Compute Length Contraction: Is ##l=1## a Possible Model?

[Office_Shredder]

TL;DR Summary

Try to go straight from invariance of spacetime intervals to length contraction

Sorry, I accidentally posted this while typing it up, then when I finished typing it I found the mistake that made me come here to make this thread. I'll still write it up just to make the thread somewhat useful, and then ask my follow-up question at the end

I'm picking units so the speed of light is 1.

Suppose a rod of length 1 in its frame is traveling at a speed of v relative to an observer, parallel to the direction of its travel. The observer sees some length distortion and measures the rod as being length $l$. The observer sets two synchronized clocks a distance of $l$ apart, and measures that the two ends of the rod hit the two clocks at exactly the same time. So the spacetime interval is
$$s^2=0^2-l^2=-l^2$$.

From the rod's perspective, it has a length of 1, and the two points the observer set out are distorted by a factor of $l$ by symmetry, so are $l^2 apart. If$l<1, then when the further point hits the front end of the rod, the closer point has a distance of $1-l^2$ left to travel to reach the back end of the rod, so the closer point hitting the back end occurs at time $\frac{1-l^2}{v}$ from the perspective of the rod, and that event is distance $1$ from the first event. If $l>1$, then when the first point hits back of the rod, the further point is a distance of $l^2-1$ from the front of the rod, so takes time $\frac{1-l^2}{v}$ to reach the front of the rod. The distance is still 1.

Squaring gets rid of the ambiguity of the sign of $1-l^2$ fortunately In either case the spacetime interval has length
$$s^2=\left( \frac{1-l^2}{v}\right)^2-1$$
Setting the two spacetime intervals equal to each other, we get that
$$\left( \frac{1-l^2}{v}\right)^2-1=-l^2$$
You can just let $u=l^2$ and use the quadratic equation to solve this, but instead I noticed we can let $u=1-l^2$, so $-l^2=u-1$ to get
$$u^2/v^2-1=u-1$$
So
$$u^2-v^2u=0$$
This has solutions $u=0$ and $u=v^2$
This corresponds to $1-l^2=0$, or $1-l^2=v^2$. Since $l$ has to be a positive number (right?...) This means $l=1$ or $l=\sqrt{1-v^2}$.

Obviously the latter solution corresponds to actual special relativity. Does the former solution correspond to anything? It makes intuitive sense that no length contraction could be a possible model of the universe, but it's weird for it to be a possible model where the thing you're assuming is spacetime invariance. One thing I haven't used here is assuming that the speed of light is constant in all frames, is there a model where that's not true and length contraction doesn't happen, or is there some other way to exclude the $l=1$ solution? I know for example if you assume the speed of light is constant in all frames then using a light clock going parallel to the motion vs perpendicular to the motion results in getting that there has to be length contraction.

 

[Nugatory]

There are two logically consistent possibilities for how the universe works. In one the speed of light is the same in all inertial frames; this leads to Lorentz transforms, length contraction, time dilation, relativity of simultaneity, and all the apparatus of SR. In the other the speed of light is (like all other speeds) not the same in all inertial frames; this leads to Galilean relativity, no length contraction or time dilation, absolute simultaneity, and all the apparatus of classical Newtonian mechanics. Either could be the right model for how our universe works, but when we do experiments we find that the SR model is the right choice.

I haven't checked your calculations, but it is not surprising that when you don't assume light speed invariance you find both possibilties.

 

[PeroK]

The Newtonian/Galilean solution is consistent with isotropy and homogeneity of space and time. There's a derivation of the two options here:

http://www2.physics.umd.edu/~yakovenk/teaching/Lorentz.pdf

 

[vela]

The invariance of the speed of light is a consequence of spacetime invariance, so it's built into your derivation.

One way to derive the Lorentz transformations is to assume a linear transformation between $(t,x)$ and $(t',x')$ and require spacetime invariance be satisfied. A trivial solution turns out to be $t'=t$ and $x'=x$, but then you can show this is simply the case where the two frames aren't moving relative to each other. I think this latter case is analogous to your $l=1$ solution, but I don't see how you can show $l=1$ implies $v=0$.

 

[Sagittarius A-Star]

and the two points the observer set out are distorted by a factor of $l$ by symmetry, so are ##l^2 apart.

What do you mean by saying, that two events are "distorted"?

 

[Office_Shredder]

What do you mean by saying, that two events are "distorted"?

Sorry, I meant the length was stretched or contracted. I'm not assuming at the start that lengths get shorter, it could go either way.

 

[Sagittarius A-Star]

Summary:: Try to go straight from invariance of spacetime intervals to length contraction

I think that is not possible without an additional assumption. Reason:
$\Delta s^2 = c^2\Delta t^2 - \Delta x^2$
$\Delta s'^2 = c^2\Delta t'^2 - \Delta x'^2$
$\Delta s^2 = \Delta s'^2$
The rod shall be at rest in the primed frame: $L' = \Delta x' = L_0$. The length measurement for $L= \Delta x$ in the unprimed frame has the condition $\Delta t = 0$. Then follows:
$0 - L^2 = c^2\Delta t'^2 - L_0^2$
That is 1 equation with 2 unknowns ($L$ and $\Delta t'$).

The additional assumption could be the Lorentz transformation for time:
$\Delta t' = \gamma (0 -v/c^2 * L$)

 

[Office_Shredder]

I think that is not possible without an additional assumption. Reason:
$\Delta s^2 = c^2\Delta t^2 - \Delta x^2$
$\Delta s'^2 = c^2\Delta t'^2 - \Delta x'^2$
$\Delta s^2 = \Delta s'^2$
The rod shall be at rest in the primed frame: $L' = \Delta x' = L_0$. The length measurement for $L= \Delta x$ in the unprimed frame has the condition $\Delta t = 0$. Then follows:
$0 - L^2 = c^2\Delta t'^2 - L_0^2$
That is 1 equation with 2 unknowns ($L$ and $\Delta t'$).

The additional assumption could be the Lorentz transformation for time:
$\Delta t' = \gamma (0 -v/c^2 * L$)

If you read the full post, you'll see that just by invariance of spacetime interval, and assuming that length distortion is symmetric for the two parties, you can reduce it down to a quartic polynomial in one variable.

The invariance of the speed of light is a consequence of spacetime invariance, so it's built into your derivation.

Thinking about this a bit more, maybe it's not. I just asserted I had some units for which $t^2-x^2$ was fixed. There's nothing used here that implies that the speed of light moves at a speed of 1, which I think is what Nugatory is getting at. An obvious way for $t^2-x^2$ to be fixed for all observers is for $t$ and $x$ to be fixed for all observers. The surprising part I guess is the *only* other option is for things to transform like Lorentz transformations.

 

[Sagittarius A-Star]

If you read the full post, you'll see that just by invariance of spacetime interval, and assuming that length distortion is symmetric for the two parties, you can reduce it down to a quartic polynomial in one variable.

I do not yet understand your calculation. But you seem to acknowledge, what I wrote: You need an additional assumption besides the invariance of the spacetime interval. Your assumtion comes from SR postulate #1 (principle of relativity).

Thinking about this a bit more, maybe it's not.

The invariance of the spacetime interval implies, that the speed c (or 1 in your units) is invariant (=SR postulate #2). Reason: You can set the spacetime interval to 0 in both frames and get the same speed.

 

[vela]

Thinking about this a bit more, maybe it's not. I just asserted I had some units for which $t^2-x^2$ was fixed. There's nothing used here that implies that the speed of light moves at a speed of 1, which I think is what Nugatory is getting at. An obvious way for $t^2-x^2$ to be fixed for all observers is for $t$ and $x$ to be fixed for all observers. The surprising part I guess is the *only* other option is for things to transform like Lorentz transformations.

Requiring $t^2-x^2$ be invariant leads to the usual Lorentz transformations and an invariant speed, or it leads to the solution $x'=x,\ t'=t$. But this latter possibility contradicts the assumption that the two frames are moving relative to each other. The point x'=0 should move with speed $v$ according to the observer at rest in the unprimed frame, but $x=x'=0$ would imply it's not.

This is different than what Nugatory and PeroK brought up: spacetime symmetries and the principle of relativity only admit two possibilities: special relativity and Galilean relativity. Galilean relativity definitely doesn't satisfy $t^2-x^2$ being invariant, so it wouldn't follow from the assumption of the invariance of spacetime intervals.

 

[Sagittarius A-Star]

Summary:: Try to go straight from invariance of spacetime intervals to length contraction
...
One thing I haven't used here is assuming that the speed of light is constant in all frames, is there a model where that's not true and length contraction doesn't happen, or is there some other way to exclude the $l=1$ solution?

Albert Einstein derived the Lorentz transformation from the invariance of the speed of light and from the symmetry of the length contraction. Then he derived the invariance of the spacetime interval.

Source (Appendix I - Simple Derivation of the Lorentz Transformation):
https://en.wikisource.org/wiki/Rela...mple_Derivation_of_the_Lorentz_Transformation

He did not get 2 solutions, because he did not solve a quadratic equation for the length contraction factor: He did not do the length contraction directly two times, forth and back.

I know for example if you assume the speed of light is constant in all frames then using a light clock going parallel to the motion vs perpendicular to the motion results in getting that there has to be length contraction.

Taylor and Wheeler derive the Lorentz transformation from the linearity requirement and the invariance of the spacetime interval: They use the time dilation formula (derived from the Minkowski metric).

Source:
https://www.eftaylor.com/spacetimephysics/04a_special_topic.pdf

They derive the length contraction from the time dilation formula (derived from the Minkowski metric) in combination with the twin paradox.

Source:
https://www.eftaylor.com/spacetimephysics/04_chapter4.pdf

 

[PeterDonis]

Albert Einstein derived the Lorentz transformation from the invariance of the speed of light and from the symmetry of the length contraction.

No, he didn't. He derived the Lorentz transformation from the invariance of the speed of light and the principle of relativity.

He did not get 2 solutions, because he did not solve a quadratic equation for the length contraction factor: He did not do the length contraction directly two times, forth and back.

I have no idea what you're talking about here.

Taylor and Wheeler derive the Lorentz transformation from the linearity requirement and the invariance of the spacetime interval

Yes.

: They use the time dilation formula (derived from the Minkowski metric).

They use it, but they don't assume it; they derive it from invariance of the interval.

They derive the length contraction from the time dilation formula (derived from the Minkowski metric) in combination with the twin paradox.

Sort of. They are just deriving obvious consequences of invariance of the interval and the Lorentz transformation.

 

[PeterDonis]

From the rod's perspective, it has a length of 1

This means you're assuming length contraction, not deriving it. I thought you were trying to derive it.

Generally speaking, whenever you use computations of intervals to try to obtain length contraction, you will always obtain time dilation and relativity of simultaneity along with it. The three things always go together, and just looking at one of them doesn't work unless you assume what you're trying to prove.

 

[PeterDonis]

Setting the two spacetime intervals equal to each other

Why are you doing that? How do you know they are both the same interval (i.e., the interval between the same pair of events)?

 

[Sagittarius A-Star]

He derived the Lorentz transformation from the invariance of the speed of light and the principle of relativity.

Yes, but he used the principle of relativity via:

Furthermore, the principle of relativity teaches us that, as judged from $K$, the length of a unit measuring-rod which is at rest with reference to $K'$ must be exactly the same as the length, as judged from $K'$, of a unit measuring-rod which is at rest relative to $K$.

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

I have no idea what you're talking about here.

I refer to the calculation in the OP.

 

[PeterDonis]

he used the principle of relativity via

That may make it seem like he's only using the "symmetry of length contraction", but he isn't. He's leaving out the fact that the two intervals he calls $\Delta x$ and $\Delta x'$ are along different directions in spacetime--$\Delta x$ is along a line of simultaneity in the unprimed frame, while $\Delta x'$ is along a line of simultaneity in the primed frame. So he is implicitly making use of more than just the length contraction aspect of the principle of relativity; he's also making use of relativity of simultaneity. (And he could make use of time dilation plus relativity of simultaneity by going through similar reasoning for deltas in the $t$ coordinate instead of the $x$ coordinate.)

I refer to the calculation in the OP.

Ah, ok. I'm not sure the calculation in the OP is correct; I posed a couple of questions to the OP in posts #13 and #14.

 

[Office_Shredder]

This means you're assuming length contraction, not deriving it. I thought you were trying to derive it.

Generally speaking, whenever you use computations of intervals to try to obtain length contraction, you will always obtain time dilation and relativity of simultaneity along with it. The three things always go together, and just looking at one of them doesn't work unless you assume what you're trying to prove.

I'm not assuming length contraction, I'm saying suppose the length contraction is some amount $l$ which could include $l=1$, let's see what values of $l$ could exist. If $l$ ends to being 1, then there is no length contraction.

Why are you doing that? How do you know they are both the same interval (i.e., the interval between the same pair of events)?

For example, suppose the observer places two synchronized clocks a distance of $l$ apart that happen to simultaneously observe the two parts of the rod passing them. We don't have to worry about how the observer did this, maybe they just guessed what $l$ should be and got lucky.. Everyone should agree on the events "the end of the rod was coincident with the clock", although obviously they won't all agree the two events occur at the same time.

 

[PeterDonis]

I'm not assuming length contraction, I'm saying suppose the length contraction is some amount $l$ which could include $l=1$, let's see what values of $l$ could exist.

That is assuming length contraction. If you're not assuming length contraction, you have no reason to allow the possibility of any value of $l$ other than $1$.

 

[Office_Shredder]

That is assuming length contraction. If you're not assuming length contraction, you have no reason to allow the possibility of any value of $l$ other than $1$.

That doesn't seem right. I'm allowing for the possibility of length contraction. All I'm assuming is that spacetime coordinates might be weird between different frames, and I'm trying to compute what it looks like like. If I compute the position vs time graph of a falling object to see if it ends up being a parabola, it would be weird to say I'm assuming it's a parabola as part of the computation.

 

[PeterDonis]

I'm allowing for the possibility of length contraction.

Yes. What makes you think it is a possibility? It isn't in Newtonian physics. That's what I mean by assuming it; you are just putting it in as a possibility without any supporting argument. Of course you already know it is present in relativity, which is why the idea occurred to you in the first place; but that doesn't change the fact that, from the standpoint of your reasoning in the OP, you are putting it in without any supporting argument or derivation from something else, i.e., as an assumption.

All I'm assuming is that spacetime coordinates might be weird between different frames

No, that's not what you're assuming. Your reasoning is entirely in terms of measurements, which are invariants, independent of any choice of coordinates. You are not even discussing coordinates at all.

 

[PeterDonis]

If I compute the position vs time graph of a falling object to see if it ends up being a parabola, it would be weird to say I'm assuming it's a parabola as part of the computation.

If you compute the position vs. time graph, you are assuming that the position varies with time. You aren't assuming that the variation takes the specific form of a parabola, but you are assuming that there is variation, i.e., that measuring the position at different times might give different results. The fact that you already know that this happens does not make it any less of an assumption from the standpoint of your calculation.

Similarly, in your argument in the OP, you are not assuming any specific form for the length contraction formula, but you are assuming that length contraction is possible, i.e, that the length measurement might give different results for different observers. The fact that you already know relativity predicts this does not make it any less of an assumption from the standpoint of your calculation.

 

[PeterDonis]

$$u^2/v^2-1=u-1$$

Note that this equation is undefined if $v = 0$. So your further steps are only valid if $v > 0$.

(In fact, your earlier equation for $s^2$ in the rod's frame is also undefined if $v = 0$.)

So
$$u^2-v^2u=0$$
This has solutions $u=0$ and $u=v^2$

No, the $u = 0$ solution corresponds to $v = 0$ (as you will see if you go back and work through how you arrived at your formula for $s^2$ in the rod's frame), which was already eliminated above.

 

[Sagittarius A-Star]

Yes. What makes you think it is a possibility? It isn't in Newtonian physics. That's what I mean by assuming it; you are just putting it in as a possibility without any supporting argument.

Einstein seems to argue in the same way as @Office_Shredder (before discussing Lorentz transformation and then length contraction):

Section 10 - On the Relativity of the Conception of Distance
...
A priori it is by no means certain that this last measurement will supply us with the same result as the first. Thus the length of the train as measured from the embankment may be different from that obtained by measuring in the train itself. This circumstance leads us to a second objection which must be raised against the apparently obvious consideration of Section 6. Namely, if the man in the carriage covers the distance $w$ in a unit of time — measured from the train, — then this distance — as measured from the embankment — is not necessarily also equal to $w$.

Source:
https://en.wikisource.org/wiki/Rela..._the_Relativity_of_the_Conception_of_Distance

 

[Office_Shredder]

Note that this equation is undefined if $v = 0$. So your further steps are only valid if $v > 0$.

Agreed. If v=0 then you don't get an event where the rod meets the clocks, the rod just kind of. .. sits there. So nothing I do would be valid.

(In fact, your earlier equation for $s^2$ in the rod's frame is also undefined if $v = 0$.)No, the $u = 0$ solution corresponds to $v = 0$ (as you will see if you go back and work through how you arrived at your formula for $s^2$ in the rod's frame), which was already eliminated above.

I don't agree, you can go back to

$$(\frac{1-l^2}{v})^2-1=-l^2$$
And observe it is satisfied by $l=1$.

Einstein seems to argue in the same way as @Office_Shredder (before discussing Lorentz transformation and then length contraction):

Source:
https://en.wikisource.org/wiki/Rela..._the_Relativity_of_the_Conception_of_Distance

I'm working through the textbook spacetime physics (which was cited above, though a chapter later than I currently am). It starts by emphasizing that constant speed of light and space-time interval lengths being the same between events are all you need, and Lorentz transformations etc are useful tools but conceptually unnecessary.

 

[PeterDonis]

Einstein seems to argue in the same way as @Office_Shredder (before discussing Lorentz transformation and then length contraction)

Yes, and his argument amounts to saying that we should consider adopting the assumption that length contraction is possible, instead of the assumption built into Newtonian mechanics that it is not.

 

[PeterDonis]

I don't agree, you can go back to

$$(\frac{1-l^2}{v})^2-1=-l^2$$
And observe it is satisfied by $l=1$.

Not if $l = 1$ requires that $v = 0$, since $v$ is in the denominator of the first term on the LHS.

 

[PeterDonis]

It starts by emphasizing that constant speed of light and space-time interval lengths being the same between events are all you need

Yes, that's correct. You can derive length contraction and time dilation (and relativity of simultaneity) from the invariance of the interval. But you don't have to make any assumptions about length and time by themselves to do so. You can proceed solely on the basis of which spacetime intervals correspond to the lengths and times you are interested in.

For example, consider the following: We have two observers, moving at a constant speed $v$ relative to each other. Each observer has a pair of clocks separated by a ruler, at rest relative to him, and has synchronized his pair of clocks using Einstein clock synchronization. Observer A, with his ruler and clocks, is moving to the right relative to observer B with his ruler and clocks.

Now consider the following events:

Event #1: Observer A's right-hand clock passes observer B's left-hand clock.

Event #2: Observer A's right-hand clock passes observer B's right-hand clock.

Event #3: Observer A's left-hand clock passes observer B's left-hand clock.

Event #4: Observer A's left-hand clock passes observer B's right-hand clock.

We can immediately deduce that:

The length of A's ruler, relative to B, is $v$ divided by the time difference, on B's left-hand clock, between events 1 and 3, or the time difference, on B's right-hand clock, between events 2 and 4. (Note that this implies that those two time differences must be the same.) These time differences are the (timelike) spacetime intervals between those pairs of events.

The length of B's ruler, relative to A, is $v$ divided by the time difference, on A's left-hand clock, between events 3 and 4, or the time difference, on A's right-hand clock, between events 1 and 2. (Note that this implies that those two time differences must be the same.) These time differences are the (timelike) spacetime intervals between those pairs of events.

So we have these relationships between intervals:

$$
\tau_{13} = \tau_{24} = \tau
$$

$$
\tau_{12} = \tau_{34} = \tau^\prime
$$

where I have used $\tau$ and $\tau^\prime$ to denote that these are intervals on B clocks (which we'll consider to be at rest in the "unprimed" frame) and A clocks (at rest in the "primed" frame) respectively. Similarly, we will call the length of A's ruler, relative to B, $L$ (since B's frame is "unprimed"), and the length of B's ruler, relative to A, $L^\prime$. Thus we have

$$
L = \frac{v}{\tau}
$$

and

$$
L^\prime = \frac{v}{\tau^\prime}
$$

We also know that events 2 and 3 must be spacelike separated (can you see why?). That means we can choose a frame in which those events happen at the same (coordinate) time, and we can set up the time coordinate so that that coordinate time is $t = 0$. So in this frame, $t_2 = t_3 = 0$.

This implies the following equations for the $t$ and $x$ coordinates of the events in the frame we just chose:

$$
\left( - t_1 \right)^2 - \left( x_3 - x_1 \right)^2 = \left( t_4 \right)^2 - \left( x_4 - x_2 \right)^2
$$

$$
\left( - t_1 \right)^2 - \left( x_2 - x_1 \right)^2 = \left( t_4 \right)^2 - \left( x_4 - x_3 \right)^2
$$

Now we make one additional assumption: that B has set up his clocks and ruler in just the right way to make events 2 and 3 happen at the same time by his clocks--in other words, the frame we just defined above, in which events 2 and 3 are simultaneous, is B's rest frame. That means $x_1 = x_3$ and $x_2 = x_4$ above (since those pairs of events both take place at the same B clock, hence the same $x$ coordinate in B's rest frame). So the first equation above becomes $t_1 = - t_4$ (note that we took a square root to get that, but we have to choose the minus square root for $t_1$--can you see why?), and the second equation becomes $x_2 - x_1 = x_4 - x_3$ (note that we took a square root to get that as well, but the square root must be the positive square root on both sides this time--can you see why?).

This now gives us a second equation for the length of A's ruler relative to B: it is the spacetime interval between events 2 and 3, which reduces to:

$$
L = x_2 - x_3 = x_4 - x_1
$$

Note carefully: this is also the length of B's ruler relative to B! (It should be obvious why this is the case.) So we can now compare the length of B's ruler relative to B (i.e., $L$) with the length of B's ruler relative to $A$ (i.e., $L^\prime$). (Note that the key additional constraint we imposed on the problem that let's us do this is that we had B set things up just right to make events 2 and 3 have the same B clock reading.) Let's make that comparison.

Equating all of the expressions for $L$ gives:

$$
L = x_2 - x_3 = x_4 - x_1 = \frac{v}{\tau} = \frac{v}{\sqrt{\left( - t_1 \right)^2 - \left( x_3 - x_1 \right)^2}} = \frac{v}{\sqrt{\left( t_4 \right)^2 - \left( x_4 - x_2 \right)^2}}
$$

The corresponding expressions for $L^\prime$ are:

$$
L^\prime = \frac{v}{\tau^\prime} = \frac{v}{\sqrt{\left( - t_1 \right)^2 - \left( x_2 - x_1 \right)^2}} = \frac{v}{\sqrt{\left( t_4 \right)^2 - \left( x_4 - x_3 \right)^2}}
$$

Substituting $x_2 = x_3 + L$ and $x_3 = x_2 - L$ in the second set of expressions gives:

$$
L^\prime = \frac{v}{\tau^\prime} = \frac{v}{\sqrt{\left( - t_1 \right)^2 - \left( x_3 - x_1 + L \right)^2}} = \frac{v}{\sqrt{\left( t_4 \right)^2 - \left( x_4 - x_2 + L \right)^2}}
$$

Taking ratios gives

$$
\frac{L^\prime}{L} = \frac{\sqrt{\left( - t_1 \right)^2 - \left( x_3 - x_1 \right)^2}}{\sqrt{\left( - t_1 \right)^2 - \left( x_3 - x_1 + L \right)^2}} = \frac{\sqrt{\left( t_4 \right)^2 - \left( x_4 - x_2 \right)^2}}{\sqrt{\left( t_4 \right)^2 - \left( x_4 - x_2 + L \right)^2}}
$$

It should be evident that these fractions are less than $1$, hence $L^\prime < L$, hence there is length contraction.

Note the difference between the above and your reasoning in the OP. In the above, we did not start with any assumptions about lengths at all. We looked at spacetime intervals and what they imply about lengths. To put it another way, we constructed lengths out of spacetime intervals. Whereas, in your OP, you assumed that "length" might be a variable at the start; you didn't construct that variable out of anything, you just assumed it was a variable.

This might seem like quibbling, but I think it's important to realize just how far you can get by constructing the things we normally take as givens--things like lengths and times--out of spacetime intervals, without assuming anything about them to start with.

 

[Sagittarius A-Star]

Whereas, in your OP, you assumed that "length" might be a variable at the start; you didn't construct that variable out of anything, you just assumed it was a variable.

I think it is O.K., that @Office_Shredder assumed that "length" might be "a variable at the start".

Reason: @Office_Shredder starts his derivation in the OP from the assumption, that the spacetime interval is invariant. If the spacetime interval is invariant, then the spatial distance cannot be invariant. From that follows, that length contraction must be possible.

So it is not a separate assumption.

What @Office_Shredder derives, is the quantitative calculation formula with the length contraction factor $\gamma$.

 

[PeterDonis]

If the spacetime interval is invariant, then the spatial distance cannot be invariant.

My post #27 amounts to a long-winded way of deducing this, yes.

 

[Sagittarius A-Star]

My post #27 amounts to a long-winded way of deducing this, yes.

Now, I apply the derivation method of the OP to Newtons physics: Try to go straight from invariance of time intervals (= absolute time) to length "contraction" (for $v \neq 0$).

$\Delta t^2 = \Delta t'^2\ \ \ \ \$(1)

The rod of length 1 is at rest in the primed frame. The observer at rest in the unprimed frame has 2 clocks at rest, with distance $l$. The observer is lucky and each end of the rod meet at the same time in the unprimed frame ($\Delta t =0$) the respective clock. Then, for symmetry reasons, the spatial distance of the clocks must be $l^2$ in the frame of the rod. As @Office_Shredder derived, in the frame of the rod, the time between one clock meets one end of the rod and the other clock meets the other end of the rod must be $\Delta t' = \frac{1-l^2}{v}$.

I put this into (1):

$0 = (\frac{1-l^2}{v})^2$
$=> l=1$

I think this shows, that the derivation method of the OP does not require before a deduction, that length contraction is possible (in case of spacetime invariance), nor in this case, that it is not possible. Instead, it deduces it from the respective invariance (for $v \neq 0$).

 

[PeterDonis]

I think this shows, that the derivation method of the OP does not require before a deduction, that length contraction is possible (in case of spacetime invariance), nor in this case, that it is not possible.

I didn't say "deduction", I said "assumption". Your argument has the same form as the OP, so whatever assumptions the OP is making, your argument is making as well.

One could, of course, also make an argument similar to the one I made in post #27, but using the Newtonian invariance property instead of the SR one. Whatever difference there is between my argument in post #27 and the OP, there would be the same difference between a Newtonian argument similar to post #27 and your argument in post #30.

 

[Sagittarius A-Star]

Your argument has the same form as the OP, so whatever assumptions the OP is making, your argument is making as well.

Newton's assumption of an absolute time enforces the length-"contraction" factor to be 1. Because the OP does not assume an absolute time, he must also drop this restiction of the length-contraction factor.

 

[PeterDonis]

Newton's assumption of an absolute time enforces the length-"contraction" factor to be 1. Because the OP does not assume an absolute time, he must also drop this restiction of the length-contraction factor.

Nothing you are saying here contradicts what I said.

 

[Sagittarius A-Star]

Summary:: Try to go straight from invariance of spacetime intervals to length contraction
...
Obviously the latter solution corresponds to actual special relativity. Does the former solution correspond to anything?

In your derivation, which is valid only for $v \neq 0$, an additional constraint is missing, that $l=1$ is only possible at $v=0$. You would need to derive from the spacetime invariance and additional equation, which contains $v$, for example the time dilation formula. Then you could derive the Lorentz transformations and get the missing constraint from the relativity of simultaneity.

But here is an easier derivation of the length contraction from spacetime invariance:

I define the length of the rod as $\Delta x' := 1$
I call the length of the rod in the unprimed frame $\Delta x = l$
The measurement events have the time difference $\Delta t = 0$.

I put this into the spacetime invariance
$\Delta t^2 - \Delta x^2 = \Delta t'^2 - \Delta x'^2$
and get
$0 - \Delta x^2 = \Delta t'^2 - \Delta x'^2$
$l^2/1 = \frac{\Delta x^2}{\Delta x'^2} = 1 - \frac{\Delta t'^2}{\Delta x'^2}$

Because of the symmetry between the temporal coordinates and spatial coordinates (in the direction of movement) in the spacetime invariance formula, it follows (in analogy to the time dilation formula):
$l^2 = 1 - v^2$

Visualization:
In an Minkowski diagram ($c=1$), the angle between the t'-axis and the t-axis is the same as the angle between the x'-axis and the x-axis.
$v=\frac{\Delta x}{\Delta t}$ if in the other frame $\Delta x' = 0\ \ \$ (relativity of "same location")
$v=\frac{\Delta t}{\Delta x}$ if in the other frame $\Delta t' = 0\ \ \ \ \$(relativity of simultaneity)