Minkowski spacetime interval's Lorentz invarianceby cefarix Tags: interval, invariance, lorentz, minkowski, spacetime 

#1
Aug505, 09:50 PM

P: 69

Maybe this is really easy, but...
Can someone show me how the sign reversal between the space and time components of Minkowski spacetime make its intervals Lorentz invariant (mathematical derivation) ? Thanks.... 



#2
Aug605, 02:08 AM

P: 142

Assuming that you are familiar with the Lorentz transformation equations themselves (object moving along x):
[tex]x'=\gamma(xvt)[/tex] [tex]t'=\gamma(tvx/c^2)[/tex] [tex]y'=y[/tex] [tex]z'=z[/tex] Lorentz invariance requires a spacetime interval between two events to have the same magnitude from any frame so: [tex]ds'^2=ds^2[/tex]. If [itex]ds^2[/itex] would have been defined Euclidean as: [tex]ds^2=c^2t^2+dx^2+dy^2+dz^2[/tex], then this equation should have hold: [tex]c^2t^2+dx^2+dy^2+dz^2=c^2t'^2+dx'^2+dy'^2+dz'^2[/tex] If you solve the primed coordinates in this equation using the Lorentz transformation equations you end up with something that is clearly nonsense (check for yourself). If on the other hand we define [itex]ds^2[/itex] Minkowskian: [tex]ds^2=c^2t^2dx^2dy^2dz^2[/tex], then solving the primed coordinates leads to a correct result. Another way to arrive at the [itex]+[/itex] is using a lightpulse that spreads with speed [itex]c[/itex] from the origin in all directions. A sphere is formed by this lightspreading according to: [tex]c^2t^2=x^2+y^2+z^2[/tex], so [tex]c^2t^2x^2y^2z^2=0[/tex]. This is also true from a moving frame: [tex]c^2t'^2x'^2y'^2z'^2=0[/tex], so [tex]c^2t^2x^2y^2z^2=c^2t'^2x'^2y'^2z'^2[/tex]. The rest of the story is the same but this last method was actually used to derive the Lorentz transformation equations in the first place. 



#3
Aug705, 04:17 PM

HW Helper
P: 4,125

I would do it this way. Take two events (t1,x1,y1,z1) and (t2,x2,y2,z2)
Using the lorentz transformations you can show that: [tex]\Delta t' = \gamma (\Delta t  v\Delta x/c^2) [/tex] [tex]\Delta x' = \gamma (\Delta x  v\Delta t)[/tex] [tex]\Delta y' = \Delta y[/tex] [tex]\Delta z' = \Delta z[/tex] To get the above just use the lorentz transformations to calculate t1',x1'....t2',x2'... and then [tex]x2x1 = \Delta x[/tex] and [tex]x2'x1' = \Delta x'[/tex] etc.... Now just calculate out: [tex]c^2*(\Delta t')^2 (\Delta x')^2(\Delta y')^2(\Delta z')^2[/tex] by substituting the above formulas. You'll see that it comes out to: [tex]c^2*(\Delta t)^2 (\Delta x)^2(\Delta y)^2(\Delta z)^2[/tex] Showing that the euclidean formula (with + instead of ) is not invariant is simple. Just take two events let's say (0,0,0,0) and (t1,0,0,0) Now [tex]c^2*(\Delta t)^2 +(\Delta x)^2+(\Delta y)^2+(\Delta z)^2[/tex] comes out to [tex]c^2*t1^2[/tex] for S' it comes out to [tex](c^2+v^2) (\gamma)^2*t1^2[/tex] The two are not equal for nonzero v and t1. 


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