Need to resort to spherical wavefront to derive the LTs?

[Ibix]

I think the OP's problem is that he thinks that to get from $$\begin{eqnarray*}&&c^2\Delta t^2-\Delta x^2-\Delta y^2-\Delta z^2\\&=&c^2\Delta t'^2-\Delta x'^2-\Delta y'^2-\Delta z'^2\end{eqnarray*}$$ to ##c^2\Delta t^2-\Delta x^2=c^2\Delta t'^2-\Delta x'^2## the derivation must (to quote from #1) "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".

But that is not a correct description of this step. The point is that we can simply choose our ##x## and ##x'## axes to be parallel to the relative motion of the frames. Then we can observe from the principle of relativity that ##\Delta y^2+\Delta z^2=\Delta y'^2+\Delta z'^2## and let those terms cancel. There's no "jump to a simpler case" here. We're still considering the general case with potentially non-zero coordinate differences in the y and z directions, but they drop out of this equation.

The rest of the thread is the OP wanting the special case where the y and z coordinate differences are zero to generalise automatically to the case where they are non-zero. But the problem there is that ##\left(c\Delta t\right)^{2n}=\left(\Delta x\right)^{2n}## for any integer ##n## and he has no way to pick ##n=1## except to bring in some extra information. So the information that ##n=1## is the part missing from his 1d derivation. But it's already present in the 3d version because that's the only case where the transverse coordinate differences drop out.

 

[Saw]

After some discussion among the mentors the thread is reopened.

Thanks indeed, because the discussion was being for me very helpful, in order to clarify and improve my ideas. I understand that you have to strike a difficult balance between allowing debate and the need to veto crank-like or ignorant discussions, but I am glad that you understood that we are not in the second case.

In the interest of moving forward @Saw please post a link to a specific derivation that you think is logically deficient.

The one that triggered my concern is this.

It is called "hyperbolic rotation", but my objection is of course not against conceiving the LT as such. My concern is precisely that I tend to believe that, when choosing the ST interval in its unanimously accepted form (no matter if you consider 1+1D or 1+3D), you are already assuming that the change of perspective between the two frames consists of a hyperbolic rotation, which by the way looks like a very sensible thing to do, given how you measure cT and X-Y-X and what the problem at hand is.

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.

They are indeed enjoyable, although they jump directly to deriving the LTs and, for this purpose, they make all necessary assumptions one by one. My concern is only with a derivation that presents the ST as the outcome of assuming just invariance of c plus pure algebraic manipulations.

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).

This is not enough if it only refers to assumptions affecting the spatial dimensions. What is missing is an assumption about how the spatial dimensions X-Y-Z *and* cT relate together to produce the invariant interval.

I will also share a couple of thoughts that I have had after our last contact:

- The discussion has been complicated by a discrepancy about whether you can legitimately derive the ST interval from 1+1D instead of from 1+3D. But let us assume we are deriving from 1+3D. There you also square both sides at a given moment (see the OP) and ...

- "Squaring both sides of an equation" is a perfectly legitimate move in order to facilitate solving for the unknown/s of the equation. However, our purpose here is not solving for an unknown, but guessing how cT and X-Y-Z combine in two frames so that their combination coincides in the outcome, even if their individual values differ. In this different context, squaring both sides is a move that you can also make, but it requires a "plus" of justification, i.e. a new assumption, precisely in the sense that I pointed out above: you square because that makes the parameters behave in the manner that you presume correct.

 

[Saw]

I think the OP's problem is that he thinks that to get from $$\begin{eqnarray*}&&c^2\Delta t^2-\Delta x^2-\Delta y^2-\Delta z^2\\&=&c^2\Delta t'^2-\Delta x'^2-\Delta y'^2-\Delta z'^2\end{eqnarray*}$$ to ##c^2\Delta t^2-\Delta x^2=c^2\Delta t'^2-\Delta x'^2## the derivation must (to quote from #1) "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".

But that is not a correct description of this step. The point is that we can simply choose our ##x## and ##x'## axes to be parallel to the relative motion of the frames. Then we can observe from the principle of relativity that ##\Delta y^2+\Delta z^2=\Delta y'^2+\Delta z'^2## and let those terms cancel. There's no "jump to a simpler case" here. We're still considering the general case with potentially non-zero coordinate differences in the y and z directions, but they drop out of this equation.

No, that is not my problem. I am fine with the fact that you can move from the general case to the simpler case, by simply choosing that (i) the ##x## and ##x'## are parallel and (ii) there is no relative motion between the two frames other than in such axes. My problem, as noted in the previous post, is only with (no matter whether you start with the general or the simple case) stating that by "squaring both sides" (i.e. through a simple algebraic arrangement) you get to the ST interval, when in fact you can only take this step based on an assumption about how the two families of dimensions (cT and X-Y-Z) relate to each other.

 

[Ibix]

by "squaring both sides" (i.e. through a simple algebraic arrangement) you get to the ST interval, when in fact you can only take this step based on an assumption about how the two families of dimensions (cT and X-Y-Z) relate to each other.

But that's just Pythagoras' theorem.

 

[Saw]

But that's just Pythagoras' theorem.

Yes! But Pythagoras' theorem relating not only spatial dimensions among themselves but linking cT, on the one hand, with X-Y-Z, on the other hand.

 

[Ibix]

Yes! But Pythagoras' theorem relating not only spatial dimensions among themselves but linking cT, on the one hand, with X-Y-Z, on the other hand.

Huh? If you have two events separated by ##\Delta x##, ##\Delta y##, ##\Delta z##, ##\Delta t## then the distance between them is ##\sqrt{\Delta x^2+\Delta y^2+\Delta z^2}##. If those events are joined by a light ray then the distance must also be the speed of light times ##\Delta t##. Thus ##c^2\Delta t^2=\Delta x^2+\Delta y^2+\Delta z^2##, or ##c^2\Delta t^2-\Delta x^2-\Delta y^2-\Delta z^2=0##. Since the same argument can be made in the primed coordinates, this implies the expression we want. This is just Pythagoras' theorem, the definition of "speed", and a bit of algebra.

We can, of course, say that ##\sqrt{c^2\Delta t^2}-\sqrt{\Delta x^2-\Delta y^2-\Delta z^2}=0## if we want. It's correct. It's just not helpful in simplifying the expression.

 

[Dale]

The one that triggered my concern is this.

It is called "hyperbolic rotation",

That derivation seems logically sound to me. As I objected above.

EDIT: I no longer believe that this derivation is logically sound. See the next post below.

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. ... It is certainly not the case that many derivations assume only the invariance of c.

I had not seen this derivation, but as I said above it is not assuming only a spherical wavefront. They assume a spherical wavefront AND linearity. That seems sufficient to me.

I don’t think that going to 1+1D implies any weakness of this derivation.

 

[Dale]

The one that triggered my concern is this.

That derivation seems logically sound to me.

Actually, I agree with you that this derivation is not logically sound as stated. It says that the two assumptions are

1) the spherical wavefront (I am going to use units where ##c=1##): $$t^2 -(x^2+y^2+z^2)=t'^2-(x'^2+y'^2+z'^2)=0$$ 2) linearity

However they throw in an unacknowledged 3rd assumption that ##y=y'## and ##z=z'##.

If we use the Voight transform $$t'=t-vx$$$$x'=x-vt$$$$y'=y/\gamma$$$$z'=z/\gamma$$ then we see that it also satisfies the stated assumptions 1) and 2) but is excluded by the unstated assumption ##y=y'## and##z=z'##.

So I agree that this specific derivation is a bad one. I think that identifying and pointing out the unstated additional assumption is easier to do using the full 1+3D transform than trying to analyze a 1+1D transform. So I disagree with your method, but agree with your claim that this specific derivation is poor.

 

[Ibix]

I'm not sure I would call linearity an assumption at all - certainly not an independent one. Any transform between inertial frames must be linear because it must map straight lines (##x=vt+x_0##) to straight lines (##x'=v't'+x'_0##). So I think it's part and parcel of the homogeneity/isotropy that we're already using in Pythagoras.

On the other hand, the principle of relativity is an assumption you need to get the Lorentz transforms. The Voight transforms don't respect the principle of relativity completely - they use different measurement scales depending on your frame's velocity with respect to some arbitrarily chosen inertial frame.

 

[Dale]

On the other hand, the principle of relativity is an assumption you need to get the Lorentz transforms. The Voight transforms don't respect the principle of relativity completely - they use different measurement scales depending on your frame's velocity with respect to some arbitrarily chosen inertial frame.

I agree. In the specific derivation of interest to the OP, the principle of relativity is added sneakily through the stipulation that ##y'=y## and ##z'=z##. It should be stated as an explicit assumption. As you say, the Voight transforms violate the 1st postulate, while accommodating the 2nd.

I'm not sure I would call linearity an assumption at all - certainly not an independent one. Any transform between inertial frames must be linear because it must map straight lines

Perhaps. I am just stating how the derivation in question identified it.

I would tend to call linearity an assumption because mapping straight lines is what an affine transform does. So linearity is a slightly stronger statement than just mapping straight lines. Anyway, it certainly doesn't hurt to be explicit.

 

[PeterDonis]

It is called "hyperbolic rotation"

A Lorentz boost does not hyperbolically rotate null vectors. It dilates them. Its action on timelike/spacelike and null vectors is fundamentally different. So, since you have been considering how boosts act on light, you need to look at dilation, not hyperbolic rotation.

 

[Dale]

A Lorentz boost does not hyperbolically rotate null vectors. It dilates them. Its action on timelike/spacelike and null vectors is fundamentally different. So, since you have been considering how boosts act on light, you need to look at dilation, not hyperbolic rotation.

That is just the name of the subsection in the Wikipedia article with the derivation in question. It isn't a statement from the OP.

 

[PeterDonis]

However they throw in an unacknowledged 3rd assumption that ##y=y'## and ##z=z'##.

It's not unacknowledged, it's stated in the text right after the spherical wave front equation.

The real problem I see with this derivation is that, as I stated in post #46 just now, a Lorentz boost does not hyperbolically rotate null vectors; it dilates them. So the "hyperbolic rotation" assumption does not apply to the spherical wavefront of light that is assumed in the derivation, making the derivation inconsistent and therefore invalid.

 

[PeterDonis]

That is just the name of the subsection in the Wikipedia article with the derivation in question. It isn't a statement from the OP.

The derivation in the Wikipedia article assumes hyperbolic rotation, as has already been remarked. It's not just the name of the subsection, it's also the next assumption introduced (because the form of the equations "suggest" it) after linearity.

 

[Dale]

It's not unacknowledged, it's stated in the text right after the spherical wave front equation.

It is not acknowledged as a separate assumption. The wording is "Next, consider relative motion along the x-axes of each frame, in standard configuration above, so that y = y′, z = z′, ". So this makes it sound like this is a "without loss of generality" convenience that follows merely from choosing the frames to be parallel with the motion along the x direction. In fact, this statement does more than that.

 

[PeterDonis]

I'm curious what the OP thinks of two other derivations given in the same Wikipedia article:

Derivation using spherical wavefronts of light:

https://en.wikipedia.org/wiki/Deriv...transformations#Spherical_wavefronts_of_light

Einstein's popular derivation:

https://en.wikipedia.org/wiki/Deriv...transformations#Einstein's_popular_derivation

 

[Dale]

The derivation in the Wikipedia article assumes hyperbolic rotation, as has already been remarked. It's not just the name of the subsection, it's also the next assumption introduced (because the form of the equations "suggest" it) after linearity.

Sure, but there is no point in complaining to the OP about it. It isn't his wording. He is complaining about this derivation too.

 

[PeterDonis]

there is no point in complaining to the OP about it. It isn't his wording.

Yes, but he said he wasn't complaining about the derivation because of that term. I think he should.

 

[PeterDonis]

when choosing the ST interval in its unanimously accepted form (no matter if you consider 1+1D or 1+3D), you are already assuming that the change of perspective between the two frames consists of a hyperbolic rotation

The interval doesn't need to be assumed. The fact that ##(ct)^2 - x^2 - y^2 - z^2 = 0## for a spherical wave front of light in any inertial frame can be derived from the fact that the speed of light is ##c## in any inertial frame. The derivation you pointed at doesn't make that explicit (which is another issue with that derivation). But the "spherical wavefronts of light" derivation that I referenced, which is later in the same Wikipedia article, does.

 

[Dale]

Yes, but he said he wasn't complaining about the derivation because of that term. I think he should.

I think then that all three of us agree that this specific derivation is not good for a variety of reasons.

 

[PeterDonis]

I think then that all three of us agree that this specific derivation is not good for a variety of reasons.

Yes, agreed.

 

[Sagittarius A-Star]

The interval doesn't need to be assumed.

Yes. According to Wikipedia, Einstein started his derivation of the LT by adding and subtracting the equations
##\begin{cases}
x' - ct' = \lambda (x - ct) \\
x' + ct' = \mu (x + ct)
\end{cases}##
From this one can also derive the invariance of the spacetime interval, by multiplying the equations. The result is:
##{x'}^2 - c^2{t'}^2 = \lambda \mu (x^2 - c^2t^2)##.
From reciprocity between both frames can be concluded: ##\lambda \mu = 1##.

 

[Saw]

Many comments since my last one, I will reply to them collectively:

- The derivation I pointed to takes as introduction at least the first part of the other one called "Spherical wavefront", so any criticism on the former applies to the latter.
- But please note that my criticism exclusively refers to the part *up to* the formulation of the ST interval. *After that* the derivation introduces linearity and, following consequences of linearity, it hints that the form of the equations suggests hyperbolic rotation, but these are new assumptions that were not deemed necessary for deriving the ST interval.
- I have by now abandoned the argument about the teachings to be drawn from a 1+1D derivation, as the same point can be made relying on the 1 +3D one.
- My criticism (as noted, circumscribed to the part until derivation of ST interval) is only this:
* based on invariance of c postulate, you get [tex]c\Delta t = \sqrt {\Delta {x^2} + \Delta {y^2} + \Delta {x^2}} [/tex]
* based on relativity postulate, you can presume that the same interval that is valid for O will be valid for O'
* but, in order to jump from the former expression to [tex]{(c\Delta t)^2} = \Delta {x^2} + \Delta {y^2} + \Delta {z^2}[/tex] and hence to the ST interval [tex]{(c\Delta t)^2} - (\Delta {x^2} + \Delta {y^2} + \Delta {z^2}) = 0[/tex] you cannot rely on the algebraic trick "squaring both sides" because that is valid only for solving equations and finding unknowns, not for guessing which form take the dimensions in an invariant interval, which in turn depends on the nature of the change of perspective that applies in the case at hand.
- Although I initially intended to restrict the thread to what the derivation misses, I have already revealed what I had in mind as to what it misses: it should admit that the fact that the difference of perspective between frames in spacetime consists of a hyperbolic rotation is a "prius" to the ST interval. In other words, the third assumption of the derivation should come first, as a sort of third postulate.
- Ok, the null vector itself would not rotate because it is the eigenvector of the rotation matrix, but the ST interval is one that assumes that the values of ct and x, y, z transform as appropriate for a change of perspective corresponding to a hyperbolic rotation.
- Now if you said that any of the last two statements is wrong, I would be disconcerted and need your guidance as to why.
PS: BTW, I noted once that the eigenvalues are the Doppler factor (so the null vector would dilate by such factor when the transformation matrix is applied to it?); but if you explain to me how and why, that'd be great.

 

[Dale]

* based on invariance of c postulate, you get [tex]c\Delta t = \sqrt {\Delta {x^2} + \Delta {y^2} + \Delta {x^2}} [/tex]

This is incorrect. ##\Delta t## can be negative, but this expression cannot have a negative ##\Delta t##. The correct form for the invariance of c postulate is ##c^2\Delta t^2=\Delta x^2+\Delta y^2+\Delta z^2## from the beginning.

you cannot rely on the algebraic trick "squaring both sides" because that is valid only for solving equations and finding unknowns, not for guessing which form take the dimensions in an invariant interval

On the contrary. When you go from the correct starting expression to yours you need to include both the positive and the negative square roots.

This is a perfectly valid algebraic operation and your restriction doesn’t make sense. If an equation is true then it remains true under correct algebraic manipulations regardless of why you are doing them. That is indeed the point of algebra.

 

[Sagittarius A-Star]

PS: BTW, I noted once that the eigenvalues are the Doppler factor (so the null vector would dilate by such factor when the transformation matrix is applied to it?); but if you explain to me how and why, that'd be great.

You can calculate the longitudinal relativistic Doppler factor by Lorentz-transforming the four-frequency of a photon, that moves in x direction.

The four-frequency of a massless particle, such as a photon, is a four-vector defined by
##N^{a}=(\nu ,\nu {\hat {\mathbf {n} }})##
where ##\nu ## is the photon's frequency and ##\hat {\mathbf {n} }## is a unit vector in the direction of the photon's motion. The four-frequency of a photon is always a future-pointing and null vector.

Source:
https://en.wikipedia.org/wiki/Four-frequency

 

[PeterDonis]

it should admit that the fact that the difference of perspective between frames in spacetime consists of a hyperbolic rotation is a "prius" to the ST interval. In other words, the third assumption of the derivation should come first, as a sort of third postulate.

No, it can't, because, as has already been pointed out and as you acknowledge, the transformation does not rotate null vectors, it dilates them.

- Ok, the null vector itself would not rotate because it is the eigenvector of the rotation matrix, but the ST interval is one that assumes that the values of ct and x, y, z transform as appropriate for a change of perspective corresponding to a hyperbolic rotation.

This is wrong. The SR (I assume you meant this instead of "ST") interval makes no such assumption. If it did, it would be wrong since the interval has to apply to null vectors.

- Now if you said that any of the last two statements is wrong, I would be disconcerted and need your guidance as to why.

Isn't it obvious? Once more: a Lorentz boost does not rotate null vectors. So you cannot base any derivation of the LT equations on hyperbolic rotation if you want it to apply to null vectors. And of course if you're starting off your derivation by looking at spherical wave fronts of light, you want your derivation to apply to null vectors.

PS: BTW, I noted once that the eigenvalues are the Doppler factor (so the null vector would dilate by such factor when the transformation matrix is applied to it?)

Yes, of course, that's what eigenvalue and eigenvector mean.

but if you explain to me how and why, that'd be great.

I'm not sure what needs explaining. Yes, the eigenvalues of the LT are the Doppler factors, and the eigenvectors are the null vectors that get dilated by the Doppler factors. This is just another way of saying that Lorentz boosts dilate null vectors and don't rotate them.

If you're asking why there are two Doppler factors, with one the reciprocal of the other, that should be obvious: they are for the two possible directions light can move relative to the direction of the Lorentz boost. Note that transformations of lightlike vectors that aren't in the same direction as the boost are more complicated; we haven't discussed those at all in this thread.

 

[PeterDonis]

you cannot rely on the algebraic trick "squaring both sides"

If this argument were valid, it would prove too much: it would prove that the Pythagorean theorem is not correct in Euclidean space.

I would suggest thinking carefully about what justifies the Pythagorean theorem (i.e., summing squares of coordinate deltas) in Euclidean space. The justification for using squares of coordinate deltas in Minkowski spacetime will be the same.

 

[Saw]

I think that we can narrow down to and organize the discussion around three issues:

1) if in order to jump:
- from what is obvious, i.e. that the spatial path traversed by light in a given frame, as composed by the Pythagorean combination of the three spatial dimensions, is equal to the time elapsed in such frame times the speed of light
[tex]c\Delta t = \sqrt {\Delta {x^2} + \Delta {y^2} + \Delta {x^2}} [/tex]
- to what is already the ST interval in the form of
[tex]{(c\Delta t)^2} = \Delta {x^2} + \Delta {y^2} + \Delta {z^2}[/tex]
the argument "I have squared both sides" is sufficient justification or you need "something else", some "added justification"
2) what this "else" would be, which in my opinion is (being very generic) hypothesizing that the new dimension, cT, is related to the old ones, X-Y-Z, in a manner that precisely justifies squaring both sides and which also has to do with the Pythagorean Theorem, but in a different manner, in order to account for the different sign
3) if in particular this "else" can be simply saying that the transformation is going to be a hyperbolic rotation, so I identify the ST interval with the equation of the hyperbola, the unit hyperbola

To me 1) is very clear, but I would not like to look stubborn. At a given time, we could agree that the matter has been discussed enough and leave the thread (without need to close it!) until someone comes with a new comment. But as of now I don't lose the hope of convincing you. Some arguments:

This is incorrect. ##\Delta t## can be negative, but this expression cannot have a negative ##\Delta t##.

As already stated, if you want to get rid of negative time you can always stipulate that what goes into the ST interval is the absolute value of cT or either square both sides but immediately squareroot them, which leads to this other (wrong) interval:
[tex]\sqrt {{{(c\Delta t)}^2}} - \sqrt {(\Delta {x^2} + \Delta {y^2} + \Delta {z^2})} = \sqrt {{{(c\Delta t')}^2}} - \sqrt {(\Delta x{'^2} + \Delta y{'^2} + \Delta z{'^2})} [/tex]
Of course, to avoid this ugly expression you must square both sides, but the subtlety that I favor is just admitting that you do it for a reason that goes beyond the pure algebraic trick.

This is a perfectly valid algebraic operation and your restriction doesn’t make sense. If an equation is true then it remains true under correct algebraic manipulations regardless of why you are doing them. That is indeed the point of algebra.

Sure. The equation remains true under the algebraic manipulation of squaring both sides. But the equation would also remain true under the algebraic manipulation of raising both sides to the 11th power, which would lead to a ST that is false. That is because what is valid for the algebraic purpose (solving for unknowns) may not be valid to guess what remains invariant under transformations. We have a set of operations that would be valid for the first purpose (squaring both sides, squaring and squarerooting to get absolute values, raising both sides to any other power...), but only the first is apt for the second purpose and that is due to "something else".

As to 2),

If this argument were valid, it would prove too much: it would prove that the Pythagorean theorem is not correct in Euclidean space.

I would suggest thinking carefully about what justifies the Pythagorean theorem (i.e., summing squares of coordinate deltas) in Euclidean space. The justification for using squares of coordinate deltas in Minkowski spacetime will be the same.

Well, it is clear to me that we can add up vectors through the Pythagorean theorem ("PT") when they are orthogonal to each other. That is what you do for X-Y-Z with the expression:
[tex]c\Delta t = \sqrt {\Delta {x^2} + \Delta {y^2} + \Delta {x^2}} [/tex]
The question is how and why the new dimension cT can join the team of dimensions "somehow" related by the PT. I have thought of two routes:
a) The resultant of combining X-Y-Z through the PT in Euclidean manner combines with cT through the hyperbola equation. In this case, cT is orthogonal to X-Y-Z in the Minkowski way (dot product with negative sign is zero).
b) As reflected in the Loedel diagram, what is (in another sense) perpendicular is cT with (simplifying) X' and cT' with X, so that the ST interval is the height of the right triangle.
How the two things match is, though, a subtle question which I find beyond my reach as of now. Comments are welcome, but really the OP was not meant to discuss this.

In any case, the fact that the orthogonality between cT and X-Y-Z is implicit or embedded in the ST interval itself seems clear to me. As a proof: look at how the Wikipedia derivation continues in the second section "Linearity". At a given point they find a term that is remininscent of the last term of the Law of Cosines and they equate it to 0 after "comparing coefficients" and noting that the term of comparison does not contain a counterpart. To be expected, because the ST interval is built by assuming orthogonality.

As to 3), what was puzzling me was that in the ST diagram the null vector remains intact, how could it be dilated by the eigenvalue = the Doppler factor?

I have now realized that the Doppler factor is also the scale at which the second frame is drawn in the Minkowski diagram, so the null vector "is" dilated in this diagram even if you dont see it on the page, right? Thanks for guiding me to this is insight (if it is correct at all).

Said this, does this mean that you are unfavoring now any derivation of the ST interval that takes as illustration a null vector? Well, the problem with deriving the ST interval is that there are three possible displays, depending on whether you choose a timelike, lightlike or spacelike vector... You have to choose one reference and then generalize the result to the others or you can repeat it with lightlike (what we have done so far) and timelike (which is the other derivation of the light clock though experiment, which Iike, although it seems not be in fashion nowadays), I would not know how to express it with spacelike, since we would present an impossible scenario of something traveling FTL... In any case, I don't see a problem in saying that we are in face of a hyperbolic rotation even if the reference is precisely the vector that acts as eigenvector of the rotation and therefore gets only dilated, without changing direction.

Ultimately, this is a question of semantics: I think that you can perfectly say that the ST interval is what remains invariant under the "hyperbolic rotation" of a vector, no matter what kind of vector we are talking about and regardless whether the effect of the rotation is change of direction or dilation. Take the example of a right triangle where the height is overlapping with the Y axis and you reflect it precisely over the Y axis. You would say that the figure as a whole has been "reflected", even if a particular side (the height) has remained untouched (not even dilated in this case) in the course of the reflection.

 

[Sagittarius A-Star]

As to 3), what was puzzling me was that in the ST diagram the null vector remains intact, how could it be dilated by the eigenvalue = the Doppler factor?

I have now realized that the Doppler factor is also the scale at which the second frame is drawn in the Minkowski diagram, so the null vector "is" dilated in this diagram even if you dont see it on the page, right? Thanks for guiding me to this is insight (if it is correct at all).

Yes, that is correct. I think a Minkowski diagram is incomplete, if the scales on each axis are omitted, as it is unfortunately often done, for example in the SR book of Leonard Susskind. It is otherwise a great book.

You will find the Doppler factor in equations (1) and (2) of the following LT derivation of Macdonald, because the ##T+X## and ##T-X## are light-cone coordinates. This LT derivation has similarities to the discussed one of Einstein.

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

The reason, why that is the Doppler factor is, that the four frequency ##\mathbf N## of a photon is a null-vector and therefore transforms in the same way. Reason, why the four frequency of a photon is a null-vector: The four-momentum of a photon is also a null-vector and, for movement in x direction, can be written as:
## \mathbf P = \frac{E}{c^2}(c, c, 0, 0) = \frac{h \nu}{c^2}(c, c, 0, 0) = \frac{h}{c}(\nu, \nu, 0, 0) = \frac{h}{c}\mathbf N ##.

 

[Sagittarius A-Star]

As already stated, if you want to get rid of negative time you can always stipulate that what goes into the ST interval is the absolute value of cT or either square both sides but immediately squareroot them, which leads to this other (wrong) interval:
[tex]\sqrt {{{(c\Delta t)}^2}} - \sqrt {(\Delta {x^2} + \Delta {y^2} + \Delta {z^2})} = \sqrt {{{(c\Delta t')}^2}} - \sqrt {(\Delta x{'^2} + \Delta y{'^2} + \Delta z{'^2})} [/tex]

The question is, if you want to write an equation, that is only valid for event pairs with a spacetime interval of zero. For arbitrary event pairs, you need the quadratic form.

I think that you can perfectly say that the ST interval is what remains invariant under the "hyperbolic rotation" of a vector, no matter what kind of vector we are talking about and regardless whether the effect of the rotation is change of direction or dilation.

In the following video, a derivation of the LT via "hyperbolic rotation" is shown, including the analogy to Euclidean rotation.
https://www.physicsforums.com/threads/videos-andrzej-dragan-course-on-relativity.1011307/

 

[Dale]

the Pythagorean combination

The Pythagorean theorem is ##a^2+b^2=c^2## to begin with. So the starting point for this proof is ##\Delta x^2+\Delta y^2+\Delta z^2=c^2\Delta t^2##. Your argument makes 0 sense. You are making a huge deal of your starting point and your starting point is incorrect. If you weren’t making a big deal of the starting point that wouldn’t matter. But it is the crux of your argument and it is wrong. The Pythagorean theorem is the squared statement, so if you insist that the starting point is important, then that is the one to use.

to avoid this ugly expression you must square both sides, but the subtlety that I favor is just admitting that you do it for a reason that goes beyond the pure algebraic trick

The reason doesn’t matter. We are doing all of the algebraic steps in any proof because we already know that they get us where we want to go in the end. We are allowed to do the algebraic "tricks" as long as we perform them correctly, regardless of our reasons for doing them. This restriction you are asserting is irrelevant.

But the equation would also remain true under the algebraic manipulation of raising both sides to the 11th power, which would lead to a ST that is false.

No, it wouldn’t be false. If you do valid algebra then you will have the same set of events as a solution. The algebraic operations are designed expressly to leave those unchanged. It won’t make it false, just ugly.

what is valid for the algebraic purpose (solving for unknowns) may not be valid to guess what remains invariant under transformations.

This is simply not true. The thing that is invariant is the set of events that forms the light cone. The algebra doesn’t change that. I don’t understand how you can have passed algebra class and believe this point you are making.

In short, I disagree with both your starting point and your reasoning from the starting point

 

[Dale]

The equation remains true under the algebraic manipulation of squaring both sides. But the equation would also remain true under the algebraic manipulation of raising both sides to the 11th power, which would lead to a ST that is false.

To show that this is wrong is rather easy. As I said above, the whole point of algebra is to leave the set of solutions unchanged.

Here is a set of events that satisfies the invariance of c (in units where c=1) based on the Pythagorean theorem. It is a standard light cone with the apex at the origin.

If I take the square root of both sides to get the corrected version of your expression then I get the exact same light cone.

I wasn't sure if you meant the 11th power of your square root formula or the 11th power of the original interval but it doesn't really matter. Here is the 11th power of the square root formula. It produces the exact same light cone.

The algebraic operations are valid. The motivation is irrelevant. Taking the 11th power does not produce a false equation, just an ugly one.

Your argument is flat out wrong from start to finish.

 

[Saw]

@Dale, thanks for your development, although I find it hard to follow, I am not sure if we are talking about the same thing.

What I meant is the following: instead of squaring both sides, another valid algebraic operation is raising both sides to the 11th power and, unless I messed up somewhere, it seems that this way we get to a ST interval that does not work:

 

[Saw]

The question is, if you want to write an equation, that is only valid for event pairs with a spacetime interval of zero. For arbitrary event pairs, you need the quadratic form.

Certainly. The question is only how to get to the quadratic form, if that can happen a little magically, through an algebraic operation, or rather it must happen, but based on a hypothesis on how ct, on the one hand, and x-y-z, on the other hand, combine together. I am thinking that this is hard to do on the basis of a null vector, because this is after all the "simple situation": no matter if you have 3 spatial dimensions, it is still a simple situation because here all you need to know is whether light can reach the second event in time and it does simply because ##c\Delta t = \Delta x. ## It is with a timelike interval that you need some additional information, the number of ticks that lapse between two events in a clock present at both, so this is a more complete and hence a better scenario to get the "full" ST interval, which can then be generalized to the simpler case (null vector) or the remaining case (spacelike vector), which is harder to take as reference for the derivation.

 

[Dale]

unless I messed up somewhere

You messed up at the beginning. As I already pointed out before your starting point is wrong. ##\Delta t## can be negative and your starting point does not reflect that.

If you use a correct starting point and correct algebra then you will get a valid result.

Also, we are currently working only with null spacetime intervals. We have not as yet made any assumption that non-null spacetime intervals are invariant. That is a separate assumption.