Variations, Euler-Lagrange, and Stokes

[haushofer]

Hi, I have some questions which I encountered during my thesis-writing, I hope some-one can help me out on this :)

First, I have some problems interpreting coordinate-transformations ( "active and passive") and the derivation of the Equations of Motion. We have

$$S = \int L(\phi, \partial_{\mu} \phi) d^{4}x$$

and

$$\delta S = 0$$

, in which the variation of the field is arbitrary. My question is: how exactly is this variation defined? One has 2 options:

$$\delta \phi = \phi^{'}(x^{'}) - \phi (x)$$
or
$$\delta \phi = \phi^{'}(x) - \phi (x)$$

where the difference lies in the argument. In notes of Aldrovandi and Pereira ( Notes for a classical course on fields ) option 1 is choosen. And in Inverno they say that option 2 is a coordinate transformation. Why is that? I tend to choose for option two, because here you actually change the field; I would say that a scalar quantity is invariant under transformation 1, so here you just state general covariance in stead of obtaining the Equations of Motion. At the other hand, you can always choose x=x'. In the end I want to look at why one is able to commute variations and partial derivatives, so I need the exact definition of the variation of the field.

Another question concerns Stokes theorem. I understand the theorem totally for n-forms ( where you define the boundary of your region with chains etc ), but why is it also valid for example in the derivation of the Euler-Lagrange equations or tensor densities?

Many thanks in forward :)

-edit My TeX-code is not working properly for some reason, but I hope it is clear.

 

[haushofer]

No-one? :(

 

[haushofer]

Ok, let me put it differently. I know about pull-backs, push-forwards, diffeomorphisms etc. I always regarded a push-forward of a vector as some sort of "generalized coordinatetransformation", but still as passively. Why is a push-forward an active transformation, and a coordinatetransformation just passive?

 

[genneth]

I suggest a good textbook -- Analysis, Manifolds and Physics is good. It covers differential geometry over finite and infinite dimensional spaces, which are what you need for the formal treatment of variational calculus.

 

[haushofer]

I suggest a good textbook -- Analysis, Manifolds and Physics is good. It covers differential geometry over finite and infinite dimensional spaces, which are what you need for the formal treatment of variational calculus.

I followed a course on differential geometry, and have the idea that I have enough knowledge of understanding this, but that I'm overlooking something.

 

[samalkhaiat]

Hi, I have some questions which I encountered during my thesis-writing, I hope some-one can help me out on this :)

First, I have some problems interpreting coordinate-transformations ( "active and passive") and the derivation of the Equations of Motion. We have

$$S = \int L(\phi, \partial_{\mu} \phi) d^{4}x$$

and

$$\delta S = 0$$

, in which the variation of the field is arbitrary. My question is: how exactly is this variation defined? One has 2 options:

$$\delta \phi = \phi^{'}(x^{'}) - \phi (x)$$
or
$$\delta \phi = \phi^{'}(x) - \phi (x)$$

where the difference lies in the argument. In notes of Aldrovandi and Pereira ( Notes for a classical course on fields ) option 1 is choosen. And in Inverno they say that option 2 is a coordinate transformation. Why is that? I tend to choose for option two, because here you actually change the field; I would say that a scalar quantity is invariant under transformation 1, so here you just state general covariance in stead of obtaining the Equations of Motion. At the other hand, you can always choose x=x'. In the end I want to look at why one is able to commute variations and partial derivatives, so I need the exact definition of the variation of the field.

Coordinate transformations and (therefore) the transformation properties (tensorial characters) of the dynamical variables play no role in the formulation of the variation principle of stationary action:

$$\delta \int_{D} d^{4}x \mathcal{L}(x) = 0$$

We obtain the E-L equation in D by varying the dynamical variables;

$$\bar{u}_{i}(x) = u_{i}(x) + \delta u_{i}(x)$$

and assuming that the variation, $\delta u_{i}$, vanishes on $\partial D$ (the surface of the 4-volume over which the integral is taken) but arbitrary in D.

In this case, it is trivial to verify that

$$\left[ \delta , \partial \right] = 0$$

The variation principle leads to an expression of the form;

$$0 = \int_{D} d^{4}x \ \frac{\delta \mathcal{L}}{\delta u_{i}} \ \delta u_{i} + \int_{D} d^{4}x \ \partial_{a} \{ F^{ai}(u) \delta u_{i} \}$$

But

$$\int_{D} d^{4}x \ \partial_{a} \{ F^{ai}(u) \delta u_{i} \} = \int_{\partial D} d \Sigma_{a} \ F^{ai}(u) \ \delta u_{i}$$

and, since

$$\delta u_{i} = 0 \ \mbox{on} \ \partial D$$

Thus

$$\int_{D} d^{4}x \ \frac{ \delta \mathcal{L}}{\delta u_{i}} \ \delta u_{i} = 0$$

Now, since the variation, $\delta u_{i}$, are arbitrary in D, hence

$$\frac{\delta \mathcal{L}}{\delta u_{i}} \equiv \frac{\partial \mathcal{L}}{\partial u_{i}} - \partial_{a} \left( \frac{\partial \mathcal{L}}{\partial \partial_{a} u_{i}} \right) = 0$$

If you want to see how the action integral changes under some group of coordinate transformations then you do need to know how the dynamical variables transform under the given group, i.e., what representation do they carry.

Consider an arbitrary infinitesimal transformations;

$$\bar{x}^{a} = x^{a} + \delta x^{a} \ \ 1$$
$$\delta x^{a} = \omega_{A} f^{Aa}(x) \ \ 1'$$

where A is a multi-index taken value in {0,1,2,...}, i.e., A = no-index, single-index, double-index, ..., $\omega_{A}$ is the parameter of the transformations, and $f^{Aa}(x)$ is an arbitrary function.

Normally, the dynamical variables belong to some finite dimensional (matrix) representation of the group;

$$\bar{u}_{i}(\bar{x}) = R_{i}{}^{j}(\omega) u_{j}(x) \ \ 2$$

The infinitesimal version of this law is obtained by expanding the transformation matrix, near the identity, to 1st order in the parameter;

$$R_{i}{}^{j} = \delta^{j}_{i} + \omega_{A} D_{i}^{Aj} \ \ 3$$

where $D^{A}$ are set of matrices satisfying the Lie algebra of the group. They act on the space of the dynamical variables. for example, if we are dealing with Lorentz group, then $D^{A}$ is the appropriate spin matrix for $u_{i}(x)$.

From (3) and (2) we find

$$\delta^{\ast} u_{i}(x) \equiv \bar{u}_{i}(\bar{x}) - u_{i}(x) = \omega_{A} D_{i}^{Aj} u_{j}(x) \ \ 4$$

So, $\delta^{\ast} u_{i}$ describes an infinitesimal change in u brought about by the coordinate transformations. Notice that $x$ and $\bar{x}$ represent the same geometrical point, i.e., $\delta^{\ast}$ compares two functions at the same point, hence the name local variation.

Since $u_{i}$ and $\partial u_{i}$ belong to different representation matrices, then $\delta^{\ast}$ does not, in general, commute with $\partial$. Indeed you can show that

$$\left[ \partial_{a} , \delta^{\ast} \right] = \omega_{A} \partial_{a} f^{Ab} \partial_{b} \ \ 5$$

This vanishes only for the translation group ( f(x) is constant).

We can also define a non-local variation symbol, $\delta$, that compares the two functions $( u_{i}, \bar{u}_{i})$ at two different points;

$$\delta_{PM} u_{i} \equiv \bar{u}_{i}(P) - u_{i}(M)$$

If the two points have the same value of the coordinate x, then

$$\delta u_{i}(x) = \bar{u}_{i}(\bar{x})|_{\bar{x} = x} - u_{i}(x) = \bar{u}_{i}(x) - u_{i}(x)$$

represents the infinitesimal change of "shape" of the dynamical variable. It is easy to see that this functional-form-variation is related to the local variation by a "drag";

$$\delta u_{i} = - \delta x^{a} \partial_{a} u_{i} + \delta^{\ast} u_{i}$$

Indeed, for scalar, vector and tensor fields, $\delta$ is nothing but a Lie derivative.
I prefer the name total variation; notice the orbital and the spin contributions in;

$$\delta u_{i}(x) = - \omega_{A} \{ f^{Aa}(x) \partial_{a} u_{i} - D_{i}^{Aj} u_{j} \}$$

One can also show that

$$\delta u = \left[ \omega_{A} G^{A} , u \right]$$

where the G's are the infinitesimal generatores of the group in question.

For more details, see post #9 and #12 in

https://www.physicsforums.com/showthread.php?t=172461

regards

sam

 

[haushofer]

Coordinate transformations and (therefore) the transformation properties (tensorial characters) of the dynamical variables play no role in the formulation of the variation principle of stationary action:

$$\delta \int_{D} d^{4}x \mathcal{L}(x) = 0$$

We obtain the E-L equation in D by varying the dynamical variables;

$$\bar{u}_{i}(x) = u_{i}(x) + \delta u_{i}(x)$$

and assuming that the variation, $\delta u_{i}$, vanishes on $\partial D$ (the surface of the 4-volume over which the integral is taken) but arbitrary in D.

In this case, it is trivial to verify that

$$\left[ \delta , \partial \right] = 0$$

The variation principle leads to an expression of the form;

$$0 = \int_{D} d^{4}x \ \frac{\delta \mathcal{L}}{\delta u_{i}} \ \delta u_{i} + \int_{D} d^{4}x \ \partial_{a} \{ F^{ai}(u) \delta u_{i} \}$$

But

$$\int_{D} d^{4}x \ \partial_{a} \{ F^{ai}(u) \delta u_{i} \} = \int_{\partial D} d \Sigma_{a} \ F^{ai}(u) \ \delta u_{i}$$

and, since

$$\delta u_{i} = 0 \ \mbox{on} \ \partial D$$

Thus

$$\int_{D} d^{4}x \ \frac{ \delta \mathcal{L}}{\delta u_{i}} \ \delta u_{i} = 0$$

Now, since the variation, $\delta u_{i}$, are arbitrary in D, hence

$$\frac{\delta \mathcal{L}}{\delta u_{i}} \equiv \frac{\partial \mathcal{L}}{\partial u_{i}} - \partial_{a} \left( \frac{\partial \mathcal{L}}{\partial \partial_{a} u_{i}} \right) = 0$$

If you want to see how the action integral changes under some group of coordinate transformations then you do need to know how the dynamical variables transform under the given group, i.e., what representation do they carry.

Consider an arbitrary infinitesimal transformations;

$$\bar{x}^{a} = x^{a} + \delta x^{a} \ \ 1$$
$$\delta x^{a} = \omega_{A} f^{Aa}(x) \ \ 1'$$

where A is a multi-index taken value in {0,1,2,...}, i.e., A = no-index, single-index, double-index, ..., $\omega_{A}$ is the parameter of the transformations, and $f^{Aa}(x)$ is an arbitrary function.

Normally, the dynamical variables belong to some finite dimensional (matrix) representation of the group;

$$\bar{u}_{i}(\bar{x}) = R_{i}{}^{j}(\omega) u_{j}(x) \ \ 2$$

The infinitesimal version of this law is obtained by expanding the transformation matrix, near the identity, to 1st order in the parameter;

$$R_{i}{}^{j} = \delta^{j}_{i} + \omega_{A} D_{i}^{Aj} \ \ 3$$

where $D^{A}$ are set of matrices satisfying the Lie algebra of the group. They act on the space of the dynamical variables. for example, if we are dealing with Lorentz group, then $D^{A}$ is the appropriate spin matrix for $u_{i}(x)$.

From (3) and (2) we find

$$\delta^{\ast} u_{i}(x) \equiv \bar{u}_{i}(\bar{x}) - u_{i}(x) = \omega_{A} D_{i}^{Aj} u_{j}(x) \ \ 4$$

So, $\delta^{\ast} u_{i}$ describes an infinitesimal change in u brought about by the coordinate transformations. Notice that $x$ and $\bar{x}$ represent the same geometrical point, i.e., $\delta^{\ast}$ compares two functions at the same point, hence the name local variation.

Since $u_{i}$ and $\partial u_{i}$ belong to different representation matrices, then $\delta^{\ast}$ does not, in general, commute with $\partial$. Indeed you can show that

$$\left[ \partial_{a} , \delta^{\ast} \right] = \omega_{A} \partial_{a} f^{Ab} \partial_{b} \ \ 5$$

This vanishes only for the translation group ( f(x) is constant).

We can also define a non-local variation symbol, $\delta$, that compares the two functions $( u_{i}, \bar{u}_{i})$ at two different points;

$$\delta_{PM} u_{i} \equiv \bar{u}_{i}(P) - u_{i}(M)$$

If the two points have the same value of the coordinate x, then

$$\delta u_{i}(x) = \bar{u}_{i}(\bar{x})|_{\bar{x} = x} - u_{i}(x) = \bar{u}_{i}(x) - u_{i}(x)$$

represents the infinitesimal change of "shape" of the dynamical variable. It is easy to see that this functional-form-variation is related to the local variation by a "drag";

$$\delta u_{i} = - \delta x^{a} \partial_{a} u_{i} + \delta^{\ast} u_{i}$$

Indeed, for scalar, vector and tensor fields, $\delta$ is nothing but a Lie derivative.
I prefer the name total variation; notice the orbital and the spin contributions in;

$$\delta u_{i}(x) = - \omega_{A} \{ f^{Aa}(x) \partial_{a} u_{i} - D_{i}^{Aj} u_{j} \}$$

One can also show that

$$\delta u = \left[ \omega_{A} G^{A} , u \right]$$

where the G's are the infinitesimal generatores of the group in question.

For more details, see post #9 and #12 in

https://www.physicsforums.com/showthread.php?t=172461

regards

sam

A late reply, but thank you very much for your effort, I will definitely look at it ! I have the feeling that this is something which is for many people quite trivial, and then it's hard to find a good explanation for it.