Suppose my action integral is $$S=\int d^4x(\nabla \times A)^2$$ and $\delta S$ gives $$\delta S =\int d^4x [2(\nabla \times A).(\nabla \times \delta A)]$$ I would like to calculate the coefficient of $\delta A$ from this action integral. But I am stuck. How can I separate the $\delta A$ from the term like this?
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- $\begingroup$ It might be useful to use index notation $(\nabla\times A)_i = \epsilon_{ijk}\partial_j A_k$ and integration by parts. $\endgroup$Heidar– Heidar2012-09-01 14:10:16 +00:00Commented Sep 1, 2012 at 14:10
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4 Let's do what Heidar says and write it with indices, and identify the Lagrangian. $$ L=\frac{1}{2}(\vec{\nabla}\times \vec{A})^2 = \frac{1}{2}\epsilon_{ijk}\partial_j A_k \epsilon_{ilm}\partial_l A_m $$ where, if you haven't heard of it yet, you pretend there is a summation symbol for each repeated index. Then since there are no bare $A_i$ sitting by themselves, only $\partial_i A_j$s the only part of Lagrange's equations that will contribute are $$ \partial_q \frac{\partial L}{\partial (\partial_q A_p)} $$ which we set equal to zero following the equations. Then $$ \frac{\partial L}{\partial (\partial_q A_p)}=\frac{1}{2}(\epsilon_{ijk}\delta_{jq}\delta_{kp}\epsilon_{ilm}\partial_l A_m+\epsilon_{ijk}\partial_j A_k \epsilon_{ilm}\delta_{lq}\delta_{mp}) $$ using $$ \frac{\partial (\partial_i A_j)}{\partial (\partial_q A_p)}=\delta_{iq}\delta_{jp}. $$ Then we have $$ \partial_q \frac{\partial L}{\partial (\partial_q A_p)}=\partial_q(\epsilon_{iqp}\epsilon_{ijk}\partial_i A_j) = \partial_q ((\delta_{qj}\delta_{pk}-\delta_{qk}\delta_{pj})\partial_{i} A_j)=\partial_q (\partial_q A_p - \partial_p A_q)=0 $$ where i used the contracted epsilon identity and changed the repeated indices as i needed them in order to combine terms. Hope this helps.
EDIT:
Well, I'll still try and help out, hopefully I don't make anything any worse. $$ \delta S = \int d^3 x \frac{1}{2}\epsilon_{ijk}\epsilon_{ilm}\delta(\partial_j A_k)\partial_l A_m+\int d^3 x\frac{1}{2}\epsilon_{ijk}\epsilon_{ilm}\partial_j A_k \delta(\partial_l A_m) $$ Now with the variations $\delta$ we can interchange the order of $\partial$ and $\delta$ $$ \delta(\partial_i A_j)=\partial_i (\delta A_k) $$ So with the two terms multiplied above we get $$ \partial_j( \delta A_k)\partial_l A_m=\partial_j (\delta A_k \partial_l A_m)-\delta A_k \partial_j \partial_l A_m $$ from the product rule. This helps isolate the variation of the field. Please (everyone) let me know if this is still confusing and/or wrong. Hope this helps.
- $\begingroup$ Sorry, your $=0$ at the very end is extremely confusing because it seems you are claiming that the previous expression is identically zero. It's surely not. Obviously, in general, the contribution is proportional to $\nabla\times B$. It would be pretty bad if if the (spatial) Maxwell term gave zero to the Maxwell equations. ;-) You shouldn't write it's zero because there are other terms in the equations from other terms in the action such as $j\cdot A$ and $(\partial_t A)^2$. $\endgroup$Luboš Motl– Luboš Motl2012-09-01 15:57:51 +00:00Commented Sep 1, 2012 at 15:57
- $\begingroup$ hmmm... but given the lagrangian from OP, are Lagrange's equations not what I wrote down in the last line? Certainly you are right if the lagrangian is more complicated. $\endgroup$kηives– kηives2012-09-01 16:27:01 +00:00Commented Sep 1, 2012 at 16:27
- $\begingroup$ Actually I have to agree with Lubos Motl because there are other terms with it in the main Lagrangian. I have already tried the procedure you have described. That causes some problem. So one of my advisor asked me to try the path integral, separate the coefficient of the terms and that might leads to the equation of motion we want. $\endgroup$aries0152– aries01522012-09-01 17:09:11 +00:00Commented Sep 1, 2012 at 17:09
- $\begingroup$ well my bad then, I thought you wanted the EOM assuming that was the whole action integral. sorry. $\endgroup$kηives– kηives2012-09-01 17:30:52 +00:00Commented Sep 1, 2012 at 17:30