Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

In physics total energy is extremal for the static solutions of equation of motions.

Can anyone explain this sentence to me?

share|improve this question
1  
Can you provide a reference so we can get some context also. –  ja72 Apr 22 '13 at 14:25

1 Answer 1

I) Here is at least a partial answer. Assume the following set-up. Let there be given a classical Lagrangian field theory in $d+1$ spacetime dimensions, with dynamical field variables $\phi^{\alpha}(x,t)$, and with no explicit time dependence.

Action $S[\phi]:=\int \! dt~ L[\phi(t,\cdot)]$.

Lagrangian functional $L:=T-V$.

Energy functional $E=T+V$.

Kinetic energy functional $T[\phi(t,\cdot)] :=\int \! d^dx ~{\cal T}(t,x)$.

Potential energy functional $V[\phi(t,\cdot)] :=\int \!d^dx~ {\cal V}(t,x)$ is a local functional$^1$ of a field $\phi(t,\cdot)$.

Kinetic energy density ${\cal T}(t,x)$ is a polynomial of $\dot{\phi}(t,x)$.

Potential energy density ${\cal V}(t,x)$ is a function of $\phi(t,x)$ and its spatial derivatives.

II) Then the Euler-Lagrange equation

$$\tag{1} \frac{\delta S}{\delta\phi^{\alpha}(t,x)} ~=~0 $$

becomes

$$\tag{2} \frac{d}{dt}\frac{\partial {\cal T}(t,x)}{\partial \dot{\phi}^{\alpha}(t,x)}~=~ -\frac{\delta V[\phi(t,\cdot)]}{\delta \phi^{\alpha}(t,x)}.$$

For a static (=time-independent) solution $\phi(t,x)$, the lhs. of (2) vanishes, i.e. a static solution is a stationary point

$$\tag{3} \frac{\delta V[\phi(t,\cdot)]}{\delta \phi^{\alpha}(t,x)}~=~0 $$

for the potential energy functional. This is our main statement. See also Ref. 1.

III) On the other hand, for a fixed time $t$, we offer the following derivation. We may consider the energy functional

$$\tag{4} E[\phi(t,\cdot),\dot{\phi}(t,\cdot)] ~=~T[\dot{\phi}(t,\cdot)]+V[\phi(t,\cdot)],$$

where $\phi$ and $\dot{\phi}$ are treated as independent fields. Note that the fixed time variable $t$ does effectively not enter this problem of section III, so in that loose sense we may, with some abuse of language, identify the field configurations with "static" field configurations. The Euler-Lagrange equation

$$\tag{5} \frac{\delta E[\phi(t,\cdot),\dot{\phi}(t,\cdot)]}{\delta \phi^{\alpha}(t,x)}~=~0~=~\frac{\delta E[\phi(t,\cdot),\dot{\phi}(t,\cdot)]}{\delta \dot{\phi}^{\alpha}(t,x)} $$

becomes

$$\tag{6} \frac{\delta V[\phi(t,\cdot)]}{\delta \phi^{\alpha}(t,x)}~=~0~=~\frac{d}{dt}\frac{\partial {\cal T}(t,x)}{\partial \dot{\phi}^{\alpha}(t,x)}.$$

If we furthermore assume that the kinetic energy functional $T[\dot{\phi}(t,\cdot)]$ is a positive definite quadratic form in $\dot{\phi}$, then a global minimum for the energy functional (4) is given by zero velocity field

$\tag{7} \dot{\phi}~=~0 \qquad \Leftrightarrow \qquad T~=~0,$

and a global minimum for the potential functional $V[\phi(t,\cdot)]$, cf. eq. (3).

References:

  1. R. Rajaraman Solitons and Instantons: An Introduction to Solitons and Instantons in Quantum Field Theory, (1987) p. 133-134.

--

$^1$ For the notion of a local functional, see also this Phys.SE answer.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.