Why is √(gμνdxμdxν) the Lagrangian for Geodesic Eq?

Click For Summary

Discussion Overview

The discussion revolves around the use of the term √(gμνdxμdxν) as the Lagrangian for the geodesic equation in the context of general relativity. Participants explore the implications of this choice, its connection to the invariance of the spacetime interval, and the nature of the action being extremized.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants note that the metric dΓ² = dt² - dx² - dy² - dz² leads to the expression dΓ² = gμνdxμdxν, suggesting a connection to the proper time.
  • One participant asks why √(gμνdxμdxν) is taken as the Lagrangian, indicating a need for clarification on this choice.
  • Another participant presents an analogy involving the expression L = -m√(1 - v²) to illustrate the relationship between the Lagrangian and the proper time.
  • Some participants discuss the concept of maximizing the interval between points in spacetime, questioning whether this choice of Lagrangian reflects that maximization.
  • There is a clarification that the action, rather than the Lagrangian itself, is what is stationary, with a distinction made between maximum and minimum values.
  • Participants express uncertainty about the implications of the term "stationary" in the context of the action and whether it refers to an extremum.
  • One participant raises a question about whether the Lagrangian is derived from a different frame of reference, while another asserts that the interval is invariant across all coordinate systems.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of the Lagrangian and its implications for the spacetime interval. There is no consensus on the nature of the action being extremized or the connection to different frames of reference.

Contextual Notes

Some statements reflect assumptions about the nature of spacetime intervals and the relationship between the Lagrangian and the action, which may depend on specific interpretations or definitions that are not fully resolved in the discussion.

Apashanka
Messages
427
Reaction score
15
From the invariance of space time interval the metric dΓ2=dt2-dx2-dy2-dz2
2=gμνdxμdxν
dΓ=√(gμνvμvμ)dt
dΓ=proper time.
Can someone please help me in sort out why the term √(gμνdxμdxν) is taken as the Lagrangian,as geodesic equation is solved by taking this to be the Lagrangian.
 
Physics news on Phys.org
Think of it like this:

√(dt2 - dx2) = √(1 - v2)dt

so -m√(dt2 - dx2) = -m√(1 - v2) dt

L = -m√(1 - v2)
 
Apashanka said:
From the invariance of space time interval the metric dΓ2=dt2-dx2-dy2-dz2
dΓ2=gμνdxμdxν
dΓ=√(gμνvμvμ)dt
dΓ=proper time.
Can someone please help me in sort out why the term √(gμνdxμdxν) is taken as the Lagrangian,as geodesic equation is solved by taking this to be the Lagrangian.
For light like interval dΓ is the maximum interval between two points in space time,
that means this choice of Lagranian actually maximises the interval between points in space time??
 
expand out the square root in L = -m√(1 - v2)

You will find a term 1/2 mv2 which is the kinetic energy T.

Then try to connect this to L = T - U
 
Apashanka said:
this choice of Lagranian actually maximises the interval between points in space time??
Yes. The requirement is that the action be extremised, not necessarily minimised.
 
Apashanka said:
For light like interval dΓ is the maximum interval between two points in space time,
that means this choice of Lagranian actually maximises the interval between points in space time??
You mean time-like. Also, it is the action that is stationary, not the Lagrangian.
 
  • Like
Likes   Reactions: Ibix
Orodruin said:
You mean time-like.
yes it is time like ,interval (dΓ)maximised
 
Orodruin said:
Also, it is the action that is stationary, not the Lagrangian.
Will you please explain this statement.
Will stationary mean here extremum.e.g either maximum or minimum??
 
  • #10
Apashanka said:
Will you please explain this statement.
Will stationary mean here extremum.e.g either maximum or minimum??
Stationary means that ##\delta S = 0##, this does not necessarily imply max or min.
 
  • #11
Apashanka said:
From the invariance of space time interval the metric dΓ2=dt2-dx2-dy2-dz2
dΓ2=gμνdxμdxν
dΓ=√(gμνvμvμ)dt
dΓ=proper time.
Can someone please help me in sort out why the term √(gμνdxμdxν) is taken as the Lagrangian,as geodesic equation is solved by taking this to be the Lagrangian.
Is this lagrangian obtained from another frame of reference (e.g other than the frame in which interval is dΓ)??
 
  • #12
Apashanka said:
(e.g other than the frame in which interval is dΓ)
Interval is invariant. It's the same for all coordinate systems.
 
  • #13
Ibix said:
Interval is invariant. It's the same for all coordinate systems.
Why I try to say is that other than the frame in which spatial separation between events is 0
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 42 ·
2
Replies
42
Views
6K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K