Acceleration of A Classical Particle

[trfrm]

Gloria in excelsis Deo.

Excuse me ... .

In sight, it is impossible that a classical particle can be accelerated without a force acting on it ... . In other word ... , the velocity of the particle cannot change if the force acting on it is zero, sightly ... .

But ... , in fact, this opinion is not always true generally ... .

Really, the Newton’s second law of motion is a differential equation

$\displaystyle\mathbf{F}=\frac{d\mathbf{p}}{dt}$

where the $\mathbf{F}$ and the $\mathbf{p}$, respectively, are the force and the momentum of the particle as function of the time $t$ ... .

If the velocity and mass of the particle, respectively, are $\mathbf{v}$ and $m$ as function of time $t$, then $\mathbf{p}=m\mathbf{v}$ ... . Thus, in general,

$\mathbf{F}=\dot{m}\mathbf{v}+m\mathbf{a}$

where $\dot{m}:=dm/dt$ is rate of the mass, and $\mathbf{a}:=d\mathbf{v}/dt$ is its acceleration ... . The last expression doesn’t only obey in non-relativistic cases, but also in relativistic cases ... .

If we set $\mathbf{F}=\mathbf{0}$, then $\mathbf{a}=-\dot{m}\mathbf{v}/m$ ... . In this case, $d\mathbf{p}/dt=\mathbf{0}$ alias $\mathbf{p}=\mathbf{p}_0$ alias $m\mathbf{v}=m_0\mathbf{v}_0$ alias $\mathbf{v}=m_0\mathbf{v}_0/m$ ... . Thus (by subtituting or by differentiating), this yields $\mathbf{a}=-\dot{m}m_0\mathbf{v}_0/m^2$ ... . [Here, the $m_0$ and the $\mathbf{v}_0$ denote the initial mass and the initial velocity, respectively ... .]

In another case, if $\mathbf{F}=\dot{m}\mathbf{v}$, then $\mathbf{a}=\mathbf{0}$ ... .

Therefore, really, the force merely changes the momentum rather than the velocity ... .

Alhamdulillah hirobbil alamin.

[trfrm]

Namo Buddhaya.

This stuff is really bad. Care to try again, given the information that $\mathbf{p}=\gamma m_0 \mathbf{v}$ . May I suggest that you forget about the nonsense $\dot{m}:=dm/dt$. This doesn't exist.
If you want to do "classical domain", just set $\gamma=1$

O.K. ... . Thanks ... .

We have known that the relativistic momentum is $\mathbf{p}=m\mathbf{v}=\gamma{m_0}\mathbf{v}$, where $m=\gamma{m_0}$ and $\gamma:=1/\sqrt{1-|\mathbf{v}|^2/c^2}$ ... . [Here, the $m_0$ is not the initial mass, but is the rest mass ... .] Then,

$\mathbf{F}=\dot{\mathbf{p}}=\dot{m}\mathbf{v}+m\mathbf{a}$

where $\dot{m}=\dot{\gamma}m_0$ and $\dot\gamma=\gamma^3(\mathbf{v}\cdot\mathbf{a})/c^2$ ... , so

$\dot{m}=m\gamma^2(\mathbf{v}\cdot\mathbf{a})/c^2$ ... .

Thus ... (by subtituting), this yields

$\mathbf{F}=m(\gamma^2(\mathbf{v}\cdot\mathbf{a})\mathbf{v}/c^2+\mathbf{a})$

which be known as the relativistic force ... .

[Note that the special relativity (without quantum aspect) is sometimes classified into the classical mechanics subject ... . I saw it in the Classical Mechanics textbook by Hebert Goldstein, 1980 ... .]

Dalam Nama Bapa dan Putera dan Roh Kudus. Amin.

[trfrm]

Horas.

Hence, if the force was zero, then the acceleration should not be zero ... . In other word (in the relativistic case without presence of force), $\mathbf{v}=m_i\mathbf{v}_i/m=\gamma_i\mathbf{v}_i/\gamma$, so $c^2|\mathbf{v}|^2={\gamma_i}^2|\mathbf{v}_i|^2(c^2-|\mathbf{v}|^2)$ alias $(c^2+{\gamma_i}^2|\mathbf{v}_i|^2)|\mathbf{v}|^2=c^2{\gamma_i}^2|\mathbf{v}_i|^2$ alias

$\displaystyle|\mathbf{v}|=\frac{c\gamma_i|\mathbf{v}_i|}{\sqrt{c^2+{\gamma_i}^2|\mathbf{v}_i|^2}}$ ... .

N.B. : The $\gamma_i$ and the $\mathbf{v}_i$, respectively, are the initial gamma and the initial velocity (at $t=0$) ... .

Because the final velocity has same direction with the initial velocity, then

$\displaystyle\mathbf{v}=|\mathbf{v}|\frac{\mathbf{v}_i}{|\mathbf{v}_i|}=\frac{c\gamma_i\mathbf{v}_i}{\sqrt{c^2+{\gamma_i}^2|\mathbf{v}_i|^2}}=\frac{c\mathbf{v}_i}{\sqrt{c^2/{\gamma_i}^2+|\mathbf{v}_i|^2}}=\mathbf{v}_i=\textrm{constant}$

Thus, in this case, $\mathbf{a}=\dot{\mathbf{v}}=\mathbf{0}$ if $\mathbf{F}=\mathbf{0}$ ... .

Nderek langkung.

[trfrm]

Terpujilah Kristus.

Oh O.K. ... . From the identity of vectors $\mathbf{A}\times(\mathbf{B}\times\mathbf{C})=(\mathbf{A}\cdot\mathbf{C})\mathbf{B}-(\mathbf{A}\cdot\mathbf{B})\mathbf{C}$, then $\mathbf{v}\times(\mathbf{v}\times\mathbf{a})=(\mathbf{v}\cdot\mathbf{a})\mathbf{v}-|\mathbf{v}|^2\mathbf{a}$ alias $(\mathbf{v}\cdot\mathbf{a})\mathbf{v}=\mathbf{v}\times(\mathbf{v}\times\mathbf{a})+|\mathbf{v}|^2\mathbf{a}$ ... .

Thus,

$\mathbf{F}=m(\gamma^2(\mathbf{v}\times(\mathbf{v}\times\mathbf{a})+|\mathbf{v}|^2\mathbf{a})/c^2+\mathbf{a})$ alias

$\mathbf{F}=m(\gamma^2\mathbf{v}\times(\mathbf{v}\times\mathbf{a})/c^2+(1+\gamma^2|\mathbf{v}|^2/c^2)\mathbf{a})$ ... .

Since $1+\gamma^2|\mathbf{v}|^2/c^2=\gamma^2$, then

$\mathbf{F}=\gamma^2m(\mathbf{v}\times(\mathbf{v}\times\mathbf{a})/c^2+\mathbf{a})$ ... .

In the last equation, we can see that there are transversal and longitudinal components of the force ... .

Haleluya.

[trfrm]

Agnus Dei, qui tollis peccata mundi.

I don't know where you are trying to get, you simply have:


$0=\mathbf{F}=m_0\gamma(\frac{\gamma^2\mathbf{v}\cdot\mathbf{a}}{c^2}\mathbf{v}+\mathbf{a})$

meaning that you get $\mathbf{a}=-\frac{\gamma^2 \mathbf{v}\cdot\mathbf{a}}{c^2}\mathbf{v}$

A little further algebraic manipulation shows that the above is impossible.

Here ... , incidentally, $\mathbf{a}=\mathbf{0}$ ... .

It's getting worse, you should really arrive to :

$\mathbf{F}=m_0\gamma(\frac{\gamma^2\mathbf{v}\cdot\mathbf{a}}{c^2}\mathbf{v}+\mathbf{a})$

It turns out that the equation $\mathbf{F}=m(\gamma^2(\mathbf{v}\cdot\mathbf{a})\mathbf{v}/c^2+\mathbf{a})$ is not different from the equation $\mathbf{F}=\gamma^2m(\mathbf{v}\times(\mathbf{v}\times\mathbf{a})/c^2+\mathbf{a})$ ... .

http://www.thescienceforum.com/physics/35197-acceleration-classical-particle-without-force-acting-particle.html#post413250

Hosana in excelcis.