Lagrangian Invariant Under Transformation

AI Thread Summary
The discussion focuses on verifying the invariance of a specific Lagrangian density under infinitesimal SO(3) rotations. Participants explore the algebraic manipulation of terms resulting from the transformation, particularly addressing the cancellation of higher-order terms and the implications of dummy indices in tensor notation. A key point raised is the necessity to consider only first-order variations due to the infinitesimal nature of the transformation, which simplifies the analysis. The conversation also touches on the calculation of Noether currents, suggesting that treating the transformation parameter as a function of position can lead to insights about conservation laws. Overall, the participants are engaged in a detailed examination of the mathematical framework underlying the invariance of the Lagrangian.
  • #51
latentcorpse said:
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.
 
Physics news on Phys.org
  • #52
fzero said:
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
 
  • #53
latentcorpse said:
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.
 
  • #54
fzero said:
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!
 
  • #55
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.
 
  • #56
fzero said:
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?
 
  • #57
latentcorpse said:
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...
 
  • #58
fzero said:
I suggested that a day or so ago...

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