Lagrangian Invariant Under Transformation

[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 because 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 because 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?

 

[fzero]

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

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

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

is this looking correct?

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

 

[latentcorpse]

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

ok so this line is correct:
$( \vec{\nabla} \phi_b ) \phi_c = \vec{\nabla} ( \phi_b \phi_c ) - \phi_b \vec{\nabla} \phi_c$

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

 

[fzero]

ok so this line is correct:
$( \vec{\nabla} \phi_b ) \phi_c = \vec{\nabla} ( \phi_b \phi_c ) - \phi_b \vec{\nabla} \phi_c$

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

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

 

[latentcorpse]

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

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

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

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

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

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

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

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

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

thanks!

 

[fzero]

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

 

[latentcorpse]

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

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

 

[fzero]

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

I suggested that a day or so ago...

 

[latentcorpse]

I suggested that a day or so ago...

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