- #1

- 129

- 0

[tex]

\delta(\sqrt{\eta^{\mu \nu}\partial_{\mu}\phi \partial_{\nu}\phi})

[/tex]

what is the variation of this quantity under variation of \phi?

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- Thread starter arroy_0205
- Start date

In summary, the conversation discusses finding the variation of the term \phi^2 under variation of \phi, and the procedure for dealing with more complicated objects containing derivatives of \phi. The rule is to go back to the definition of the derivative and use the chain rule. In the specific example given, the result is zero due to the dependence on the derivative of \phi instead of \phi itself.

- #1

- 129

- 0

[tex]

\delta(\sqrt{\eta^{\mu \nu}\partial_{\mu}\phi \partial_{\nu}\phi})

[/tex]

what is the variation of this quantity under variation of \phi?

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- #2

- 129

- 0

Is the rule here

[tex]

\delta(\sqrt{\eta^{\mu \nu}\partial_{\mu}\phi \partial_{\nu}\phi})=\frac{\partial(\sqrt{\eta^{\mu \nu}\partial_{\mu}\phi \partial_{\nu}\phi})}{\partial \phi}\delta(\eta^{\mu \nu}\partial_{\mu}\phi \partial_{\nu}\phi)

[/tex]

But then the derivative part creates problem. The part under square root depends on derivative of \phi and not on \phi itself, so the result is zero. I am confused.

[tex]

\delta(\sqrt{\eta^{\mu \nu}\partial_{\mu}\phi \partial_{\nu}\phi})=\frac{\partial(\sqrt{\eta^{\mu \nu}\partial_{\mu}\phi \partial_{\nu}\phi})}{\partial \phi}\delta(\eta^{\mu \nu}\partial_{\mu}\phi \partial_{\nu}\phi)

[/tex]

But then the derivative part creates problem. The part under square root depends on derivative of \phi and not on \phi itself, so the result is zero. I am confused.

Last edited:

- #3

- 410

- 3

[tex]\delta F[\phi] = F[\phi + \delta \phi] - F[\phi][/tex].

In any case, in your case the chain rule applies:

[tex]

\delta(\sqrt{\eta^{\mu \nu}\partial_{\mu}\phi \partial_{\nu}\phi})=

\frac{1}{ 2\sqrt{\eta^{\mu \nu}\partial_{\mu}\phi \partial_{\nu}\phi} }

\delta(\eta^{\mu \nu}\partial_{\mu}\phi \partial_{\nu}\phi)

[/tex]

The Variational Technique for Derivatives of $\phi$ is a mathematical method used to find the first and second derivatives of a function $\phi$ with respect to its variables. It is based on the calculus of variations and involves finding the critical points of a functional, which is an integral of the function $\phi$ and its derivatives.

The Variational Technique for Derivatives of $\phi$ works by finding the critical points of a functional, which is an integral of the function $\phi$ and its derivatives. This is achieved by setting the functional's first variation to zero and solving for the variables. The resulting equations are then solved to find the first and second derivatives of $\phi$.

The Variational Technique for Derivatives of $\phi$ has many applications in physics, engineering, and other fields where optimization problems arise. It is commonly used to find solutions to equations of motion, such as the Euler-Lagrange equations, and to analyze the stability of systems.

The Variational Technique for Derivatives of $\phi$ has several advantages, including its ability to handle complex and nonlinear problems, its use of the calculus of variations, which provides a powerful tool for optimization, and its application to a wide range of systems and equations.

While the Variational Technique for Derivatives of $\phi$ is a powerful method, it does have some limitations. It is often difficult to implement and requires a good understanding of the underlying mathematics. It also may not provide solutions in all cases, as some problems may not have a well-defined functional or may be too complex to solve using this technique.

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