Parallel axes between inertial frames in Special Relativity

Question

In "CLASSICAL ELECTRODYNAMICS" by J.D.Jackson, 3rd Edition , $\S$ 11.3, the author gives in equation (11.19) a generalization of Lorentz transformation as follows :

If the axes in K and K' remain parallel, but the velocity $\:\mathbf{v}\:$ of the frame K' in frame K is in an arbitrary direction, the generalization of (11.16) is

$$ \begin{align} x'_{0} & =\gamma\left(x_{0}-\boldsymbol{\beta}\boldsymbol{\cdot}\mathbf{x}\right)\\ \mathbf{x}^{\prime} & = \mathbf{x} +\dfrac{\left(\gamma-1\right)}{\beta^{2}}\left(\boldsymbol{\beta}\boldsymbol{\cdot}\mathbf{x} \right)\boldsymbol{\beta}-\gamma\boldsymbol{\beta}x_{0} \end{align} \Biggr\} \tag{11.19} $$ where $$ \begin{align} \boldsymbol{\beta} & = \dfrac{\mathbf{v}}{c}\; \qquad \beta=|\boldsymbol{\beta}| \\ \gamma &=\left(1-\beta^2 \right)^{-1/2} \end{align} \tag{11.17} $$

and $$ \begin{align} x'_{0} & =\gamma\left(x_{0}-\beta x_{1}\right)\\ x'_{1} & =\gamma\left(x_{1}-\beta x_{0}\right)\\ x'_{2} & =x_{2}\\ x'_{3} & =x_{3} \end{align} \Biggr\} \tag{11.16} $$ the Lorentz Transformation with the velocity $\:\mathbf{v}\:$ parallel to the common $\:x-x'\:$ axis.

In case (11-16) it's permissible to talk about parallel axes. But in the generalized case (11-19) is it permissible to talk about parallel axes ? What is the meaning of parallelism in this later case ?

Answer 1

In general, Lorentz (or rather Poincare) transformation are those transformations that keep the speed of light in any reference frame the same. They can be decomposed into the following:

1.) Translations
2.) Rotations and
3.) Boosts

Translations and Rotations are defined just like in the Galileo case and don't make up the "interesting" physics of special relativity. Therefore, one often only really talks about the boosts when talking about Lorentz transformations. These boosts can be done in some coordinate direction, most easily in the direction of an coordinate axis, for example the $x$-axes. The corresponding transformation is written in your equation (11.16). Now Jackson talks about a generalization, by that he means here to write down the transformation for a boost in $\overrightarrow{v}$-direction which he does in (11.19).
In principle it's the same kind of transformation. So in the same fashion as for the boost in $x_1$ direction, the spatial coordinate axes are parallel, i.e. the transformation does not include any rotation. If you were to shoot an arrow in $\overrightarrow{x}_i$-direction, $i=1,2,3$, the observer in the reference frame with primed coordinates would also observe the arrow going in $\overrightarrow{x}'_i$-direction.
The parallelity refers to the spatial coordinates only.

Answer 2

See: wikipedia:....

$\mathbf{r}=\mathbf {r} _{\perp }+ \mathbf {r} _{\parallel }\;\;, \;\mathbf{r'}=\mathbf {r'} _{\perp }+ \mathbf {r'} _{\parallel }$

then the transformations are:$$\begin{cases} t'=\gamma(t-\frac{\mathbf {r}_{\parallel }.\mathbf{v} }{c^{2}} )\\\mathbf {r'}_{\parallel }=\gamma(\mathbf {r}_{\parallel }-\mathbf{v}t)\\\mathbf{r'}_{\perp } =\mathbf {r} _{\perp } \end{cases}$$

....

Introducing a unit vector $\mathbf{n}=\mathbf{v}/v=\boldsymbol{\beta}/ \beta \;$ in the direction of relative motion, the relative velocity is $\mathbf{v} = v\mathbf{n}$ with magnitude $v$ and direction $\mathbf{n}$, and vector projection and rejection give respectively $${\displaystyle \mathbf {r} _{\parallel }=(\mathbf {r} \cdot \mathbf {n} )\mathbf {n} \,,\quad \mathbf {r} _{\perp }=\mathbf {r} -(\mathbf {r} \cdot \mathbf {n} )\mathbf {n} }\;\;\;\;$$(*)

Accumulating the results gives the full transformations,

$$\begin{align} t' & =\gamma\left(t-\frac{v\boldsymbol{n}\boldsymbol{\cdot}\mathbf{r}}{c^{2}}\right)\\ \mathbf{r}^{\prime} & = \mathbf{r} +\left(\gamma-1\right)\left(\boldsymbol{r}\boldsymbol{\cdot}\mathbf{n} \right)\boldsymbol{n}-\gamma vt\boldsymbol{n} \;\;\end{align} \Biggr\} $$ with $x_{0}=ct\;,\;x'_{0}=ct'\;,\;\mathbf{r}\equiv \mathbf{x}$

$$\begin{align} x'_{0} & =\gamma\left(x_{0}-\boldsymbol{\beta}\boldsymbol{\cdot}\mathbf{x}\right)\\ \mathbf{x}^{\prime} & = \mathbf{x} +\dfrac{\left(\gamma-1\right)}{\beta^{2}}\left(\mathbf{x}\boldsymbol{\cdot}\boldsymbol{\beta} \right)\boldsymbol{\beta}-\gamma\boldsymbol{\beta}x_{0} \end{align} \Biggr\} \tag{11.19}$$

Answer 3

A Lorentz boost is just linear acceleration in a spatial direction.

An amazing fact that a lot of folks don't know: all linear acceleration in N dimensions is rotation in N + 1 dimensions.

Think of a candle in a dark room sitting on the edge of a rotating turntable. Look at it from the side. The candle appears to be accelerating, and then slowing down and reversing direction. This is in one dimension.

But seen from above, in 2 dimensions, a more complete view reveals simple rotation.

Linear acceleration in N dimensions is rotation in N + 1 -- and that's true in any number of dimensions.

Exercise for the student: what is an example in two and three dimensions?

So what does that have to do with the question?

Well, you can accelerate an object in the X direction, and its path is parallel and colinear to its previous path. This is accurate in 3 dimensions, but incomplete.

Seen as it really is, in 4 dimensions, the two world lines (4-momentum vectors) are NOT parallel.

The object that is moving faster is pointed slightly away from the direction of future time. Another name for this is time dilation. If the object accelerates all the way to c, its momentum vector would rotate 90 degrees, and it wouldn't be moving in the time direction at all.

The situation is symmetric, because the coordinate system is arbitrary. Each object will consider the other one as having veered off the "straight and narrow" path in which time passes at the normal rate -- i.e., the path that takes the least amount of proper time.

You're welcome! [curtsies]

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WRETCHED GEEKS ONLY:

If you understand that, but just barely, skip this part. But if you want to expend a little effort, there's a way to look at it that makes it much easier to see what's going on.

The reason this all seems so strange is that the universe is not euclidean. Momentum through time and space is contravariant. That means that when you increase your speed through one, you necessarily, always move slower through the other. This is a consequence of the fact that distance in one of the directions you can move (i.e, time) is negative relative to the other 3.

Space like this is called "pseudometric," and Einstein said that spacetime is like that.

This is his equation that defines the absolute (Lorentz-invariant 4-dimensional) distance between two events. Note that:

===[ Elapsed time is SUBTRACTED from the spatial distance between the events.

An easy way to look at it is to think of yourself as driving your car along the equator of a globe. Everyone moving at the same speed as you is also moving along the equator. The passage of time is motion in that direction.

People moving faster than you are also moving parallel to the equator, but they are on a parallel path slightly north of yours.

As they drive past you, you peek through their window and see that their clock is running slow.

If they suddenly accelerate to c, you would see them veer straight north. The North Pole is reachable at the speed of light.

Here's the thing:

When you accelerate, you're taking an exit on the highway that's just slightly more northward. But after you finish accelerating, you look out the window and you're amazed to see that you're still driving along the equator and the North Pole is still 90 degrees away from you.

This is actually what's really going on. The globe is a Riemann sphere, the North Pole represents infinite momentum, and the South Pole represents an object with zero mass-energy and zero momentum. Such an object would, by definition, not exist.

The Lorentz Transformation is just a map between Euclidean space and the surface of the Riemann sphere.

...That probably just makes it more confusing.

Oh well.