Why the square-root of the metric

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In summary, the Lagrangian density for a free scalar particle is scaled by \sqrt{-g} because it is a tensor density and this factor is necessary to maintain invariance under a change of coordinate charts. This is explained in more detail in Ben Crowell's online book, and the reason for this factor is that it represents the volume element in a set of arbitrary coordinates.
  • #1
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In a paper about field theory in curved spacetime, an author says that the Lagrangian density for a free scalar particle is

[itex]L = \sqrt{-g} ((\nabla_\mu \Phi)(\nabla^\mu \Phi) - m^2 \Phi^2)[/itex]

Is there a simple explanation for why this is scaled by [itex]\sqrt{-g}[/itex]?
 
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The simple explanation is the Lagrangian is a tensor density. You want to map a volume to a number (a Lorentz scalar), but representing the volume requires the factor of sqrt(g).

http://www.lightandmatter.com/html_books/genrel/ch04/ch04.html [Broken] (Ben Crowell's online book) talks a little about tensor densities in an informal way.
 
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  • #3
Because the Lagrangian density is integrated over spacetime to obtain the action, and without the factor of [itex]\sqrt{-g}[/itex], the integral isn't invariant under a change of coordinate charts. [Edit: This is another way of saying what pervect said.]
 
  • #4
Because [itex]\sqrt{-g}=det(J)[/itex], where J is the Jacobian matrix transforming from Cartesian coordinates to the coordinates being used in the particular problem.

The volume element in a set of arbitrary coordinates is given by dV=det(J)dx1dx2...dxn. For example, in Cartesian coordinates in 3D space dV=dxdydz. The determinant of the Jacobian transforming to spherical coordinates is r2sinθ, so the volume element in spherical coordinates is dV=r2sinθdrdϕdθ.

You need the factor in front of dnx in order to keep things invariant under a coordinate transformation.
 
  • #5


The square-root of the metric, also known as the determinant of the metric, is an important factor in the Lagrangian density for a free scalar particle in a curved spacetime. This is because the Lagrangian density is a measure of the energy of a system, and in a curved spacetime, the metric itself is affected by the presence of matter and energy. The square-root of the metric takes into account the curvature of the spacetime, which in turn affects the dynamics of the scalar particle.

One way to understand this is through the concept of the action principle, which states that the path taken by a particle between two points in spacetime is the one that minimizes the action, a quantity that is related to the Lagrangian density. In a flat spacetime, the action is simply the integral of the Lagrangian density over time. However, in a curved spacetime, the metric itself is changing and therefore the action must also take into account the variation of the metric. This is where the square-root of the metric comes in - it is a way to incorporate the changing metric into the action and therefore the Lagrangian density.

Furthermore, in general relativity, the metric is a fundamental quantity that describes the geometry of spacetime. The square-root of the metric is a way to account for the non-Euclidean nature of curved spacetime, and it is necessary for accurately describing the dynamics of particles in this type of spacetime.

In summary, the inclusion of the square-root of the metric in the Lagrangian density for a free scalar particle is essential for taking into account the curvature of spacetime and accurately describing the dynamics of the particle in a curved spacetime.
 

1. Why is the square-root of the metric used in Einstein's theory of general relativity?

The square-root of the metric is used in Einstein's theory of general relativity because it is a fundamental component of the theory's mathematical framework. The metric tensor, which is a mathematical object that describes the geometry of spacetime, is raised to the power of 1/2 in order to properly account for the curvature of spacetime and the effects of gravity.

2. How does the square-root of the metric relate to the concept of spacetime curvature?

The square-root of the metric is crucial in understanding the concept of spacetime curvature in Einstein's theory of general relativity. This is because the metric tensor, when raised to the power of 1/2, represents the spacetime curvature at a given point. This curvature is what causes objects to move in a curved path in the presence of massive objects, such as stars or planets.

3. Can you explain the significance of the square-root of the metric in the Schwarzschild solution?

The Schwarzschild solution is a mathematical solution to Einstein's field equations, which describe the curvature of spacetime due to a massive, non-rotating object. In this solution, the square-root of the metric plays a crucial role in determining the event horizon of a black hole. The event horizon is the boundary at which the gravitational pull of a black hole becomes so strong that even light cannot escape, and the square-root of the metric is used to calculate this boundary.

4. What is the relationship between the square-root of the metric and the speed of light?

In Einstein's theory of general relativity, the speed of light is a fundamental constant that is used to define the units of measurement for both space and time. The square-root of the metric is also used in these units and is directly related to the speed of light. This relationship allows for the consistent treatment of both space and time in the theory, and it is essential in accurately describing the effects of gravity.

5. How does the square-root of the metric affect measurements of time and distance in general relativity?

The square-root of the metric is a crucial factor in accurately measuring time and distance in general relativity. This is because the metric tensor, when raised to the power of 1/2, represents the spacetime interval between two events. This interval is used to measure both time and distance in the theory, and the square-root of the metric ensures that these measurements are consistent and correctly account for the effects of gravity.

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