Occasionally you will hear that the Lorentz transformation is comparable to a rotation. In this post I will demonstrate how that is so, but also how all boosts in general are also mere rotations. First consider the active velocity boosting of a test mass and look at the inner product of the velocity four-vector with itself \(g_{\mu \nu}U^{\mu}U^{\nu} = c^{2}\) This is a statement that the four-d speed of anything through four-dimensional spacetime is c. In three dimensional space the local speed of massive objects is always less than c, but they also travel through time. With this time component of the speed considered in the velocity four-vector the entire length of the vector calculated in that manner is always exactly c. As it is a constant the proper time derivative of this is zero. \(\frac{D(g_{\mu \nu}U^{\mu}U^{\nu})}{d\tau} = 0\) Since the covariant divergence of the metric tensor is zero the product rule for general relativity yields \(g_{\mu \nu}\frac{DU^{\mu}}{d\tau}U^{\nu} + g_{\mu \nu}\frac{DU^{\nu}}{d\tau}U^{\mu} = 0\) The metric is symmetric so \(g_{\mu \nu}\frac{DU^{\mu}}{d\tau}U^{\nu} + g_{\nu \mu }\frac{DU^{\nu}}{d\tau}U^{\mu} = 0\) Rename indexes on the second term \(\mu \shortrightarrow \nu\) \(\nu \shortrightarrow \mu\) \(g_{\mu \nu}\frac{DU^{\mu}}{d\tau}U^{\nu} + g_{\mu \nu}\frac{DU^{\mu}}{d\tau}U^{\nu} = 0\) \(2g_{\mu \nu}\frac{DU^{\mu}}{d\tau}U^{\nu} = 0\) \(g_{\mu \nu}\frac{DU^{\mu}}{d\tau}U^{\nu} = 0\) Next note that the covariant proper time derivative of the velocity four-vector is the acceleration four-vector. \(g_{\mu \nu}A^{\mu}U^{\nu} = 0\) Since this says that the inner product of four-acceleration and four-velocity is zero, these vectors are orthogonal. Also multiplying through by the mass. \(g_{\mu \nu}mA^{\mu}U^{\nu} = 0\) From the relation between four-force and four-acceleration \(F^{\mu} = mA^{\mu}\)this yields \(g_{\mu \nu}F^{\mu}U^{\nu} = 0\) This says that four-force is always orthogonal to four-velocity. Now these results should make since because the four-d speed never changes as it is always c. Real forces then never change the four-speed, they merely cause deflections in direction in four dimensions. That's what you get when force is orthogonal to velocity, a rotation of direction. All active boosts then in four-d are merely rotations of direction in spacetime. Now this has all been about forced boosts of actual test masses moving about in spacetime, but the principle holds as well for passive boosts between frames of observers. This is particularly easy to show for the Lorentz transformation. Consider for example in three space the matrix expression for rotation of the x,y coordinates about an angle \(\theta\) to a new coordinate system using x',y' and for simplicity leave z = z' \( \begin{pmatrix}x' \\ y' \\ z' \end{pmatrix}= \begin{pmatrix}\cos{\theta} & \sin{\theta} & 0 \\ -\sin{\theta} & \cos{\theta} & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix}x \\ y \\ z \end{pmatrix}\) Please Register or Log in to view the hidden image! Just compare that to a Lorentz transformation. Start with the Lorentz transformation as a matrix expression for a boost along x, \( \begin{pmatrix}ct' \\ x' \\ y' \\ z' \end{pmatrix}= \begin{pmatrix}\gamma & -\gamma \beta & 0 & 0 \\ -\gamma \beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix}ct \\ x \\ y \\ z \end{pmatrix}\) Now switch variables from velocity to rapidity \(v = c\tanh{\theta} \) The Lorentz transformation becomes \( \begin{pmatrix}ct' \\ x' \\ y' \\ z' \end{pmatrix}= \begin{pmatrix}\cosh{\theta} & -\sinh{\theta} & 0 & 0 \\ -\sinh{\theta} & \cosh{\theta} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix}ct \\ x \\ y \\ z \end{pmatrix}\) Now you should be able to see that as far as the linear algebra is concerned, the Lorentz transformation is dealt with the same way as a rotation. The difference is merely the matter of "complexity" in hyperbolic trig and trig function relationships and a sign, and the sign part is merely due to the fact that in this case one axis is being rotated in the opposite direction of the other instread of in the same direction as before. (see image) Please Register or Log in to view the hidden image! I suppose that I should footnote that the rapidity as the matrix angle isn't equal to the drawing angle at which you place the primed axis on paper(rapidity \(\theta\) isn't the image authors \(\theta\)), but I hope that the analog of Lorentz transformation to rotation will be the the point understood by readers.
I have personally carefully retraced Lorentz's derivation of the transformation you are refering to here. Therefore I am closely and exactly familiar with the resulting apparition ( or lack of it ) of a rotation. Before indulging in a discussion of a "rotation", I believe it is fair to ask you if you personally have carefully reproduced Lorentz's derivation process, or, if you have perhaps only read of it without carefully doing it yourself? What you have displayed here is the result, alledgedly, of the derivation. What I am writing of is the derivation itself. Have you carefully gone through the DERIVATION?
IM WITH TRILARIAN..... if you edited my post..... id be really pissed off. discuss... thats fine.. but to change anothers work... is just wrong. -MT
Editing posts to remove irrelevant insults/provocations is nothing new. No meaningful post content was changed. We can discuss this further in the open government forum if you like.
I think the majority of us guessed that was what the edit was about. It might be a good idea to state what edits are deleting or changing when they are done, so some members don't draw incorrect conclusions. I think both you and James R do a great job moderating Physics & Math, no complaints from me.
Being in the batter's box when the post(s) were edited, I know exactly what was edited, on the part of both parties. The pitcher threw a spitball, which has been illegal for umpteen years in Major League play, at the batter's head. The pitcher was surprised and disapointed when I turned on it and drove it at least 700 feet over the centerfield fence. However, the umpire, in correct interpretation of the official rulebook, called the pitch void and therefore called the home run void. The batter was extremely agravated to lose one of the longest home runs on record, but can only admit that the ump made the right call. The pitcher should have never thrown the ball like that in the first place. Like Merlin told Arthur; " You must burn the Queen. Or else you must burn the Law.". In this forum it really can be to all our advantage to not burn our rules.
