Categories: Mathematics, Physics.

Noether’s theorem

Consider the following general functional J[f]J[f], where LL is a known Lagrangian density, f(x,t)f(x, t) is an unknown function, and fxf_x and ftf_t are its first-order derivatives:

J[f]=(x0,t0)(x1,t1)L(f,fx,ft,x,t)dxdt\begin{aligned} J[f] = \iint_{(x_0, t_0)}^{(x_1, t_1)} L(f, f_x, f_t, x, t) \dd{x} \dd{t} \end{aligned}

Then the calculus of variations states that the ff which minimizes or maximizes J[f]J[f] can be found by solving this Euler-Lagrange equation:

0=Lfddx(Lfx)ddt(Lft)\begin{aligned} 0 = \pdv{L}{f} - \dv{}{x} \Big( \pdv{L}{f_x} \Big) - \dv{}{t} \Big( \pdv{L}{f_t} \Big) \end{aligned}

Now, the first steps are similar to the derivation of the Beltrami identity (which is a special case of Noether’s theorem): we need to find relations between the explicit dependence of LL on the free variables (x,t)(x, t), and its implicit dependence via ff, fxf_x and ftf_t.

Let us start with xx. The “hard” (explicit + implicit) derivative dL/dx\idv{L}{x} is given by the chain rule:

dLdx=Lfdfdx+Lfxdfxdx+Lftdftdx+Lx\begin{aligned} \dv{L}{x} &= \pdv{L}{f} \dv{f}{x} + \pdv{L}{f_x} \dv{f_x}{x} + \pdv{L}{f_t} \dv{f_t}{x} + \pdv{L}{x} \end{aligned}

Inserting the Euler-Lagrange equation into the first term and using that dft/dx=dfx/dt\idv{f_{t}}{x} = \idv{f_{x}}{t}:

dLdx=(ddx(Lfx)+ddt(Lft))fx+Lfxdfxdx+Lftdftdx+Lx=ddx(Lfxfx)+ddt(Lftfx)+Lx\begin{aligned} \dv{L}{x} &= \bigg( \dv{}{x} \Big( \pdv{L}{f_x} \Big) + \dv{}{t} \Big( \pdv{L}{f_t} \Big) \bigg) f_x + \pdv{L}{f_x} \dv{f_x}{x} + \pdv{L}{f_t} \dv{f_t}{x} + \pdv{L}{x} \\ &= \dv{}{x} \Big( \pdv{L}{f_x} f_x \Big) + \dv{}{t} \Big( \pdv{L}{f_t} f_x \Big) + \pdv{L}{x} \end{aligned}

Leading to the following expression for the “soft” (explicit only) derivative L/x\ipdv{L}{x}:

Lx=ddx(LfxfxL)+ddt(Lftfx)\begin{aligned} - \pdv{L}{x} &= \dv{}{x} \Big( \pdv{L}{f_x} f_x - L \Big) + \dv{}{t} \Big( \pdv{L}{f_t} f_x \Big) \end{aligned}

And then by going through the same process for the other variable tt, we arrive at:

Lt=ddx(Lfxft)+ddt(LftftL)\begin{aligned} - \pdv{L}{t} &= \dv{}{x} \Big( \pdv{L}{f_x} f_{t} \Big) + \dv{}{t} \Big( \pdv{L}{f_t} f_t - L \Big) \end{aligned}

Now we define the so-called stress-energy tensor TνμT_\nu^\mu as a useful abbreviation (the name comes from its application in relativity), where δνμ\delta_\nu^\mu is the Kronecker delta:

TνμLfμfνδνμL\begin{aligned} T_\nu^\mu &\equiv \pdv{L}{f_\mu} f_\nu - \delta_\nu^\mu L \end{aligned}

Such that the two relations we just found can be written as follows:

Lx=ddxTxx+ddtTxtLt=ddxTtx+ddtTtt\begin{aligned} \boxed{ \begin{aligned} - \pdv{L}{x} &= \dv{}{x} T_x^x + \dv{}{t} T_x^t \\ - \pdv{L}{t} &= \dv{}{x} T_t^x + \dv{}{t} T_t^t \end{aligned} } \end{aligned}

And with this definition of TνμT_\nu^\mu we can also rewrite the Euler-Lagrange equation in the same way, noting that ff=f/f=1f_f = \ipdv{f}{f} = 1:

Lf=ddxTfx+ddtTft\begin{aligned} \boxed{ - \pdv{L}{f} = \dv{}{x} T_f^x + \dv{}{t} T_f^t } \end{aligned}

These three equations are the framework we need. What happens if LL does not explicitly contain xx, so L/x\ipdv{L}{x} is zero? Then the corresponding equation clearly turns into:

ddtTxt=ddxTxx\begin{aligned} \dv{}{t} T_x^t &= - \dv{}{x} T_x^x \end{aligned}

Such continuity relations are very common in physics. This one effectively says that if TxtT_x^t increases with tt, then TxxT_x^x must decrease with xx by a certain amount. Yes, this is very abstract, but when you apply this technique to a specific physical problem, TxxT_x^x and TxtT_x^t are usually quantities with a clear physical interpretation.

For L/t=0\ipdv{L}{t} = 0 and L/f=0\ipdv{L}{f} = 0 we get analogous continuity relations, so there seems to be a pattern here: if LL has a continuous symmetry (i.e. there is a continuous transformation with no effect on the value of LL), then there exists a continuity relation specific to that symmetry. This is the qualitative version of Noether’s theorem.

In general, for L(f,fx,x,t)L(f, f_x, x, t), a continuous transformation (not necessarily a symmetry) consists of shifting the coordinates (f,x,t)(f, x, t) as follows:

ff+εαfxx+εαxtt+εαt\begin{aligned} f \:\:\to\:\: f + \varepsilon \alpha^f \qquad\qquad x \:\:\to\:\: x + \varepsilon \alpha^x \qquad\qquad t \:\:\to\:\: t + \varepsilon \alpha^t \end{aligned}

Where ε\varepsilon is the amount of shift, and (αf,αx,αt)(\alpha^f, \alpha^x, \alpha^t) are parameters controlling the direction of the shift in (f,x,t)(f, x, t)-space. Given a specific LL, suppose we have found a continuous symmetry, i.e. a direction (αf,αx,αt)(\alpha^f, \alpha^x, \alpha^t) such that the value of LL is unchanged by the shift, meaning:

0=dLdεε=0=Lfdfdε+Lxdxdε+Ltdtdε\begin{aligned} 0 &= \dv{L}{\varepsilon} \bigg|_{\varepsilon = 0} = \pdv{L}{f} \dv{f}{\varepsilon} + \pdv{L}{x} \dv{x}{\varepsilon} + \pdv{L}{t} \dv{t}{\varepsilon} \end{aligned}

Where we set ε=0\varepsilon = 0 to get rid of it. Negating and inserting our three equations yields:

0=LfαfLxαxLtαt=(ddxTfx+ddtTft)αf+(ddxTxx+ddtTxt)αx+(ddxTtx+ddtTtt)αt\begin{aligned} 0 &= - \pdv{L}{f} \alpha^f - \pdv{L}{x} \alpha^x - \pdv{L}{t} \alpha^t \\ &= \bigg( \dv{}{x} T_f^x + \dv{}{t} T_f^t \bigg) \alpha^f + \bigg( \dv{}{x} T_x^x + \dv{}{t} T_x^t \bigg) \alpha^x + \bigg( \dv{}{x} T_t^x + \dv{}{t} T_t^t \bigg) \alpha^t \end{aligned}

This is a continuity relation! Let us make this clearer by defining some current densities:

JxTfxαf+Txxαx+TtxαtJtTftαf+Txtαx+Tttαt\begin{aligned} J^x &\equiv T_f^x \alpha^f + T_x^x \alpha^x + T_t^x \alpha^t \\ J^t &\equiv T_f^t \alpha^f + T_x^t \alpha^x + T_t^t \alpha^t \end{aligned}

So that the above equation can be written in the standard form of a continuity relation:

0=ddxJx+ddtJt\begin{aligned} \boxed{ 0 = \dv{}{x} J^x + \dv{}{t} J^t } \end{aligned}

This is the quantitative version of Noether’s theorem: for every symmetry (αf,αx,αt)(\alpha^f, \alpha^x, \alpha^t) we can find, Noether gives us the corresponding continuity relation. This result is easily generalized to more variables x1,x2,...x_1, x_2, ... and/or more unknown functions f1,f2,...f_1, f_2, ....

Continuity relations tell us about conserved quantities. Of the free variables (x,t)(x, t), we choose one as the dynamic coordinate (usually tt) and then all others are transverse coordinates. Let us integrate the continuity relation over all transverse variables:

0=x0x1 ⁣(ddxJx+ddtJt)dx=[Jx]x0x1+ddtx0x1 ⁣Jtdx\begin{aligned} 0 &= \int_{x_0}^{x_1} \! \bigg( \dv{}{x} J^x + \dv{}{t} J^t \bigg) \dd{x} \\ &= \big[ J^x \big]_{x_0}^{x_1} + \dv{}{t} \int_{x_0}^{x_1} \! J^t \dd{x} \end{aligned}

Usually the problem’s boundary conditions ensure that [Jx]x0x1=0[J^x]_{x_0}^{x_1} = 0, in which case x0x1Jtdx\int_{x_0}^{x_1} J^t \dd{x} is a conserved quantity (i.e. a constant) with respect to the dynamic coordinate tt.

In the 1D case L(f,ft,t)L(f, f_t, t) (i.e. if LL is a Lagrangian rather than a Lagrangian density), the current density JxJ^x does not exist, so the conservation of the current JtJ^t is clearly seen:

ddtJt=0\begin{aligned} \dv{}{t} J^t &= 0 \end{aligned}


  1. O. Bang, Nonlinear mathematical physics: lecture notes, 2020, unpublished.