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

• I
• Pony
In summary, the spacetime interval is unchanged by a Lorentz transformation, and the following formulas are equivalent:1) Speed of light in frame S: c = Δx/Δt2) Speed of light in frame S': c = Δx'/Δt' (i.e. same as in S)3) 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
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!

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

malawi_glenn
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.

malawi_glenn and Dale
Pony said:
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##

malawi_glenn
Dale said:
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.

Dale and malawi_glenn
robphy said:
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.

robphy said:
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!

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.

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Pony said:
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.

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.

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malawi_glenn
Sagittarius A-Star said:
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, physicsworks and 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
robphy said:
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.

robphy said:
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.

Sagittarius A-Star said:
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.

Sagittarius A-Star said:
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.

Sagittarius A-Star
Sagittarius A-Star said:
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).

PeterDonis said:
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.

PeterDonis said:
Reciprocity isn't "demanded", it's proven using Eq. (4).
Reciprocity is demanded by SR postulate 1, it's proven using Eq. (4).

Sagittarius A-Star said:
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##.

Sagittarius A-Star
Sagittarius A-Star said:
Reciprocity is demanded by SR postulate 1
If there is such a "demand", the proof in the referenced paper makes no use of it.

PeterDonis said:
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##.

Sagittarius A-Star said:
It makes use of it via:

... resulting in a formula using the same ##\gamma##.
Reciprocity has already been proven at that point.

PeterDonis said:
Reciprocity has already been proven at that point.
The reciprocity of ##v## has already been proven at that point, not that of ##\gamma##.

Sagittarius A-Star said:
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 said:
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)##.

PeterDonis said:
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

Sagittarius A-Star said:
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.

PeterDonis said:
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.

Did we loose OP?

PeterDonis said:
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)##.

Sagittarius A-Star said:
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##.

Sagittarius A-Star
malawi_glenn said:
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.

Pony said:
You can write other proofs, if you know.
Pony said:
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:
$$dx'^2 - c^2dt'^2 = dx^2 - c^2dt^2$$

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Pony
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.

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