Lagrangian Invariant Under Transformation

RHS is evaluated to FIRST order in \delta\phi_a. In the current form, the expression for L has a lot of second order terms in \theta that you wouldn't have if you had expanded the fields to first order. If you were expanding the fields to second order, then you'd have those terms, but you'd also have to include all the terms from expanding the first order variations of the fields to second order.So you have a couple options: you can either compute the first order variation of the fields and then expand to second order, or you can do the expansion as you have it, but then you have to expand the derivatives on the fields to second order. In either case you have
  • #1
latentcorpse
1,444
0
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.
 
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  • #2
latentcorpse said:
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.
 
  • #3
fzero said:
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?
 
  • #4
latentcorpse said:
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.
 
  • #5
fzero said:
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!
 
  • #6
latentcorpse said:
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.
 
  • #7
fzero said:
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 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 [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!
 
  • #8
latentcorpse said:
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 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 [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.
 
  • #9
fzero said:
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?
 
  • #10
latentcorpse said:
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]?
 
  • #11
fzero said:
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.
 
  • #12
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.
 
  • #13
latentcorpse said:
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.
 
  • #14
vela said:
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?
 
  • #15
fzero said:
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.
 
  • #16
latentcorpse said:
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.

latentcorpse said:
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.
 
  • #17
fzero said:
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
 
  • #18
latentcorpse said:
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.
 
  • #19
fzero said:
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?
 
  • #20
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?
 
Last edited:
  • #21
latentcorpse said:
[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?

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

[tex]\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 ).[/tex]

latentcorpse said:
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?

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)

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

where I've stripped off the parameters of the transformation so that the Noether current(s) [tex]j_a^\mu[/tex] do not depend on them. Incidentally, I was sloppy about this in my previous expression for the Noether current, where I pulled out [tex]\theta[/tex] but not [tex]n_a[/tex]. Since we are dealing with an [tex]SO(3)[/tex] rotation, there are 3 currents since the rank of SO(3) is 3.
however this formula
 
Last edited:
  • #22
fzero said:
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)

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

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

[itex]L=\frac{1}{2} \partial_\mu \phi_a \partial^\mu \phi_a + \frac{1}{2} m^2 \phi_a{}^2[/itex]

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

and so using your formula we get

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

am i allowed to rearrange for j by dividing by theta and n_a. surely not as n_a is a vector?
 
  • #23
latentcorpse said:
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


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

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

[itex]L=\frac{1}{2} \partial_\mu \phi_a \partial^\mu \phi_a + \frac{1}{2} m^2 \phi_a{}^2[/itex]

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

and so using your formula we get

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

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 [tex]n_a[/tex], the RHS side will be proportional to [tex]n_a[/tex] when you put in the form of [tex]\delta\phi_a[/tex] (after redefining dummy indicies). You identify the current with the rest of the expression.
 
  • #24
fzero said:
There was a typo, I meant to write


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

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 [tex]n_a[/tex], the RHS side will be proportional to [tex]n_a[/tex] when you put in the form of [tex]\delta\phi_a[/tex] (after redefining dummy indicies). You identify the current with the rest of the expression.



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

so that would mean

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

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

and so [itex]j_a{}^\mu = \frac{1}{2} \epsilon_{abc} \phi_c \partial^\mu \phi_a[/itex]
 
  • #25
latentcorpse said:
ok so [itex]\delta \phi_a = \theta \epsilon_{abc} n_b \phi_c[/itex] presumably since [itex]\phi_a \rightarrow \phi_a + \delta \phi_a[/itex]

so that would mean

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

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

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

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.
 
  • #26
fzero said:
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 [itex]\delta \phi_a[/itex] then my indices will match up on the n's but my indices won't match up in my expression for [itex]\delta \phi_a[/itex] as i'll have:

[itex]\delta \phi_a = \theta \epsilon_{bac} n_a \phi_c[/itex] 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

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

how does that look now?
 
  • #27
latentcorpse said:
yes but say i relabel a to b in my epsression for [itex]\delta \phi_a[/itex] then my indices will match up on the n's but my indices won't match up in my expression for [itex]\delta \phi_a[/itex] as i'll have:

[itex]\delta \phi_a = \theta \epsilon_{bac} n_a \phi_c[/itex] 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

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

how does that look now?

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


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

and relabel the index on the LHS:

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

Reorder the terms a bit to show the structure

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

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

[itex]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 ,[/itex]

This is the same as

[itex]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 .[/itex]

where in the last part we have just moved the derivative term to the right for aesthetics and relabeled the indicies [tex]b\leftrightarrow c[/tex] 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.
 
  • #28
fzero said:
I'm not sure why you have [tex]\partial^\mu \theta[/tex] appearing in the expression now. Take the first line


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

and relabel the index on the LHS:

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

Reorder the terms a bit to show the structure

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

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

[itex]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 ,[/itex]

This is the same as

[itex]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 .[/itex]

where in the last part we have just moved the derivative term to the right for aesthetics and relabeled the indicies [tex]b\leftrightarrow c[/tex] 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?
 
  • #29
latentcorpse said:
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 [tex]\partial^\mu[/tex] on the [tex]\phi_b[/tex].
 
  • #30
fzero said:
Yes they agree if you put the [tex]\partial^\mu[/tex] on the [tex]\phi_b[/tex].

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

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

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

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

anyway so i have the EOMs but i first need to deduce that the [itex]Q_a[/itex] 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.
 
  • #31
latentcorpse said:
the final thing i have to do is to deduce that the three quantities

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

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

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

anyway so i have the EOMs but i first need to deduce that the [itex]Q_a[/itex] 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

[itex]\partial_\mu (\frac{\partial L}{\partial (\partial_\mu {\phi_a})})=\frac{\partial L}{\partial \phi_a}.[/itex]

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 [tex]Q_a[/tex] are conserved, note that

[tex]Q_a = \int d^3 x j_{a 0}.[/tex]

Try to relate [tex]\dot{Q}_a[/tex] to the conservation of the Noether currents.
 
  • #32
fzero said:
Those are the Euler-Lagrange equations for a 1d theory, in a field theory the EL equations are

[itex]\partial_\mu (\frac{\partial L}{\partial (\partial_\mu {\phi_a})})=\frac{\partial L}{\partial \phi_a}.[/itex]

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 [tex]Q_a[/tex] are conserved, note that

[tex]Q_a = \int d^3 x j_{a 0}.[/tex]

Try to relate [tex]\dot{Q}_a[/tex] to the conservation of the Noether currents.

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

Now the Noether current [itex]j_a{}^\mu[/itex] will be conserved by definition. This means that [itex]Q_a[/itex] 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:

[itex]\partial_\mu \partial^\mu \phi_a + m^2 \phi_a = 0 [/itex]
 
  • #33
latentcorpse said:
ok. i find that [itex]Q_a = \int d^3x \epsilon_{abc} \dot{\phi_b} \phi_c = - \frac{1}{2} \int j_a{}^0[/itex]

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

Now the Noether current [itex]j_a{}^\mu[/itex] will be conserved by definition. This means that [itex]Q_a[/itex] 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:

[itex]\partial_\mu \partial^\mu \phi_a + m^2 \phi_a = 0 [/itex]

If the relative signs are correct, that's the right equation of motion. Did you use that to show that [tex]\dot{Q}_a=0[/tex]?
 
  • #34
fzero said:
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?

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

ok. i gave this a shot:

one of the 4 eqns looks like [itex]\ddot{\phi_a}+m^2 \phi_a = 0 \Rightarrow \phi_a = A \cos{mt} + B \sin{mt}[/itex]
i was then going to get an expression for [itex]\dot{\phi_a}[/itex] and plug that back into the expression for [itex]Q_a[/itex] but I don't think this is right since i have no indices in my expression for [itex]\phi_a[/itex].
 
  • #35
latentcorpse said:
ok. i have no idea how to do this. any advice?

Write down an expression for [tex]\dot{Q}_a[/tex] in terms of [tex] j_{a0}[/tex] and use [tex]\partial^\mu j_{a\mu}=0[/tex] 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 [itex]\ddot{\phi_a}+m^2 \phi_a = 0 \Rightarrow \phi_a = A \cos{mt} + B \sin{mt}[/itex]
i was then going to get an expression for [itex]\dot{\phi_a}[/itex] and plug that back into the expression for [itex]Q_a[/itex] but I don't think this is right since i have no indices in my expression for [itex]\phi_a[/itex].

Here you want to use the expression for the [tex]Q_a[/tex] in terms of the [tex]\phi_a[/tex] and relate the expressions to terms that either vanish because of the equation of motion or because they're total derivatives.
 
<h2>1. What is a Lagrangian invariant under transformation?</h2><p>A Lagrangian invariant under transformation is a physical quantity that remains unchanged when the coordinates used to describe the system are transformed. This means that the equations of motion derived from the Lagrangian will still be valid, regardless of the coordinate system used.</p><h2>2. Why is it important for a Lagrangian to be invariant under transformation?</h2><p>It is important for a Lagrangian to be invariant under transformation because it ensures that the physical laws and equations of motion derived from it will be valid in any coordinate system. This allows for a more comprehensive and accurate understanding of the system being studied.</p><h2>3. How is the Lagrangian invariant under transformation calculated?</h2><p>The Lagrangian invariant under transformation is calculated by using a mathematical tool called the Euler-Lagrange equation. This equation takes into account the transformation of coordinates and ensures that the Lagrangian remains unchanged.</p><h2>4. Can the Lagrangian be invariant under all types of transformations?</h2><p>No, the Lagrangian can only be invariant under certain types of transformations. These include translations, rotations, and Lorentz transformations. Other types of transformations may result in a different Lagrangian, which would lead to different physical laws and equations of motion.</p><h2>5. How does the concept of invariance under transformation relate to symmetries in physics?</h2><p>Invariance under transformation is closely related to symmetries in physics. A symmetry is a transformation that leaves a physical system unchanged. When a Lagrangian is invariant under a particular transformation, it means that the system has a symmetry under that transformation. This connection between invariance and symmetry allows for the use of powerful mathematical techniques, such as Noether's theorem, to derive important physical laws and conservation principles.</p>

1. What is a Lagrangian invariant under transformation?

A Lagrangian invariant under transformation is a physical quantity that remains unchanged when the coordinates used to describe the system are transformed. This means that the equations of motion derived from the Lagrangian will still be valid, regardless of the coordinate system used.

2. Why is it important for a Lagrangian to be invariant under transformation?

It is important for a Lagrangian to be invariant under transformation because it ensures that the physical laws and equations of motion derived from it will be valid in any coordinate system. This allows for a more comprehensive and accurate understanding of the system being studied.

3. How is the Lagrangian invariant under transformation calculated?

The Lagrangian invariant under transformation is calculated by using a mathematical tool called the Euler-Lagrange equation. This equation takes into account the transformation of coordinates and ensures that the Lagrangian remains unchanged.

4. Can the Lagrangian be invariant under all types of transformations?

No, the Lagrangian can only be invariant under certain types of transformations. These include translations, rotations, and Lorentz transformations. Other types of transformations may result in a different Lagrangian, which would lead to different physical laws and equations of motion.

5. How does the concept of invariance under transformation relate to symmetries in physics?

Invariance under transformation is closely related to symmetries in physics. A symmetry is a transformation that leaves a physical system unchanged. When a Lagrangian is invariant under a particular transformation, it means that the system has a symmetry under that transformation. This connection between invariance and symmetry allows for the use of powerful mathematical techniques, such as Noether's theorem, to derive important physical laws and conservation principles.

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