KAMP
Fisika => Mekanika Klasik => Topik dimulai oleh: trfrm pada Juli 20, 2018, 12:44:43 PM
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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 ... .
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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 ... .]
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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 $\mathbf{F}=\mathbf{0}$ ... .
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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 ... .
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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
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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
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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 ... .