For the relativistic charged particle in EM field we have the following equation for the hamiltonian
$$H\left( {\vec r,\vec P,t} \right) = c\sqrt {{m^2}{c^2} + {p^2}} + e\varphi = c\sqrt {{m^2}{c^2} + {{\left( {\vec P – e\vec A\left( {\vec r,t} \right)} \right)}^2}} + e\varphi. $$
Then the hamiltonian's equations of motion can be written as
$$\frac{{d\vec r}}{{dt}} = \frac{{\partial H}}{{\partial \vec P}} = \frac{{c\vec p}}{{\sqrt {{m^2}{c^2} + {p^2}} }}$$
and
$$\frac{{d\vec P}}{{dt}} = – \frac{{\partial H}}{{\partial \vec r}} = \vec ve\frac{{\partial \vec A}}{{\partial \vec r}} – e\frac{{\partial \phi }}{{\partial \vec r}}$$
Where ${\vec P}$ is the generalised momentum. I don't understand why the generalized momentum is used in the last equation instead of the ordinary momentum.
Futhermore, the expression for the ordinary momentum ${\vec p}$ is obtained and has such a form
$$\frac{{d\vec p}}{{dt}} = – e\frac{{\partial \vec A}}{{\partial t}} – e\frac{{\partial \phi }}{{\partial \vec r}} + e\left( {\vec v\frac{{\partial \vec A}}{{\partial \vec r}} – \frac{{\partial \vec A}}{{\partial \vec r}}\vec v} \right)$$
And it is not clear for me why the last term in this equation $e\left( {\vec v\frac{{\partial \vec A}}{{\partial \vec r}} – \frac{{\partial \vec A}}{{\partial \vec r}}\vec v} \right)$ is not
zero and what ${\frac{{\partial \vec A}}{{\partial \vec r}}}$ means? If ${\vec A}$ was a scalar it would be just a gradient value but the vector quantity confuses me
Best Answer
Short answer
The generalized momentum $\vec P$ is used because Hamilton's equations of motion relate the time-derivative of the generalized momentum $d \vec P/dt$ -- not the time-derivative of the kinetic momentum $d \vec p/dt$ -- to the negative partial derivatives of the Hamiltonian with respect to the generalized position $-\partial H/\partial \vec r$.
Correct. If $\vec A$ were instead a scalar field, that term would denote the gradient of a scalar field. It turns out that we can apply the concept of a gradient not only to scalars but also to vectors and, more generally, tensors, a class of geometric objects to which scalars, or zeroth-order tensors, and vectors, or first-order tensors, belong. As the gradient of a zeroth-order tensor yields a first-order tensor, you might guess that the gradient of a first-order tensor yields a second-order tensor, a geometric object which can be represented using an $n \times n$ matrix; you'd be correct, and while it is not as clear in the notation you have chosen, the fact that $\partial \vec A/\partial \vec r$ -- the gradient of the vector field $\vec A$ -- is a second-order tensor is exactly the reason why $\vec v (\partial \vec A/\partial \vec r) - (\partial \vec A/\partial \vec r)\vec v \neq 0$.
In fact, just by inspection, you should be able to at least convince yourself that since
$$- e\frac{\partial \vec A}{\partial t} - e\frac{\partial \phi}{\partial \vec r} = e \left(-\frac{\partial \vec A}{\partial t} - \frac{\partial \phi}{\partial \vec r}\right) = e \vec E$$
it must be that $\vec v (\partial \vec A/\partial \vec r) - (\partial \vec A/\partial \vec r)\vec v = \vec v \times \vec B$, since the right-hand side of the equation for $d \vec p/dt$ should yield the correct expression for the Lorentz force.
Long answer
Now let's prove our conviction.
Tensors in $\mathbb R^3$
Consider a vector basis $\{\mathbf e_i\}$, where the index $i$ ranges from 1 to 3, and for simplicity, assume that this vector basis is Euclidean; in other words, the inner product of basis vectors $\mathbf e_i$ and $\mathbf e_j$ gives,
$$\mathbf e_i \cdot \mathbf e_j = \delta_{ij} \tag{1}$$
where
$$\left(\delta_{ij}\right) = \left(\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right) \tag{2}$$
is the identity matrix.
A vector $\mathbf f$ may be expressed in this coordinate basis as,
$$\mathbf f = \sum_{i=1}^3 f_i\, \mathbf e_i = \left(\begin{matrix} f_1 & f_2 & f_3 \end{matrix}\right)_{\{\mathbf e_i\}} \tag{3}$$
while a second-order tensor $\mathbf F$ may be expressed as,
$$\mathbf F = \sum_{i=1}^3 \sum_{j=1}^3 F_{ij}\, \mathbf e_i \otimes \mathbf e_j = \left(\begin{matrix} F_{11} & F_{12} & F_{13} \\ F_{21} & F_{22} & F_{23} \\ F_{31} & F_{32} & F_{33} \end{matrix}\right)_{\{\mathbf e_i \otimes \mathbf e_j\}} \tag{4}$$
where $\mathbf e_i \otimes \mathbf e_j$ is called the outer product of $\mathbf e_i$ and $\mathbf e_j$. The outer product is defined such that, given vectors $\mathbf a$, $\mathbf b$, $\mathbf c$, and $\mathbf d$,
$$(\mathbf a \otimes \mathbf b) \cdot (\mathbf c \otimes \mathbf d) = (\mathbf b \cdot \mathbf c)(\mathbf a \otimes \mathbf d) \tag{5}$$
or, equivalently,
$$(\mathbf a \otimes \mathbf b) \cdot \mathbf c = (\mathbf b \cdot \mathbf c)\mathbf a$$ $$\mathbf c \cdot (\mathbf a \otimes \mathbf b) = (\mathbf c \cdot \mathbf a)\mathbf b \tag{6}$$
The transpose of $\mathbf a \otimes \mathbf b$, denoted as $(\mathbf a \otimes \mathbf b)^T$, is defined as,
$$(\mathbf a \otimes \mathbf b)^T = \mathbf b \otimes \mathbf a \tag{7}$$
and so (6) may be rewritten as,
$$(\mathbf a \otimes \mathbf b) \cdot \mathbf c = \mathbf c \cdot (\mathbf a \otimes \mathbf b)^T = (\mathbf b \cdot \mathbf c)\mathbf a$$ $$\mathbf c \cdot (\mathbf a \otimes \mathbf b) = (\mathbf a \otimes \mathbf b)^T \cdot \mathbf c = (\mathbf c \cdot \mathbf a)\mathbf b \tag{8}$$
Additionally, applying (7) to (4), the transpose of $\mathbf F$, denoted $\mathbf F^T$, is,
$$\mathbf F^T = \sum_{i=1}^3 \sum_{j=1}^3 F_{ij}\, \mathbf e_j \otimes \mathbf e_i = \sum_{i=1}^3 \sum_{j=1}^3 F_{ji}\, \mathbf e_i \otimes \mathbf e_j = \left(\begin{matrix} F_{11} & F_{21} & F_{31} \\ F_{12} & F_{22} & F_{32} \\ F_{13} & F_{23} & F_{33} \end{matrix}\right)_{\{\mathbf e_i \otimes \mathbf e_j\}} \tag{9}$$
If the second-order tensor $\mathbf F^T = \mathbf F$, then $\mathbf F$ is said to be symmetric; if $\mathbf F^T = -\mathbf F$, then $\mathbf F$ is said to be antisymmetric.
A third-order tensor $\mathbf \Phi$ may be expressed as,
$$\mathbf \Phi = \sum_{i=1}^3 \sum_{j=1}^3 \sum_{k=1}^3 \Phi_{ijk}\, \mathbf e_i \otimes \mathbf e_j \otimes \mathbf e_k \tag{10}$$
One commonly-encountered third-order tensor known as the alternating tensor, denoted as $\mathbf \epsilon$, is defined as,
$$\mathbf \epsilon = \sum_{i=1}^3 \sum_{j=1}^3 \sum_{k=1}^3 \epsilon_{ijk}\, \mathbf e_i \otimes \mathbf e_j \otimes \mathbf e_k \tag{11}$$
such that $\mathbf \epsilon$ is antisymmetric under an exchange of any two indices (e.g $\epsilon_{jik} = -\epsilon_{ijk}$) and $\epsilon_{123} = 1$. Note that this implies that any component $\epsilon_{ijk}$ that has two or more indices set to the same value is equal to zero (e.g $\epsilon_{112} = 0$).
A common practice in many publications is to omit the summation symbols found in the expressions for the tensors given above; this is known as the Einstein summation convention, and this convention will be used from this point onward. The rules of the Einstein summation convention are as follows:
Relevant Tensor Operations
The scalar product of two vectors $\mathbf a$ and $\mathbf b$, denoted $\mathbf a \cdot \mathbf b$, is given by,
$$\mathbf a \cdot \mathbf b = a_i b_j\, \mathbf e_i \cdot \mathbf e_j = a_i b_j \delta_{ij} = a_i b_i \tag{12}$$
Similarly, the inner product for a vector $\mathbf a$ and a second-order tensor $\mathbf B$ is given by,
$$(\mathbf B \cdot \mathbf a)_i = B_{ij} \delta_{jk} a_k = B_{ij}a_j $$ $$(\mathbf a \cdot \mathbf B)_i = a_k \delta_{kj} B_{ji} = B_{ji}a_j \tag{13}$$
where $(\mathbf B \cdot \mathbf a)_i$ and $(\mathbf a \cdot \mathbf B)_i$ are, respectively, the $i$-th components of the vectors $\mathbf B \cdot \mathbf a$ and $\mathbf a \cdot \mathbf B$.
The vector product of two vectors $\mathbf a$ and $\mathbf b$, denoted $\mathbf a \times \mathbf b$, is given by,
$$(\mathbf a \times \mathbf b)_i = \epsilon_{ijk} a_j b_k \tag{14}$$
where $(\mathbf a \times \mathbf b)_i$ is the $i$-th component of $\mathbf a \times \mathbf b$. Similarly, the curl of a vector $\mathbf a$ is given by,
$$(\nabla \times \mathbf a)_i = \epsilon_{ijk} \frac{\partial a_k}{\partial x_j} \tag{15}$$
where $x_j$ is the $j$-th component of the position vector $\mathbf r$ relative to the origin of our coordinate basis.
The gradient of a vector $\mathbf a$, a second-order tensor denoted $\nabla \mathbf a$, is defined as,
$$(\nabla \mathbf a)_{ij} = \frac{\partial a_i}{\partial x_j} \tag{16}$$
where $(\nabla \mathbf a)_{ij}$ is the component of $\nabla \mathbf a$ associated with the outer product $\mathbf e_i \otimes \mathbf e_j$.
A Note About Antisymmetric Second-Order Tensors
If a second-order tensor $\mathbf F$ is antisymmetric (i.e. $F_{ji} = -F_{ij}$), then there exists some scalars $f_k$ such that,
$$F_{ij} = \epsilon_{ijk} f_k \tag{17}$$
or, equivalently,
$$f_k = \frac{1}{2}\epsilon_{ijk}F_{ij} \tag{18}$$
and the vector $\mathbf f$ for which $f_k$ is the $k$-th component is called the axial vector of $\mathbf F$. Note that the definition for the vector product in (14) also involves the components of the alternating tensor. This is no accident, as the vector product in $\mathbb R^3$ between two vectors $\mathbf a$ and $\mathbf f$ is actually the result of an inner product between an antisymmetric tensor $\mathbf F$ and a vector $\mathbf a$,
$$F_{ij}a_j = \epsilon_{ijk} a_j f_k \tag{19}$$
The Lorentz Force
Now we are in a position to consider the equation you have written for $d \vec p/dt$ using clearer notation,
$$\begin{align} \frac{d\mathbf p}{dt} & = e\left[\left(-\frac{\partial \mathbf A}{\partial t} - \nabla \phi \right) + \mathbf v \cdot \nabla \mathbf A - \nabla \mathbf A \cdot \mathbf v\right] \\ & = e\left(\mathbf E + \mathbf v \cdot \nabla \mathbf A - \nabla \mathbf A \cdot \mathbf v\right) \end{align} \tag{20}$$
Let's rewrite this in summation notation:
$$\begin{align} \frac{dp_i}{dt} & = e\left(E_i + v_l \delta_{lj}\frac{\partial A_j}{\partial x_i} - \frac{\partial A_i}{\partial x_j} \delta_{jl} v_l\right) \\ & = e\left(E_i + \left(\frac{\partial A_j}{\partial x_i} - \frac{\partial A_i}{\partial x_j}\right) v_j\right) \end{align} \tag{21}$$
The term $(\partial A_j/\partial x_i) - (\partial A_i/\partial x_j)$ is clearly the component $F_{ij}$ of an antisymmetric second-order tensor $\mathbf F$, and so the components of its corresponding axial vector $f_k$ are,
$$\begin{align} f_k & = \frac{1}{2} \epsilon_{ijk} \left(\frac{\partial A_j}{\partial x_i} - \frac{\partial A_i}{\partial x_j}\right) \\ & = \frac{1}{2} \left(\epsilon_{ijk}\frac{\partial A_j}{\partial x_i} - \epsilon_{ijk}\frac{\partial A_i}{\partial x_j}\right) \\ & = \frac{1}{2} \left(\epsilon_{kij}\frac{\partial A_j}{\partial x_i} + \epsilon_{kji}\frac{\partial A_i}{\partial x_j}\right) \\ & = \frac{1}{2} \left[\left(\nabla \times \mathbf A\right)_k + \left(\nabla \times \mathbf A\right)_k\right] \\ & = \left(\nabla \times \mathbf A\right)_k \\ & = B_k \end{align} \tag{22}$$
Thus,
$$\begin{align} \frac{dp_i}{dt} & = e\left(E_i + \left(\frac{\partial A_j}{\partial x_i} - \frac{\partial A_i}{\partial x_j}\right) v_j\right) \\ & = e\left(E_i + \epsilon_{ijk} B_k v_j\right) \\ & = e\left(E_i + \left(\mathbf v \times \mathbf B\right)_i \right) \\ \end{align} \tag{23}$$
A Final Note on the Tensor Formulation of Electromagnetism
When studying electromagnetic phenomena in the context of relativity, due to the non-Euclidean nature of the geometry of spacetime, we cannot simplify our mathematical analysis by working in a 3-dimensional Euclidean basis; however, since tensors are geometric objects that exist independently of any coordinate basis used to describe them, a similar analysis can be performed to yield the correct result, and in Minkowski spacetime you will end up constructing a $4 \times 4$ antisymmetric tensor of the form,
$$F_{\mu \nu} = \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu} = \left(\begin{matrix} 0 & -E_x/c & -E_y/c & -E_z/c \\ E_x/c & 0 & B_z & -B_y \\ E_y/c & -B_z & 0 & B_x \\ E_z/c & B_y & -B_x & 0 \\\end{matrix}\right) \tag{24}$$
more commonly know as the electromagnetic field tensor, and the Lorentz Force Law for a charge $q$ will take the form of
$$\frac{dp_\mu}{d \tau} = qF_{\mu\nu}U^{\nu} \tag{25}$$
where $p_\mu$ are the covariant components of the charge's four-momentum, $\tau$ is the proper time experienced by the charge, and $U^{\nu}$ are the contravariant components of the charge's four-velocity.