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**but I do not know why there is a square root -g . Could you give me the proof of this integral???? I mean How is this integral constructed??? What is the logic of this???? Thanks in advance....**

**I = ∫( − g ( x))^1/2 L ( x) d 4 x,**You are using an out of date browser. It may not display this or other websites correctly.

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ChrisVer

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The [itex]\int d^4 x \sqrt{-g}[/itex] is invariant...

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George Jones

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http://www.physics.uoguelph.ca/poisson/research/agr.pdf

or pages 51 - 53 of Carroll's lecture notes,

http://xxx.lanl.gov/abs/gr-qc/9712019

Both of these sets of lecture notes evolved into books, and the books are better than the notes.

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It might be worth looking at how it works in 2 spacelike dimensions. There, you can think of an integral over all space as the limit of a sum, where the sum is over little rectangles with sides [itex]\vec{\delta x}[/itex] and [itex]\vec{\delta y}[/itex]. The issue is: how to compute the area of the rectangle (in 3-D, it would be computing the volume of a parallelepiped, in 4-D, it would be computing the hypervolume of some 4-dimensional cell). In Cartesian coordinates, the displacement vectors [itex]\vec{\delta x}[/itex] and [itex]\vec{\delta y}[/itex] are orthogonal, and the lengths are just [itex]|\delta x|[/itex] and [itex]|\delta y|[/itex], respectively, so you just use the usual formula for area of a rectangle: [itex]A = |\delta x| |\delta y|[/itex]. But if you're using curvilinear coordinates, the two displacements are not necessarily orthogonal. In that case, how do you compute the area?

Well, here's a heuristic argument: In good-old Euclidean space, the area of a parallelogram with sides [itex]\vec{U}[/itex] and [itex]\vec{V}[/itex] is given by:

[itex]A = \vec{U} \times \vec{V} = |U||V| sin(\theta)[/itex] where [itex]\theta[/itex] is the angle between the two sides. We also know that:

[itex]\vec{U} \cdot \vec{V} = |U||V| cos(\theta)[/itex]

So we can relate the cross product to the dot product as follows:

[itex] |\vec{U} \times \vec{V}|^2 = |U|^2 |V|^2 sin^2(\theta) = |U|^2 |V|^2 - |U|^2 |V|^2 cos^2(\theta) = (\vec{U} \cdot \vec{U}) (\vec{V} \cdot \vec{V}) - (\vec{U} \cdot \vec{V})^2[/itex]

The dot-product can be written in terms of the metric:

[itex]\vec{U} \cdot \vec{V} = g_{\mu \nu} U^\mu V^\nu[/itex]

So we can rewrite the cross-product in terms of the metric:

[itex] |\vec{U} \times \vec{V}|^2 = (g_{\mu \nu} g_{\alpha \beta} - g_{\mu \alpha} g_{\nu \beta}) U^\mu U^\nu V^\alpha V^\beta[/itex]

Now, let's specialize to the case [itex]\vec{U} = \vec{\delta x}[/itex] and [itex]\vec{V} = \vec{\delta y}[/itex]. In that case, [itex]U^x = \delta x, U^y = 0, V^x = 0, V^y = \delta y[/itex]. So we have:

[itex] |\vec{\delta x} \times \vec{\delta y}|^2 = (g_{x x} g_{y y} - g_{x y} g_{x y}) \delta x^2 \delta y^2[/itex] (All other terms are zero)

Note that the expression involving the metric is just the determinate of the 2x2 matrix:

[itex]\left( \begin{array} \\ g_{x x} & g_{x y} \\ g_{y x} & g_{y y} \end{array} \right)[/itex]

So,

[itex] |\vec{\delta x} \times \vec{\delta y}|^2 = det(g) \delta x^2 \delta y^2[/itex]

So,

[itex] |\vec{\delta x} \times \vec{\delta y}| = \sqrt{det(g)} \delta x \delta y[/itex]

It would take a lot more work to see that this generalizes to more than 2 dimensions and to the case of a pseudo-Euclidean metric, but it gives you a little bit of the flavor for why something like [itex]\sqrt{det(g)}[/itex] might show up.

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Not really, the main change to make for n dimensions is to use the completely anti-symmetric scalar n-tuple product. You can define it using the Levi-Civita symbol in Cartesian coordinates. It essentially boils down to finding a volume spanned by a parallelepiped.It would take a lot more work to see that this generalizes to more than 2 dimensions

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Stevendarly Thanks for your explanatory answer, but how do We obtain square root -g ????????? I can not see any -g term in your answer...

It might be worth looking at how it works in 2 spacelike dimensions. There, you can think of an integral over all space as the limit of a sum, where the sum is over little rectangles with sides [itex]\vec{\delta x}[/itex] and [itex]\vec{\delta y}[/itex]. The issue is: how to compute the area of the rectangle (in 3-D, it would be computing the volume of a parallelepiped, in 4-D, it would be computing the hypervolume of some 4-dimensional cell). In Cartesian coordinates, the displacement vectors [itex]\vec{\delta x}[/itex] and [itex]\vec{\delta y}[/itex] are orthogonal, and the lengths are just [itex]|\delta x|[/itex] and [itex]|\delta y|[/itex], respectively, so you just use the usual formula for area of a rectangle: [itex]A = |\delta x| |\delta y|[/itex]. But if you're using curvilinear coordinates, the two displacements are not necessarily orthogonal. In that case, how do you compute the area?

Well, here's a heuristic argument: In good-old Euclidean space, the area of a parallelogram with sides [itex]\vec{U}[/itex] and [itex]\vec{V}[/itex] is given by:

[itex]A = \vec{U} \times \vec{V} = |U||V| sin(\theta)[/itex] where [itex]\theta[/itex] is the angle between the two sides. We also know that:

[itex]\vec{U} \cdot \vec{V} = |U||V| cos(\theta)[/itex]

So we can relate the cross product to the dot product as follows:

[itex] |\vec{U} \times \vec{V}|^2 = |U|^2 |V|^2 sin^2(\theta) = |U|^2 |V|^2 - |U|^2 |V|^2 cos^2(\theta) = (\vec{U} \cdot \vec{U}) (\vec{V} \cdot \vec{V}) - (\vec{U} \cdot \vec{V})^2[/itex]

The dot-product can be written in terms of the metric:

[itex]\vec{U} \cdot \vec{V} = g_{\mu \nu} U^\mu V^\nu[/itex]

So we can rewrite the cross-product in terms of the metric:

[itex] |\vec{U} \times \vec{V}|^2 = (g_{\mu \nu} g_{\alpha \beta} - g_{\mu \alpha} g_{\nu \beta}) U^\mu U^\nu V^\alpha V^\beta[/itex]

Now, let's specialize to the case [itex]\vec{U} = \vec{\delta x}[/itex] and [itex]\vec{V} = \vec{\delta y}[/itex]. In that case, [itex]U^x = \delta x, U^y = 0, V^x = 0, V^y = \delta y[/itex]. So we have:

[itex] |\vec{\delta x} \times \vec{\delta y}|^2 = (g_{x x} g_{y y} - g_{x y} g_{x y}) \delta x^2 \delta y^2[/itex] (All other terms are zero)

Note that the expression involving the metric is just the determinate of the 2x2 matrix:

[itex]\left( \begin{array} \\ g_{x x} & g_{x y} \\ g_{y x} & g_{y y} \end{array} \right)[/itex]

So,

[itex] |\vec{\delta x} \times \vec{\delta y}|^2 = det(g) \delta x^2 \delta y^2[/itex]

So,

[itex] |\vec{\delta x} \times \vec{\delta y}| = \sqrt{det(g)} \delta x \delta y[/itex]

It would take a lot more work to see that this generalizes to more than 2 dimensions and to the case of a pseudo-Euclidean metric, but it gives you a little bit of the flavor for why something like [itex]\sqrt{det(g)}[/itex] might show up.

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haushofer

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You can also consult Zwiebach's string theory book, chapter 6.

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DrGreg

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##\sqrt{-g}## is shorthand for ##\sqrt{-\text{det}(g)} = \sqrt{|\text{det}(g)|}##.Stevendarly Thanks for your explanatory answer, but how do We obtain square root -g ????????? I can not see any -g term in your answer...

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Hi,ChrisVer you said [itex]\int d^4 x \sqrt{-g}[/itex] is invariant, but I see that integration of multiplication of lagrangian density and infinite small volume element is actually invariant, so I wonder if there is a proof or derivation of this situation. Could you provide me it's derivation ???

The [itex]\int d^4 x \sqrt{-g}[/itex] is invariant...

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This is just a regular change of coordinates for a multivariable integral.Hi,ChrisVer you said [itex]\int d^4 x \sqrt{-g}[/itex] is invariant, but I see that integration of multiplication of lagrangian density and infinite small volume element is actually invariant, so I wonder if there is a proof or derivation of this situation. Could you provide me it's derivation ???

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pardon orodruin, would you mind spelling it out, or explain in more detail ????This is just a regular change of coordinates for a multivariable integral.

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Thanks for your valuable responses...

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ChrisVer

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http://www.nyu.edu/classes/tuckerman/stat.mech/lectures/lecture_2/node2.html

I call it fun because it can have interesting analogies from other 'fields' (like the conservation of volumes in phase-space diagrams)...

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hi, I was looking some equations related to invariance of lagrangian. In my attachment you can see that there are 4 equations, and while the we proceed to 4. equations from 3. equations (green box) you can see that first term and third term of 3. equation vanishes. I would like to ask how these terms vanish mathematically. Could you explain this situation using some mathematical demonstrations??????

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ChrisVer

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they are total derivatives within the integral, they will vanish eventually...

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haushofer

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ChrisVer

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I mean at the end of the day, in every variation formalism you go to, you will eventually get some EOM that will have:

[itex]S = \int L(x ,\dot{x} ;t ) dt [/itex]

Correct?

then the concept is that the Lagrangian is not uniquely defined, but there is a whole set of Lagrangians that could in general produce the same equations of motions. Those lagrangians in the above case are:

[itex] L + \frac{d}{dt} K [/itex]

with the last term being a total derivative... The total derivatives don't survive into the equations of motion after the Lagrangian variation. The [itex]dK/dt[/itex] will by default satisfy the Euler-Lagrange equation.

Another way to see that is by looking at the integrals you have; they will eventually give you 'surface' terms, which in general can lead to some "constraints" over your fields... where? at infinity...

What is this integral equal to:

[itex] \int_{a}^{b} \frac{dA(x)}{dx} dx = ?[/itex]

[itex]A(x)[/itex] is a parametrized line, but the result of this int is 1 number... the generalization to more dimension's why these type of things are called 'surface' terms....

Another example from the 'classical case' of a relativistic string is that those "surface" constraints will generally result to the known "boundary" conditions for the oscillating string (like Neumann or Dirichlet ones).

[itex]S = \int L(x ,\dot{x} ;t ) dt [/itex]

Correct?

then the concept is that the Lagrangian is not uniquely defined, but there is a whole set of Lagrangians that could in general produce the same equations of motions. Those lagrangians in the above case are:

[itex] L + \frac{d}{dt} K [/itex]

with the last term being a total derivative... The total derivatives don't survive into the equations of motion after the Lagrangian variation. The [itex]dK/dt[/itex] will by default satisfy the Euler-Lagrange equation.

Another way to see that is by looking at the integrals you have; they will eventually give you 'surface' terms, which in general can lead to some "constraints" over your fields... where? at infinity...

What is this integral equal to:

[itex] \int_{a}^{b} \frac{dA(x)}{dx} dx = ?[/itex]

[itex]A(x)[/itex] is a parametrized line, but the result of this int is 1 number... the generalization to more dimension's why these type of things are called 'surface' terms....

Another example from the 'classical case' of a relativistic string is that those "surface" constraints will generally result to the known "boundary" conditions for the oscillating string (like Neumann or Dirichlet ones).

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ChrisVer

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