What is the relation between velocity and momentum in Hamiltonian mechanics?

[mathmari]

Hey!

We have an electromagnetic field. The relation between energy and momentum of a particle with charge $q$ is $$\left (\frac{E-q\phi}{c}\right )^2=m^2c^2+\|p-\frac{q}{c}A\|^2$$ where $c$ is the velocity of light and $\phi=\phi (x,t), A=A(x,t)$ are the scalar and vector potential of the field with values in $\mathbb{R}$ and $\mathbb{R}^3$ respectively.
Find the relation between the velocity and momentum and the equation of motion.

So, to get the relation between the velocity and momentum, we have to expresss the energy in respect of the velocity, right? (Wondering)

I have done the following:

The momentum is defined as $p=m\cdot v$, where $v$ is the velocity, $p$ is the momentum, and $m$ is the mass. Solving for $m$ we get $m=\frac{p}{v}$

The relationship between mass and energy is $E=m\cdot c^2$. So, we get $E= \frac{p}{v}\cdot c^2$.

Therefore, we get the following:
$$\left (\frac{E-q\phi}{c}\right )^2=m^2c^2+\|p-\frac{q}{c}A\|^2\Rightarrow \left (\frac{\frac{p}{v}\cdot c^2-q\phi}{c}\right )^2=m^2c^2+\|p-\frac{q}{c}A\|^2$$

Is this correct? (Wondering)

 

[I like Serena]

Hey mathmari! (Smile)

Yep. I believe that is correct - assuming non-relativistic speeds.
That is, if $v \ll c$, so that $p=\gamma_v m_0 v \approx m_0 v$ and $E=\sqrt{(m_0c^2)^2 + (pc)^2} \approx m_0c^2$. (Nerd)

 

[mathmari]

Yep. I believe that is correct - assuming non-relativistic speeds.
That is, if $v \ll c$, so that $p=\gamma_v m_0 v \approx m_0 v$ and $E=\sqrt{(m_0c^2)^2 + (pc)^2} \approx m_0c^2$. (Nerd)

So, do we have to assume thatwe have non-relativistic speeds? Or do we have to rake cases? (Wondering)

 

[I like Serena]

So, do we have to assume thatwe have non-relativistic speeds? Or do we have to rake cases? (Wondering)

Not really. We can include relativistic speeds, meaning it becomes generally true without any conditions.
It's just that the formula becomes more complicated.
But as long as we're only looking for a relation between speed and momentum that should be fine.

For completeness,
$$\gamma_v = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$
which is the Lorentz factor associated with the speed $v$.
And nowadays $m_0$ is usually denoted as just $m$.
($m$ used to denote the relativistic mass, but modern physics has abandoned that notion and only refers to rest mass.) (Thinking)

 

[mathmari]

Not really. We can include relativistic speeds, meaning it becomes generally true without any conditions.
It's just that the formula becomes more complicated.
But as long as we're only looking for a relation between speed and momentum that should be fine.

For completeness,
$$\gamma_v = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$
which is the Lorentz factor associated with the speed $v$.
And nowadays $m_0$ is usually denoted as just $m$.
($m$ used to denote the relativistic mass, but modern physics has abandoned that notion and only refers to rest mass.) (Thinking)

Ah ok.. I see!

 

[mathmari]

For the equation of motion do we use the fact that $x=u\cdot t$ ? (Wondering)

 

[I like Serena]

At a second look, I'm realizing I was wrong before. (Tmi)

I'm concluding that the energy $E$ is the total energy of the system, which is constant, so it's not dependent on $v$.

We can see that by seeing what happens for $\phi(\mathbf x,t)=0$ and $\mathbf A(\mathbf x, t) = \mathbf 0$:
$$\left(\frac Ec\right)^2=m^2c^2+p(t)^2$$
This is the same as energy equation for relativistic speeds that I just mentioned! (Nerd)
The original equation is more general, since it includes the effect of an electromagnetic field with components $\phi(\mathbf x,t)$ and $\mathbf A(\mathbf x,t)$.

If we try to find the equation of motion for this simpler equation, we get:
$$p(t)=\sqrt{\left(\frac Ec\right)^2-m^2c^2}=\text{constant}
\quad \Rightarrow\quad v(t)=\text{constant}$$
I believe this is as close as we can get to an equation of motion, since we don't have sufficient information to find the vector functions $\mathbf v(t)$ or $\mathbf x(t)$.

That is, unless we are constrained to 1 dimension (are we?), or unless we use Newton's first law, in which case we have $\mathbf x(t)=\mathbf v_0 \cdot t$. (Thinking)

For the equation of motion do we use the fact that $x=u\cdot t$ ? (Wondering)

Where did you get that? How is it a fact? (Wondering)

 

[mathmari]

At a second look, I'm realizing I was wrong before. (Tmi)

I'm concluding that the energy $E$ is the total energy of the system, which is constant, so it's not dependent on $v$.

We can see that by seeing what happens for $\phi(\mathbf x,t)=0$ and $\mathbf A(\mathbf x, t) = \mathbf 0$:
$$\left(\frac Ec\right)^2=m^2c^2+p(t)^2$$
This is the same as energy equation for relativistic speeds that I just mentioned! (Nerd)
The original equation is more general, since it includes the effect of an electromagnetic field with components $\phi(\mathbf x,t)$ and $\mathbf A(\mathbf x,t)$.

So, is the following wrong?

The momentum is defined as $p=m\cdot v$, where $v$ is the velocity, $p$ is the momentum, and $m$ is the mass. Solving for $m$ we get $m=\frac{p}{v}$
The relationship between mass and energy is $E=m\cdot c^2$. So, we get $E= \frac{p}{v}\cdot c^2$.
Therefore, we get the following:
$$\left (\frac{E-q\phi}{c}\right )^2=m^2c^2+\|p-\frac{q}{c}A\|^2\Rightarrow \left (\frac{\frac{p}{v}\cdot c^2-q\phi}{c}\right )^2=m^2c^2+\|p-\frac{q}{c}A\|^2$$

Does this only hold when $\phi(\mathbf x,t)=0$ and $\mathbf A(\mathbf x, t) = \mathbf 0$ ? (Wondering)

 

[I like Serena]

So, is the following wrong?

Does this only hold when $\phi(\mathbf x,t)=0$ and $\mathbf A(\mathbf x, t) = \mathbf 0$ ? (Wondering)

Yes, and we need $\mathbf p(t)=\mathbf 0$ as well.
More specifically, $p\ne mv$ (at relativistic speeds), and $E_{total}\ne mc^2$, so we cannot substitute those. (Sweating)

$mc^2$ is the contribution to the total energy of a mass $m$ at rest (the term $m^2c^2$ corresponds to it).
$pc = \|\mathbf p\|c$ is the contribution of momentum to the total energy (the classical kinetic energy $\frac 12mv^2$ is hidden in there).
$q\phi$ is the contribution of electric energy
$q\mathbf A$ is the contribution of magnetic energy.

The total energy $E$ is the combination of all these quantities. (Nerd)

 

[mathmari]

Yes, and we need $\mathbf p(t)=\mathbf 0$ as well.
More specifically, $p\ne mv$ (at relativistic speeds), and $E_{total}\ne mc^2$, so we cannot substitute those. (Sweating)

$mc^2$ is the contribution to the total energy of a mass $m$ at rest (the term $m^2c^2$ corresponds to it).
$pc = \|\mathbf p\|c$ is the contribution of momentum to the total energy (the classical kinetic energy $\frac 12mv^2$ is hidden in there).
$q\phi$ is the contribution of electric energy
$q\mathbf A$ is the contribution of magnetic energy.

The total energy $E$ is the combination of all these quantities. (Nerd)

Ah ok. (Thinking)

But how can we find then a relation between the velocity and momentum? Could you give me a hint? (Wondering)

 

[I like Serena]

Ah ok. (Thinking)

But how can we find then a relation between the velocity and momentum? Could you give me a hint? (Wondering)

The relation between velocity and momentum is:
$$\mathbf p=\gamma_v m \mathbf v$$
where $\gamma_v$ is what I wrote before.
It doesn't depend on the given equation. (Thinking)

We can substitute it in the given equation, and together with $\mathbf v(t) = \mathbf{\dot x}(t)$, we have an equation that we could conceivable solve to find an equation of motion that satisfies the energy equation.

 

[mathmari]

The relation between velocity and momentum is:
$$\mathbf p=\gamma_v m \mathbf v$$
where $\gamma_v$ is what I wrote before.
It doesn't depend on the given equation. (Thinking)

We can substitute it in the given equation, and together with $\mathbf v(t) = \mathbf{\dot x}(t)$, we have an equation that we could conceivable solve to find an equation of motion that satisfies the energy equation.

Ah ok!

So, we get the following:
\begin{align*}\left (\frac{E-q\phi}{c}\right )^2=m^2c^2+\|\mathbf p-\frac{q}{c}A\|^2&\Rightarrow \left (\frac{E-q\phi}{c}\right )^2=m^2c^2+\|\gamma_v m \mathbf v-\frac{q}{c}A\|^2 \\ & \Rightarrow \left (\frac{E-q\phi}{c}\right )^2=m^2c^2+\|\gamma_v m \mathbf{\dot x}(t)-\frac{q}{c}A\|^2 \\ & \Rightarrow \|\gamma_v m \mathbf{\dot x}(t)-\frac{q}{c}A\|^2=\left (\frac{E-q\phi}{c}\right )^2-m^2c^2\end{align*}
We cannot bring $\mathbf{\dot x}(t)$ out of the norm and so we cannot solve this equation for $\mathbf{\dot x}(t)$, can we? (Wondering)

Do we just say that the equation motion $\mathbf{x}(t)$ satisfies the above equation? (Wondering)

 

[I like Serena]

Ah ok!

So, we get the following:
\begin{align*}\left (\frac{E-q\phi}{c}\right )^2=m^2c^2+\|\mathbf p-\frac{q}{c}A\|^2&\Rightarrow \left (\frac{E-q\phi}{c}\right )^2=m^2c^2+\|\gamma_v m \mathbf v-\frac{q}{c}A\|^2 \\ & \Rightarrow \left (\frac{E-q\phi}{c}\right )^2=m^2c^2+\|\gamma_v m \mathbf{\dot x}(t)-\frac{q}{c}A\|^2 \\ & \Rightarrow \|\gamma_v m \mathbf{\dot x}(t)-\frac{q}{c}A\|^2=\left (\frac{E-q\phi}{c}\right )^2-m^2c^2\end{align*}

More specifically, we have:
$$ \|\gamma_v(\mathbf{\dot x}(t)) m \mathbf{\dot x}(t)-\frac{q}{c}\mathbf A(\mathbf x(t), t)\|^2
=\left (\frac{E-q\phi(\mathbf x(t), t)}{c}\right )^2-m^2c^2$$
(Nerd)

We cannot bring $\mathbf{\dot x}(t)$ out of the norm and so we cannot solve this equation for $\mathbf{\dot x}(t)$, can we?

Do we just say that the equation motion $\mathbf{x}(t)$ satisfies the above equation?

I don't think so.
At best we can study some special cases, and see if we can come up with a function $\mathbf x(t)$ that satisfies the equation.
Such as when $\phi=\text{constant}$ and $\mathbf A =\textbf{constant}$, then $\mathbf x(t)=\mathbf v_0 t$ is a solution. (Thinking)

Or alternatively, we can indeed say that $\mathbf{x}(t)$ satisfies the above equation.

 

[mathmari]

I don't think so.
At best we can study some special cases, and see if we can come up with a function $\mathbf x(t)$ that satisfies the equation.
Such as when $\phi=\text{constant}$ and $\mathbf A =\textbf{constant}$, then $\mathbf x(t)=\mathbf v_0 t$ is a solution. (Thinking)

What special cases do we have to take? Do we maybe consider if $\phi=\text{constant}$ is constant or not and if $\mathbf A =\textbf{constant}$ is constant or not? (Wondering)

 

[I like Serena]

What special cases do we have to take? Do we maybe consider if $\phi=\text{constant}$ is constant or not and if $\mathbf A =\textbf{constant}$ is constant or not? (Wondering)

We might consider the case of a constant uniform electric field in the absence of a magnetic field.
Or an electric field as caused by a point charge $Q$.
Or a uniform magnetic field.

According to wiki we have:
(Please note the distinction between scalar energy $E$ and electric field vector $\mathbf E$.)
$$\mathbf E = -\nabla \phi- \pd {\mathbf A}t\\
\mathbf B = \nabla\times \mathbf A
$$
So we would get a constant uniform electric field in the x-direction if $\phi(x,y,z,t) = \phi_0 + Cx$ and/or $\mathbf A(x,y,z,t) = \mathbf A_0 + Dt \mathbf{\hat x}$.
Consequently, the solution should be that a particle with charge $q$ gets linearly accelerated in the x-direction.
That is, the solution for its momentum should be $\mathbf p(t) = \mathbf p_0 + q\mathbf E_0 t$. (Thinking)Suppose we pick $\phi = Cx$ and $\mathbf A = \mathbf 0$...
And suppose we approximate $\mathbf p(t) = m\mathbf{\dot x}(t)$ as it is for non-relativistic speeds, what would we get...? (Thinking)

 

[mathmari]

So we would get a constant uniform electric field in the x-direction if $\phi(x,y,z,t) = \phi_0 + Cx$ and/or $\mathbf A(x,y,z,t) = \mathbf A_0 + Dt \mathbf{\hat x}$.
Consequently, the solution should be that a particle with charge $q$ gets linearly accelerated in the x-direction.
That is, the solution for its momentum should be $\mathbf p(t) = \mathbf p_0 + q\mathbf E_0 t$. (Thinking)Suppose we pick $\phi = Cx$ and $\mathbf A = \mathbf 0$...
And suppose we approximate $\mathbf p(t) = m\mathbf{\dot x}(t)$ as it is for non-relativistic speeds, what would we get...? (Thinking)

We get then the following:
\begin{align*}\|m \mathbf{\dot x}(t)\|^2
=\left (\frac{E-qCx}{c}\right )^2-m^2c^2&\Rightarrow m^2\|\mathbf{\dot x}(t)\|^2
=\left (\frac{E-qCx}{c}\right )^2-m^2c^2\\ & \Rightarrow \|\mathbf{\dot x}(t)\|^2
=\left (\frac{E-qCx}{mc}\right )^2-c^2\end{align*}
or not? (Wondering)

Can we do something further here? (Wondering)

 

[I like Serena]

We get then the following:
\begin{align*}\|m \mathbf{\dot x}(t)\|^2
=\left (\frac{E-qCx}{c}\right )^2-m^2c^2&\Rightarrow m^2\|\mathbf{\dot x}(t)\|^2
=\left (\frac{E-qCx}{c}\right )^2-m^2c^2\\ & \Rightarrow \|\mathbf{\dot x}(t)\|^2
=\left (\frac{E-qCx}{mc}\right )^2-c^2\end{align*}
or not? (Wondering)

Can we do something further here? (Wondering)

Sure.
As a next step, we have:
$$\dot x(t)^2+\dot y(t)^2+\dot z(t)^2=\left (\frac{E-qCx(t)}{mc}\right )^2-c^2$$

For starters we want to find a single solution, later on we can try to generalize.
So let's start with $y(t)=z(t)=0$, and see if we can find a solution.
Then:
$$\dot x(t)^2=\left (\frac{E-qCx(t)}{mc}\right )^2-c^2$$
(Thinking)

 

[mathmari]

Then:
$$\dot x(t)^2=\left (\frac{E-qCx(t)}{mc}\right )^2-c^2$$
(Thinking)

According to Wolfram the solutions are the following:
$$x(t) = \frac{-m^2c^4e^{qCk_1-\frac{qCt}{mc}}-e^{\frac{qCt}{mc}-qCk_1}+2E}{2qC}$$ and $$x(t) = \frac{-m^2c^4e^{\frac{qCt}{mc}+qCk_1}-e^{-\frac{qCt}{mc}-qCk_1}+2E}{2qC}$$
So, these are in this case the motion equations, aren't they? (Wondering)

 

[I like Serena]

According to Wolfram the solutions are the following:
$$x(t) = \frac{-m^2c^4e^{qCk_1-\frac{qCt}{mc}}-e^{\frac{qCt}{mc}-qCk_1}+2E}{2qC}$$ and $$x(t) = \frac{-m^2c^4e^{\frac{qCt}{mc}+qCk_1}-e^{-\frac{qCt}{mc}-qCk_1}+2E}{2qC}$$
So, these are in this case the motion equations, aren't they? (Wondering)

That looks strange does it not? (Sweating)

And we were expecting something like a linearly accelerated particle.
Let's check by simplifying further.

Suppose we pick:
\begin{array}{}\mathbf A &= \mathbf 0 &\quad\Rightarrow\quad \mathbf B = \nabla \mathbf A = \mathbf 0\\
\phi &= -(1 \text{ N/C})x &\quad\Rightarrow\quad \mathbf E= -\nabla\phi - \pd {\mathbf A}t = (1 \text{ N/C})\mathbf{\hat x} \\
m&=1\text{ kg} \\
q &= 1\text{ C} \\
c &= 3\cdot 10^8 \text{ m/s} \\
\mathbf x(0) &= 0 \\
\mathbf{\dot x}(0) &= 0 \\
\end{array}
Since the mass is initially at rest, and has electric potential $\phi_0=0$, it follows that:
$$E=mc^2$$
And classically we have:
$$m\mathbf{\ddot x} =\mathbf F = q(\mathbf E + \mathbf v \times \mathbf B) \quad\Rightarrow\quad \ddot x = 1
\quad\Rightarrow\quad \dot x=t
\quad\Rightarrow\quad x=\frac 12 t^2$$
So we should find a solution that is close to this one - certainly for non-relativistic speeds. Do we? (Thinking)

 

[I like Serena]

Hey!

We have an electromagnetic field. The relation between energy and momentum of a particle with charge $q$ is $$\left (\frac{E-q\phi}{c}\right )^2=m^2c^2+\|p-\frac{q}{c}A\|^2$$ where $c$ is the velocity of light and $\phi=\phi (x,t), A=A(x,t)$ are the scalar and vector potential of the field with values in $\mathbb{R}$ and $\mathbb{R}^3$ respectively.
Find the relation between the velocity and momentum and the equation of motion.

So, to get the relation between the velocity and momentum, we have to expresss the energy in respect of the velocity, right?

Hi again mathmari! (Smile)

I've just realized that we're supposed to use Hamilton's equations of motion here.

[box=blue]

In Hamiltonian mechanics, a classical physical system is described by a set of canonical coordinates $\boldsymbol r = (\boldsymbol q, \boldsymbol p)$, where each component of the coordinate $q_i$, $p_i$ is indexed to the frame of reference of the system.

The time evolution of the system is uniquely defined by Hamilton's equations:
$$ \frac{\mathrm{d}\boldsymbol{p}}{\mathrm{d}t} = -\frac{\partial \mathcal{H}}{\partial \boldsymbol{q}}\quad,\quad \frac{\mathrm{d}\boldsymbol{q}}{\mathrm{d}t} = +\frac{\partial \mathcal{H}}{\partial \boldsymbol{p}}$$
where $\mathcal H = \mathcal H(\boldsymbol q, \boldsymbol p, t)$ is the Hamiltonian, which often corresponds to the total energy of the system.​

[/box]

In our case the total energy $E=E(x,p,t)$ is the Hamiltonian $\mathcal H$, and $x$ is the generalized coordinate $\boldsymbol q$.
So Hamilton's equations become:
[box=yellow]
$$\dot p = -\pd E x \quad,\quad \dot x = \pd E p$$
[/box]
The requested relation between velocity and momentum follows from the second equation.
And with the first equation the equation of motion follows as well.

So:
$$\left (\frac{E-q\phi}{c}\right )^2=m^2c^2+\|p-\frac{q}{c}A\|^2 \quad\Rightarrow\quad
\pd {}p \left[\left (\frac{E-q\phi}{c}\right )^2\right]=\pd{}p \left[m^2c^2+\|p-\frac{q}{c}A\|^2\right] \\ \quad\Rightarrow\quad
2\left (\frac{E-q\phi}{c}\right )\cdot \frac 1c \pd E p = \pd {}p\left( \|p-\frac{q}{c}A\|^2\right) \quad\Rightarrow\quad
\frac 2c\left (\frac{E-q\phi}{c}\right )\dot x = 2(p-\frac{q}{c}A) \\ \quad\Rightarrow\quad
\dot x = \frac{(p-\frac{q}{c}A)c}{\left (\frac{E-q\phi}{c}\right )}
= \frac{(p-\frac{q}{c}A)c}{\sqrt{m^2c^2+\|p-\frac{q}{c}A\|^2}}
$$

This is the relation between velocity $\dot x$ and momentum $p$.

We can solve it for $p$ and rewrite it as:
$$
p = \frac{m\dot x}{\sqrt{1-\frac{\dot x^2}{c^2}}} + \frac qc A
$$
And indeed, this looks like the familiar $p=mv$ or rather the relativistic version $p=m\gamma v$.We still have to evaluate $\dot p = -\pd E x$, after which we can complete the equations of motion and find $x(t)$. (Sweating)