Tell me what I'm doing wrong (velocity addition in 3+1 dimensions)

1. Mar 19, 2009

Fredrik

Staff Emeritus
A 3+1-dimensional Lorentz transformation can be written as

$$\Lambda=\gamma\begin{pmatrix}1 & -v^T\beta \\ -v & \beta\end{pmatrix}$$

where v is a 3×1 matrix representing the velocity difference, $\gamma=1/\sqrt{1-v^2}$, and $\beta$ is a 3×3 matrix that's orthogonal when v=0. When $\beta=I$, $\Lambda$ is a pure boost.

The product of two boosts with arbitrary velocities is

$$\Lambda(u)\Lambda(v)=\gamma_u\gamma_v(1+u^Tv)\begin{pmatrix}1 & -w^T \\ -w & \beta\end{pmatrix}$$

where

$$w=\frac{u+v}{1+u^Tv}$$

and

$$\beta=\frac{I+uv^T}{1+u^Tv}\neq I$$

so this isn't another boost, but it's a Lorentz transformation with velocity w. (If you noticed that the upper right component of the matrix on the right-hand side is $$-w^T$$ even though the formula for a general Lorentz transformation says that we should have $$-w^T\beta$$ in that position, don't worry about it. It's not a mistake. This w happens to be an eigenvector with eigenvalue 1 of $$\beta^T$$).

My question is this: Why doesn't this mean that the velocity addition rule in 3+1 dimensions is

$$u\oplus v=\frac{u+v}{1+u^Tv}$$

The correct result is supposed to be

$$\vec u\oplus\vec v=\frac{1}{1+\vec u\cdot\vec v}\bigg(\vec u+\vec v+\frac{\gamma_{\vec u}}{1+\gamma_{\vec u}}\vec u\times(\vec u\times\vec v)\bigg)$$

according to a book mentioned in another thread of this forum. (That book didn't include a derivation).

I'm pretty sure my algebra is correct. That's why I didn't bother posting all the details. If that last formula is the correct velocity addition rule, then there's something wrong with my whole approach to this.

Last edited: Mar 19, 2009
2. Mar 19, 2009

Fredrik

Staff Emeritus
That part was actually wrong. I thought I had verified this, but what I had verified is that $v$ is an eigenvector of $\beta^T$ with eigenvalue 1:

$$\beta^Tv=\frac{I+vu^T}{1+u^Tv}v=\frac{v+vu^Tv}{1+u^Tv}=\frac{v(1+u^Tv)}{1+u^Tv}=v$$

$u$ on the other hand, is an eigenvector of $\beta$ with eigenvalue 1:

$$\beta u=\frac{I+uv^T}{1+u^Tv}u=\frac{u+uv^Tu}{1+u^Tv}=\frac{u(1+v^Tu)}{1+u^Tv}=u$$

This is rather frustrating. I don't see a mistake anywhere in my calculations, but the results seem to contradict each other. I guess I'll have to take another look at this tomorrow.

3. Mar 19, 2009

DrGreg

I think you'll be kicking yourself...

For a pure boost $\beta$ is not I, it is

$$\beta = \gamma^{-1} I + (1 - \gamma^{-1})\frac{vv^T}{v^Tv}$$

It acts as the identity on v, but as $\gamma^{-1} I$ on the orthogonal complement.

4. Mar 19, 2009

Fredrik

Staff Emeritus
Thanks. I realized that there was something wrong with my $\beta$ right after I posted #2. I tried multplying two "boosts" (defined by $\beta=I$) together and got a result that wasn't even a Lorentz transformation). Unfortunately I didn't have time to look into it further right away.

How did you obtain that result for $\beta$?

The only formula I've found that describes a property of $\beta$ is

$$\beta^{-1}=\beta^T\frac{I-vv^T}{1-v^Tv}$$

which implies that $\beta$ is orthogonal when v=0.

Last edited: Mar 19, 2009
5. Mar 20, 2009

Fredrik

Staff Emeritus
I almost understand this now. The first hint is that in 1+1 dimensions we have

$$\Lambda=\gamma\begin{pmatrix}1 & -v\\ -v & 1\end{pmatrix}$$

so a 3+1-dimensional pure boost in the x direction should be

$$\Lambda=\begin{pmatrix}\gamma & -\gamma v & 0 & 0\\ -\gamma v & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\end{pmatrix}=\gamma\begin{pmatrix}1 & -v & 0 & 0\\ -v & 1 & 0 & 0\\ 0 & 0 & 1/\gamma & 0\\ 0 & 0 & 0 & 1/\gamma\end{pmatrix}$$

and this is a Lorentz transformation with

$$\beta=\begin{pmatrix}1 & 0 & 0\\ 0 & 1/\gamma & 0\\ 0 & 0 & 1/\gamma\end{pmatrix}$$

Every (3-)vector x can be split up into a component that's parallel to the velocity vector v and a component that's perpendicular to it:

$$x=x_\parallel+x_\perp$$

$$x_\parallel=\Big\langle\frac{v}{\|v\|},x\Big\rangle\frac{v}{\|v\|}=\frac{\langle v,x\rangle}{\|v\|^2}v=\frac{v^Txv}{v^Tv}=\frac{vv^T}{v^Tv}x$$

$$x_\perp=x-x_\parallel$$

It seems plausible that it's going to be true in general that when $\Lambda$ is a pure boost, $\beta$ is going to act as $I$ on $x_\parallel$ and as $\gamma^{-1}I$ on $x_\perp$:

$$\beta x=x_\parallel+\frac{1}{\gamma}x_\perp=\frac{vv^T}{v^Tv}x+\frac{1}{\gamma}\bigg( x-\frac{vv^T}{v^Tv}x\bigg)=\bigg(\frac{1}{\gamma}I+\bigg(1-\frac{1}{\gamma}\bigg)\frac{vv^T}{v^Tv}\bigg)x$$

so if what I called "plausible" is in fact true, then a pure boost $\beta$ is what you said.

I'm still not sure how to prove that it's true in general, or even if it's something that we should prove. Maybe we should just take this as a definition of what we mean by a "pure boost". Is there another way to define "pure boost" that can be used as the starting point of a proof that the "plausible" statement is true?

Edit: Sweet! I was finally able to prove that velocity addition law. I'm not going to LaTeX all the details, at least not right now, but the method suggested in #1, with the correct $\beta$ gives us

$$u\oplus v=\frac{u+\beta_uv}{1+u^T\beta_u v}$$

It's easy to show that $u^T\beta_u v=u^T v$, and a bit messy but still not too hard to show the rest. (I found it easier to start with the final result and work backwards).

Last edited: Mar 20, 2009
6. Mar 20, 2009

DrGreg

I'm claiming it's "geometrically obvious" that every boost is isomorphic to a boost along the x-axis, you just need to isometrically rotate your spatial axes. And the $\beta$ I quoted is the result of that rotation combined with an x-boost. It must the the right $\beta$ because it has the desired effect applied to v or anything orthogonal to v.

Now does this "geometrically obvious" claim require a more rigorous proof?

7. Mar 22, 2009

Fredrik

Staff Emeritus
I don't think it's obvious enough. It's intuitive enough, at least if we consider the physical motivation for the Lorentz transformation, but I'd rather show that it's just a straightforward calculation from the condition $\Lambda^T\eta\Lambda=\eta$ and a definition of the term "pure boost". A pure boost is supposed to be a Lorentz transformation without a rotation, but how do we express that formally? It would be easy if we could show that in general $\beta$ can be expressed in the form f(v)R+g(v) where R is an orthogonal matrix that's independent of v, f(0)=1 and g(0)=0. Then we could define a pure boost as one with R=I. But I haven't figured out the general form of $\beta$. The closest I've been able to get is the condition

$$\beta\beta^T=\frac{I}{\gamma^2}+vv^T$$

Maybe it's as simple as replacing the I in the formula you showed me with R:

$$\beta=\frac{R}{\gamma}+\left(1-\frac{1}{\gamma}\right)\frac{vv^T}{v^Tv}$$

8. Mar 24, 2009

DrGreg

(I've done a back-of-envelope calculation that I haven't double-checked, never mind latexed, so this reply is a bit vague but I hope you'll be able to fill in the details yourself. I worked backwards from the physics & 4d-geometry to get the algebraic result I wanted.)

Yes, you can put

$$\beta = f(v) R + g(v)$$​

Put $g(v) = 0$ and

$$f(v) = \gamma^{-1} I + (1 - \gamma^{-1})\frac{vv^T}{v^Tv}$$​

In fact, given $\beta$, use these equations to define R. It's not difficult to invert the matrix $f(v)$ (consider the special case $\textbf{v} = (v, 0, 0)^T$) -- it takes the same form but with different coefficients in front of I and $\textbf{v} (\textbf{v}^T \textbf{v})^{-1} \textbf{v}^T$. Then you ought to be able to show that $R^TR = I$ from your definition of a Lorentz transform.

Then, as you say, define a pure boost to be one with R=I. As a by-product, we've also decomposed a Lorentz transform into a spatial rotation followed by a boost.

9. Mar 24, 2009

Fredrik

Staff Emeritus
So you're suggesting that the most general form of $\beta$ is

$$\beta= \left(\gamma^{-1} I + (1 - \gamma^{-1})\frac{vv^T}{v^Tv}\right)R$$

where R is a rotation. I think you're right, but I still don't feel that we have proved it. Do you see a way to prove that every $\beta$ that satisfies

$$\beta\beta^T=\frac{I}{\gamma^2}+vv^T$$

can be expressed in the form above?

10. Mar 24, 2009

Fredrik

Staff Emeritus
I worked through the details of what you suggested, but I'm not going to LaTeX all of it. Let $\beta$ be given, and define $R=X^{-1}\beta$, where X is what we called f(v) above, i.e.

$$X=\gamma^{-1} I + (1 - \gamma^{-1})\frac{vv^T}{v^Tv}=\frac{I}{\gamma} +\frac{\gamma}{\gamma+1}vv^T$$

$$X^{-1}=\gamma I-\frac{\gamma^2}{\gamma+1}vv^T$$

The R constructed this way is orthogonal, as we can see by multiplying this stuff together:

$$RR^T=X^{-1}\beta\beta^T(X^{-1})^T=\left(\gamma I-\frac{\gamma^2}{\gamma+1}vv^T\right)\left(\frac{I}{\gamma^2}+vv^T\right)\left(\gamma I-\frac{\gamma^2}{\gamma+1}vv^T\right)=\dots=I$$

(Yes, I verified it. Finding $X^{-1}$, verifying that result, and verifying that R is orthogonal, made me feel like a random equation generator. It took like, 500 tries and I seemed to get a different result each time). So no matter what $\beta$ is (as long as it consists of the ij components of a Lorentz transformation), we can always express it as XR, with X defined as above and R orthogonal.

I'm not sure what all this means though. I'm going to have to think about it. For example, should it bother us that R appears to depend explicitly on v? Also, it seems that we have defined a "pure boost" to be a Lorentz transformation with a $\beta$ that has that particular form. Shouldn't we define the concept in a more intuitive way and derive the form of $\beta$ from that?

I'm pretty tired right now and it's possible that these things will be perfectly clear to me tomorrow, but right now they aren't.

11. Mar 25, 2009

DrGreg

I started to draft an answer to post #9 (fleshing out the details of #8), but once again you've worked it out yourself. (Unfortunately I have lots of other things to do as well as contribute to this forum so my response rate can be slow.)

Note that we have now proved that every Lorentz transform is the product of a boost and a spatial rotation (as we'd expect on geometrical grounds). The two are independent, so you can, if you wish, hold the rotation operator constant as you vary the boost velocity.

You say that, in the argument of #10, R depends on v, but both $\beta$ and X-1 depend on v and, I would say, "the v's cancel out" when constructing R.

And our reason for saying $\beta = X$ defines a pure boost does have an intuitive motivation: if, given v, we choose our spatial x-axis parallel to v, our definition of pure boost reduces to the standard boost matrix for "standard configuration".

I, too, got entangled in algebra when trying to prove the result. I eventually found it easier to introduce the notation

$$E(\lambda) = I + (\lambda - 1)\frac{\textbf{v}\textbf{v}^T}{\textbf{v}^T\textbf{v}}$$​

where $\lambda$ is an arbitrary scalar parameter and v is regarded as constant.

It is not hard to show that the function E is then a group homomorphism ($E(\lambda\mu) = E(\lambda)E(\mu)$) and this lemma makes the algebra a bit easier. (Lemma motivated by the canonical form of $E(\lambda)$ when $\textbf{v}=(v,0,0)^T$.)

(Of course your X is $\gamma^{-1}E(\gamma)$ with inverse $\gamma E(\gamma^{-1})$.)

Last edited: Mar 25, 2009
12. Mar 25, 2009

Fredrik

Staff Emeritus
Not sure about the validity of the cancellation argument, but I found another one. If we want to prove that any given $\beta$ that satisifes

$$\beta\beta^T=\frac{I}{\gamma^2}+vv^T$$

can be expressed as XR, where R is orthogonal, we can prove this by explicitly constructing such an R. That's what we just did. The components of this R will depend on the components of v and three more independent parameters (since the condition $$\Lambda^T\eta\Lambda=\eta[/itex] implies that $\Lambda$ has 6 independent components). If we change the values of those three in a way that keeps R orthogonal, it will change our $\beta$, but the condition above will still be satisfied. We can prove that my last sentence above is true by proving that $\beta=XR$ satisfies the condition above for any given orthogonal matrix R. I just verified that it does, so I'm satisfied. I feel that this is what we need to be able to think of v and R as independent. Yes, I think I understand this now. Another way of looking at it is to note that [tex]\beta=\begin{pmatrix}1 & 0 & 0\\ 0 & 1/\gamma & 0\\ 0 & 0 & 1/\gamma\end{pmatrix}$$

defines a valid Lorentz transformation when v is in the x direction, and that this is a) exactly what we expect a pure boost to look like if we have studied 1+1-dimensional Lorentz transformations, and b) the only choice of $\beta$ that satisfies

$$\beta x=x_\parallel+\frac 1 \gamma x_\perp$$

for arbitrary x. Since the decomposition of x into parts that are parallel and perpendicular to v is unique, this equation can be taken as the definition of $\beta$ when v is in the x direction, and it obviously works when v is in the y or z directions too. So it seems extremely natural to define a pure boost for arbitrary v to be a Lorentz transformation with a $\beta$ that satisfies this condition. The stuff I did in #5 then proves that $\beta$ must take the form we've been using.

Well that's unacceptable. . (OK, not really ). You should at least make another 118 posts quickly, so we can nominate you for a science advisor medal. (Really). Thanks for your help. I really appreciate it.

Last edited: Mar 26, 2009