Acceleration of A Classical Particle

[trfrm]

Terpujilah Kristus.

If $\mathbf{F}=0$ this means $\mathbf{p}=$const, i.e. $\mathbf{p}=\gamma(v_0)m_0 \mathbf{v_0}$ (i.e $\mathbf{v}=\mathbf{v_0}$) so $\mathbf{a}=0$. This is the correct chain of inference, what you posted so far, isn't.

In this context (this relativistic case) ... , it is right ... .

But, in general case, if $\mathbf{F}=\mathbf{0}$, then $\mathbf{p}=\mathbf{p}_i$ (where the index $i$ denotes the initial state) ... , then ... , $m\mathbf{v}=m_i\mathbf{v}_i$ alias $\mathbf{v}=m_i\mathbf{v}_i/m$ ... .

Because in general, the mass is not constant, then

$\mathbf{a}:=\dot{\mathbf{v}}=-m_i\mathbf{v}_i\dot{m}/m^2$

which must not to be zero ... .

But, however, in the relativistic case, $\dot{m}=\gamma^2m(\mathbf{v}\cdot\mathbf{a})/c^2$, so the last equation becomes

$\mathbf{a}=-m_i\mathbf{v}_i\gamma^2(\mathbf{v}\cdot\mathbf{a})/(mc^2)$ ... .

The possible solution of the last equation is $\mathbf{a}=\mathbf{0}$ ... .

I think it is important to distinguish pure mathematics from physics in this context. Mathematically it is certainly possible to come up with all sorts of relations affecting speed and acceleration even in the absence of a force, but that does not necessarily mean those relations are physically meaningful. In the context of physics, changing the state of motion of an object requires changing its momentum, and, at least classically, that will require a force.

This seemly comes from the origin definition of the force as derivative of the linear momentum respect to time, that is the Newton’s second law of motion as a differential equation [Goldstein, Classical Mechanics, 1980, Chapter 1] ... .

http://www.thescienceforum.com/physics/35197-acceleration-classical-particle-without-force-acting-particle.html#post413388

Sampai jumpa lagi.

[trfrm]

Agnus Dei, qui tollis peccata mundi.

I don't know why you insist on using the antiquate notion of relativistic mass, this is the source of confusion. Using the modern notion of invariant mass makes things a lot more straightforward. And yes, $\mathbf{a}=0$ is the only acceptable solution for the equation:

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

I’m sorry ... . There, $m_i=\gamma_im_0$ and $m=\gamma{m_0}$ are the initial and final relativistic mass, respectively ... .

Dalam Nama Bapa dan Putera dan Roh Kudus. Amin.