I Need to resort to spherical wavefront to derive the LTs?

[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.

 

[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 c\Delta t = \sqrt {\Delta {x^2} + \Delta {y^2} + \Delta {x^2}}
* 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 {(c\Delta t)^2} = \Delta {x^2} + \Delta {y^2} + \Delta {z^2} and hence to the ST interval {(c\Delta t)^2} - (\Delta {x^2} + \Delta {y^2} + \Delta {z^2}) = 0 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 c\Delta t = \sqrt {\Delta {x^2} + \Delta {y^2} + \Delta {x^2}}

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