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Office_Shredder

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- 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.

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.

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