Gradient theorem for time-dependent vector field

[dEdt]

Let's say we have some time-independent scalar field $\phi$. Obviously $$\phi\left(\mathbf{q}\right)-\phi\left(\mathbf{p}\right) = \int_{\gamma[\mathbf{p},\,\mathbf{q}]} \nabla\phi(\mathbf{x})\cdot d\mathbf{x}.$$
This is of course still true if the path $\gamma$ is the trajectory of a particle moving through space. But let's say we have a time-dependent field instead, with $\gamma$ still being the trajectory of the particle. Will
$$\phi(\mathbf{q},t_2)-\phi(\mathbf{p},t_1)=\int_{\gamma[\mathbf{p},\,\mathbf{q},t]} \nabla\phi(\mathbf{x} (t))\cdot d\mathbf{x}?$$

 

[Mandelbroth]

Let's say we have some time-independent scalar field $\phi$. Obviously $$\phi\left(\mathbf{q}\right)-\phi\left(\mathbf{p}\right) = \int_{\gamma[\mathbf{p},\,\mathbf{q}]} \nabla\phi(\mathbf{x})\cdot d\mathbf{x}.$$
This is of course still true if the path $\gamma$ is the trajectory of a particle moving through space. But let's say we have a time-dependent field instead, with $\gamma$ still being the trajectory of the particle. Will
$$\phi(\mathbf{q},t_2)-\phi(\mathbf{p},t_1)=\int_{\gamma[\mathbf{p},\,\mathbf{q},t]} \nabla\phi(\mathbf{x} (t))\cdot d\mathbf{x}?$$

Consider that the gradient theorem is basically a thinly disguised form of Stokes' Theorem, id est,

$$\int\limits_{\partial\gamma} \phi = \int\limits_{\gamma} d\phi$$

So, let's consider evaluating the exterior derivative of $\phi(\vec{x}(t))$.

$$d\left(\phi(\vec{x}(t))\right) = \sum_{i=1}^{n}\frac{\partial\phi}{\partial x^i}dx^i = \langle\nabla\phi(\vec{x}(t)), \cdot\rangle$$
where the angle brackets denote the inner product of phi with something.

Do you agree with this? There are other ways to do this, but this method seems easiest.

(I'm fairly sure I did this right, but if someone from the upper math-y places wants to correct me, they should.)

If this works how I thinks it does,

$$\int\limits_{\partial\gamma}\phi(\vec{x}(t)) = \phi(\vec{q}(t))-\phi(\vec{p}(t)) = \int\limits_{\gamma} \langle\nabla\phi(\vec{x}(t)), d\vec{x}\rangle$$

where $\vec{q}(t)$ is the vector q at time t.

 

[dEdt]

I'm sorry, but you're using a lot of terminology and notation that I'm unfamiliar with. What does $$\int_{\partial \gamma}\phi$$ mean? What's an exterior derivative?

 

[Mandelbroth]

I'm sorry, but you're using a lot of terminology and notation that I'm unfamiliar with. What does $$\int_{\partial \gamma}\phi$$ mean? What's an exterior derivative?

Oh...sorry. That may be out of what you've learned before. My apologies.

$\int\limits_{\partial\gamma}\phi$ would be the oriented sum of phi at the endpoints of $\gamma$. Because $\gamma$ is 1-dimensional (it's a curve), its boundary is a finite set of points (namely, its endpoints).

The exterior derivative is a a generalization of the differential. See here.

 

[HallsofIvy]

Let's say we have some time-independent scalar field $\phi$. Obviously $$\phi\left(\mathbf{q}\right)-\phi\left(\mathbf{p}\right) = \int_{\gamma[\mathbf{p},\,\mathbf{q}]} \nabla\phi(\mathbf{x})\cdot d\mathbf{x}.$$
This is of course still true if the path $\gamma$ is the trajectory of a particle moving through space. But let's say we have a time-dependent field instead, with $\gamma$ still being the trajectory of the particle. Will
$$\phi(\mathbf{q},t_2)-\phi(\mathbf{p},t_1)=\int_{\gamma[\mathbf{p},\,\mathbf{q},t]} \nabla\phi(\mathbf{x} (t))\cdot d\mathbf{x}?$$

The left side is evaluated at two different values of t. What value of t are you using on the right?