Minkowski spacetime interval's Lorentz invariance

  • Thread starter cefarix
  • Start date
  • #1
63
0

Main Question or Discussion Point

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.... :wink:
 

Answers and Replies

  • #2
142
0
Assuming that you are familiar with the Lorentz transformation equations themselves (object moving along x):
[tex]x'=\gamma(x-vt)[/tex]
[tex]t'=\gamma(t-vx/c^2)[/tex]
[tex]y'=y[/tex]
[tex]z'=z[/tex]
Lorentz invariance requires a space-time 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^2-dx^2-dy^2-dz^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^2-x^2-y^2-z^2=0[/tex].
This is also true from a moving frame:
[tex]c^2t'^2-x'^2-y'^2-z'^2=0[/tex],
so
[tex]c^2t^2-x^2-y^2-z^2=c^2t'^2-x'^2-y'^2-z'^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
learningphysics
Homework Helper
4,099
5
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]x2-x1 = \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.
 

Related Threads on Minkowski spacetime interval's Lorentz invariance

  • Last Post
Replies
17
Views
6K
  • Last Post
Replies
2
Views
675
  • Last Post
Replies
10
Views
4K
Replies
4
Views
2K
Replies
24
Views
4K
Replies
20
Views
35K
Replies
4
Views
8K
Replies
6
Views
358
Replies
4
Views
1K
Replies
1
Views
3K
Top