Deriving General Lorentz Boost Equation

[HJ Farnsworth]

Greetings,

I have been having trouble deriving the equation for the general Lorentz boost for velocity in an arbitrary direction. It seems to me that given the 1D Lorentz transformations...

matrix for Lorentz transformation in x-direction, X:
{{1/sqrt(1-v^2), -v/sqrt(1-v^2), 0, 0},
{-v/sqrt(1-v^2), 1/sqrt(1-v^2), 0, 0},
{0, 0, 1, 0},
{0, 0, 0, 1}}

matrix for Lorentz transformation in y-direction, Y:
{{1/sqrt(1-v^2), 0, -v/sqrt(1-v^2), 0},
{0, 1, 0, 0},
{-v/sqrt(1-v^2), 0, 1/sqrt(1-v^2), 0},
{0, 0, 0, 1}}

matrix for Lorentz transformation in z-direction, Z:
{{1/sqrt(1-v^2), 0, 0, -v/sqrt(1-v^2)},
{0, 1, 0, 0},
{0,0,1,0},
{-v/sqrt(1-v^2), 0, 0, 1/sqrt(1-v^2)}}

...all I should have to do is replace each v in X with v_x, each v in Y with v_y, and each v in Z with v_z, and then multiply the matrices to get ZYX as the general Lorentz boost equation. This should sequentially compute the Lorentz transformation in the x-direction with the x-velocity, then the y-direction with the y-velocity, then the z-direction with the z-velocity. However, when I do that, the answer I get only vaguely resembles the general Lorentz boost equation found on Wikipedia (http://en.wikipedia.org/wiki/Lorentz_transformation#Matrix_form).

I can't find the derivation for the general boost matrix anywhere, and more importantly, I have no idea why the simple math I used is wrong - it seems conceptually correct to me.

Please tell me what the flaw in my logic is, and how to correct it.

Thanks for any help you can give.

-HJ Farnsworth

 

[HJ Farnsworth]

By the way, in case this wasn't clear in my previous post, the v_x, v_y, and v_z in the previous post were for 3-velocity vector v = v_xe_x + v_ye_y + v_ze_z, where the e's are orthonormal basis vectors in the x-, y-, and z-directions.

 

[George Jones]

Please tell me what the flaw in my logic is, and how to correct it.

In general, the product of two non-colinear boosts is not a boost. To find a boost in an arbitrary direction from a boost B in the x direction, compute
R B R^{-1},
where R is the rotation matrix that rotates the x direction into the arbitrary direction.

 

[pervect]

Offhand, I'd say that what you should do to get the general boost is to perform some rotation R, so that the rotated coorinates xr, yr, zr have the boost going along one of the axes - say the x-axis for specificity.

Then you do the boost.

Then you perform the inverse rotation R^(-1) such that R(-1)R = a unit matrix, and you should have the general boost along an arbitary direction.

It's important to get the right order, and I'm not totally sure I've gotten that right in my text description. Basically, I expect it to go like this:

<br /> R^{-1} \Lambda R<br />

where R is the rotation matrix. The composition process works from right to left, first you apply R, then do the boost \Lambda, then you do the inverse of R, see http://en.wikipedia.org/w/index.php?title=Transformation_matrix&oldid=431791983

"Composing and inverting transformations"There's probably an easier way.

[add] I see George snuck an answer in ahead of me, they seem to be equivalent except that the matrix R is defined diffrently.

 

[HJ Farnsworth]

Hey,

Thanks for the quick responses . Your answers make perfect sense - in fact I can't believe I didn't think of that. Ah well, we all have off days I guess.

I still have one question though. I understand why RBR^-1 works - it rotates the coordinate axes to be in the easiest direction, does the standard Lorentz transformation, then rotates the coordinate axes back so we have the transformed coordinates in our original coordinate system. However, I am still not quite clear on why a series of orthogonal boosts does not work. How come the product of two non-colinear boosts is not a boost? Please be explicit/detailed in telling me why my original thought process in my first post was wrong.

Again, thanks very much.

-HJ Farnsworth

 

[Ben Niehoff]

More precisely, the product of two non-parallel boosts is not purely a boost. Generically, the result will be a boost and a rotation. For example, if you boost in the X direction and then boost in the Y direction, the result will be a boost in some direction in the XY plane, combined with an additional rotation about the Z axis.

This can be seen from the Lorentz algebra:

\begin{gather*}[J_a, J_b] = \varepsilon_{abc} J_c \\ [K_a, K_b] = -\varepsilon_{abc} J_c \\ [J_a, K_b] = \varepsilon_{abc} K_c \end{gather*}

where J_a are the rotations, and K_a are the boosts. In particular, two boosts do not generically commute (as per second line).

 

[3nTr0pY]

Hey,

Thanks for the quick responses . Your answers make perfect sense - in fact I can't believe I didn't think of that. Ah well, we all have off days I guess.

I still have one question though. I understand why RBR^-1 works - it rotates the coordinate axes to be in the easiest direction, does the standard Lorentz transformation, then rotates the coordinate axes back so we have the transformed coordinates in our original coordinate system. However, I am still not quite clear on why a series of orthogonal boosts does not work. How come the product of two non-colinear boosts is not a boost? Please be explicit/detailed in telling me why my original thought process in my first post was wrong.

Again, thanks very much.

-HJ Farnsworth

Sorry for digging this thread up, but why does RBR^-1 work?

Is B the v_x boost matrix? If so, why does it cause it go in the 'easiest direction' when it acts on the rotation matrix. I can see why you then have to do the inverse to go back to the original coordinate basis, but why is the resulting matrix now in a more general form? As you can guess, my understanding of matrices is pretty shaky.
EDIT: Using that method I can't seem to get the right result anyway... Is the rotation matrix supposed to be: {{coshx, -sinhx, 0, 0}, {-sinhx, coshx, 0, 0}, {0, 0, 1, 0}, {0, 0, 0, 1}} ?

 

[DrGreg]

Given a 3D velocity v = (v_x, v_y, v_z) with v = |v|, <br /> \textbf{B} = \begin{bmatrix}<br /> \gamma &amp; -\gamma v &amp; 0 &amp; 0\\ <br /> -\gamma v &amp; \gamma &amp; 0 &amp; 0\\ <br /> 0 &amp; 0 &amp; 1 &amp; 0\\ <br /> 0 &amp; 0 &amp; 0 &amp; 1<br /> \end{bmatrix}<br />
Choose two unit vectors f and g so that v, f and g are all orthogonal to each other.
<br /> \textbf{R} = \begin{bmatrix}<br /> 1 &amp; 0 &amp; 0 &amp; 0\\ <br /> 0 &amp; v_x/v &amp; f_x &amp; g_x\\ <br /> 0 &amp; v_y/v &amp; f_y &amp; g_y\\ <br /> 0 &amp; v_z/v &amp; f_z &amp; g_z<br /> \end{bmatrix}<br />

Now work out RBR^T

 

[3nTr0pY]

Now work out RBR^T

Thanks for the help. I did what you said but without understanding why it works.

Comparing it to the required result, I get something very similar but which requires equations such as the following to be true:

v_x²/v² + f_x² + g_x² = 1

and

-(v_xv_y)/v² = f_xf_y + g_xg_y Why are these true? (And why does the method even work at all?)

 

[DrGreg]

Thanks for the help. I did what you said but without understanding why it works.

Comparing it to the required result, I get something very similar but which requires equations such as the following to be true:

v_x²/v² + f_x² + g_x² = 1

and

-(v_xv_y)/v² = f_xf_y + g_xg_y

Why are these true?

The way I defined R ensures that its columns are orthonormal. It is a fact about square matrices, maybe not immediately obvious but nevertheless true, that orthonormal columns imply orthonormal rows. Or, to put it another way, \mathbf{R}^T\mathbf{R} = \mathbf{I} implies \mathbf{RR}^T = \mathbf{I}. Your equations above follow from this.

(And why does the method even work at all?)

The matrix R was chosen as an orthogonal matrix satisfying<br /> \textbf{R}^T \, \begin{bmatrix} 0 \\ v_x \\ v_y \\ v_z \end{bmatrix}<br /> = \begin{bmatrix} 0 \\ v \\ 0\\ 0 \end{bmatrix}<br />or equivalently
<br /> \textbf{R} \, \begin{bmatrix} 0 \\ v \\ 0\\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ v_x \\ v_y \\ v_z \end{bmatrix}<br />(and, of course, leaving the time dimension unmodified).

 

[3nTr0pY]

Great. Thanks again for your help.