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Qmechanic
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TL;DR: Generically$^1$ an action principle gets destroyed if we apply EOMs in the action.

Examples:

  1. This is particularly clear if we try to vary wrt. a dynamical variable that no longer appears in the action after substituting an EOM.

  2. The 1D static model $$V(q)~=~\frac{k}{2}q^2+{\cal O}(q^3), \qquad k~\neq ~0,$$ has a trivial stationary point $q\approx 0$. (We ignore here possible non-trivial stationary points for simplicity.) We can replace the potential $V$ with a new potential $$\tilde{V}(q)~=~a+bq+\frac{c}{2}q^2+{\cal O}(q^3), \qquad \qquad c~\neq ~0,\qquad b~=~0,$$ without changing the trivial stationary point $q\approx 0$. Note that it is crucial that $b=0$, i.e. it only works for a zero-measure set.

  3. For the 2D kinetic term $L=T = \frac{m}{2}\left(\dot{r}^2+r^2\dot{\theta}^2\right)$ in polar coordinates, if we substitute the angular variable $\theta$ with its EOM, the remaining Lagrangian for the radial variable $r$ gets a wrong sign in one of its terms! See e.g. this & this Phys.SE posts for an explanation.

  4. Specifically, we can not derive EFE from OP's action (2).

--

$^1$ The word generically means here generally modulo a zero-measure set of exceptions.

TL;DR: Generically$^1$ an action principle gets destroyed if we apply EOMs in the action.

Examples:

  1. This is particularly clear if we try to vary wrt. a dynamical variable that no longer appears in the action after substituting an EOM.

  2. The 1D static model $$V(q)~=~\frac{k}{2}q^2+{\cal O}(q^3), \qquad k~\neq ~0,$$ has a trivial stationary point $q\approx 0$. (We ignore here possible non-trivial stationary points for simplicity.) We can replace the potential $V$ with a new potential $$\tilde{V}(q)~=~a+bq+\frac{c}{2}q^2+{\cal O}(q^3), \qquad \qquad c~\neq ~0,\qquad b~=~0,$$ without changing the trivial stationary point $q\approx 0$. Note that it is crucial that $b=0$, i.e. it only works for a zero-measure set.

  3. For the 2D kinetic term $L=T = \frac{m}{2}\left(\dot{r}^2+r^2\dot{\theta}^2\right)$ in polar coordinates, if we substitute the angular variable $\theta$ with its EOM, the remaining Lagrangian for radial variable $r$ gets a wrong sign in one of its terms! See e.g. this & this Phys.SE posts for an explanation.

  4. Specifically, we can not derive EFE from OP's action (2).

--

$^1$ The word generically means here generally modulo a zero-measure set of exceptions.

TL;DR: Generically$^1$ an action principle gets destroyed if we apply EOMs in the action.

Examples:

  1. This is particularly clear if we try to vary wrt. a dynamical variable that no longer appears in the action after substituting an EOM.

  2. The 1D static model $$V(q)~=~\frac{k}{2}q^2+{\cal O}(q^3), \qquad k~\neq ~0,$$ has a trivial stationary point $q\approx 0$. (We ignore here possible non-trivial stationary points for simplicity.) We can replace the potential $V$ with a new potential $$\tilde{V}(q)~=~a+bq+\frac{c}{2}q^2+{\cal O}(q^3), \qquad \qquad c~\neq ~0,\qquad b~=~0,$$ without changing the trivial stationary point $q\approx 0$. Note that it is crucial that $b=0$, i.e. it only works for a zero-measure set.

  3. For the 2D kinetic term $L=T = \frac{m}{2}\left(\dot{r}^2+r^2\dot{\theta}^2\right)$ in polar coordinates, if we substitute the angular variable $\theta$ with its EOM, the remaining Lagrangian for the radial variable $r$ gets a wrong sign in one of its terms! See e.g. this & this Phys.SE posts for an explanation.

  4. Specifically, we can not derive EFE from OP's action (2).

--

$^1$ The word generically means here generally modulo a zero-measure set of exceptions.

Added explanation
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Qmechanic
  • 206.6k
  • 48
  • 566
  • 2.3k

TL;DR: Generically,$^1$ an action principle gets destroyed if we apply EOMs in the action. This is particularly clear if a dynamical variable no longer appears in the action after substituting an EOM. Specifically, we can not derive EFE from OP's action

Examples:

  1. This is particularly clear if we try to vary wrt. a dynamical variable that no longer appears in the action after substituting an EOM.

  2. The 1D static model $$V(q)~=~\frac{k}{2}q^2+{\cal O}(q^3), \qquad k~\neq ~0,$$ has a trivial stationary point $q\approx 0$. (We ignore here possible non-trivial stationary points for simplicity.) We can replace the potential $V$ with a new potential $$\tilde{V}(q)~=~a+bq+\frac{c}{2}q^2+{\cal O}(q^3), \qquad \qquad c~\neq ~0,\qquad b~=~0,$$ without changing the trivial stationary point $q\approx 0$. Note that it is crucial that $b=0$, i.e. it only works for a zero-measure set.

  3. For the 2D kinetic term $L=T = \frac{m}{2}\left(\dot{r}^2+r^2\dot{\theta}^2\right)$ in polar coordinates, if we substitute the angular variable $\theta$ with its EOM, the remaining Lagrangian for radial variable $r$ gets a wrong sign in one of its terms! See e.g. this & this Phys.SE posts for an explanation.

  4. Specifically, we can not derive EFE from OP's action (2).

--

$^1$ The word (2).generically means here generally modulo a zero-measure set of exceptions.

Generically, an action principle gets destroyed if we apply EOMs in the action. This is particularly clear if a dynamical variable no longer appears in the action after substituting an EOM. Specifically, we can not derive EFE from OP's action (2).

TL;DR: Generically$^1$ an action principle gets destroyed if we apply EOMs in the action.

Examples:

  1. This is particularly clear if we try to vary wrt. a dynamical variable that no longer appears in the action after substituting an EOM.

  2. The 1D static model $$V(q)~=~\frac{k}{2}q^2+{\cal O}(q^3), \qquad k~\neq ~0,$$ has a trivial stationary point $q\approx 0$. (We ignore here possible non-trivial stationary points for simplicity.) We can replace the potential $V$ with a new potential $$\tilde{V}(q)~=~a+bq+\frac{c}{2}q^2+{\cal O}(q^3), \qquad \qquad c~\neq ~0,\qquad b~=~0,$$ without changing the trivial stationary point $q\approx 0$. Note that it is crucial that $b=0$, i.e. it only works for a zero-measure set.

  3. For the 2D kinetic term $L=T = \frac{m}{2}\left(\dot{r}^2+r^2\dot{\theta}^2\right)$ in polar coordinates, if we substitute the angular variable $\theta$ with its EOM, the remaining Lagrangian for radial variable $r$ gets a wrong sign in one of its terms! See e.g. this & this Phys.SE posts for an explanation.

  4. Specifically, we can not derive EFE from OP's action (2).

--

$^1$ The word generically means here generally modulo a zero-measure set of exceptions.

Added explanation
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Qmechanic
  • 206.6k
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  • 566
  • 2.3k

Generically, an action principle gets destroyed if we apply EOMs in the action. This is particularly clear if a dynamical variable no longer appears in the action after substituting an EOM. Specifically, we can no longernot derive EFE from OP's action (2).

Generically, an action principle gets destroyed if we apply EOMs in the action. Specifically, we can no longer derive EFE from OP's action (2).

Generically, an action principle gets destroyed if we apply EOMs in the action. This is particularly clear if a dynamical variable no longer appears in the action after substituting an EOM. Specifically, we can not derive EFE from OP's action (2).

Source Link
Qmechanic
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