I Derivation of SR's time-dilatation in 1d?

[Pony]

Hey, I am looking for a derivation of time-dilatation or some trivially equivalent formulas (Lorentz-transformation, conservation of 4-distance (edit: invariance of spacetime interval) etc) in 1 dimension, using that c is observer independent.

I only can find the one that uses a light-clock moving perpendicularly to it's axis, but I don't like that one. I am looking for an other approach.

Thank you!

 

[Dale]

Start with the spacetime interval: $$c^2 d\tau^2=c^2 dt^2-dx^2$$ $$\frac{c^2 d\tau^2}{c^2 dt^2}=\frac{c^2 dt^2}{c^2 dt^2}-\frac{dx^2}{c^2 dt^2}$$ $$\frac{d\tau}{dt}=\sqrt{\frac{c^2 dt^2}{c^2 dt^2}-\frac{dx^2}{c^2 dt^2}}$$ $$\frac{1}{\gamma}=\frac{d\tau}{dt}=\sqrt{1-\frac{v^2}{c^2 }}$$

 

[robphy]

To keep the focus on 1D, use a “longitudinal light clock” where the light signals are parallel to the direction of relative motion.

In my opinion, the clearest derivation that is physically motivated with operational definitions (radar measurements) is due to Bondi (Relativity and Common Sense).

If you are impatient and only care about standard textbook formulas, then don’t use Bondi.

 

[Sagittarius A-Star]

Hey, I am looking for a derivation of time-dilatation or some trivially equivalent formulas (Lorentz-transformation, conservation of 4-distance etc) in 1 dimension, using that c is observer independent.

Consider two ticks of a clock at rest in the primed frame ($\Delta x' = 0$). Inverse Lorentz transformation for time:
$\Delta t = \gamma (\Delta t' + v\Delta x'/c^2) = \gamma \Delta \tau$

 

[Pony]

Start with the spacetime interval: $$c^2 d\tau^2=c^2 dt^2-dx^2$$ $$\frac{c^2 d\tau^2}{c^2 dt^2}=\frac{c^2 dt^2}{c^2 dt^2}-\frac{dx^2}{c^2 dt^2}$$ $$\frac{d\tau}{dt}=\sqrt{\frac{c^2 dt^2}{c^2 dt^2}-\frac{dx^2}{c^2 dt^2}}$$ $$\frac{1}{\gamma}=\frac{d\tau}{dt}=\sqrt{1-\frac{v^2}{c^2 }}$$

I claimed in my opening post that this set of equations are equivalent, and my goal is to derive them, using the fact that c is constant.

The common approach goes like (for example on the wiki https://en.wikipedia.org/wiki/Time_dilation#Simple_inference but everywhere else that I can find) : you move a light clock with velocity v, you get a right triangle with sides x, ct, cτ (the green triangle on the wiki page), Pythagoras gives you
$$ c^2\tau^2 + x^2 = c^2t^2 $$
then you do what you wrote. I am not satisfied that it uses the fact that perpendicular length contraction is zero, and I am looking for an another approach.

 

[Pony]

To keep the focus on 1D, use a “longitudinal light clock” where the light signals are parallel to the direction of relative motion.

Can you do that? I can't.

In my opinion, the clearest derivation that is physically motivated with operational definitions (radar measurements) is due to Bondi (Relativity and Common Sense).

If you are impatient and only care about standard textbook formulas, then don’t use Bondi.

It seems he uses Doppler effect, I played with it before, I will try that too, thanks!

 

[Nugatory]

The appendix of Einstein’s popular book “Relativity: The Special and General Theory” has a largely algebraic derivation of the Lorentz transformations. The basic approach is to look for a coordinate transformation between inertial frames that keeps the coordinate speed of light finite and unchanged, so seems to be what you are looking for.

Because such a transformation must preserve the spacetime interval, this derivation is equivalent to @Dale's, but is much messier and harder to follow.

 

[malawi_glenn]

@Pony You mean you want do derive the Lorentz-transformation and after that time dilation?

You should be able to find the steps below in almost every introductury book on relativity.

Speed of light in frame S: c = Δx/Δt
Frame S' moves at constant speed v in S.
Speed of light in S': c = Δx'/Δt' (i.e. same as in S)

Ansatz for linear transformatio: x' = γ(x-vt) and x = γ(x'+vt') to resemble Gallilean transformation, but we do not assume t = t' and we have a constant γ that should only depend on v. Furthermore, we want γ → 1 and t → t' as v/c → 0 to obtain Gallilean transformation as a low speed case. Goal is to determine γ.

a beam of light is emitted. In S we measure the time the light beam has travelled as Δt = Δx/c and in S' we have Δt' = Δx'/c.
We obtain with some algebra: Δx' = γ(Δx-vΔx/c) = γ(1 - v/c)Δx and Δx = γ(Δ'x+vΔx'/c) = γ(1 - v/c)Δx'.
The ratio Δx' / Δx = γ(1 - v/c) = 1/(γ(1 + v/c)) then solve for γ (I think you can do this).
After you have obtain γ, figure out the transformation rules for t and t'. When you have done this, show that (cΔt)² - (Δx)² = (cΔt')² - (Δx')² and from there the time dilation formula follows.

 

[Dale]

I claimed in my opening post that this set of equations are equivalent

Maybe it seemed that way to you, but I certainly didn’t see anything in your OP that indicated you did not want a derivation starting from the spacetime interval.

Anyway, “guess my secret requirements” isn’t a fun game to play.

 

[Sagittarius A-Star]

Hey, I am looking for a derivation of time-dilatation or some trivially equivalent formulas (Lorentz-transformation

A derivation of Lorentz transformation and time dilation together:
http://www.faculty.luther.edu/~macdonal/LorentzT/LorentzT.html

 

[Ibix]

Note that the transforms must be linear from the definition of an inertial frame. Thus they can be written as$$\left(\begin{array}{c}x'\\t'\end{array}\right)=
\left(\begin{array}{cc}A(v)&B(v)\\C(v)&D(v)\end{array}\right)
\left(\begin{array}{c}x\\t\end{array}\right)$$For convenience I'll call the matrix $\Lambda$. Note that

1. $\Lambda(vt,t)^T$ must map to $(0,t')^T$ because that's what "a frame moving with velocity $v$" means.
2. $\Lambda^{-1}(vt',t')^T$ must likewise map to $(0,t)^T$.
3. If you set $t=T$ in the first equation and $t'=T$ in the second, the output $t'$ and $t$ coordinates must be equal from the principle of relativity.
4. $\Lambda(ct,t)^T$ must map to $(ct',t')^T$ by the invariance of the speed of light.

I think those four equations should let you solve for the components of $\Lambda$. It's a purely one dimensional calculation.

The other approach is to note that transverse length contraction cannot happen from the principle of relativity and just use a transverse light clock.

 

[robphy]

A derivation of Lorentz transformation and time dilation together:
http://www.faculty.luther.edu/~macdonal/LorentzT/LorentzT.html

Eqs (1) and (2) is the Lorentz Boost transformation (essentially) in light cone coordinates. This is what Bondi’s method gives… but with more physical motivation (why write X+T?)….

Oh look what this apparently-magical system of equations does.
It’s cute… and curious.
But I think one is left not understanding what is going on physically….
But yep, you got the equations.

 

[Sagittarius A-Star]

Eqs (1) and (2) is the Lorentz Boost transformation (essentially) in light cone coordinates.

Yes. The appendix of Einstein’s popular book “Relativity: The Special and General Theory”, mentioned in #7 by @Nugatory, used also light cone coordinates.

https://en.wikisource.org/wiki/Rela...mple_Derivation_of_the_Lorentz_Transformation

 

[Pony]

Thank you all! That Macdonald paper is much shorter and uses much less things than I expected.

You can write other proofs, if you know.

 

[Sagittarius A-Star]

Eqs (1) and (2) is the Lorentz Boost transformation (essentially) in light cone coordinates.

Eqs (1) and (2) is not yet the Lorentz Boost transformation, because it contains only information included from SR postulate 2 (invariance of the speed of light). It is yet missing information from SR postulate 1 (principle of relativity). Therefore, two yet independent constants $\gamma$ and $v$ must be used.

Only after essentially transforming from light cone coordinates in Eqs (1) and (2) to normal coordinates in Eqs (3) and (4), SR postulate 1 gets included, by demanding the reciprocity of the time-dilation factor $\gamma$ and of $v$, with a sign reversal. From this, the function $\gamma(v)$ is derived to connect the two originally independent constants and get the Lorentz Boost transformation.

This is what Bondi’s method gives… but with more physical motivation (why write X+T?)….

At least according to Wikipedia, differently to the above, Bondi derived the longitudinal Doppler factor $k$ from both postulates and used it then in normal coordinates instead of in terms of $v$. I don't know his book.
https://en.wikipedia.org/wiki/Bondi_k-calculus

The reasoning for X+T stands in the paragraph below Eqs (1) and (2) to show from postulate 2, that the shown "transformation" is valid for "arbitrary" events, not only those on a certain world-line.

 

[PeterDonis]

Eqs (1) and (2) is not yet the Lorentz Boost transformation, because it contains only information included from SR postulate 2 (invariance of the speed of light). It is yet missing information from SR postulate 1 (principle of relativity).

No, it isn't. The relativity principle is used in the second paragraph, before Eqs (1) and (2), to derive the properties of $d \tau / dt$ in all inertial frames.

Therefore, two yet independent constants $\gamma$ and $v$ must be used.

Those two constants are not independent in Eqs (1) and (2) of the referenced paper; $\gamma$ is defined at the end of the second paragraph to be a function of $v$. The precise functional form of $\gamma$ is not yet determined, but just the fact that $\gamma$ is a function of $v$ means it's not an independent constant.

 

[PeterDonis]

demanding the reciprocity of the time-dilation factor $\gamma$ and of $v$, with a sign reversal.

Reciprocity isn't "demanded", it's proven using Eq. (4).

 

[Sagittarius A-Star]

No, it isn't. The relativity principle is used in the second paragraph, before Eqs (1) and (2), to derive the properties of $d \tau / dt$ in all inertial frames.

Those two constants are not independent in Eqs (1) and (2) of the referenced paper; $\gamma$ is defined at the end of the second paragraph to be a function of $v$. The precise functional form of $\gamma$ is not yet determined, but just the fact that $\gamma$ is a function of $v$ means it's not an independent constant.

That's correct. But those two constants must be treated as independent in Eqs (1) and (2), as long it is not shown, that there is a dependency, see footnote 4:

⁴This does not assume time dilation: as far as (B) is concerned, γ could be identically 1.

 

[Sagittarius A-Star]

Reciprocity isn't "demanded", it's proven using Eq. (4).

Reciprocity is demanded by SR postulate 1, it's proven using Eq. (4).

 

[PeterDonis]

those two constants must be treated as independent in Eqs (1) and (2), as long it is not shown, that there is a dependency

No, this is not correct. The function $\gamma(v)$ being $1$ for all $v$ does not mean $\gamma$ is an independent constant. It just means the function $\gamma(v)$ turns out to have a constant value of $1$ for all $v$. No other constant value is possible because we already know that $\gamma = 1$ when $v = 0$; that in itself means $\gamma$ is not independent of $v$.

 

[PeterDonis]

Reciprocity is demanded by SR postulate 1

If there is such a "demand", the proof in the referenced paper makes no use of it.

 

[Sagittarius A-Star]

If there is such a "demand", the proof in the referenced paper makes no use of it.

It makes use of it via:

Thus, switching I and I′ ...

... resulting in a formula using the same $\gamma$.

 

[PeterDonis]

It makes use of it via:

... resulting in a formula using the same $\gamma$.

Reciprocity has already been proven at that point.

 

[Sagittarius A-Star]

Reciprocity has already been proven at that point.

The reciprocity of $v$ has already been proven at that point, not that of $\gamma$.

 

[PeterDonis]

The reciprocity of $v$ has already been proven at that point, not that of $\gamma$.

Hm, yes, I see, one would have to know that $\gamma$ can only be a function of $v^2$ to know that $-v$ would give the same $\gamma$ as $v$; just knowing that it's a function of $v$ isn't enough.

 

[PeterDonis]

Hm, yes, I see, one would have to know that $\gamma$ can only be a function of $v^2$ to know that $-v$ would give the same $\gamma$ as $v$; just knowing that it's a function of $v$ isn't enough.

Actually, on thinking this over, I'm not so sure a separate assumption of reciprocity for $\gamma$ is required here. The second paragraph of the referenced paper says "speed" $v$, and the reasoning there is based on spatial homogeneity and isotropy. If $\gamma$ only depends on the speed $v$, not on the direction, which in 1d means $\gamma$ can only be a function of the absolute value of $v$ (in 3d it would mean that $\gamma$ could only be a function of the norm of the vector $\vec{v}$), that is sufficient to know that $\gamma(-v) = \gamma(v)$.

 

[Sagittarius A-Star]

Hm, yes, I see, one would have to know that $\gamma$ can only be a function of $v^2$ to know that $-v$ would give the same $\gamma$ as $v$; just knowing that it's a function of $v$ isn't enough.

Yes. The author could have already written it as a function of $v^2$ in the following quote. But in the end, the reciprocity of $\gamma$ comes from postulate 1 together with (B) "Inertial frames are homogeneous and spatially isotropic":

By (B), dτ/dt is the same for all inertial clocks with speed v in I. Then by the relativity principle, dτ/dt is the same for inertial clocks with speed v in other inertial frames. Set γ = γ(v) = dτ/dt.⁴

 

[PeterDonis]

in the end, the reciprocity of $\gamma$ comes from postulate 1 together with (B) "Inertial frames are homogeneous and spatially isotropic"

Yes, agreed. That brings up an interesting point, though: spatial homogeneity and isotropy is actually a stronger assumption than reciprocity of $\gamma$ and $v$. It would be interesting to see how far one could get if one only assumed reciprocity and did not assume spatial homogeneity and isotropy.

 

[Sagittarius A-Star]

Yes, agreed. That brings up an interesting point, though: spatial homogeneity and isotropy is actually a stronger assumption than reciprocity of $\gamma$ and $v$. It would be interesting to see how far one could get if one only assumed reciprocity and did not assume spatial homogeneity and isotropy.

Then you would also have to assume linearity of the LT, what the author doesn't.

³We do not assume that the Lorentz transformation is linear.

 

[malawi_glenn]

Did we loose OP?

 

[Sagittarius A-Star]

If $\gamma$ only depends on the speed $v$, not on the direction, which in 1d means $\gamma$ can only be a function of the absolute value of $v$ (in 3d it would mean that $\gamma$ could only be a function of the norm of the vector $\vec{v}$), that is sufficient to know that $\gamma(-v) = \gamma(v)$.

Yes. Then still for the reciprocity of the time-dilation factor $\gamma$ the 1st postulate is needed for $\gamma'(v^2) = \gamma(v^2)$.

 

[PeterDonis]

Yes. Then still for the reciprocity of the time-dilation factor $\gamma$ the 1st postulate is needed for $\gamma'(v^2) = \gamma(v^2)$.

You don't need $\gamma'(v^2) = \gamma (v^2)$ as a separate item the way the paper does it. You already have that there is a single function $\gamma$ that is the same for all frames (the paper derives this from homogeneity and isotropy and the principle of relativity). So all you need in addition to that is that $\gamma$ only depends on the speed, not on the direction, of $v$.

 

[Vanadium 50]

Did we loose OP?

Did you mean "did he go away", no, he was here yesterday.

If you mean "did we leave him behind in the dust"? Well, maybe. It's a rather odd thing to ask for ("Derive this, but don't use the usual way") so I would hope that there's some interest in why this is not the usual way.

 

[Sagittarius A-Star]

You can write other proofs, if you know.

Hey, I am looking for a derivation of time-dilatation or some trivially equivalent formulas ... edit: invariance of spacetime interval ... in 1 dimension, using that c is observer independent.

Einstein wrote in the appendix of his popular book from 1916 transformation formulas, that satisfy the invariance of the speed of light signals, that move through the origin in (+x) direction and (-x) direction. The disappearance of the left sides involves the disappearance of the right sides:
$$(x' - ct') = \lambda (x - ct)\ \ \ \ \ \ \ \ \ \ (3)$$$$(x' + ct') = \mu (x + ct)\ \ \ \ \ \ \ \ \ \ (4)$$From this stage, I continue differently than Einstein did. He derived the LT, I want to derive directly the invariance of the spacetime interval.

I multiply equations (3) and (4) and get:
$$(x'^2 - c^2t'^2) = \lambda\mu (x^2 - c^2t^2)$$From the 1st postulate (and assuming inertial frames are homogeneous and spatially isotropic), from the invariance of causality [edited] and when having a fixed unit of time follows reciprocity: $\lambda\mu = +1$. Then using differentials:
$$dx'^2 - c^2dt'^2 = dx^2 - c^2dt^2$$

 

[vanhees71]

This is a bit too quick, since from the invariance of light-like vectors alone you don't get only the Lorentz (or Poincare) group but a larger symmetry group including dilations, the conformal group.

 

[Sagittarius A-Star]

This is a bit too quick, since from the invariance of light-like vectors alone you don't get only the Lorentz (or Poincare) group but a larger symmetry group including dilations, the conformal group.

I don't understand this. I didn't derive the LT. That is, what Einstein did with additionally invoking postulate 1.

When deriving the invariance of the spacetime interval, I did not refer only to the invariance of the speed of light, but I used additional assumptions to get $\lambda = \frac{1}{\mu}$.

Are assumptions missing?
Would it solve the issue, if I delete the word "transformation" in my first sentence in my posting #34?

 

[Pony]

You should elaborate how did you get $\lambda = 1 / \mu$.

 

[Sagittarius A-Star]

You should elaborate how did you get $\lambda = 1 / \mu$.

[edited] I take the definition of the second of the SI and argue that this must be the same for any inertial observer at rest relative to the Cs atom used to define it.

From multiplying equations (3) and (4) I got:
$(x'^2 - c^2t'^2) = \lambda\mu (x^2 - c^2t^2)\ \ \ \ \ \ \ \ \ \ (5)$

Since both polynomials are quadratic, they do not depend on the orientation of the axes. Therefore, the proportionality factor $\lambda\mu$ cannot depend on the sign of $v$. Together with the mention assumptions, I can switch the primed and unprimed coordinates and get the inverse relation:
$(x^2 - c^2t^2) = \lambda\mu (x'^2 - c^2t'^2)\ \ \ \ \ \ \ \ \ \ (6)$
I substitute equation (6) into equation (5):
$(x'^2 - c^2t'^2) = \lambda\mu \lambda\mu (x'^2 - c^2t'^2)$
$\Rightarrow$
$(\lambda\mu)^2 = 1$.
That the sign of $\lambda\mu$ must be positive, can be derived from the invariance of causality or from the fact, that $\lambda\mu$ must be $1$ for $v=0$ and therefore, from continuity, also for all other $v$.

 

[vanhees71]

As I said, if you only look for the transformations, which leave the light cone invariant, you get the conformal group, i.e., a larger group than the (proper orthochronous) Lorentz group. I think the right argument is that there must be a fixed unit of time or length, i.e., you can take the definition of the second of the SI and argue that this must be the same for any inertial observer at rest relative to the Cs atom used to define it. This together with the independence of $c$ of the motion of light sources (Einstein's 2nd postulate) also defines a length unit, as it is realized in the SI by fixing the speed of light to a defined value. This breaks the scale invariance and reduces the conformal to the Lorentz group as symmetry group of SR spacetime.

 

[Sagittarius A-Star]

there must be a fixed unit of time or length, i.e., you can take the definition of the second of the SI and argue that this must be the same for any inertial observer at rest relative to the Cs atom used to define it.

Yes. I think Einstein addressed this in his popular book from 1916 by:

Under these conditions we understand by the "time" of an event the reading (position of the hands) of that one of these clocks which is in the immediate vicinity (in space) of the event.
...
It has been assumed that all these clocks go at the same rate if they are of identical construction.

Source:
https://en.wikisource.org/wiki/Rela..._I#Section_8_-_On_the_Idea_of_Time_in_Physics

 

[pervect]

I'm going to put another plug in for Bondi's k-calculus. Wiki has a treatment at https://en.wikipedia.org/wiki/Bondi_k-calculus, which I haven't read in detail. Bondi's book, also mentioned, "relativity and common sense", is a much longer treatment - probably longer than necessary, the basic idea doesn't need a whole book to explain IMO. But if the Wiki article is too short, the book may be helpful.

On other observation. In spite of the name, k-calculus does not involve calculus, just algebra.

The basic idea is simple. If we have two co-located observers moving at some relative velocity v, one or both of which is emitting light pulses, the relative motion causes a doppler shift. The doppler shift can be characterized by some number k, the ratio of the frequency of emission to the frequency of reception - or to the ratio of the period of reception to the period of emission.

The other thing one needs to assume is that the doppler shift between any two observers not moving relative to each other is unity, so that the periods and frequencies for the observer and reciever are identical in this case. This will be true in the context of special relativity, and it allows one to conclude that the doppler shift factor k is in general a constant of motion - it does not vary with time as an object moves away from the source object. One basically uses the property that the k-factor doesn't depend on distance for a stationary observer to conclude that it can't depend on distance for a moving observer. One can view this as a result of imagining a co-located stationary observer that is present at the location of any moving observer at any desired time.

Thus if a source object emits pulses with a regular spacing of T, the destination object will receive pulses with a spacing of kT.

Furthermore, by symmetry, given that the destination object recieves pulses at intervals of kT, and reflects or re-emits them to the source object, the source object will recieve the radar pulses at time k^2 T.

This is all one need to set up a basic "radar" system, since one knows the arrival time and reception time of radar pulses, and that the fact that the speed of light is constant.

From this, one can derive that a light signal, emitted at time T, arrives at time kT. So the event of reception, according to the radar set and calculation of the emitter, is that the signal arrives at time (k^2+1)T/2.

However, we know from the previous argument that the time of reception of the signal is kT, which is not the same number.

If k=2, for example, one can conlulde that the time (k^2+1)/2 = (4+1)/2 = 2.5 in the emission frame corresponds to the time 2 in the reception frame.

As well as this direct illustration of time dilation, a bit more math allows one to express v in terms of k, and invert the result to find k in terms of v. One can also proceed further to derive the 1space+1time Lorentz transform. This isn't super hard, but I'll leave it to the wiki article and/or Bondi's book as it requires some illustrating diagrams and is a bit longer than I want to write a post about. The goal of this post is to simply motivate the OP to take the effort to read up on the topic on their own, by providing an overview of the process and conclusion.

 

[robphy]

I'm going to put another plug in for Bondi's k-calculus. Wiki has a treatment at https://en.wikipedia.org/wiki/Bondi_k-calculus, which I haven't read in detail. Bondi's book, also mentioned, "relativity and common sense", is a much longer treatment - probably longer than necessary, the basic idea doesn't need a whole book to explain IMO. But if the Wiki article is too short, the book may be helpful.

[snip]

There's a PF Insight for that: (my) https://www.physicsforums.com/insights/relativity-using-bondi-k-calculus/ .

 

[vanhees71]

The idea of this socalled "k calculus" is pretty simple. One introduces light-cone coordinates $x^{\pm}=x^0 \pm x^3$ instead of $x^0$ and $x^3$ and the two "perpendicular" space-like coordinates $x^1$ and $x^2$ (with $x^{\mu}$ usual Lorentzian coordinates).

Now if we consider only boosts in the $x^0 x^3$ plane, we have to fulfill only the constraint
$$x^+ x^-=(x^0)^2-(x^3)^2=\text{const}.$$
For all $(x^0,x^3)$. This implies that the linear transformations leaving this product of the light-cone coordinates invariant must be of the form
$$x^{\prime +} =\exp (-\eta) x^+, \quad x^{\prime -}=\exp( \eta) x^{-}, \quad \eta \in \mathbb{R}.$$
Then you have
$$x^{\prime 0}=\frac{1}{2} (x^{\prime +} + x^{\prime -}) = \cosh \eta x^0 - \sinh \eta x^3,$$
$$x^{\prime 3}=\frac{1}{2} (x^{\prime +}-x^{\prime -})=-\sinh \eta x^0 + \cosh \eta x^3,$$
which is the corresponding Lorentz boost with rapidity $\eta$. Since the origin of the frame $\Sigma'$ moves with velocity $\beta=\sinh \eta /\cosh \eta=\tanh \eta$ we have $\cosh \eta=\gamma=1/\sqrt{1-\beta^2}$ and $\sinh \eta=\beta \gamma$.

It's an elegant way to derive the Lorentz boost. That's all behind it. I never understood, why one made an entire philosophy around this ;-).

 

[robphy]

The idea of this socalled "k calculus" is pretty simple. One introduces light-cone coordinates $x^{\pm}=x^0 \pm x^3$ instead of $x^0$ and $x^3$

[snip]

It's an elegant way to derive the Lorentz boost. That's all behind it. I never understood, why one made an entire philosophy around this ;-).

Probably the simplest description is that one works in the eigenbasis of the Lorentz boost transformation.
Those light-cone coordinates are coordinates for this eigenbasis,
and the lightlike directions are the eigenvectors.
The "k" is the Doppler factor (the exponential function of the rapidity).

---

I'm glad this "philosophy" (way of thinking) about relativity exists and was accessible to me
because its presentation of the above mathematics starts from
physical principles, operational experiments, and "thinking in terms of spacetime diagrams"
that tell a story of WHY we are able to write some of those equations
in a language that is more accessible to a more general audience.

It helped me see that physical meaning that I couldn't see in the standard textbook treatment,
and it emphasizes relativistic principles (including causal structure) and spacetime geometry
(rather than merely algebraic formulae and group transformations).

In Bondi's storyline, the "light-cone coordinates" are essentially represented by
elapsed times on the inertial observer's wristwatch for a radar experiment.
etc...
The usual "relativistic effects" are developed without explicit use of the Lorentz transformations.
The Doppler k-factor (with its simpler properties and simpler physical interpretation) is given precedence over the time-dilation $\gamma$ factors.

---

From https://www.physicsforums.com/threa...rstand-special-relativity.999183/post-6451991

In my opinion, the Lorentz Transformations are given too much emphasis for a beginner.

Consider this letter by Mermin
"Lapses in relativistic pedagogy"
American Journal of Physics 62, 11 (1994); https://doi.org/10.1119/1.17728

...Lorentz transformation doesn't belong in a first exposure to special relativity. Indispensable as it is later on, its very conciseness and power serve to obscure the subtle interconnnectedness of spatial and temporal measurements that makes the whole business work. Only a loonie would start with real orthogonal matrices to explain rotations to somebody who had never heard of them before, but that's how we often teach relativity. You learn from the beginning how to operate machinery that gives you the right answer but you acquire little insight into what you're doing with it.

By the way, Mermin later published
"Space–time intervals as light rectangles"
American Journal of Physics 66, 1077 (1998) https://doi.org/10.1119/1.19047 ,
which, together with Bondi's k-calculus, inspired my "Relativity on Rotated Graph Paper" idea.
(The area invariance was already known, but not well-known.)

---

So, consider me a k-calculus evangelist :) .

 

[vanhees71]

Well, I'm an "algebraist", but it's in a way right. The Lorentz transformation is overemphasized in teaching relativity in comparison to the use of the Galilei transformation in Newtonian mechanics. On the other hand to get to the spacetime geometry in terms of Minkowski space it's may be unavoidable in some sense.

I don't see that the Doppler effect of light is very intuitive. I think it's better to use gedanken experiments a la Einstein (1905) about light signals. The light-cone coordinates are also natural when thinking about "radar timing", which I think is the approach in Bondi's book.

On the other hand, how else do you want to introduce rotations than via orthogonal transformations? In the plane you can simply read them off from elementary trigonometry, which leads you directly to orthogonal transformations in the 2D real vector space. The generalization to 3D is then most simple by using orthogonal transformations right away.

 

[Pony]

[edited] I take the definition of the second of the SI and argue that this must be the same for any inertial observer at rest relative to the Cs atom used to define it.

From multiplying equations (3) and (4) I got:
$(x'^2 - c^2t'^2) = \lambda\mu (x^2 - c^2t^2)\ \ \ \ \ \ \ \ \ \ (5)$

Since both polynomials are quadratic, they do not depend on the orientation of the axes. Therefore, the proportionality factor $\lambda\mu$ cannot depend on the sign of $v$. Together with the mention assumptions, I can switch the primed and unprimed coordinates and get the inverse relation:
$(x^2 - c^2t^2) = \lambda\mu (x'^2 - c^2t'^2)\ \ \ \ \ \ \ \ \ \ (6)$
I substitute equation (6) into equation (5):
$(x'^2 - c^2t'^2) = \lambda\mu \lambda\mu (x'^2 - c^2t'^2)$
$\Rightarrow$
$(\lambda\mu)^2 = 1$.
That the sign of $\lambda\mu$ must be positive, can be derived from the invariance of causality or from the fact, that $\lambda\mu$ must be $1$ for $v=0$ and therefore, from continuity, also for all other $v$.

I think that works! Putting together with the beginning of the Macdonald paper, I think I have a proof that I actually can remember!

Let A,B be two frames, $X,T,X',T' : E \longrightarrow R$ denote the coordinate functions for the events in A and B. Let their origins coincide: when $(X,T)=(0,0),$ also $(X',T')=(0,0).$

Let H denote the trajectory of the origin of A. H is a set of events.

step 1 : the points of H suffer a linear transformation from A to B.

It is true, since X,T,X',T' are both linear functions of T on H. (Checking one-by-one:
$T' = k_1 T$ for some constant k1 is because we assume that time changes a constant factor. $X' = vT' = k_2 T$ because A have a constant speed in B. $T = T$, and $X = 0$.)

Thus on H $$ X + T = e_1 (X'+T') $$ $$ X - T = e_2 (X'-T') $$ for some constants $e_1,e_2.$ If I could show that $e_1 e_2 = 1$ then that would be enough, sadly I can't do that at this point.

Now assume that c, the speed of light is 1 in both frames.

step 2 : $$ X + T = e_1 (X' + T') $$ $$ X - T = e_2 (X' - T') $$ hold not just on H, but for every event (with $e_1, e_2$ only depending on A,B).

For an event e, we can send and receive lightrays to H that transport the values (X+T) and (X-T), thus this holds (see the figure on the Macdonald paper).

Now the frames have a symmetric role. Multiply the equations, and you get $$ X^2 - T^2 = e_1 e_2 (X'^2 - T'^2)$$ where $e_1 e_2$ must be 1, thus this quantity (=length of spacetime intervals) remains constant in different frames QED.

It's basically the Macdonald paper, but without the need to remember the ugly looking equations. I am a bit disappointed that I had to use that c=1 in both frames.

 

[Sagittarius A-Star]

If I could show that $e_1 e_2 = 1$ then that would be enough, sadly I can't do that at this point.

If you want, then you can invoke already at this stage the 1st postulate together with a symmetry argument. See section "Lorentz Transformation" in the Insights article:
https://www.physicsforums.com/insights/relativity-using-bondi-k-calculus/

I am a bit disappointed that I had to use that c=1 in both frames.

You can change it, if you substitute in MacDonad's formulas $T$ by $cT$, $T'$ by $cT'$ and $v$ by $v/c$.

 

[Pony]

If you want, then you can invoke already at this stage the 1st postulate together with a symmetry argument. See section "Lorentz Transformation" in the Insights article:
https://www.physicsforums.com/insights/relativity-using-bondi-k-calculus/

Without the second postulate I don't think I can show that $X^2-T^2 = X'^2-T'^2$ on H. Wouldn't that be too strong? What are the forms of these equations (LT, time dilatation, spacetime interval) if we only assume the first postulate (Galilean relativity principle), but not the second postulate?

edit: or you meant that I should invoke 2nd postulate at that point, the constant speed of light?

 

[vanhees71]

If you only assume the 1st postulate + assumption that any inertial observers considers his "space" as Euclidean (including all the symmetries, i.e., isotropy and homogeneity) + homogeneity of time (for any inertial observer) + that the transformations between inertial reference frames form a group, there are only two spacetime models left: the Galilei-Newton spacetime (a fiber bundle) and Einstein-Minkowski spacetime (a pseudo-Euclidean affine manifold with a fundamental form of signature (1,3) or (3,1)). See, e.g.,

https://doi.org/10.1063/1.1665000

 

[Pony]

I know that one can enumerate them in a group theoretic way, but if one only works with equations, then the equations for LT, time dilatation and spacetime interval will have an unified look that work in both cases I think.