Physical meaning of spatial part of Lorentz transform is, obviously, rotations in R-space. Does anybody have a nice physical interpretation of boosts though? I can understand to some extent off-diagonal terms, but what do diagonal ones represent? And, yes, I can apply the transform matrix to a vector and say that whatever comes out is expected. I am wondering if there is a more elegant interpretation than just stating that Lorentz transform behaves like Lorentz transform.

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    $\begingroup$ What, beyond, 'a change in velocity' do you expect? $\endgroup$ Jun 16, 2016 at 23:47
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    $\begingroup$ I do not like alienating time component and try to avoid it as much as possible. I was trying to think about that as a weird analogue of rotation. $\endgroup$
    – MsTais
    Jun 16, 2016 at 23:54
  • $\begingroup$ If it looks like a rotation it looks like a rotation... even though it is not a rotation. Is there a deeper meaning to that? I don't think so. $\endgroup$
    – CuriousOne
    Jun 16, 2016 at 23:57
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    $\begingroup$ Not a physical interpretation, but there is, for me at least, a very strong physical justification. It has to be something like a rotation, but it can't be exactly like a rotation because that would violate causality. So it's a signatured rotation. See my answer. $\endgroup$ Jun 17, 2016 at 8:22

1 Answer 1


A boost is "the only other" plausible choice for transformations that mix spacetime co-ordinates, aside from rotations. The two notions in this sense are complementary. See for example:

Palash B. Pal, "Nothing but Relativity", Eur. J. Phys. 24, pp315-319,2003

Given certain "reasonable" assumptions about our Universe; roughly

  1. The first relativity postulate; together with
  2. Assumptions of homogeneity of space and time and
  3. Symmetry of the relationships between two inertial observers

one deduces linearity of the co-ordinate transformations between them.

Now we make the assumption that:

  1. The transformation should be a continuous function of the velocity parameter. There should be a smooth path, parameterized by a velocity / rapidity joining each boost to the identity transformation.

So our transformation must be of the form (this isn't immediately obvious, you need to do some work to prove this): $X\mapsto \exp(\eta\,K)\,X$, where $X$ is our spacetime co-ordinate column vector, $K$ a $2\times2$ constant matrix and $\eta$ our velocity / rapidity parameter that encodes the smoothness of the path through the matrix exponential $\exp$.

If you then assume, further to the above, that time is the same for all observers, then your transformation is uniquely defined: you've got Galilean relativity.

If you then relax the assumption of universal time and then say that time itself is an observer-dependent co-ordinate, then there are more possibilities.

Basically, modulo co-ordinate scalings, there end up being two possibilities consistent with the three assumptions above: rotations and boosts; the former corresponding choosing our normalized $K = \left(\begin{array}{cc}0&-1\\1&0\end{array}\right)$ for rotations, and $K = \left(\begin{array}{cc}0&+1\\1&0\end{array}\right)$ for the other possibility. Rotations of course manifest themselves in transformations that mix the spatial co-ordinates alone.

So could rotations mix space and time? Well, not if we demand causality stays true. You can always find a rotation that would invert the direction of the time difference between two events.

The only other possibility is a boost where we replace our $K$ matrix with $K = \left(\begin{array}{cc}0&+1\\1&0\end{array}\right)$. This has real eigenvalues as opposed to the purely imaginary ones for the rotation $K$ above. It also has the crucial property that, as long as the speed parameter does not exceed $c$, the sign of the time component stays the same. Causality is upheld. Indeed, you can take this as the reason why we say there can be no faster than light travel: we're forced to this postulate if we demand the orders of irreversible physical processes to be observer independent. Also, no transformation with greater than light speed is connected to the identity by the path $X(\eta) = \exp(\eta\,K) X$. No finite sequence of finite boosts will get you to a relative speed of $c$.

So, in summary, when looked at in the above light, I believe the Lorentz transformation is the most natural thing in the world. It's one of a small number of possibilities that are all like rotations. When it comes to the mixing of space and time co-ordinates as opposed to the mixing of space co-ordinates, the transformation can't be exactly like a rotation, because we must have a mechanism to uphold causality. This mechanism is the Lorentz transformation's nontrivial signature.

More Material and Background

Demonstration of Linearity of the Transformations

Assumptions (1), (2) and (3) show that the coordinate transformation must be affine: JoshPhysics does this here (his answer to Physics SE "Homogeneity of space implies linearity of Lorentz transformations") and Mark H does it beautifully here (his answer to "Why do we write the lengths in the following way? Question about Lorentz transformation").

We can then choose a spacetime translation to make the transformation linear and homogeneous.

Proof that the One-Parameter Lie Group Follows from the Continuity / Monotonicity Postulate (4)

This is a very first baby step in the greatly more difficult solution of Hilbert's famous fifth problem. We first begin with a more precise statement of our group operation continuity and monotonicity postulate suitable for mathematical inferences.

Continuity and Monotonicity of Group Composition Postulate (Version 2): The matrix group of transformations between inertial frames whose relative motion is defined by a velocity in one direction characterizing the group is the continuous image of the real line (the line of generalized velocities) i.e. $\sigma:\mathbb{R}\to\mathcal{M}(N,\,\mathbb{R})$ is the whole group and defines a continuous ($C^0$) path through the set $\mathcal{M}(N,\,\mathbb{R})$ of $N\times N$ real matrices. Furthermore:

  1. We define $\sigma(0)=\mathrm{id}$ and $\sigma(\eta)$ for $\eta>0$ always defines relative motion in the same direction whilst $\eta<0$ always defines motion in the opposite direction ($\eta$ has the same sign as the relative velocity along the chosen direction);
  2. The group operations are continuous so that the function $p:\mathbb{R}\times\mathbb{R}\to\mathbb{R}$ defined by $\sigma(\eta_1)\,\sigma(\eta_2) \stackrel{def}{=}\sigma(p(\eta_1,\,\eta_2))$ is a continuous function of $\eta_1$ and $\eta_2$ and the function $\iota:\mathbb{R}\to\mathbb{R}$ defined by $\sigma(\eta)^{-1} = \sigma(\iota(\eta))$ is also a continuous function of $\eta$;
  3. The mapping $\sigma$ is monotonic, in the sense that if $\sigma(\eta_1)\,\sigma(\eta_2)=\sigma(p(\eta_1,\,\eta_2)=\eta_3)$ and if $\eta_1,\,\eta_2>0$, then $\eta_3>\eta_1$ and $\eta_3>\eta_2$. Likewise, if $\eta_1,\,\eta_2<0$, then $\eta_3<\eta_1$ and $\eta_3<\eta_2$. Intuitively: the composition of two relative motions in the same direction is a ``swifter'' relative motion in the same direction.\newline

We now work a "trick" that Henry Briggs used to calculate his 1624 Arithmetica Logarithmica tables of logarithms with (See Feynman Lectures, Vol 1. Ch 22) and which was broadened to closed matrix groups by von Neumann in 1929. We rescale the argument of the function $\sigma:\mathbb{R}\to\mathcal{M}(N,\,\mathbb{R})$ so as to choose a "unit" relative motion to be slow enough that the all the transformation matrices in the set $\{\sigma(\eta)|\,\eta\in[0,\,1]\}=\sigma([0,\,1])$, i.e. the continuous path of transformation matrices in the group linking the identity $\sigma(0)=\mathrm{id}$ and $\sigma(1)$, all fulfill the bound $\|\sigma(\eta) - \mathrm{id}\| < 1;\,\eta\in[0,\,1]$. Such a "unit" motion exists by dint of continuity of $\sigma$. The significance of this bound is that, when it is fulfilled, the matrix logarithm Taylor series converges, so that every $\sigma(\eta)$ in the path segment $\sigma([0,\,1])$ has a logarithm defined by $\log(\sigma(\eta)) = K = (\sigma(\eta) - \mathrm{id}) -\frac{1}{2} (\sigma(\eta) - \mathrm{id})^2 + \frac{1}{3}(\sigma(\eta) - \mathrm{id})^3 - \cdots$. Now we consider the function $\mathrm{sqr}:\mathbb{R}\to\mathbb{R}$ defined by $\sigma(\eta)\,\sigma(\eta)=\sigma(\mathrm{sqr}(\eta))$. By the group composition continuity postulate, this is a continuous function of a real variable, with $\mathrm{sqr}(0)=0$ and, by the monotonicity axiom in that postulate we see that $\mathrm{sqr}(1)>1$. By the intermediate value theorem, therefore, there is a $\eta_{\frac{1}{2}}\in[0,\,1]$ such that $\sigma(\eta_{\frac{1}{2}})\,\sigma(\eta_{\frac{1}{2}}) = \sigma(1)$. That is, $\sigma(1)$ has a square root in the path segment $\sigma([0,\,1])$. But now, by dint of the convergence of the logarithm series, every matrix within the ball defined by $\mathcal{V}=\{\gamma|\,\|\gamma-\mathrm{id}\|<1\}$ has a unique square root inside that ball (Although it may very well have other square roots outside the ball) defined by $\sqrt{\sigma(\eta)} = \exp\left(\frac{1}{2}\log(\sigma(\eta))\right)$ because the logarithm is defined and maps the ball $\mathcal{V}$ into a neighborhood $\mathcal{U}=\log(\mathcal{V})=\{K|\,\exp(K)\in\mathcal{V}\}$ and both the matrix exponential, defined by the universally convergent matrix exponential matrix Taylor series and logarithm are bijective maps between $\mathcal{U}$ and $\mathcal{V}$. Therefore, if there were two square roots $\varsigma_1,\,\varsigma_2$ inside the ball $\mathcal{V}$, then both have logarithms so their squares are $\sigma(1)=\exp(2\,\log\varsigma_1) = \exp(2\,\log\varsigma_1)$. Then because $\sigma(1)$ also belongs to $\mathcal{V}$, it has a unique logarithm so that $\log\varsigma_1=\log\varsigma_2$, whence $\varsigma_1=\varsigma_2$. So we now know that the path segment contains a square root $\sigma(\eta_{\frac{1}{2}})$ of $\sigma(1)$ where $\eta_{\frac{1}{2}}\in[0,\,1]$ and that square root must be the unique square root $\exp\left(\frac{1}{2}\,K(1)\right)$, where $K(1)=\log(\sigma(1))$, since the whole path segment $\sigma([0,\,1])$ lies in the ball $\mathcal{V}$.

Now we repeat this trick with $\sigma(\eta_{\frac{1}{2}})=\exp\left(\frac{1}{2}\,K(1)\right)$ to show that there is a $\sigma(\eta_{\frac{1}{4}}) = \exp\left(\frac{1}{4}\,K(1)\right)$ with $\eta_{\frac{1}{4}}\in[0,\,\eta_{\frac{1}{2}}]$. It also follows, by the monotonicity axiom in our group composition continuity postulate, that $\sigma(\eta_{\frac{3}{4}}) = \exp\left(\frac{3}{4}\,K(1)\right)$ with $\eta_{\frac{3}{4}}\in[\eta_{\frac{1}{2}},\,1]$ lies in our path segment. This is because $\sigma(\eta_{\frac{3}{4}}) = \sigma(\eta_{\frac{1}{4}})\,\sigma(\eta_{\frac{1}{2}})$, so $\eta_{\frac{3}{4}}>\eta_{\frac{1}{2}}$, whilst $\sigma(1) = \sigma(\eta_{\frac{1}{4}})\,\sigma(\eta_{\frac{3}{4}})$, so $\eta_{\frac{3}{4}}<1$. Next we repeat this trick for $\sigma(\eta_{\frac{1}{4}})$, thus showing $\{\exp\left(\frac{k}{8} K(1)\right)|\,k\in0,\,1,\,\cdots\,8\}\subset\sigma([0,\,1])$. Repeating this process inductively shows that the matrix $\exp\left(n\,2^{-m}\,K(1)\right)$ for any integer $n$ and $m$ belongs to our group, that is, every power $\exp\left(q\,K(1)\right)$ of $\exp\left(K(1)\right)$ lies in the path segment $\sigma([0,\,1])$ where $q$ is a rational number in the interval $[0,\,1]$ with a finite binary expansion. The path segment $\sigma([0,\,1])$ being a closed subset of the set of square matrices, it must contain the closure of the set of transformation matrices we have shown to belong to it, to wit, $\{\exp\left(n\,2^{-m}\,K(1)\right)|\,n,\,m\in\mathbb{N};\,n\,2^{-m}\in[0,\,1]\}$. But the set of numbers of the form $n\,2^{-m}$ is {\it dense} in the interval $[0,\,1]$, therefore the path segment contains path segment $\{e^{\eta\,K(1)}|\,\eta\in[0,\,1]\}$ which also links $\sigma(0)$ and $\sigma(1)$. It follows that our path segment must be the path segment $\{e^{\eta\,K(1)}|\,\eta\in[0,\,1]\}$ because the function $\eta\mapsto e^{\eta\,K(1)}$ is the unique possible continuous function of one real variable whose values coincide with the values determined above on the dense subset of $\{n\,2^{-m}|\,n,\,m\in\mathbb{Z}\}$ of the real line. Thus we are lead to the conclusion in \eqref{LieGroup} and the understanding that some constant $4\times 4$ matrix $K$ wholly characterizes our transformation group of inertial frames in our World and which group acts linearly on the spacetime co-ordinates, at least if our four postulates hold good.

There are several ways to argue the form of \eqref{LieGroup}; another simple way is to understand that any closed matrix group with elements arbitrarily near to the identity (i.e. there is a member $\gamma(\epsilon)$ with $\|\gamma(\epsilon)-\mathrm{id}\|<\epsilon$ for any $\epsilon>0$) must contain at least one ray of the form $\{\exp(\eta\,K)|\,\eta\in\mathbb{R}\}$ for some square matrix $K$. von Neumann began exploring such thoughts in 1929\cite{vonNeumann1929}; a most excellent, well readable, sophomore level account of these ideas is given in Chapter 7 of Stillwell's "Naïve Lie Theory".


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