The quaternion Lorentz boost $v'=hvh^*+ 1/2( (hhv)^*-(h^*h^*v)^*)$ where $h$ is $(\cosh(x),\sinh(x),0,0)$ was derived by substituting the hyperbolic sine and cosine for the sine and cosine in the usual unit quaternion rotation $v'=hvh^*$ and then subtracting out unwanted factors in Lorentz boosts with Quaternions.

Does this transformation form a group? Do two transformations of the form $v'=hvh^*+ 1/2( (hhv)^*-(h^*h^*v)^*)$ make another transformation of the form $v'=fvf^*+ 1/2( (ffv)^*-(f^*f^*v)^*)$?

Is this transformation useful?

For more information, see his blog, The Stand-Up Physicist.

  • $\begingroup$ This Phys.SE post is also discussed in the second half of this entry on Doug Sweetser's blog. $\endgroup$
    – Qmechanic
    Jun 29, 2012 at 18:58

3 Answers 3


I) The $x$-boost formula of The Stand-up Physicist (also known as Doug Sweetser) can be simplified to

$$\tag{1} q^{\prime}-bq\bar{b}~=~ \bar{q} \frac{\bar{b}^2-b^2}{2} $$ $$~=~ - \bar{q} i {\rm Im} (b^2)~=~ - \bar{q} i\sinh(2\alpha)~=~(q_{\perp}-\bar{q}_{\parallel}) i\sinh(2\alpha), $$


$$b~:=~\cosh(\alpha)+i \sinh(\alpha)~\in~ \mathbb{C}, \qquad b^2~=~ 1+ i \sinh(2\alpha),\qquad |b|^2~=~\cosh(2\alpha), $$

$$q~:=~q_{\parallel}+q_{\perp}~\in~ \mathbb{H}~\cong~M(1,3;\mathbb{R}), \qquad q_{\parallel}~:=~t+ix , \qquad q_{\perp}~:=~jy+kz. $$

Formula $(1)$ indeed reproduces the well-known Lorentz transformation formulas for $x$-boosts with rapidity $2\alpha$,

$$ t^{\prime}~=~t\cosh(2\alpha) - x \sinh(2\alpha), \qquad x^{\prime}~=~x\cosh(2\alpha) - t \sinh(2\alpha),$$ $$ y^{\prime}~=~y,\qquad z^{\prime}~=~z, $$

or equivalently,

$$ q^{\prime}_{\parallel} ~=~q_{\parallel}\cosh(2\alpha)-\bar{q}_{\parallel} i \sinh(2\alpha) , \qquad q^{\prime}_{\perp}~=~q_{\perp}. $$

This is because

$$bq\bar{b}~=~b(q_{\parallel}+q_{\perp})\bar{b}~=~q_{\parallel}|b|^2+ q_{\perp}\bar{b}^2~=~q_{\parallel}\cosh(2\alpha)+ q_{\perp}(1- i \sinh(2\alpha)).$$

II) As is well-known, the $x$-boosts $b$ form an Abelian non-compact $U(1)$ subgroup of the Lorentz group $SO(1,3;\mathbb{R})$. Since the pertinent group acts freely on Minkowski space, the $x$-boost formula $(1)$ must also respect this group structure. In terms of rapidity $2\alpha$, this Abelian subgroup is just the additively written group $(\mathbb{R},+)$, cf. this and this Phys.SE post.

III) There are many important descriptions of the Minkowski space $M(1,3;\mathbb{R})$ and the Lorentz group (meaning both rotations and boosts), but as far as I can tell, the quaternions $\mathbb{H}$ are not too useful in this respect. For instance, the set of Hermitian matrices $u(2,\mathbb{C})$ seems to be a much more powerful description of Minkowski space $M(1,3;\mathbb{R})$, cf. this Phys.SE post.

In contrast, the quaternions $\mathbb{H}$ play an important role for the Euclidean compact counterpart $SO(4,\mathbb{R})$ of the Lorentz group $SO(1,3;\mathbb{R})$, because the Lie group $U(1,\mathbb{H})\times U(1,\mathbb{H})$ is (a double cover of) $SO(4,\mathbb{R})$.

IV) The Lie group

$$U(1,\mathbb{H})~:=~\{r\in \mathbb{H} \mid \bar{r}r=1\}~\cong~ SU(2,\mathbb{C})$$

is (a double cover of) the rotation group

$$SO(3,\mathbb{R})~\cong~ SO({\rm Im}(\mathbb{H}),\mathbb{R}),$$

which in turn is both a subgroup of $SO(4,\mathbb{R})$ and $SO(1,3;\mathbb{R})$. A rotation $r$ is implemented as

$$\tag{2} \tilde{q}~=~rq\bar{r}, \qquad q\in \mathbb{H}, \qquad r\in U(1,\mathbb{H}).$$


$$ \qquad \tilde{q}^{\prime}~=~rq^{\prime}\bar{r}, \qquad \tilde{b}~=~rb\bar{r}, \qquad \tilde{i}~=~ri\bar{r}. $$

Hence, the formula $(1)$ behaves covariantly under rotations, i.e. formula $(1)$ holds not just for boosts $b$ in the $x$-direction, but for boosts $b$ in any direction!

V) Potential problems lie elsewhere. Is this formula $(1)$ interesting? I fail to see how formula $(1)$ could be useful (as compared to more potent standard approaches). It is apparently not Lorentz covariant. Why have a formula that works for boosts, but not for rotations, cf. formula $(2)$? Note that two successive boosts in different directions do in general not constitute a pure boost, i.e. boosts in generic directions do not form a proper subgroup.

  • $\begingroup$ For rotations specifically, you need to consider another sort of b or h in the original post. If $b = (0, a, b, c)$ and b has a norm of unity, then the second and third terms of the original formula cancel leaving the rotational equation first used by Rodrigues. The b in your post is for any and all boosts which also might involve rotations. If the scalar is zero, that cannot be about the rapidity, but could be about pure rotations. $\endgroup$
    – sweetser
    Nov 21, 2019 at 16:59

I worked it out, mostly just reproducing Qmechanic, but I wanted to explain why this embedding works in a conceptual way, and how to generalize it to the whole Lorentz group. But the construction is an interesting way to show how boosts work in a Dirac representation of the Gamma matrices, it doesn't give something new.

The fact that you can represent Lorentz transformations using quaternion operations is not surprising by itself, because given any quaternion v, you can find it's components by pure quaternionic operations:

$$2 Re v = v + \bar{v}$$ $$2 Im v = -2 Re iv $$ $$2 JP v = -2 Re jv $$ $$2 KP v = -2 Re kv $$

Where JP and KP mean the "j-part" and the "k-part" of the quaternion. If you work out the individual components, they end up being commutors of v and its conjugate with i. So you can write any linear transformation you like in terms of v and commutators with i,j,k.

But this formula doesn't do this. The formula is (or should be--- you screwed up the stuff in the second parentheses, it should be $(h^*h^*v)^*$, both terms need to be proportional to the conjugate of v)

$$ h v \bar{h} + ({h^2\over 2} - {\bar{h}^2\over 2}) \bar{v} $$

(as Qmechanic say) where $h= \cosh(x)+i\sinh(x)$. This doesn't do any specific coordinate projections, yet it makes a linear transformation on v which is a Lorentz transformation on the components, and it looks like it should mean something, because it only acts on v and v-bar in a simple way.

From this you immediately see that the 1-d transformations form a group, because (as Qmechanic writes) if you do two linear transformations one after another, you multiply the matrices. Since this reproduces the Lorentz transformation, you end up multiplying the Lorentz transformation matrices. So to verify that it forms a group, you just need to verify that it indeed does a Lorentz transformation.

Verifying that it works

You describe a quaternion as a pair of complex numbers as follows:

$$ v = z_1 + z_2 j $$

Where multiplying $z_2$ by j gives the j and k terms. Then note the following formal algebraic relations, for any complex number $z$:

$$ zj = j\bar{z}$$

So that you can slide a complex number past a j by conjugating--- this is the quaternion algebra. The quaternionic conjugate is simple but you should know that $(ab)^*=b^*a^*$ (I use * and bar interchangeably).

$$ \bar{v} = \bar{z}_1 + \bar{j}\bar{z}_2 = \bar{z}_1 - z_2 j $$

Also the algebra is simple--- you just conjugate a complex number whenever you move it past a j. . Then you never need to write down real components, you can make do with complex ones. The complex numbers commute, so all you have to worry about is the placement of the j, and you can always rearrange any expression to put the j last, by moving it past the complex numbers, conjugating them as you go.

From this, and knowing that $h = \cosh(x) + i \sinh(x) $, you can see that

$$ h v\bar{h} = h\bar{h} z_1 + h^2 z_2 j $$

using the relations , $ \bar{h}h = \cosh(2x)$, $h^2 = 1 + 2i \sinh(2x) $ so that $h^2 + \bar h^2 = 2$, you can add the above to the thing below:

$$ ({h^2 - \bar{h}^2\over 2} ) (\bar{z}_1 - z_2 j) $$

to find

$$ h\bar{h} z_1 + {h^2 - \bar{h}^2\over 2i} i \bar{z}_1 + {h^2 + \bar{h}^2\over 2} z_2 $$

And this is the Lorentz transformation (since $h\bar{h}$ is $\cosh(2x)$ while $h^2-\bar{h}^2=2i\sinh(2x)$, while the thing multiplying $z_2$ is 1. This is repeating Qmechanic with a little more detail (and originally typos, sorry).

Why it works and embedding the whole Lorentz group

Why it works

The reason it works is because of the spinor representations of the Lorentz group. This is usually written in terms of gamma-matrices, not quaternions, but i-times the gamma matrices gives the quaternion algebra.

For convenience, it is useful to consider the usual Pauli matrices as a complex quaternions: I,J,K are i-times the Pauli matrices, and they make the quaternions, but there is also lowercase "i", which commutes with all three and squared to -1. This is not a division algebra anymore, but it's the full space of 2 by 2 complex matrices.

In terms of the complex quaternions, a standard Dirac representation is

$$\gamma^0 = (1,0;0,-1)$$ $$\gamma^i = (0,i\sigma_i;i\sigma_{i_0},0)$$

Another one, the chiral one, is

$$\gamma^0 = (0,i;i,0)$$ $$\gamma^i = (0,i\sigma_i;-i\sigma_i,0,0)$$

(you should always multiply Dirac matrices as 2 by 2 quaternionic matrices. This is how everyone does it in their heads anyway). The point is that v-slash in a chiral representation is very close to a quaternion.

$$ v\cdot\gamma^0 =( 0 , i v_0 + v_1 I + v_2 J + v_3 J, -i v_0 + v_1 I + v_2 J + v_3 K) $$

define the quaternion $V$ to be $v_0 + v_1 I + v_2 J + v_3 K$. All the Dirac stuff is complex quaternions with an imaginary time-component, which you can do formally by writing $V-(1+i)(V+\bar{V})$ The boost generator is the product of $\gamma_0$ and $\gamma$ in the direction of the boost, and what it does in a Dirac rep is mix up components, and in a Chiral rep, it multiplies by

$$ (\cosh(x) + i \sinh x I) V (cosh(x) + i \sinh(x) I ) $$

This is very close to the form of the stand-up physicist. To make it pure quaternionic, you just have to get rid of the "i" factors, and I don't know any elegant way to do this.

The reason the final thing is interesting is only because it is expressed with both left multiplications and right multiplications (left and right on V, only left for $\bar{V}$), so that it isn't a matrix acting on the quaternion on the left and on the right. The algebra involved is exactly the same as representing a Dirac spinor, but the algebra has been rearranged in a slightly more elegant way. I think this is kind of nifty, and it might be useful for something.

  • $\begingroup$ THANKS! There is alot of information to process there. I saw that it formed a boost group when I realized that, after the first boost to (t',x',y',z'), second boost acts on (t',x',y',z') in exactly the same way and therefore the standard trig identites like sinh(x+y)=sinh(x)cosh(y)+sinh(y)cosh(x) apply. $\endgroup$ May 25, 2012 at 4:08
  • $\begingroup$ @Barrett: yeah--- that's not a big deal. It's a linear transformation, so composition is like composition of ordinary Lorentz transformations. The major thing is to understand why the quaternions have this strange Lorentz group action, and this is explained by the standard (1/2,0) representation of the Lorentz group, and the map between the helicity $\gamma$ matrices and the quaternion algebra. I have some typos and the answer is incomplete. I'll fix it up. $\endgroup$
    – Ron Maimon
    May 25, 2012 at 6:06
  • $\begingroup$ I've really just started to learn group theory and I know a reasonable amount, but I haven't solidified my understanding yet. So this is a lot for me to take in even though each individual step seems straightforward. Given that it appears to represent the full Lorentz Group, do you think this is an important result? Is this useful? $\endgroup$ May 25, 2012 at 7:02
  • $\begingroup$ @Barrett: It does represent the whole Lorentz group, I could see it by piddling around, but my argument was wrong! It doesn't work the way I said it, in a Chiral representation. You don't need any group theory to work this out, just familiarity with Dirac matrices and how they work to represent the Lorentz group. The correct representation that gives this is Dirac's original representation where $\gamma_0$ is diagonal. $\endgroup$
    – Ron Maimon
    May 25, 2012 at 9:55
  • $\begingroup$ Comment to v4: There seems to be a typo in the second term of the formula for $v'$. Here $\bar{v}$ should be shifted to the left of ${h^2 - \bar{h}^2\over 2}$ and with opposite sign. $\endgroup$
    – Qmechanic
    May 25, 2012 at 18:51

Rotations in 3D space are done with a transformation known to Rodrigues in the 1840, $v' = h v h^* $ where $h$ has a norm of one. The transformation written in the question, $v' = h v h^* + \frac{1}{2}((v v h)^* - (v^* v^* h)^*) $, is a generalization of the spatial rotation. Notice that if the scalar of a quaternion $h$ is equal to zero, then the two vvh terms cancel leaving only the Rodrigues transformation. Thus all rotations are possible. Lorentz boosts use a different parameter $h$, one where the scalar is unity or greater, $h=(\cosh(x), \vec{I} \sinh(x))$. The Lie group SO(3, 1) of the Lorentz group has a Lie algebra so(3, 1) that has 6 dimensions. That appears to be in conflict with this quaternion function that only has one quaternion parameter and thus 4 or less degrees of freedom. Notice however that rotations are ratios of distances and thus are elements of space-time. The boosts are are ratios of changes in distance over time. Velocities live in the tangent space of space-time. The three rotations in space-time and three boosts in the tangent space of space-time constitute six degrees of freedom.

  • $\begingroup$ self-disclosure: I am the Stand-Up Physicist. I only noticed the h=(0, I) case in 2013. Dr. Kharinov independently found the Loretnz boost transformation. He too thinks it can only do a boost. $\endgroup$
    – sweetser
    Oct 9, 2019 at 14:45

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