Kinetics

Work and Kinetic Energy§

Put simply, we can follow the same definition

$$W=\int \mathbf F\cdot d\mathbf r$$

But, notice that our defintiion of $\mathbf p$ have changed, it is now $p=\gamma m_0v$. By noticing the inner product is invariant under rotation, let's make it into a direction where only $x$ direction is relavent.

$$\begin{align*} W &= \int F\dd x \\ &= \int \frac{\dd p}{\dd t}\dd x \\ &= \int v\dd p \\ &= m_0\int v(v\dd \gamma + \gamma\dd v) \\ \end{align*}$$

Evaluating this is possible but messy, let's have a trick. Notice that

$$\begin{align*} \gamma^2(1-\beta^2) &= 1 \\ \gamma^2c^2 - \gamma^2v^2 &= c^2 \\ \dd(\gamma^2c^2 - \gamma^2v^2) &= 0 \\ c^2 \gamma\dd\gamma - v^2\gamma\dd\gamma - \gamma^2v\dd v &= 0 \\ c^2 \dd\gamma - v^2\dd\gamma - \gamma v\dd v &= 0 \\ v(v\dd\gamma + \gamma\dd v) &= c^2 \dd\gamma \end{align*}$$

So,

$$W=m_0\int c^2\dd\gamma = m_0 c^2 (\gamma - 1) = K$$

or

$$\begin{equation} K = m_0c^2(\gamma-1) \tag{\htmlId{eq-eq-kinetic-energy}{1}}\end{equation}$$

We chose to integrate from $\gamma = 1$ which implies a rest frame.

Notice what happens if we take $\dd W$, we get

$$\dd W = \dd (m_0 c^2\gamma) = m_0c^2\dd\gamma = \dd K$$

Or then we get ( if define $m = \gamma m_0$ )

$$\dd m = \frac{\dd K}{c^2}$$

This means nothing for us now because we can say the mass changes due to velocity changes which is already not a news when we define $m=\gamma m_0$. But what if we are able to gain or drop the mass irrelavant to the velocity, then we have the motivation to literally write $E = m_0c^2$ as the rest energy. Which indeed is the case as we will see.

Rest Energy§

Consider a box in its rest frame. Then emits light symetrically to left and right to conserves momentum. Yet, by energy convservation, the energy of the box must be dropped. For now, we haven't proof this energy must comes from the mass because the box might and can have variations of internal energy insides of it. So, just to quantize it, energy conservation gives us

$$E_\text{initial} = E_\text{after} + L$$

Where $E$ is the energy of the box and $L$ is the total energy of both light. And

$$p_\text{initial} = p_\text{after} + p_L - p_L = 0$$

Well, now consider the same setup in a different frame. for example now the box is moving in velocity $v$ in thsi frame, then by doopler shift, the energy of the light in this new frame will be

$$L' = \gamma L \qquad \text{will be derived later}$$

Then

$$E'_\text{initial} = E'_\text{after} + \gamma L$$

Since now the box is moving, $E' - E = K$ is the kinetic energy

$$K_\text{initial} = K_\text{after} + (\gamma - 1)L$$

Then

$$\Delta K = 2(\gamma - 1)L \underbrace{=}_\text{approximation} \frac{v^2}{2c^2}L$$

And since we are already under approximation $\Delta K = \dfrac{1}{2}\Delta m_0v^2$ since we already know from the rest frame the velocity of the box should not change.

$$\Delta m_0=\frac{L}{c^2}$$

The mass of the box literally droped new to energy dropped, irrelavent to its internal energy. Or we should now say, internal energy will be part of its mass. Mass is energy.

Now then we will have the motivation to say

$$E_\text{rest} = m_0c^2$$

Total relativistic energy§

Now, from $\href{#eq-eq-kinetic-energy}{(1)}$, we see

$$K = m_0c^2(\gamma-1) = m_0c^2\gamma - m_0c^2$$

Then we can rearrange and define the total relativistic energy as

$$E = \gamma m_0c^2 = K + \underbrace{m_0c^2}_\text{rest energy}$$

And remeber from Four vectors. We have

$$P^\mu = (\gamma_um_0c, \mathbf p)$$

Which we then substitude,

$$P^\mu = (\frac{E}{c}, \mathbf p)$$

So, if we calculate the invariant this time this way,

$$P^2 = -\frac{E^2}{c^2}+p^2$$

Which we already know should be

$$P^2 = -\frac{E^2}{c^2}+p^2 = -m_0^2c^2$$

Thus rearranging, we get

$$E^2 = m_0^2c^4 + p^2c^2$$

which for anyone who is actaully new, note that here $p=\gamma m_0 v$. And rest energy can be recovered by taking $p = 0$. And the energy of light can be taken as $m_0 = 0$. Which should be the expected result from electrodynamics.

Lorentz Force§

Field strength Tensor§

Although technically not part of kinetic. But it is related, we need field strength tensor to express the lorentz force, so let's expand out the matrix element now for the field strength tensor from Four vectors.

$$F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$$

Notice that $F^{\mu\nu} = -F^{\nu\mu}$. Automatically, all diagonals will be $0$ and we will only have independant variables in $\mu < \nu$ ( total of 6 ).

For $\mu = 0$ and $\nu = i$

$$F^{0i} = -\frac{1}{c}\partial_t A^i - \frac{1}{c}\partial_i\phi$$

This, by definition in Lorentz transformation. By definition is the elctric field. So

$$F^{0i} = \frac{E^i}{c}$$

For $\mu = i$ and $\nu = j$

$$F^{ij} = \partial^i A^j - \partial^j A^i$$

This is the curl of $A$, which then by definition is the magnetic field, again looking from Lorentz transformation. So

$$\begin{equation} F^{ij} = \epsilon^{ijk}B_k \tag{\htmlId{eq-eq-field-strength-tensor-magnetic}{2}}\end{equation}$$

Putting it togather, we get

$$F^{\mu\nu} = \begin{pmatrix} 0 & E^x/c & E^y/c & E^z/c \\ -E^x/c & 0 & B^z & -B^y \\ -E^y/c & -B^z & 0 & B^x \\ -E^z/c & B^y & -B^x & 0 \\ \end{pmatrix}$$

Lorentz Force in Strength Tensor§

Now, we see, the lorentz force is

$$\mathbf F = q\mathbf v\times \mathbf B$$

Which, using the Strength tensor and in index notation we get ($f$ is force, $F$ is field strenght tensor )

$$f^i = q\epsilon^{ijk}v_jB_k$$

And from $\href{#eq-eq-field-strength-tensor-magnetic}{(2)}$ we get

$$f^i = qv_jF^{ij}$$

Energy of light wave packet§

We will explain why this make sense later. But using the field strength tensor, we can see that $E$ and $B$ are one object in a whole and they mix after lorentz transformation

$$F'^{\alpha\beta} = F^{\mu\nu}\Lambda^{\alpha}_{\mu}\Lambda^{\beta}_{\nu}$$

And let's us only boost in the $x$ direction. And we only care about $E^y, B^z$ and $E^z, B^y$. Thus

$$\begin{align*} E'^y/c &= F'^{02} \\ &= F^{\mu\nu}\Lambda^{0}_{\mu}\Lambda^2_{\nu} \\ &= F^{\mu2}\Lambda^{0}_{\mu}\Lambda^2_{2} \\ &= F^{02}\Lambda^{0}_{0}\Lambda^2_{2} + F^{12}\Lambda^{0}_{1}\Lambda^2_{2} \\ E'^y &= \gamma (E^y - v B^z) \\ \end{align*}$$

The same goes to the rest

$$E'^z = \gamma(E^z + vB^y)\qquad B'^z = \gamma(B^z - \frac{v}{c^2}E^y)\quad B'^y = \gamma(B^y + \frac{v}{c^2}E^z)$$

Assuming there is no $x$ component in the field, we get

$$E'^2 = (E'^y)^2 + (E'^z)^2 = \gamma^2(E^2 + v^2B^2 + 2v(\mathbf E\times \mathbf B)^x)$$ $$B'^2 = (B'^y)^2 + (B'^z)^2 = \gamma^2(B^2 + \left(\frac{v}{c^2}\right)^2E^2 - 2\frac{v}{c^2}(\mathbf E\times \mathbf B)^x)$$

Adding them up, we will get the the energy density

$$u'=\frac{1}{2}\epsilon_0E'^2 + \frac{1}{2\mu_0}B'^2 = \gamma\left(u(1+\beta^2)-\frac{2v}{c^2}S_x\right)$$

Again if the light is directly traveling down $x$. Then, $(\mathbf E\times \mathbf B)^x = EB$. Then we will get

$$u' = \gamma^2 u (1-\beta)^2$$

Thus, for energy after transformation

$$U' = u'V' = \gamma^2u(1-\beta)^2\cdot \frac{V}{\gamma}=\gamma(1-\beta)U$$

We see, energy is frame dependant, although this is not a news for us when we already know kinetic energy is frame dependant.

Well, then, if we add up light going in both direction in the rest frame, but now in the moving frame, we have both $+\beta$ and $-\beta$. They then will add up to $0$ thus in the rest energy section we get

$$L' = \gamma L$$

And it can be easily seen as a combination of two effect. First, is doopler effect, although irrelavent to the frequency ( at least without quantum mechanics ), field cannot physically disappear, thus their amplitude must mush togather to form a higher amplitude.

Well, we can easily proof this using the transformation of $P^\mu$ ( assuming it is justified ). But that's the issue, we are justifing the time component of the four vector really is energy with either light and matter.