Lagrangian Invariant Under Transformation

[latentcorpse]

Verify that the Lagrangian density
L= \frac{1}{2} \partial_\mu \phi_a \partial^\mu \phi_a - \frac{1}{2} m^2 \phi_a{}^2
for a triplet of real fields \phi_a (a=1,2,3) is invariant under the infinitesimal SO(3) rotation by \theta

\phi_a \rightarrow \phi_a + \theta \epsilon_{abc} n_b \phi_c

plugging this in i get:

L= \frac{1}{2} \partial_\mu ( \phi_a + \theta \epsilon_{abc} n_b \phi_c) \partial^\mu ( \phi_a + \theta \epsilon_{abc} n_b \phi_c ) - \frac{1}{2} m^2 ( \phi_a{}^2 + 2 \phi_a \theta \epsilon_{abc} n_b \phi_c + \theta^2 \epsilon_{abc} \epsilon_{abc} n_b{}^2 \phi_c{}^2)

but now i don't know how to get rid of anything. any ideas?

thanks.

 

[fzero]

Verify that the Lagrangian density
L= \frac{1}{2} \partial_\mu \phi_a \partial^\mu \phi_a - \frac{1}{2} m^2 \phi_a{}^2
for a triplet of real fields \phi_a (a=1,2,3) is invariant under the infinitesimal SO(3) rotation by \theta

\phi_a \rightarrow \phi_a + \theta \epsilon_{abc} n_b \phi_c

plugging this in i get:

L= \frac{1}{2} \partial_\mu ( \phi_a + \theta \epsilon_{abc} n_b \phi_c) \partial^\mu ( \phi_a + \theta \epsilon_{abc} n_b \phi_c ) - \frac{1}{2} m^2 ( \phi_a{}^2 + 2 \phi_a \theta \epsilon_{abc} n_b \phi_c + \theta^2 \epsilon_{abc} \epsilon_{abc} n_b{}^2 \phi_c{}^2)

but now i don't know how to get rid of anything. any ideas?

thanks.

Some of these terms are going to be too confusing if you use the same indices too many times. For instance,

\theta^2 \epsilon_{abc} \epsilon_{abc} n_b{}^2 \phi_c{}^2 = \theta^2 \epsilon_{abc} \epsilon_{ade} n_b n_d \phi_c \phi_e.

Now you want to use an identity for contraction of \epsilon symbols:

\epsilon_{abc} \epsilon_{ade} = \delta_{bd}\delta_{ce} - \delta_{be} \delta_{cd}.

Other terms should be closely examined keeping in mind the antisymmetry of \epsilon_{abc}, like

2 \phi_a \theta \epsilon_{abc} n_b \phi_c .

All terms in the expansion can be evaluated using one or both of these ideas.

 

[latentcorpse]

Some of these terms are going to be too confusing if you use the same indices too many times. For instance,

\theta^2 \epsilon_{abc} \epsilon_{abc} n_b{}^2 \phi_c{}^2 = \theta^2 \epsilon_{abc} \epsilon_{ade} n_b n_d \phi_c \phi_e.

Now you want to use an identity for contraction of \epsilon symbols:

\epsilon_{abc} \epsilon_{ade} = \delta_{bd}\delta_{ce} - \delta_{be} \delta_{cd}.

Other terms should be closely examined keeping in mind the antisymmetry of \epsilon_{abc}, like

2 \phi_a \theta \epsilon_{abc} n_b \phi_c .

All terms in the expansion can be evaluated using one or both of these ideas.

hi thanks for your reply.
i want to first of all ask you how you know you can just randomly change the c index to and e index in order to make the algebra easier? Is it just because it's dummy?

anyway i can work out the first bit using the contraction identity for epsilons. But I'm having a little bit of trouble evaluating the second term using the antisymmetry of epsilon. Surely I am trying to get something that will cancel the contribution form the other term i just derived?

 

[fzero]

hi thanks for your reply.
i want to first of all ask you how you know you can just randomly change the c index to and e index in order to make the algebra easier? Is it just because it's dummy?

Yes it's because they're dummy indicies. Since the expression has the indicies summed in pairs, you should never have the same index appearing more than twice.

anyway i can work out the first bit using the contraction identity for epsilons. But I'm having a little bit of trouble evaluating the second term using the antisymmetry of epsilon. Surely I am trying to get something that will cancel the contribution form the other term i just derived?

Note that you can write

<br /> 2 \phi_a \theta \epsilon_{abc} n_b \phi_c = \theta \epsilon_{abc} n_b (\phi_a \phi_c+ \phi_c\phi_a) <br />

and it should be obvious.

 

[latentcorpse]

Yes it's because they're dummy indicies. Since the expression has the indicies summed in pairs, you should never have the same index appearing more than twice.
Note that you can write

<br /> 2 \phi_a \theta \epsilon_{abc} n_b \phi_c = \theta \epsilon_{abc} n_b (\phi_a \phi_c+ \phi_c\phi_a) <br />

and it should be obvious.

ok so using the antisymmetry, that term will go to 0 because if we swap a and c indices we pick up a minus sign from the epsilon, right?.

so i have

( \phi_a{} + \theta \epsilon_{abc} n_b \phi_c )( \phi_a{} + \theta \epsilon_{abc} n_b \phi_c )
= \phi_a{}^2 + \theta \epsilon_{abc} n_b ( \phi_a \phi_c + \phi_c \phi_a ) + \theta^2 ( \delta_{bd} \delta_{ce} - \delta_{be} \delta_{cd} ) n_b n_d \phi_c \phi_e
= \phi_a{}^2 + 0 + \theta^2 ( n_b{}^2 \phi_c{}^2 - n_b n_c \phi_c \phi_b )

hopefully that's correct algebra-wise. I'm a bit concerned because I'm trying to show the lagrangian is invariant so surely i only want a \phi_a{}^2 term? will all the other crap i have there cancel with stuff from the first part of the lagrangian? and do you have any advice on how to go about evaluating those derivatives?

thanks again for your help!

 

[fzero]

ok so using the antisymmetry, that term will go to 0 because if we swap a and c indices we pick up a minus sign from the epsilon, right?.

so i have

( \phi_a{} + \theta \epsilon_{abc} n_b \phi_c )( \phi_a{} + \theta \epsilon_{abc} n_b \phi_c )
= \phi_a{}^2 + \theta \epsilon_{abc} n_b ( \phi_a \phi_c + \phi_c \phi_a ) + \theta^2 ( \delta_{bd} \delta_{ce} - \delta_{be} \delta_{cd} ) n_b n_d \phi_c \phi_e
= \phi_a{}^2 + 0 + \theta^2 ( n_b{}^2 \phi_c{}^2 - n_b n_c \phi_c \phi_b )

hopefully that's correct algebra-wise. I'm a bit concerned because I'm trying to show the lagrangian is invariant so surely i only want a \phi_a{}^2 term? will all the other crap i have there cancel with stuff from the first part of the lagrangian? and do you have any advice on how to go about evaluating those derivatives?

thanks again for your help!

Oh I should have realized you'd run into trouble. The problem is that this is an O(\theta^2) term and we're only varying the fields to first order. Unless you consider higher orders in the variation, you won't be able to cancel this term. What you really want to compute is

L(\phi_a+\delta\phi_a) = L(\phi_a) + \frac{\delta L}{\delta\phi_a} \delta\phi_a + O((\delta\phi_a)^2)

and show that \delta L/\delta\phi_a, which comes from only first-order terms, vanishes.


As for the dealing with the derivative terms, you can usually use integration by parts to make the symmetry more transparent.

 

[latentcorpse]

Oh I should have realized you'd run into trouble. The problem is that this is an O(\theta^2) term and we're only varying the fields to first order. Unless you consider higher orders in the variation, you won't be able to cancel this term. What you really want to compute is

L(\phi_a+\delta\phi_a) = L(\phi_a) + \frac{\delta L}{\delta\phi_a} \delta\phi_a + O((\delta\phi_a)^2)

and show that \delta L/\delta\phi_a, which comes from only first-order terms, vanishes.


As for the dealing with the derivative terms, you can usually use integration by parts to make the symmetry more transparent.

ok. well that sorts out the second part of the lagrangian. how did you know we are working only to first order though? i copied the question out word for word and nowhere does it mention first order - does it?

the derivative terms give (to first order):

\partial_\mu \phi_a \partial^\mu \phi_a + \partial_\mu \phi_a \partial^\mu \theta \epsilon_{abc} n_b \phi_c + \partial_\mu \theta \epsilon_{abc} n_b \phi_c \partial^\mu \phi_a

now the first term is what i want so i need to cancel the 2nd and 3rd terms:

\partial_\mu \phi_a \partial^\mu \theta \epsilon_{abc} n_b \phi_c + \partial_\mu \theta \epsilon_{abc} n_b \phi_c \partial^\mu \phi_a

not really sure how to integrate these by parts or how that is going to help me?? i thought about trying a similar trick as last time and "moving" the \partial \phi_a terms about and trying to pick up a minus sign from the antisymmetry of the levi-civita but i couldn't get it to work (there was no relabelling of indices involved)

thanks again.

thanks!

 

[fzero]

ok. well that sorts out the second part of the lagrangian. how did you know we are working only to first order though? i copied the question out word for word and nowhere does it mention first order - does it?

What you wrote was "infinitesimal rotation." As I said, if you took the rotation out to 2nd order you would find that the Lagrangian was invariant up to 3rd order terms.

the derivative terms give (to first order):

\partial_\mu \phi_a \partial^\mu \phi_a + \partial_\mu \phi_a \partial^\mu \theta \epsilon_{abc} n_b \phi_c + \partial_\mu \theta \epsilon_{abc} n_b \phi_c \partial^\mu \phi_a

now the first term is what i want so i need to cancel the 2nd and 3rd terms:

\partial_\mu \phi_a \partial^\mu \theta \epsilon_{abc} n_b \phi_c + \partial_\mu \theta \epsilon_{abc} n_b \phi_c \partial^\mu \phi_a

not really sure how to integrate these by parts or how that is going to help me?? i thought about trying a similar trick as last time and "moving" the \partial \phi_a terms about and trying to pick up a minus sign from the antisymmetry of the levi-civita but i couldn't get it to work (there was no relabelling of indices involved)

thanks again.

thanks!

We're assuming that \theta is a constant, what's called a "global" transformation. As you can see, the Lagrangian is not invariant under "local" transformations where \theta(x) is a function. In order for the Lagrangian to be invariant under local transformations, we would need to couple the theory to a gauge field.

 

[latentcorpse]

We're assuming that \theta is a constant, what's called a "global" transformation. As you can see, the Lagrangian is not invariant under "local" transformations where \theta(x) is a function. In order for the Lagrangian to be invariant under local transformations, we would need to couple the theory to a gauge field.

ok. but how do i show those two terms with the derivatives in them cancel out?

 

[fzero]

ok. but how do i show those two terms with the derivatives in them cancel out?

If \theta is a constant, what is \partial_\mu \theta?

 

[latentcorpse]

If \theta is a constant, what is \partial_\mu \theta?

oh that is rather obvious isn't it!

thanks!

do you know anything about Noether currents? The next thing I'm supposed to do is find the Noether current j^\mu. We have actually spent a couple of lectures on these but I don't really grasp how to go about calculating them.

Thanks.

 

[vela]

It's not that obvious.

\partial_\mu \phi_a \rightarrow \partial_\mu (\phi_a+\theta \epsilon_{abc} n_b \phi_c) = \partial_\mu \phi_a + \partial_\mu (\theta \epsilon_{abc} n_b \phi_c)

You're not just differentiating the θ.

 

[fzero]

oh that is rather obvious isn't it!

thanks!

do you know anything about Noether currents? The next thing I'm supposed to do is find the Noether current j^\mu. We have actually spent a couple of lectures on these but I don't really grasp how to go about calculating them.

Thanks.

Well one way to compute the Noether currents is to actually consider the transformation parameter to be a function of x instead of just throwing those terms away. We actually do this within the action, so we're considering

\delta S = \int d^4x~ \delta L = \int d^4x ~ \partial_\mu \theta (\cdots )

If we can write this term as

\int d^4x ~ \theta ~ \partial_\mu j^\mu ,

then we conclude that j^\mu is the conserved Noether current.

 

[latentcorpse]

It's not that obvious.

\partial_\mu \phi_a \rightarrow \partial_\mu (\phi_a+\theta \epsilon_{abc} n_b \phi_c) = \partial_\mu \phi_a + \partial_\mu (\theta \epsilon_{abc} n_b \phi_c)

You're not just differentiating the θ.

good point.

so id have to do a product rule on that 2nd term you've written and show that it all vanishes. i get:

\theta ( \partial_\mu \epsilon_{abc} ) n_b \phi_c + \theta \epsilon_{abc} n_b ( \partial_\mu \phi_c )

as teh derivative of \theta is zero as its just a constant and the derivative of n_b is zero as it is also a constant.

how do i get those two remaining terms to go away though?

 

[latentcorpse]

Well one way to compute the Noether currents is to actually consider the transformation parameter to be a function of x instead of just throwing those terms away. We actually do this within the action, so we're considering

\delta S = \int d^4x~ \delta L = \int d^4x ~ \partial_\mu \theta (\cdots )

If we can write this term as

\int d^4x ~ \theta ~ \partial_\mu j^\mu ,

then we conclude that j^\mu is the conserved Noether current.

is this always the case? i.e. is this the defn of noether current or have you adapted it for this situation?

anyway, if i can write L in the form \theta \partial_\mu j^\mu then j^\mu will be my noether current, correct? only problem is, if i use the unmodified L, there is no theta involved and if i use the modified L, i struggle to put it into the desired form as there are terms that i can't get to go into that format.

 

[fzero]

good point.

so id have to do a product rule on that 2nd term you've written and show that it all vanishes. i get:

\theta ( \partial_\mu \epsilon_{abc} ) n_b \phi_c + \theta \epsilon_{abc} n_b ( \partial_\mu \phi_c )

as teh derivative of \theta is zero as its just a constant and the derivative of n_b is zero as it is also a constant.

how do i get those two remaining terms to go away though?

You have to include the whole expression, not just part of it. Look at\partial^\mu \phi _a\left( \theta ( \partial_\mu \epsilon_{abc} ) n_b \phi_c + \theta \epsilon_{abc} n_b ( \partial_\mu \phi_c )\right)

The first term vanishes because \epsilon_{abc} is a constant.

is this always the case? i.e. is this the defn of noether current or have you adapted it for this situation?

You can always calculate the Noether current by letting the parameter depend on position ("gauging" the parameter), but in a more general case you might have to consider various subtleties that can occur if there is an accompanying transformation of coordinates \delta x^\mu or if the fields are not scalar fields.

anyway, if i can write L in the form \theta \partial_\mu j^\mu then j^\mu will be my noether current, correct? only problem is, if i use the unmodified L, there is no theta involved and if i use the modified L, i struggle to put it into the desired form as there are terms that i can't get to go into that format.

What appears in the expression is

\delta L = L(\phi +\delta \phi) - L(\phi)

If L is truly invariant under \delta\phi, the variation of the action will have the stated form.

 

[latentcorpse]

You have to include the whole expression, not just part of it. Look at


\partial^\mu \phi _a\left( \theta ( \partial_\mu \epsilon_{abc} ) n_b \phi_c + \theta \epsilon_{abc} n_b ( \partial_\mu \phi_c )\right)

The first term vanishes because \epsilon_{abc} is a constant.

right. i see that. but the 2nd one remains surely as \partial^\mu \phi_a , \partial_\mu \phi_c aren't necessarily zero

 

[fzero]

right. i see that. but the 2nd one remains surely as \partial^\mu \phi_a , \partial_\mu \phi_c aren't necessarily zero

You need to consider the symmetries as well.

 

[latentcorpse]

You need to consider the symmetries as well.

hmmmm. I'm still not getting this to work out. also, have you reversed the order of your \partial^\mu and \partial_\mu by accident or deliberately?


i get

\partial_\mu \phi_a \partial^\mu \phi_a \rightarrow \partial_\mu ( \phi_a + \theta \epsilon_{abc} n_b \phi_c ) \partial^\mu ( \phi_a + \theta \epsilon_{abc} n_b \phi_c )

= \partial_\mu \phi_a \partial^\mu \phi_a + \partial_\mu \phi_a \partial^\mu ( \theta \epsilon_{abc} n_b \phi_c ) + \partial_\mu ( \theta \epsilon_{abc} n_b \phi_c ) \partial^\mu \phi_a + \mathcal{O}( \theta^2)

now hopefully it's correct up to there

now the first term we can leave alone as it's correct

the 2nd on gives \partial_\mu \phi_a \partial^\mu ( \theta \epsilon_{abc} n_b \phi_c ) = \partial_\mu \phi_a ( \partial^\mu \theta ) \epsilon_{abc} n_b \phi_c + \partial_\mu \phi_a \theta ( \partial^\mu \epsilon_{abc} ) n_b \phi_c + \partial_\mu \phi_a \theta \epsilon_{abc} ( \partial^\mu n_b ) \phi_c + \partial_\mu \phi_a \theta \epsilon_{abc} n_b \partial&amp;\mu \phi_c

now as \theta, \epsilon_{abc} and n_b are all constant, this gives

\partial_\mu \phi_a \theta \epsilon_{abc} n_b \partial&amp;\mu \phi_c

and the final term in the original expression gives \partial_\mu ( \theta \epsilon_{abc} n_b \phi_c ) \partial^\mu \phi_a = \theta \epsilon_{abc} n_b \partial_\mu \phi_c \partial^\mu \phi_a = - \theta \epsilon_{abc} n_b \partial_\mu \phi_a \partial^\mu \phi_c

and that cancels the previous bit leaving only \partial_\mu \phi_a \partial^\mu \phi_a as desired and hence the lagrangian is invariant. is this ok?

 

[latentcorpse]

ok so i think that's the lagrangian invariance taken care of at last. i amen't getting anywher with the noether current though...

i have the following formula in my notes (although I'm not sure if it can be applied generally or if it has been tailored for a specific example in which case it may well be irrelevant):

(j^\mu)_\nu = \frac{\partial L}{\partial (\partial_\mu \phi_a)} \partial_\nu \phi_a - \delta^\mu{}_\nu L

anyway this gives

(j^\mu)_\nu - \frac{1}{2} \partial^\mu \phi_a \partial_\nu \phi_a - \frac{1}{2} \partial_\mu \phi_a \partial_\nu \phi_a = \frac{1}{2} \partial_\nu \phi_a (\partial^\mu - \partial_\mu) \phi_a

is this correct or am i barking up completely the wrong tree?

 

[fzero]

\partial_\mu \phi_a \theta \epsilon_{abc} n_b \partial&amp;\mu \phi_c

and the final term in the original expression gives \partial_\mu ( \theta \epsilon_{abc} n_b \phi_c ) \partial^\mu \phi_a = \theta \epsilon_{abc} n_b \partial_\mu \phi_c \partial^\mu \phi_a = - \theta \epsilon_{abc} n_b \partial_\mu \phi_a \partial^\mu \phi_c

and that cancels the previous bit leaving only \partial_\mu \phi_a \partial^\mu \phi_a as desired and hence the lagrangian is invariant. is this ok?

It's fine, but the terms vanish separately too, since

\epsilon_{abc} \partial_\mu \phi_a \partial^\mu \phi_c = \frac{1}{2} \epsilon_{abc} (\partial_\mu \phi_a \partial^\mu \phi_c + \partial_\mu \phi_c \partial^\mu \phi_a ).

ok so i think that's the lagrangian invariance taken care of at last. i amen't getting anywher with the noether current though...

i have the following formula in my notes (although I'm not sure if it can be applied generally or if it has been tailored for a specific example in which case it may well be irrelevant):

(j^\mu)_\nu = \frac{\partial L}{\partial (\partial_\mu \phi_a)} \partial_\nu \phi_a - \delta^\mu{}_\nu L

anyway this gives

(j^\mu)_\nu - \frac{1}{2} \partial^\mu \phi_a \partial_\nu \phi_a - \frac{1}{2} \partial_\mu \phi_a \partial_\nu \phi_a = \frac{1}{2} \partial_\nu \phi_a (\partial^\mu - \partial_\mu) \phi_a

is this correct or am i barking up completely the wrong tree?

That looks like a formula for the stress-energy tensor, though I don't think you computed it correctly. That is the Noether current associated to spacetime translations. There is a related expression for the Noether current under a general transformation of the fields (but not coordinates)

\theta n_a j_a^\mu = \frac{\partial L}{\partial (\partial_\mu \phi_a)} \delta \phi_a,

where I've stripped off the parameters of the transformation so that the Noether current(s) j_a^\mu do not depend on them. Incidentally, I was sloppy about this in my previous expression for the Noether current, where I pulled out \theta but not n_a. Since we are dealing with an SO(3) rotation, there are 3 currents since the rank of SO(3) is 3.
however this formula

 

[latentcorpse]

It's fine, but the terms vanish separately too, since

\epsilon_{abc} \partial_\mu \phi_a \partial^\mu \phi_c = \frac{1}{2} \epsilon_{abc} (\partial_\mu \phi_a \partial^\mu \phi_c + \partial_\mu \phi_c \partial^\mu \phi_a ).
That looks like a formula for the stress-energy tensor, though I don't think you computed it correctly. That is the Noether current associated to spacetime translations. There is a related expression for the Noether current under a general transformation of the fields (but not coordinates)

\theta n_a j_a^\mu = \frac{\partial L}{\partial (\partial_\mu \phi_a)} \delta \phi_\mu,

where I've stripped off the parameters of the transformation so that the Noether current(s) j_a^\mu do not depend on them. Incidentally, I was sloppy about this in my previous expression for the Noether current, where I pulled out \theta but not n_a. Since we are dealing with an SO(3) rotation, there are 3 currents since the rank of SO(3) is 3.
however this formula

ok thanks for that.

is this formula for the noether current suitable for any example or have you tailored it to this one? also, do the indices on the LHS and RHS of your expression for noether current balance properly?

anyway, we have

L=\frac{1}{2} \partial_\mu \phi_a \partial^\mu \phi_a + \frac{1}{2} m^2 \phi_a{}^2

\Rightarrow \frac{\partial L}{\partial ( \partial_\mu \phi_a )} = \frac{1}{2} \partial^\mu \phi_a

and so using your formula we get

\theta n_a j_a{}^\mu =\frac{1}{2} \partial^\mu \phi_a \delta \phi_\mu

am i allowed to rearrange for j by dividing by theta and n_a. surely not as n_a is a vector?

 

[fzero]

ok thanks for that.

is this formula for the noether current suitable for any example or have you tailored it to this one? also, do the indices on the LHS and RHS of your expression for noether current balance properly?

There was a typo, I meant to write


<br /> \theta n_a j_a^\mu = \frac{\partial L}{\partial (\partial_\mu \phi_a)} \delta \phi_a, <br />

This is specific to the example because of the form of the transformation parameters. In a general case, you'd have to figure out the appropriate way to deal with those.

anyway, we have

L=\frac{1}{2} \partial_\mu \phi_a \partial^\mu \phi_a + \frac{1}{2} m^2 \phi_a{}^2

\Rightarrow \frac{\partial L}{\partial ( \partial_\mu \phi_a )} = \frac{1}{2} \partial^\mu \phi_a

and so using your formula we get

\theta n_a j_a{}^\mu =\frac{1}{2} \partial^\mu \phi_a \delta \phi_\mu

am i allowed to rearrange for j by dividing by theta and n_a. surely not as n_a is a vector?

You don't really divide by n_a, the RHS side will be proportional to n_a when you put in the form of \delta\phi_a (after redefining dummy indicies). You identify the current with the rest of the expression.

 

[latentcorpse]

There was a typo, I meant to write


<br /> \theta n_a j_a^\mu = \frac{\partial L}{\partial (\partial_\mu \phi_a)} \delta \phi_a, <br />

This is specific to the example because of the form of the transformation parameters. In a general case, you'd have to figure out the appropriate way to deal with those.



You don't really divide by n_a, the RHS side will be proportional to n_a when you put in the form of \delta\phi_a (after redefining dummy indicies). You identify the current with the rest of the expression.

ok so \delta \phi_a = \theta \epsilon_{abc} n_b \phi_c presumably since \phi_a \rightarrow \phi_a + \delta \phi_a

so that would mean

\theta n_a j_a{}^\mu = \frac{1}{2} \partial^\mu \phi_a \theta \epsilon_{abc} n_b \phi_c

(slight problem in that my n indices don't match up)

and so j_a{}^\mu = \frac{1}{2} \epsilon_{abc} \phi_c \partial^\mu \phi_a

 

[fzero]

ok so \delta \phi_a = \theta \epsilon_{abc} n_b \phi_c presumably since \phi_a \rightarrow \phi_a + \delta \phi_a

so that would mean

\theta n_a j_a{}^\mu = \frac{1}{2} \partial^\mu \phi_a \theta \epsilon_{abc} n_b \phi_c

(slight problem in that my n indices don't match up)

and so j_a{}^\mu = \frac{1}{2} \epsilon_{abc} \phi_c \partial^\mu \phi_a

I mentioned that you would have to relabel dummy indices. Whatever you do you need to match the free index on the RHS with the index on the LHS.

 

[latentcorpse]

I mentioned that you would have to relabel dummy indices. Whatever you do you need to match the free index on the RHS with the index on the LHS.

yes but say i relabel a to b in my epsression for \delta \phi_a then my indices will match up on the n's but my indices won't match up in my expression for \delta \phi_a as i'll have:

\delta \phi_a = \theta \epsilon_{bac} n_a \phi_c and so there are 2 a indices on RHS and only 1 on LHS plus the b's and c's are also out of sync?

however, on the other hand, if I am free to sub it in and THEN relabel it, i get

\theta n_a j_a{}^\mu = \frac{1}{2} \partial^\mu \phi_a \theta \epsilon_{abc} n_b \phi_c
\Rightarrow \theta n_a j_a{}^\mu = \frac{1}{2} \partial^\mu \phi_b \theta \epsilon_{bac} n_a \phi_c
\Rightarrow j_a{}^\mu = - \frac{1}{2} \partial^\mu \theta \epsilon_{abc} \phi_b \phi_c

how does that look now?

 

[fzero]

yes but say i relabel a to b in my epsression for \delta \phi_a then my indices will match up on the n's but my indices won't match up in my expression for \delta \phi_a as i'll have:

\delta \phi_a = \theta \epsilon_{bac} n_a \phi_c and so there are 2 a indices on RHS and only 1 on LHS plus the b's and c's are also out of sync?

however, on the other hand, if I am free to sub it in and THEN relabel it, i get

\theta n_a j_a{}^\mu = \frac{1}{2} \partial^\mu \phi_a \theta \epsilon_{abc} n_b \phi_c
\Rightarrow \theta n_a j_a{}^\mu = \frac{1}{2} \partial^\mu \phi_b \theta \epsilon_{bac} n_a \phi_c
\Rightarrow j_a{}^\mu = - \frac{1}{2} \partial^\mu \theta \epsilon_{abc} \phi_b \phi_c

how does that look now?

I'm not sure why you have \partial^\mu \theta appearing in the expression now. Take the first line


\theta n_a j_a{}^\mu = \frac{1}{2} \partial^\mu \phi_a \theta \epsilon_{abc} n_b \phi_c

and relabel the index on the LHS:

\theta n_d j_d{}^\mu = \frac{1}{2} \partial^\mu \phi_a \theta \epsilon_{abc} n_b \phi_c.

Reorder the terms a bit to show the structure

\theta n_d j_d{}^\mu = \theta n_b \left( \frac{1}{2} \epsilon_{abc} \partial^\mu \phi_a \phi_c \right) .

Looking at the free index within the parentheses, we see that we have have

j_b{}^\mu = \frac{1}{2} \epsilon_{abc} \partial^\mu \phi_a \phi_c = - \frac{1}{2} \epsilon_{bac} \partial^\mu \phi_a \phi_c ,

This is the same as

j_a{}^\mu = - \frac{1}{2} \epsilon_{abc} \partial^\mu \phi_b \phi_c = \frac{1}{2} \epsilon_{abc} \phi_b \partial^\mu \phi_c .

where in the last part we have just moved the derivative term to the right for aesthetics and relabeled the indicies b\leftrightarrow c accordingly, picking up another minus sign from the epsilon symbol.

You might as well practice a bit with relabeling and rearranging indices like this because you'll be seeing it a lot in any field theory course.

 

[latentcorpse]

I'm not sure why you have \partial^\mu \theta appearing in the expression now. Take the first line


\theta n_a j_a{}^\mu = \frac{1}{2} \partial^\mu \phi_a \theta \epsilon_{abc} n_b \phi_c

and relabel the index on the LHS:

\theta n_d j_d{}^\mu = \frac{1}{2} \partial^\mu \phi_a \theta \epsilon_{abc} n_b \phi_c.

Reorder the terms a bit to show the structure

\theta n_d j_d{}^\mu = \theta n_b \left( \frac{1}{2} \epsilon_{abc} \partial^\mu \phi_a \phi_c \right) .

Looking at the free index within the parentheses, we see that we have have

j_b{}^\mu = \frac{1}{2} \epsilon_{abc} \partial^\mu \phi_a \phi_c = - \frac{1}{2} \epsilon_{bac} \partial^\mu \phi_a \phi_c ,

This is the same as

j_a{}^\mu = - \frac{1}{2} \epsilon_{abc} \partial^\mu \phi_b \phi_c = \frac{1}{2} \epsilon_{abc} \phi_b \partial^\mu \phi_c .

where in the last part we have just moved the derivative term to the right for aesthetics and relabeled the indicies b\leftrightarrow c accordingly, picking up another minus sign from the epsilon symbol.

You might as well practice a bit with relabeling and rearranging indices like this because you'll be seeing it a lot in any field theory course.

ok. i see what you've done in that last line - that's neat! but isn't it basically the same as what i had in my last post except i forgot to cancel my theta? your way is of course much more clearly explained but are the answers not the same?

 

[fzero]

ok. i see what you've done in that last line - that's neat! but isn't it basically the same as what i had in my last post except i forgot to cancel my theta? your way is of course much more clearly explained but are the answers not the same?

Yes they agree if you put the \partial^\mu on the \phi_b.

 

[latentcorpse]

Yes they agree if you put the \partial^\mu on the \phi_b.

great. thanks. the final thing i have to do is to deduce that the three quantities

Q_a = \int d^3x \epsilon_{abc} \dot{\phi_b} \phi_c are all conserved and verify that using the field equations satisfied by \phi_a

well we can use the Euler Lagrange equation to get the field equations:
\frac{d}{dt}(\frac{\partial L}{\partial \dot{\phi_a}})=\frac{\partial L}{\partial \phi_a}
\Rightarrow \frac{d}{dt} ( \frac{1}{2} \partial^0 \phi_a ) = - \frac{1}{2} m^2 \phi_a
where I have noted that only the \partial^0 derivative will survive as \frac{\partial L}{\partial \dot{\phi_a}} will only be non zero when \mu=0

so this gives \frac{1}{2} \frac{\partial^2 L}{\partial t^2} = - m^2 \phi_a
\Rightarrow \ddot{\phi_a}+2m^2 \phi_a=0

anyway so i have the EOMs but i first need to deduce that the Q_a are conserved - any adivce on how to do that? do i just take a time derivative? i don't see how i can get anything useful out of that.

 

[fzero]

the final thing i have to do is to deduce that the three quantities

Q_a = \int d^3x \epsilon_{abc} \dot{\phi_b} \phi_c are all conserved and verify that using the field equations satisfied by \phi_a

well we can use the Euler Lagrange equation to get the field equations:
\frac{d}{dt}(\frac{\partial L}{\partial \dot{\phi_a}})=\frac{\partial L}{\partial \phi_a}
\Rightarrow \frac{d}{dt} ( \frac{1}{2} \partial^0 \phi_a ) = - \frac{1}{2} m^2 \phi_a
where I have noted that only the \partial^0 derivative will survive as \frac{\partial L}{\partial \dot{\phi_a}} will only be non zero when \mu=0

so this gives \frac{1}{2} \frac{\partial^2 L}{\partial t^2} = - m^2 \phi_a
\Rightarrow \ddot{\phi_a}+2m^2 \phi_a=0

anyway so i have the EOMs but i first need to deduce that the Q_a are conserved - any adivce on how to do that? do i just take a time derivative? i don't see how i can get anything useful out of that.

Those are the Euler-Lagrange equations for a 1d theory, in a field theory the EL equations are

\partial_\mu (\frac{\partial L}{\partial (\partial_\mu {\phi_a})})=\frac{\partial L}{\partial \phi_a}.

The calculation should be similar, but you will probably need integrate by parts to find that certain terms vanish because they are the integral of a total spatial derivative.

As for why the Q_a are conserved, note that

Q_a = \int d^3 x j_{a 0}.

Try to relate \dot{Q}_a to the conservation of the Noether currents.

 

[latentcorpse]

Those are the Euler-Lagrange equations for a 1d theory, in a field theory the EL equations are

\partial_\mu (\frac{\partial L}{\partial (\partial_\mu {\phi_a})})=\frac{\partial L}{\partial \phi_a}.

The calculation should be similar, but you will probably need integrate by parts to find that certain terms vanish because they are the integral of a total spatial derivative.

As for why the Q_a are conserved, note that

Q_a = \int d^3 x j_{a 0}.

Try to relate \dot{Q}_a to the conservation of the Noether currents.

ok. i find that Q_a = \int d^3x \epsilon_{abc} \dot{\phi_b} \phi_c = - \frac{1}{2} \int j_a{}^0

Now the Noether current j_a{}^\mu will be conserved by definition. This means that Q_a is constant and therefore conserved - i don't think this can be sufficient though!

as for the E-L eqns - that was stupid of me to use 1D case. anyway i get:

\partial_\mu \partial^\mu \phi_a + m^2 \phi_a = 0

 

[fzero]

ok. i find that Q_a = \int d^3x \epsilon_{abc} \dot{\phi_b} \phi_c = - \frac{1}{2} \int j_a{}^0

Right, I think you even fixed a minus sign that I got wrong.

Now the Noether current j_a{}^\mu will be conserved by definition. This means that Q_a is constant and therefore conserved - i don't think this can be sufficient though!

There's a line or two of computation to verify this, but it's an important result that there is an associated conserved charge to every conserved current.

as for the E-L eqns - that was stupid of me to use 1D case. anyway i get:

\partial_\mu \partial^\mu \phi_a + m^2 \phi_a = 0

If the relative signs are correct, that's the right equation of motion. Did you use that to show that \dot{Q}_a=0?

 

[latentcorpse]

There's a line or two of computation to verify this, but it's an important result that there is an associated conserved charge to every conserved current.

ok. i have no idea how to do this. any advice?

If the relative signs are correct, that's the right equation of motion. Did you use that to show that \dot{Q}_a=0?

ok. i gave this a shot:

one of the 4 eqns looks like \ddot{\phi_a}+m^2 \phi_a = 0 \Rightarrow \phi_a = A \cos{mt} + B \sin{mt}
i was then going to get an expression for \dot{\phi_a} and plug that back into the expression for Q_a but I don't think this is right since i have no indices in my expression for \phi_a.

 

[fzero]

ok. i have no idea how to do this. any advice?

Write down an expression for \dot{Q}_a in terms of j_{a0} and use \partial^\mu j_{a\mu}=0 to show that it vanishes. It's kind of simple, I'm not sure what else I could say without giving it completely away.

ok. i gave this a shot:

one of the 4 eqns looks like \ddot{\phi_a}+m^2 \phi_a = 0 \Rightarrow \phi_a = A \cos{mt} + B \sin{mt}
i was then going to get an expression for \dot{\phi_a} and plug that back into the expression for Q_a but I don't think this is right since i have no indices in my expression for \phi_a.

Here you want to use the expression for the Q_a in terms of the \phi_a and relate the expressions to terms that either vanish because of the equation of motion or because they're total derivatives.

 

[latentcorpse]

Write down an expression for \dot{Q}_a in terms of j_{a0} and use \partial^\mu j_{a\mu}=0 to show that it vanishes. It's kind of simple, I'm not sure what else I could say without giving it completely away.

ok so we had Q_a = - \frac{1}{2} \int j_a{}^0

i notice you have the 0 index "down" - is this important?

anyway \dot{Q_a} = - \frac{1}{2} \int \partial_0 j_a{}^0 (where i hope my indices are correct since i want to contract over the 0 index and \partial_0 = \frac{\partial}{\partial t}

but \partial_\mu j_a{}^\mu = 0 \Rightarrow \partial_0 j_a{}^0 = - \partial_i j_a{}^i

giving \dot{Q_a} = \frac{1}{2} \int \partial_i j_a{}^i

how does that go to 0 though?

 

[fzero]

ok so we had Q_a = - \frac{1}{2} \int j_a{}^0

i notice you have the 0 index "down" - is this important?

In the Minkowski metric there's a minus sign between j_{a0} = - j_a^{0}. Put the index wherever you find convenient, as I think I've made a couple of sign errors already because of it.

anyway \dot{Q_a} = - \frac{1}{2} \int \partial_0 j_a{}^0 (where i hope my indices are correct since i want to contract over the 0 index and \partial_0 = \frac{\partial}{\partial t}

but \partial_\mu j_a{}^\mu = 0 \Rightarrow \partial_0 j_a{}^0 = - \partial_i j_a{}^i

giving \dot{Q_a} = \frac{1}{2} \int \partial_i j_a{}^i

how does that go to 0 though?

It's an integral of a total derivative, which will vanish by Stokes' theorem as long as the fields \phi are "well-behaved." For smooth solutions like plane waves, etc. it holds.

 

[latentcorpse]

In the Minkowski metric there's a minus sign between j_{a0} = - j_a^{0}. Put the index wherever you find convenient, as I think I've made a couple of sign errors already because of it.



It's an integral of a total derivative, which will vanish by Stokes' theorem as long as the fields \phi are "well-behaved." For smooth solutions like plane waves, etc. it holds.

ok. so we got to \dot{Q_a} = \int \vec{\nabla} \cdot \vec{j_a} d^3x
how is this zero by stokes? stokes is to do with the curl of a vector, no?

sorry if I am being a total idiot here!

 

[fzero]

ok. so we got to \dot{Q_a} = \int \vec{\nabla} \cdot \vec{j_a} d^3x
how is this zero by stokes? stokes is to do with the curl of a vector, no?

sorry if I am being a total idiot here!

The curl theorem is one form of a more general theorem that also includes the Gauss or divergence theorem:

\int_\Sigma \nabla \cdot \vec{F} ~dV = \oint_{\partial \Sigma} \vec{F}\cdot d\vec{S}.

The surface integral vanishes because we're in R^3 and the boundary vanishes.

 

[latentcorpse]

The curl theorem is one form of a more general theorem that also includes the Gauss or divergence theorem:

\int_\Sigma \nabla \cdot \vec{F} ~dV = \oint_{\partial \Sigma} \vec{F}\cdot d\vec{S}.

The surface integral vanishes because we're in R^3 and the boundary vanishes.

ok. but how do we know the boundary vanishes?

 

[fzero]

ok. but how do we know the boundary vanishes?

Like I said, \Sigma = R^3. It's also important that the integrand is generally well-defined.

 

[latentcorpse]

Like I said, \Sigma = R^3. It's also important that the integrand is generally well-defined.

so it's basically because there is no boundary to \mathbb{R}^3?

and then for the very last bit, i have

Q_a = \int d^3x \epsilon_{abc} \dot{\phi_b} \phi_c

and the Euler Lagrange eqns give an EOM \partial^\mu \phi_a = 2 m^2 \phi_a

so am i trying to show that Q_a is going to be zero? that's what is osunds like when you were talking about using terms that vanish because of the EOM or becuase they are total derivatives, but surely if Q_a is conserved, i just have to show that it is a constant?

either way, I'm struggling to get anything productive from subbing from the EOM into my equation - what would you recommend substituting and where?

thanks.

 

[fzero]

so it's basically because there is no boundary to \mathbb{R}^3?

and then for the very last bit, i have

Q_a = \int d^3x \epsilon_{abc} \dot{\phi_b} \phi_c

and the Euler Lagrange eqns give an EOM \partial^\mu \phi_a = 2 m^2 \phi_a

The EOM should be \partial_\mu\partial^\mu \phi_a = m^2 \phi_a (up to signs depending on the signature of your metric), since it's just the statement that p^2 = m^2.

so am i trying to show that Q_a is going to be zero? that's what is osunds like when you were talking about using terms that vanish because of the EOM or becuase they are total derivatives, but surely if Q_a is conserved, i just have to show that it is a constant?

You want to show that \dot{Q}_a=0.

either way, I'm struggling to get anything productive from subbing from the EOM into my equation - what would you recommend substituting and where?

thanks.

When you compute \dot{Q}_a=0, you will get a term in the integrand that depends on \ddot{\phi_b}. You will use the EOM to rewrite this term. After integrating a new term by parts, you'll find that everything vanishes for one reason or another.

 

[latentcorpse]

The EOM should be \partial_\mu\partial^\mu \phi_a = m^2 \phi_a (up to signs depending on the signature of your metric), since it's just the statement that p^2 = m^2.
You want to show that \dot{Q}_a=0.
When you compute \dot{Q}_a=0, you will get a term in the integrand that depends on \ddot{\phi_b}. You will use the EOM to rewrite this term. After integrating a new term by parts, you'll find that everything vanishes for one reason or another.

hmmmm. so i find that

\partial_\mu \left( \frac{ \partial L}{ \partial{ ( \partial_\mu \phi_a ) }} \right) = \frac{\partial L}{ \partial \phi_a} \Rightarrow \frac{1}{2} \partial_\mu \partial^\mu \phi_a = - m^2 \phi_a

(note that i have a factor of 2 kicking about that you didn't?)

giving
\ddot{\phi_a} - \nabla^2 \phi_a = - 2 m^2 \phi_a

and then

\dot{Q_a} = \int d^3x \epsilon_{abc} \left( \ddot{\phi_b} \phi_c + \dot{\phi_b} \dot{\phi_c} \right) = \int d^3x \epsilon_{abc} \left( \left( \nabla^2 \phi_b - 2 m^2 \phi_b \right) \phi_c + \dot{\phi_b} \dot{\phi_c} \right)

i get the feeling this is going massively off track as there's no obvious cancellations...

 

[fzero]

hmmmm. so i find that

\partial_\mu \left( \frac{ \partial L}{ \partial{ ( \partial_\mu \phi_a ) }} \right) = \frac{\partial L}{ \partial \phi_a} \Rightarrow \frac{1}{2} \partial_\mu \partial^\mu \phi_a = - m^2 \phi_a

(note that i have a factor of 2 kicking about that you didn't?)

Both terms in the Lagrangian are morally sums of squares, so they carry the same factors of two after taking the derivative.

giving
\ddot{\phi_a} - \nabla^2 \phi_a = - 2 m^2 \phi_a

and then

\dot{Q_a} = \int d^3x \epsilon_{abc} \left( \ddot{\phi_b} \phi_c + \dot{\phi_b} \dot{\phi_c} \right) = \int d^3x \epsilon_{abc} \left( \left( \nabla^2 \phi_b - 2 m^2 \phi_b \right) \phi_c + \dot{\phi_b} \dot{\phi_c} \right)

i get the feeling this is going massively off track as there's no obvious cancellations...

The derivative term can be integrated by parts. Use symmetry considerations to deal with the other terms.

 

[latentcorpse]

Both terms in the Lagrangian are morally sums of squares, so they carry the same factors of two after taking the derivative.

but surely not?

L=\frac{1}{2} \partial_\mu \phi_a \partial^\mu \phi_a - \frac{1}{2}m^2 \phi_a{}^2

so we get

\frac{\partial L}{\partial \phi_a} = -\frac{1}{2} m^2 2 \phi_a = - m^2 \phi_a

and

\frac{\partial L}{ \partial ( \partial_\mu \phi_a ) } = \frac{1}{2} \partial^\mu \phi_a
and then
\partial_\mu \left( \frac{\partial L}{ \partial ( \partial_\mu \phi_a ) } \right) = \frac{1}{2} \partial_\mu \partial^\mu \phi_a

no? what am i doing wrong here?

The derivative term can be integrated by parts. Use symmetry considerations to deal with the other terms.

ok. so i assume u mean the nabla term by the derivative term. we can use divergence theorem on that to give

\int d^3x \nabla^2 \phi_b = \int d^3x \vec{\nabla \phi_b} \cdot d \Sigma
and I am assuming this is going to be zero again for the same reason as the last surface integral was - because we are in \mathbb{R}^3 and so d \Sigma vanishes, correct?

this leaves me with \dot{Q_a} = \int d^3x \epsilon_{abc} \left( \dot{\phi_b} \dot{\phi_c} - 2m^2 \phi_b \right)
can i use the EOM on that? my EOM uses 4-derivatives instead of time derivatives though?

 

[fzero]

but surely not?

\frac{\partial L}{ \partial ( \partial_\mu \phi_a ) } = \frac{1}{2} \partial^\mu \phi_a
and then
\partial_\mu \left( \frac{\partial L}{ \partial ( \partial_\mu \phi_a ) } \right) = \frac{1}{2} \partial_\mu \partial^\mu \phi_a

no? what am i doing wrong here?

\partial^\mu \phi_a is not independent of \partial_\mu \phi_a, so you get the same factor of 2 you did in the mass term.

ok. so i assume u mean the nabla term by the derivative term. we can use divergence theorem on that to give

\int d^3x \nabla^2 \phi_b = \int d^3x \vec{\nabla \phi_b} \cdot d \Sigma
and I am assuming this is going to be zero again for the same reason as the last surface integral was - because we are in \mathbb{R}^3 and so d \Sigma vanishes, correct?

You left out part of the integrand, so you're missing an extra term after integrating by parts.

this leaves me with \dot{Q_a} = \int d^3x \epsilon_{abc} \left( \dot{\phi_b} \dot{\phi_c} - 2m^2 \phi_b \right)
can i use the EOM on that? my EOM uses 4-derivatives instead of time derivatives though?

You're missing a \phi_c in the second term. I already said that you need to consider the symmetry of these terms, not the EOM.

 

[latentcorpse]

\partial^\mu \phi_a is not independent of \partial_\mu \phi_a, so you get the same factor of 2 you did in the mass term.

how does this work?

You left out part of the integrand, so you're missing an extra term after integrating by parts.

You're missing a \phi_c in the second term. I already said that you need to consider the symmetry of these terms, not the EOM.

so i get
\dot{Q_a} = \int d^3x \epsilon_{abc} ( \ddot{\phi_b} \phi_c + \dot{\phi_b} \dot{\phi_c} )
we know:
\ddot{\phi_a}=\nabla^2+m^2 \phi_a
giving:
\dot{Q_a} = \int d^3x \epsilon_{abc} \left( \nabla^2 \phi_b \phi_c + m^2 \phi_b \phi_c + \dot{\phi_b} \dot{\phi_c}

but I am not sure if i can use the divergence theorem to get rid of the \nabla^2 term because i now have a \phi_c multiplying it and i can't take that outside the integral, can i?

assuming i get rid of that term, I'm left with:
\dot{Q_a}=\int d^3x \epsilon_{abc} \left( \dot{\phi_b} \dot{\phi_c} + m^2 \phi_b \phi_c \right) = \int d^3x \epsilon_{acb} \left( \dot{\phi_b} \dot{\phi_c} + m^2 \phi_b \phi_c \right) =-\int d^3x \epsilon_{abc} \left( \dot{\phi_b} \dot{\phi_c} + m^2 \phi_b \phi_c \right)
by relabelling b to c
but then we deduce
\dot{Q_a} = - \dot{Q_a} \Rightarrow \dot{Q_a}=0 as required!

thanks!

 

[fzero]

how does this work?

If you can't make sense of it in index form, just expand into components:

\partial_\mu \phi_a \partial^\mu \phi_a = (\dot{\phi}_a)^2 - (\nabla\phi_a)^2.

so i get
\dot{Q_a} = \int d^3x \epsilon_{abc} ( \ddot{\phi_b} \phi_c + \dot{\phi_b} \dot{\phi_c} )
we know:
\ddot{\phi_a}=\nabla^2+m^2 \phi_a
giving:
\dot{Q_a} = \int d^3x \epsilon_{abc} \left( \nabla^2 \phi_b \phi_c + m^2 \phi_b \phi_c + \dot{\phi_b} \dot{\phi_c}

but I am not sure if i can use the divergence theorem to get rid of the \nabla^2 term because i now have a \phi_c multiplying it and i can't take that outside the integral, can i?

Remember that

\nabla(fg) = (\nabla f)g + f (\nabla g)

so you will get two terms when you rearrange the derivative.

 

[latentcorpse]

If you can't make sense of it in index form, just expand into components:

\partial_\mu \phi_a \partial^\mu \phi_a = (\dot{\phi}_a)^2 - (\nabla\phi_a)^2.



Remember that

\nabla(fg) = (\nabla f)g + f (\nabla g)

so you will get two terms when you rearrange the derivative.

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?