Partial integration vs total integration and time-dependent force

[Happiness]

Integration as antiderivative

Question: Why isn't a distinction made between anti-total-derivatives and anti-partial-derivatives in common usage of integration?

Consider the functions $G_1(x, t)$ and $F(x, t)$ such that $F(x, t)=\frac{d}{dx}G_1(x, t)=\frac{\partial G}{\partial x}+\frac{dt}{dx}\frac{\partial G}{\partial t}=\frac{\partial G}{\partial x}+\frac{1}{v}\frac{\partial G}{\partial t}$, where $v=\frac{dx}{dt}$ is the velocity.

Then we define $G_1(x, t)$ as the anti-total-derivative (also known as the total-integration) of $F(x, t)$ with respect to $x$. $F(x, t)=\frac{d}{dx}G_1(x, t)\iff G_1(x, t)=\int_a^xF(x', t(x'))d_{full}x'+c$, where $a$ and $c$ are constants.

Next, consider the functions $G_2(x, t)$ and $F(x, t)$ such that $F(x, t)=\frac{\partial}{\partial x}G_2(x, t)$.

Then we define $G_2(x, t)$ as the anti-partial-derivative (also known as the partial-integration) of $F(x, t)$ with respect to $x$. $F(x, t)=\frac{\partial}{\partial x}G_2(x, t)\iff G_2(x, t)=\int_{a(y)}^xF(x', t(x'))d_{partial}x'+c(y)$.

Example: Partial integration

Question: Is equation (2) below correct?

Consider a force $F$ that varies with displacement $x$ and time $t$. In general, the acceleration $a$ and velocity $v$ of a particle subjected to such a force also vary with $x$ and $t$.

$a=\frac{\partial v}{\partial x}\frac{dx}{dt}+\frac{\partial v}{\partial t}=v\frac{\partial v}{\partial x}+\frac{\partial v}{\partial t}$

Since $F=ma=mv\frac{\partial v}{\partial x}+m\frac{\partial v}{\partial t}$,

$mv\frac{\partial v}{\partial x}=F-m\frac{\partial v}{\partial t}.\,\,\,\,\,$ -------- (1)

Integrating (partial-integration) both sides wrt $x$,

$\int mv\frac{\partial v}{\partial x} d_{partial}x=\int(F(x, t)-m\frac{\partial v}{\partial t})d_{partial}x$

$\frac{1}{2}mv^2=g(x, t)+h(t)\,\,\,\,\,$ -------- (2) Is this correct?

where $\frac{\partial}{\partial x}g(x, t)=F(x, t)-m\frac{\partial v(x, t)}{\partial t}$.

Example: Total differential

Question: I obtain below $\frac{\partial v}{\partial t}=0$, which is inconsistent with $F(x, t)$. What's wrong?

Multiplying $dx$ to both sides of (1),

$mv\frac{\partial v}{\partial x}dx=(F-m\frac{\partial v}{\partial t})dx$.

Adding $mv\frac{\partial v}{\partial t}dt$ to both sides,

$mv(\frac{\partial v}{\partial x}dx+\frac{\partial v}{\partial t}dt)=(F-m\frac{\partial v}{\partial t})dx+mv\frac{\partial v}{\partial t}dt.\,\,\,\,\,$ -------- (3)

Since $dv=\frac{\partial v}{\partial x}dx+\frac{\partial v}{\partial t}dt$ and $vdt=dx$,

$mvdv=(F-m\frac{\partial v}{\partial t})dx+m\frac{\partial v}{\partial t}dx=Fdx$.

Integrating both sides,

$\frac{1}{2}mv^2=\int F(x, t)dx$.

I believe the integration on the RHS should be a partial-integration. (But why?) Then

$\frac{1}{2}mv^2=G(x, t)+H(t)$ (This is consistent with equation (2).)

where $\frac{\partial}{\partial x}G(x, t)=F(x, t)$.

But simplifying the RHS of (3), we have

$mv\frac{\partial v}{\partial x}dx+mv\frac{\partial v}{\partial t}dt=Fdx+0dt$

By comparing terms with $dt$, we deduce

$\frac{\partial v}{\partial t}=0$.

But this is inconsistent with a time-dependent force $F(x, t)$. What's wrong?

 

[Charles Link]

Integration as antiderivative

Question: Why isn't a distinction made between anti-total-derivatives and anti-partial-derivatives in common usage of integration?

Consider the functions $G_1(x, t)$ and $F(x, t)$ such that $F(x, t)=\frac{d}{dx}G_1(x, t)=\frac{\partial G}{\partial x}+\frac{dt}{dx}\frac{\partial G}{\partial t}=\frac{\partial G}{\partial x}+\frac{1}{v}\frac{\partial G}{\partial t}$, where $v=\frac{dx}{dt}$ is the velocity.

Then we define $G_1(x, t)$ as the anti-total-derivative (also known as the total-integration) of $F(x, t)$ with respect to $x$. $F(x, t)=\frac{d}{dx}G_1(x, t)\iff G_1(x, t)=\int_a^xF(x', t(x'))d_{full}x'+c$, where $a$ and $c$ are constants.

Next, consider the functions $G_2(x, t)$ and $F(x, t)$ such that $F(x, t)=\frac{\partial}{\partial x}G_2(x, t)$.

Then we define $G_2(x, t)$ as the anti-partial-derivative (also known as the partial-integration) of $F(x, t)$ with respect to $x$. $F(x, t)=\frac{\partial}{\partial x}G_2(x, t)\iff G_2(x, t)=\int_{a(y)}^xF(x', t(x'))d_{partial}x'+c(y)$.

Example: Partial integration

Question: Is equation (2) below correct?

Consider a force $F$ that varies with displacement $x$ and time $t$. In general, the acceleration $a$ and velocity $v$ of a particle subjected to such a force also vary with $x$ and $t$.

$a=\frac{\partial v}{\partial x}\frac{dx}{dt}+\frac{\partial v}{\partial t}=v\frac{\partial v}{\partial x}+\frac{\partial v}{\partial t}$

Since $F=ma=mv\frac{\partial v}{\partial x}+m\frac{\partial v}{\partial t}$,

$mv\frac{\partial v}{\partial x}=F-m\frac{\partial v}{\partial t}.\,\,\,\,\,$ -------- (1)

Integrating (partial-integration) both sides wrt $x$,

$\int mv\frac{\partial v}{\partial x} d_{partial}x=\int(F(x, t)-m\frac{\partial v}{\partial t})d_{partial}x$

$\frac{1}{2}mv^2=g(x, t)+h(t)\,\,\,\,\,$ -------- (2) Is this correct?

where $\frac{\partial}{\partial x}g(x, t)=F(x, t)-m\frac{\partial v(x, t)}{\partial t}$.

Example: Total differential

Question: I obtain below $\frac{\partial v}{\partial t}=0$, which is inconsistent with $F(x, t)$. What's wrong?

Multiplying $dx$ to both sides of (1),

$mv\frac{\partial v}{\partial x}dx=(F-m\frac{\partial v}{\partial t})dx$.

Adding $mv\frac{\partial v}{\partial t}dt$ to both sides,

$mv(\frac{\partial v}{\partial x}dx+\frac{\partial v}{\partial t}dt)=(F-m\frac{\partial v}{\partial t})dx+mv\frac{\partial v}{\partial t}dt.\,\,\,\,\,$ -------- (3)

Since $dv=\frac{\partial v}{\partial x}dx+\frac{\partial v}{\partial t}dt$ and $vdt=dx$,

$mvdv=(F-m\frac{\partial v}{\partial t})dx+m\frac{\partial v}{\partial t}dx=Fdx$.

Integrating both sides,

$\frac{1}{2}mv^2=\int F(x, t)dx$.

I believe the integration on the RHS should be a partial-integration. (But why?) Then

$\frac{1}{2}mv^2=G(x, t)+H(t)$ (This is consistent with equation (2).)

where $\frac{\partial}{\partial x}G(x, t)=F(x, t)$.

But simplifying the RHS of (3), we have

$mv\frac{\partial v}{\partial x}dx+mv\frac{\partial v}{\partial t}dt=Fdx+0dt$

By comparing terms with $dt$, we deduce

$\frac{\partial v}{\partial t}=0$.

But this is inconsistent with a time-dependent force $F(x, t)$. What's wrong?

Just a quick comment or two: Your force F you can write as a field that is a function of x and t, but the velocity is simply a function of t or a function of x where x=x(t). You can not separate the velocity with v=v(x,t). a=a(x,t)=F(x,t)/m seems to be ok, but a(x,t) is a two parameter function and really should not be confused with the actual acceleration a(t)=a(x(t),t) that occurs... editing... Your "force" F=F(x,t) is really a field function and not an actual force. The actual force experienced is F(t)=F(x(t),t). It would be good to even subscript the $F_p(t)$ and $a_p(t)$ so that it doesn't get confused with F(x,t) and a(x,t). I would also subscript the particle position $x_p(t)$ and write e.g. $F_p(t)=F(x_p(t),t)$ so that it doesn't get confused with the parameter x that is used in the field function $F(x,t)$.

 

[Charles Link]

Besides my response in post #2, you may find it of interest that a $\vec{v}(\vec{x},t)$ formalism is used in plasma physics where the particles (e.g. electrons) have a density $n(\vec{x},t)$ and a velocity $\vec{v}(\vec{x},t)$ in a field $E(\vec{x},t)$. Derivatives (vector derivatives, etc.) are done on the $n(\vec{x},t)$ and $\vec{v}(\vec{x},t)$ in some applications. Sometimes, the density is written $f(\vec{x},\vec{v},t)$ where the density function includes both position and velocity. Equations then can be written for $f(\vec{x},\vec{v},t)$ in an electromagnetic field and look similar to those of a charged particle in an electromagnetic field, but are somewhat different including the sign of the terms. Anyway, you may find this application of interest...editing... I just looked this up in one of my plasma physics textbooks (author Ichimaru). The equation that describes the behavior of $f(\vec{x},\vec{v},t)$ in an electromagnetic field is called the Vlasov equation.

 

[Happiness]

a $\vec{v}(\vec{x},t)$ formalism is used in plasma physics

So I guess equation (2) is correct?

$\frac{1}{2}mv^2=g(x(t), t)+h(t)$

That can be interpreted as the kinetic energy (or rather, the change in kinetic energy) depends on $x(t)$, which is the path the particle takes.

But I'm still confused with why $\frac{\partial v}{\partial t}=0$ and how do we know if the integration $\int dx$ in the steps following equation (3) is a partial-integration. What if we express $F(x, t(x))$ as a function $J(x)$ of $x$ and then perform the integration $\int dx$? That is,

$\frac{1}{2}mv^2=\int J(x)dx=K(x)+c$.

 

[Charles Link]

So I guess equation (2) is correct?

$\frac{1}{2}mv^2=g(x(t), t)+h(t)$

That can be interpreted as the kinetic energy (or rather, the change in kinetic energy) depends on $x(t)$, which is the path the particle takes.

But I'm still confused with why $\frac{\partial v}{\partial t}=0$ and how do we know if the integration $\int dx$ in the steps following equation (3) is a partial-integration. What if we express $F(x, t(x))$ as a function $J(x)$ of $x$ and then perform the integration $\int dx$? That is,

$\frac{1}{2}mv^2=\int J(x)dx=K(x)+c$.

The difficulty I think is your use of v(x,t). In the plasma physics, (and also in fluid flow), v(x,t) describes the velocity of the flow at (x,t). In the case of a single particle, it is incorrect to use this formalism. It seems you may be eager to use this mathematics in some form=it really is an interesting formalism. e.g. The derivative $dn(\vec{x},t)/dt=\vec{v} \cdot \nabla n(\vec{x},t)+\frac{\partial{n(\vec{x},t)}}{\partial{t}}$ describes the rate of change of $n$ with time when moving along with the fluid. If you want to research it further, you can find it being applied in some form in most introductory plasma physics textbooks.

 

[wrobel]

Consider a force FF that varies with displacement xx and time tt. In general, the acceleration aa and velocity vv of a particle subjected to such a force also vary with xx and tt.

nonsense

 

[Charles Link]

nonsense

It is not uncommon for an electric field to be described by $\vec{E}=\vec{E}(\vec{x},t)$. The force $\vec{F}$ for a particle of charge $Q$ is given by $\vec{F}=Q \vec{E}(\vec{x},t)$. The OP's mistake is to use equations that are designed for a fluid flow, such as $\vec{v}=\vec{v}(\vec{x},t)$ to describe a single particle in which case it must be written $\vec{v}=\vec{v}(t)$.

 

[wrobel]

It is not uncommon for an electric field to

what exactly is not uncommon? to claim that the velocity of the particle depends upon its position in the space?

 

[Charles Link]

what exactly is not uncommon? to claim that the velocity of the particle depends upon its position in the space?

I've been trying to explain to the OP how using v=v(x,t) is incorrect for a single particle, but it can be used in applications such as fluid flow It appears the OP is still on a learning curve with some of these concepts... v=v(t) can also be written as v=v(x) since x=x(t) for the particle (where x is the location of the particle at time t), but it is incorrect to write the velocity v=v(x,t) as a field equation except in the case of a fluid. I do think the OP, with a little further study, will be able to see why what he attempted to do simply is incorrect.