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LaTeX practice


Lorentz Jr

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c = 10 m/s, v/c = .866, [math]\gamma [/math] = 2, and L = 10m.

To calculate the difference between the two clocks on the train, we'll read them simultaneously at a time t in the ground frame.

The time on the clock in back is [math]\displaystyle{t_{B}' = \gamma (t - \frac{v x_B}{c^2})}[/math].

The time on the clock in front is [math]\displaystyle{t_{F}' = \gamma (t - \frac{v x_F}{c^2})}[/math].

So [math]c^2(t_{B}'-t_{F}') = \gamma v(x_F - x_B)[/math].

Now we need to calculate [math]x_F[/math] and [math]x_B[/math].

[math]x_B = \gamma(x_{B}' - v t_{B}') = \gamma v t_{B}'[/math].

[math]x_F = \gamma(x_{F}' - v t_{F}') = L + \gamma v t_{F}'[/math].

So [math]x_F - x_B = \gamma(L + v(t_{F}' - t_{B}'))[/math].

Now we get the desynchronization [math]\Delta = t_{B}'-t_{F}'[/math]:

[math]c^2(t_{B}'-t_{F}') = \gamma v(\gamma(L + v(t_{F}' - t_{B}')))[/math]

[math]c^2\Delta = \gamma^2 v(L - v\Delta)[/math]

[math](1 - \beta^2)\Delta = \beta(L/c - \beta\Delta)[/math]

[math]\Delta = \beta L/c = vL/c^2[/math]

Edited by Lorentz Jr
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[math]\displaystyle{ x = \gamma L \left(1 + \sqrt{2} \frac{v}{c}\right) = 2(10m)(1 + .866\sqrt{2}) = 44.5m  }[/math]

[math]\displaystyle{ t= \gamma \frac{L}{c}\left(\sqrt{2} + \frac{v}{c}\right) = \frac{2(10m)}{1m/s}\left(\sqrt{2} + .866\right) = 45.6s  }[/math]

Edited by Lorentz Jr
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  • 3 weeks later...

[math]\displaystyle{ dt' = dt_s / \gamma } [/math]

[math]\displaystyle{ t_{r1} = \frac{y + dy}{c} }[/math]

[math]\displaystyle{  t_{r2} = \frac{ds}{v} + \frac{y}{c} }[/math]

[math]\displaystyle{   dt_r = \frac{ds}{v} - \frac{dy}{c} }[/math]

[math] dy = ds \sin \theta [/math]

[math] dt' = dt_r [/math]

[math] \displaystyle{ \frac{dt_s }{\gamma} = \frac{ds}{v} - \frac{dy}{c} }[/math]

[math] \displaystyle{ \frac{ds }{\gamma v }= \frac{ds}{v} - \frac{ds \sin \theta}{c}} [/math]

[math] c = \gamma c - \gamma v \sin\theta [/math]

[math] \displaystyle{  \sin\theta = \frac{c}{v} \left( 1 - \frac{1}{\gamma}\right) } [/math]

 

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