tags:

This post (and perhaps a few to follow) is prompted by this thread over on SciForums. It’s a discussion about Lorentz transforms and, surprise surprise, someone doesn’t get it. In this case Magneto.

The thread started with Rpenner going over the form of a Lorentz transform, specifically a boost. He did what most textbooks do and selected the boost in the x axis direction. This is mathematically simpler than a general one and helps to show that the directions orthogonal to the velocity vector do not get changed by the boost. While the choice of velocity to go in the x axis might sound like not a general case it is actually equivalent to it. Unfortunately Magneto doesn’t understand this (despite claiming to be well read in this stuff, so much so he thought he was in the running to head up a research group at CalTech, which he wasn’t). It’s problem I can imagine others having, though most people wouldn’t be as dense as Magneto when they are given the explanation which I’ll do now.

Bases of $\mathbb{R}^{3}$

For the sake of simple examples I will do everything in the familiar realm of 3 dimensional space but all of this can be generalised to any number of dimensions (even infinite in some cases!), usually by just changing 3 to N and repeating a few things enough times.

So what’s a basis of $\mathbb{R}^{3}$? A basis is a list of 3 vectors in terms of which all vectors can be written. If $\{\mathbf{e}_{1},\mathbf{e}_{2},\mathbf{e}_{3}\}$ is a basis then it means any vector $\mathbf{v}$ can be written as a linear combination of them, $\mathbf{v} = A_{1}\mathbf{e}_{1}+A_{2}\mathbf{v}_{2}+A_{3}\mathbf{v}_{3}$. We call $A_{i}$ the components of $\mathbf{v}$ in the basis $\{\mathbf{e}_{1},\mathbf{e}_{2},\mathbf{e}_{3}\}$. If we change basis to $\{\mathbf{f}_{1},\mathbf{f}_{2},\mathbf{f}_{3}\}$ then we get new components because now we have $\mathbf{v} = B_{1}\mathbf{f}_{1}+B_{2}\mathbf{f}_{2}+B_{3}\mathbf{f}_{3}$. A simple example is that if $\mathbf{f}_{i} = 2\mathbf{e}_{i}$ then obviously we have $B_{i} = \frac{1}{2}A_{i}$ but more complicated combinations are allowed.

So how do we know if a set of vectors form a basis? They must be linearly independent, which means you can’t write one in terms of the others. A quick way to check this is to compute the triple vector product, $\mathbf{e}_{1} \cdot (\mathbf{e}_{2} \times \mathbf{e}_{3})$. If it is zero then they are not a basis. If it is non-zero then you’re good to go. It’s often convenient to have the vectors be mutually orthogonal, which means they are at right angles to one another so $\mathbf{e}_{i}\cdot \mathbf{e}_{j} = 0$ if $i \neq j$.

All of these things are true for the x,y,z axis directions, $\mathbf{e}_{x},\mathbf{e}_{y},\mathbf{e}_{z}$. If we take $\mathbf{e}_{1} = \mathbf{e}_{x}$ etc then the components of $\mathbf{e}_{x}$ are (1,0,0) because obviously $\mathbf{e}_{x} = 1\mathbf{e}_{1} + 0\mathbf{e}_{2}+0\mathbf{e}_{3}$ and likewise for the rest. This is the standard basis everyone is familiar with from school.

Rotations

Rotations are things most people should be familiar with. In linear algebra a rotation can be represented as a matrix acting on vectors. Again, as for vectors, the components of the matrix depend upon the choice of basis. Suppose we want to consider a rotation about $\mathbf{e}_{1}$. This will leave $\mathbf{e}_{1}$ invariant and will mix the other two directions. It will depend on the angle, call it $\theta$. Through some simple geometry (see Wikipedia) you find the matrix has the following form,

$R(\theta,\mathbf{e}_{1}) = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & \cos \theta & -\sin \theta \\ 0 & \sin \theta & \cos \theta \end{array}\right)$

The effect of this on the basis is the following,

$\begin{array}{lcrcrcl} \mathbf{e}_{1} &\to& \mathbf{e}_{1} &&&\equiv& \mathbf{f}_{1} \\ \mathbf{e}_{2} &\to& \cos\theta\, \mathbf{e}_{2} &-& \sin \theta\, \mathbf{e}_{3} &\equiv& \mathbf{f}_{2} \\ \mathbf{e}_{3} &\to& \sin\theta \,\mathbf{e}_{2} &+& \cos \theta \,\mathbf{e}_{3} &\equiv& \mathbf{f}_{3} \end{array}$.

Similar things occur for $R(\theta,\mathbf{e}_{2})$ and $R(\theta,\mathbf{e}_{3})$, just shuffle around the 1,2,3 labels. So what’s nice about rotations? They don’t change the dot product between two vectors is the really important thing. This is a special case of what has previously been talked about for Lorentz transforms, the group SO(N) leaves the Euclidean metric $\delta$ invariant.

$\delta(\mathbf{X},\mathbf{Y})=\delta(R\cdot\mathbf{X},R\cdot\mathbf{Y}) \quad \Leftrightarrow \quad \delta = R^{\top} \cdot \delta \cdot R$

In a less formal way it means rotations don’t change the length of vectors and they don’t change the angles between vectors and you can write the dot product between vectors in terms of those things,

$\mathbf{X}\cdot \mathbf{Y} = \Vert \mathbf{X} \Vert \Vert \mathbf{Y} \Vert \cos \alpha$,

where $\alpha$ is the angle between the vectors. So what does that mean for a basis? Well we used the rotation $R(\theta,\mathbf{e}_{1})$ to define a set of new vectors $\mathbf{f}_{i}$ above. Well let’s consider their dot products,

$\mathbf{f}_{i} \cdot \mathbf{f}_{j} = (R(\theta,\mathbf{e}_{1}) \cdot \mathbf{e}_{i}) \cdot (R(\theta,\mathbf{e}_{1})\cdot \mathbf{e}_{j}) = \mathbf{e}_{i} \cdot \mathbf{e}_{j}$ .

Another thing about rotations is that if two vectors have the same length then there exists a rotation (in more than 2 dimensions there are infinitely many!) which will turn one into the other.

$\Vert \mathbf{X} \Vert = \Vert \mathbf{Y} \Vert \quad \Rightarrow \quad \exists \, R_{\mathbf{X},\mathbf{Y}} \; \textrm{such that} \; \mathbf{X} = R_{\mathbf{X},\mathbf{Y}} \cdot \mathbf{Y}$

For example, if $\mathbf{v} = (v_{1},v_{2},v_{3})$ then there is a rotation which changes it to $(v,0,0)$ where $v^{2} = v_{1}^{2}+v_{2}^{2}+v_{3}^{2}$ and vice versa.

Lorentz Boosts

So let’s return to Rpenner’s Lorentz boost, where the velocity was $(v,0,0)$. It results in the y and z directions being unchanged, ie the boost warps only 1 direction of 3. This can be seen from the matrix representation, which is the identity in those two directions.

$\Lambda(v,0,0) = \left( \begin{array}{cccc} \gamma & -v \gamma & 0 & 0 \\ -v \gamma & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{array} \right)\quad \Rightarrow \quad \Lambda(v,0,0) \cdot \left( \begin{array}{c} A_{0} \\ A_{1} \\ A_{2} \\ A_{3} \end{array} \right) = \left( \begin{array}{c} \gamma A_{0} - v A_{1} \\ \gamma A_{1} - vA_{0} \\ A_{2} \\ A_{3} \end{array} \right)$

Acting the above on a space-time vector will not change the last two components. Magneto’s Lorentz boost is in the $\mathbf{v} = (v_{1},v_{2},v_{3})$ direction. He claims it is ‘more general’ in the sense that all the components change,

$\Lambda(v_{1},v_{2},v_{3}) = \left( \begin{array}{cccc} \gamma&-v_{1}\,\gamma&-v_{2}\,\gamma&-v_{3}\,\gamma\\ -v_{1}\,\gamma&1+(\gamma-1)\dfrac{v_{1}^2}{v^2}&(\gamma-1)\dfrac{v_{1} v_{2}}{v^2}&(\gamma-1)\dfrac{v_{1} v_{3}}{v^2}\\ -v_{2}\,\gamma&(\gamma-1)\dfrac{v_{2} v_{1}}{v^2}&1+(\gamma-1)\dfrac{v_{2}^2}{v^2}&(\gamma-1)\dfrac{v_{2} v_{3}}{v^2}\\ -v_{3}\,\gamma&(\gamma-1)\dfrac{v_{3} v_{1}}{v^2}&(\gamma-1)\dfrac{v_{3} v_{2}}{v^2}&1+(\gamma-1)\dfrac{v_{3}^2}{v^2}\\ \end{array} \right)$

where $\gamma = \gamma(v) = (1-v^{2})^{-\tfrac{1}{2}}$ and we have used the notation $v^{2} = v_{1}^{2}+v_{2}^{2}+v_{3}^{2}$ (this is not a coincidence). Clearly acting that on a general vector will change all 3 spatial components, so is Magneto correct in claiming all 3 directions are warped, rather than 1?

As just explained, if two vectors are of the same length there is a rotation which turns one into the other. Since $v^{2} = v_{1}^{2}+v_{2}^{2}+v_{3}^{2}$ the two vectors used by Rpenner and Magneto are equal in size and since rotations are special cases of Lorentz transforms there’s a Lorentz transform which converts Rpenner’s vector into Magneto’s and vice versa. Now notice how the direction which is warped in Rpenner’s case includes a contradiction factor $\gamma$ and is time dependent. Clearly it’s length has changed and changes in time. So the question is can we find a direction (or even 2) in Magneto’s version which isn’t changed?

Remember that rotations don’t change inner products and that the size of a vector is expressed in terms of an inner product, $\Vert \mathbf{X} \Vert^{2} = \mathbf{X} \cdot \mathbf{X}$. Remember the vectors $\mathbf{e}_{y}$ and $\mathbf{e}_{z}$ don’t change their length under Rpenner’s boost. But we also have the fact we can rotate Rpenner’s boost vector to become Magneto’s. So we can do the following things,

• Rotate Magneto’s setup to be the same as Rpenner’s setup
• Do Rpenner’s boost
• Rotate Rpenner’s setup back to Magneto’s setup

There is some vector $\mathbf{Y}$ in Magneto’s setup which is rotated into $\mathbf{e}_{y}$ in Rpenner’s setup, will not be changed by the boost and then is rotated back to $\mathbf{Y}$. Since $\mathbf{e}_{y}$ is orthogonal to latex $\mathbf{e}_{x}$ it follows that latex $\mathbf{Y}$ is orthogonal to latex $\mathbf{v}$. So we have determined there is a direction which Magneto’s boost doesn’t warp! Similarly with $\mathbf{e}_{z}$ and so there is actually 2 directions! If we did this in N dimensional space there would be N-1 such independent directions.

In fact this method of rotate, boost, unrotate is precisely why Rpenner’s method is just as general as Magneto’s. The rotation is equivalent to picking a basis set of vectors and it’s logical to pick one which makes your life easy. It’s easy to compute rotations and it’s easy to do boosts in the x direction. When you combine the rotation matrices with Rpenner’s boost matrix you end up with Magneto’s.

Of course some of you might be thinking that I should actually construct the direction in question, else my logic might not be valid. Firstly, the logic stands, regardless of whether I construct the vector. Secondly that’s precisely what I’m going to do now!

Explicit Construction

The x,y,z directions form an orthogonal basis, they are all at right angles to one another and that makes life easier in general so the first thing to do is to make an orthogonal basis which includes $\mathbf{v} = (v_{1},v_{2},v_{3})$. With a moment’s thought a vector orthogonal to this can be found, $\mathbf{u} = (-v_{2},v_{1},0)$. To get the third vector we use the cross product,

$\mathbf{w} = \mathbf{v} \times \mathbf{u} = (-v_{2}v_{3} , -v_{1}v_{3} , v_{1}^{2}+v_{2}^{2})$.

Let’s call this set $\{\mathbf{v},\mathbf{u},\mathbf{w}\} = \{\mathbf{f}_{1},\mathbf{f}_{2},\mathbf{f}_{3}\}$. Since they are orthogonal they are a basis and any vector can be written in terms of them, $\mathbf{X} = A_{1}\mathbf{f}_{1} + A_{2}\mathbf{f}_{2} + A_{3}\mathbf{f}_{3}$. We can make them orthonormal by normalising but what is important are the directions. One gripe which someone might have is that if they represent velocities why does $\mathbf{w}$  have squared terms. You can view them as dimensionless coefficients, with the [tex]\mathbf{e}_{i}\$ carrying the units.

In the previous section I said that because the x and y directions are orthogonal a direction orthogonal to $\mathbf{f}_{1}$ should be unchanged by Magneto’s boost. So let’s check that by acting Magneto’s boost on each one. Remember, at the end the only changes which should occur should be in the $\mathbf{v} = \mathbf{f}_{1}$, the coefficients of the other vectors shouldn’t change. We start by acting $\Lambda(v_{1},v_{2},v_{3})$ on the components of $\mathbf{f}_{2}$. Since this is space-time we have to pick a time, so we use t=T for the time component. The time component is allowed to be altered, as its common to all vectors.

$\begin{array}{rcl}\Lambda(v_{1},v_{2},v_{3})\cdot \mathbf{f}_{2} &=& \left( \begin{array}{cccc} \gamma&-v_{1}\,\gamma&-v_{2}\,\gamma&-v_{3}\,\gamma\\ -v_{1}\,\gamma&1+(\gamma-1)\dfrac{v_{1}^2}{v^2}&(\gamma-1)\dfrac{v_{1} v_{2}}{v^2}&(\gamma-1)\dfrac{v_{1} v_{3}}{v^2}\\ -v_{2}\,\gamma&(\gamma-1)\dfrac{v_{2} v_{1}}{v^2}&1+(\gamma-1)\dfrac{v_{2}^2}{v^2}&(\gamma-1)\dfrac{v_{2} v_{3}}{v^2}\\ -v_{3}\,\gamma&(\gamma-1)\dfrac{v_{3} v_{1}}{v^2}&(\gamma-1)\dfrac{v_{3} v_{2}}{v^2}&1+(\gamma-1)\dfrac{v_{3}^2}{v^2}\\ \end{array} \right) \left( \begin{array}{c} T \\ -v_{2} \\ v_{1} \\ 0 \end{array}\right) \\ &=& \left( \begin{array}{c} \gamma T \\ -\gamma T v_{1}-v_{2} \\ -\gamma T v_{2}+ v_{1} \\ -\gamma T v_{3} \end{array}\right)\end{array}$

So how has the spatial part changed? Clearly there’s some T dependent terms and some non-T dependent terms. Dropping the time component we have the 3-vectors

$\left( \begin{array}{c} -\gamma T v_{1}-v_{2} \\ -\gamma T v_{2}+ v_{1} \\ -\gamma T v_{3} \end{array}\right) = -\gamma T \left( \begin{array}{c} v_{1} \\ v_{2} \\ v_{3} \end{array}\right) + \left( \begin{array}{c} -v_{2} \\ v_{1} \\ 0 \end{array}\right) = -\gamma T \mathbf{f}_{1} + \mathbf{f}_{2}$

So it turns out the coefficient of $\mathbf{f}_{2}$ doesn’t change, only a shift in the $\mathbf{f}_{1}$ direction occurs. Doing the same calculation for $\mathbf{f}_{3}$ gives the same result. Overall we have the effect of Magneto’s Lorentz boost being

$\Lambda(v_{1},v_{2},v_{3}) \;:\; A_{1}\mathbf{f}_{1} + A_{2} \mathbf{f}_{2} + A_{3} \mathbf{f}_{3} \to \tilde{A}_{1}\mathbf{f}_{1} + A_{2} \mathbf{f}_{2} + A_{3} \mathbf{f}_{3}$.

The $A_{1}$ coefficient is altered, gaining time dependence, while $A_{2},A_{3}$ are untouched. We can therefore conclude that Magneto’s assertions were incorrect.

The one thing he did get right is that this is general in the sense of being able to recover specific cases, you can just slap in any values for the $v_{i}$ components you want or define the $\mathbf{f}_{j}$ appropriately. In Rpenner’s example just set $\mathbf{f}_{1} = \mathbf{e}_{x}$ etc,  this is nothing more than just applying a rotation to align your vectors nicely before you start churning through the algebra. This method generalises to $\mathbb{R}^{N}$, where a boost in a given direction warps that direction and leaves the N-1 perpendicular directions unchanged. None of this material is particularly complicated, it’s just something you get a handle on by working through examples and gaining some experience. Unfortunately, despite his years spend writing a book on what he thinks relativity is Magneto has little experience with this stuff. He even confuses vectors and scalars, despite such things being the fundamental concepts upon which relativity is built. Maybe one day he’ll stop writing books about relativity and actually read some instead….