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Proof that Classical Action is Invariant in a Galilean Transformation
by Miles Mathis
Abstract In their paper entitled "Symmetries and Conservation Laws: Consequences of Noether's Theorem" [American Journal of Physics, accepted 23 May 2003] authors Jozef Hanc, Slavomir Tuleja and Martina Hancova provide equations to support current wisdom that classical action is not invariant in a Galilean transformation. They present this lack of invariance as being "surprising," and ask, "Is the Galilean transformation incorrect?" I prove that the specific transformation used in their paper and in most other 20th and 21st century papers is not correct. I then provide Galilean transforms that are both correct and invariant.
The Galilean Transformation.
In their paper entitled “Symmetries and Conservation Laws: Consequences of Noether’s Theorem” [American Journal of Physics, accepted 23 May 2003] authors Jozef Hanc, Slavomir Tuleja and Martina Hancova present an introduction to action, symmetry and other related topics. It is their exposition of action that concerns me here, specifically their claim that classical action is not invariant under a Galilean transformation.
I do not claim that these authors are in any way unusual or iconoclastic in their explanations, either mathematically or conceptually. They lucidly set down what is now accepted by most of those interested in the problem. In fact, I chose their paper to critique for this very reason. It is beautifully transparent where most other explanations of action are not. The reader is presented with simple algebra, the Lagrangian being put into common variables, and integrals simplified into averages. This presentation I have no quarrel with. In many ways I can imagine that it is quite helpful to students. It will also help me in showing the simple algebraic error in the Galilean transform used by the authors and by many others.
They begin the section in question by offering the basic equation for the action S of a particle when no potential exists:
S = ½ m (x_{2} – x_{1})^{2}/(t_{2} – t_{1})
Then we do a Galilean transform. The authors imagine a rocket flying by slowly (slowly so that the transform does not become relativistic) and we want S' from the point of view of the rocket. We are told that the transforms are these:
x' = x  vt t' = t
This makes the solution
S' = ½ m (x'_{2}  x'_{1})^{2}/(t'_{2}  t'_{1})  vm(x'_{2}  x'_{1}) + ½ mv^{2}(t'_{2}  t'_{1})
I am not concerned with their solution except to notice that the action is not invariant. S ≠ S'. The problem is with the first transform and may be stated briefly. The authors are likely to have imported the equation from a paper of Einstein, or to have taken it from a paper that did so itself. Their acceptance of it is therefore easy to understand. Einstein himself told us in "On the Electrodynamics of Moving Bodies" [Annalen der Physik, 1905] that this equation is the Galilean transform. Special Relativity is derived from this equation in that paper. Unfortunately this equation, as used, is not a Galilean transform. It is assumed to be for this reason: the transform that would actually have been used by Galileo, Newton, or even a young Maxwell, is
x' = x  a
or some simple variation of that. The variable a is the distance between the origins of the two coordinate systems. Since a is a distance, it seems logical to assume that it may also be expressed vt. But it can't. The Galilean transform x' = (x  a) allows you to transform a point in one system to a point in another. Neither of those two x variables is a Δx. Therefore velocity is no part of the transform. The Galilean transform is not concerned with motion. It is not concerned with time or velocity. It only transforms points on a graph. To be even more rigorous, notice that in the equation x' = (x  a) the two x variables must be found at the same time. They are two points, one in the primed field, one in the unprimed field. They are separated by a distance a but they are separated by no time. In Galilean transforms, distance separation does not imply a necessary time separation, since the speed of light was thought to be infinite. Light could travel an x in no t. Therefore Galilean transforms do not take into account relativistic transforms, obviously. What this means is that the equation x' = (x  a) has no time component. It is an equation at an instant. You therefore cannot substitute vt for a, since a velocity requires a time. This is an algebraic equation, not a calculus equation. You cannot have a velocity at an instant in an algebraic equation, since there is no way to expand it. It is meaningless. Furthermore, the equation x' = x  vt is not a sensible equation since it proposes to show that x ≠ x' . But if t = t' , then x must equal x' . There can be no length contraction if there is no time dilation. We are not in a relativistic situation here. The rocket cannot measure a greater Δx than the object itself. That is why we had the rocket go slowly in the first place. A Galilean transform is Galilean (instead of Lorentzian) because in the time of Galileo light was thought to have an infinite speed. If light has an infinite speed, then the rocket will see everything at exactly the same time as the object itself. There can be no transform due to velocity. Therefore a Galilean transform with a velocity variable in it is a contradiction in terms. It is impossible. The equation x' = x  vt has also been called Newton's Principle of Relativity. But it is not that either. Neither Newton nor Galileo could have had any possible use for a transform caused by velocity, since constant velocity was not believed to make any difference in measurement, no matter how great the velocity was. A transform between coordinate systems was made necessary by distance only; velocity was not a factor.
I have shown that x' = x  vt cannot be true if the x variables stand for points, since a velocity cannot have anything to do with transforming a point on one graph to a point on another graph when these graphs are Galilean. But the equation also cannot be thought of as Δx' = Δx  vΔt, since an observer in a Galilean system cannot possibly find any length contraction. In all Galilean systems with constant velocities, Δx' = Δx.
You can see that what the authors mean by t = t' is really Δt = Δt' . We are being given a period equivalence. Time is going at the same rate in both coordinate systems. If this is true, then the analogous transform for distance between the same two systems should also be in terms of delta variables. That is, Δx = Δx' . The points in the two systems may be different, if we have different origins. But distances are the same. We can make the points equal, too, just by making the origins equal. Then the rocket and the object are in the same system and we don't even need a transform at all. This is the way that Galileo or Newton or anyone before Maxwell would really have solved. Put both objects in the same graph and do direct subtractions of points. Then you can see that action is truly invariant, since it becomes a tautology. The authors ask if the Galilean transform is incorrect. The Galilean transform x' = x  vt is incorrect. But we can provide Galilean transforms that are correct and that provide invariance. They are: Δt = Δt' Δx = Δx' Δy = Δy' Δz = Δz' x' = (x  a) is also true and provides invariance, if and only if the x variables are understood to be points measured at the same time. Of course, this means that S' = ½ m (Δx')^{2}/(Δt') = ½ m (Δx)^{2}/(Δt) = S
This paper was sent to the American Journal of Physics in early 2005. In July I received this reply from the editors:
The author brings up an interesting point: Did Galileo (or Newton, "or even a young Maxwell") make use of what we call the Galilean transformation? The author says "the transform actually used by scientists before relativity is x' = (x  a) or some simple variation of that." Here the symbol "a" stands for a distance under the assumption of simultaneity. There are no references in this paper, so we are given no evidence for these historical statements. He continues, saying that time cannot enter this equation and therefore neither can velocity, "since a velocity cannot have anything to do with transforming a point on one graph to a point on another graph when these graphs are Cartesian or when the spaces are Galilean." The entire paper concerns points in space.
We are talking here about a description of motion, and motion is described by EVENTS marking the sequential locations of, say, a particle in spacetime, most simply x and t. Events also dominate quantum physics and general relativity. Events are the nails on which physics hangs. The equation x' = x  a, where "a" is a distance and time is simultaneous has no possible chance of describing motion. I cannot wordsearch this pdf document, but have not found a single use of the word EVENT in this present manuscript.
This paper is disqualified for publication in AJP because the formalism the author proposes has no chance of describing motion. If the author wishes to convince us that the Galilean transformation would have been foreign to Galileo, Newton, and possibly even the young Maxwell, let him make the case with a study of historical context and references. The editors can then decide if this is a useful footnote on the history of physics and appropriate to the AJP.
AMERICAN JOURNAL of PHYSICS Jan Tobochnik, Editor Harvey Gould, Associate Editor Martin Ligare, Assistant Editor Jennifer Perry, Assistant to the Editor Anonymous Referee's review: This manuscript argues that the majority of postEinsteinian physicists are misinterpreting the concept of a Galilean transformation. The current interpretation as the limit of the Lorentz transformation as the velocity of light approaches infinity (or, as some see it, a contraction of the Lorentz transformation) is incorrect because the action of a free particle is not invariant under such a transformation. The author argues that the correct transformation is a static translation in space rather than one that moves with the frame of a moving inertial observer. The content is directly applicable to the undergraduate classroom and the manuscript is written so that this audience may understand the arguments.
I would recommend against publication on three grounds. I disagree with the premise. Firstly, the action of a free particle should not be invariant under what we currently call a Galilean transformation. The measurements of energy will differ since one frame has kinetic energy relative to the other. However, a variational principle on both actions will yield the same equations of motion (or force laws), as pointed out by [Hanc, et al., Am. J. Phys. 72 (2004), 428 435], the article the author critiques for committing the misinterpretation. Secondly, the proposed replacement Galilean transformation is not physically interesting. It's just a translation in space.
I am not an historian of science so I cannot verify how Galilean transformations were viewed prior to the turn of the twentieth century. If, indeed, we all have it wrong, this manuscript should contain documentation to that effect.* That said, such a manuscript with historical backing is more appropriate as a brief article in a history of science journal, a feature in an educational journal such as Physics Teacher, or a letter to the editor of a trade publication, such as Physics Today.
This was my reply:
Dear Editors,
I have to admit I was astonished at the form your dismissal of my paper took. It seems to me that you and your referee have not really addressed the content of my paper. Your referee's main point is that he disagrees with my premise, that premise being that action is invariant in this Galilean equation. But that is not my premise, that is my conclusion, a conclusion I argue for with simple mathematical and logical steps. He does not address any of those steps or even attempt to refute them. Nor does he refute my actual premises, which are 1) that the x variable in the equation in question must be either a point or a length, 2) that in Galilean equations the speed of light must be infinite, 3) that there is no possibility of length contraction or time dilation in Galilean equations. His only substantive argument is that kinetic energy cannot be invariant across a Galilean transformation, an argument that betrays a serious misunderstanding of kinematics and of the variables in this equation. This equation does not even allow one to calculate a transform for kinetic energy, since it only has one velocity variable in it. Galilean equations do not allow for mass increase, by definition. So to find a difference in kinetic energy, you must find a difference in velocity. Those are the only two variables in a kinetic energy equation. To find a kinetic energy transform, you must be given or be able to assign two velocities. I encourage you to simply do a kinetic energy transform with classical coordinate systems. It is very simple. But you will find that you need two velocities. This is only to be expected, since you have to transform from one velocity to the other. But the equation we are talking about here has only one velocity variable in it. How are you going to transform velocity? The only way you can get two velocities is if you let the particle measure its own velocity relative to the observer. Then you would have the particle's measurement and the observer's measurement. But in this case, you are already into the realm of Relativity. Before Relativity, you could not even pose the question in this way. Before Relativity there was no possibility of any difference between the particle's measurements and the observer's measurements. That is what Relativity is. Before Einstein, the velocities must be equal in this situation. There is no possibility of a transform. The particle and the observer both see everything the same, since they both see with light, and the light is travelling infinitely fast. Before Einstein, no one ever thought that the observer and the particle could possibly see things differently, so they didn't even think to include that in their math. I do not need to prove this by giving footnotes to Galileo or Maxwell. It is or should be something every schoolboy knows. All I am saying is that before Relativity there was no Relativity. Where is the necessity of footnotes here? It is not an historical question, it is a question of logic. A Galilean transform cannot assume or imply length differences or time differences. That is why it is Galilean in the first place. Therefore, if you have a Galilean transform like this, and you want to transform kinetic energy only between the observer's field and the particle's field, then the equation truly is invariant, by a tautology, since according to all preEinstein math they were equivalent. There could be no difference, by definition. Before Einstein, kinetic energy transforms were done between two moving systems, both of which were moving against the universal background. We do not have that situation here, since the Galilean transform I am critiquing in this paper only has one velocity in it. The Lorentz transforms can find kinetic energy differences with only one given velocity, since in Relativity the difference is between the particle and the observer. But in a Galilean transform this is impossible. There is no difference between the particle's field and the observer's field. Therefore you must be given two velocities to calculate a kinetic energy transform.
Next your referee says, "the proposed replacement Galilean transformation is not physically interesting. It's just a translation in space." That is just my point. A real Galilean transformation was only a translation. It dealt with points, not distances. Whether that is "physically interesting" is not pertinent. The point of science is not to be provocative or sexy, it is to be consistent and correct. The current interpretation of the Galilean transform is inconsistent because the x variables in the equation must be either points or distances. If they are points, then we just have a translation in space, a translation your referee calls uninteresting. If they distances, then the equation must mean that x is not equal to x' . If x is not equal to x' , then you have length contraction. If you have length contraction then you do not have a Galilean transformation. It is that simple.
Finally, in your own response, you say that I offer no evidence to support my contention. You imply that historical quotes would be more convincing than simple mathematics and logic. As a scientist I find this amazing. I have not written a historical treatise here and don't intend to make it into one. I am making a mathematical and logical argument, which argument is science. My postulates don't require footnotes, since they are either common sense or things we were all taught in high school. I should not have to footnote an assertion that kinematic equations before Einstein did not contain relativistic assumptions. Then you critique my paper for not using the word "event." I don't see how this is to the point, since the paper I am responding to also does not center around events, and yet it was published. Classically, an event takes place at a time and a place, and my equations have variables for time and place. Besides that, in what sense is velocity an event? All we have in this problem is body moving by at constant velocity. Where is the event? I am aware that you can always speak in terms of events if you like, but I don't see how that that would benefit this paper. You say, "The equation x' = (x  a), where 'a' is a distance and time is simultaneous has no possible chance of describing motion," and you dismiss my paper on that basis. But in saying so, you have chosen to ignore my point, that being that because this equation has no chance of describing motion, it cannot be analogous to the Lorentz transform and cannot be the equation that the Lorentz transform resolves to classically. I suspect that you can see this, and that the real reason you have dismissed my paper in such a strange way is that this implication frightens you. You can see that this paper has implications for Relativity and you do not want to go there. I assure you that I am not a classicist and that I have no intention of arguing against the theory of Relativity. I believe in time dilation and length contraction and mass increase. But this misunderstanding about the Galilean transform must be cleared up. Its misuse has caused serious problems in kinematics, and its continued misuse a century after its birth is a mathematical and physical embarrassment to history. Your publication of this paper would be a great benefit to the field. The subject must be reopened. I agree that you may take some hits in the short term from those in the mainstream who do not like to be contradicted, but in the long term it is papers like this that make the name of journals like yours. I beg you to try again to comprehend my thesis. You are missing something crucial here, and you are mistaken to accept the review of this referee.
PS: As proof that the equation x' = (x  a) is universal knowledge mentioned without footnotes to this day I refer you to Richard Feynman's Six NotsoEasy Pieces, equation 1.2. There he uses it as a Newtonian transform in precisely the way I do, and he assigns it to Newton and to classical mechanics without footnote or quote. The same equation or a simple variation of it would be found in most high school physics textbooks in the 20th century, and I assume still is.
The Editors reviewed my appeal in this way:
Dear Prof. Mathis,
Below and attached you will find copies of the reviewers' reports on your manuscript "WHY CLASSICAL ACTION MUST BE INVARIANT IN A GALILEAN TRANSFORMATION," our manuscript #18452. I agree with the reviewers. Either your manuscript is just plain wrong or so misleading that we cannot consider it for publication in AJP. Thus, we cannot consider any further appeals.
Sincerely,
Jan Tobochnik
Review#1 Jan: I hold these truths to be selfevident to any practicing physicist:
1. The Lorentz transformation transforms events and can be used to compare the description of the motion of a particle as observed with respect to two reference frames in uniform relative motion.
2. The conventional Galilean transformation used in all modern texts, whatever its historical source, is the limit of the Lorentz transformation when all relative speeds are very much smaller than c. In that limit it also can transform events and describe motion of a particle as observed with respect to two reference frames in uniform relative motion.
3. The translation of a point in space described by the equation x' = x  a with no reference to time has no ability to describe motion in any frame, much less the transformation of that motion between frames in relative motion.
4. Any discussion of the transformation properties of action is necessarily a discussion of motion and therefore cannot be carried out using the transformation x' = x  a for points in space.
Review #2
Referee's response to appeal: Manuscript 18452 In my report I described the main conclusion of the manuscript as a "premise," (as when a barrister begins an argument with a premise that is to be substantiated). I agree that in scientific writing this is poor usage, as "premise" (in the first definition supplied by the OED) is also the correct term for A in the propositional sentence "A implies B." I apologize for being sloppy.
To get to the heart of the argument, let's do an example. There is a large rock of mass one lmru (Large Rock Mass Unit) in the garden that does not move. There are two observers, one standing on the rock and one running past the rock at 10 m/s. When the running observer passes the rock they both take measurements and simultaneously start the clock. One second later both observers take second measurements. The stationary observer notes that the rock has zero velocity and thus zero kinetic energy. The moving observer sees the rock moving backwards relative to him at a speed of 10 m/s and thus measures a kinetic energy of 50lmru·m^{2}/s^{2}. Both the author and I can agree that the kinetic energy is not conserved when transforming from the frame of the stationary observer to that of the running observer. The issue may be whether the transformation from stationary frame to moving frame represents a Galilean transformation. The author never supplies a definition but he does demand that a Galilean transformation satisfy 1) that the x variable must be a point, 2) that the speed of light is infinite, and 3) that there is no possibility of length contraction nor time dilation. The example above meets all three criteria. Furthermore, if the primes represent the coordinates of the rock in the moving observer's reference frame, then t_{1} = t'_{1} = 0, t_{2} = t'_{2} = 1 s, x_{1} = x'_{1} = 0, x_{2} = 0, and x'_{2} = −10 m, which satisfy the equations x'_{i} = x_{i} − vt_{i} t'_{i} = t_{i} where i = 1, 2 and v is the relative velocities of the frames. Furthermore they satisfy Δx' = Δx − vΔt Since my choice of one second is arbitrary, I can replace it with any t'_{2} = t_{2} = t. Thus the equations in question (a) represent a Galilean transformation 1 and (b) cannot be adequately described if I set vt to a constant a. So I see no problem with the current interpretation of Galilean transformations. Also in contradiction to the manuscript, we have a transformation that meets his criteria yet x' ≠ x and Δx' ≠ Δx. (Note that these inequalities do not imply in any way that space has been contracted.)
Again, I am no historian of science, but I know that both Euler and Lagrange considered such transformations at least in the context of moving rigid bodies. They even had relative rotations in their transformations. (To ascertain whether they actually wrote down the equations as we do is not germane.) Thus the "moving frame" velocitydependent Galilean transformation dates back well before the advent of special relativity.
The author answers his own appeal when he admits that his replacement Galilean transformation is not physically interesting. That alone indicates that the manuscript does not meet the standards of the American Journal of Physics.
If this reply again misses the mark, as the author has indicated of my original report, then I would argue that the manuscript as it stands is sufficiently unclear as to warrant rejection.
To which I answered,
Dear Editors and Reviewers at AJP,
You may consider this a simple reply rather than a further appeal if you like. First of all, I appreciate your reviewer #2 taking the time to develop a little thought problem and to try to get to the bottom of this in a way that made more sense to him. It may help us come to an understanding. Using the example of a rock and an observer passing it by at velocity v, he shows that points can be transformed from one system to another with this equation: x' = x  vt. But I reply that he must be aware that the equation in this form cannot be the equation the Lorentz transform resolves to, since in the Lorentz transformation the x variables do not stand for points. They stand for lengths or distances.* You cannot claim that a relativistic equation where the x variables are distances resolves to a Galilean equation where the x variables are points.
The reviewer then says that Δx' = Δx  vΔt is also true, though his argument here is faulty. He is right in one sense: that equation can be made to work, but only if Δx is zero. By his way of making it work in his thought problem, Δx must always be zero, since as he showed with the kinetic energy equation, there is no motion of the rock relative to the rock. Therefore his equation always resolves to Δx' = vΔt. But we already knew that. That is not a transformation, that is just a definition of velocity. Besides, if Δx = 0, then it is still a point. He may say we can express zero as Δx if we want to, in order to create the necessary equation, but that is not what Einstein intended to do. Einstein was trying to transform a length in one system to a length in another. That is what the Lorentz transform is able to do, as we all know. Therefore, the Lorentz transform should resolve to a Galilean transform where the x variables stand for the same length seen by different observers.
Let this reviewer's rock be flat and let us put a meter stick on it, then run his experiment as before. According to the observer standing on the rock with the stick, the length is Δx = 1 m; it is not zero. According to the observer running by, the length is also Δx. In Galilean transforms lengths do not and cannot transform since there is no length contraction. In a Galilean transform, Δx = Δx'. This falsifies the current Galilean transform, which is quite simply a mathematical mirage. The Lorentz transform resolves to Δx = Δx', which is a tautology, not a true Galilean transform. You can call it a Galilean transform if you like, but historically, a Galilean transform was x' = x  a, as everyone up to Feynman used it [see my email on Six notsoeasy Pieces]. At a stretch you can put a = vt, as the reviewer has, but that does not make the equation a transform. In that form it is really just a definition of velocity, since it resolves like this: x' = x  vt x  x' = vt Δx = vt The bottom line: there is no Galilean transformation with a velocity variable in it. It never existed. The Lorentz transformation @ c = ∞ resolves to Δx = Δx' There is no transformation of lengths in Galilean systems, by definition of Galilean. My paper is interesting not because x' = x  a is a sexy new bit of math, but because I have solved a simple albegraic mistake that has lain in the open for a century at least. If correcting hundredyearold problems elegantly is not your cup of tea at AJP, then that is certainly your decision to make, and may you rest well with it in the future.
* The easiest way to prove this beyond a doubt is to show that Einstein and Lorentz and everyone else since has derived the Lorentz transforms by substituting the light equations x = ct and x' = ct' into the Galilean transform and solving for gamma. The x variables in the light equations obviously cannot be thought of as points, therefore the same variables in the Lorentz transformation cannot be points. As I said in my previous reply, this does not doom the transforms. Length, time, velocity and mass transforms can be derived without the Galilean transform. But it is important to settle this question once and for all. I can't stand seeing it claimed a hundred years after the fact that the Galilean transform is not invariant.
Miles Mathis
As an addendum to this exchange I would like to add some comments on the review #1 in the editors' reply to my appeal. This review has no content beyond the fact that the reviewer does not want to be bothered with anything that questions accepted theory in any way. Current theory is called by him or her "a selfevident truth." That is a very strange way to define science. It is a good thing that Einstein did not accept that Newton's theory was "selfevident truth," as he was no doubt assured by all that it was. This reviewer's points 3 and 4 completely miss my point, since my point is that the correct Galilean transform, understood either historically or logically, has nothing to do with motion. It cannot transform length, period of time, or velocity, since in a Galilean situation these variables do not change or transform. Velocity has nothing to transform, therefore a transform in the current form is absurd. The only thing that "transforms" with velocity is points, and that "transform" can be found with a simple velocity equation: x = vt, where x is the distance between the points in question.
Reviewer #2, who was the original referee, also concludes his review with more unnecessarily heavyhanded dismissals. The first responses to my paper were strangely antagonistic, and this reviewer keeps up the attitude here. Because he does not comprehend my argument, he assume the fault must be all mine. I think it is clear by now that the fault is due to his very poor knowledge of both classical and relativistic kinematics. All his conceptions are sloppy and he fails to make any of the important distinctions. Furthermore, he fails to understand me when I make them for him, although I am using high school algebra. I suppose I shouldn't be too hard on him, since current physics is populated by people just like him. If anybody understood basic kinematics, then Einstein's mistakes, including the misassignment of the "Galilean transformation", would not have already lasted 100 years. We are clearly living in an era where all physicists have skipped basic classical instruction to race into specialization or esoterica. They never took the time to learn fundamental concepts such as how variables are assigned in kinematics. Physicists now do math without ever attaching it firmly to underlying concepts. The math has become freefloating.
As another example of this, look at his continued insistence that there is no invariance across a classical kinetic energy transform. He does this by showing that the rock and its observer will get different numbers for the rock's kinetic energy. It is beyond belief that he would assert it, except that this is current wisdom. He is just stating what everyone accepts. The absurdity does not reveal itself to current physicists at all. The reason that it is absurd is that there is no transform. He is saying there is no invariance, or conservation of kinetic energy; but the reason there is no invariance is that there is no transform. Invariance is a concept that applies to transforms. There either is or is not invariance across a transform. But of course this assumes you have a transform. The stone cannot possibly measure any kinetic energy, by definition. You cannot transform from a zero constant. A transform requires two sets of variables. Zero is not a variable. This reviewer's example is therefore just silly.
The reviewer hasn't shown invariance, he has just shown the definition of kinetic energy. I showed above that his Galilean transforms didn't work because they resolved to the equation x = vt, which is the definition of velocity. Likewise, his example of kinetic energy transformed from the rock to the observer simply resolves to the definition of kinetic energy, K = mv^{2}/2. This is not a real transform. It is a fake and manufactured transform, so that variance across it is proof of absolutely nothing.
It is like claiming that there is no invariance of velocity across the velocity "transform" x = vt. The object itself measures no velocity and I measure it to have a velocity, therefore I have discovered a velocity variance. Goggledygook, obviously. There is no transform and therefore all talk of variance is illogical.
Beyond that, the whole argument about conservation of kinetic energy is a nonstarter. Kinetic energy is never conserved, not in classical events, not in relativistic events, and not across Lorentz transforms. Momentum and total energy are conserved in systems; they are also invariant under a Lorentz transformation. Conservation of energy and invariance are not the same thing anyway. And, Lorentz transforms are not invariant in the case he gives either. Let me give examples to show all the things I have just claimed. Last one first. If we let the reviewer's observer passing by his rock go near the speed of light, in order to make the "transform" a Lorentz transform, what will we find? All the changes to numbers will be to the numbers of the observer. He will calculate the mass of the stone to be increasing simply because he will be using this equation: m = γm_{r}. But the stone's numbers will not be changing. The stone still finds its mass to be m_{r} and its velocity to be zero. Therefore, mass is not conserved under the "transform" and neither is kinetic energy. The observer will find some large number for kinetic energy and the stone will still find a zero kinetic energy. By the reviewer's definition, no conservation and no invariance. This contradicts the current dogma, so something must have gone wrong. What is wrong is that the reviewer's example is not an example of a transform.
In relativity, kinetic energy is never invariant under a Lorentz transform. Momentum is. Momentum and kinetic energy can never be invariant at the same time, due to the form of the equation for each. Momentum is mv and kinetic energy is mv^{2}/2. If you use relativistic equations, my point is the same, since you do not square the velocity in a relativistic momentum equation, but in a kinetic energy equation you square both v and c. Therefore you cannot "conserve" both at the same time. If one is invariant, the other varies, and vice versa.
Finally, according to the reviewer's example, not even total energy would be conserved in a Lorentz transformation. Let us find the total energy of our rock using relativity. From the point of view of the speeding observer, the total energy of the rock is m_{r}c^{2} + γm_{r}c^{2}. From the point of view of the rock, the total energy is just m_{r}c^{2}. Total energy is not conserved across this "transform". Why? Because we have no transform. We just have the calculation of total energy. To find a conservation of energy, we must find it in a system. A rock by itself is not a system. When the observer speeding by calculates a total energy, that energy applies to a certain situation. It means that if the rock hits a wall that is stationary relative to the observer, that much energy will be released (supposing the rock is completely annihilated). It seems that the ghost of the rock must be quite surprised to see that much energy released by its anniliation, since it thought it had no kinetic energy. The reason that energy is conserved and the reason that the ghost of the rock would really not be surprised is that the rock would measure the approaching wall to have kinetic energy. The system of wall + rock conserves total energy. There is a conservation of energy measured before the collision and energy measured after the collision.
By the same token, momentum is conserved through the collision. Momentum is also invariant, but only if you have a transform. There is no transform from the momentum of the rock measured by the rock, to the rock measured by the wall or the observer. The rock measures zero momentum and the others do not. You will have a transform and invariance in this situation: let a bug crawl along the rock at a constant velocity. Let the bug measure its own velocity by using mile markers or something like that, just as a car does. Then measure the velocity and momentum of the bug from a stationary point on the rock. Even if the bug is going .99c, both measurements of momentum will be the same. That is the invariance that is maintained across a Lorentz transform. It is maintained by definition. Meaning that it is maintained because Einstein assumed momentum was invariant and built the transforms up from that assumption. That is the whole point of the transforms.
*As a footnote, you may be interested to know that I later tripped historical justification for my assumption here. I was eventually able to discover Woldemar Voigt's 1887 (preLorentz) derivation of x' = x  vt from x' = x  a, proving that my analysis is correct both as math and as history. You can read the whole story here.
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