notes for 普通物理（1）

Lecture 1

Chapter 1

Vectors: \(\vec{A}=(x,y,z)\).

Operations on vectors:

• Addition \(\vec{A}+\vec{B}\)
• Scalar product \(c\vec{A}\)
• Dot product \(\vec{A}·\vec{B}=x_1x_2+y_1y_2+z_1z_2=|\vec{A}||\vec{B}|\cos\theta\)
• Cross product \(\vec{A}\times\vec{B}\)

Vector calculus: functions in the form \(\vec{A}(t)=(A_x(t),A_y(t),A_z(t))=A_x(t)\hat{x}+A_y(t)\hat{y}+A_z(t)\hat{z}\).

\(\dfrac{\mathrm d\vec{A}(t)}{\mathrm dt}=\dfrac{\mathrm dA_x(t)}{\mathrm dt}\hat{x}+\dfrac{\mathrm dA_y(t)}{\mathrm dt}\hat{y}+\dfrac{\mathrm dA_z(t)}{\mathrm dt}\hat{z}\) (assumption: \(\hat{x},\hat{y},\hat{z}\) are constant)

\(\nabla\equiv\dfrac{\partial}{\partial x}\hat{x}+\dfrac{\partial}{\partial y}\hat{y}+\dfrac{\partial}{\partial z}\hat{z}\) (an operator)

\(\mathrm d\phi=\dfrac{\partial\phi}{\partial x}\mathrm dx+\dfrac{\partial\phi}{\partial y}\mathrm dy+\dfrac{\partial\phi}{\partial z}\mathrm dz=\nabla\varphi\mathrm·d\vec{r}\)

Units in physics:

• Length(m)
• Mass(kg)
• Time(s)
• Temperature(K)

Chapter 2

Basis concepts for kinematics: coordinate system, reference frame, relativity of motion, point particle (size can be ignored, translational motion)

Velocity \(\vec{v}=\dfrac{\mathrm d\vec{r}}{\mathrm dt}\)

Acceleration: \(\vec{a}=\dfrac{\mathrm d\vec{v}}{\mathrm dt}\)

Lecture 2

Polar coordinate: \(x=r\cos\theta,y=r\sin\theta,r=\sqrt{x^2+y^2},\theta=\tan^{-1}\dfrac{y}{x}\)

\(\hat{r}=\cos\theta\hat{x}+\sin\theta\hat{y},\hat{\theta}=-\sin\theta\hat{x}+\cos\theta\hat{y}\)

\(\hat{x}=\cos\theta\hat{r}-\sin\theta\hat{\theta},\hat{y}=\sin\theta\hat{r}+\cos\theta\hat{\theta}\)

Polar coordinate with physics:

• Position \(\vec{r}(t)=r(t)\hat{r}\)
• Velocity \(\vec{v}=\dfrac{\mathrm d}{\mathrm dt}\vec{r}(t)=\dfrac{\mathrm d}{\mathrm dt}r(t)·\hat{r}+r(t)\dfrac{\mathrm d}{\mathrm dt}\hat{r}=\dot{r}\hat{r}+r\dot{\hat{r}}\), since \(\hat{r}=\cos\theta\hat{x}+\sin\theta\hat{y}\), \(\dfrac{\mathrm d\hat{r}}{\mathrm dt}=\dfrac{\mathrm d\theta}{\mathrm dt}·(-\sin\theta\hat{x}+\cos\theta\hat{y})=\dot{\theta}\hat{\theta}\), thus \(\vec{v}=\dot{r}\hat{r}+r\dot{\theta}\hat{\theta}\).
• Acceleration \(\vec{a}=\dfrac{\mathrm d}{\mathrm dt}\vec{v}=(\ddot{r}\hat{r}+\dot{r}\dot{\hat{r}})+(\dot{r}\dot{\theta}\hat{\theta}+r\ddot{\theta}\hat{\theta}+r\dot{\theta}\dot{\hat{\theta}})\), since \(\dot{\hat{r}}=\dot{\theta}\hat{\theta},\dot{\hat{\theta}}=-\dot{\theta}\hat{r}\), so \(\vec{a}=(\ddot{r}\hat{r}+\dot{r}\dot{\theta}\hat{\theta})+(\dot{r}\dot{\theta}\hat{\theta}+r\ddot{\theta}\hat{\theta}-r\dot{\theta}^2\hat{r})=(\ddot{r}-r\dot{\theta}^2)\hat{r}+(2\dot{r}\dot{\theta}+r\ddot{\theta})\hat{\theta}\).

Circular motion: angular velocity \(\omega=\dfrac{v}{r}\), \(\vec{\omega}=\omega\hat{z}\).

Let \(\vec{\theta}(t)=\theta\hat{z}\), then \(\vec{\omega}=\dfrac{\mathrm d\vec{\theta}}{\mathrm dt}\).

Relative motion: \(x'=x-ut,y'=y,z'=z,t'=t\).

Lecture 3

Angular velocities are vectors (it is ok to add them together under the rule of vectors)

Common forces in classical mechanics: gravity, supporting force, elastic force, friction.

Classification of forces: gravitation, electromagnetic force, weak force, strong force

Hooke's law: \(F=-k(x-x_{eq})\)

Two springs with coefficient \(k\) hanging together, \(k_p=2k,k_s=\dfrac{k}{2}\).

Newton's universal law of gravitation: \(\vec{F_{1,2}^G}=-G\dfrac{m_{1,2}}{r_{1,2}^3}\vec{r_{1,2}}\), where \(G=6.67\times 10^{-11}N\cdot m^2\cdot kg^{-2}\).

Newton's first law is only true for inertia reference frames (reference frame has no acceleration).

Newtonian Damping: A particle moves along \(x\)-axis with initial velocity \(v_0\) experiencing a frictional force \(F=\gamma mv^2\). Then \(\dfrac{\mathrm dv}{\mathrm dt}=\gamma v^2\), \(\gamma\mathrm dt=-\dfrac{\mathrm dv}{v^2}=\mathrm d(\dfrac{1}{v})\), \(\gamma t=\dfrac{1}{v}-\dfrac{1}{v_0}\), \(v=\dfrac{1}{\gamma t+\frac{1}{v_0}}=\dfrac{v_0}{1+v_0\gamma t}\).

Translational non-inertial reference frame: inertial force \(\vec{F_a}=-m\vec{a_0}\), where \(a_0\) is the acceleration of the non-inertial reference frame relative to some inertial reference frame.

Lecture 4

Cross product: let \(\vec{A}\times\vec{B}=\vec{C}\), then \(|\vec{C}|=|\vec{A}||\vec{B}|\sin\theta\), \(C\perp A,C\perp B\) (\(C=0\) if \(A\) and \(B\) are parallel), the direction is determined by right hand rule.

Another definition for cross product: \(\vec{A}\times\vec{B}=\det\begin{bmatrix}\hat{x}&\hat{y}&\hat{z}\\x_1&y_1&z_1\\x_2&y_2&z_2\end{bmatrix}\)

Rotational non-inertial reference frame:

• \(r=r',\theta=\theta'+\omega t,\hat{r}=\hat{r}',\hat{\theta}=\hat{\theta'}\).
• Since \(\vec{r}=\vec{r'}=x\hat{x}+y\hat{y}=x'\hat{x'}+y'\hat{y'}\), \(\vec{v}=\dot{x}\hat{x}+\dot{y}\hat{y}=\dot{x'}\hat{x'}+\dot{y'}\hat{y'}+x'\dot{\hat{x'}}+y'\dot{\hat{y'}}\), since \(\dot{x'}\hat{x'}+\dot{y'}\hat{y'}=\vec{v'}\), \(\dot{\hat{x'}}=\omega y'\), \(\dot{\hat{y'}}=y'(-\omega\hat{x'})\), so \(\vec{v}=\vec{v'}+\vec{\omega}\times\vec{r'}\).
• Since \(r=r',\dot{r}=\dot{r'}\), we have \(\ddot{r}=\ddot{r'}\). Since \(\theta=\theta'+\omega t\), \(\dot{\theta}=\dot{\theta'}+\omega\), \(\theta=\ddot{\theta}\), so \(\vec{a_r}=(\ddot{r}+r\dot{\theta}^2)\hat{r}=[\ddot{r'}+r'(\dot{\theta'}+\omega)^2]\hat{r'}=[\ddot{r'}+r'\dot{\theta'}-2r'\dot{\theta'}\omega-r'\omega^2]\hat{r'}=\vec{a'_r}-2\vec{v_{\theta'}}\times\vec{\omega}-\omega^2\vec{r}\), \(\vec{a_{\theta}}=r'\ddot{\theta'}+2r'(\dot{\theta'}+\omega)\vec{\theta'}=\vec{a'_{\theta}}-2\vec{v_{r'}}\times\vec{\omega}\), \(\vec{a}=\vec{a'}-2\vec{v'}\times\vec{\omega}-\omega^2\vec{r'}\). \(\vec{F}_{\text{inertial}}=2m\vec{v'}\times\vec{\omega}+m\omega^2\vec{r'}\), \(\vec{F_{\text{real}}}+\vec{F_{\text{inertial}}}=m\vec{a'}\).

Lecture 5

\(\vec{F_c}=m\omega^2\vec{r'},\vec{F_{cor}}=2m\vec{v'}\times\vec{\omega}\).

Momentum: \(\vec{p}=m\vec{v}\). If \(m\) is constant, then \(\dfrac{\mathrm d}{\mathrm dt}\vec{p}=\dfrac{\mathrm d}{\mathrm dt}(m\vec{v})=\vec{F}\).

Impulse \(\vec{I}=\Delta\vec{p}=\int_{t_1}^{t_2}\vec{F}\mathrm dt\).

For a system of particles, \(\vec{p}_{sys}=\sum\limits_{i}\vec{p_i},\vec{F_i}=\vec{F_i^{ext}}+\sum\limits_{j\ne i}\vec{F_{j,i}},\vec{F}_{sys}=\sum\limits_{i}\vec{F_i}=\sum\limits_{i}\vec{F^{ext}_i}\). Since \(\vec{F_i}=\dfrac{\mathrm d}{\mathrm dt}\vec{p_i}\), \(\vec{F_{sys}}=\sum\limits_{i}\dfrac{\mathrm d}{\mathrm dt}\vec{p_i}=\dfrac{\mathrm d}{\mathrm dt}\vec{p_{sys}}\). So if \(\vec{F}_{sys}=0\), then \(\vec{p_{sys}}\) is constant.

Aerospace 101:

• rocket velocity is \(v(t)\).
• ejecting velocity relative to rocket \(-u\).
• rocket mass \(m_r(t)\)
• mass of ejected fuel at between time \(t\) and \(t+\mathrm dt\) is \(\mathrm d m_f\)
• external force \(F_{ext}\)

Then \(p_{sys}(t)=m_r(t)v(t)\), \(p_{sys}(t+\mathrm dt)=(v(t)+\mathrm dv)m_r(t+\mathrm dt)+(v(t)-u)\mathrm dm_f\). Since \(\mathrm dm_r=-\mathrm dm_f\), \(m_r(t+\mathrm dt)=m_r(t)+\mathrm d m_r\). Thus \(F_{ext}=\dfrac{\mathrm dp_{sys}(t)}{\mathrm dt}=v(t)\dfrac{\mathrm dm_r}{\mathrm dt}+m_r(t+\mathrm dt)\dfrac{\mathrm dv}{\mathrm dt}-(v(t)-u)\dfrac{\mathrm dm_r}{\mathrm dt}=m_r(t+\mathrm dt)\dfrac{\mathrm dv}{\mathrm dt}+u\dfrac{\mathrm dm_r}{\mathrm dt}\).

In gravity-free space, \(F_{ext}=0\), \(-\dfrac{\mathrm dm_r}{\mathrm dt}u=m_r(t)\dfrac{\mathrm dv(t)}{\mathrm dt}\), \(\dfrac{\mathrm dv(t)}{\mathrm dt}=-\dfrac{u}{m_r(t)}\cdot\dfrac{\mathrm dm_r(t)}{\mathrm dt}\), so \(v(t)-v(0)=u\ln(\dfrac{m_{r}(0)}{m_{r}(t)})\).

In constant gravitational field, \(\vec{F}_{ext}=m_r\vec{g}\). Assume that the positive of the \(x\)-axis is upwards, then \(-m_r(t)g=m_r(t+\mathrm dt)\dfrac{\mathrm dv}{\mathrm dt}+u\dfrac{\mathrm dm_r}{\mathrm d_t}\). So \(\mathrm dv=-g-\dfrac{\mathrm dm_r}{m_r}u\), \(v(t)-v(0)=u\ln(\dfrac{m_{r}(0)}{m_{r}(t)})-gt\).

Center of mass: \(\vec{r_c}=\dfrac{1}{m_{sys}}\sum\limits_{i=1}^Nm_i\vec{r_i}\), where \(m_{sys}=\sum\limits_{i=1}^Nm_i\).

Lecture 6

Center of mass reference frame: \(\vec{v_c}=\vec{r_c}=\vec{p_c}=0\), in general not inertial.

2-body problem:

• \(F_{ext}=\vec{F_1}+\vec{F_2}\)
• \(\vec{a_c}=\dfrac{F_{ext}}{m_1+m_2}=\vec{F_1+F_2}\)
• Inertial forces: \(\vec{F_1^{in}}=-m_1\vec{a_c}=-\dfrac{m_1}{m_1+m_2}(\vec{F_1}+\vec{F_2})\), \(\vec{F_2^{in}}=-m_2\vec{a_c}=-\dfrac{m_2}{m_1+m_2}(\vec{F_1}+\vec{F_2})\).
• \(\vec{F'_1}=\vec{F_1}+\vec{F_1^{in}}=\dfrac{m_2\vec{F_1}-m_1\vec{F_2}}{m_1+m_2}\), \(\vec{F'_2}=\vec{F_2}+\vec{F_2^{in}}=\dfrac{m_1\vec{F_2}-m_2\vec{F_1}}{m_1+m_2}\).

Since \(\vec{F'_1}=m_1\vec{a'_1},\vec{F_2}=m_2\vec{a'_2}\), \(\vec{a'}=\vec{a'_1}-\vec{a'_2}=(\dfrac{1}{m_1}+\dfrac{1}{m_2})\vec{F'_1}\).

Chapter 5

Energy and work

\(W=\vec{F}\cdot\Delta\vec{r}\), \(\mathrm dW=\vec{F}\cdot\mathrm dr\).

\(\bar{P}=\dfrac{W}{\Delta t}\), \(P=\dfrac{\mathrm d W}{\mathrm dt}=\vec{F}\cdot\vec{v}\).

\(P=\vec{F}\cdot\vec{v}=m\vec{a}\cdot\vec{v}=m\dfrac{\mathrm d\vec{v}}{\mathrm dt}\cdot\vec{v}=m\cdot\dfrac{1}{2}\cdot\dfrac{\mathrm d(\vec{v}\cdot\vec{v})}{\mathrm dt}=\dfrac{\mathrm d(\frac{1}{2}mv^2)}{\mathrm dt}\).

Lecture 7

The work \(W\) is not a state function.

\(K=\dfrac{1}{m}m\vec{v}^2=\dfrac{1}{2}m\vec{v}\cdot\vec{v}\).

In center of mass reference frame moving with velocity \(\vec{u}\)

• \(K=\sum K_i=\sum\dfrac{1}{2}m_iv_i^2\)
• \(\vec{v_i}=\vec{v'_i}+\vec{u}\), where \(\vec{v'_i}\) is the velocity of particle \(i\) in the C.O.M R.F.
• \(K_{COM}=\sum\dfrac{1}{2}m_i(v'_i)^2\)

So

\[\begin{aligned} K&=\dfrac{1}{2}\sum m_i|\vec{v'_i}+\vec{u}|^2\\ &=\dfrac{1}{2}\sum m_i((v'_i)^2)+u^2+2\vec{v_i}\cdot\vec{u}\\ &=\dfrac{1}{2}\sum m_i(v'_i)^2+\dfrac{1}{2}\sum m_iu^2+(\sum m_i\vec{v_i})\cdot\vec{u}\\ &=K_{com}+\dfrac{1}{2}Mu^2 \end{aligned} \]

Potential energy (only conservative forces can define potential energy)

Definition of conservative force: \(\oint\vec{F}\cdot\vec{r}=0\) for any closed curve.

\(U_B-U_A=-\int_A^B\vec{F}\cdot\mathrm d\vec{r}=-W_c\) (path independent)

For gravitational force, \(\vec{F}=-G\dfrac{m_Am_B}{r_{AB}^3}\vec{r_{AB}}\).

So

\[\begin{aligned} W&=\int\vec{F}\cdot\mathrm dr\\ &=-Gm_Am_B\int\dfrac{\vec{r}\cdot\mathrm d\vec{r}}{r^3}\\ &=-Gm_Am_B\int\dfrac{\mathrm dr}{r^2}\\ &=-Gm_Am_B(\dfrac{1}{r})_i^f\\ &=-Gm_Am_B(\dfrac{1}{r_f}-\dfrac{1}{r_i}) \end{aligned} \]

So \(U=\dfrac{-Gm_Am_B}{r}\).

Escape velocity: \(\dfrac{1}{2}mv^2=\dfrac{GMm}{R}\), so \(v=\sqrt{\dfrac{2GM}{R}}\) (for earth, the value is \(11.2\text{km/s}\))

The change of the mechanical energy equals to the work done by non-conservative forces.

Bernoulli equation: \(\dfrac{p}{\rho g}+\dfrac{v^2}{2g}+h=0\) (\(p\): pressure, \(\rho\): density, \(h\): height)

Proof: \(\mathrm dE=(\dfrac{1}{2}v_2^2\mathrm dm+gh_2\mathrm dm)-(\dfrac{1}{2}v_1^2\mathrm dm+gh_1\mathrm dm)\), \(\mathrm dW=P_1A_1v_1\mathrm dt-P_2A_2v_2\mathrm dt\), \(\mathrm dm=\rho(v_2A_2\mathrm dt-v_1A_1\mathrm dt)\). Since \(\mathrm dE=\mathrm dW\), so \((P_1-P_2)(A_1v_1\mathrm dt-A_2v_2\mathrm dt)=\rho(v_1A_2\mathrm dt-v_1A_1\mathrm dt)(\dfrac{1}{2}v_2^2+gh_2-\dfrac{1}{2}v_1^2-gh_1)\), so \(P_1+\rho(\dfrac{1}{2}v_1^2+gh_1)=P_2+\rho(\dfrac{1}{2}v_2^2+gh_2)\).

Lecture 8

Mechanical energy \(E=K+U\).

\(\Delta E=W_{n.c}\)

Friction: some part of the energy turn into heat

The change of energy of the system plus the change of energy of surrounding is \(0\), i.e., \(\Delta E_{sys}+\Delta_{sur}=0\) (1st law)

Angular momentum: \(\vec{L_s}=\vec{r_s}\times\vec{p}\), where \(\vec{p}\) is the momentum and \(s\) is the reference point.

Magnitude of angular momentum: \(|\vec{L}|=|\vec{r_s}||\vec{p}|\sin\theta\).

Let \(v=v_r\hat{r}+v_{\theta}\hat{\theta}\), then \(\vec{L}=m\vec{r}\times\vec{r}=mrv_{\theta}(\hat{r}\times\hat{\theta})=mr^2\dot{\theta}(\vec{r}\times\vec{\theta})=mr^2\vec{\omega}\).

\(\dfrac{\mathrm d\vec{L}}{\mathrm dt}=\dfrac{\mathrm d}{\mathrm dt}=\vec{v}\times\vec{p}+\vec{r}\times\vec{F}=\vec{r}\times\vec{F}=\vec{\tau}\) (torque)

Meteor flyby: A meteor with mass \(m\) is moving at speed \(v_0\), let \(\theta\) be the angle between \(\vec{v_0}\) and \(\vec{r}\), \(h=d\sin\theta\), the radius of the Earth is \(R\), then if the meteor will exactly hit Earth, then:

• \(\vec{F}=-\dfrac{GMm}{r^2}\hat{r}\)
• \(\vec{L_i}=mhv_0\vec{k},L_f=m_fv_fR\vec{k}\)
• Since \(\vec{r}\times\vec{F}=0\), \(\vec{L_f}=\vec{L_i}\), so \(v_0h=v_fR\).
• \(\dfrac{1}{2}v_f^2-\dfrac{1}{2}mv_0^2=0+\dfrac{GMm}{R}\), so \(v_f=v_0\sqrt{1+\dfrac{2GM}{Rv_0^2}}\), \(h=R\sqrt{1+\dfrac{2GM}{Rv_0^2}}\).

So if \(h\le R\sqrt{1+\dfrac{2GM}{Rv_0^2}}\) the meteor will hit Earth.

Kepler's second law: \(\mathrm dA=\dfrac{1}{2}(v\mathrm dt)r\sin\theta=\dfrac{1}{2}|\vec{r}\times\vec{v}|\mathrm dt\), so \(\dfrac{\mathrm dA}{\mathrm dt}=\dfrac{1}{2}|\vec{r}\times\vec{v}|=\dfrac{\vec{L}}{2m}\). Since \(\vec{r}\) is parallel with \(\vec{F}\), \(L\) is constant, so \(\dfrac{\mathrm dA}{\mathrm dt}\) is constant.

Lecture 9

According to the definition of angular momentum, we have \(\vec{\omega}=\dfrac{\vec{L}}{mr^2}\), so \(mr^2\) can be viewed as "rotational inertia".

Leverage: \(\tau=l_2F-l_1G\ge 0\), so in order to lift the mass, \(l_2F\ge l_1G\).

For a system of particles: \(\vec{L_i}=\vec{r_i}\times\vec{p_i}\), \(\vec{L}=\sum\vec{L_i}=\sum\vec{r_i}\times\vec{p_i}\).

The change of angular momentum of a system:

\[\begin{aligned} &\dfrac{\mathrm d}{\mathrm dt}\sum\limits_{i}\vec{r_i}\times\vec{p_i}\\ =&\sum\limits_{i}(\dfrac{\mathrm d\vec{r_i}}{\mathrm dt}\times\vec{p_i}+\vec{r_i}\times\dfrac{\mathrm d}{\mathrm dt}\vec{p_i})\\ =&\sum\limits_{i}(\vec{v_i}\times\vec{p_i}+\vec{r_i}\times\vec{F_i})\\ =&\sum\limits_{i}\vec{\tau_i} \end{aligned} \]

If for all \(i,j\), \(\vec{F}_{i,j}\) is parallel to \(\vec{r_i}\), then \(\vec{r_i}\times\vec{F_{j,i}}+\vec{r_j}\times\vec{F_{i,j}}=\vec{r_{j,i}}\times\vec{F_{j,i}}=0\), so \(\sum\limits_{i}\vec{\tau_i}=\sum\limits_{i}\vec{\tau_i^{ext}}\). (contact forces)

Reference points: Consider an object \(i\) and two reference points \(A,B\), \(\vec{L_A}=\vec{r_{A,i}}\times\vec{p_i}\), \(\vec{L_B}=\vec{r_{B,i}}\times\vec{p_i}\), so \(\vec{L_A}-\vec{L_B}=(\vec{r_{A,i}}-\vec{r_{B,i}})\times\vec{p_i}=\vec{r_{A,B}}\times\vec{p_i}\). Now consider replacing the object with a system of particles, then \(\vec{L_A}-\vec{L_B}=\sum\limits_{i}(\vec{L_A^i}-\vec{L_B^i})=\sum\limits_{i}\vec{r_{A,B}}\times\vec{p_i}=\vec{r_{A,B}}\times\vec{p_{sys}}\). So in the C.O.M R.F of the system, \(\vec{p_{sys}}=0\), \(\vec{L_A}=\vec{L_B}\).

For a system of particles,

\[\begin{aligned} &\vec{L_{sys}}-\vec{L_{COM}}\\ =&\sum\limits_{i}(m_i\vec{r_i}\times\vec{v_i}-m_i\vec{r'_i}\times\vec{v'_i})\\ =&\sum\limits_{i}m_i((\vec{r'_i}+\vec{r_{COM}})\times(\vec{v'_i}+\vec{v_{COM}})-\vec{r'_i}\times\vec{v'_i})\\ =&\sum\limits_{i}m_i(\vec{r_{COM}}\times\vec{v_{COM}}+\vec{r'_i}\times\vec{v_{COM}}+\vec{r_{COM}}\times\vec{v'_i}) \end{aligned} \]

Since \(\sum\vec{r'_i}=0,\sum\vec{v'_i}=0\), \(\vec{L_{sys}}=\vec{L_{COM}}+M\vec{r_{COM}}\times\vec{v_{COM}}=\vec{L_{COM}}+\vec{L}\), where \(\vec{L}\) is the angular momentum of the system if we treat the whole system as a point particle.

This is similar to kinetic energy (\(K=K_{com}+\dfrac{1}{2}Mu^2\))

Lecture 10

Momentum:

\(\vec{p}=m\vec{v}\)

\(\vec{F}=\dfrac{\mathrm d}{\mathrm dt}\vec{p}\)

\(\vec{F^{ext}}=0\Rightarrow \sum\limits_{i}\vec{p_i}=\text{const}\)

Under C.O.M R.F, \(\vec{p}_{COM}=0\), \(\vec{p}=M\vec{v}_{COM}\).

Angular Momentum:

\(\vec{L}=\vec{r}\times\vec{p}\)

\(\tau=\dfrac{\mathrm d}{\mathrm dt}\vec{L}=\vec{r}\times\vec{F}\)

\(\vec{\tau^{ext}}=0\Rightarrow \sum\limits_{i}\vec{L_i}=0\)

Under C.O.M R.F, \(\vec{L}=\vec{r}_{COM}\times\vec{p}_{COM}+\vec{L}_{COM}\).

Energy:

\(K=\dfrac{1}{mv^2}\)

\(P=\dfrac{\mathrm dW}{\mathrm dt}=\vec{F}\cdot\vec{v}\)

Internal forces can change the energy.

Under C.O.M R.F, \(K=\dfrac{1}{2}v_{COM}^2+K_{COM}\).

Collision.

Elastic collision:

• \(m_{A}v_A^i+m_Bv_B^i=m_Av_A^f+m_Bv_B^f\)

• \(\dfrac{1}{2}m_A(v_A^i)^2+\dfrac{1}{2}m_B(v_B^i)^2=\dfrac{1}{2}m_A(v_A^f)^2+\dfrac{1}{2}m_B(v_B^f)^2\).

Solve it and we get

• \(v_A^f=\dfrac{m_A-m_B}{m_A+m_B}v_A^i+\dfrac{2m_A}{m_A+m_B}v_B^i\)
• \(v_B^f=\dfrac{m_B-m_A}{m_A+m_B}v_B^i+\dfrac{2m_B}{m_A+m_B}v_A^i\)

\(v_A^i+v_A^f=v_B^i+v_B^f\), \(v_i^{rel}=-v_f^{rel}\).

If \(m_A>>m_B\), \(v_A^f\approx v_B^i\), \(v_B^f\approx -v_B^i+2v_A^i\)

Let \(\Delta K=K_f-K_i\), in inelastic collision, \(\Delta K<0\); in super-elastic collision: \(\Delta K>0\).

Totally inelastic: after collision \(v_A^f=v_B^f\) (two objects stick together). \(\Delta K=\dfrac{1}{2}m_A(v_A^i)^2+\dfrac{1}{2}m_B(v_B^i)^2-\dfrac{(m_Av_A+m_Bv_B)^2}{2(m_1+m_2)}\) (maximal kinetic energy change).

Lecture 12

Rigid body: a system of particles such that \(\forall i,j\), \(|\vec{r_i}-\vec{r_j}|\) is constant over time.

Theorem: At instant time \(t\), any motion of a rigid body can be treated as a translational motion + fixed axis rotation.

Rolling wheel: \(\vec{v_p}=(v_{COM}+R\omega_c\sin\theta)\hat{i}-R\omega_c\cos\theta\hat{j}\).

• \(v_{COM}=R\omega_c\): rolling without slipping
• \(v_{COM}>R\omega_C\): skidding
• \(v_{COM}<R\omega_C\): slipping

Moment of inertia: \(I_{axis}=\int\rho\mathrm dm=\sum\limits_{i}m_i\rho_i^2\), where \(\rho_i\) is the distance of particle \(i\) to the axis.

Angular momentum \(\vec{L_S}=I_S\omega_z\hat{k}\), total torque \(M=I\alpha\).

Parallel axis theorem: \(I=I_{COM}+md^2\), where \(I_{COM}\) is the rotational inertia if we displace the axis to let it pass through the C.O.M and \(d\) is the displacement from the C.O.M.

Lecture 13

Kinetic energy of a rigid body:

\[\begin{aligned} K&=\sum\limits_{i}\omega(L_z)_i\\ &=\sum\limits_{i}\dfrac{1}{2}m_i(\omega r_i)^2\\ &=\dfrac{1}{2}(\sum\limits_{i}m_ir_i^2)\omega^2\\ &=\dfrac{1}{2}I\omega^2 \end{aligned} \]

A cylinder with mass \(m\) and radius \(R\) rolling down the slope: \(mgh=\dfrac{1}{2}mv_{COM}^2+\dfrac{1}{2}I_{COM}\omega^2\). Since \(\omega=\dfrac{v_{COM}}{R}\), \(I=\dfrac{1}{2}MR^2\), \(v_{COM}=\sqrt{\dfrac{2gh}{1+\frac{I}{MR^2}}}=\sqrt{\dfrac{4}{3}gh}\).

Lecture 15

Simple harmonic motion: \(x(t)=A\cos(\omega t+\phi_0)\).

• \(A\): amplitude
• \(\omega\): angular frequency
• \(\phi_0\): initial phase
• \(T=\dfrac{2\pi}{\omega}\): period
• \(f=\dfrac{1}{T}\): frequency

\(\dot{x}(t)=-A\omega\sin(\omega t+\phi_0),\ddot{x}(t)=A\omega^2\cos(\omega t+\phi_0)\), so \(\ddot{x}(t)+\omega^2x(t)=0\).

Dynamics: \(F=ma=-m\omega^2x\). For Hooke's law in the form of \(F=-kx\), we have \(k=m\omega^2\), so \(\omega=\sqrt{\dfrac{k}{m}},T=2\pi\sqrt{\dfrac{m}{k}}\).

Energy: \(U(t)=\dfrac{1}{2}kx^2=\dfrac{1}{2}kA^2\cos^2(\omega t+\phi_0)\), \(K(t)=\dfrac{1}{2}m(-\omega A\sin(\omega t+\phi_0))^2=\dfrac{1}{2}kA^2\sin^2(\omega t+\phi_0)\), so \(E(t)=U(t)+K(t)=\dfrac{1}{2}kA^2\).

Composition of simple harmonic motions:

• \(x_1=A_1\cos(\omega t),x_2=A_2\cos(\omega t+\phi)\). Then \(x=A\cos(\omega+\theta)\), where \(A=\sqrt{A_1^2+A_2^2+2A_1A_2\cos\phi}\), \(\tan\theta=\dfrac{A_2\sin\phi}{A_1+A_2\cos\phi}\).
  • \(\phi=0\): \(A=A_1+A_2,\theta=0\)
  • \(\phi=\pi\): \(A=|A_1-A_2|,\theta=0\).
• \(x_1=A\cos(\omega_1t),x_2=A\cos(\omega_2t+\phi)\). Then \(x=2A\cos(\dfrac{\omega_1-\omega_2}{2}t-\dfrac{\phi}{2})\cos(\dfrac{\omega_1+\omega_2}{2}+\dfrac{\phi}{2})\), not a SHM function, but if \(\omega_1-\omega_2<<\omega_1+\omega_2\), \(\cos(\dfrac{\omega_1-\omega_2}{2}t-\dfrac{\phi}{2})\) will change very slowly (beats).

Composition of simple harmonic motions in perpendicular directions: \(x=A\cos(\omega_xt),y=B\cos(\omega_yt+\phi)\):

• \(\omega_x=\omega_y\): \(\dfrac{x^2}{A^2}+\dfrac{y^2}{B^2}-\dfrac{2xy\cos\phi}{AB}=\sin^2\phi\).
• \(\omega_x\ne \omega_y\): Lissajous curve

Damped oscillation: \(F_d=-bv\), so \(-bv-kx=ma\). Let \(\omega_0^2=\dfrac{k}{m},\beta=\dfrac{b}{2m}\), then \(\ddot{x}+2\beta\dot{x}+\omega_0^2x=0\). The general solution can be written as \(x(t)=c_+e^{\lambda_+t}+c_-e^{\lambda_-t}\), where \(\lambda_{\pm}=-\beta\pm\sqrt{\beta^2-\omega_0^2}\).

• Under damping: \(\beta<\omega_0\), \(x(t)=Ae^{-\beta t}\cos(\omega t+\phi)\), where \(\omega=\sqrt{\omega_0^2-\beta^2}\).
• Over damping: \(\beta>\omega_0\), \(x(t)=c_+e^{\lambda_+t}+c_-e^{\lambda_-t}\).
• Critical damping: \(\beta=\omega_0\), \(x(t)=e^{bt/2m}(c_0+c_1t)\) (fastest return to the equilibrium position)

Force oscillation: \(F(t)=F_0\cos(\omega t)\), \(F_0\cos(\omega t)-kx-b\dot{x}=m\ddot{x}\). Let \(\omega_0^2=\dfrac{k}{m},\beta=\dfrac{b}{2m},f=\dfrac{F_0}{m}\), then \(x(t)=A\cos(\omega t+\phi)\), where \(A=\dfrac{f}{\sqrt{(\omega_0-\omega^2)^2+4\beta\omega^2}}\). When \(\omega=\omega_0\), \(A\) is maximal (resonance)

Lecture 16

Transverse wave: a wave whose oscillations are perpendicular to the direction of the wave’s propagation direction. (contrast: longitudinal wave)

For transverse wave, we have \(y(x,t)=A\cos(kx-\omega t+\phi)\).

Wavelength: \(\lambda=\dfrac{2\pi}{k}\), period: \(T=\dfrac{2\pi}{\omega}\), frequency: \(f=\dfrac{1}{T}=\dfrac{\omega}{2\pi}\), speed: \(v=\dfrac{\omega}{k}\). (phase velocity, possible to \(>c\))

Energy: Let \(\mathrm dm=\mu\mathrm dx\), \(\dfrac{\mathrm dK}{\mathrm dx}_{avg,t}=\dfrac{1}{4}\mu\omega^2A^2\) (average line density of kinetic energy over time), \(\dfrac{\mathrm dE}{\mathrm dx}_{avg,t}=\dfrac{1}{2}\mu\omega^2A^2\).

Dynamic equation of transverse wave: since \(y(x,t)=A\cos(kx-\omega t+\phi)\), \(F=ma=m\dfrac{\mathrm d^2y}{\mathrm dt^2}=-m\omega^2y\), \(\dfrac{\mathrm d^2y}{\mathrm d x^2}=-k^2y\), so we have \(\dfrac{\mathrm d^2y}{\mathrm d x^2}=-\dfrac{1}{v^2}\cdot \dfrac{\mathrm d^2y}{\mathrm dt^2}\).

Interference of waves: \(y(x,t)=y_1(x,t)+y_2(x,t)\).

If \(y_1=A\cos(\omega t-kx),y_2=A\cos(\omega t-kx+\phi)\), then \(y=2A\cos\dfrac{\phi}{2}\cos(\omega t-kx+\dfrac{\phi}{2})\).

• \(\phi=0\): a new wave with amplitude \(2A\).
• \(\phi=\pi\): \(0\).

If \(y_1=A\cos(\omega t-kx),y_2=A\cos(\omega t+kx)\), then \(y=2A\cos kx\cos\omega t\). (standing wave equation)

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Lecture 17

Lagrangian mechanics

Coordinate: \(q\), momentum: \(p\)

Lagrangian \(L(q,\dot{q})=T(q,\dot{q})-V(q)\), where \(T\) and \(V\) denote the kinetic energy and potential energy, respectively.

For a single particle, we have \(T=\dfrac{1}{2}m\dot{q}^2\) and \(-\dfrac{\mathrm dV}{\mathrm dq}\) is the force.

Path: \(q(\tau)\), action: \(S[q(\tau)]=\int_0^tL(q(\tau),\dot{q}(\tau))\mathrm d\tau\).

Principle of least action: \(q_{\text{actual}}(\tau)=\arg\min_{q(\tau)}S[q(\tau)]\).

Consider a free particle with \(V(q)=0\) everywhere, then since \((\int\dot{q}\mathrm d\tau)^2\le(\int\dot{q}^2\mathrm d\tau)\cdot(\int\mathrm d\tau)\), so the \(\arg\min\) is obtained at \(\dot{q}(\tau)=\text{const}\) (Newton 1st law)

We slightly change the path by \(\delta q\) and see whether the action will go down:

\[\begin{aligned} \delta S&=\int_0^t \delta L\text d\tau\\ &=\int_0^t \left(\frac{\partial L}{\partial q}\delta q+\frac{\partial L}{\partial \dot q}\delta \dot{q}\right)\text d\tau\\ &=\int_0^t \left(\frac{\partial L}{\partial q}\delta q-\frac{\text d}{\text d\tau}\left(\frac{\partial L}{\partial \dot q}\right)\delta q\right)\text d\tau +\left.\frac{\partial L}{\partial \dot q}\delta q\right|_{0}^t\\ \end{aligned} \]

If we know the correct endpoint of the path, then \(\delta q(0)=\delta q(t)=0\), so \(\delta S[q(\tau)]=\int_0^t(\dfrac{\partial L}{\partial q}\delta q-\dfrac{\mathrm d}{\mathrm d\tau}(\dfrac{\partial L}{\partial\dot{q}})\delta q)\mathrm d\tau\).

Since \(\delta q\) is arbitrary, to let \(\partial S[q(\tau)]=0\), we should have \(\dfrac{\partial L}{\partial q}-\dfrac{\mathrm d}{\mathrm dt}(\dfrac{\partial L}{\partial\dot{q}})=0\). (Euler-Lagrange equation)

Plug in the definition of \(L\):

• \(\dfrac{\partial L}{\partial q}=-\dfrac{\partial V}{\partial q}=F\).
• \(\dfrac{\partial L}{\partial\dot{q}}=m\dot{q}=p\), \(\dfrac{\mathrm d}{\mathrm dt}(\dfrac{\partial L}{\partial\dot{q}})=\dfrac{\mathrm dp}{\mathrm dt}\).

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Holonomic constraints: \(f_{\alpha}(q_1,q_2,\cdots,q_n)\).

Constrained form of the principle of the least action: \(q_{actual}(\tau)=\arg\min\limits_{q(\tau)}\max\limits_{\mu_{\alpha}}\int_0^t(L-\sum\limits_{\alpha}\mu_{\alpha}f_{\alpha}(q_1,q_2,\cdots,q_n))\mathrm d\tau\).

Constrained form of EL-equation: \(\dfrac{\partial L}{\partial q_i}-\dfrac{\mathrm d}{\mathrm dt}(\dfrac{\partial L}{\partial\dot{q_i}})-\sum\limits_{\alpha}\mu_{\alpha}\dfrac{\partial}{\partial q_i}f_{\alpha}(q_1,q_2,\cdots,q_n)=0\), \(f_{\alpha}(q_1,q_2,\cdots,q_n)=0\).

Lecture 18

Generalized momentum: \(p=\dfrac{\partial L}{\partial\dot{q}}\)。

E-L equation can be rewritten as \(\dot{p}=\dfrac{\partial L}{\partial q}\).

Continuous symmetries: \(\dfrac{\partial}{\partial s}L(Q(s,t),\dot{Q}(s,t))=0\).

Noether's Theorem: For each continuous symmetry, there exists a conserved quantity (\(\dfrac{\partial L}{\partial \dot{q}}\dfrac{\partial Q}{\partial s}\)).

Proof:

\[\begin{aligned} 0=\dfrac{\partial L}{\partial s}|_{s=0}&=\sum\limits_{i}\dfrac{\partial L}{\partial Q_i}\dfrac{\partial Q_i}{\partial s}|_{s=0}+\dfrac{\partial L}{\partial\dot{Q_i}}\dfrac{\partial\dot{Q_i}}{\partial s}|_{s=0}\\ &=\sum\limits_{i}\dfrac{\partial L}{\partial q_i}\dfrac{\partial Q_i}{\partial s}|_{s=0}+\dfrac{\partial L}{\partial\dot{q_i}}\dfrac{\partial\dot{Q_i}}{\partial s}|_{s=0}\\ &=\sum\limits_{i}\dfrac{\mathrm d}{\mathrm dt}(\dfrac{\partial L}{\partial \dot{q_i}})\dfrac{\partial Q_i}{\partial s}|_{s=0}+\dfrac{\partial L}{\partial\dot{q_i}}\dfrac{\partial\dot{Q_i}}{\partial s}|_{s=0}\\ &=\sum\limits_{i}\dfrac{\mathrm d}{\mathrm dt}(\dfrac{\partial L}{\partial\dot{q_i}}\dfrac{\partial Q_i}{\partial s})|_{s=0} \end{aligned} \]

So \(\sum\limits_{i}\dfrac{\partial L}{\partial\dot{q_i}}\dfrac{\partial Q_i}{\partial s}|_{s=0}\) is a conserved quantity.

If \(Q(s,t)=q(t)+s\), then \(\dfrac{\partial L}{\partial \dot{q_i}}\dfrac{\partial Q_i}{\partial s}=p\) which is the momentum.

Hamiltonian mechanics:

In Lagrangian mechanics, \(\mathrm dL=\dfrac{\partial L}{\partial q}\mathrm dq+\dfrac{\partial L}{\partial\dot{q}}\mathrm d\dot{q}\), since \(\dfrac{\partial L}{\partial\dot{q}}=p\), \(\mathrm dL=\dfrac{\partial L}{\partial q}\mathrm dq+p\mathrm d\dot{q}\). Consider \(p\dot{q}-L\), we have \(\mathrm d(p\dot{q}-L)=\dot{q}\mathrm dp+p\mathrm d\dot{q}-\mathrm dL=\dot{q}\mathrm dp-\dfrac{\partial L}{\partial q}\mathrm dq\).

So we define \(H=p\dot{q}-L\), then \(\dfrac{\partial H}{\partial p}=\dot{q}\), \(\dfrac{\partial H}{\partial q}=-\dot{p}\).

Lecture 19

Relativity:

Assumption:

1. The laws of physics are the same for all observers in all inertia reference frames.
2. The speed of light in vacuum has the same value \(c\) in all directions and in all inertia reference frames

Loss of simultaneity: two events happen simultaneously in RF \(S\) may not happen simultaneously in RF \(S'\).

Relativity of time: \(\Delta t=\gamma\Delta t_{proper}\), where \(\gamma=\dfrac{1}{\sqrt{1-(\frac{v}{c})^2}}\).

Relativity of length:

Alice: \(t_A=\dfrac{2l_{prop}}{c}\).

Bob: \(t_B=\dfrac{l}{c-v}+\dfrac{l}{c+v}=\dfrac{2lc}{c^2-v^2}=\dfrac{2l\gamma^2}{c}\), \(t_B=\gamma t_A\), so \(l=\dfrac{l_{prop}}{\gamma}\).

Lorentz transformation: \(\begin{cases}x'=\gamma(x-vt)\\t'=\gamma(t-vx/c^2)\\y'=y\\z'=z\end{cases}\)

Lecture 20

Relativity of velocities: \(u_x'=\dfrac{u_x-v}{1-vu_x/c^2}\), \(u_y'=\dfrac{u_y}{\gamma(1-vu_x/c^2)}\), \(u_z'=\dfrac{u_z}{\gamma(1-vu_x/c^2)}\).

Invariant interval: For any event, \(s^2=c^2t^2-x^2-y^2-z^2\) is invariant under Lorentz transformation.

• Timelike separation: \(s^2>0\), there exists a R.F. such that the \(2\) events happen at the same place.
• Spacelike separation: \(s^2<0\), there exists a R.F. such that the \(2\) events happen at the same time.
• Lightlike separation: \(s^2=0\), in every R.F. a photon emitted at one of the events will arrive at the other.

Dynamics of special relativity:

• Mass: \(m(v)=\dfrac{1}{\sqrt{1-\frac{v^2}{c^2}}}m_0=\gamma m_0\), where \(m_0\) is the rest mass.
• Kinetic energy \(K\) is the work done by an external force to accelerate the object from rest to a given velocity. \(K=\int_0^lF\text ds=\int_0^l \dfrac{\text dp}{\text dt}\text ds=\int_0^l\text d(mv)\cdot \dfrac{\text ds}{\text dt}=\gamma m_0c^2-m_0c^2\).
• Mass energy \(E_0=m_0c^2\), total energy \(E=E_0+K=mc^2\).

\(E^2=(pc)^2+(m_0c)^2\).

Lecture 21

Systems:

• Isolated: no energy change, no matter change
• Closed: has energy change, no matter change
• Open: has energy change, has matter change

Extensive: volume, mass, energy

Intensive: temperature, pressure, density

Zeroth law of thermodynamics: If bodies \(A\) and \(B\) are each in thermal equilibrium with a third body \(T\), then \(A\) and \(B\) are in thermal equilibrium with each other.

Linear expansion: \(\Delta L=L\alpha\Delta T\)

Volume expansion: \(\Delta V=V\beta\Delta T\)

Heat capacity: \(Q=C\Delta T\)

Specific heat: \(Q=cm\Delta T\)

Conduction: \(P_{cond}=\dfrac{Q}{t}=kA\dfrac{T_H-T_C}{L}\)

Radiation: \(P_{rad}=\sigma\epsilon AT^4\).

Lecture 22

First law of thermodynamics: \(\Delta E=Q-W\), \(W\) is the amount of work the system done on the environment.

Adiabatic process: \(Q=0\)

Constant-volume process: \(W=0\)

Cyclical process: \(\Delta E=0\)

Free expansion: \(Q=W=0\)

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Avogadro's number: \(1\text{mol}=6.02\times 10^{23}\)

Ideal gas: \(pV=nRT\), where \(R=8.31J/(\text{mol}\cdot K)\) (macroscopic)

\(pV=NkT\), where \(k=\dfrac{R}{N_A}=1.38\times 10^{-23}J/K\) (microscopic)

Isothermal expansion: \(W=\int_{v_i}^{v_f}p\mathrm dV=\int_{v_i}^{v_f}\dfrac{nRT}{V}\mathrm dV=nRT\ln\dfrac{v_f}{v_i}\).

Constant volume: \(W=0\), \(\mathrm dE=nC_V\mathrm dT\).

Constant pressure: \(W=P\mathrm dV\), \(\mathrm dE=nC_P\mathrm dT-P\mathrm dV\).

So \(nC_V\mathrm dT=nC_P\mathrm dT-P\mathrm dV\), and since \(\mathrm P\mathrm dV=nR\mathrm dT\), \(C_P-C_V=R\). (for ideal gas)

Adiabatic process: \(pV^{\gamma}=\text{const}\), where \(\gamma=\dfrac{C_P}{C_V}\). So then \(W=\dfrac{p_iV_i^{\gamma}}{1-\delta}V^{1-\delta}|_i^f\).

Proof: For adiabatic process, \(\mathrm dE=-\mathrm dW=-P\mathrm dV\), and since \(E=nC_VT\), so \(n\mathrm dT=-\dfrac{P}{C_V}\mathrm dV\). Since \(PV=nRT\), \(P\mathrm dV+V\mathrm dT=nR\mathrm dT\). And since \(R=C_P-C_V\), \(\dfrac{P\text dV+V\text dP}{C_P-C_V}=-\dfrac{P}{C_V}\text dV\implies \dfrac{\text dP}{P}+\dfrac{C_P}{C_V}\dfrac{\text dV}{V}=0\implies PV^{\gamma}=const\).

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When a gas molecule collides with the shaded wall, \(\Delta p_x=-2mv_x\), \(F=\dfrac{\Delta p}{\Delta t}=\dfrac{2mv_x}{\Delta t}\). The time between collisions is the time the molecule takes to travel to the opposite wall and back again, i.e., \(\dfrac{2L}{v_x}\), so \(F=\dfrac{mv_x^2}{L}\).

So the pressure \(p=\dfrac{F_x}{L^2}=\dfrac{m}{L^3}\sum v_i^2=\dfrac{nmN_A}{L^3}(v_x^2)_{avg}\).

Assume that on average \(v_x^2=\dfrac{1}{3}v^2\), so \(p=\dfrac{nMv_{rms}^2}{3V}\), where \(M\) is the molar mass of the gas.

Combining the equation and ideal gas law we get \(v_{rms}=\sqrt{\dfrac{3RT}{M}}\).

\(E=\dfrac{1}{2}Mv_{rms}^2=\dfrac{3}{2}RT\), \(K_{avg}=\dfrac{3}{2}kT\). (for monatomic gas)

The RMS of air molecule is \(500m/s\), close to the speed of sound.

Lecture 23

Mean free path: the average distance traversed by a molecule between collisions.

For ideal gas, \(\lambda=\dfrac{1}{\sqrt{2}\pi d^2N/V}\)

• \(He, Ar\): \(C_V\approx\dfrac{3}{2}R\) (monatomic)
• \(N_2,O_2\): \(C_V\approx\dfrac{5}{2}R\) (diatomic)
• \(NH_4,CO_2\): \(C_V\approx 3R\).

For a molecule, each degree of freedom associates an energy of \(\dfrac{1}{2}kT\) per molecule.

As the temperature \(T\) increases, the molar specific heats changes, as the rotational and oscillatory degree of freedom are turned on successfully.

Maxwell's speed distribution law: \(P(v)=4\pi(\dfrac{M}{2\pi RT})^{\frac{3}{2}}v^2e^{-\frac{Mv^2}{2RT}}\).

Characteristic speeds: \(v_{avg}=\sqrt{\dfrac{8RT}{\pi M}},v_{rms}=\sqrt{\dfrac{3RT}{M}}\).

Boltzmann distribution: \(P(E)\approx e^{-E/kT}\).

Van der Waals equation: modified ideal gas law: \((P+\dfrac{a}{V_m^2})(V_m-b)=RT\).

Lecture 25

Only in quasi-static process we can define the thermodynamic quantities of the system, such as pressure \(P\), temperature \(T\), volume \(V\).

A reversible process is a process which can be reversed back to its original state, with infinitesimal changes introduced to some properties of the system via its environment.

Irreversible processes: heat conduction, free expansion, work transferring to heat (such as friction)

Proof that the three statements are equivalent:

• Heat conduction \(\to\) conversion from work to heat: If heat conduction is reversible, then combining it with a Carnot engine we can produce a perfect engine that transform heat to work with \(100\%\) efficiency.

• Conversion from work to heat \(\to\) heat conduction: If heat to work conversion can be perfect, then we can extract work from high temperature object and transverse work to heat on the low temperature object.

• Conversion from work to heat \(\to\) Free adiabatic expansion: If heat to work conversion can be perfect, then first make the gas to experience reversible isothermal compression, then apply the heat-to-work

  conversion, showing that adiabatic free expansion is reversible.

• Free adiabatic expansion \(\to\) Conversion from work to heat: If the gas can be adiabatically compressed perfectly, then substitute the adiabatic compression in he Carnot cycle to this process, we can obtain a perfect heat-work engine.

Entropy: If an irreversible process occurs in an isolated system, the entropy \(S\) of the system always increases, it never decreases. A state function.

\(\Delta S_{i\to f}=\int_i^f\dfrac{\bar{\mathrm{d}}Q}{T}\) (for reversible processes)

For irreversible processes, to calculate \(\Delta S\), we need to construct an equivalent reversible path between the same initial and final states.

Second law: For an isolated system, \(\Delta S\ge 0\).

For any cyclic quasi-static process, \(\oint\dfrac{\bar{\mathrm{d}}Q}{T}\le 0\) (real process): Consider the isolated system formed by the system and the environment, we have \(\Delta S_{sys}+\Delta S_{env}\ge 0\), since the process is cyclic, \(\Delta S_{sys}=0\), so \(\Delta S_{env}=-\oint\dfrac{\bar{\mathrm{d}}Q}{T}\ge 0\).

For ideal gas:

• \(\text dE=\bar dQ-p\text dV=T\text dS-p\text dV\).

• \(T \text dS=p\text dV+nc_V\text dT\).

• \(\Delta S=\int\dfrac{nR}{V}\text dV+\dfrac{nc_V}{T}\text dT=nR\ln\dfrac{V_f}{V_i}+nc_V\ln\dfrac{T_f}{T_i}\).

For phase change: \(\Delta S=\dfrac{mL}{T}\).

Enthalpy: \(H=U+PV\).

Helmholtz free energy: \(F=U-TS\).

Gibbs free energy: \(G=U+PV-TS\).

Lecture 26

Boltzmann entropy: \(S=k\ln\Omega\), \(\Omega\) is the number of microscopic states in the macroscopic state.

Microcanonical Ensemble: all possible microstates of an isolated system that the real system might be in, where an isolated system means \(N\) (the number of particles), \(E\) (the energy of all particles) and \(V\) (the volume) are fixed.

Principle of Maximum Entropy: The system described by a microcanonical ensemble stays on the macroscopic state with maximum Boltzmann entropy.

Particle statistical distribution:

\(\epsilon_i\): energy level. \(g_i\): degeneracy of the \(i\)-th energy level. \(n_i\): the number of particles in the \(i\)-th energy level. Assume \(\sum n_i=N,\) \(\sum n_i\epsilon_i=E\).

• Maxwell (distinguishable): \(\Omega(\{n_i\})=N!\prod\dfrac{g_i^{n_i}}{n_i!}\)
• Bosons (indistinguishable, allows multiple particle in the same microstate): \(\Omega(\{n_i\})=\prod\dbinom{n_i+g_i-1}{g_i-1}\)
• Fermions (indistinguishable, does not allow multiple particle in the same microstate): \(\Omega(\{n_i\})=\prod\dbinom{g_i}{n_i}\).

Distributions:

• Maxwell: \(n_i=g_ie^{-\alpha-\beta\epsilon_i}\), where \(\alpha=-\dfrac{\mu}{kT},\beta=\dfrac{1}{kT}\), \(\mu\) is chemical potential.
• Bose-Einstein: \(n_i=\dfrac{g_i}{e^{\alpha+\beta\epsilon_i}-1}\)
• Fermi-Dirac: \(n_i=\dfrac{g_i}{e^{\alpha+\beta\epsilon_i}+1}\).