Lagrangian Invariant Under Transformation

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

[itex]\phi_a \rightarrow \phi_a + \theta \epsilon_{abc} n_b \phi_c[/itex]

plugging this in i get:

[itex]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)[/itex]

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

thanks.
 

fzero

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

[itex]\phi_a \rightarrow \phi_a + \theta \epsilon_{abc} n_b \phi_c[/itex]

plugging this in i get:

[itex]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)[/itex]

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,

[itex]\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. [/itex]

Now you want to use an identity for contraction of [tex]\epsilon[/tex] symbols:

[tex]\epsilon_{abc} \epsilon_{ade} = \delta_{bd}\delta_{ce} - \delta_{be} \delta_{cd}.[/tex]

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

[itex] 2 \phi_a \theta \epsilon_{abc} n_b \phi_c .[/itex]

All terms in the expansion can be evaluated using one or both of these ideas.
 
Some of these terms are going to be too confusing if you use the same indices too many times. For instance,

[itex]\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. [/itex]

Now you want to use an identity for contraction of [tex]\epsilon[/tex] symbols:

[tex]\epsilon_{abc} \epsilon_{ade} = \delta_{bd}\delta_{ce} - \delta_{be} \delta_{cd}.[/tex]

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

[itex] 2 \phi_a \theta \epsilon_{abc} n_b \phi_c .[/itex]

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

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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

[itex]
2 \phi_a \theta \epsilon_{abc} n_b \phi_c = \theta \epsilon_{abc} n_b (\phi_a \phi_c+ \phi_c\phi_a)
[/itex]

and it should be obvious.
 
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

[itex]
2 \phi_a \theta \epsilon_{abc} n_b \phi_c = \theta \epsilon_{abc} n_b (\phi_a \phi_c+ \phi_c\phi_a)
[/itex]

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

[itex]( \phi_a{} + \theta \epsilon_{abc} n_b \phi_c )( \phi_a{} + \theta \epsilon_{abc} n_b \phi_c )[/itex]
[itex]= \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[/itex]
[itex]= \phi_a{}^2 + 0 + \theta^2 ( n_b{}^2 \phi_c{}^2 - n_b n_c \phi_c \phi_b )[/itex]

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 [itex]\phi_a{}^2[/itex] 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

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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

[itex]( \phi_a{} + \theta \epsilon_{abc} n_b \phi_c )( \phi_a{} + \theta \epsilon_{abc} n_b \phi_c )[/itex]
[itex]= \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[/itex]
[itex]= \phi_a{}^2 + 0 + \theta^2 ( n_b{}^2 \phi_c{}^2 - n_b n_c \phi_c \phi_b )[/itex]

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 [itex]\phi_a{}^2[/itex] 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 [tex]O(\theta^2)[/tex] 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

[tex]L(\phi_a+\delta\phi_a) = L(\phi_a) + \frac{\delta L}{\delta\phi_a} \delta\phi_a + O((\delta\phi_a)^2) [/tex]

and show that [tex]\delta L/\delta\phi_a[/tex], 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.
 
Oh I should have realized you'd run into trouble. The problem is that this is an [tex]O(\theta^2)[/tex] 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

[tex]L(\phi_a+\delta\phi_a) = L(\phi_a) + \frac{\delta L}{\delta\phi_a} \delta\phi_a + O((\delta\phi_a)^2) [/tex]

and show that [tex]\delta L/\delta\phi_a[/tex], 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):

[itex] \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 [/itex]

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

[itex]\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[/itex]

not realy 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 [itex]\partial \phi_a[/itex] 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

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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):

[itex] \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 [/itex]

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

[itex]\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[/itex]

not realy 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 [itex]\partial \phi_a[/itex] 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 [tex]\theta[/tex] is a constant, what's called a "global" transformation. As you can see, the Lagrangian is not invariant under "local" transformations where [tex]\theta(x)[/tex] 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.
 
We're assuming that [tex]\theta[/tex] is a constant, what's called a "global" transformation. As you can see, the Lagrangian is not invariant under "local" transformations where [tex]\theta(x)[/tex] 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

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ok. but how do i show those two terms with the derivatives in them cancel out?
If [tex]\theta[/tex] is a constant, what is [tex]\partial_\mu \theta[/tex]?
 
If [tex]\theta[/tex] is a constant, what is [tex]\partial_\mu \theta[/tex]?
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 [itex]j^\mu[/itex]. We have actually spent a couple of lectures on these but I don't really grasp how to go about calculating them.

Thanks.
 

vela

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It's not that obvious.

[tex]\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)[/tex]

You're not just differentiating the θ.
 

fzero

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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 [itex]j^\mu[/itex]. 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

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

If we can write this term as

[tex] \int d^4x ~ \theta ~ \partial_\mu j^\mu ,[/tex]

then we conclude that [tex]j^\mu[/tex] is the conserved Noether current.
 
It's not that obvious.

[tex]\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)[/tex]

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:

[itex]\theta ( \partial_\mu \epsilon_{abc} ) n_b \phi_c + \theta \epsilon_{abc} n_b ( \partial_\mu \phi_c )[/itex]

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

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

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

If we can write this term as

[tex] \int d^4x ~ \theta ~ \partial_\mu j^\mu ,[/tex]

then we conclude that [tex]j^\mu[/tex] 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 [itex]\theta \partial_\mu j^\mu[/itex] then [itex]j^\mu[/itex] 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

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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:

[itex]\theta ( \partial_\mu \epsilon_{abc} ) n_b \phi_c + \theta \epsilon_{abc} n_b ( \partial_\mu \phi_c )[/itex]

as teh derivative of [itex]\theta[/itex] is zero as its just a constant and the derivative of [itex]n_b[/itex] 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


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

The first term vanishes because [tex]\epsilon_{abc}[/tex] 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 [tex]\delta x^\mu[/tex] or if the fields are not scalar fields.

anyway, if i can write L in the form [itex]\theta \partial_\mu j^\mu[/itex] then [itex]j^\mu[/itex] 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

[tex]\delta L = L(\phi +\delta \phi) - L(\phi)[/tex]

If L is truly invariant under [tex]\delta\phi[/tex], the variation of the action will have the stated form.
 
You have to include the whole expression, not just part of it. Look at


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

The first term vanishes because [tex]\epsilon_{abc}[/tex] is a constant.
right. i see that. but the 2nd one remains surely as [itex]\partial^\mu \phi_a , \partial_\mu \phi_c[/itex] aren't necessarily zero
 

fzero

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right. i see that. but the 2nd one remains surely as [itex]\partial^\mu \phi_a , \partial_\mu \phi_c[/itex] aren't necessarily zero
You need to consider the symmetries as well.
 
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 [itex]\partial^\mu[/itex] and [itex]\partial_\mu[/itex] by accident or deliberately?


i get

[itex]\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 )[/itex]

[itex]= \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)[/itex]

now hopefully it's correct up to there

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

the 2nd on gives [itex]\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&\mu \phi_c[/itex]

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

[itex]\partial_\mu \phi_a \theta \epsilon_{abc} n_b \partial&\mu \phi_c[/itex]

and the final term in the original expression gives [itex]\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[/itex]

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

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

anyway this gives

[itex] (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[/itex]

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