=0 (4.10)
In the next section we will derive, using the Lagrangian formalism, one of the most important theorems of modern physics. We need this theorem to see the deep connection between symmetries and conserved quantities. The conserved quantities are the appropriate
quantities20to describe nature and from this theorem we learn how 20We can see them as anchors in an otherwise extremely complicated world. While everything changes, the conserved quantities stay the same.
we can work with them in a theoretical context.
4.5 Noether’s Theorem
Noether’s theorem shows that each symmetry of the Lagrangian is directly related to one conserved quantity. Or formulated slightly different: The notions physicists commonly use to describe nature (the conserved quantities) are directly connected to symmetries.
This certainly is one of the most beautiful insights in the history of science.
4.5.1 Noether’s Theorem for Particle Theories
Let’s have a look at what we can say about conserved quantities in particle theories. We will restrict to continuous symmetries, because then we can look at infinitesimal changes. As we have seen in earlier chapters, we can built up finite changes by repetition of infinitesimal changes. The invariance of the Lagrangian under an infinitesimal
transformation21q→q=q+δq can be expressed mathematically 21The symbol for a small variationδ may not be confused with the symbol for partial derivatives∂.
δL = L
q,dq
dt, t
− L
q+δq,d(q+δq) dt , t
= L
q,dq
dt, t
− L
q+δq,dq dt+ dδq
dt , t
=! 0. (4.11)
Demanding that the Lagrangian is invariant can be too restrictive.
What really needs to be invariant for the dynamics to stay the same is the action and not the Lagrangian. Of course, if the Lagrangian is invariant, the action is automatically invariant:
δS= dtL How can the Lagrangian change, while the action stays the same? It turns out we can always add the total time derivative of an arbitrary function G to the Lagrangian
L → L + dG dt without changing the action because22
22UsingδG= ∂G∂qδq, because G=G(q) and we vary q. Therefore the variation of G is given by the rate of change∂G∂q
In the last step, we use that the variationδq vanishes at the initial and final moments of time (t1, t2). We conclude, there is no need for us to demand that the variation of the LagrangianδLvanishes, rather we have a less restrictive condition
δL=! dG
dt. (4.13)
This means the Lagrangian can change without changing the ac-tion and therefore the equaac-tion of moac-tion, as long as the change can be written as total derivative of some function dGdt. Rewriting Eq. 4.11, now with dGdt instead of 0 on the right hand side yields
δL = L
We expand the second term as Taylor series, keep only terms of first order inδq, because δq is infinitesimal and use the notation dqdt = ˙q:
→δL = L − L −∂L
We can rewrite Eq. 4.15 by using the Euler-Lagrange equation23. This
23Eq. 4.7: ∂∂qL =dtd ∂∂ ˙qL
This can be rewritten, using the product rule24 24If you’re unsure about the product rule, have a look at appendix B.1.
→δL = −d dt
∂L
∂ ˙qδq
= dG dt
→ d dt
∂L
∂ ˙qδq+G
≡J
=0. (4.16)
Therefore, we have found a quantity J that is conserved in time:
J= ∂L
∂ ˙qδq+G, (4.17)
because we have
d
dtJ=0→J =const.
To illustrate this, we will use one later result from Sec. 10.2:
New-ton’s second law for a free particle with constant mass is25 25If q denotes the position of some ob-ject,dqdt = ˙q is the velocity of the object and dtddqdt =ddt22q =¨q the acceleration.
m ¨q=0. (4.18)
The Lagrangian that reproduces this famous equation of motion is26 26We work now with more than one spatial dimension, which means instead of q and a, we use vectorsq anda.
L = 1
2m ˙q2 (4.19)
as you can check, by putting it into the Euler-Lagrange equation (Eq. 4.7).
Let us compute for this Lagrangian for different symmetries the corresponding conserved quantities.
Our Lagrangian is invariant (δL = 0) under spatial translations
q → q = q+a, wherea denotes some constant vector, because L = 12m ˙q2does not depend onq. The corresponding conserved quantity reads, with G = 0, which certainly fulfils the condition Eq. 4.13
Jtrans = ∂L
∂ ˙qa=m ˙qa= pa, (4.20) wherep = m ˙q is what we usually call momentum in classical mechanics. The equation dtdJ = 0 holds for arbitrarya and therefore the momentum is conserved, because the Lagrangian is invariant under spatial translations.
We now want to look at rotations and therefore need more than one dimension, because a rotation in one dimension makes no sense.
Instead of working with vectors likeq, we can use an index nota-tion, which is quite useful here. We can simply rewrite all equations
with q → qi. Let’s now have a look at an infinitesimal rotation27 27Here we write the generator of rotations by using the Levi-Civita symbol. This was explained in the text below Eq. 3.63.
qi → qi =qi+ ijkqjakand thereforeδqi= ijkqjak. Our Lagrangian is
invariant under such transformations, because again,L = 12m ˙q2does not depend on q, and the corresponding conserved quantity is28
28This was derived in Eq. 4.16, again we work here with G=0. In the last step we rewrite our term in vector notation, where×is called the cross product andL is what we usually call angular mo-mentum in classical mechanics. Therefore, invariance under rota-tions leads us to conservation of angular momentum.
Next let’s have a look at invariance under time translations29. An
29This means that physics doesn’t care about if we perform an experiment yesterday, today or in 50 years, given the same initial conditions, the physical laws stay the same.
infinitesimal time displacement t→t=t+ has the effect δL = L
The left hand side is exactly the total derivative
−dL
dt , (4.22) which tells us that in generalδL =0, but the condition in Eq. 4.13 is fulfilled anyway with G= −L.
We can put this into Eq. 4.16, which yields d
The conserved quantityHis called the Hamiltonian and represents the total energy of the system. For our example Lagrangian, we have
H = ∂L
which is exactly the kinetic energy of our system. The kinetic energy is the total energy, because we worked without a potential/external force. The Lagrangian that leads to Newton’s second law for a parti-cle in an external potential: m ¨q= dVdq is
which is the correct total energy=kinetic energy+potential energy.
The conserved quantity that follows from boost invariance is rather strange and the corresponding computation can be found in the appendix Sec. 4.6. The resulting conserved quantity is
˜Jboost =pt−1 2mvt
≡ ˜pt
−mq= ˜pt−mq. (4.25)
We can see that this quantity depends on the starting time and by choosing the starting time appropriately we can make it zero. Be-cause this quantity is conserved, this conservation law tells us that zero stays zero for all times.
To summarize for particle theories we have the following connec-tions:
• Translational invariance in space⇒conservation of momentum
• Boost invariance30⇒conservation of ˜pt−mq 30Another name for a boost is transla-tion in momentum space, because the transformation q→q+vt, changes the momentum to m ˙q→m(˙q+v).
• Rotational invariance⇒conservation of angular momentum
• Translational invariance in time⇒conservation of energy
Noether’s theorem shows us why those notions31are used in every 31Except for the conserved quantity following from boost invariance.
physical theory of nature in one or another form. As long as we have the usual spacetime symmetries of our physical laws we have momentum, energy and angular momentum as conserved quantities.
It is instructive to have a look at how those notions occur in field theories. For field theories we have two kinds of symmetries. On the one hand, our Lagrangian can be invariant under transformations of spacetime, which means a transformation like a rotation. On the other hand, we can have invariance under transformations of the field itself, which are called internal symmetries.