# Education and Crank Claims: Special Relativity

Discussion in 'Physics & Math' started by rpenner, Oct 5, 2011.

1. ### DonQuixoteRegistered Senior Member

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That's an interesting answer. I will have to study it in order to see if I think the "qualification" makes the "yes" into a "no". But it will have to wait, I'm afraid.

For my purpose, I can not presuppose length contraction, time dilation or relativity of simultaneity. I am trying to discover these phenomena through Einsteins descriptions of the experiments and the two postulates. If I put the conclusion among my premises, I am doing circular logic, and I would only be confusing myself. I note that Einstein says nothing about length contraction in this experiment. He is apparently satisfied that he can explain the relativity of simultaneity to me without it.

This does not mean that I deny length contraction, I have just not discovered it yet.

I have found a slowdown of clocks, but all of it is meassured fom only one frame of reference, and I have been told it is really only the Doppler effect, and that there is an additional, relativistic slowdown of the clocks. I have still to find that one, too.

I haven't looked into length contraction yet.

I have re-read the paragraph of "Relativity" you pointed me to. I have not changed my mind. I will come back to you with a post where I explain my reading of it, so that you may point out faults in my reasoning.

Last edited: Oct 11, 2011

3. ### OnlyMeValued Senior Member

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rpenner, I'll second the above correction. I also understood his post as referring to "measuring" speed based on one point and my response was intended as agreement. But I did not go back and read either post, as I am now on the road with very slow data transfer rate.

5. ### OnlyMeValued Senior Member

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3,914
It would be a good idea to separate the concepts of time dilation and length contraction from the relativity of simultaneity, though they are not completely separated in practice.

To understand simultaneity requires only an understanding of how the observation of events is affected by the transmission of light, as viewed from different locations. What appears to happen at the same time from one location may appear to happen in a different order from a different location. Examples involving light are hard to make based on everyday experience. However, sound which travels far slower than light offers a real world experience that may be helpful.

During a lightening storm if you are near a lightening strike the flash and thunder happen at the same time or very close together. If you are some distance away from where the lightening strikes you will see the flash "essentially" when the lightening strikes, while the thunder will be delayed.

We know the thunder happened at the same time but there is a time delay based on distance for observes who are some distance apart. This is a time of sound delay.

In the case of special relativity it is the "time of light" delay that affects when different observers observe things as happening. When two events happen an observer between the events may observe them as happening at the same time while another observer in another location would observe them one before (or after) the other.

This is what I was referring to in the early posts when in mentioned time of light delays and it does not involve the Doppler effect.

The Lorentz Transformations come into play after this in two ways that become more complex and begin to deal with time dilation, length contraction and the twin paradox thought experiments. But it will be very helpful to work out a good understanding of how the time of light delay affects how events are observed to occur from separate locations (frames of reference, FoR), before diving into the more involved aspects involving time dilation, length contraction and the Lorentz Transformations (LT)

Just one more note. Once you grasp the above the LT can help to translate and reconcile observations from differing FoR. They help one observer understand what another observer "sees" when they are in different FoR.

*If I messed anything up here I am sure rpenner or one of the other knowledgeable posters can correct me. My access to the Internet will be limited and at times less than functional for a while yet.

7. ### DonQuixoteRegistered Senior Member

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66
The relativity of simultaneity

Janus58 an I are having a dispute about what can be agreed upon between different frames of reference. We disagree about what Einstein actually says. Janus58 has kindly pointed me in the direction of where I might find the right answer, but the problem remains, because it seems we are reading it very differently. The terms in which we frame(!) our disagreemet are very different, though. Janus58 takes length contraction as a given, but I am trying to discover it form Einsteins examples. Therefore, I decided to provide you all with my reading of the relevant passage from "Relativity", so that you can examine it, and tell me where I'm wrong in my considerations.

I have copied Einsteins text from the link supplied by Janus58. The only alteration I have made is to include a footnote in the running text (clearly marked with "footnote") and delete a number "3" that seemed to be a strange artifact of the copying operation.

So here goes Einstein:

This is undisputed

E does state that M' is at the midpoint of the distance A -> B on the train, although he does it in the form of a definition (Let M' be...). He never uses this definition for anything.
The point M' coincides with M. This will be true for M'in the moving frame, and for M on the embankment frame.It is not, however, stated that all the points A, B and M' coincides simultaneously with the points A, B and M on the embankment, considered from the reference of the train. At least I hope so, for if the observer at M' could assume that in his frame, A and B on the train coincides simultaneously with A and B on the embankment, and that M' is the midpoint, he would be justified to expect the light from A and B to reach him simultaneously, according to the postulates of relativity. Observing something else would be a breach of the postulates.

Note: I'm saying that Einstein does not say this.

All this is explicitly concidered from the frame of the embankment. (The "In reality" bit, I consider just to be a slip of the tongue).

This appears to be unwarranted unless you add: "as determined from the frame of the embankment".

This seems to be similarly unjustified.

Note: I am not saying it is wrong. It might very well be correct, I don't know. All I'm sayin' is that it doesn't follow from the experiment.

The simultaneity as meassured from the embankment frame of the correspondence of A, M' and B of the train with A, M and B on the embankment, is in itself an observation of simultaneity (at different points, I might add). If the same simultaneous correspondance could be determined from the moving frame, that would in itself be a demonstration that the same phenomenon is simultaneous in both frames. But this is not what is claimed.

However, Einstein reverts to using the result obtained from the embankment frame without carrying that caveat over to the conclusion. That is not justified in my book.

This is obviously bordering on crank territory, so please, shoot it down. Maybe I omitted something important?

8. ### CptBorkRobbing the Shalebridge CradleValued Senior Member

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Progress update​

Well I have some great news for those interested. As I mentioned, I've seen many objections to the conventionally-taught derivation of the Lorentz transformations, as first enunciated by Einstein (with some presumed influence/guidance from Poincare, Lorentz etc.). Although most of those objections are crap, there have been some reasonable objections that even though Maxwell's equations require that all inertial observers see light pulses in vacuum propagating at $c$, perhaps different observers are seeing different light pulses constituting different manifestations of the same EM field, and that's why they both see light always moving at the same speed.

I vaguely remembered some comments and deductions from my undergrad advanced mechanics class about the Lorentz invariance of Maxwell's equations, and I looked through a few books and lots of online notes and papers to try and find more info on the matter. Unfortunately, I wasn't able to find what I was looking for (I'm sure it's out there, but after failing to find it in Jackson Electrodynamics, which graduates often treat as the Bible of electromagnetism, I gave up on the search and decided to do it myself).

I have my old notes from the relevant class lying around somewhere, but I was able to find material on the web that looked pretty similar to what I remember seeing. Unfortunately, all of this material appears to merely assume the Lorentz transformations and then show how the EM fields must transform in order to keep Maxwell's equations invariant, and that's probably all we did in class too.

So like I said, I decided to take care of things myself, and the initial work is now finished. Firstly I used some simple arguments to show that the spacetime coordinates of events in the "moving" frame must depend on a linear combination of the corresponding coordinates in the "rest" frame, and similarly I showed that the relationship between EM fields in the "moving" and "rest" frames must also be linear, and exclusively dependent upon the content of those fields and the relative velocity between the two frames.

Next I used a series of simple symmetry arguments (rotating the axes in various ways), as well as the reciprocity between the two reference frames (either frame can be taken as the "rest" frame with equal validity), and the result cleaned up my equations dramatically. I then derived the most general possible coordinate transform (it reduces to Galilean transforms when certain classical assumptions are used, and reduces to Lorentz transforms when relativistic assumptions are used instead), and the most general possible transformations between the EM fields.

Finally, by requiring that Maxwell's equations in vacuum must hold in both frames, I was able to determine all the remaining undetermined constants, yielding both the correct form of the Lorentz transformations as well as the correct transformation rules for EM fields. Interestingly enough, I didn't even have to use all of Maxwell's equations in both frames- I used the four equations in the rest frame, and three of the equations in the moving (primed) frame, and the final condition $\nabla'\times\vec{B}'=\frac{1}{c^2}\frac{\partial \vec{E}'}{\partial t'}$ popped out as an automatic consequence without any a priori requirements.

So this portion with my original goals is completed, and I'll probably start my own thread going through everything in detail so as not to clutter this one. I'm not so sure anymore if this is what Lorentz and Poincare actually did. From what I've read, it sounds like they really struggled to produce something suitable, and although they derived the correct transformations in the end, it also sounds like they had to make all kinds of ad-hoc assumptions in order to do so. I'm sure many others have performed this derivation before- if not in Lorentz's time, then certainly since then- but like I say, I couldn't find it on the web or in the books I searched, so I figured it would be quickest and easiest to do from scratch, and then I'll post it on these forums for archival purposes (or maybe write up a complete LaTeX document eventually, when I've finished the thesis I'm supposed to be doing instead).

For those who don't like mathematics, all I can say is please submit your complaints and queries to nature- electrodynamics is a mathematical theory, and if one wants to argue for/against it, they will have to argue based on what the math actually says.

Plans for part 2​

Having deduced the rules for coordinate and EM field transforms directly from Maxwell's equations in vacuum, we can then apply Maxwell's equations in the presence of charges and currents in order to determine how electric charge and current densities transform between different frames. Next, given these transformation rules, we will be able to show that the total electric charge is invariant when switching between reference frames, even though the charge and current densities might vary. I already worked it out for the case where the "rest" frame observes a static charge density with no currents, but I've been doing some mental calculations and I think I'll be able to show it for general charge/current configurations too, although it'll be somewhat tricky.

Once we've established the invariance of total electric charge between different reference frames, we can then apply the Lorentz force law $\vec{F}=q\vec{E}+q\vec{u}\times\vec{B}$
($\vec{u}$ is the observed velocity of the charge, $q$) to determine how electromagnetic forces acting on massive particles transform between different reference frames, and thus how arbitrary forces acting on massive particles transform in general. We can then define the momentum of a moving particle in terms of these forces, determine the transformation rules for the momentum of EM fields and massive particles, and show that momentum conservation in the rest frame implies momentum conservation in all other frames.

At this point we can also show that the kinetic energy of a particle with rest mass $m_0$ must equal $\left(\gamma-1\right)m_0c^2$, where of course as will be shown in the writeup of part 1, $\gamma=\frac{1}{\sqrt{1-v^2/c^2}}$. Einstein interpreted $m_0c^2$ as the particle's total energy at rest, and $\gamma m_0c^2$ as its total energy in motion, but there were many objections to his interpretation. It was subsequently shown that this interpretation can be deduced by assuming that a system's centre-of-mass has a constant, uniform motion in the absence of any external forces, but this itself is still a questionable assumption.

Einstein then found another derivation of $E=mc^2$ which only requires momentum conservation (which we will establish first), and uniform motion of the centre-of-mass pops out as a secondary consequence. You can view his derivation here: http://en.wikipedia.org/wiki/Mass-energy_equivalence#First_correct_derivation_.281905.29

My problem with this derivation is that it only works in the near-classical limit, and further deductive steps are required to give a completely rigorous Relativistic derivation. I've already worked out such a derivation, and it's surprisingly not much more difficult than the near-classical derivation. However, to complete this rigorous derivation, we will not only need to know how the momentum of an EM field transforms, but also how its energy transforms- this is easy to do for an EM plane wave by applying the relativistic Doppler shift, but I'm hoping I'll be able to do it for general cases.

Summary of the overall objective​

The ultimate idea is to show that all the results of Special Relativity can be derived directly from basic physical reasoning, together with Maxwell's equations and the Lorentz electromagnetic force law. By assuming that Newton's laws apply to objects in any frame where such objects are seen at rest, and that the laws of physics (including Maxwell's equations) are the same in all inertial reference frames, we can derive the entire Special Theory of Relativity. Basically, with the proper foresight, Maxwell could have done it all by himself.

No need for "magical" assumptions like momentum conservation in arbitrary reference frames, or uniform motion of the centre-of-mass in the absence of external forces, or that anything moving at lightspeed must be moving at lightspeed in all other inertial frames. Instead, this can all be derived from more basic, fundamental postulates, such that if you want to argue with any of these results, you basically need to argue with Maxwell himself (or else argue that his equations only work on Earth). Modern physics education makes a lot of unjustified assumptions in order to expedite the learning process. Don't agree with the derivation they give for $E=\gamma m_0c^2$? Many physicists will probably tell you that it's been experimentally proven as a requirement for energy conservation in particle collisions and decays, but the point is that you don't need to know about any such experiments in order to rigorously deduce it, it all comes out as a necessary consequence of good ole' fashioned electromagnetism.

So I've finished the first part which is deriving the Lorentz transformations and the transformation rules for electromagnetic fields, and hopefully I'll be able to write them up in a separate post very soon. The rest can follow afterwards, and I'm totally open to suggestions, corrections, (reasonable) objections, and other ideas for improvement or self-doubt.

Last edited: Oct 12, 2011
9. ### DonQuixoteRegistered Senior Member

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66
Yes, I think they're very different beasts. But light signals is involved in all of them, so that's the similarity.

It seems to me that the logical form is very different between the length contraction and time dilation on one hand, and relativity of simultaneity on the other. The "vice versa" that is attached to all of them are different.

I really appreciate your good intentions when you try to explain this to me. I'm sorry to be so picky about what I will accept.

Last edited: Oct 11, 2011
10. ### TachBannedBanned

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5,265
No, they aren't different at all. All three are direct consequences of the Lorentz transforms. If you accept the Lorentz transforms as correct, the three are immediate consequences. To understand that you need to learn a little calculus, last I remember, you were going to learn math, right?

11. ### DonQuixoteRegistered Senior Member

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Yes. I am not there yet.

12. ### TachBannedBanned

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5,265
Then, you aren't ready for a simple, straightforward explanation.

13. ### DonQuixoteRegistered Senior Member

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That may, or may not be true. I have no way of establishing it. I have shown my line of reasoning. If your answer is "you're an idiot", I am not going to argue with you. If you don't want to engage in the discussion, that's all right with me.

Last edited: Oct 11, 2011
14. ### TachBannedBanned

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5,265
Words are not the appropriate language for physics explanations. The language of physics is math, so, if you want to converse in physics , you will need to learn math. For example, you have been exchanging lengthy posts with others without managing to reach any conclusion. The answers to your questions are expressible in a handful of formulas. Why are you so adverse to learning math?

15. ### DonQuixoteRegistered Senior Member

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I asked politely (my first post) if non-mathematical discussions were in order. I did not receive a confirmation that it was, but I did not hear a "no" either.

I am not adverse to learning math. But I haven't learned much so far. I think it may have someting to do with notation. I simply cannot remember what represents what. I could never learn to read music either. I do agree that it would be a *huge* advantage to know math.

The same goes for logical notation. I find it cumbersome. That doesn't mean, in my experience, that I am incapable of logical reasoning. But then your mileage may vary.

16. ### Janus58Valued Senior Member

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1,890
It is not required for A', M' and B' to all align simultaneously with A, M and B in the frame of the train. All that is required is that M' is at the midpoint between A' and B' when considered from the frame of the Train. And this is something that both frames agree upon. As long as M' is half way between A' and B' then the observer at M can expect light traveling to him from either point to take the same amount of time to reach him.
It is perfectly justified.

M is midway between A and B
M' is midway between A' and B'

At the moment that M align with M', A' aligns with A and B' aligns with B, when considered from the frame of the embankment, lightning strikes points A'A and B'B leaving marks at A A', B, and B' .

The light from said flashes reach M at the same time leading the observer at M to conclude that the strikes took place at the same time.

Also, when considered from the embankment, M' is moving towards B and away from A. M' intercepts the flash from B at some point between M and B (call it point C. ) and then sometime later, the flash from A catches up to M' when M' is at some point closer to B (call it point D).

Now consider things from the frame of the Train. when does the observer at M' see the flashes? He has to agree with the embankment frame that he sees one flash when he is next to point C of the embankment and Sees the other when he is next to Point D. Since C and D are at different points of the embankment, and it takes some time between when M is next to C and when it is next to D, as considered from the frame of the train, the Observer at M sees the flashes at different times.

M' is also halfway between A' and B', which are where the lightning strikes left char marks on the train. Thus, as considered from the train, the lightning struck points A' and B' of the train. If the flash reaches M' at different times and the distance to their sources are equal, then the only conclusion that can be made in the train frame is that the lightning strikes were not simultaneous.

The simultaneity as meassured from the embankment frame of the correspondence of A, M' and B of the train with A, M and B on the embankment, is in itself an observation of simultaneity (at different points, I might add). If the same simultaneous correspondance could be determined from the moving frame, that would in itself be a demonstration that the same phenomenon is simultaneous in both frames. But this is not what is claimed.

However, Einstein reverts to using the result obtained from the embankment frame without carrying that caveat over to the conclusion. That is not justified in my book.

This is obviously bordering on crank territory, so please, shoot it down. Maybe I omitted something important?[/QUOTE]

17. ### Neddy BateValued Senior Member

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1,589
According to the embankment frame, the simultaneous correspondence of A, M' and B of the train with A, M and B on the embankment is essentially the same observation of simultaneity as the lightning strikes. If there were no lightning strikes, and the observer at M wanted to know whether those points all corresponded at the same time, he would have to wait for the light showing their correspondence to reach him.

Thus, according to the train frame, the observation of non-simultaneity of the lightning strikes is also essentially the same observation as the non-correspondence of A, M' and B of the train with A, M and B on the embankment. If there were no lightning strikes, and the observer at M' wanted to know whether those points all corresponded at the same time, he would have to wait for the light showing their correspondence to reach him. In this case, M' would conclude that those points did not all correspond at the same time.

18. ### CptBorkRobbing the Shalebridge CradleValued Senior Member

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Ok, just wanted to update that I considered the relativistic transformation laws for completely general charge and current densities, and was able to show that the total electric charge $\int \mathrm{d}^3x\, \rho\left(\vec{x},0\right)$ is a Lorentz invariant (in addition to being conserved as a function of time in each frame). Was a bit tricky because you end up having to integrate over charge and current densities appearing at different positions and times, but by employing the conservation relationship between charge and current densities as a function of time, and a little fancy manipulation with multivariable calculus, things end up cancelling out in just the right way, as with many other things in physics. That will be useful when I want to start deriving things from the Lorentz force law, because I need the charges $q$ to be the same in both frames. I'll post that separate thread I promised, it's on the way!

19. ### DonQuixoteRegistered Senior Member

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66
You are of course right. M' must only be able to do the same meassurement in his frame as M does on the embankment.

This is one thing it seems I just couldn't get my head around. It seemed to me as just something E said, but never did anything with. I see now that it is crucial for M' to be able to know he is in the middle of A' B'.

All in all, I will conclude that my reading of the paragraph was misguided.
I suspected as much.

Thank you very much for taking the time pointing it out to me.

20. ### DonQuixoteRegistered Senior Member

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That is what I realised, but it seems this realisation did me no good.

21. ### rpennerFully WiredRegistered Senior Member

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4,833
Education: What fixes the form of the given Lorentz transform?

1) All Lorentz transforms must preserve the speed of light in any direction. (Second postulate of special relativity, consistency of the speed of light)
2) Because of the principle of relativity, it shouldn't matter if we are are the first or second observer, so an inverse Lorentz transform should be be in the same form as the second.

The next parts come because we wish to parametrize the transform a certain way in terms of v in the +x direction.

3) The given Lorentz transform must cancel a velocity of v in the +x direction.
4) The inverse transform should simply have -v substituted for v
5) Therefore the Lorentz transform should transform zero motion into velocity v in the -x direction.
6) This transform should be the identity transform in the case v = 0
7) The transform should be a smooth function of v

Since speed and the coordinate difference between two events are related we have:
$\begin{pmatrix} \Delta x' \\ \Delta y' \\ \Delta z' \\ \Delta t' \end{pmatrix} = \Delta t' \begin{pmatrix} u'_x \\ u'_y \\ u'_z \\ 1 \end{pmatrix} = { \Huge \Lambda} \begin{pmatrix} \Delta x' \\ \Delta y' \\ \Delta z' \\ \Delta t' \end{pmatrix} = \Delta t \begin{pmatrix} \Lambda_{xx} & \Lambda_{xy} & \Lambda_{xz} & \Lambda_{xt} \\ \Lambda_{yx} & \Lambda_{yy} & \Lambda_{yz} & \Lambda_{yt} \\ \Lambda_{zx} & \Lambda_{zy} & \Lambda_{zz} & \Lambda_{zt} \\ \Lambda_{tx} & \Lambda_{ty} & \Lambda_{tz} & \Lambda_{tt} \end{pmatrix} \begin{pmatrix} u_x \\ u_y \\ u_z \\ 1 \end{pmatrix}$

Consistency of the speed of light requires that $u_x ^2 + u_y ^2 + u_z ^2 = c^2$ implies $u'_x ^2 + u'_y ^2 + u'_z ^2 = c^2$ or

$\left( c \begin{pmatrix} \Lambda_{xx} & \Lambda_{xy} & \Lambda_{xz} \\ \Lambda_{yx} & \Lambda_{yy} & \Lambda_{yz} \\ \Lambda_{zx} & \Lambda_{zy} & \Lambda_{zz} \end{pmatrix} \begin{pmatrix} \cos \theta \\ \sin \theta \cos \phi \\ \sin \theta \sin \phi \end{pmatrix} + \begin{pmatrix} \Lambda_{xt} \\ \Lambda_{yt} \\ \Lambda_{zt} \end{pmatrix} \right)^2 = \left( c^2 \begin{pmatrix} \Lambda_{tx} \\ \Lambda_{ty} \\ \Lambda_{tz} \end{pmatrix} \cdot \begin{pmatrix} \cos \theta \\ \sin \theta \cos \phi \\ \sin \theta \sin \phi \end{pmatrix} + c \Lambda_{tt} \right)^2$

To cancel a velocity of v in the +x direction, we require
$v \begin{pmatrix} \Lambda_{xx} & \Lambda_{xy} & \Lambda_{xz} \\ \Lambda_{yx} & \Lambda_{yy} & \Lambda_{yz} \\ \Lambda_{zx} & \Lambda_{zy} & \Lambda_{zz} \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} v \Lambda_{xx} \\ v \Lambda_{yx} \\ v \Lambda_{zx} \end{pmatrix} = - \begin{pmatrix} \Lambda_{xt} \\ \Lambda_{yt} \\ \Lambda_{zt} \end{pmatrix}$
Thus $\Lambda_{xt} = -v \Lambda_{xx} , \; \Lambda_{yt} = -v \Lambda_{yx} , \;\Lambda_{zt} = -v \Lambda_{zx}$

Likewise to transform zero velocity to -v in the +x direction we should have
$\begin{pmatrix} \Lambda_{xt} \\ \Lambda_{yt} \\ \Lambda_{zt} \end{pmatrix} = \begin{pmatrix} -v \\ 0 \\ 0 \end{pmatrix} \Lambda_{tt}$
Thus $\Lambda_{xt} = -v \Lambda_{tt} = -v \Lambda_{xx} , \; \Lambda_{yt} = 0 , \;\Lambda_{zt} = 0$

So consistency of the speed of light is now:
$\left( c \begin{pmatrix} \Lambda_{xx} & \Lambda_{xy} & \Lambda_{xz} \\ 0 & \Lambda_{yy} & \Lambda_{yz} \\ 0 & \Lambda_{zy} & \Lambda_{zz} \end{pmatrix} \begin{pmatrix} \cos \theta - \frac{v}{c} \\ \sin \theta \cos \phi \\ \sin \theta \sin \phi \end{pmatrix} \right)^2 = \left( c^2 \begin{pmatrix} \Lambda_{tx} \\ \Lambda_{ty} \\ \Lambda_{tz} \end{pmatrix} \cdot \begin{pmatrix} \cos \theta \\ \sin \theta \cos \phi \\ \sin \theta \sin \phi \end{pmatrix} + c \Lambda_{tt} \right)^2$

For $\theta = 0$ this is: $(c - v)^2 ( \Lambda_{xx}^2 ) = c^2 ( c \Lambda_{tx} + \Lambda_{tt})^2$
For $\theta = \pi$ this is: $(c + v)^2 ( \Lambda_{xx}^2 ) = c^2 ( -c \Lambda_{tx} + \Lambda_{tt})^2$
So we can say that $\left( \Lambda_{tt} + v \Lambda_{tx} \right) \left(c^2 \Lambda_{tx} + v \Lambda_{tt} \right) = 0$ Since we want v=0 to be the identity matrix we now know $\Lambda_{tx} = -\frac{v}{c^2} \Lambda_{tt}$

For $\theta = \frac{\pi}{2}$ we have:

$\left( c \begin{pmatrix} \Lambda_{xx} & \Lambda_{xy} & \Lambda_{xz} \\ 0 & \Lambda_{yy} & \Lambda_{yz} \\ 0 & \Lambda_{zy} & \Lambda_{zz} \end{pmatrix} \begin{pmatrix} - \frac{v}{c} \\ \cos \phi \\ \sin \phi \end{pmatrix} \right)^2 = \left( \begin{pmatrix} -v \Lambda_{xx} \\ c^2 \Lambda_{ty} \\ c^2 \Lambda_{tz} \end{pmatrix} \cdot \begin{pmatrix} 0 \\ \cos \phi \\ \sin \phi \end{pmatrix} + c \Lambda_{xx} \right)^2$
For $\phi = 0$ this is : $\left( -v \Lambda_{xx} + c \Lambda_{xy} )^2 + c^2 \Lambda_{yy}^2 + c^2 \Lambda_{zy}^2 = ( c \Lambda_{xx} + c^2 \Lambda_{ty} )^2$
For $\phi = \pi$ this is : $\left( v \Lambda_{xx} + c \Lambda_{xy} )^2 + c^2 \Lambda_{yy}^2 + c^2 \Lambda_{zy}^2 = ( c \Lambda_{xx} - c^2 \Lambda_{ty} )^2$
Therefore $\Lambda_{ty} = - \frac{v}{c^2} \Lambda_{xy}$
Working with $\phi = \pm \frac{\pi}{2}$ we come to the analogous $\Lambda_{tz} = - \frac{v}{c^2} \Lambda_{xz}$, and we have:
${ \Huge \Lambda}(v) = \begin{pmatrix} \Lambda_{xx} & \Lambda_{xy} & \Lambda_{xz} & -v \Lambda_{xx} \\ 0 & \Lambda_{yy} & \Lambda_{yz} & 0 \\ 0 & \Lambda_{zy} & \Lambda_{zz} & 0 \\ \frac{-v}{c^2} \Lambda_{xx} & \frac{-v}{c^2} \Lambda_{xy} & \frac{-v}{c^2} \Lambda_{xz} & \Lambda_{xx} \end{pmatrix}$
And by solving the inverse we have:
${ \Huge \Lambda^{-1}}(v) = \begin{pmatrix} \frac{ \Lambda_{xx}^{-1} }{1 - \frac{v^2}{c^2}} & \Lambda_{xx}^{-1} \frac{ \Lambda{xz} \Lambda{zy} - \Lambda_{xy} \Lambda{zz}}{ \Lambda_{yy} \Lambda{zz} - \Lambda{yz} \Lambda{zy} } & \Lambda_{xx}^{-1} \frac{ \Lambda{xy} \Lambda{yz} - \Lambda_{xz} \Lambda{yy}}{ \Lambda_{yy} \Lambda{zz} - \Lambda{yz} \Lambda{zy} } & v \frac{ \Lambda_{xx}^{-1} }{1 - \frac{v^2}{c^2}} \\ 0 & \frac{ \Lambda{zz}}{ \Lambda_{yy} \Lambda{zz} - \Lambda{yz} \Lambda{zy} } & - \frac{ \Lambda{yz}}{ \Lambda_{yy} \Lambda{zz} - \Lambda{yz} \Lambda{zy} } & 0 \\ 0 & - \frac{ \Lambda{zy}}{ \Lambda_{yy} \Lambda{zz} - \Lambda{yz} \Lambda{zy} } & \frac{ \Lambda{yy}}{ \Lambda_{yy} \Lambda{zz} - \Lambda{yz} \Lambda{zy} } & 0 \\ \frac{v}{c^2} \frac{ \Lambda_{xx}^{-1} }{1 - \frac{v^2}{c^2}} & 0 & 0 & \frac{ \Lambda_{xx}^{-1} }{1 - \frac{v^2}{c^2}} \end{\pmatrix} = { \Huge \Lambda}(-v)$
So we have $\Lambda_{xx} = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} , \; \Lambda{xy} = \Lambda{xz} = 0, \; \Lambda_{yy} = \Lambda_{zz} = \cos \psi(v) , \; \Lambda_{yz} = - \Lambda{zy} = \sin \psi(v)$ where $\psi(v)$ is any odd function of v. Notably, $\psi(v) = 0$ gives us our standard Lorentz tranformation, and $\psi(v) = k v$ gives a corkscrew transformation that rotates the y-z plane at progressively higher velocities.

So if we add the following constraint on geometry
8) Where possible, we want the x, y, and z directions for both observers to correspond
then we fix $\psi(v) = 0$ and recover the standard Lorentz transform for all velocities.

This freedom suggests that a Lorentz transformation in the X direction commutes with any rotation about the X axis, but I haven't yet prepared to discuss the general concept of commuting transforms.

Last edited: Oct 13, 2011
22. ### rpennerFully WiredRegistered Senior Member

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Education: So, since it is more special than the postulates of special relativity alone, does the Lorentz transform, as given in the x direction, actually have different invariants than special relativity?

Yes. Special relativity ( 1 + 2 ) preserves $I = c^2 (\Delta t)^2 - (\Delta x)^2 - (\Delta y)^2 - (\Delta z)^2$, while adding ( 3, 4, 5, 6, 7 ) which constrains the direction of movement preserves $J = (\Delta y)^2 + (\Delta z)^2$. Adding ( 8 ) constrains the relations even more and constrains $K = \tan^{-1} \frac{\Delta z}{\Delta y}$.
If $I$ and $J$ are invariants, so is $I + J = c^2 (\Delta t)^2 - (\Delta x)^2$. So if we write:
$\begin{pmatrix} \Delta x' \\ \Delta y' \\ \Delta z' \\ \Delta t' \end{pmatrix} = \begin{pmatrix} \sqrt{I + J} \sinh \alpha ' \\ \sqrt{J} \cos K \\ \sqrt{J} \sin K \\ \frac{1}{c} \sqrt{I + J} \cosh \alpha ' \end{pmatrix} = \begin{pmatrix} \cosh \tanh^{-1} \frac{v}{c} & 0 & 0 & -c \sinh \tanh^{-1} \frac{v}{c} \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ - \frac{1}{c} \sinh \tanh^{-1} \frac{v}{c} & 0 & 0 & \cosh \tanh^{-1} \frac{v}{c} \end{pmatrix} \begin{pmatrix} \sqrt{I + J} \sinh \alpha \\ \sqrt{J} \cos K \\ \sqrt{J} \sin K \\ \frac{1}{c} \sqrt{I + J} \cosh \alpha \end{pmatrix} = { \Huge \Lambda } \begin{pmatrix} \Delta x' \\ \Delta y' \\ \Delta z' \\ \Delta t' \end{pmatrix}$
This reduces to : $\alpha' = \alpha - \tanh^{-1} \frac{v}{c}$ which is one of the many ways to justify the term hyperbolic rotation when describing the Lorentz transform.

$\frac{(\sinh a \, \pm \, \tanh b \, \cosh a)}{\sqrt{1 - \tanh^2 b }} = \frac{\sinh a \, \pm \, \cosh a \tanh b }{ \textrm{sech} b } = \sinh (a \pm b)$
$\frac{(\cosh a \, \pm \, \tanh b \, \sinh a)}{\sqrt{1 - \tanh^2 b }} = \frac{\cosh a \, \pm \, \sinh a \tanh b }{ \textrm{sech} b } = \cosh (a \pm b)$

23. ### rpennerFully WiredRegistered Senior Member

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4,833
Education: So what does a general Lorentz transform look like?

A general Lorentz transform does not constrain the boost to be in the x direction and it doesn't constrain the x,y,z directions in one frame to point in the same directions as the x,y,z directions in another.
Symbolically, ${\huge \Lambda } ( \Theta, \Phi, \Psi, v, \theta, \phi ) = { \huge R }(\Theta, \Phi, \Psi ) { \huge R } (\theta, \phi, 0 ) { \huge L }(v) { \huge R^{-1} } (\theta,\phi, 0 ) = { \huge R }(\Theta, \Phi, \Psi ) { \huge R } (\theta, \phi, 0 ) { \huge L }(v) { \huge R } (0,-\phi, -\theta )$

$R$ is a Euclidean rotation, realigning the axes any which way. There are many ways to parameterize this, but all require 3 parameters. One example is:
$R(\Theta, \Phi, \Psi) = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & \cos \Theta & \sin \Theta & 0 \\ 0 & - \sin \Theta & \cos \Theta & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \cos \Phi & 0 & -\sin \Phi & 0 \\ 0 & 1 & 0 & 0 \\ \sin \Phi & 0 & \cos \Phi & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & \cos \Psi & \sin \Psi & 0 \\ 0 & - \sin \Psi & \cos \Psi & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$
$L$ is just the Lorentz transform in the x direction.
$L = \begin{pmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & \frac{1}{c} \end{pmatrix} \begin{pmatrix} \quad \cosh \, \tanh^{-1} \, \frac{v}{c} \quad & 0 & 0 & \quad -\sinh \, \tanh^{-1} \, \frac{v}{c} \quad \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \quad -\sinh \, \tanh^{-1} \, \frac{v}{c} \quad & 0 & 0 & \quad \cosh \, \tanh^{-1} \, \frac{v}{c} \quad \end{pmatrix} \begin{pmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & c \end{pmatrix}$

$\Lambda = {\small \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & \cos \Theta & \sin \Theta & 0 \\ 0 & - \sin \Theta & \cos \Theta & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \cos \Phi & 0 & -\sin \Phi & 0 \\ 0 & 1 & 0 & 0 \\ \sin \Phi & 0 & \cos \Phi & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & \cos \Psi & \sin \Psi & 0 \\ 0 & - \sin \Psi & \cos \Psi & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & \cos \theta & \sin \theta & 0 \\ 0 & - \sin \theta & \cos \theta & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \cos \phi & 0 & -\sin \phi & 0 \\ 0 & 1 & 0 & 0 \\ \sin \phi & 0 & \cos \phi & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \quad \cosh \, \tanh^{-1} \, \frac{v}{c} \quad & 0 & 0 & \quad -c \, \sinh \, \tanh^{-1} \, \frac{v}{c} \quad \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \quad - \frac{1}{c} \, \sinh \, \tanh^{-1} \, \frac{v}{c} \quad & 0 & 0 & \quad \cosh \, \tanh^{-1} \, \frac{v}{c} \quad \end{pmatrix} \begin{pmatrix} \cos \phi & 0 & \sin \phi & 0 \\ 0 & 1 & 0 & 0 \\ - \sin \phi & 0 & \cos \phi & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & \cos \theta & - \sin \theta & 0 \\ 0 & \sin \theta & \cos \theta & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} }$

This is painful to multiply out but $R (\theta, \phi, 0 ) L(v) R (0,-\phi, -\theta )$ is fully equivalent to the general formula which keeps the axes aligned which Tach has been linked to since post 90, perhaps earlier.