Spacetime Inverval Invariance using Lorentz Transformations

maccyjj
Messages
2
Reaction score
0

Homework Statement


Prove that the spacetime interval
-(ct)^{2} + x^{2} + y^{2} + z^{2}
is invariant.

[/itex]<br /> <h2>Homework Equations</h2><br /> Lorentz transformations<br /> \Deltax&amp;#039; = \gamma(\Deltax-u\Deltat)<br /> \Deltay&amp;#039; = \Deltay<br /> \Deltaz&amp;#039; = \Deltaz<br /> \Deltat&amp;#039; = \gamma(\Deltat-u\Deltax/c^{2})<br /> <br /> <br /> <h2>The Attempt at a Solution</h2><br /> I have tried to prove that \Delta S = \Delta S&amp;#039;<br /> So first I said that \Delta S&amp;#039; = - \Delta (ct&amp;#039;)^{2} + \Delta (x&amp;#039;)^{2} + \Delta (y&amp;#039;)^{2} + \Delta (z&amp;#039;)^{2}<br /> <br /> And inserted all the Lorentz Transformations above into the above formula.<br /> <br /> I end up simplyfying it to get<br /> <br /> \gamma^{2} (x^{2} + u^{2}t^{2} - c^{2}t^{2} - \frac{u^{2}x^{2}}{c^{2}}) + y^{2} + z^{2}<br /> <br /> How does this equal S = - \Delta (ct)^{2} \Delta (x)^{2} + \Delta (y)^{2} + \Delta (z)^{2} ? I can&#039;t see a way to get rid of the extra terms to get this simple function.<br /> <br /> Any help would be really really great!
 
Physics news on Phys.org
If you collect the terms, you'll see that the coefficient of x2 is \gamma^2(1-u^2/c^2). Use the definition of \gamma to simplify that.
 
Oh of course! How did I miss that?

Thank you so much I got it out now!
 
Back
Top