A simple boost in the $x$ direction is given by: $$ \Lambda = \begin{pmatrix} \cosh(\rho) & \sinh(\rho) & 0 & 0 \\ \sinh(\rho) & \cosh(\rho) & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix} $$

Which get linearized to the following transformation: $$ x^0 \mapsto x^0 , \quad x^1 \mapsto x^1 + \frac vc x^0 $$

How come the zeroth component is not linearized to $x^0 \mapsto x^0 + \frac vc x^1$? Is that because there is another factor $c$ in the time components? Since $x^0 = ct$, that would mean the time is transformed like $$ t \mapsto t + \frac v{c^2} x,$$ and $c^{-2}$ is just so small that is ignored?

Or is it just to fit the Galilei transformation?

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    $\begingroup$ You can only linearise the expression when $v/c$ is small. That implies $v/c^2$ is very small. $\endgroup$ Jun 22, 2014 at 20:43
  • $\begingroup$ I see, that explains it. $\endgroup$ Jun 22, 2014 at 21:01
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    $\begingroup$ Unless I am missing something how you define your coordinates, there is an error in your post - the space transformations should be $ x^1 \mapsto x^1 + v x^0 $ or $ x^1 \mapsto x^1 + v t $. Otherwise it is dimensionally incorrect. $\endgroup$
    – DJBunk
    Jun 23, 2014 at 16:10
  • $\begingroup$ @SimonLyons - $v/c$ and $v/c^2$ have different dimensions, there is no meaning to comparing them. $\endgroup$
    – DJBunk
    Jun 23, 2014 at 16:13
  • $\begingroup$ That's true, and I agree there's probably a mistake in his post. I stand by what I said though - If $v/c$ is of order $\epsilon$, whatever pops out of his calculation will be of order $\epsilon^2$. $\endgroup$ Jun 24, 2014 at 16:55

2 Answers 2


The Lorentz transformations : \begin{pmatrix} \cosh(\rho) & \sinh(\rho) \\ \sinh(\rho) & \cosh(\rho) \\ \end{pmatrix} form a group.

The galilean transformations : \begin{pmatrix} 1 & 0 \\ v/c & 1 \\ \end{pmatrix} form a group

But the transformations :

\begin{pmatrix} 1 & v/c \\ v/c & 1 \\ \end{pmatrix} do not form a group.

If you restrict your transformations to a group structure, which is the simplest hypothesis, you cannot keep the last example of transformations.


The Lorentz boost has two different low-velocity limits: the Galilean transformation appropriate for transforming ultra-timelike four-vectors, which is usually what we're interested in if we want to recover low-velocity kinematics, and the whimsically named "Carroll transformation" appropriate for transforming ultra-spacelike four-vectors. Your proposed time transformation is part of the latter.

Intuitively, the effect of the Lorentz transformation is a rotation along curves of constant spacetime intervals:

Spacetime wheel Image from here.

If we zoom in on the upper vertices of the hyperbola, the low-velocity Lorentz boosts of timelike vectors with $ct\gg|\mathbf{x}|$ are approximated by: $$\begin{eqnarray*}ct' \approx ct\text{,}&\quad&\mathbf{x}' \approx \mathbf{x} - \mathbf{v}t\text{.}\end{eqnarray*}$$ Treating this low-velocity approximation as a transformation in its own right, we get the Galilean boost: $$\begin{eqnarray*}ct' = ct\text{,}&\quad&\mathbf{x}' = \mathbf{x} - \mathbf{v}t\text{.}\end{eqnarray*}$$ This is sensible, because the tangent lines to the hyperbolas on ultra-timelike vectors are horizontal, so a low-velocity boost should not change the time coordinate by an appreciable amount.

If instead we zoom in on the right vertices, and repeat the above procedure on ultra-spacelike ($ct\ll|\mathbf{x}|$) vectors, we get the Carroll transformation: $$\begin{eqnarray*}ct' = ct - (\mathbf{v}\cdot\mathbf{x})/c^2\text{,}&\quad&\mathbf{x}' = \mathbf{x}\text{.}\end{eqnarray*}$$ Note that this reflects the fact that a low-velocity boost of an ultra-spacelike vector does not change the spatial coordinates appreciably, as the tangent lines to the hyperbolas are vertical there.

On a purely formal level, they are equally valid low-velocity limits. The Galilean transformation is much more physically significant because material objects should follow timelike vectors, not spacelike ones, so we can interpret it as a possible transformation between inertial frames formed by ideal clocks and rulers (or some other means). In contrast, the Carroll transformation does not allow a sensible interpretation as a transformation between physical inertial frames.

“My dear, here we must run as fast as we can, just to stay in place. And if you wish to go anywhere you must run twice as fast as that.” ― Lewis Carroll, Alice in Wonderland

  • $\begingroup$ "If we zoom in on the upper vertices of the hyperbola, we get the 'ultra-timelike' limit that recovers the Galilean boost: $ct'=ct, x'=x−vt$". The first equation boils down to $x'=x$ (if $c$ is still constant in all frames), which contradicts the second equation, and also it is exactly what you put here: "If instead we zoom in on the right vertices, we get the 'ultra-spacelike' limit that is the Carroll transformation: $ct′=ct−(v⋅x)/c^2, x′=x$". Also, if $x'=x$ in the second case, then $v=0$, and therefore $ct′=ct−(v⋅x)/c^2$ boils down to $ct′=ct$, i.e. first case. Or I am missing something? $\endgroup$ Jun 23, 2014 at 11:49
  • $\begingroup$ @brightmagus: In spacetime, Lorentz boosts go along hyperbolas, while Galilean boosts go along (constant-time) lines. If we consider the latter to be the tangent-line approximation of the former at low speed ('zooming in' flattens the hyperbola), we can also do the same thing at the other branches. This gets us an unphysical but mathematically valid transformation for time that the OP has. In their respective regimes, a low-v Lorentz boost is approximated by the G/C transformations, but they aren't supposed to keep $c$ invariant. $\endgroup$
    – Stan Liou
    Jun 23, 2014 at 13:15
  • $\begingroup$ "This gets us an unphysical but mathematically valid transformation for time .... they aren't supposed to keep c invariant." But I just showed it is not valid mathematically. Perhaps because you claim that speed of light is not invariant, and yet you still call it c in both cases, which makes it invariant. For instance: $ct′=ct$ is it the same $c$ on both sides of the equation? And is $c=x/t$ and $c=x'/t'$ here? If not, then how is it defined? $\endgroup$ Jun 23, 2014 at 14:07
  • $\begingroup$ @brightmagus: it's mathematically valid, and even a group. $x'=x$ does not imply $v = 0$ because we're not dealing with Lorentz trns. anymore. The CT is just an approximation to low-v Lorentz boosts for highly spacelike vectors, which we can then take on its own right (just as we can take a tangent-line approximation to any differentiable function and then consider the line itself). The CT cannot serve as a transformation between inertial frames, and it won't have sensible causal relations if we pretend that it does. But I've called it unphysical already, so what's the problem? $\endgroup$
    – Stan Liou
    Jun 23, 2014 at 14:37
  • $\begingroup$ What is $c$ in your equations? What does it mean? How is it defined? Does $ct$ equal $x$, and does $ct'$ equal $x'$? If not, then what does it equal to? $\endgroup$ Jun 23, 2014 at 14:44

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