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Let's take two events [itex]P_1,P_2[/itex] in a Minkowsky spacetime, and let's choose them both lying on the [itex]\omega[/itex] axis, separated by a certain distance.

Now, I want to calculate the world line of the particle which experiences the least proper time during its trip between the two points.

The proper time, taking the velocity of the particle costant in its modulus (but not in its direction!!) can be written [itex]d\tau=dt\sqrt{1-v^2/c^2}=dt\sqrt{dt^2-dx^2/c^2}=\frac{1}{c}\sqrt{dw^2-dx^2}[/itex]; so it's easy to set up a variational calculus, minimizing the integral[tex]\tau=\frac{1}{c}\int_{P_1}^{P_2}\sqrt{1-\left(\frac{dx}{d\omega}\right)^2}d\omega[/tex](the quantity under the square root can't be negative, you all know why).

So by taking [itex]L=\sqrt{1-{x'}^2(\omega)}[/itex] with [itex]x'(\omega)=\frac{dx}{d\omega}[/itex] we need to calculate the euler-lagrange equation:[tex]\frac{\partial L}{\partial x(\omega)}-\frac{d}{d\omega}\frac{\partial L}{\partial x'(\omega)}=0[/tex]but since L doesn't depend on x(w) but only on x'(w) we get to[tex]\frac{d}{d\omega}\frac{\partial L}{\partial x'(\omega)}=\frac{d}{d\omega}\frac{x'(\omega)}{ \sqrt{1-{x'}^2(\omega)} }=0[/tex]and so [tex]\frac{x'(\omega)}{ \sqrt{1-{x'}^2(\omega)} }=G\Rightarrow x'=\frac{G}{\sqrt{1+G^2}}[/tex] where G is a constant, determinable by the position of the two point on the spacetime.

So we end up with[tex]x(\omega)=\frac{G}{\sqrt{1+G^2}}\omega[/tex]

This is supposed to be the path between the two point on the spacetime of the particle which experiences the least proper time.

In the case considered in the beginning I took both point lying on the w axis; so the constant G turns out to be 0, so the path I'm looking for is [itex]x=0[/itex], which corresponds to a particle at rest.

But this is impossible, since it's know that this path it's the one with the longest proper time!!

I found various solution for this:

1) I did not understood nothing of relativity

2) I've done various mistake

3) The variational method I've used gave me effectively an extremal path, but the longest, not the shortest. If so, does this means that there are not a minimum path? (I don't think so)ORI have to introduce some lagrange multiplier-style constraint on the calculation (like the fact that I cant'get back in time)??

Can you help me to solve this riddle??

Thanks for your disponibility!!!

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# Easy question on proper time

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