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

[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.

 

[Sagittarius A-Star]

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?

I didn't write "without the second postulate". MacDonald wrote in his first line "Assume: (A) The speed of light is the same in all inertial frames. (Take c = 1.)". The following equation are already based on postulate 2.

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.

I should better re-write them without $c=1$:
$x + ct = e_1 (x'+ct')$
$x - ct = e_2 (x'-ct')$
That the 2nd postulate was invoked can be seen from the fact, that they contain $ct'$ instead of $c't'$.

You could now take the first equation, invoke the 1st postulate to switch primed and unprimed coordinates, if you also describe for symmetry reasons a light beam, that moves in the opposite direction:
$x'-ct' = e_1 (x - ct)$
By putting the unprimed quantities on the left side, you get your new second equation in:
$$ x + ct = e_1 (x'+ct') $$$$ x - ct = \frac{1}{e_1} (x'-ct') $$

 

[Pony]

I didn't write "without the second postulate". MacDonald wrote in his first line "Assume: (A) The speed of light is the same in all inertial frames. (Take c = 1.)". The following equation are already based on postulate 2.

Not how I wrote it, I think I never used postulate 2 before I said it. I only used postulate 1/Galilean relativity. So does the Macdonald paper BTW, it uses postulate 2 first when it refers it. (In a relativist mindset it is hard to imagine that time and space could exist without something relating them. But then Newtonian kinematics, where time is measured with wristwatches and distance is measured with measuring tapes, and just assuming that light follows newtonian mechanics too, is consistent.)

That the 2nd postulate was invoked can be seen from the fact, that they contain $ct'$ instead of $c't'$.

You can't change what I write and point out that you use postulate 2 in that.

It doesn't really matter tho.

 

[Sagittarius A-Star]

Not how I wrote it, I think I never used postulate 2 before I said it.

How are the units in your $X+T$ in posting #46 consistent without $c=1$?
And after you say that you define $c=1$ you show the same equations.

 

[Pony]

Both are real numbers.

And after you say that you define you show the same equations.

This may seem confusing, but I believe it works the way I wrote it, that is, we can not use an equation, and derive some results in a more general way, then restrict ourselves later. "Now we assume c=1, which means some additional restrictions for our T and X functions, but the previous facts still remain true, and we even can keep using the same symbols."

So those equations are true in the Newtonian setting as well (I think at least), and I hoped for a nice looking spacetime interval formula that is agnostic whether we assume postulate 2 or don't.

 

[Sagittarius A-Star]

This may seem confusing

Yes. I think it would be less confusing to use from begin to end the same unit system. But yes, I also think it works the way you wrote it.

 

[Pony]

Well, okay. I am sorry about that.

I am not interested talking about aesthetics anymore.

Yes. You didn't write to which value you defined $c$ before you defined $c=1$.
You are adding a distance to a time interval, which makes no sense with an undefined $c$.

The way I wrote does not need it, and is a bit more general. But you can assume that Galileo worked with seconds and light-seconds, and so do I until step 1. Then Einstein came and noticed that light moves with velocity = 1 lightsec/sec in every frame, and so I postulated this as well before step 2. (But not sooner.)

But it really is super agnostic to measurement units, and later only assumes that c=1.

 

[DAH]

The way I see it in SR $c=1$ but it still has units $\text{m/s}$ so that $ct$ has units of meters, the same as the spatial coordinates. The relative velocities in the transformations will always be less than 1 or a fraction.

For time dilation in one dimension, if two ticks of a watch are separated by $(\delta t, 0)$ in the frame of the watch, where $\delta t=1$ second. Then in a frame that is moving at $v=\text{d}x^{\prime}/\text{d}t^{\prime}$ relative to the watch, the same ticks are separated by $(\delta t^{\prime}, \delta x^{\prime})$.

We know the spacetime separations are $\delta s=\delta s^{\prime}$, so then $(c\delta t)^2=(c\delta t^{\prime})^2-(\delta x^{\prime})^2$.
If we divide by $(c\delta t^{\prime})^2$ we get:
\begin{align*}
\frac{c^2(\delta t)^2}{c^2(\delta t^{\prime})^2}&=1-\frac{(\delta x^{\prime})^2}{c^2(\delta t^{\prime})^2}\\\\
\left(\frac{\delta t}{\delta t^{\prime}}\right)^2&=1-\left(\frac{\text{d}x^{\prime}}{c\text{d}t^{\prime}}\right)^2=1-\frac{v^2}{c^2}\\\\
\frac{\delta t}{\delta t^{\prime}}&=\sqrt{1-\frac{v^2}{c^2}}
\end{align*}
Which can be rearranged to give $$\delta t^{\prime}=\gamma \delta t$$
This is basically the same as @Dale 's post #2

 

[Pony]

I don't think that the symmetry approach in #38 is working anymore. It is too vague to point out a mistake, but I don't think that the assumptions imply the conclusion.

A better approach can be orienting the x axes against each other from the start, then we get $$ T+X = e_1 (T'-X')$$ $$ T-X = e_2 (T'+X')$$ equations which are true for every event ( a light ray has +1 and -1 velocities in the frames when we extend these equations from the trajectories to every event ) and in both of the two frames, with the same $e_1,e_2$ constants (because the frames now have a truly symmetric role). So $$ T'+X' = e_1 (T-X)$$ $$ T'-X' = e_2 (T+X)$$ also hold, which means
$$ T + X = e_1(T'-X') = e_1e_2 (T+X) $$ which means $e_1 e_2 = 1.$ Thus $T^2 - X^2 = T'^2 -X'^2$ and SR follows easily.

 

[Sagittarius A-Star]

A better approach can be orienting the x axes against each other from the start
$$T+X = e_1 (T'-X')$$

Then you can define time as that, what a standardized clock shows and directly conclude with the two postulates and symmetry from ... $$T+X = e_1 (T'-X')$$$\Rightarrow$ (swap frames and satisfy descriptions in both frames of a light beam in opposite direction)$$T'+X' = e_1 (T-X)$$$\Rightarrow$$$\begin{cases}T+X = e_1 (T'-X')\\
T- X = \frac{1}{e_1} (T'+X')\end{cases}$$Then multiply those.

These are essentially calculations in light-cone coordinates. If you rotate the axes of a normal spacetime coordinate system $(X, T)$ counter-clockwise by $45°$, than you get the axes of a light-cone coordinate system $({L_+}, {L_-})$.$$\begin{cases}{L_+} = X \cos(45°) + T \sin(45°) = (T+X) \frac{1}{\sqrt{2}}\\
{L_-} = -X \sin(45°) + T \cos(45°) = (T- X) \frac{1}{\sqrt{2}}\end{cases}$$Lorentz transformation (with x axes in same direction):$$\begin{cases}{L_+}=k {L_+}' \\{L_-}=\frac{1}{k} {L_-}'\end{cases}$$

Calculating $k$ on the worldline of an object at $X' = 0 \ \ \ \Rightarrow \ \ \ {L_+}'={L_-}'$:​

$$k^2 = \frac{{L_+}}{{L_-}} = \frac{1+v}{1-v} = \gamma^2 (1+v)^2 = \frac{1}{\gamma^2 (1-v)^2}$$$$\begin{cases}T+X = \gamma (1+v) (T'+X') \ \ \ \ \ (1)\\
T- X = \gamma (1-v) (T'-X') \ \ \ \ \ (2)\end{cases}$$

Add and subtract Eqs. (1) and (2):​

$$\begin{cases}T= \gamma (T'+v X') \ \ \ \ \ (3)\\X = \gamma (v T'+X') \ \ \ \ \ (4)\end{cases}$$​

Invariance of the spacetime interval:
$${L_+}{L_-} = {L_+}'{L_-}'$$

 

[Pony]

Who's clock shows, where, what do you mean 'shows'? What does the first equation mean, and how do I conclude that?

 

[Sagittarius A-Star]

Who's clock shows, where, what do you mean 'shows'?

See posting #40. That definition of time separates for example SR from Lorentz ether theory.

What does the first equation mean, and how do I conclude that?

I copied the first equation from your posting #58.