Light clock moving to demonstrate time dilation

[JesseM]

You keep accusing me of “playing semantic games”. Again, as if semantics were a sort of idle good-for-nothing and “boring” activity, instead of what it is, the art of understanding what the heck it means what you are saying…!

No, I have already explained my position on semantics in physics: it's important insofar as people want to make sure they understand each other on issues that are relevant to physics, namely mathematical facts about the mathematical models used in physics, and facts about what these models predict about quantitative empirical observations. But I have further said that if two people already understand what the other is saying about these kinds of issues, and if they in fact agree on all such issues, then to continue to debate what words are best to describe certain thing is a waste of time (unless they are debating pedagogy, how best to explain these things to other people in a way that will be most easily understood). Do you disagree with me about this? Do you think there is something useful about debating word-definitions even if we already agree on all the mathematical/predictive aspects of relativity, and even if we are not having a pedagogical debate about how to explain things to some other people beyond ourselves?

In any case, I've said that I'm fine with you defining words any way you want to. If you define what you mean by "real-life and measurable invariant speed"--whether "invariant speed" is being defined in a way that presupposes it's finite, and whether this phrase requires that there be actual objects moving at that speed, for example--then I'll be happy to accept that definition and the proof may or may not say that the first postulate implies such an invariant speed. But even this seems not to satisfy you--is it your position that any definition other than the one you have in your head is unacceptable, and that I am therefore wrong to suggest it's just a matter of arbitrary definition, that "invariant speed" denotes some important concept that has only one "true" definition?

- Does not lead to a super-transformation that contains both the LT and the GT. It gives out the LT with some make-up, in the form of a constant (“aunque la mona se vista de seda, mona se queda”, see translation above) plus a mutilated and unrecognizable GT.

Is there any mathematical sense in which this is true? I think the answer is "obviously not", since from a mathematical point of view it's indisputable that if you plug in K=0 you get the GT, while if you plug in K=1/c^2 you get the LT; mathematically they are both just special cases of the generalized transformation. If it's not mathematical, what is it? You have a reaction of aesthetic distaste that the general transformation "looks more like" the LT or something? Do you disagree that what's "relevant to physics" is just mathematics and empirical predictions?

- And that is so because the derivation has made an extra implicit assumption from the outset: it has chosen that t≠t’

Hopefully you would agree that this is simply false mathematically, that the proof that the first postulate alone implies the final generalized transformation is fine on a purely mathematical level. So if you're no longer talking about mathematics you need to explain what new domain of discourse you are talking about. For example, as I asked earlier, are you maybe trying to suggest something about the motivations or thoughts of the authors of the proof, perhaps saying that they never would have come up with the idea for the proof if they hadn't already known that t≠t' in the Lorentz transformation?

and that is why it simply gives out the LT (where the existence of an invariant speed is a must) and it only accommodates the GT in a very exceptional case (where such invariant speed is infinite).

Again, it seems you are speaking non-mathematically here, since the phrase "very exceptional case" has no obvious mathematical meaning. Why is K=0 more "exceptional" than K=1/c^2?

So we are not in face of any valuable contribution. In fact, the derivation is based on two postulates. By pretending, however, that it uses only one, it lies and only creates confusion.

Do you claim it is "based on two postulates" in a purely mathematical sense? If you agree it is valid in a purely mathematical sense, what criteria are you using to judge whether it is "valuable" or not? Do you imagine the authors of the paper would have claimed it was "valuable" in the sense of somehow showing Einstein used a poor approach, as suggested by the imaginary "author #1" in my previous post?

But I am not going to convince you and you are not going to convince me.

But the reason you aren't convincing me is not that I understand what you are saying but disagree with you, it's that I honestly have no idea what you are trying to say in the first place.

Well, of course Einstein didn’t explore what would follow from the first postulate alone. That is precisely what I am arguing: He knew that with the aid of the 1st postulate alone, you derive nothing

The proof shows that, mathematically, you can derive a generalized coordinate transformation from the first postulate--do you disagree that this result is valid in a purely mathematical sense? And Einstein couldn't have known what would or wouldn't follow mathematically from a certain initial assumption without actually doing the math, there isn't any historical evidence that he tried to look at the consequences of the first postulate alone. Math is like that, you can't magically "know" what will follow from certain axioms without doing the math.

Again, this is not to suggest that Einstein somehow made a mistake in his approach. If you look at the opinions of author #1 and author #2 in my previous post, my own opinion about the historical significance would be much closer to that of author #2.

But leaving historical Einstein aside, what was important in my comment is that for any transformation to be constructed you need an invariant something. I don’t think that is mathematical “nonsense”, as you said. I think that is how a transformation operates.

Is "that is how a transformation operates" meant to be a mathematical claim? If so, can you come up with a general proof that any transformation must include an "invariant something"? (and 'something' is of course hopelessly vague mathematically, you would need to specify the precise range of mathematical entities that could qualify).

Do you think there is an "invariant something" in the following coordinate transformation?

x' = 38x + 15vt
t' = (17t²v/x) + (6x/v)

...or were you talking specifically about situations where the laws of physics were invariant under a coordinate transformation where each coordinate system is moving at constant speed v relative to every other, satisfying the first postulate? In that case the proof shows that the above coordinate transformation is inconsistent with the first postulate, but still, I wonder how you would show that in any universe where the first postulate holds the coordinate transformation must include an "invariant something" without making use of the proof we've been discussing.

Take the example I gave many posts ago about two observers, two coordinate systems at rest with each other, but whose origins are in different positions. S measures that the origin of S’ is 30 m away from the origin of S. Now S’ declares that a third point S’’ is situated at x’ = 20 mjxfs. What is x, the distance from S to S’’ in S frame? S argues that S’ should clarify what “mjxfs” means, whether it is units of length measured with a 1-metre rod or what, so as to be able to determine whether the 30 m that he has measured as distance from S to S’ can be combined or not with the 20 mjxfs. Unfortunately, S’ refuses to answer: “we don’t need to share any invariant concept, your m does not have to be equal to my mjxfs, that is mathematical nonsense”. How do you build then the transformation?

You simply haven't given enough initial assumptions to derive a coordinate transformation in this word-problem. I don't understand what S' means by the phrase "mathematical nonsense" anyway--S doesn't seem to have made anything resembling a mathematical claim that S' can declare as nonsense, he's only asked about the relation between their systems of units! In contrast, you seemed to be trying to make some mathematical claim about a property that all coordinate transformations must share, but you gave no mathematical justification for why they should all have this property.

Edit: Well, I reviewed what I said here and admit that it's to a good extent imprecise. I'll try to give some more detail. I had said before that I understand that the relativity principle has two components:

1) given two experiments E and E', result of experiment E with ball B wrt frame S = result of identical experiment E' with object B' wrt frame S' and
2) in a single experiment E, the coordinates of ball B in S can be related through a transformation with the coordinates of that same B in S'.

It's my impression that 1) is the condition for 2) to exist, since it permits to establish common measurement units which are used in the transformation.

I think I may see what you're getting at with 2) but if so you aren't stating it well. If each frame is using some physical ruler at rest in that frame to establish position coordinates, then 1) implies that they can all use some common physical procedure to build these rulers (likewise with clocks and with the synchronization procedure for clocks). Then I would write 2) as:
2) In a single experiment E, if each observer uses these identically-constructed systems of rulers and clocks to assign position and time coordinates to events in the experiment, then the coordinate transformation should accurately relate the coordinates found by different observers.

On the other hand, when talking about general coordinate transformations without assuming that the laws of physics must look the same in all the coordinate systems (first postulate), there is no requirement that different coordinate systems be defined in terms of identically-constructed physical rulers and clocks at rest in that system. If you know how the laws of physics work in one coordinate system, you can write down any arbitrary coordinate transformation you like and then use the transformation itself to determine how they will look in other coordinate systems given by that transformation. And if you have the coordinate transformation in hand, then your version of 2) is basically a tautology, since in any single experiment E you can always use the coordinate transformation to relate the coordinates of events that happened in that experiment as seen in different coordinate systems given by the transformation.

Frame S decomposes a problem in two component vectors and it turns out that for a certain component the measurements of S' are valid for S's purposes, due precisely to 1). What may vary, however, in different models of relativity is the vector that is considered as usable inter-frames. In the t=t' model, it is time. In the c=c' model, it is a speed.

Why should 1) imply that any vector is frame-invariant, if you don't use the results of the proof?

 

[yuiop]

Galileo essentially came up with the principle of relativity, sometime paraphrased as "the laws of physics are the same in all inertial reference frames". In this thread we have learned there are two main transformation systems that are consistent with that principle. In Galileo's time the technology was not available to experimentally determine whether the GT or LT transformations were the correct description of nature, but it is interesting to conjecture when it would have first been possible to choose the LT transformations based on the available experimental data. Most people think the Michelson Morley experiment was the pivotal experiment that ruled out the Galilean transformations but I would argue that is not the case. As discussed earlier in this thread a ballistic theory of light would be completely consistent with a null result of the MM experiment and the GT can not be ruled out without first ruling out the ballistic theory of the speed of light. When Maxwell came up with his equations for the propagation of electromagnetic waves in 1865, which indicated that the speed of light is independent of the speed of the emitter, that was the first opportunity to rule out the GT and come up with the LT of Special Relativity. In fact Einstein claims that he was unaware of the Michelson Morley experiment and based his second postulate on Maxwell's equations. I guess if you classify Maxwell's equations as "a law of physics", then the first postulate alone is enough to derive Special Relativity.

 

[Saw]

To me it seems like you are guilty of a physics analogue of the intentional fallacy in literary criticism--instead of just focusing on the text of the proof itself, looking at whether it's valid mathematically and so forth,

This accusation is groundless and unfair.

There are two things to analyze when one is confronted with a mathematical proof: (i) the axioms or assumptions or postulates that the proof starts with and (ii) the mechanics of the subsequent reasoning, ie. the algebra. As to the algebra, I have often said that I do not dispute it. As to the axioms, precisely the question is that the proof holds that it starts with just one axiom, whereas I point out that it starts with an extra-assumption, namely t ≠ t’. In order to prove this assertion, I do look at the text of the “proof” and beg you to do the same:

First line: the author writes t and t’. But t = t’, on the one hand, and t≠t’, on the other hand, are “mutually exclusive” options. They are both possible and compatible with the principle of relativity, but exclusive of one another. Hence writing one of them at the outset of a mathematical proof equates to excluding the other. Writing, in particular, t and t’ is tantamount to postulating that t is not equal to t’. (Well, you hold the contrary, but NOT on the basis of the TEXT of the derivation, which is uncontestable! You hold that on the basis of a funny conception about the right to be arbitrary in the choice of English words that I later comment!)

Second, do we have to look at the final line? Actually, we don’t have to. What we have seen so far is enough to show that the proof makes a second assumption and so its claim that it derives anything out of the principle of relativity ALONE is a lie. But we can still look at the final line just for confirmation purposes. The final line is the LT with one particularity: instead of 1/s^2, the author has written K. But since this K (i) MUST have units of inverse squared speed and (ii) MUST be the same in the direct and reverse transformation, it follows that s is an “invariant speed”. Mathematically this means that there is no “substantial” difference between the LT and this so called generalized transform. The difference is purely in the arbitrary choice of symbols: the author has written K where she could have perfectly written 1/s^2.

I could accept, yes, that the derivation differs from Einstein’s original approach in that, instead of assuming that light must travel at an invariant speed, it assumes that t≠t’ and, together with it (both things are logically inter-connected) an invariant speed that does not need to be associated to light. That is all it proves. It’s not too much though it’s quite true: relativity + t≠t' is = to an invariant speed. But it’d be a blunder to state that it mathematically proves that the principle of relativity alone requires an invariant speed. The principle of relativity can do perfectly well with an invariant time and without any invariant speed, just as it can do with an invariant speed and without any invariant time.

All these are mathematical reasons. You can say they are simplistic: a six-year-old infant could have utilized them. They do not show a magnificent display of mathematical knowledge. But it cannot be said, in good faith, that they are not mathematical.

you are coming up with some extra-mathematical ideas about the "intentions" behind the proof, like that the proof shows Einstein took the "wrong approach" somehow in deriving the LT (even though his approach was perfectly valid mathematically), that he "could have been much more aggressive" and that it was a mistake for him to make the assumptions he did. But the proof is just a proof, in assessing it's validity I don't care about the author's secret motivations in coming up with it.

You call it the “proof” because you think that it proves something that Einstein himself did not prove. Don’t you? Well, that is what I was highlighting: you and others think that this derivation proves something that was not proven before. In particular, let us repeat it again, you think that this way you prove that the principle of relativity implies an invariant speed. That is not a secret motivation. It is the crux of the argument. I simply note it and point out that I disagree with it.

At a deeper level, yes, some authors may have deeper motivations, a certain mental framework that leads them to think that way. It’d be quite interesting to hypothesize about that, but it’s not my wish to do so now.

What does "super-expression of both GT and LT" mean? It must mean something other than "an equation which reduces to the GT if you plug in one value for an arbitrary constant, and reduces to the LT if you plug in a different value for that constant", since mathematically it is undeniable that the "generalized transformation" does have that property!

Obviously, “super-expression of both GT and LT” is another expression for “generalized transform”. However, I am afraid that there is a difference between the objective meaning of this phrase (which is the one you are handling, I think) and the mathematical properties of the equation under consideration (which do not stand up to such high expectations).

As to the objective meaning of the expression “generalized transform”, we just have to remember the crux of the argument: it is an equation that comes out by making a single assumption, the principle of relativity, and therefore it comprises all valid relativistic transformations. In mathematical and logical terms, it corresponds to the “class of all relativistic transformations”.

As to the mathematical properties, I agree with your definition: the equation contains a constant; if you plug in one value for that constant (an infinite speed), you get the GT; if you plug in another value (any finite speed > v), you get the LT.

Do these properties justify the claim that we are in front of a generalized transform, in the sense of the “class of all relativistic transformations”?

Again and again and again I will note that we must not forget a crucial detail: in order to get the GT, you must plug in the constant an INFINITE speed. However, without this derivation and still respecting the principle of relativity, I can get the GT without the need to assume any invariant speed, least of all an infinite one, by assuming that t = t’. In view of this, there are two logical possibilities:

- One is what most authors seem to be saying: that the GT is only possible if you admit an infinite invariant speed. That’s wrong: it only appears to be mathematically true if you look at this derivation and buy that it is the only possible derivation that the principle of relativity admits, but it turns out to be wrong as soon as you realize that such principle also permits another proof by which you get the GT without any restriction (and why not the LT as a special case of the latter? Have you tried at doing that seriously?).

- The other possibility is what you say here with your queer conception of the meaning of conventional (English, French or Spanish) language:

The english phrase "invariant speed" appears nowhere in the proof. You are free to define "invariant speed" in a way that rules out all but finite speeds, and then say that the proof can accommodate t=t' without an "invariant speed". Mathematically the proof accommodates t=t' in the case of K=0, the language you use as English shorthand for the K=0 case is totally irrelevant as long as we both understand the mathematical meaning of any such shorthand. For example, if we both agree in advance to denote the K=0 case with the English shorthand "cows develop a sudden hunger for human flesh", then it would be perfectly correct to say that "the generalized transform can only accommodate the t=t' case if we also assume that cows develop a sudden hunger for human flesh". We would both understand what this means mathematically, and if the domain of discourse is math, then semantics is only important insofar as it's important that we all have the same understanding of what any English phrases used are supposed to denote mathematically.

I'm happy to admit anything you like about English phrases like "invariant speed" and "invariant time" as long as you explain in advance what they are supposed to denote on a purely mathematical level. But if you think that it matters whether we choose to define English phrases in one way as opposed to another way, then apparently you have left the "domain of discourse" of mathematics, and are using some non-mathematical criteria to judge which definitions are "better", in which case you need to actually explain the details of what these criteria are, and why they are relevant to a discussion about physics.

Again, you have a wrong conception of language. You confuse the choice of specific symbols with the choice of meanings. For the choice of symbols, one is relatively free to act arbitrarily. That is so in mathematical and conventional language. But once a certain symbol is chosen to denote a particular idea or meaning, one must be consistent and keep the same symbol for ever, both on mathematical and conventional language. There is no possibility to change the symbol a posteriori, for convenience in order to puzzle your opponent in a discussion. In this case, clearly the s contained in K is, in both mathematical and conventional languages, what is universally alluded to as an “invariant speed”. Ok, long ago, our ancestors could have chosen other symbols to denote that meaning. As you suggest, they could have resolved to choose the shorthand "cows develop a sudden hunger for human flesh", but they didn’t. They could even have chosen the symbols “JesseM is totally wrong and is perfectly aware of that” and maybe that would have eased our discussion. Unfortunately they didn’t. They chose to call this meaning “invariant speed”, in technical (mathematical!) jargon, “a speed that is measured with the same value by all observers”, in more day-to-day language. Similarly, they chose to call the value of s that permits the GT to show up an “infinite” speed. They could have used other symbols, but they didn´t. Both things (invariant speed and infinite) have universally accepted meanings, so it would not be wise to choose now other alternative symbols.

Conclusion: whether you like it or not, in a rigorous mathematical analysis, in English and even in Basque language, the “proof” only contains a special case GT, not the full-fledged GT. Why so? Because it starts with an extra-assumption (t≠t’)!

Again I don't know how to interpret this, you certainly aren't saying anything that I can translate into mathematically meaningful statements. What does "the Lorentz Transformation with some cosmetics" mean?

Whatever your basis for attacking it, it seems like it's coming from a mode of thinking that's alien to the mode of thinking physicists normally use to discuss results in physics, which is perhaps why no one here can understand what you are trying to say.

All that looks like arguments ad hominem. For whatever reasons, you are not fond of and don’t find it easy to understand metaphoric language. It’s a pity because the whole of language is metaphoric, all concepts are analogies, all measurements are mirrors or similes of the observed reality... But I am sure others do appreciate it, physicists included, and even find it enlightening. At least my wife does.

To focus on one example: Yes, I claim that the proof is just the LT with some “cosmetic ornament”. This is a very simple figure of speech that I have seen successfully used thousands of times. The mental process goes as follows. As I explained above, if you open up the constant K you find 1/s^2, s being a speed. Then you realize that you are just in front of the LT. The fact of writing K instead of 1/s^2 has not changed the essence of the equation, only its external appearance, just like cosmetics do not change the essence of a woman or an actor, only their looks. Normally this analogue rings a bell inside people’s heads and helps to convey the meaning in question. It may not be your case, but you should not try to patronize me into forsaking this kind of language, which others, including physicists, utilize as well and appreciate.

Anyhow, I am now talking to you, ‘cause I like it, and I wish to continue to do so, that is why I try to reduce the metaphoric tone, though I sometimes lose control…

Finally, your comments about the meaning of transformations look very helpful. I will address them later.

 

[JesseM]

There are two things to analyze when one is confronted with a mathematical proof: (i) the axioms or assumptions or postulates that the proof starts with and (ii) the mechanics of the subsequent reasoning, ie. the algebra. As to the algebra, I have often said that I do not dispute it. As to the axioms, precisely the question is that the proof holds that it starts with just one axiom, whereas I point out that it starts with an extra-assumption, namely t ≠ t’.

The distinction you're making doesn't make sense. A proof involves starting with some axioms, then in a step-by-step manner showing that the statements you have so far (axioms plus implications derived in later steps) imply some further statements, finally ending in some final statement which is what you wanted to prove. Do you disagree that this is how all mathematical proofs work?

If you don't disagree, then if you think the axioms do not logically imply the final theorem, then there must be some mistake in the intermediate steps, i.e a step where they say something like "since we have shown A, B, and C, this implies D according to mathematical rule X" when in fact A, B, and C do not imply D on their own without some extra (unstated) assumption. So if you want to show the proof doesn't work mathematically you can't just make broad conceptual arguments, you have to point to a specific step where they made a mathematical statement that wasn't actually justified by what came before. This is just the nature of proofs in math, whether they stand or fall is completely determined by whether each step is valid in isolation!

In order to prove this assertion, I do look at the text of the “proof” and beg you to do the same:

First line: the author writes t and t’. But t = t’, on the one hand, and t≠t’, on the other hand, are “mutually exclusive” options. They are both possible and compatible with the principle of relativity, but exclusive of one another. Hence writing one of them at the outset of a mathematical proof equates to excluding the other.

What do you mean "writing one of them at the outset"? The don't write either t=t' or t≠t' at the outset, they just write that t must be some function of x,t,v and then use the principle of relativity (along with the 'isotropy of space' and the 'homogeneity of space and time', I'm not sure if these qualify as extra initial axioms or if they can be derived from the principle of relativity) to derive various constraints on the form of the function. Again, if you understand the basic nature of mathematical deduction and proof, you will understand that to find a flaw in one you need to find a specific step that is not justified in terms of previous steps or axioms, the vague conceptual argument above doesn't qualify.

Writing, in particular, t and t’ is tantamount to postulating that t is not equal to t’.

"Is tantamount to" is not anything resembling a mathematical/logical argument! And anyway this makes no sense, "writing" two variables without specifying the function that relates them cannot in itself imply anything about the relation between them!

(Well, you hold the contrary, but NOT on the basis of the TEXT of the derivation, which is uncontestable! You hold that on the basis of a funny conception about the right to be arbitrary in the choice of English words that I later comment!)

Huh? English words are irrelevant to the validity of a mathematical proof, or to specific mathematical questions like whether writing down t' = T(x,t,v), where T is an arbitrary function, in itself implies anything about whether t=t' or t≠t'. Mathematically it doesn't, and I don't "hold that" because of any conception about the free use of English words in informal descriptions of mathematical results, I hold that for mathematical reasons having nothing to do with informal English descriptions.

Second, do we have to look at the final line? Actually, we don’t have to. What we have seen so far is enough to show that the proof makes a second assumption and so its claim that it derives anything out of the principle of relativity ALONE is a lie. But we can still look at the final line just for confirmation purposes. The final line is the LT with one particularity: instead of 1/s^2, the author has written K. But since this K (i) MUST have units of inverse squared speed and (ii) MUST be the same in the direct and reverse transformation, it follows that s is an “invariant speed”. Mathematically this means that there is no “substantial” difference between the LT and this so called generalized transform. The difference is purely in the arbitrary choice of symbols: the author has written K where she could have perfectly written 1/s^2.

Sure, the author (actually a he not a she) could have written 1/S^2 instead of K, but there would be no grounds in any of the previous steps for saying S must take a finite value, so one could still show that the Galilei transformations are one valid case of the general transformation.

All these are mathematical reasons. You can say they are simplistic: a six-year-old infant could have utilized them. They do not show a magnificent display of mathematical knowledge. But it cannot be said, in good faith, that they are not mathematical.

Sorry, but your arguments above don't make any sense as mathematical reasoning. It's just totally illogical to say that the act of writing down symbols for two variables somehow implies that they are unequal!

You call it the “proof” because you think that it proves something that Einstein himself did not prove. Don’t you?

Of course, it proves that the first postulate (possibly plus some assumptions about the homogeneity of space and time) implies that the laws of physics must be invariant under a certain general coordinate transformation equation.

In particular, let us repeat it again, you think that this way you prove that the principle of relativity implies an invariant speed. That is not a secret motivation. It is the crux of the argument.

No, I seriously don't care about how you define nomathematical terms like "invariant speed". If you want to know what I think the first postulate implies, look at the equation they end up with for the coordinate transformation.

What you call the “generalised transformation” (as if it were a super-expression of both GT and LT)

What does "super-expression of both GT and LT" mean? It must mean something other than "an equation which reduces to the GT if you plug in one value for an arbitrary constant, and reduces to the LT if you plug in a different value for that constant", since mathematically it is undeniable that the "generalized transformation" does have that property!

Obviously, “super-expression of both GT and LT” is another expression for “generalized transform”.

So when you said "as if it were a super-expression of both GT and LT", the "as if" wasn't meant to imply any skepticism about equating the two phrases? That seems pretty weird, like if I were to say, during the course of a discussion "what you call 'synonyms' (as if they were two words that meant exactly the same thing)" but then were to later clarify that yes, I agreed the two words did mean exactly the same thing!

However, I am afraid that there is a difference between the objective meaning of this phrase (which is the one you are handling, I think) and the mathematical properties of the equation under consideration (which do not stand up to such high expectations).

As to the objective meaning of the expression “generalized transform”, we just have to remember the crux of the argument: it is an equation that comes out by making a single assumption, the principle of relativity, and therefore it comprises all valid relativistic transformations. In mathematical and logical terms, it corresponds to the “class of all relativistic transformations”.

As to the mathematical properties, I agree with your definition: the equation contains a constant; if you plug in one value for that constant (an infinite speed), you get the GT; if you plug in another value (any finite speed > v), you get the LT.

Do these properties justify the claim that we are in front of a generalized transform, in the sense of the “class of all relativistic transformations”?

No, the proof itself is what justifies the claim that the equation represents the class of all relativistic transforms.

Again and again and again I will note that we must not forget a crucial detail: in order to get the GT, you must plug in the constant an INFINITE speed. However, without this derivation and still respecting the principle of relativity, I can get the GT without the need to assume any invariant speed, least of all an infinite one, by assuming that t = t’.

So? You can certainly use a different set of axioms that go beyond just the first postulate to derive the GT, but does that somehow disprove the claim that the first postulate alone can be used to derive a generalized transformation which includes the GT as a special case?

In view of this, there are two logical possibilities:

- One is what most authors seem to be saying: that the GT is only possible if you admit an infinite invariant speed.

DEFINE YOUR TERMS PLEASE. If you don't explicitly define what you mean by the term "invariant speed"--then this is just meaningless verbiage. And none of the definitions I've suggested so far--whether definitions which presuppose an invariant speed must be finite, or definitions which presuppose that there must be actual physical objects moving at this speed--would in any way imply that the answer would depend on how you derive the GT! It shouldn't matter whether you arrive at the GT by first deriving the generalized transform and then plugging in K=0/S=infinity or if you arrive at the GT by some completely different route, either way there must be a single answer to whether the GT have an "invariant speed"! If you are switching between two different definitions of "invariant speed" depending on how the GT were derived, then this would be an equivocation fallacy.

but it turns out to be wrong as soon as you realize that such principle also permits another proof by which you get the GT without any restriction (and why not the LT as a special case of the latter? Have you tried at doing that seriously?).

"Seriously" because I don't know what the hell you're talking about. The GT cannot be a special case of the LT, and the LT cannot be a special case of GT, because they are different, distinct, not the same. They can both be special cases of a generalized transformation that includes a constant that is allowed to take any value, though.

Again, you have a wrong conception of language. You confuse the choice of specific symbols with the choice of meanings. For the choice of symbols, one is relatively free to act arbitrarily. That is so in mathematical and conventional language. But once a certain symbol is chosen to denote a particular idea or meaning, one must be consistent and keep the same symbol for ever, both on mathematical and conventional language.

In a specific domain, you must indeed be consistent. But the point is that when using English phrases to make precise technical claims about math or physics, one must give these phrases technical definitions which are naturally going to be separate from their imprecise "conventional language" definitions. And when dealing with an English phrase that hasn't yet been assigned any specific technical definition, like "invariant speed", you're free to define it in whatever way is most convenient, you don't have to worry about whether it corresponds too closely with the conventional-language meaning of the same phrase (look at the physics definition of action, for example). Once you have chosen a technical meaning, of course you should stick to the same meaning, I certainly wouldn't dispute that!

The problem is that you consistently refuse to actually define what technical meaning you want to assign to "invariant speed", and then bring up new arguments seemingly based on some implicit meaning which exists in your head and which seems to be based on your understanding of the conventional meaning of the words, like your sudden invocation of the notion of a "REAL-LIFE and MEASURABLE infinite invariant speed" in post #102. My point is just that you can't treat phrases in technical discussions the same way you treat phrases in conventional language--if there is even an ounce of ambiguity about the precise technical meaning of a phrase, you have to be willing to give an explicit definition if you want to continue using it in a technical context!

There is no possibility to change the symbol a posteriori, for convenience in order to puzzle your opponent in a discussion. In this case, clearly the s contained in K is, in both mathematical and conventional languages, what is universally alluded to as an “invariant speed”.

What does "invariant speed" mean in mathematical language? Can you actual define it? If you're defining it in a purely mathematical way, why did you previously talk as though it needed to be "real-life and measurable"? It seems like you are the one who is changing the symbol a posteriori, or more likely just not not having a clear definition in your mind to begin with.

Ok, long ago, our ancestors could have chosen other symbols to denote that meaning. As you suggest, they could have resolved to choose the shorthand "cows develop a sudden hunger for human flesh", but they didn’t.

I'm not talking about our ancestors. I'm talking about the fact that phrases like "invariant speed" and "cows develop a sudden hunger for human flesh" have no technical meaning now, their conventional meaning is not sufficient to give them any clear technical meaning. Sometimes in a technical discussion you can use a new phrase and trust that the context and the conventional meaning will make it reasonably clear what the technical meaning is supposed to be, as with the paper's own use of the phrase "invariant speed". But it is not similarly clear from the context what you mean, and so much of your argument revolves around this ill-defined phrase that you really need to spell it out explicitly.

Unfortunately they didn’t. They chose to call this meaning “invariant speed”, in technical (mathematical!) jargon

Who did? Are you talking about the paper's use of this term? Like I said, it seems reasonably clear what the author meant based on the context, but it doesn't seem to correspond to how you use the term. For example, I would say that based on how the author uses it, he would presumably say that the Galilei transformation by definition has an invariant speed of infinity regardless of how you derive it or of whether any object in a universe with Galilei-invariant laws actually travels at infinite speed.

 

[JesseM]

(continued from previous post)

“a speed that is measured with the same value by all observers”, in more day-to-day language.

What do you mean by "measured"? I asked before whether this actually required any objects that move at this speed and you didn't answer:

I also don't understand what it would mean to say the invariant speed should be "real-life and measurable", isn't the invariant infinite speed just a mathematical property of the Galilei transformation?

Consider an analogy: suppose we lived in a universe with Lorentz-symmetric fundamental universe, but unlike in the real universe, in this all particles have nonzero rest mass and it is impossible for any real physical effect to travel at c. Would you say in this case that the invariant speed c is not "real-life and measurable"? But it is still perfectly measurable in the sense that if we write down the equations of the laws of physics in one inertial frame, we can see that they will remain unchanged under a particular unique coordinate transformation which involves the constant c (whose value can be determined from the equations), one which has the velocity addition law w = (u + v)/(1 + uv/c^2). We will also still observe that a clock moving at velocity v in a given frame is slowed down by a factor of sqrt(1 - v^2/c^2), which would be another empirical way to determine the value of c.

I suppose it is again just a semantic matter whether you choose to define "real-life and measurable invariant speed" in such a way that you need to be able to find actual physical entities that travel at that speed, or whether you define it in such a way that if one can empirically determine the coordinate transformation that the laws of physics are invariant under, and mathematically speaking this coordinate transformation includes an invariant speed, then that counts as a "real-life and measurable invariant speed". I would find the second definition more natural (and more reflective of how physicists would likely talk about these issues), but if you prefer the first definition, then under that definition we can say that the Galilei transformation does not include a "real-life and measurable invariant speed" if no actual physical object or effect (like instantaneous action-at-a-distance) travels at infinite speed.

But as I said before, regardless of what definition you choose, it makes no sense that somehow if one author derives the Galilei transformation as a special case of a more general transformation, and another derives the Galilei transformation in a more unique way, that is somehow going to cause them to disagree about whether the GT implies an "invariant speed"!

Conclusion: whether you like it or not, in a rigorous mathematical analysis, in English and even in Basque language, the “proof” only contains a special case GT, not the full-fledged GT. Why so? Because it starts with an extra-assumption (t≠t’)!

I don't know what "special case" vs. "full-fledged" GT means. Mathematically this is the GT:

x'=x-vt
t'=t

If you just mean "the proof derives the GT (a phrase that has a single unique meaning, namely the coordinate transform I just wrote down) as a special case of a more general transformation, whereas some other proof might derive the GT (with the phrase here having exactly the same meaning) as a unique outcome of some different postulates", then I'd agree. But if you mean to imply that "the GT" does not imply the unique coordinate transformation above then you'll have to define what you mean by "the GT", since that's what everyone else means by the phrase.

And obviously I disagree that the proof starts with an extra assumption, since your argument that the mere act of writing down two variables implies that they cannot be equal is completely absurd, and you have not shown any step of the proof where the assumption of t≠t' is needed. Moreover the idea that a proof which starts with t≠t' and derives the GT as a special case is also completely absurd, since the GT requires that t=t'; it's logically impossible to start with an axiom A and derive the possibility of ~A as a special case of your conclusion!

Again I don't know how to interpret this, you certainly aren't saying anything that I can translate into mathematically meaningful statements. What does "the Lorentz Transformation with some cosmetics" mean?

Whatever your basis for attacking it, it seems like it's coming from a mode of thinking that's alien to the mode of thinking physicists normally use to discuss results in physics, which is perhaps why no one here can understand what you are trying to say.

All that looks like arguments ad hominem.

Neither are ad hominems because they address your style of argument, not anything about you personally.

For whatever reasons, you are not fond of and don’t find it easy to understand metaphoric language.

I like metaphoric language just fine in everyday speech, I just don't like it in cases where it's being used to talk about a precise technical subject like mathematics and the technical meaning of the metaphors is completely ambiguous (especially when the person using it refuses to ever state clearly what the technical meaning is supposed to be when asked, and is using these metaphoric arguments to try to refute mainstream technical conclusions).

But I am sure others do appreciate it, physicists included, and even find it enlightening.

Pretty sure no physicist would understand or agree with the argument you are making here.

To focus on one example: Yes, I claim that the proof is just the LT with some “cosmetic ornament”. This is a very simple figure of speech that I have seen successfully used thousands of times. The mental process goes as follows. As I explained above, if you open up the constant K you find 1/s^2, s being a speed. Then you realize that you are just in front of the LT.

No, I am in front of a general equation that reduces to the LT in the specific case of s=c, and reduces to the GT in the specific case of s=infinity. The equation itself is not identical to either one.

The fact of writing K instead of 1/s^2 has not changed the essence of the equation, only its external appearance, just like cosmetics do not change the essence of a woman or an actor, only their looks.

I don't know what "the essence of the equation" means. Clearly in a purely mathematical sense the equation is not identical to the LT, because the equation involves a variable which can (consistent with the assumptions of the proof) take any value from 0 to infinity, whereas the LT is a different equation that has the constant c=299792458 meters/second in place of the variable. Do you think that the equation y=x^N, where N is defined to be a variable whose domain consists of the positive integers, has the same "essence" as the equation y=x²? This talk of "essence" appears to be every bit as metaphoric as "cosmetics", so defining one metaphor in terms of another metaphor does not help me understand what the actual mathematical meaning of your statements is supposed to be.

 

[Saw]

I’d agree to dedicating some time to definitions:

Definition of “invariant speed”

With regard to this issue, I see that you fluctuate between two positions:

(a) Sometimes you consider that this is a technical, mathematical expression, which you ask me earnestly to define, instead of handling the informal and conventional meaning that would be lurking in my head:

DEFINE YOUR TERMS PLEASE. If you don't explicitly define what you mean by the term "invariant speed"--then this is just meaningless verbiage.

The problem is that you consistently refuse to actually define what technical meaning you want to assign to "invariant speed

(b) But sometimes you find it useful to say that this is NOT a technical expression and declare that you do not really care about its definition.

No, I seriously don't care about how you define nomathematical terms like "invariant speed".

And when dealing with an English phrase that hasn't yet been assigned any specific technical definition, like "invariant speed", you're free to define it in whatever way is most convenient

Which is your final choice? Is it a technical expression or not? Can we agree on a definition or not?

If this is an expression that is actually technical (it really seems so!) and we can agree on a definition, you should remember that we had almost reached that point, when I said:

Thus I don’t define "speed" in a way that presupposes it must be finite = a speed can be finite or infinite = an infinite speed is not a contradiction in terms. And if the speed in question is frame-invariant, that doesn’t change the conclusion = an invariant speed can be infinite.

(By the way, as to what “frame-invariant” means… I suppose we have agreed on this: We say that the speed of an object is invariant if the *same* object in the *same* experiment is measured to have the same speed in all frames. For example, light in SR model, where the speed of a single light beam is said to be identical for all observers)

And you had agreed:

Right, that's what I mean when I say "invariant speed".)

Certainly, you complain about

some implicit meaning which exists in your head and which seems to be based on your understanding of the conventional meaning of the words, like your sudden invocation of the notion of a "REAL-LIFE and MEASURABLE infinite invariant speed" in post #102.

Well, hopefully you agree that there are several levels here:

(1) If you take the principle of relativity (PR) as an axiom, any valid transformation must have an invariant “something”: an invariant speed, in your opinion; an invariant speed or alternatively an invariant time, in my opinion.

(2) Does that mean that such invariant value must necessarily be measured, that there must exist some physical object permitting the actual measurement of a frame-invariant value? Obviously, no. Did you think I had that idea in my head? If so, that is not the case.

(3) If actually experiments prove that the invariant value is never measured, would that prove that the PR is, to some extent, wrong? Here I would tend to answer, yes, but this may be problematic. Let us go slowly now. Please let me know your opinion so far.

 

[JesseM]

I’d agree to dedicating some time to definitions:

Definition of “invariant speed”

With regard to this issue, I see that you fluctuate between two positions:

(a) Sometimes you consider that this is a technical, mathematical expression, which you ask me earnestly to define, instead of handling the informal and conventional meaning that would be lurking in my head

I don't consider that it is already a technical expression, I just ask that if you continue to use this expression, you need to assign it a precise technical meaning. And I also say that as long as you assign it some precise meaning, it doesn't matter to me what meaning you do assign it, consistent with this:

(b) But sometimes you find it useful to say that this is NOT a technical expression and declare that you do not really care about its definition.

Yes, as you are using it so far it does not seem to correspond to any precise technical meaning, which is why I ask you to assign it some precise meaning. I don't care in advance whether you choose to assign it meaning X or meaning Y, but I do care that we settle on some meaning and then stick to it afterwards.

If this is an expression that is actually technical (it really seems so!) and we can agree on a definition, you should remember that we had almost reached that point, when I said:

Thus I don’t define "speed" in a way that presupposes it must be finite = a speed can be finite or infinite = an infinite speed is not a contradiction in terms. And if the speed in question is frame-invariant, that doesn’t change the conclusion = an invariant speed can be infinite.

(By the way, as to what “frame-invariant” means… I suppose we have agreed on this: We say that the speed of an object is invariant if the *same* object in the *same* experiment is measured to have the same speed in all frames. For example, light in SR model, where the speed of a single light beam is said to be identical for all observers)

And you had agreed:

Right, that's what I mean when I say "invariant speed".)

What I had in mind at that point was a purely theoretical definition based on the mathematical form of the transform; the idea is that if (hypothetically, regardless of whether this is physically possible or not) something is moving at a speed S in one frame, with S allowed to take infinite as well as finite values, then according to the transform it will also be traveling at S in another. I was thinking of a purely kinematical definition which ignores the question of whether the dynamical laws actually allow things to travel at S; to bring the latter question into the definition hadn't even occurred to me at that point, so I thought that as long as you allowed the "invariant speed" to be infinite, that meant we were on the same page (from the context I'm fairly confident this sort of kinematical definition of 'invariant speed' was being assumed implicitly by both the author of the paper and bcrowell, by the way). But then you subsequently brought up the "real-life and measurable" issue which made me realize we actually weren't.

Well, hopefully you agree that there are several levels here:

(1) If you take the principle of relativity (PR) as an axiom, any valid transformation must have an invariant “something”: an invariant speed, in your opinion; an invariant speed or alternatively an invariant time, in my opinion.

I don't agree there is any reason a priori to think that the principle of relativity implies an invariant anything before you have done an actual derivation of what conclusions follow from this principle like the one in the proof. If you think there is some simple argument for why the principle of relativity implies some variable must be invariant between frames, I'd like to see that argument laid out in precise terms.

Also, like I said, I am happy to agree to any definition of "invariant speed" that you choose, so long as you choose some clear definition. It may well be that given your definition, I would say the principle of relativity does not necessarily imply an invariant speed. Likewise, I would say that if we use the kinematical definition I talked about above (one which allows 'speed' to be infinite), then the proof shows definitively that the principle of relativity does imply an "invariant speed"; if you disagree that this is true even given the definition I gave, I'd like to know your argument for that.

(2) Does that mean that such invariant value must necessarily be measured, that there must exist some physical object permitting the actual measurement of a frame-invariant value? Obviously, no. Did you think I had that idea in my head? If so, that is not the case.

Well, in post #102 you stressed the difference between "the THEORETICAL possibility of an infinite speed" and a "REAL-LIFE and MEASURABLE infinite speed". What types of measurements would you say demonstrate a "measurable infinite speed" if no actual physical objects or effects travel at infinite speed? The only answer that occurred to me was that if we can perform measurements which allow us to determine the dynamical laws of physics in one frame, and we find that the equations for these laws are invariant under the Galilei transformation, then since the GT features an invariant infinite speed at the theoretical kinematical level, this could qualify as a sort of "measurement" of an invariant infinite speed. That's why I made the analogy to imaginary Lorentz-symmetric laws of physics which don't actually allow any particle or effect to travel at c:

Consider an analogy: suppose we lived in a universe with Lorentz-symmetric fundamental universe, but unlike in the real universe, in this all particles have nonzero rest mass and it is impossible for any real physical effect to travel at c. Would you say in this case that the invariant speed c is not "real-life and measurable"? But it is still perfectly measurable in the sense that if we write down the equations of the laws of physics in one inertial frame, we can see that they will remain unchanged under a particular unique coordinate transformation which involves the constant c (whose value can be determined from the equations), one which has the velocity addition law w = (u + v)/(1 + uv/c^2). We will also still observe that a clock moving at velocity v in a given frame is slowed down by a factor of sqrt(1 - v^2/c^2), which would be another empirical way to determine the value of c.

So, please tell me whether this sort of measurement of the equations of the laws of physics, showing that they were invariant under a coordinate transformation which at a theoretical/kinematical level features an invariant infinite speed, qualifies as a "real-life and measurable" invariant infinite speed in your way of thinking. If not, can you think of any other way in which an invariant infinite speed could be "measurable" in terms of how you are defining this word, besides actually measuring an object or effect that travels with infinite speed?

(3) If actually experiments prove that the invariant value is never measured, would that prove that the PR is, to some extent, wrong?

Again, depends on what you mean by "measured". But as understood by physicists, the principle of relativity just means that the equations of the laws of physics are invariant under a certain type of coordinate system, one which involves a family of coordinate systems moving at constant coordinate velocity relative to one another.

 

[Saw]

Hmmm… Yes, we can agree that the S that appears in the final equation of the derivation we comment can be called an “invariant speed” with this meaning:

- It is a speed.
- It is inside a constant, which is the same in the two frames whose values are being compared in the transformation = it is a frame-invariant speed.
- It can take finite or infinite values.
- It does not necessarily correspond to the speed of any object that physically exists.

With regard to the latter point, paraphrasing a little, I am thinking that this invariant speed, in the sense that you seemed to have in mind and that now becomes apparent to me, belongs to the type of idealized concepts that abound in physics and are so helpful. For example, inertia. In reality, no object in the universe is ever totally free from the influence of forces, since at least there will always be gravity, no matter how miniscule due to long distance from the closest gravity source. In spite of that, the idea of inertia pervades physics and luckily so, because it’s most useful. Other examples: perfect rigidity or elasticity of an object… Even infinity could be classified within this lot of “idealized useful” concepts…

This brings us to the issue of the utility of the invariant speed. You link it to measurement, with these comments:

Consider an analogy: suppose we lived in a universe with Lorentz-symmetric fundamental universe, but unlike in the real universe, in this all particles have nonzero rest mass and it is impossible for any real physical effect to travel at c. Would you say in this case that the invariant speed c is not "real-life and measurable"? But it is still perfectly measurable in the sense that if we write down the equations of the laws of physics in one inertial frame, we can see that they will remain unchanged under a particular unique coordinate transformation which involves the constant c (whose value can be determined from the equations), one which has the velocity addition law w = (u + v)/(1 + uv/c^2). We will also still observe that a clock moving at velocity v in a given frame is slowed down by a factor of sqrt(1 - v^2/c^2), which would be another empirical way to determine the value of c.

Well, yes, I suppose the “measurement” can be indirect: even if you cannot find any object that travels at an “invariant speed” (as ascertained by direct measurements with clocks and rods over such hypothetical object), you can somehow say that the “invariant speed” has been indirectly measured with this procedure:

- Clock A has ticked time = t in A frame,
- But in B frame, where clock A is moving at velocity v, two clocks have registered for that interval a longer time = t’.
- So t’ = t / sqrt(1 - v^2/s^2). By solving for s, we get a finite value for that invariant speed…

And if instead, like the t=t’ model claims, the two observers measure the same time intervals, if they solve for s they find an infinite speed.

Given this, with this definition of “invariant speed” as a pure mathematical object not necessarily existing physically, we would have made a great progress and I would admit that the equation in question is actually a “generalized transform” that perfectly accommodates both the LT (if the “invariant speed” is finite) and the GT (if it’s infinite).

If so, my only concern would be only whether this expression is somehow a little “biased in its external form” (sorry for the imprecision, I cannot be more precise now) or at least whether it’d be possible to derive an equally generalized transform with a different “look”. But please don’t pay attention to that now. I think that I could make the question clearer after a similar definition effort applied to the principle of relativity. Therefore, if you agree on the former definition of “invariant speed”, I would attempt in the next post to define with my words the “principle of relativity”, for your correction.

 

[Saw]

Although you may still want to comment th previous post, I try to make more progress in the target of agreeing on definitions.

As to the t=t’ model that is at the same time relativist, that is what you describe physically as one with moveable aether or with light traveling at constant speed relative to the emitter. I don’t know how it came to appear we disagreed on that.

As to the Principle of Relativity (PoR), we agreed it means that:

If we have two identical experiments carried out in two different frames (identical apparatuses, identical initial coordinate values), you get identical results in both frames (final coordinate values in S = final coordinate values in S’) = if a law of physics applies in S, the same law holds good in S’ = there is no way for experimenters to discern whether their lab is stationary or in relative motion.

Then I mentioned a second issue, indicating that this is a sort of second requirement of the PoR:

Second requirement: for a given unique experiment carried out in one frame, the two frames may obtain different coordinates; however, you can relate the coordinates in frame S with the coordinates in S’ for the same object… through a transformation equation.

About the first phrase,

Second requirement: for a given unique experiment carried out in one frame, the two frames may obtain different coordinates

you commented:

No, not if the coordinate description of the initial setup was identical (including details like the initial distance of various components of the apparatus from the origin). In this case the coordinate description of the results must be identical too.

I suppose there was a misunderstanding here. I was not repeating the same thing (comparison between the coordinates of object A in S and the coordinates of object A’ in S’) but talking about another thing (comparison between the coordinates of the same object A in S and S’). I imagine that you agree that, for a single experiment, even if the initial coordinate description is identical (x=x’= 0 m and t = t’ = 0 s), the final coordinates in each frame may differ. For example, in the t = t’ model, after “some time”, the time coordinate will be the same, but the x coordinate will be different. And in a t≠t’ model, both the t and the x coordinates will differ.


On the second phrase

however, you can relate the coordinates in frame S with the coordinates in S’ for the same object… through a transformation equation.

you commented:

No! If that was all that was required, then even in a Galilean aether model where light moved at c in all directions in a preferred frame but moved at c+v in one direction and c-v in the other direction in some other inertial frame, then your version of the "first postulate" would still be satisfied since we could related the coordinates back to the preferred frame via a coordinate transformation. In fact this would make the first postulate into a tautology, since all coordinate transformations are just different ways of labeling the same physical events, so if you have the description of some results in coordinate system A, and you transform these events into coordinate system B and look at the description in B, it's automatically going to be true that if you perform the inverse transformation on these results in B you'll get back the original description you had in A.

Again I suppose here there was another misunderstanding. I didn’t say that that (transformation) was all that is required in the PoR. What I said was that this was a second requirement of the PoR, not that it was its only requirement. Is that wrong? Isn’t that what is meant when it’s stated that the laws of physics are invariant “under a transformation”? For example, you analyze a single experiment in two frames, one with an opening and a closing event. The coordinates in each frame for the closing event differ, but there is an equation relating them, so that if you know (i) the coordinates in S and (ii) the velocity of other frames wrt to S, you can figure out the coordinates in all other frames for such event. That is inherent to any relativist model, insn't it?

To sum up, I was saying that all relativist models are transformational. What you pointed out is that a model can be transformational without being relativist, but –in spite of the “No!” with which you start your paragraph- that does not contradict my assertion (does it?), since the set of relativists models can be a subset of a wider group of transformational models, where there would also exist transformational but not relativist models…

However, on further thinking I have doubts that it is possible to have transformational models that are not relativists. You mention the case of the immovable aether model, in which light only travels at c in the aether frame, but it can travel at other speeds in other frames. No problem with stating that such model is not relativist. What I doubt, exclusively, is that you can still deem it “transformational”, if it is not relativist. Why? Let us take a ship in frame S’ traveling at +v wrt the aether frame = S. A light beam is projected towards the ship. From S we can guess that in S’ light travels at (c – v). And inversely from S’ we can guess that light travels wrt S at (c-v)+v= c. So far, so good. But if we actually measure the speed of the light beam in S’, with clocks and rulers…, would we get c-v? For that purpose, don’t we need that in S and S’ time intervals are the same and in general t = t’? But if time is what is measured with clocks and the measurement of a time interval in clocks S and S’ is nothing but two physical experiments… isn’t t = t’ the same as saying that the results of the clock experiments are identical? I am not totally sure, but I tend to think that all transformational models must be relativist. What is your opinion?

 

[yuiop]

Section 7.1.2 of this Wikipedia article http://en.wikipedia.org/wiki/Lorent...mation_matrices_consistent_with_group_axioms" it claims to derive the Lorentz transformation from the first postulate only without having to explicitly require the constancy of the speed of light. Any thoughts?

In the http://en.wikipedia.org/wiki/Lorentz_transformation#Lorentz_transformations" section of the same Wikpedia article it states "the Michelson–Morley experiment showed that it [the speed of light] is an absolute speed" which I think is a little misleading because the MM experiment does not by itself rule out the ballistic theory of light. It is only when you establish that the velocity of light is independent of the velocity of the light source that the MM experiment has any descriminating value. In the very next section it does acknowledge that "Another condition is that the speed of light must be independent of the reference frame, in practice [independent] of the velocity of the light source".

 

[Saw]

section 7.23 of the same article, http://en.wikipedia.org/wiki/Lorentz_transformation#Principle_of_relativity" it claims to derive the Lorentz transformation from the first postulate only without having to explicitly require the constancy of the speed of light. Any thoughts?

To me, it seems clear that you cannot derive the LT only on the basis of the PoR. However, the way in which light, and the speed of light, and its constancy or not constancy, may play a role in that derivation is a tricky subject. This thread has been very revealing in that respect and I’ll tell you how I see things now.

The idea that the speed of light is “constant” (always the “same” in different situations) can have three different meanings:

(1) Always the same, no matter the source, only in a preferred frame, not in any other frame:

This will happen if (a) light is a perturbation in a medium (the aether) that is immovable and (b) time is frame-invariant.

(2) Always the same, in the frame where the source is at rest, not in any other frame:

This will happen if:

2.a) either light is a perturbation in a medium (the aether) that moves with the source
2.b) or what you call the ballistic model: light is like any massive particle: somehow the photon travels by inertia with the source and when accelerated its acquired state of motion is added to the state of motion of the source.

and also time is frame-invariant.

(3) Always the same, in any frame, the source frame and any other observing frame.

This requires that time is frame-variant. This model doesn’t make any particular physical assumption, although it is compatible with the idea that light is a perturbation in a medium (the aether) that is immovable (I think it’s not compatible with the moveable aether or the ballistic theory).

Coming back to your point, the teaching of this thread for me is that derivations in the style of the http://arxiv.org/PS_cache/physics/pdf/0302/0302045v1.pdf" show that light plays a role before and after the derivation, though not inside the derivation itself.

If you start with the PoR as unique assumption:

First, you start by ruling out model (1), because it is not relativist. I think that this is in line with what Einstein said when he said he appreciated a clash between the PoR and the “constancy of the speed of light”. The clash was with meaning (1) and not with meaning (2), as you point out.

Second, admitting models (2) and (3) as both complying with the PoR, you get a generalised transformation which becomes the GT or the LT, depending on the value of a certain constant, which in turn contains an invariant speed. The latter is a mathematical construct that does not necessarily corresponding to the speed of any physical object and in particular does not have to be the speed of light.

Given this:

- If time is experimentally proved to be frame-invariant, then the invariant speed turns out to be infinite, the constant is zero and model (2) and the GT are successful.

- If instead time is proved to be relative, then the invariant speed is finite, the constant is not zero and the LT succeeds.

So in the end to reach the LT you do need experiment and, as you say, not the MM experiment (which is also compatible with the GT) but others.

And what is the physical reason, the dynamical (and not merely kinematical) explanation why experiments may prove time to be relative and the LT true?

Well, to start with, the fact that light, as you point out, does not take the motion of the source (you cannot hold a photon in your hands as if it were a pie, as in Einstein’s teenager’s dream) rules out model (2). Taking one step further and choosing model (3) seems to require that the speed of light is precisely the invariant speed that appears in the derivation. It seems that commentators suggest that it’s one of these two: either (i) speed of light = the invariant speed, if photons are perfectly massless or (ii) even if photons were not perfectly massless, the LT would still be true but its invariant speed would be… (now I am asking) some number > than the speed of the physical object called light?

 

[yuiop]

Hi Saw,

I think that is nice summary in your last post. I would just like to add a couple of additional observations.

The idea that the speed of light is “constant” (always the “same” in different situations) can have three different meanings:

(1) Always the same, no matter the source, only in a preferred frame, not in any other frame:

This will happen if (a) light is a perturbation in a medium (the aether) that is immovable and (b) time is frame-invariant.

(2) Always the same, in the frame where the source is at rest, not in any other frame:

This will happen if:

2.a) either light is a perturbation in a medium (the aether) that moves with the source
2.b) or what you call the ballistic model: light is like any massive particle: somehow the photon travels by inertia with the source and when accelerated its acquired state of motion is added to the state of motion of the source.

and also time is frame-invariant.

(3) Always the same, in any frame, the source frame and any other observing frame.

This requires that time is frame-variant. This model doesn’t make any particular physical assumption, although it is compatible with the idea that light is a perturbation in a medium (the aether) that is immovable (I think it’s not compatible with the moveable aether or the ballistic theory).

I think when mentioning the aether it is important to mention that are basically two types of aether. There is the pre-Lorentz Newtonian aether which is imovable and does not affect the clock rates and lengths of material moving relative to it and there is the Lorentz Ether which has the property that time dilates and length contracts for any matter moving relative to the Lorentz Ether. In other words the Lorentz Ether physically interacts with matter (massive particles) whereas the classic aether just provides a medium for light waves to travel in and has no interaction with massive particles.

https://www.physicsforums.com/showthread.php?t=382163&page=8
Well, to start with, the fact that light, as you point out, does not take the motion of the source (you cannot hold a photon in your hands as if it were a pie, as in Einstein’s teenager’s dream) rules out model (2). Taking one step further and choosing model (3) seems to require that the speed of light is precisely the invariant speed that appears in the derivation. It seems that commentators suggest that it’s one of these two: either (i) speed of light = the invariant speed, if photons are perfectly massless or (ii) even if photons were not perfectly massless, the LT would still be true but its invariant speed would be… (now I am asking) some number > than the speed of the physical object called light?

Your last question is very tricky and I am not totally sure of the answer to it, but it was being disussed in this recent thread https://www.physicsforums.com/showthread.php?t=387039

 

[Saw]

even if photons were not perfectly massless, the LT would still be true but its invariant speed would be… (now I am asking) some number > than the speed of the physical object called light?

Your last question is very tricky and I am not totally sure of the answer to it, but it was being disussed in this recent thread https://www.physicsforums.com/showthread.php?t=387039

Thanks, I looked at that thread. So, as to the question whether it’s theoretically possible that light has vanishingly rest mass, there are diverging opinions. But it seems you are answering the question posed here: if light had some miniscule mass, it would travel at speed less than c, at a speed that would not be perfectly invariant…

Anyhow, coming back to the meaning of the http://arxiv.org/PS_cache/physics/pdf/0302/0302045v1.pdf" , I have been thinking that the idea that such derivation gives a generalized transform comprising both the GT and the LT may be judged under this analogy:

There is a general concept of what we could call the “closed circuit”, which can be:

- A rectilinear circuit around a centre, with an invariant radius.
- A circumference = a curve around a centre, also with an invariant radius.
- An ellipse = a curve around a centre, but flattened, so that it has a minor axis (2b) ≠ major axis (2a), although with another invariant element, which is here (instead of the radius) the sum of the two radio-vectors (distance from a point of the curve to one focus + distance to the other, which is always = 2a). The degree of flattening of an ellipse is its eccentricity:

$$e = \frac{c}{a} = \sqrt {1 - \frac{{b^2 }}{{a^2 }}}$$


where c = distance from centre to each focus, a = semi-major axis and b = semi-minor axis

Then, well, we can choose to consider that the most complicated case, i.e. the ellipse, is the “general” case and the other two are its special and “degenerate” cases, because in them the nature of the figure is changing:

- In the case of the circumference, there is no flatness, no eccentricity (e=0), because a = b = radius.
- In the case of the rectilinear circuit, there is total flatness, the maximum possible eccentricity (e = 1), because a = c… and in the second formula, could we also say that e = 1 because either b = 0… or a = ∞?

Applying this reasoning by analogy to the derivation of the http://arxiv.org/PS_cache/physics/pdf/0302/0302045v1.pdf" , I tend to think, now again, that what such derivation obtains is really the LT, although one could say that it also accommodates the GT as a “degenerate case”, i.e. one where the figure changes nature and really becomes something else. Is all this very far from truth?