Four vectors

Four position§

Worldline§

We have been using the word "world line", "four position", "four space" for a long time without explaning it. For anyone that wondering what it is, anything with a four should and will obey lorentz transformation. And worldline, is the trajectory traced out by 4-position, which is just the $(ct, \mathbf x)$ space. So, if you wish, replace all the word "world line" with "trajectory" in all previous articles and you will get the same meaning at least up until now. And we already see how it behave under lorentz transformation on the last article, so we will not be doing it here.

Notation§

Rather, we will be formalizing the notation we have been using. $X^\mu$ is the component of the 4-position, where when $\mu = 0$ we get $X^0 = ct$ and $X^1 = x, X^2 = y, X^3 = z$. Or, if we write the index using latin rather than greek, we only get $1,2,3$ rather than $0,1,2,3$. So $X^i$ meant $X^i=\mathbf x$. And if $X^i\equiv 0$, we call the time component here $X^0=ct = c\tau$ being the proper time, as it is the time measured from the frame of the target object itselves. And following the same nameing logic, the length measured from the frame of the target object is proper length.

Spacelike, timelike, lightlike§

We alrady know, the lorentz transformation is a hyperbolic rotation, which we also know, it have the asymptote at $|\mathbf x|=ct$. Which we call that position vector $|\mathbf x|=ct$ lightlike vector. And anything slower than light a timelike vector and anything faster than light a spacelike vector. timelike and spacelike are topologically seperated, we cannot transform a vector from timelike to spacelike or vis versa. Which, then now we know, any object that is travelling slower than light can never go faster than light, even with infinite energy.

We already seen an example of lightlike vector, which is literally light themselves. Well, what about timelike vectors and spacelike vectors. For timelike vectors, it is literally any everyday object. And for spacelike vectors, we already have seen one, which is the length vector where we set $\dd t' = 0$ with non-zero $\dd x$.

And using the invariant to detect if a vector is spacelike or timelike or lightlike ( as, using velocity is only valid for four-position ), we use it's invariant. See, we first use the four-position to build the intuition. Note that rotation is part of the lorentz group transformation, so let's rotate into a frame that $\dd y = \dd z = 0$

$$\dd s^2 = -c^2\dd t^2 + \dd x^2$$

If it is time-like, which its velocity should be less then the speed of light. We have $\dd x = v\dd t$.

$$\dd s^2 = -c^2\dd t^2 + v^2\dd t^2 = (v^2 - c^2)\dd t^2 < 0$$

We will have an imaginary proper length ( since it is not actually a length, it is a trajectory ).

If it is light-like, then we should have $\dd x = c\dd t$. Then we can see, $\dd s^2 = 0$. And for spacelike it is $\dd s^2 > 0$. And we actually, see, a length is measured by a spacelike vector. Which, from now on, just like a length is an invariant under isotropy and homogeneity of space. we will call this invariant the length of the four-vector.

Four Velocity§

Now, let's get formal, a 4-velocity, is by definition

$$U^\mu = \frac{\dd{}}{\dd{\tau}}X^\mu$$

Where again, $\tau$ is the proper time ( the clock of te moving object itselves, not ours ).

Then

$$U^\mu = (\frac{\dd{}}{\dd{\tau}}ct,\frac{\dd{}}{\dd{\tau}}\mathbf x) = \gamma_u(c,\mathbf u)$$

So, note that if $u$ is the velocity they measured in our frame. As in the rest frame itselves it will always measure $\mathbf u = 0$.

And again, we will have the invariant ( the length ) as

$$-\gamma_u^2(c^2 - u^2) = -c^2\gamma_u^2 ( 1 - \beta^2 ) = -c^2$$

We see, no matter what velocity we are on, the length of the velocity vector must be a constant $-c^2$ which, we have seen, a negative length is a timelike vector. velocity vector are generally timelike.

Four acceleration§

By defintion

$$A^\mu := \frac{\dd{}}{\dd\tau}U^\mu = \gamma_u\frac{\dd{}}{\dd t}U^\mu$$

We can actaully calculate some of it's properties without brute force expansion. Take note that the length of a four-velocity is a universal constant. We will have

$$\begin{align*} \eta_{\mu\nu}U^\mu U^\nu &= -c^2 \\ \frac{\dd{}}{\dd\tau}(\eta_{\mu\nu}U^\mu U^\nu) &= 0 \\ \eta_{\mu\nu}\frac{\dd{}}{\dd\tau}(U^\mu)U^\nu &= 0 \\ \eta_{\mu\nu}A^\mu U^\nu &= 0 \\ \end{align*}$$

This means that the acceleration vector must be forever orthogonal to the velocity vector ( a timelike vector ) making the acceleration vector generally a spacelike vector ( orthogonality is not a 90 degree, but rather a mirror image from the lightlike line ). Well, do the same for

$$\begin{align*} \frac{\dd{}}{\dd t}\mathbf u \cdot \mathbf u &= 2\mathbf u\cdot \mathbf a \\ \frac{\dd{}}{\dd t} u^2 &= 2\mathbf u\cdot \mathbf a \\ u\frac{\dd u}{\dd t} &= \mathbf u\cdot \mathbf a \\ \end{align*}$$

Then we can see from $\href{#eq-eq-derivative-of-gamma}{(1)}$,

$$\frac{\dd{}}{\dd t}\gamma = \gamma_u^3\frac{1}{c^2}\mathbf u \cdot \mathbf a$$

Then from the definition

$$\begin{align*} A^\mu &= \gamma_u\left[(c, \mathbf u)\frac{\dd{}}{\dd t}\gamma_u + \gamma_u\frac{\dd{}}{\dd t}(c, \mathbf u )\right] \\ &= \gamma_u\left[(1, \boldsymbol \beta) \gamma_u^3\boldsymbol \beta \cdot \mathbf a + \gamma_u(0, \mathbf a )\right] \\ &= \gamma_u^4\left[(\boldsymbol \beta \cdot \mathbf a, (\boldsymbol \beta \cdot \mathbf a)\boldsymbol \beta + \frac{1}{\gamma_u^2}\mathbf a)\right] \\ \end{align*}$$

We can double check all properties ( space like, and orthogonality ) from this general form, but we didn't need to. Instead, we will calculate the proper accelaration.

$$\begin{align*} A^2 &= \gamma_u^8\left[-(\boldsymbol \beta \cdot \mathbf a)^2 + (\boldsymbol \beta \cdot \mathbf a)^2\boldsymbol\beta^2 + \frac{1}{\gamma_u^4}\mathbf a^2 + \frac{2}{\gamma_u^2}(\boldsymbol \beta \cdot \mathbf a)^2\right] \\ &= \gamma_u^8\left[-\frac{1}{\gamma_u^2}(\boldsymbol \beta \cdot \mathbf a)^2 + \frac{1}{\gamma_u^4}\mathbf a^2 + \frac{2}{\gamma_u^2}(\boldsymbol \beta \cdot \mathbf a)^2\right] \\ &= \gamma_u^6\left[\frac{1}{\gamma_u^2}\mathbf a^2 + (\boldsymbol \beta \cdot \mathbf a)^2\right] \\ \end{align*}$$

Let's be in the instaneteous rest frame, we will be $\beta = 0$ then $\gamma_u = 1$ we get

$$A^2 = a^2$$

So, the proper accelaration is the accelaration in the rest frame itselves. Well, from now on, we will stop explicitly stating out what's the meaning of that invariant. Because, it is already clear that whenever we call something invariant, we can also use the rest frame and call it the physical meaning. So

Invariant is the measurement of that quantity in the rest frame.

Four momentum§

$$P^\mu := m_0U^\mu$$

Then

$$P^\mu = \gamma_u (m_0c, \mathbf p)$$

And thus, our momentum must also be carefully defined to be $p=\gamma mu$ ( since momentum is the measurement of movement, and we can never exceed speed of light, therefore the closer we are to the speed of light we should actually have more movement ). We get

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

And then calculating its invariant, we have

$$\begin{align*} P^2 &= \gamma_u^2(p^2 - m_0^2c^2) \\ &= -m_0^2c^2\gamma_u^2(1 - \beta^2) \\ &= -m_0^2c^2 \\ \end{align*}$$

We, see, momentum is as well a timelike constant vector just like velocity

Four Force§

By relativity of simulteneity, and also the limit of the speed of light. A force actually looses meaning. Therefore, although we are able to define a four-force, it is actaully not a useful concept anymore. Specifically, the framework of forces in newtonian mechanics actually makes force causaily instetanous, like the force of gravity acts instantly without needing any information to travel, which has infinite velocity. And also, newton's third law state that the force and reaction force happens simultaneously, where simulteneity is already relative under lorentz transformation.

But, anyhow, this is the definition

$$F^\mu = :\frac{\dd{}}{\dd\tau}P^\mu$$

Well, by the same trick from accelaration ( since we are assuming mass doesn't change w.r.t time ).

$$\begin{align*} \eta_{\mu\nu}P^\mu P^\nu &= -m_0^2c^2 \\ \eta_{\mu\nu}(\frac{\dd{}}{\dd\tau} P^\nu)P^\mu &= 0 \\ \eta_{\mu\nu}F^\nu P^\mu &= 0 \\ \end{align*}$$

$F$ and $P$ are orthogonal. So, force is a spacelike vector. We will get

$$\begin{align*} F^\mu &= \gamma_u \left[(m_0c, m_0\mathbf u)\frac{\dd{}}{\dd t}\gamma_u + \gamma_u\frac{\dd{}}{\dd t}(m_0c, m_0\mathbf u)\right] \\ &= \gamma_u \left[(m_0c, m_0\mathbf u)\frac{\gamma_u^3}{c}\boldsymbol \beta\cdot \mathbf a + \gamma_u(0, \mathbf f)\right] \\ &= \gamma_u^4 \left[((\boldsymbol \beta\cdot \mathbf a)m_0, (\boldsymbol \beta\cdot \mathbf a)m_0\boldsymbol\beta + \frac{1}{\gamma_u^2}\mathbf f)\right] \\ \end{align*}$$

Where now we should redefine $f = \gamma m_0 a$

$$F^\mu = \gamma^3\left[((\boldsymbol\beta\cdot\mathbf f), (\boldsymbol\beta\cdot\mathbf f)\boldsymbol\beta + \frac{1}{\gamma_u^2}\mathbf f)\right]$$

Which, in the rest frame $F^2=f^2$.

Current Density§

$$J^\mu := \rho_0 U^\mu$$

With the proper density $\rho_0$ We then will get

$$J^\mu = \gamma_u(\rho_0 c, \mathbf J)$$

Where, by length contraction we already know exist we should have the lab density $\rho = \gamma_u\rho_0$. And $j = \gamma \rho_0 u$, We get

$$J^\mu = (\rho c, \mathbf J)$$

Vector potential§

Recall, in Lorentz transformation, we see that the D'lambert operator is lorentz invariant, and we concluded therefore maxwell equation must be lorentz covariant. The logic is loose and now is the time to tightnen it up. See that the source term

$$\square \phi = -\frac{\rho}{\epsilon_0} \qquad \square \mathbf A = -\mu_0\mathbf J$$

• We already seen that $(\rho c, \mathbf J)$ is a 4-vector which already means they are lorentz covariant
• We already known that by definition $\square$ is Lorentz invariant
• We demand principle of relativity ( physics law should not depend on reference frame )
• We fix Lorenz Gauge

Then, therefore we can conclude

$$A^\mu = (\frac{\phi}{c}, \mathbf A)$$

Will be a four-vector and

$$\square A^\mu = (\square \frac{\phi}{c}, \square \mathbf A) = -\mu_0(\rho c, \mathbf J) = -\mu_0J^\mu$$

And therefore, again maxwell equation will be and must be lorentz covariant.

Field Strength Tensor§

Now, let's find a tensor that can help us recover the maxwell equations. Which it is more like a tool rather than a mandatory object. Without it, at most recovering maxwell equation is frusrating but doable.

Let's take note on maxwell equations structure where it is full of differential operators. Thus, from what we have here, and from the condition we said above that makes $A$ a four vector. We see, in index notation that we have. $\eta_{\mu\nu}\partial^\mu\partial^\nu = \partial_\mu\partial^\mu$ is the dlambert operator. Where be careful $\partial_0 = \dfrac{\partial}{\partial ct}$

$$\partial_\mu\partial^\mu A^\nu = -\mu_0J^\nu$$

With the lorentz gauge condition $\partial_\mu A^\mu = 0$. Then we can create

$$\partial^\nu\partial_\mu A^\mu = 0$$

Then, subtract them, because we want an actual tensor that is gauge invariant or invariant over gauge transformation ( $A \to A + \grad \chi$ ) .

$$\partial_\mu\partial^\mu A^\nu - \partial^\nu\partial_\mu A^\mu = -\mu_0J^\nu$$

Then swapping the order of differential, we will end up with the curl

$$\begin{align*} \partial_\mu\partial^\mu A^\nu - \partial_\mu\partial^\nu A^\mu &= -\mu_0J^\nu \\ \partial_\mu(\partial^\mu A^\nu - \partial^\nu A^\mu) &= -\mu_0J^\nu \\ \end{align*}$$

See whats inside, we defined it as $F^{\mu\nu}$

$$\partial_\mu F^{\mu\nu} = -\mu_0 J^\nu$$

This $F$ will be a gauge invariant term which since it is trivial we will not check it explicitly. But, since it is gauge invariant, just like $B$ and $E$. We should suspect that $F$ have something to do with $E$ and $B$, and historically speaking, this is what motivate physist to find $F$, a tensor that represent $E$ and $B$ after they realise that $E$ and $B$ arent 4-vectors. We will not expand it here to find its element, ( which will directly be $E$ and $B$ ), rather we leaves it to later articles.

4-wave-vector§

Remember what a wave vector is: It is "a vector that should do a measurement" to find a $2\pi$ phase on a wave. So we call $k^\mu$, that will measure a phase of $2\pi$ on the creast of the wave ( seperated in $\Delta x^\mu$ ) And we know the phase of a wave should be $\mathbf k\cdot\Delta \mathbf x - \omega \Delta t$

$$\eta_{\mu\nu}k^\mu \Delta x^\nu = k^i\Delta x^i - \omega \Delta t = 2\pi$$

Which means $k^0 = \omega/c$ since $x^0 = ct$.

Therefore, then making it a vector, we have

$$k^\mu = (\omega/c, \mathbf k)$$