Lagrangian Invariant Under Transformation

[fzero]

ok. so i now understand why that factor of 2 comes down.

as for rearranging the derivative, if we look just at the integrand, we have:

\nabla^2 \phi_b \phi_c + m^2 \phi_b \phi_c + \dot{\phi_b} \dot{\phi_c}
= \vec{\nabla} \cdot \vec{\nabla} \phi_b \phi_c + m^2 \phi_b \phi_c + \dot{\phi_b} \dot{\phi_c}
now as for rearranging the derivative, i get:
( \vec{\nabla} \phi_b ) \phi_c = \vec{\nabla} ( \phi_b \phi_c ) - \phi_b \vec{\nabla} \phi_c
which means
\vec{\nabla} \cdot ( \vec{\nabla} \phi_b ) \phi_c = \nabla^2 ( \phi_b \phi_c ) - \vec{\nabla} \cdot ( \phi_b \vec{\nabla} \phi_c )

is this looking correct?

The \nabla\cdot on the left acts only on \phi_b, i.e. the term is (\nabla^2 \phi_b) \phi_c. This contradicts the RHS of the last line that you wrote.

 

[latentcorpse]

The \nabla\cdot on the left acts only on \phi_b, i.e. the term is (\nabla^2 \phi_b) \phi_c. This contradicts the RHS of the last line that you wrote.

ok so this line is correct:
<br /> ( \vec{\nabla} \phi_b ) \phi_c = \vec{\nabla} ( \phi_b \phi_c ) - \phi_b \vec{\nabla} \phi_c<br />

but i don't know how to rearrange to get \nabla^2 \phi_b without dot producting with \vec{\nabla} \cdot on the LHS

 

[fzero]

ok so this line is correct:
<br /> ( \vec{\nabla} \phi_b ) \phi_c = \vec{\nabla} ( \phi_b \phi_c ) - \phi_b \vec{\nabla} \phi_c<br />

but i don't know how to rearrange to get \nabla^2 \phi_b without dot producting with \vec{\nabla} \cdot on the LHS

It seems easier in index notation, since the dot is giving you trouble. Start with (\partial_i \partial_i \phi_b) \phi_c.

 

[latentcorpse]

It seems easier in index notation, since the dot is giving you trouble. Start with (\partial_i \partial_i \phi_b) \phi_c.

right I am just going round in circles with this and not getting anywhere.
best I've got is the following:

\nabla^2 (fg) = \vec{\nabla} \cdot \vec{\nabla} (fg) = \vec{\nabla} \cdot \left( \left( \vec{\nabla} f \right) g + f \vec{\nabla} g \right) = ( \nabla^2 f ) g + 2 \vec{\nabla} f \cdot \vec{\nabla} g + f \nabla^2 g

now because of the antisymmetry of levi civita and the symmetry of the \vec{\nabla} f \cdot \vec{\nabla} g term, these will vanish.

this means that if we let f=\phi_b,g=\phi_c we have

(\nabla^2 \phi_b) \phi_c = \nabla^2 ( \phi_b \phi_c) - \phi_b \nabla^2 \phi_c

again, the \nabla^2 ( \phi_b \phi_c ) term is symmetric so will vanish by antisymmetry of levi civita, leaving

(\nabla^2 \phi_b) \phi_c =- \phi_b \nabla^2 \phi_c

i'm not sure if this is correct and even if it is, how does it help?

thanks!

 

[fzero]

What you did will work if you do some symmetry operations and relabeling of idices, but it's going off on a tangent by quite a bit. Just integrate by parts once: (\partial_i \partial_i \phi_b) \phi_c = \partial_i ( (\partial_i \phi_b) \phi_c) - \partial_i \phi_b\partial_i\phi_c.

 

[latentcorpse]

What you did will work if you do some symmetry operations and relabeling of idices, but it's going off on a tangent by quite a bit. Just integrate by parts once: (\partial_i \partial_i \phi_b) \phi_c = \partial_i ( (\partial_i \phi_b) \phi_c) - \partial_i \phi_b\partial_i\phi_c.

well if i substitue what you've written above, won't the 2nd term vanish by symmetry and the first term vanish by the same divergence theorem argument we used earlier?

 

[fzero]

well if i substitue what you've written above, won't the 2nd term vanish by symmetry and the first term vanish by the same divergence theorem argument we used earlier?

I suggested that a day or so ago...

 

[latentcorpse]

I suggested that a day or so ago...

lol. thanks a million. I'm sorry for being so slow...