\left(\frac{1}{2}(a+b)\right)}\cos{\left(\frac{1}{2}(a-b)\right)}\right]\right>\\ &=2A\sin{\left(\frac{(kx+\omega{t})+(kx-\omega{t})}{2}\right)}\cos{\left(\frac{(kx+\omega{t})-(kx-\omega{t})}{2}\right)}\\ &=2A\sin{(kx)}\cos{(\omega{t})} \end{align*}$$

$$\left\{ \begin{align*} kx&=n\pi;\\ \left(\frac{2\pi}{\lambda}\right)L&=n\pi;\\ \lambda&=\frac{2L}{n}; \end{align*} \right\}$$

vibration of standing wave

$$\begin{align*} f&=\frac{v}{\lambda}=\left(\frac{n}{2L}\right)v=\frac{nv}{2L}\\ f_{c}&\underset{\text{EM Wave (1D)}}{=\left(\frac{c}{2L}\right)n};\\ f_{c}&\underset{\text{EM Wave (3D)}}{=\left(\frac{c}{2L}\right)\sqrt{(n_x)^2+(n_y)^2+(n_z)^2}};\\ &\left<\begin{matrix}n:\text{mode \#}&&L:\text{size of capacity}\end{matrix}\right> \end{align*}$$

For EM wave

BLAKC BODY

https://en.m.wikipedia.org/wiki/Black_body

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이찬희 (MarkiiimarK)

Never Stop Learning.