A person travelling through a geodesic.

  • Context: Graduate 
  • Thread starter Thread starter Kevin_spencer2
  • Start date Start date
  • Tags Tags
    Geodesic
Click For Summary
SUMMARY

A person traveling along a geodesic experiences no acceleration, as described by the geodesic equation Δ_u u = 0. In Euclidean space, a particle's trajectory is linear, represented by X(u) = au + b, indicating no acceleration. While an individual in free fall along a geodesic will not feel any acceleration and an accelerometer will read zero, a global coordinate system reveals nonzero second time derivatives of position coordinates, particularly evident in scenarios like circular orbits around a central mass.

PREREQUISITES
  • Understanding of geodesic equations in differential geometry
  • Familiarity with Newtonian mechanics and free fall concepts
  • Knowledge of local versus global coordinate systems
  • Basic principles of orbital mechanics
NEXT STEPS
  • Study the implications of the geodesic equation in general relativity
  • Explore the concept of inertial frames in curved spacetime
  • Learn about the mathematical formulation of circular orbits in gravitational fields
  • Investigate the relationship between local and global coordinates in physics
USEFUL FOR

Physicists, mathematicians, and students of general relativity or differential geometry seeking to deepen their understanding of geodesics and their implications in motion and acceleration.

Kevin_spencer2
Messages
29
Reaction score
0
A person traveling through a geodesic. Does experiment some kind of acceleration?, since the geodesic equation analogue to Newton one is:

[tex]\Delta _{u} u =0[/tex] and in an Euclidean space there's no acceleration for a particle line [tex]X(u)=au+b[/tex]
 
Physics news on Phys.org
Someone following a geodesic will be in "free fall". They won't feel any acceleration - if they have an accelerometer, it will also read zero. If you set up a global coordinate system, however, the second time derivatives of the position coordinates will in general be nonzero.

Think, for example, of an object orbiting a central mass in a circular orbit. Such an object is locally in free fall, and could set up a local coordinate system that is nearly inertial over small distances.

In a global coordinate system anchored to the central mass, the second time derivative of the spatial postion coordinates will be nonzero, however.
 

Similar threads

  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 35 ·
2
Replies
35
Views
3K
  • · Replies 30 ·
2
Replies
30
Views
2K
  • · Replies 0 ·
Replies
0
Views
1K
  • · Replies 25 ·
Replies
25
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 27 ·
Replies
27
Views
7K