Do variations obey the product rule?












3












$begingroup$


I have been trying to derive the Einstein equation from the Einstein-Hilbert action
$$ S[g_{mu nu}] = frac{1}{16 pi} int_M text{d}^4x sqrt{-g}R $$
The standard derivation states that the variation $delta S =0$ when we vary the metric components. In this derivation, we use the fact that $delta (sqrt{-g}R)= delta sqrt{-g} R + sqrt{-g}delta R$. To me this seems quite obvious but I thought I would try and prove this.



My understanding is that if we have an action
$$ S[q] = int text{d}t L(q,dot{q},t)$$
we vary it as
$$ delta S[q] = int text{d}t delta L(q,dot{q},t)$$
so
$$ delta L= frac{partial L}{partial q} delta q + frac{partial L}{partial dot{q}} delta dot{q} \
= left( frac{partial L}{partial q} - frac{text{d}}{text{d}t} frac{partial L}{partial dot{q}} right) delta q + frac{text{d}}{text{d}t} left( frac{partial L}{partial dot{q}} delta q right)$$
As we integrate over $delta L$, we ignore the last term as this produces a boundary term which we can take to vanish, therefore I say



$$ delta L = left( frac{partial L}{partial q} - frac{text{d}}{text{d}t} frac{partial L}{partial dot{q}} right) delta q $$



Okay, now let's say my Lagrangian is a product: $L(q,dot{q},t) = f(q,dot{q},t)g(q,dot{q},t)$. Plugging this into the above formula for the variation, I have



$$ frac{partial L}{partial q} = frac{partial f}{partial q} g + frac{partial g }{partial q} f $$



$$ frac{partial L}{partial dot{q}} = frac{partial f}{partial dot{q}}g + frac{partial g}{partial dot{q}} f$$



$$ frac{text{d}}{text{d}t} frac{partial L}{partial dot{q}} = left( frac{text{d}}{text{d}t} frac{partial f}{partial dot{q}} right) g + left( frac{text{d}}{text{d}t}frac{partial g}{partial dot{q}} right) f + frac{partial f}{partial dot{q}} frac{text{d}g}{text{d}t} + frac{partial q}{partial dot{q}} frac{text{d}f}{text{d}t}$$



so the variation of this Lagrangian is



$$ delta L = left( frac{partial f}{partial q} - frac{text{d}}{text{d}t} frac{partial f}{partial dot{q}} right)g delta q + left( frac{partial g}{partial q} - frac{text{d}}{text{d}t} frac{partial g}{partial dot{q}} right)f delta q - frac{partial f}{partial dot{q}} frac{text{d}g}{text{d}t}delta q - frac{partial q}{partial dot{q}} frac{text{d}f}{text{d}t}delta q$$



or



$$ delta L = g delta f + f delta g - frac{partial f}{partial dot{q}} frac{text{d}g}{text{d}t}delta q - frac{partial g}{partial dot{q}} frac{text{d}f}{text{d}t}delta q $$



Now I can't seem to get rid of those horrible extra terms - I can't see how they would produce a boundary term when integrated. Maybe my understanding was incorrect and variations do not obey the product rule? Many standard resources suggest that varying the Einstein-Hilbert action obeys the product rule... is this just an exception?




My question:



How can I show that variations obey $delta (fg) = f delta g + g delta f$






calculus-of-variations euler-lagrange-equation







share|cite|improve this question











$endgroup$












  • $begingroup$
    en.wikipedia.org/wiki/Functional_derivative
    $endgroup$
    – caverac
    Dec 27 '18 at 23:16
















3












$begingroup$


I have been trying to derive the Einstein equation from the Einstein-Hilbert action
$$ S[g_{mu nu}] = frac{1}{16 pi} int_M text{d}^4x sqrt{-g}R $$
The standard derivation states that the variation $delta S =0$ when we vary the metric components. In this derivation, we use the fact that $delta (sqrt{-g}R)= delta sqrt{-g} R + sqrt{-g}delta R$. To me this seems quite obvious but I thought I would try and prove this.



My understanding is that if we have an action
$$ S[q] = int text{d}t L(q,dot{q},t)$$
we vary it as
$$ delta S[q] = int text{d}t delta L(q,dot{q},t)$$
so
$$ delta L= frac{partial L}{partial q} delta q + frac{partial L}{partial dot{q}} delta dot{q} \
= left( frac{partial L}{partial q} - frac{text{d}}{text{d}t} frac{partial L}{partial dot{q}} right) delta q + frac{text{d}}{text{d}t} left( frac{partial L}{partial dot{q}} delta q right)$$
As we integrate over $delta L$, we ignore the last term as this produces a boundary term which we can take to vanish, therefore I say



$$ delta L = left( frac{partial L}{partial q} - frac{text{d}}{text{d}t} frac{partial L}{partial dot{q}} right) delta q $$



Okay, now let's say my Lagrangian is a product: $L(q,dot{q},t) = f(q,dot{q},t)g(q,dot{q},t)$. Plugging this into the above formula for the variation, I have



$$ frac{partial L}{partial q} = frac{partial f}{partial q} g + frac{partial g }{partial q} f $$



$$ frac{partial L}{partial dot{q}} = frac{partial f}{partial dot{q}}g + frac{partial g}{partial dot{q}} f$$



$$ frac{text{d}}{text{d}t} frac{partial L}{partial dot{q}} = left( frac{text{d}}{text{d}t} frac{partial f}{partial dot{q}} right) g + left( frac{text{d}}{text{d}t}frac{partial g}{partial dot{q}} right) f + frac{partial f}{partial dot{q}} frac{text{d}g}{text{d}t} + frac{partial q}{partial dot{q}} frac{text{d}f}{text{d}t}$$



so the variation of this Lagrangian is



$$ delta L = left( frac{partial f}{partial q} - frac{text{d}}{text{d}t} frac{partial f}{partial dot{q}} right)g delta q + left( frac{partial g}{partial q} - frac{text{d}}{text{d}t} frac{partial g}{partial dot{q}} right)f delta q - frac{partial f}{partial dot{q}} frac{text{d}g}{text{d}t}delta q - frac{partial q}{partial dot{q}} frac{text{d}f}{text{d}t}delta q$$



or



$$ delta L = g delta f + f delta g - frac{partial f}{partial dot{q}} frac{text{d}g}{text{d}t}delta q - frac{partial g}{partial dot{q}} frac{text{d}f}{text{d}t}delta q $$



Now I can't seem to get rid of those horrible extra terms - I can't see how they would produce a boundary term when integrated. Maybe my understanding was incorrect and variations do not obey the product rule? Many standard resources suggest that varying the Einstein-Hilbert action obeys the product rule... is this just an exception?




My question:



How can I show that variations obey $delta (fg) = f delta g + g delta f$






calculus-of-variations euler-lagrange-equation







share|cite|improve this question











$endgroup$












  • $begingroup$
    en.wikipedia.org/wiki/Functional_derivative
    $endgroup$
    – caverac
    Dec 27 '18 at 23:16














3












3








3





$begingroup$


I have been trying to derive the Einstein equation from the Einstein-Hilbert action
$$ S[g_{mu nu}] = frac{1}{16 pi} int_M text{d}^4x sqrt{-g}R $$
The standard derivation states that the variation $delta S =0$ when we vary the metric components. In this derivation, we use the fact that $delta (sqrt{-g}R)= delta sqrt{-g} R + sqrt{-g}delta R$. To me this seems quite obvious but I thought I would try and prove this.



My understanding is that if we have an action
$$ S[q] = int text{d}t L(q,dot{q},t)$$
we vary it as
$$ delta S[q] = int text{d}t delta L(q,dot{q},t)$$
so
$$ delta L= frac{partial L}{partial q} delta q + frac{partial L}{partial dot{q}} delta dot{q} \
= left( frac{partial L}{partial q} - frac{text{d}}{text{d}t} frac{partial L}{partial dot{q}} right) delta q + frac{text{d}}{text{d}t} left( frac{partial L}{partial dot{q}} delta q right)$$
As we integrate over $delta L$, we ignore the last term as this produces a boundary term which we can take to vanish, therefore I say



$$ delta L = left( frac{partial L}{partial q} - frac{text{d}}{text{d}t} frac{partial L}{partial dot{q}} right) delta q $$



Okay, now let's say my Lagrangian is a product: $L(q,dot{q},t) = f(q,dot{q},t)g(q,dot{q},t)$. Plugging this into the above formula for the variation, I have



$$ frac{partial L}{partial q} = frac{partial f}{partial q} g + frac{partial g }{partial q} f $$



$$ frac{partial L}{partial dot{q}} = frac{partial f}{partial dot{q}}g + frac{partial g}{partial dot{q}} f$$



$$ frac{text{d}}{text{d}t} frac{partial L}{partial dot{q}} = left( frac{text{d}}{text{d}t} frac{partial f}{partial dot{q}} right) g + left( frac{text{d}}{text{d}t}frac{partial g}{partial dot{q}} right) f + frac{partial f}{partial dot{q}} frac{text{d}g}{text{d}t} + frac{partial q}{partial dot{q}} frac{text{d}f}{text{d}t}$$



so the variation of this Lagrangian is



$$ delta L = left( frac{partial f}{partial q} - frac{text{d}}{text{d}t} frac{partial f}{partial dot{q}} right)g delta q + left( frac{partial g}{partial q} - frac{text{d}}{text{d}t} frac{partial g}{partial dot{q}} right)f delta q - frac{partial f}{partial dot{q}} frac{text{d}g}{text{d}t}delta q - frac{partial q}{partial dot{q}} frac{text{d}f}{text{d}t}delta q$$



or



$$ delta L = g delta f + f delta g - frac{partial f}{partial dot{q}} frac{text{d}g}{text{d}t}delta q - frac{partial g}{partial dot{q}} frac{text{d}f}{text{d}t}delta q $$



Now I can't seem to get rid of those horrible extra terms - I can't see how they would produce a boundary term when integrated. Maybe my understanding was incorrect and variations do not obey the product rule? Many standard resources suggest that varying the Einstein-Hilbert action obeys the product rule... is this just an exception?




My question:



How can I show that variations obey $delta (fg) = f delta g + g delta f$






calculus-of-variations euler-lagrange-equation







share|cite|improve this question











$endgroup$




I have been trying to derive the Einstein equation from the Einstein-Hilbert action
$$ S[g_{mu nu}] = frac{1}{16 pi} int_M text{d}^4x sqrt{-g}R $$
The standard derivation states that the variation $delta S =0$ when we vary the metric components. In this derivation, we use the fact that $delta (sqrt{-g}R)= delta sqrt{-g} R + sqrt{-g}delta R$. To me this seems quite obvious but I thought I would try and prove this.



My understanding is that if we have an action
$$ S[q] = int text{d}t L(q,dot{q},t)$$
we vary it as
$$ delta S[q] = int text{d}t delta L(q,dot{q},t)$$
so
$$ delta L= frac{partial L}{partial q} delta q + frac{partial L}{partial dot{q}} delta dot{q} \
= left( frac{partial L}{partial q} - frac{text{d}}{text{d}t} frac{partial L}{partial dot{q}} right) delta q + frac{text{d}}{text{d}t} left( frac{partial L}{partial dot{q}} delta q right)$$
As we integrate over $delta L$, we ignore the last term as this produces a boundary term which we can take to vanish, therefore I say



$$ delta L = left( frac{partial L}{partial q} - frac{text{d}}{text{d}t} frac{partial L}{partial dot{q}} right) delta q $$



Okay, now let's say my Lagrangian is a product: $L(q,dot{q},t) = f(q,dot{q},t)g(q,dot{q},t)$. Plugging this into the above formula for the variation, I have



$$ frac{partial L}{partial q} = frac{partial f}{partial q} g + frac{partial g }{partial q} f $$



$$ frac{partial L}{partial dot{q}} = frac{partial f}{partial dot{q}}g + frac{partial g}{partial dot{q}} f$$



$$ frac{text{d}}{text{d}t} frac{partial L}{partial dot{q}} = left( frac{text{d}}{text{d}t} frac{partial f}{partial dot{q}} right) g + left( frac{text{d}}{text{d}t}frac{partial g}{partial dot{q}} right) f + frac{partial f}{partial dot{q}} frac{text{d}g}{text{d}t} + frac{partial q}{partial dot{q}} frac{text{d}f}{text{d}t}$$



so the variation of this Lagrangian is



$$ delta L = left( frac{partial f}{partial q} - frac{text{d}}{text{d}t} frac{partial f}{partial dot{q}} right)g delta q + left( frac{partial g}{partial q} - frac{text{d}}{text{d}t} frac{partial g}{partial dot{q}} right)f delta q - frac{partial f}{partial dot{q}} frac{text{d}g}{text{d}t}delta q - frac{partial q}{partial dot{q}} frac{text{d}f}{text{d}t}delta q$$



or



$$ delta L = g delta f + f delta g - frac{partial f}{partial dot{q}} frac{text{d}g}{text{d}t}delta q - frac{partial g}{partial dot{q}} frac{text{d}f}{text{d}t}delta q $$



Now I can't seem to get rid of those horrible extra terms - I can't see how they would produce a boundary term when integrated. Maybe my understanding was incorrect and variations do not obey the product rule? Many standard resources suggest that varying the Einstein-Hilbert action obeys the product rule... is this just an exception?




My question:



How can I show that variations obey $delta (fg) = f delta g + g delta f$






calculus-of-variations euler-lagrange-equation




calculus-of-variations euler-lagrange-equation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 27 '18 at 23:47







Matt0410

















asked Dec 27 '18 at 23:12









Matt0410Matt0410

29319




29319












  • $begingroup$
    en.wikipedia.org/wiki/Functional_derivative
    $endgroup$
    – caverac
    Dec 27 '18 at 23:16


















  • $begingroup$
    en.wikipedia.org/wiki/Functional_derivative
    $endgroup$
    – caverac
    Dec 27 '18 at 23:16
















$begingroup$
en.wikipedia.org/wiki/Functional_derivative
$endgroup$
– caverac
Dec 27 '18 at 23:16




$begingroup$
en.wikipedia.org/wiki/Functional_derivative
$endgroup$
– caverac
Dec 27 '18 at 23:16










1 Answer
1






active

oldest

votes


















4












$begingroup$

It is true that
$$ delta L = f delta g + g delta f,$$
but the expressions for variations are:
begin{align} delta f&=frac{partial f}{partial q} delta q+ frac{partial f}{partial dot q}delta dot q = left[ frac{partial f}{partial q} - frac{d}{dt}left(frac{partial f}{partial dot q} right) right] delta q + frac{d}{dt}left( frac{partial f}{partial dot q} delta qright)\ delta g&=frac{partial g}{partial q} delta q+ frac{partial g}{partial dot q}delta dot q = left[ frac{partial g}{partial q} - frac{d}{dt}left(frac{partial g}{partial dot q} right) right] delta q + frac{d}{dt}left( frac{partial g}{partial dot q} delta qright)\delta L&=frac{partial L}{partial q} delta q+ frac{partial L}{partial dot q}delta dot q = left[ frac{partial f}{partial q}g+frac{partial g}{partial q}f - frac{d}{dt}left(frac{partial f}{partial dot q}g + frac{partial g}{partial dot q}f right) right] delta q + frac{d}{dt}left( frac{partial f}{partial dot q}g delta q + frac{partial g}{partial dot q}f delta qright)end{align}
In each case, the variation is equal to the Euler-Lagrange equation times $delta q$, plus a total derivative. The total derivatives are important!






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The OP wants to know why $delta(fg) = fdelta g + gdelta f$. You just cannot start with "it is true that $delta(fg) = fdelta g + gdelta f$"
    $endgroup$
    – caverac
    Dec 27 '18 at 23:59










  • $begingroup$
    @caverac I think the OP wants to verify that $delta(fg) = fdelta g + g delta f$ holds in the example given. And this is manifestly true from the expressions for $delta f$, $delta g$ and $delta L$ that I wrote down.
    $endgroup$
    – Kenny Wong
    Dec 28 '18 at 0:01










  • $begingroup$
    @KennyWong Yes this is what I was looking for, thank you.
    $endgroup$
    – Matt0410
    Dec 28 '18 at 0:12











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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

It is true that
$$ delta L = f delta g + g delta f,$$
but the expressions for variations are:
begin{align} delta f&=frac{partial f}{partial q} delta q+ frac{partial f}{partial dot q}delta dot q = left[ frac{partial f}{partial q} - frac{d}{dt}left(frac{partial f}{partial dot q} right) right] delta q + frac{d}{dt}left( frac{partial f}{partial dot q} delta qright)\ delta g&=frac{partial g}{partial q} delta q+ frac{partial g}{partial dot q}delta dot q = left[ frac{partial g}{partial q} - frac{d}{dt}left(frac{partial g}{partial dot q} right) right] delta q + frac{d}{dt}left( frac{partial g}{partial dot q} delta qright)\delta L&=frac{partial L}{partial q} delta q+ frac{partial L}{partial dot q}delta dot q = left[ frac{partial f}{partial q}g+frac{partial g}{partial q}f - frac{d}{dt}left(frac{partial f}{partial dot q}g + frac{partial g}{partial dot q}f right) right] delta q + frac{d}{dt}left( frac{partial f}{partial dot q}g delta q + frac{partial g}{partial dot q}f delta qright)end{align}
In each case, the variation is equal to the Euler-Lagrange equation times $delta q$, plus a total derivative. The total derivatives are important!






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The OP wants to know why $delta(fg) = fdelta g + gdelta f$. You just cannot start with "it is true that $delta(fg) = fdelta g + gdelta f$"
    $endgroup$
    – caverac
    Dec 27 '18 at 23:59










  • $begingroup$
    @caverac I think the OP wants to verify that $delta(fg) = fdelta g + g delta f$ holds in the example given. And this is manifestly true from the expressions for $delta f$, $delta g$ and $delta L$ that I wrote down.
    $endgroup$
    – Kenny Wong
    Dec 28 '18 at 0:01










  • $begingroup$
    @KennyWong Yes this is what I was looking for, thank you.
    $endgroup$
    – Matt0410
    Dec 28 '18 at 0:12
















4












$begingroup$

It is true that
$$ delta L = f delta g + g delta f,$$
but the expressions for variations are:
begin{align} delta f&=frac{partial f}{partial q} delta q+ frac{partial f}{partial dot q}delta dot q = left[ frac{partial f}{partial q} - frac{d}{dt}left(frac{partial f}{partial dot q} right) right] delta q + frac{d}{dt}left( frac{partial f}{partial dot q} delta qright)\ delta g&=frac{partial g}{partial q} delta q+ frac{partial g}{partial dot q}delta dot q = left[ frac{partial g}{partial q} - frac{d}{dt}left(frac{partial g}{partial dot q} right) right] delta q + frac{d}{dt}left( frac{partial g}{partial dot q} delta qright)\delta L&=frac{partial L}{partial q} delta q+ frac{partial L}{partial dot q}delta dot q = left[ frac{partial f}{partial q}g+frac{partial g}{partial q}f - frac{d}{dt}left(frac{partial f}{partial dot q}g + frac{partial g}{partial dot q}f right) right] delta q + frac{d}{dt}left( frac{partial f}{partial dot q}g delta q + frac{partial g}{partial dot q}f delta qright)end{align}
In each case, the variation is equal to the Euler-Lagrange equation times $delta q$, plus a total derivative. The total derivatives are important!






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The OP wants to know why $delta(fg) = fdelta g + gdelta f$. You just cannot start with "it is true that $delta(fg) = fdelta g + gdelta f$"
    $endgroup$
    – caverac
    Dec 27 '18 at 23:59










  • $begingroup$
    @caverac I think the OP wants to verify that $delta(fg) = fdelta g + g delta f$ holds in the example given. And this is manifestly true from the expressions for $delta f$, $delta g$ and $delta L$ that I wrote down.
    $endgroup$
    – Kenny Wong
    Dec 28 '18 at 0:01










  • $begingroup$
    @KennyWong Yes this is what I was looking for, thank you.
    $endgroup$
    – Matt0410
    Dec 28 '18 at 0:12














4












4








4





$begingroup$

It is true that
$$ delta L = f delta g + g delta f,$$
but the expressions for variations are:
begin{align} delta f&=frac{partial f}{partial q} delta q+ frac{partial f}{partial dot q}delta dot q = left[ frac{partial f}{partial q} - frac{d}{dt}left(frac{partial f}{partial dot q} right) right] delta q + frac{d}{dt}left( frac{partial f}{partial dot q} delta qright)\ delta g&=frac{partial g}{partial q} delta q+ frac{partial g}{partial dot q}delta dot q = left[ frac{partial g}{partial q} - frac{d}{dt}left(frac{partial g}{partial dot q} right) right] delta q + frac{d}{dt}left( frac{partial g}{partial dot q} delta qright)\delta L&=frac{partial L}{partial q} delta q+ frac{partial L}{partial dot q}delta dot q = left[ frac{partial f}{partial q}g+frac{partial g}{partial q}f - frac{d}{dt}left(frac{partial f}{partial dot q}g + frac{partial g}{partial dot q}f right) right] delta q + frac{d}{dt}left( frac{partial f}{partial dot q}g delta q + frac{partial g}{partial dot q}f delta qright)end{align}
In each case, the variation is equal to the Euler-Lagrange equation times $delta q$, plus a total derivative. The total derivatives are important!






share|cite|improve this answer









$endgroup$



It is true that
$$ delta L = f delta g + g delta f,$$
but the expressions for variations are:
begin{align} delta f&=frac{partial f}{partial q} delta q+ frac{partial f}{partial dot q}delta dot q = left[ frac{partial f}{partial q} - frac{d}{dt}left(frac{partial f}{partial dot q} right) right] delta q + frac{d}{dt}left( frac{partial f}{partial dot q} delta qright)\ delta g&=frac{partial g}{partial q} delta q+ frac{partial g}{partial dot q}delta dot q = left[ frac{partial g}{partial q} - frac{d}{dt}left(frac{partial g}{partial dot q} right) right] delta q + frac{d}{dt}left( frac{partial g}{partial dot q} delta qright)\delta L&=frac{partial L}{partial q} delta q+ frac{partial L}{partial dot q}delta dot q = left[ frac{partial f}{partial q}g+frac{partial g}{partial q}f - frac{d}{dt}left(frac{partial f}{partial dot q}g + frac{partial g}{partial dot q}f right) right] delta q + frac{d}{dt}left( frac{partial f}{partial dot q}g delta q + frac{partial g}{partial dot q}f delta qright)end{align}
In each case, the variation is equal to the Euler-Lagrange equation times $delta q$, plus a total derivative. The total derivatives are important!







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 27 '18 at 23:48









Kenny WongKenny Wong

19k21440




19k21440












  • $begingroup$
    The OP wants to know why $delta(fg) = fdelta g + gdelta f$. You just cannot start with "it is true that $delta(fg) = fdelta g + gdelta f$"
    $endgroup$
    – caverac
    Dec 27 '18 at 23:59










  • $begingroup$
    @caverac I think the OP wants to verify that $delta(fg) = fdelta g + g delta f$ holds in the example given. And this is manifestly true from the expressions for $delta f$, $delta g$ and $delta L$ that I wrote down.
    $endgroup$
    – Kenny Wong
    Dec 28 '18 at 0:01










  • $begingroup$
    @KennyWong Yes this is what I was looking for, thank you.
    $endgroup$
    – Matt0410
    Dec 28 '18 at 0:12


















  • $begingroup$
    The OP wants to know why $delta(fg) = fdelta g + gdelta f$. You just cannot start with "it is true that $delta(fg) = fdelta g + gdelta f$"
    $endgroup$
    – caverac
    Dec 27 '18 at 23:59










  • $begingroup$
    @caverac I think the OP wants to verify that $delta(fg) = fdelta g + g delta f$ holds in the example given. And this is manifestly true from the expressions for $delta f$, $delta g$ and $delta L$ that I wrote down.
    $endgroup$
    – Kenny Wong
    Dec 28 '18 at 0:01










  • $begingroup$
    @KennyWong Yes this is what I was looking for, thank you.
    $endgroup$
    – Matt0410
    Dec 28 '18 at 0:12
















$begingroup$
The OP wants to know why $delta(fg) = fdelta g + gdelta f$. You just cannot start with "it is true that $delta(fg) = fdelta g + gdelta f$"
$endgroup$
– caverac
Dec 27 '18 at 23:59




$begingroup$
The OP wants to know why $delta(fg) = fdelta g + gdelta f$. You just cannot start with "it is true that $delta(fg) = fdelta g + gdelta f$"
$endgroup$
– caverac
Dec 27 '18 at 23:59












$begingroup$
@caverac I think the OP wants to verify that $delta(fg) = fdelta g + g delta f$ holds in the example given. And this is manifestly true from the expressions for $delta f$, $delta g$ and $delta L$ that I wrote down.
$endgroup$
– Kenny Wong
Dec 28 '18 at 0:01




$begingroup$
@caverac I think the OP wants to verify that $delta(fg) = fdelta g + g delta f$ holds in the example given. And this is manifestly true from the expressions for $delta f$, $delta g$ and $delta L$ that I wrote down.
$endgroup$
– Kenny Wong
Dec 28 '18 at 0:01












$begingroup$
@KennyWong Yes this is what I was looking for, thank you.
$endgroup$
– Matt0410
Dec 28 '18 at 0:12




$begingroup$
@KennyWong Yes this is what I was looking for, thank you.
$endgroup$
– Matt0410
Dec 28 '18 at 0:12


















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