# Lagrangian density

1. May 2, 2014

### Geometry_dude

Now this is a bit of a mix of a math and a physics question, but I think it is best asked here.

Assume we are given a Lorentzian manifold $(Q, g)$ together with a metric connection $\nabla$. Naturally we define geodesics $\gamma$ via
$$\nabla_{\dot \gamma} \dot \gamma = 0 \quad ,$$
$$0=\ddot \gamma^k + \Gamma^k{}_{(ij)} \, \dot \gamma^i \dot \gamma^j \quad ,$$
where $\dot \gamma = \frac{d \gamma}{d \tau}$.
Alternatively, we could have defined the Lagrangian $L \in C^\infty(TQ,\mathbb{R})$, i.e. a function on the tangent bundle, by
$$L(q,\dot q) = \frac{1}{2} g_{q}(\dot q, \dot q)$$
for $(q,\dot q) \in TQ$ and then varied the action
$$S(\gamma) = \int_{\gamma} L (\gamma, \dot \gamma) \, d t$$
leading to the same local geodesic equation.

Now, in physics, one makes the step to "field theory" by considering a function $\phi \in C^\infty(Q, \mathbb R)$ and then taking a new functional
$$\tilde S (\phi) = \int_Q \mathcal L (\phi, d \phi) \, \mu$$
where $\mu \in \Omega^n(Q)$ is the canonical volume form. $\mathcal L$ is a function from some space of smooth functions cartesian product with their exterior derivative to $\mathbb R$ and is called the Lagrange density.
My question is:
How do the two formalisms connect? What justifies us in calling $\tilde S$ the action as well? Is there a mathematical discipline dealing with such systems $(Q,g, \mathcal L)$?

Last edited: May 2, 2014
2. May 2, 2014

### Ben Niehoff

Careful! These two definitions are not always equivalent. The first condition

$$\nabla_{\dot \gamma} \dot \gamma = 0 \quad ,$$
defines "autoparallels", or curves whose tangent vector is parallel-transported along itself.

The second condition minimizing the action

$$S(\gamma) = \int_{\gamma} L (\gamma, \dot \gamma) \, d t$$
defines a "locally length-extremizing curve". If the connection is Levi-Civita (i.e., both metric and torsion-free --- these are independent conditions!), then these two definitions for "geodesics" coincide. But for other connections, they need not define the same sets of curves.

This question has basically nothing to do with the first half of your post, so it's not clear to me why you've put these things together.

An action is any scalar quantity that is a functional from some function space $\mathcal{F}$ into $\mathbb{R}$. Extremizing the action with respect to its arguments is a standard variational problem.

In the length-minimizing curve problem, the function space $\mathcal F$ is the space of functions $\gamma : \mathbb{R} \to M$ where $M$ is the manifold.

In the field theory problem, the function space $\mathcal F$ is the space of functions $\phi : M \to \mathbb{R}$. One can also write actions involving functions from $M$ into $TM$ or $T^*M$, or tensor products of those (i.e., particles with integer spin), or functions into the spin bundle, if it exists* (particles with half-integer spin), or functions into Lie algebras (gauge theories), or even functions into manifolds (sigma models).

* Not every manifold admits a spin structure.

3. May 2, 2014

### Matterwave

I'm not sure what you mean with your question...$\tilde{S}$ is called the action if it is the correct action leading to the correct equations of motion. It is the action for the field, given you have the correct Lagrange density. This is why we call any S "THE action". The $S$ you wrote earlier is, up to a constant prefactor (-mc), the action for a free particle (but I think your Lagrangian is meant to have a square-root, instead of a 1/2?). If your action $\tilde{S}$ is the action for a free field, then one needs one more action term describing the interaction of the field and the particles. For the electro-magnetic field, for example, this action term is:

$S_{int}=-\frac{e}{c}\int A_\mu dx^\mu$

Varying the total action with respect to the fields only gives you the field equations, and varying the total action with respect to the matter particles only gives you the equations of motion of the particles.

I'm not sure if this was your question though.

4. May 3, 2014

### Geometry_dude

The torsion does not have an effect on the shape of geodesics.
Proof: Let $(.)$ and $[.]$ denote the anti-symmetrization w.r.t. enclosed indicies.
$$\Gamma^k{}_{ij} \, \dot \gamma^i \, \dot \gamma^j = \Gamma^k{}_{ij} \, \dot \gamma^{(i} \, \dot \gamma^{j)} = \Gamma^k{}_{(ij)} \, \dot \gamma^i \, \dot \gamma^j$$
Since
$$\Gamma^k{}_{ij}= \Gamma^k{}_{(ij)} + \Gamma^k{}_{[ij]}$$
and
$$T^k{}_{ij}= 2 \Gamma^k{}_{[ij]}$$
by definition of torsion, it is only relevant for parallel transport of vectors, not for geodesics.

Hence the definition of the connection being metric is enough to uniquely define geodesics. The metric assumption gives the conceptual link to straight lines.

The reason is that they are conceptually related and somehow (the how is my question) the second follows by considering "infinite degrees of freedom". For example, a free Klein-Gordon field has the density
$$\mathcal L (q, \phi, d \phi) = \frac{1}{2} g^{-1}_q (d \phi, d \phi) - (m^2+ \xi \mathcal S) \phi^2 (q)$$
where $m$ is the mass (a constant), $\mathcal S$ the Ricci scalar and $\xi$ some parameter. The first summand seems very similar to the free particle Lagrangian I mentioned above indicating a deeper relation.

In the Wikipedia article on Lagrangians they do everything in a Newtonian formulation (take $Q= \mathbb R ^3$ and $g= \delta$, the standard inner product with respect to the frame induced by the standard chart) and s.t. $t= \tau$ (the parameter of $\gamma$ is ordinary Newtonian time $t$) and then they simply set
$$L(t) = \int_{\mathbb R^3} \mathcal L (x,y,z,t) \, dx dy dz$$
but this seems really strange and false from a conceptual point of view to me.
So my question is, what is the conceptual relation between $L$ and $\mathcal L$?

Yes, this makes sense. Then a general "classical (relativistic) field" in physics is just a section of a smooth fiber bundle, where the base space is a "spacetime" (that is a manifold $Q$ equipped with a Lorentzian metric $g$ and a metric connection as discussed above) and a field action is just a real functional on those sections. So what you are suggesting is that the field action and the "curve action" just carry this name, because they are both functionals on spaces of (smooth) functions, is that correct?

Sorry, I forgot to mention that I hid the mass in the metric. If you don't like that, you get
$$L(q,\dot q)= \frac{m}{2} g_q (\dot q, \dot q)$$
for the motion of a mass point on straight lines. When g is unitless, then this has unit energy, as it should have.
What you probably mean is the proper time $\Delta \tau$ for a curve $\gamma$ parametrized by $s$ from $s_0$ to $s_1$
$$\Delta \tau (\gamma) = \frac{1}{c^2}\int_{s_0}^{s_1} \sqrt{g_{\gamma(s)}(\dot \gamma (s), \dot \gamma (s))} \, d s$$
where I wrote $\dot \gamma = \frac{d \gamma}{d s}$ and $s$ has physical dimension of time.
Note that I use the $(+,-,-,-)$ convention.
If you vary this integral and let the variation vanish, you should also get the geodesic equation, but I haven't checked this.
$$d F=0$$
am I correct?

5. May 3, 2014

### Ben Niehoff

The relation between fields and particles is the "eikonal approximation". A particle is a localized excitation of a field that travels along the characteristics (I think) of the field equation. This is the relationship between wave and ray optics.

If you assume a localized ("delta function") excitation of the field and pullback to its worldline, then the field action becomes the particle action. Try it with the free particle.

Yes.

You do. Classically it doesn't matter if you use the length functional, the squared length functional, or even the length-raised-to-any-power functional. However, quantum mechanically, the specific form of the action tells us how the system must be quantized, and these are not necessarily equivalent (even the constant out front makes a difference).

No, he gave the interaction part of the action which should give the Lorentz force. But his integral needs to be pulled back to the worldline of some particle. Try it, it's instructive.

$F$ is typically defined as $dA$, so you won't get that out of an action. The action for Maxwell's equations is

$$S[A] = \int_\mathcal{M} \bigg( -\frac12 \, F \wedge \star F - \frac12 \, A \wedge \star J \bigg),$$
where $F \equiv d A$. If you vary this action with respect to $A$, you should get the other half of Maxwell's equations, that couple $F$ to the external source $J$.

$\star$ is the Hodge dual, and note that I use $(-,+,+,+)$ signature (so all kinetic terms come in with a coefficient of $-1/2$). I have also set the speed of light and the elementary charge to 1.

6. May 3, 2014

### Matterwave

The usual free-particle action is simply:

$$S=-mc\int ds$$

This is good for a free particle in special relativity, general relativity, or classical mechanics, just take the appropriate form of ds. In terms of the metric, for a massive particle, then this action is:

$$S=-mc\int \sqrt{g(T,T)}dt$$

Where T is the tangent vector to the particle's world line, with t being the parametrization of that worldline. Varying this action and setting it equal to 0 will yield the geodesic equations because free particles travel along geodesics.

Now if there is a background E&M field as well as the gravitational field, we have to add the interaction term I showed you earlier to the action, and vary both this term and the interaction term together with respect to the particle's coordinates. This will give you the motion of this particle, which will no longer be geodesic motion due to the presence of the EM field. We have to vary the action:

$$S=-mc\int ds-\frac{e}{c}\int A_\mu dx^\mu$$

It will result in the equations for the 4-velocity of this particle:

$$\nabla_u u=\frac{e}{m}F(\quad,u)$$

At this point, we have only cared about the motion of the (test) particle due to a gravitational field, described by the metric, and a E&M field, described by the field tensor. We have NOT talked at all about how the fields interact! To do that, we look at the Field Lagrangians. The free EM field Action is:

$$S_{em}=-\frac{1}{16\pi c}\int F_{\mu\nu} F^{\mu\nu} \sqrt{-g}d^4 x$$

This is the same equation as Ben gave, but in different units (Gaussian units) and his action included the interaction term I mentioned before. This free EM field action by itself, when varied with respect to it's degrees of freedom (the 4-potential) will give the vacuum Maxwell equations. Incorporating the interaction term gives the non-vacuum Maxwell equations. The last thing we have to add is the action of the gravitational field itself, given by the Einstein-Hilbert action:

$$S_{g}=\frac{c^4}{16\pi G}\int R\sqrt{-g}d^4 x$$

For the gravitational field, we don't have to include some such "coupling" terms as for the EM field, since the couplings were already done by changing the geometry. So our total action is:

$$S=-\sum_{particles}mc\int ds-\sum_{particles}\frac{e}{c}\int A_\mu dx^\mu--\frac{1}{16\pi c}\int F_{\mu\nu} F^{\mu\nu} \sqrt{-g}d^4 x+\frac{c^4}{16\pi G}\int R\sqrt{-g}d^4 x$$

Variations of this total action with respect to all degrees of freedom should yield, 1) the particles' equation of motion in the EM and gravitational fields, 2) the non-vacuum Maxwell's equations in curved spacetime, and 3) the Einstein field equations with stress-energy from the EM field as well as the particles. Do I actually want to try to do this all in one go? Heck no. Heeeeck no.

The main point is. There are actions for the particles, and for the fields, one will give you E.O.M. of the particles, and the other will give you the field equations. For the former, you have that the action is the integral of a Lagrangian. For the latter, the action is the integral of a Lagrangian, which itself is the integral of a Lagrange density.

7. May 4, 2014

### Geometry_dude

While I don't see why one would need quantum theory for this, I think I understand what you mean. The basic idea is taking the step to distributions.
So I was wrong in saying that $\phi$ is a function, but it should be a distribution acting on test functions $f \in C^\infty_{\text{test}}(Q, \mathbb R )$ on the Lorentzian manifold. So the first step is doing this for $1$ particle:
$$\gamma (\tau) \to \phi :=\delta_{\gamma(\tau)}$$
with
$$\delta_{\gamma(\tau)}(f) = f \circ \gamma \, (\tau)$$.
Then $\mathcal L$ becomes a functional on the space of distributions and I somehow need to choose $\mathcal L$ such that whatever I get out of
$$\partial_i \frac{\partial \mathcal L}{\partial(\partial_i \phi)}= \frac{\partial \mathcal L}{\partial \phi}$$
becomes equivalent to
$$\frac{d}{d \tau} \frac{\partial L}{\partial \dot q^i}= \frac{\partial L}{\partial q^i}$$
for the above $\delta$-distribution, right? This would directly give me the continuum generalization of the Lagrangian to the Lagrange density.
How do I get the exterior derivative on the space of distributions? The fact that all this is happening of a general manifold further complicates the matter.

I think I do get the general idea now, though. Now I only have problems with the implementation. Thank you for your help.

Matterwave, thank you for this exposition of the Lagrangian formulation of field theory in classical mechanics. Do you know whether there have there been any successful attempts to include also the self-fields of the particles?

8. May 4, 2014

### Matterwave

It sort of depends on what you mean by including the self-fields. For example, in E&M, one can calculate the effect of a particle's field on itself by energy and momenta considerations. An accelerating particle emits radiation, which carry away energy and momentum. By integrating over the field's energy and momentum, and by conservation of energy and momentum, one can then look at how that particle will react (to its own field).

However, as the field itself usually diverges at the point of the particle's location, it's quite a hairy situation one finds oneself in if one wants to, for example, use the Lorentz force law on the particle for its own field. It is beyond my knowledge of if there are successful ways of consistently removing these infinities.