Why is invariant interval invariant?

Bob_for_short

It makes sense to speak of a(V) if ds ≠ 0. And you started from ds = 0.

Concerning the group character of LT, it was first written in H. Poincaré seminal article.
He required the mechanical equations to be also invariant under LT, i.e., introduced ds ≠ 0.

See http://www.univ-nancy2.fr/poincare/bhp/hp1906rpen.xml

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lalbatros

It makes sense to speak of a(V) if ds ≠ 0. And you started from ds = 0.
kof9595995 was asking about why ds² should be invariant.
Therefore I was implicitely discussing the transformation from ds² to ds'² in the general case.
I skipped a few obvious steps to go directly to the consequence of the equivalence of ds²=0 and ds'²=0.

To be more explicit, the general form of the transformation could be this:

ds'² = f(x,y,z,t,x',y',z',t',Vx,Vy,Vz,ds²)

Obvious arguments (same as those listed in Einstein 1905) lead to the simplification:

ds'² = f(V,ds²)

Developping this in series, assuming ds² is small enough, leads to

 . . . ds'² = b(V) + a(V) ds²

This relation  should be valid for any interval, even along light rays.
Since we should have ds²=0 => ds'²=0, we can conclude that:

b(V) = 0

And this leads us to:

 . . . ds'² = a(V) ds²

This is were I started my https://www.physicsforums.com/showpost.php?p=2440322&postcount=20".

But https://www.physicsforums.com/showpost.php?p=2437046&postcount=3", this was indeed an easy question.

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kof9595995

It's not exactly intuitive, but you might find this post relevant.
Hi DrGreg, you proved dedr=de'dr', and denote it as ds^2. But I don't think it's good enough, you've shown according to first postulate, ds^2=dedr is invariant, but you didn't show further that $$\Delta {s^2} = \Delta {x^2} + \Delta {y^2} + \Delta {z^2} - {c^2}\Delta {t^2}$$
So the invariance of dedr is not so satisfactory because I can also argue that dedr=dt^2, which bring us back to Galilean transformation

DrGreg

Gold Member
Hi DrGreg, you proved dedr=de'dr', and denote it as ds^2. But I don't think it's good enough, you've shown according to first postulate, ds^2=dedr is invariant, but you didn't show further that $$\Delta {s^2} = \Delta {x^2} + \Delta {y^2} + \Delta {z^2} - {c^2}\Delta {t^2}$$
So the invariance of dedr is not so satisfactory because I can also argue that dedr=dt^2, which bring us back to Galilean transformation
Carry on reading my post. I found e and r in terms of x and t and said "Substitute (4) and (5) into (3) to get the standard equation for the Lorentz interval." I didn't give all the details (an exercise for the reader!) and I ignored y and z, but once you've got the result in 2D spacetime, it's not hard to generalise to 4D spacetime.

If you can't work out the missing details, ask and I'll help.

Just in case you're confused, in my version I'm using the convention that

$$ds^2 = dt^2 - \frac{dx^2}{c^2}$$​

If you want consistency with your version I'd have to redefine $ds^2 = -c^2\,de\,dr$.

P.S. the link in my original post is now dead. Site moved to http://autotheist.synthasite.com/bondi1.php

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kof9595995

Sorry I did not read your post carefully enough. Your post do seem to be the nicest answer to my question so far, though also as you said, "not exactly intuitive".
Thanks

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