Need to resort to spherical wavefront to derive the LTs?

[Saw]

TL;DR Summary

Possible weakness of the derivation of the ST interval based on the spherical wavefront

I have been reading Wikipedia’s derivations of the Lorentz Transformations. Many of them start with the equation of a spherical wavefront and this reasoning:

- We are asked to imagine two events: light is emitted at 1 and absorbed somewhere else at 2. For a given reference frame, the distance traversed by the light in 3D is given by the Pythagorean Theorem:

$$\sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2} + {{({z_2} - {z_1})}^2}}$$

- But this distance can also be expressed as the product of the speed of light by the time elapsed between events 1 and 2:

$$c({t_2} - {t_1}) = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2} + {{({z_2} - {z_1})}^2}}$$

- Putting the intervals in short form and squaring both sides, we get:

$${(c\Delta t)^2} = \Delta {x^2} + \Delta {y^2} + \Delta {z^2}$$

- And rearranging:

$${(c\Delta t)^2} - (\Delta {x^2} + \Delta {y^2} + \Delta {z^2}) = 0$$

- If the speed of light is constant, another primed frame should obtain the same result, so we get:

$${(c\Delta t)^2} - (\Delta {x^2} + \Delta {y^2} + \Delta {z^2}) = {(c\Delta t')^2} - (\Delta x{'^2} + \Delta y{'^2} + \Delta z{'^2}) = 0$$

- Then the derivations jump to the simpler case where the relative velocity between the two frames is along an overlapping X axis, so you can neglect the Y and Z dimensions and continue the derivation of the Lorentz Transformation with this expression:

$${(c\Delta t)^2} - \Delta {x^2} = {(c\Delta t')^2} - \Delta x{'^2} = 0$$

But if this reasoning is correct (or at least complete), then one must also be able to replicate it by considering from the start a situation where only the X axis is relevant, instead of introducing it at the end. However, if you repeat the same train of thought with this start, what you get is this:

- The distance traversed by light between events 1 and 2 is:

$$\begin{array}{l} c({t_2} - {t_1}) = {x_2} - {x_1}\\ c\Delta t = \Delta x \end{array}$$

- Rearranging:

$$c\Delta t - \Delta x = 0$$

- If the speed of light is constant, another primed frame should obtain the same result, so we get:
$$c\Delta t - \Delta x = 0 = c\Delta t' - \Delta x'$$

Ok, it is true that, when we were confronted with the equality between the c * time interval and space interval, we could have squared both sides, but we could also have raised both sides to the 11th power (in fact, we could also have done that when playing with the wavefront formula)… and would that mean that the spacetime interval must include that power?

 

[PeterDonis]

if this reasoning is correct (or at least complete), then one must also be able to replicate it by considering from the start a situation where only the X axis is relevant

This is not correct. Spacetime is 4 dimensional, not 2 dimensional. So you can't do a valid derivation in just 2 dimensions. You have to look at the 4d case and see what it requires, before you restrict to 2d to get a simpler form for that special case.

 

[Ibix]

The point about dropping the $\Delta y^2$ and $\Delta z^2$ terms is that one can reason that $\Delta y^2+\Delta z^2=\Delta y'^2+\Delta z'^2$. They aren't dropped just because we'd like a 1d problem - there is a formal physical argument that those terms cancel out. So it is not that we're using a 1d case; rather that we've shown that the other two dimensions don't turn out to be relevant.

 

[Dale]

TL;DR Summary: Possible weakness of the derivation of the ST interval based on the spherical wavefront

The distance traversed by light between events 1 and 2 is …
we could have squared both sides

In addition to the other responses, you are forgetting that light travels in both directions. So even for 1+1D you need to square your intervals. Squaring both sides isn’t optional.

 

[Ibix]

In addition to the other responses, you are forgetting that light travels in both directions. So even for 1+1D you need to square your intervals.

Although that's strictly only an argument against odd powers - $\Delta t^4-\Delta x^4=0$ also satisfies this. I'm not sure you can eliminate higher even powers from consideration except by including a second spatial dimension and appealing to Pythagoras (as the original proof quoted in the OP does) or introducing elements from other proofs such as requiring linearity of the transforms.

 

[Saw]

This is not correct. Spacetime is 4 dimensional, not 2 dimensional. So you can't do a valid derivation in just 2 dimensions. You have to look at the 4d case and see what it requires, before you restrict to 2d to get a simpler form for that special case.

Hmmm... I thought that might be the reason, at some time. But in the end I was not convinced. Spacetime is what the problem at hand requires. It will certainly require at least two dimensions, namely ct and one spatial dimension, say X, but if I decide to live without Y and Z (i.e. restrict my problems to frames that share X), why should I not be able to make a proper derivation?

In addition to the other responses, you are forgetting that light travels in both directions. So even for 1+1D you need to square your intervals. Squaring both sides isn’t optional.

I did not catch that. Why should considering that light travels in both directions require the squares?

Although that's strictly only an argument against odd powers - $\Delta t^4-Delta x^4=0$ also satisfies this. I'm not sure you can eliminate higher even powers from consideration except by including a second spatial dimension and appealing to Pythagoras (as the original proof quoted in the OP does) or introducing elements from other proofs such as requiring linearity of the transforms.

Maybe you can elaborate on this. What is what makes the difference between odd and even powers?

 

[Dale]

I'm not sure you can eliminate higher even powers from consideration except by including a second spatial dimension and appealing to Pythagoras (as the original proof quoted in the OP does) or introducing elements from other proofs such as requiring linearity of the transforms.

Yes, I agree. Or more importantly for physics, through experiment.

 

[PeterDonis]

if I decide to live without Y and Z (i.e. restrict my problems to frames that share X), why should I not be able to make a proper derivation?

Because you can't just handwave away that spacetime has four dimensions, and the general laws of physics, which is what you're talking about here (since those laws have to be Lorentz invariant), don't care that you are only interested in a two dimensional special case.

 

[Ibix]

but if I decide to live without Y and Z (i.e. restrict my problems to frames that share X), why should I not be able to make a proper derivation?

You can do - just square your coordinate deltas. Your un-squared form must be wrong anyway because of the sign issues, and I rather suspect that if you try higher powers you'll find that your transforms can't be linear and thus you can't have inertial frames.

Maybe you can elaborate on this. What is what makes the difference between odd and even powers?

If you swap the direction of your x axis (or the direction of propagation of light) then $\Delta t$ doesn't change but $\Delta x$ swaps sign. Thus $c\Delta t\neq\Delta x$ in one direction or other. The sign goes away if the deltas are raised to an even power.

 

[Dale]

Why should considering that light travels in both directions require the squares?

$$c\Delta t - \Delta x =0$$$$c(t-t_0)-(x-x_0)=0$$$$x=ct+( x_0 - c t_0)$$This describes something moving at $c$ in the $+x$ direction and excludes anything moving at $c$ in the $-x$ direction. So it is insufficient to describe light.

 

[pervect]

One other problem with $c \Delta t - \Delta x = 0$ is that it's only true for light propagating in the +x direction. Light propagating in the opposite direction satisfies $c \Delta t + \Delta x = 0$. But $c^2 \Delta t^2$ is always equal to $\Delta x^2$, regardless of the direction of propagation. So only the statement with the squares is always true. Furthermore, the solution for $c^2 \Delta t^2 = \Delta x^2$ is just either case, i.e. either $c \Delta t = \Delta x$ or $c \Delta t = - \Delta x$. Writing it as one equation removes the need to consider the direction of propagation.

 

[Saw]

the general laws of physics, which is what you're talking about here (since those laws have to be Lorentz invariant)

Well, that the laws of physics are Lorentz invariant is out of question. The question is just whether deriving this Lorentz invariance based on the logic reproduced in the OP is enough or you must make some further assumptions.

If you swap the direction of your x axis (or the direction of propagation of light) then $\Delta t$ doesn't change but $\Delta x$ swaps sign. Thus $c\Delta t\neq\Delta x$ in one direction or other. The sign goes away if the deltas are raised to an even power.

I see, understood. But answering also Dale's later comment (also Pervect's), what you all mean then is that the invariant spacetime interval must consider the absolute values... Would it be enough in that case, in order to derive the interval in 1D for both directions, to stipulate the following expression?

$$\sqrt {{{(c\Delta t)}^2}} - \sqrt {\Delta {x^2}} = \sqrt {{{(c\Delta t')}^2}} - \sqrt {\Delta x{'^2}}$$

It would still be wrong, of course, but it would still be the ST interval flowing from not making further assumptions...

I rather suspect that if you try higher powers you'll find that your transforms can't be linear

But then you are making a further assumption consisting of linearity. The thing is, however, that the derivations that dissatisfy me only make such assumption about the linearity of the transformation at a further stage. And if you do introduce linearity in the deduction of the ST interval, is it linearity with regard to space or, as it seems logical, also with regard to time (rather, c*time)?

 

[Ibix]

But then you are making a further assumption consisting of linearity

...but only because you've decided to pretend space is one dimensional for some reason.

 

[PeterDonis]

The question is just whether deriving this Lorentz invariance based on the logic reproduced in the OP is enough or you must make some further assumptions.

You're not deriving Lorentz invariance in the OP. You're assuming it, by assuming that the speed of light is the same in both frames, and then exploring what that means. And to do that correctly, you must take into account that spacetime has 4 dimensions. You cannot ignore two of them just because you want to consider the special case of motion in a single direction, because Lorentz invariance applies to all cases, not just that special case. Your claim that since you are only interested in the 2d case you should be able to do a derivation in 2d is simply wrong.

 

[Saw]

Before replying to other comments, would we agree that the objection about the fact that light moves also in the negative direction is duly answered by my reply to Ibix in post #12?

 

[PeterDonis]

would we agree that the objection about the fact that light moves also in the negative direction is duly answered by my reply to Ibix in post #12?

No. Light moving in the negative $x$ direction is moving in the negative $x$ direction. You can't just take absolute values and ignore that.

And all that is only applicable in any case if you adopt the claim that you should be able to do a derivation restricted to 2d if you are only interested in motion in one direction, which, as I have already pointed out, is wrong.

 

[Saw]

No. Light moving in the negative $x$ direction is moving in the negative $x$ direction. You can't just take absolute values and ignore that.

Hmm… I am not ignoring that light is moving in the negative direction. Instead of that, thanking and accommodating the comment of those who made it, I proposed an alternative form of ST interval in 1 spatial direction that would be valid for light traveling at an invariant speed either in the negative or the positive axis. By the way, probably such improvised form could dispense with the square root of the square of ct and only keep it for the x interval.

If this is ok, of which certainly I am not sure, we can move on to continue discussing on whether it is wrong or not the claim that a valid derivation should be possible with just X axis, only with further assumptions that may be implicit in the customary 3D derivation.

 

[PeterDonis]

If this is ok, of which certainly I am not sure, we can move on to continue discussing on whether it is wrong or not the claim that a valid derivation should be possible with just X axis, only with further assumptions that may be implicit in the customary 3D derivation.

You've got it backwards. If your claim that a valid derivation is possible with just $t$ and $x$ is wrong--which it is--then everything else you are saying is simply irrelevant. That claim has to be decided first, before it even makes sense to investigate what kind of derivation with just $t$ and $x$ might work.

 

[PeterDonis]

probably such improvised form could dispense with the square root of the square of ct

You can't, because the Lorentz transformation also has to work for intervals where $t_2 < t_1$. You can't handwave that away any more than you can handwave away that spacetime has four dimensions, not two.

 

[Dale]

Would it be enough in that case, in order to derive the interval in 1D for both directions, to stipulate the following expression?

That expression would work for 0, but it would not be compatible with experiment for other values.

Edit: @PeterDonis has a good point too

 

[Saw]

You can't, because the Lorentz transformation also has to work for intervals where $t_2 < t_1$. You can't handwave that away any more than you can handwave away that spacetime has four dimensions, not two.

Edit: @PeterDonis has a good point too

Then we come back to the original expression =

$$\sqrt {{{(c\Delta t)}^2}} - \sqrt {\Delta {x^2}} = \sqrt {{{(c\Delta t')}^2}} - \sqrt {\Delta x{'^2}}$$
which (let me highlight it) is not meant to be a valid one: it is just what you get by applying the same (incomplete) logic that is applied in the customary derivation of ST interval with 3 spatial dimensions.

That expression would work for 0, but it would not be compatible with experiment for other values.

Sure. That is the problem. If you apply the (incomplete) logic in the simple case, you don't arrive at the LTs, which is what is compatible with experiments. My point is only that the customary derivation of the ST interval, which is based on 3 spatial dimensions and does get to the right destination, only does it because it contains other non-declared assumptions and that if you did apply such assumptions in the simple case (1 spatial dimension), you would also arrive at the LTs and agreement with experiment.

You can't (...) handwave away that spacetime has four dimensions, not two.

I am not denying that, if you want to solve all problems that reality may confront you with, you will need to consider four dimensions, if that is what you mean.

But I do claim that the same logic playing in the more "advanced" or "complicated" situation must also shine out in the "elementary" or "simple" case, even if it takes a simpler form.

Far from being "handwaving", this is just the orthodox and habitual line of thought in mathematics.

Take the example of finding the modulus of a vector in ordinary physical space. Here the logic is "get rid of direction and focus on quantity". If the components of the vector (x, y) are expressed in an orthonormal base, you can obtain such an outcome by applying the Pythagorean Theorem, because this does in a quick manner what you could slowly do with these steps:

- Multiply the vector by itself in full form:
$$\left( {x*{\bf{i}} + {\rm{ }}y*{\bf{j}}} \right){\rm{ }}\left( {x*{\bf{i}} + {\rm{ }}y*{\bf{j}}} \right){\rm{ }} = x*{\bf{i}}\;x*{\bf{i}}\; + x*{\bf{i}}*y*{\bf{j}} + y*{\bf{j}}{\rm{ }}*x*{\bf{i}} + y*{\bf{j}}{\rm{ }}*y*{\bf{j}}$$
- Note that the two terms in the middle vanish out because you are multiplying orthogonal vectors, whose dot product is 0.
- Note that the two terms at the extremes reduce to scalar multiplication because you are multiplying a unitary vector by itself, this product being 1.
- Thus you get an expression where you have actually dispensed with direction and obtained only modulus:
$$x*x + {\rm{ }}y*y = {x^2} + {\rm{ }}{y^2}$$
- But you realize that this way you have overshot since what you actually got was the square of the modulus, so you apply the square root:
$$\sqrt {{x^2} + {\rm{ }}{y^2}}$$

What if you are faced now with a simpler situation where you need the modulus of what is only a scalar, say a real number, without direction? It still may have a negative sign, so you multiply the number by itself to get rid of the sign and finally apply the square root, thus obtaining the so-called "absolute value". You could say, however, that also this absolute value is the result of applying the Pythagorean Theorem, only with y = 0.

Within the realm of scalars, the same logic applies in the comparison between real and complex numbers: to get the magnitude of a complex number, you multiply it by its complex conjugate (thus getting rid of the phase) and then apply the square root (also because you had overshot); what you do in order to obtain the absolute value of a real number is the same, based on the same logic, it is just that the imaginary part is 0.

Ok, you realize this a posteriori, once you are familiar with the logic of the full form, but the important thing is that by applying the same (complete) logic in the advanced and elementary cases, you do get equal and correct solutions in both cases. The complete form, after simplification (y=0), reduces to the simpler form that can be directly inferred in the simpler case by applying an identical rationale.

In the case at hand, however, the complete form reduces (after simplification) to:

$${(c\Delta t)^2} - \Delta {x^2} = {(c\Delta t')^2} - \Delta x{'^2}$$
which is not the same expression that we had derived by applying the same logic starting at the simpler scenario:
$$\sqrt {{{(c\Delta t)}^2}} - \sqrt {\Delta {x^2}} = \sqrt {{{(c\Delta t')}^2}} - \sqrt {\Delta x{'^2}}$$

To sum up, the customary derivation of the ST interval, by applying an incomplete logic in the advanced case, does get the right answer, only because it plays with implicit assumptions, but applying the same incomplete logic in the simple case leads to failure, only because we have inadvertently dropped the implicit assumptions. Give me the complete logic, i.e. the hidden assumptions acting in the derivation made in spatial 3D with cT and we should also be capable of deriving the ST interval in 1D with cT.

 

[Ibix]

But I do claim that the same logic playing in the more "advanced" or "complicated" situation must also shine out in the "elementary" or "simple" case, even if it takes a simpler form.

That hope would seem to be false by your own example. To me, the most you can be showing is that this derivation wouldn't work if the universe were one dimensional. So what? It isn't.

The full derivation includes an explanation for why the $\Delta y$ and $\Delta z$ terms cancel with their primed counterparts (can't speak for whatever you read, of course). It isn't simply waving its hands and reducing to 1d. You are throwing away two dimensions without any justification, and then seem surprised when this has consequences for your reasoning.

 

[Saw]

That hope would seem to be false by your own example.

Really? Why? I think it is hard to deny that, in the modulus example, the same complete logic leads to fine solutions both in the complicated and the simple case.

To me, the most you can be showing is that this derivation wouldn't work if the universe were one dimensional. So what? It isn't.

The derivation can perfectly work in a one dimensional (rather two dimensional: X and cT) world if it speaks out all its assumptions.

You are throwing away two dimensions without any justification, and then seem surprised when this has consequences for your reasoning.

No, I am not throwing away the Y and Z dimensions. They are still there, also for me, even if I have the legitimate expectation to be able to derive what happens with X and cT in the very real and plausible case where Y and Z are irrelevant. Same way that, when I deduce -based on full logic- how to calculate the absolute value of a real number, I am not throwing away complex numbers or vectors, I am not claiming that the universe does not have them.

A different thing is that a trick for finding the deep rationale of a rule, which is very appreciated by mathematicians, is looking at the more complicated case. But the rationale is the same and works the same in both cases, simple and complex.

 

[Ibix]

Really? Why?

How are you going to get from $\sqrt{\left(c^2\Delta t^2\right)}-\sqrt{\left(\Delta x^2\right)}=\sqrt{\left(c^2\Delta t'^2\right)}-\sqrt{\left(\Delta x'^2\right)}$ to the Lorentz transforms?

The derivation can perfectly work in a one dimensional (rather two dimensional: X and cT) world if it speaks out all its assumptions.

I don't understand what assumptions you think are missing in the 3d case. In the 1d case you have to add extra stuff to get to the squared coordinate differences being the correct things to equate, sure, but that's what I mean about dropping dimensions without justification having consequences.

They are still there, also for me,

Fair enough. Then you know you need to use Pythagoras to calculate moduli of coordinate differences and your trying to equate the root-square of the coordinate differences is obviously not a general rule. So it's hardly surprising you don't get the correct general transformation rule.

 

[Dale]

That is the problem. If you apply the (incomplete) logic in the simple case, you don't arrive at the LTs, which is what is compatible with experiments.

That is not a problem. That is science. Needing compatibility with experiment is always required for science, by design. It’s a feature, not a bug

Give me the complete logic,

The formula needs to accurately predict the outcome of experiments.

My point is only that the customary derivation of the ST interval, which is based on 3 spatial dimensions and does get to the right destination, only does it because it contains other non-declared assumptions

I don't think that this conclusion follows from your logic. Why would trouble in your 1+1D approach indicate the existence of a hidden assumption in the usual 1+3D approach? If there is an assumption in the 1+3D case then that should be visible directly and not require investigation of a 1+1D case. And if a 1+1D case fails how can you attribute that to an assumption in a 1+3D derivation rather than just a problem from the reduced richness of the geometry.

 

[Sagittarius A-Star]

Give me the complete logic, i.e. the hidden assumptions acting in the derivation made in spatial 3D with cT and we should also be capable of deriving the ST interval in 1D with cT.

Do you have a link to such a a derivation, which uses hidden assumptions?

The following linked paper shows a derivation of the LT in 1D with cT. It proves, that, if equations (1) and (2) are true at $F$ (on the world line through the origin), then they are also true at $E$ (not on this world line).

Adding and subtracting equations (1) and (2) yields the LT.
Multiplying equations (1) and (2) yields the ST interval invariance.

Source ("World’s Fastest Derivation of the Lorentz Transformation"):
http://www.faculty.luther.edu/~macdonal/LorentzT/LorentzT.html

 

[Saw]

Ibix, we are going in circles...
It is obvious that the above expression is wrong. Our discrepancy is just that:
- You say it is wrong because I get it after dropping dimensions.
- I say that it is wrong because it is derived based on the incomplete assumption that c is invariant, which is the only declared assumption of the 3D customary derivation of the ST interval.
Which assumptions I am missing? We could talk about that maybe in another thread, but I wanted to discuss here just two things:
A) If the customary derivation shows in itself some degree of mathematical inconsistency, thus betraying that its logic is incomplete. For example, the "squaring both sides" issue. You yourself admitted that squaring to odd powers could be a problem. I made an adaptation of the wrong formula to show that there could be an issue also with even powers. What to do with that? This inconsistency could be solved if you make some further assumption in the sense that you *must* use the squares because this or that.
B) The statement that you should be able to derive right equations in a simplified scenario (eg: only X and cT) but with complete assumptions (an example about modulus of a vector as a generalization of the concept of the absolute value of a scalar; this is pure illustration; don't confuse it with the wrong derivation of the ST, which -I repeat- I am not advocating ). I made an effort in #21 to prove that this is day-to-day business in math. You seem to have totally missed that argument. When I asked "Really? Why" it was because I thought that you were dismissing the said attempt as a failure. I would kindly ask you to re-read that post with attention!

 

[Dale]

Which assumptions I am missing?

Maybe just the assumption that a problem in a 1+1D derivation implies a problem in a 1+3D derivation. It certainly doesn’t seem like a justified claim to me.

 

[Saw]

That is not a problem. That is science. Needing compatibility with experiment is always required for science, by design. It’s a feature, not a bug

The formula needs to accurately predict the outcome of experiments.

Did I say by chance that needing agreement with experiment is a problem? No, I did not say that. The problem is only that deriving the ST interval with only one assumption (c is invariant) gives an incorrect answer in a context (2 dimensions: X and cT) where such an assumption should also work, if it were in fact complete. By the way, I gave specific examples where (as it is a must) the correct logic (sufficient assumption) working in the advanced scenario also works (with the same assumptions) in the simple one, even if (obviously) under a simpler form. That is in post # 21. What is your answer to that?

Why would trouble in your 1+1D approach indicate the existence of a hidden assumption in the usual 1+3D approach?

It is not "my" trouble. It is the trouble of those who try to derive something without sufficient basis.

If there is an assumption in the 1+3D case then that should be visible directly and not require investigation of a 1+1D case.

It is visible. I did mention that the customary derivation takes a dubious step when "squaring both sides". Why not raising both sides to the 3rd power? Ibix accepted that this is a problem with all *odd* powers. Besides, I slightly modified the derivation (based on incomplete assumptions) in 1+1D case, not because I advocate such derivation, but to show that the problem also arises with *even* powers. Is that visible enough?

 

[Saw]

Do you have a link to such a a derivation, which uses hidden assumptions?

The following linked paper shows a derivation of the LT in 1D with cT. It proves, that, if equations (1) and (2) are true at $F$ (on the light world line through the origin), then they are also true at $E$ (not on this light world line).

Adding and subtracting equations (1) and (2) yields the LT.
Multiplying equations (1) and (2) yields the ST interval invariance.

Source ("World’s Fastest Derivation of the Lorentz Transformation"):
http://www.faculty.luther.edu/~macdonal/LorentzT/LorentzT.html

I will have a look at that paper, thanks, but maybe I should say that by "hidden assumptions" I don't mean anything mysterious. It may be something as simple as assuming that the cT dimension is orthogonal to the space dimension(s), as it is widely accepted. Based on this, you could postulate that the ST interval is what it is, either from the 1+1D scenario or the 1+3D one. Without it (or something alike; indeed discussing the missing assumptions, as I said, is not the object of the thread), the usual derivation seems to be making jumps in the air.

 

[Sagittarius A-Star]

I will have a look at that paper

I recommend this. It contains only one page and I think it is, what you are looking for, if you multiply the equations (1) and (2).

The $T+X$ and $T-X$ are light-cone coordinates.

 

[Dale]

The problem is only that deriving the ST interval with only one assumption (c is invariant)

I don't know anyone who does that. Einstein used relativity and invariance of c. Pal used relativity and experiment. Robertson used linearity and experiment. Minkowski started from the invariance of the spacetime interval. Susskind used the invariance of c and an explicit ansatz. It is certainly not the case that many derivations assume only the invariance of c.

Since the 1+3D Voight transforms have the invariance of c but differ from the Lorentz transforms, it is clear in 1+3D that the invariance of c is insufficient. My preference is to directly assume the spacetime interval and check it against experiment.

It is not "my" trouble. It is the trouble of those who try to derive something without sufficient basis.

It is your trouble because, as far as I can tell, you are the only one doing that.

I did mention that the customary derivation takes a dubious step when "squaring both sides". Why not raising both sides to the 3rd power? Ibix accepted that this is a problem with all *odd* powers. Besides, I slightly modified the derivation (based on incomplete assumptions) in 1+1D case, not because I advocate such derivation, but to show that the problem also arises with *even* powers.

It is not a dubious step. That choice is the one that is consistent with experiment.

 

[PeterDonis]

we are going in circles...

Yes, because you are refusing to listen to the responses you are getting. We can't have a discussion if you insist on repeating wrong things even after you have been shown they are wrong.

Thread closed.

 

[Dale]

After some discussion among the mentors the thread is reopened. In the interest of moving forward @Saw please post a link to a specific derivation that you think is logically deficient.

 

[strangerep]

Give me the complete logic, i.e. the hidden assumptions acting in the derivation made in spatial 3D with cT and we should also be capable of deriving the ST interval in 1D with cT.

OK, here is the piece you seem to be missing...

In the 1+3D derivation, one tacitly assumes spatial isotropy (invariance under rotation of coordinates in 3-space). That rules out, e.g., the taxicab metric for 3-space in favour of the usual Euclidean metric.

In 1+1D, the analogue of this is to assume parity invariance (no change under reversal of spatial coordinates).

If you haven't studied it already, you might enjoy the group theoretic derivation(s) of the Lorentz transformations. See section 8 in this Wikipedia page.