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

[Apashanka]

From the invariance of space time interval the metric dΓ²=dt²-dx²-dy²-dz²
dΓ²=g_μνdx^μdx^ν
dΓ=√(g_μνv^μv^μ)dt
dΓ=proper time.
Can someone please help me in sort out why the term √(g_μνdx^μdx^ν) is taken as the Lagrangian,as geodesic equation is solved by taking this to be the Lagrangian.

 

[Ibix]

https://www.physicsforums.com/threads/action-for-a-relativistic-free-particle.963274/

 

[PGaccount]

Think of it like this:

√(dt² - dx²) = √(1 - v²)dt

so -m√(dt² - dx²) = -m√(1 - v²) dt

L = -m√(1 - v²)

 

[Apashanka]

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??

 

[PGaccount]

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

 

[Ibix]

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.

 

[Orodruin]

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.

 

[Apashanka]

You mean time-like.

yes it is time like ,interval (dΓ)maximised

 

[Apashanka]

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??

 

[Orodruin]

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.

 

[Apashanka]

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)??

 

[Ibix]

(e.g other than the frame in which interval is dΓ)

Interval is invariant. It's the same for all coordinate systems.

 

[Apashanka]

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