Show that Geodesic is space-like everywhere

  • Thread starter Thread starter wam_mi
  • Start date Start date
  • Tags Tags
    Geodesic
Click For Summary
SUMMARY

The discussion centers on proving that if a geodesic is space-like at one point, it remains space-like everywhere. The geodesic equation is given as \(\ddot{X}^{\mu}+\Gamma^{\mu}_{\alpha \beta}\dot{X}^{\alpha}\dot{X}^{\beta} = 0\), and the metric is defined as \(ds^{2} = g_{\alpha \beta} \dot{X}^{\alpha} \dot{X}^{\beta} = + 1\). The participant concludes that using parallel transport preserves the inner product, thereby maintaining the space-like character of the geodesic throughout its entirety.

PREREQUISITES
  • Understanding of geodesic equations in differential geometry
  • Familiarity with the concept of metrics in curved spacetime
  • Knowledge of parallel transport and its properties
  • Basic grasp of inner product spaces in the context of physics
NEXT STEPS
  • Study the implications of the geodesic equation in general relativity
  • Explore the properties of metrics in 4-dimensional spacetime
  • Learn about parallel transport and its role in preserving geometric properties
  • Investigate the relationship between space-like, time-like, and null geodesics
USEFUL FOR

Students and researchers in theoretical physics, particularly those focusing on general relativity and differential geometry, will benefit from this discussion.

wam_mi
Messages
78
Reaction score
1

Homework Statement



If the geodesic is space-like somewhere, show that the geodesic is space-like everywhere.

Homework Equations



Geodesic equation: \ddot{X}^{\mu}+\Gamma^{\mu}_{\alpha \beta}\dot{X}^{\alpha}\dot{X}^{\beta} = 0


The Attempt at a Solution



I looked at the metric

ds^{2} = g_{\alpha \beta} \dot{X}^{\alpha} \dot{X}^{\beta} = + 1,

where g_{\alpha \beta} is the general curved metric in 4 dimensions of space-time. I try to write it in the form

g_{\alpha \beta} \dot{X}^{\alpha} \dot{X}^{\beta} = g_{\alpha \beta} \dot{X'}^{\alpha} \dot{X'}^{\beta}

where X is in one frame while X' is in another.

What exactly do I need to do now? I'm confused...

Thanks
 
Physics news on Phys.org
I think I got it. Is it correct to say if one puts the inner product into the parallel transport expression, one finds that the expression vanishes as parallel transport preserves the inner product such that the character of the geodesic never changes.

Thanks
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
0
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
4K