0
$\begingroup$

Consider,again (Question about derivation of four-velocity vector) the following:

For a massive particle with position $x^{\mu}(t) = (x^{0},x^{1},x^{2},x^{3}) \equiv (x^{0},\vec{x})$ we define the coodinate velocity as:$$ v^{\mu} := \frac{dx^{\mu}}{dt} \equiv (c,\vec{v})\tag{1}$$ Where the spatial components of $(1)$ coincide with classical velocity vector and t is the coordinate time.

But, $(1)$ is not a vector object indeed, because the components didn't transforms as vectors under a lorentz transformation:

$$\frac{dx'^{\mu}}{dt'} = \Lambda^{\mu'}_{\nu}\frac{dx^{\nu}}{dt} \frac{dt}{dt'} = \frac{\Lambda^{\mu'}_{\nu}}{\Lambda^{0'}_{\nu}x^{\nu}}\frac{dx^{\nu}}{dt} \neq \Lambda^{\mu'}_{\nu}\frac{dx^{\nu}}{dt}\tag{2}$$

So, I'm still in doubt about $(2)$. We know that $(1,0)-$tensors in Minkowski space transforms like:

$$ A'^{\mu} = \Lambda^{\mu'}_{\nu} A^{\nu}$$

where $\Lambda^{\mu'}_{\nu}$ is the boost matrix:

$$\Lambda^{\mu'}_{\nu} = \begin{bmatrix} \gamma & -v\gamma & 0 & 0 \\ -v\gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} $$

So, firstly, Einstein have derived the kinematics of Relativity in terms of 3D Euclidean geometry and with the postulates of special relativity, showing the very fundamental concepts of physics near to speed of light (e.g. time dilation). Then, Minkowski introduced the spacetime and all the whole of a invariant interval, i.e. the metric formalism of special relativity. Furthermore, with Minkowski's formalism a new type of object was needed to ensamble the kinematical quantities into spacetime formalism. One of these quantities was the 4-velocity vector, which required a concept of the 4-vector. But even if the one defines a true 4-vector, this vector must transform properly under a lorentz transformation; that's the case of four-velocity.

So, consider ,then, an attempt to 4-four velocity, defined as:

$$v^{\mu} := (\frac{dx^{0}}{dt},\frac{dx^{1}}{dt},\frac{dx^{2}}{dt},\frac{dx^{3}}{dt}) \tag{3}$$

Where $t$, is the coodinate time in $S$ reference frame [with $x^{\mu} = (ct,x(t),y(t),z(t))$ coordintes]. Then, if we want a true Minkowski we must to check the tensor character:

$$\frac{dx'^{\mu}}{dt'} = \Lambda^{\mu'}_{\nu}\frac{dx^{\nu}}{dt} \tag{4}$$

As you might expect this isn't right (and the true behaviour is by introducing the proper time). And that is the point, I do not understand this, what is $t'$? Propertime or just coodinate time in $S'$?Well, actually I'm lost in what really is happening in $(2)$, why we use the chain rule, what is the dependence of the times [ I mean if is $t(t')$ or $t'(t)$]and so on.

$\endgroup$
  • 1
    $\begingroup$ I find it useful to think of four-velocity as the derivative of the coordinate positions with respect to the arc length on the world line. "You're not thinking four-dimensionally" -Doc Brown $\endgroup$ – R. Rankin Jan 8 at 0:17
2
$\begingroup$

what is $t′$, Propertime or just coodinate time in $S′$?

Answer: It is the latter. It is the time as measured in frame $S'$. However, it can be made proper time, depending on the event. By definition, the proper time is the time measured by an observer comoving with the event being described.

So if the $event$ under examination is something far away from both of them, say, some space ship flying by and both $S$ and $S'$ measure how long it takes, then neither $t$ nor $t'$ would be the proper time. The proper time in this case $\Delta \tau$ would be the time it took for somebody on the space ship to fly by.

why we use the chain rule, what is the dependence of the time?

For two observers in standard configuration, their coordinates are related by a standard Lorentz transformation. That is, they are related by

$$ {\begin{aligned}t'&=\gamma \left(t-{\frac {vx}{c^{2}}}\right)\\x'&=\gamma \left(x-vt\right)\\y'&=y\\z'&=z\end{aligned}} $$

This can be written in matrix form as

$$ \begin{bmatrix} ct' \\ x' \\ y'\\ z' \end{bmatrix} = \underbrace{ \begin{bmatrix}\gamma &-\gamma \beta &0&0\\-\gamma \beta &\gamma &0&0\\0&0&1&0\\0&0&0&1\\\end{bmatrix} }_{\Lambda^\mu_{\ \ \nu}} \begin{bmatrix} ct \\ x \\ y\\ z \end{bmatrix} $$

The matrix form is not necessary to address your main concern. The dependence you are inquiring about is exactly

$$t'=\gamma \left(t-{\frac {vx}{c^{2}}}\right) $$

so unless we take $t$ do be the proper time we see that $\frac{dt'}{dt}$ is somewhat complicated.

However it is always the case that

$$ t' = \gamma \tau \implies dt' = \gamma d\tau $$

when the time $dt'$ refers to the time passed on the space ship as observed in $S'$. Note that this is just the usual formula for time dilation. This is one reason why it's beneficial to work with proper time instead.

$\endgroup$
  • $\begingroup$ I just saw your second question.. I'll add that to my answer now $\endgroup$ – InertialObserver Jan 7 at 21:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.