0
$\begingroup$

The following notation is used below:

  • d: exterior derivative
  • $\delta$: codifferential (adjoint of d)
  • $\times$: skew-symmetric operator of a $\mathbb{R}^3$-vector
  • $\nabla\times$: Curl operator in vector calculus
  • $\langle x,y \rangle_A$: Inner product with weight matrix $A$
  • $\iota_{(\bullet)}$: interior product of a $k$-form with argument vector

Lorentz force law:

Let us consider a unity charge particle with configuration $q\in\mathcal{Q}\subset\mathbb{R}^3$, the motion of which is affected by the time-invariant scalar potential $\Phi(q):\mathcal{Q}\mapsto \mathbb{R}$ and a time-invariant vector (gauge) potential $A(q):\mathcal{Q}\mapsto T^*\mathcal{Q}$. For this system, the Lagrangian is given by $L = \frac{1}{2}\langle \dot{q},\dot{q} \rangle_M - \Phi + \langle \dot{q}, A \rangle$ (as shown here). Correspondingly, the Euler-Lagrange equations give:

$M\ddot{q} = -d\Phi -(\nabla \times A)\times \dot{q} \tag{1}$

Considering the Hamiltonian, $\mathcal{H} = \frac{1}{2}\langle \dot{q},\dot{q} \rangle_M + \Phi(q)$, we obtain $\dot{\mathcal{H}} = 0$, implying the conservation of energy due to the skew-symmetric property of the matrix operator $(\nabla \times A)\times$.

Let us now consider in higher dimensions, $q\in\mathcal{Q}\subset\mathbb{R}^n$. In this case, for the gauge potential $A$, we obtain the antisymmetric electromagnetic tensor $B = dA$, which yields through the interior product the magnetic force, i.e., $f_\mu = \dot{q}^\nu B_{\mu\nu} \Rightarrow f = \iota_{\dot{q}}B$ (as explained here). In this case too, we obtain the Lorentz force law as,

$M\ddot{q} = -d\Phi - \mathcal{B}(q)\dot{q} \tag{2}$ where the operator $\mathcal{B}$ is obtained from $\mathcal{B}(q)\dot{q}= \iota_{\dot{q}}B$, is skew-symmetric and, hence, we also obtain $\dot{\mathcal{H}}=0$.

Hodge Theorem:

From the Hodge theorem (Sec 5.3), we know that, if $\mathcal{Q}$ is compact, we can decompose a force ($n$-vector of 1-forms) as

$F(q) = d\alpha(q) + \delta \beta(q) + \gamma(q) \tag{3}$

where $F,\gamma \in\Omega^1(\mathcal{q})$, $\alpha\in \Omega^0(\mathcal{q})$, $\beta \in \Omega^2(\mathcal{q})$. In this case, $\alpha$ is the scalar potential $0$-form (similar to $\Phi$), and I see $\beta$ as the $2$-form electromagnetic tensor (similar to $B$). I am trying to analyze an arbitrary force $F(q)$ using the Hodge theorem in terms of properties like, positive-definiteness off $\alpha$, degeneracy of equilibrium, etc. However, I am unable to draw any relation between the magnetic component of the Lorentz force, $\iota_{\dot{q}}B$ and $\delta \beta(q)$, where both $B,\beta$ are $2$-forms (electromagnetic tensors). From both these $2$-forms, we obtain a resultant $1$-form using two different operations. It would be helpful if you could guide me to some theorem, which might relate the interior product and the codifferential. Are there some special conditions, in which we can choose $\delta \beta = \iota{\dot{q}}B$ for the Hodge decomposition?

I apologize if the question is trivial or if I have made some notation errors. I come from a robotics background, so I am not extremely well equipped with concepts of electromagnetism and differential forms.

$\endgroup$

1 Answer 1

0
$\begingroup$

I have investigated this topic more thoroughly now, and I would like to post it here as it might be helpful for others.

Formerly, I only looked at the Euler-Lagrange equations for the particle. When the particle is moving in the presence of an EM field, the total system will additionally have the Euler-Lagrange equations for the field. In particular, we get two equations:

$\delta dV = \rho(q)$, which is Gauss' Law

$\delta \mathcal{B} = J$, which is Ampere's law,

where $\rho$ and $J$ refer to the charge density and the current density, respectively. We also get $d^2 V =0$ and $d\mathcal{B} = d^2A = 0$, due to the property of the exterior derivative. The above two equations (laws) describe the spatial flow of the EM field (not the particle). Now, the current density, $J$, is nothing but the rate of charge flow in an infinitesimal volume, i.e., $J = \rho(q)\dot{q}$. Thus, from Ampere's law, we obtain: $\delta \mathcal{B} = \rho(q)\dot{q}$.

Furthermore, the EM field is represented by a tensor $\mathcal{T} \equiv (dV,\mathcal{B})$, and the force density, i.e., the force on the particle by virtue of the EM field in an infinitesimal volume is given by $f= \iota_{(\rho,J)}\mathcal{T} = \iota_{\rho}dV + \iota_{J}\mathcal{B} = \rho(q) (dV + \mathcal{B}\dot{q})$, which integrated over the volume of the point charge identically gives the force acting on it.

Thus, the total force on the particle due to the EM field is written as, $f = \iota_{\delta dV}dV + \iota_{\delta \mathcal{B}}\mathcal{B}$. Keeping this mind, the total flow density ($4$-current) in the EM field is $(\delta dV,\delta\mathcal{B})$.

Now coming to the Hodge decomposition, the decomposition is not for the force acting on the particle, but for the current density $(\delta dV,\delta\mathcal{B})$ due to the EM field $(dV,\mathcal{B})$. So, corresponding to a force (n-vector of 1-forms), the total current density decomposition can be given using Hodge theorem.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.