Alternative Lagrange's Equations

1. Aug 2, 2010

CGR_JAMA

The Lagrangian function $$L(\mathop{q}\limits_{(x)} ,\partial _{\bullet } \mathop{q}\limits_{(x)} )$$ on a field theory's action-integral is a generally covariant scalar-density so the resulting field equations obtained by using the standard Lagrange's expresions will inevitably involve tensor-densities:

$$I\equiv k.\int _{D}L(\mathop{q}\limits_{(x)} ,\partial _{\bullet } \mathop{q}\limits_{(x)} ).d\Omega$$

$$\delta I\equiv 0$$      $$\leftrightarrow$$      $$\partial _{k} \left(\frac{\partial L}{\partial \partial _{k} \mathop{q}\limits_{(x)} } \right)-\frac{\partial L}{\partial \mathop{q}\limits_{(x)} } =0$$            (tensor densitie's equations)

The following method avoids working with them: let's assume the Lagrangian density can be represented as the product of a reference density (which can never be null) and a scalar function both depending on the dynamic variables:

$$L(\mathop{q}\limits_{(x)} ,\partial _{\bullet } \mathop{q}\limits_{(x)} )\equiv \sigma (\mathop{q}\limits_{(x)} ,\partial _{\bullet } \mathop{q}\limits_{(x)} ).\Lambda (\mathop{q}\limits_{(x)} ,\partial _{\bullet } \mathop{q}\limits_{(x)} )$$

Based on them the Lagrange's equations can be modified for directly returning tensor expressions:

$$$\partial _{k} \left(\frac{\partial \Lambda }{\partial \partial _{k} \mathop{q}\limits_{(x)} } \right)-\frac{\partial \Lambda }{\partial \mathop{q}\limits_{(x)} } +\left(\partial _{k} (\frac{\partial \psi }{\partial \partial _{k} \mathop{q}\limits_{(x)} } )-\frac{\partial \psi }{\partial \mathop{q}\limits_{(x)} } +\frac{\partial \psi }{\partial \partial _{k} \mathop{q}\limits_{(x)} } .\partial _{k} \psi \right).\Lambda +\frac{\partial \psi }{\partial \partial _{k} \mathop{q}\limits_{(x)} } .\partial _{k} \Lambda +\frac{\partial \Lambda }{\partial \partial _{k} \mathop{q}\limits_{(x)} } .\partial _{k} \psi =0$$$

/      $$\psi \equiv \ln (\sigma )$$      ,      $$\sigma \ne 0$$

On a curved Riemann space (also on a flat one when working with curvilinear coordinates) there is already a reference density based on the metric tensor:

$$\sigma (g_{\bullet \bullet } )\equiv \sqrt{-\left|g_{\bullet \bullet }$$

$$\psi \equiv \ln (\sqrt{-\left|g_{\bullet \bullet } \right|} )$$      $$\to$$      $$\frac{\partial \psi }{\partial g^{ij} } =-{\tfrac{1}{2}} .g_{ij}$$      ,      $$\partial _{i} \psi =\Gamma _{i}$$

For a metric-affine dependent scalar function the equivalent Lagrange's equations become:

$$\Lambda \equiv \Lambda (g^{\bullet \bullet } ,{\Gamma }^{\bullet } _{\bullet \bullet } ,\partial _{\bullet } {\Gamma }^{\bullet } _{\bullet \bullet })$$

$$\to$$      $$\partial _{k} \left(\frac{\partial \Lambda }{\partial \partial _{k} \mathop{q}\limits_{(x)} } \right)-\frac{\partial \Lambda }{\partial \mathop{q}\limits_{(x)} } +\Gamma _{k} .\frac{\partial \Lambda }{\partial \partial _{k} \mathop{q}\limits_{(x)} } =0$$            /            $$\mathop{q}\limits_{(x)} \ne g^{ij}$$

$$\frac{\partial \Lambda }{\partial g^{ij} } -{\tfrac{1}{2}} .g_{ij} .\Lambda =0$$                              /            $$\mathop{q}\limits_{(x)} =g^{ij}$$

This can be applied directly to General Relativity's action for obtaining the known Einstein's equations:

$$$I_{G} \equiv \int _{D}\sqrt{-\left|g_{\bullet \bullet } \right|} .g^{kr} .R_{kr}. d\Omega$$$

$$\to$$      $$\nabla _{i} g^{jk} =0$$

$$$R_{ij} -{\tfrac{1}{2}} .R.g_{ij} =0$$$

I hope this helps.

CGR_JAMA

Last edited: Aug 3, 2010
2. Aug 3, 2010

Mentz114

Thanks, that is interesting. I'll try it next time I have a difficult EL equation.

3. Aug 4, 2010

Altabeh

Hi

Please correct all the mathematical/typing errors so I can check if your method works and is more helpful than using straightly EL equations.

AB

4. Aug 4, 2010

CGR_JAMA

What may be confussing is the $$\mathop{q}\limits_{(x)}$$ which I used for representing dynamic variables. The $$(x)$$ lower-index is used for referencing many of them, nothing to do with coordinates.

5. Aug 4, 2010

Altabeh

How do you assure that this assumption can be made for a general Lagrangian? Beside this, if there is one, there must be lots of them because by definition your scalar function can be anything depending on the choice of your reference density. The mathematical and physical properties of such sub-decompositions have to be clarified. For example, linearity of a vector
field $$\textb{f}(x)$$ calls for the following two conditions:

$$k.\textb{f}(x)=\textb{f}(kx);$$
$$\textb{f}(x)+\textb{f}(y)=\textb{f}(x+y),$$

for $$k=const.$$ Now I can write $$\textb{f}(x)=a\textb{h}(x)$$ so $$\textb{h}$$ is itself a linear vector field and if it exists, we can make such decompositions countlessly. In this example, one cannot ever put a restriction on either of $$a$$ or $$\texb{h}(x),$$ but if this was some other kind of multiplicative decomposition, we'd define a restriction.

How did you manage to get the above expression? For example, the first term must appear with a $$\sigma$$.

AB

6. Aug 4, 2010

CGR_JAMA

The Lagrangian is a scalar-density and by multiplying or dividing it by any scalar function another scalar-density is obtained. There is not physical assumption taken here:

$$.\ \ \ \ \ \ L=(L / \Lambda).\Lambda=\sigma.\Lambda=\sigma.(L / \sigma )$$

Yes, you are right, the choice of density function can be anything and that allows you to choose the one you think will simplify the desired calculations. About the physical properties, this is a mathematical procedure, no physics is being considered and no physics will be extracted out of the selected choice. You will end with the same field equations no matter if you follow Lagrange's equations or this method.

Just sustitute $$L$$ by $$( \sigma.\Lambda)$$ in Lagrange's equation, apply Leibnitz rule, divide all by $$\sigma$$ and identify the $$\psi \equiv \ln (\sigma )$$ and its derivatives in the resulting expression.

By the way, if you obtain the field equations following Lagrange and work with them for obtaining tensor ones, you will be doing this same thing. This method is just a shortcut to that procedure.

7. Aug 5, 2010

Altabeh

I noticed something here. If you set $$\sigma=\sqrt{-g}$$ then by definition $$\Lambda=R=g^{ik}R_{ik}$$ which in turn means $$\frac{\partial \Lambda}{\partial g{^ik}}=R_{ik}$$; providing the Einstein field equations. This is not only a shortcut to variational method, but has this advantage that it is almost * excluding the role of density-making part of Lagrangian so that you're not going to be bored by

1- using extraordinarily cumbersome calculations regarding scalar densities such as the equivalent Lagrangian;

2- reappearance of density-weight factors when being differentiated.

Actually as you also said, reducing variational method to a simple tensorial calculation could be so helpful because of the very annoying conditions this method would generate. All you've done here is a decomposition of quantities forming a general Lagrangian based on density-scalar separation and applying this to EL equations to get both parts appearing together then deviding all terms by one to at the very least lessen the strong role of it. Nice to give it a shot.

* Yet the presence of $$\Gamma _{k}$$ does not let the equations be completely density-weight independent. However the action is maximally optimized in such a way that the appearance of the repeated density-weight factors in all terms in the ordinary variational method has really been narrowed down to only one term in the action integral.

AB

Last edited: Aug 5, 2010
8. Aug 5, 2010

Altabeh

Yes, you're right.

No, I mean when it comes to physical (dynamical) properties when finding the equations of motion it all falls upon the Lagrangian to do all of this. If decomposition were to be applied by taking into consideration such properties, then your method would be more pithy and innovative.

AB

9. Aug 5, 2010

CGR_JAMA

Thanks AB for your appreciations. Let me point out just one more thing:

In the equivalent equation only survive derivatives of the logarithm of the scalar-density. This really stands for the ration between two densities which returns a scalar function, so the Jacobian won't show up at all.

$$.\ \ \ \ \ \ \ \ \partial \psi = \partial \ln (\sigma ) =\partial \sigma / \sigma$$

Last edited: Aug 5, 2010