2
$\begingroup$

Can it be shown that the limiting velocity of a mass is the speed of light under any nature of external force? Specifically with the example of say, a linearly increasing force F(1+bx).

I tried some cases but the integrals seemed to get too tedious for a leisurely nab at physics :p

$\endgroup$
  • $\begingroup$ Does the relativistic mass formula help you? If you take that formula and put v=c, you will obtain a singularity, so the mass goes to infinity. Which in some sense, provides the speed limit of a mass particle. $\endgroup$ – Robert Poenaru Jul 26 '17 at 13:19
  • $\begingroup$ I was looking for something from the newton's law's perspective. Something like the force F(1+bx) itself goes to infinity, but does the velocity do so? $\endgroup$ – Derangedmuon Jul 26 '17 at 13:28
  • $\begingroup$ Well...if you want to study a relativistic particle, I guess you have to drop the newton's law :D feelsbadman $\endgroup$ – Robert Poenaru Jul 26 '17 at 13:31
  • $\begingroup$ But is there any way to do so? The classic example of a constant force on a relativistic particle is in most standard books. But none involve a force which itself goes to infinity like the one above. I was curious if the same treatment could be carried out. $\endgroup$ – Derangedmuon Jul 26 '17 at 13:35
  • $\begingroup$ Take a look in my answer as user82794 herein : Inertia on relativistic mass when particle is near speed of light. Look at this as an exercise of non-relativistic classical mechanics. It answers your question for constant force in one dimension (b=0). $\endgroup$ – Frobenius Jul 26 '17 at 21:23
1
$\begingroup$

If you're interested in seeing what happened for some specific variation of force, I'm not going to say anything useful. My approach would be to use the general result (established in a few lines in any good SR textbook) that $$\int_{u=0}^{v} \frac{d [m \gamma(u)\ \vec{u}]}{dt}. d \vec{r} = m \gamma (v) c^2 -mc^2.$$ This equates the work done by any force (defined as rate of change of relativistic linear momentum) to kinetic energy acquired. $v$ is the 'final' velocity of the body due to the continued action of the force. But, as we know, $\gamma (v)$ goes to infinity as $v$ approaches $c$, so the force would have to do an infinite amount of work to get the body up to the speed of light!

$\endgroup$
  • $\begingroup$ If the OP wants a dynamical argument, then a simpler one is that $m^2=E^2-p^2$ (in natural units), where $m$ is a fixed property of the particle. Therefore we always have $E>|p|$, which implies subluminal velocity. But more fundamentally this is really a kinematic fact. Combining two Lorentz transformations (which by definition have $|v|<c$) results in another Lorentz transformation. Therefore no continuous process of acceleration can bring a particle from subluminal to superluminal speed. $\endgroup$ – Ben Crowell Jul 26 '17 at 16:26
  • $\begingroup$ Yes. I was just trying to preserve something of Derangedmuon's approach. $\endgroup$ – Philip Wood Jul 26 '17 at 16:55
0
$\begingroup$

enter image description here

Let a 1-dimensional force continuous, bounded and integrable
\begin{equation} 0<f\left(x\right)<+\infty\:, \qquad 0<\int\limits_{\xi=0}^{\xi=x}f\left(\xi\right)\mathrm{d}\xi<+\infty\:, \qquad\forall x \in \mathbb{R}^{+} \tag{01} \end{equation} acting on a particle initially at rest on the origin with rest mass $\;m_{0}$, see Figure. On one hand we have \begin{equation} \mathrm{d}\left(mc^2\right)=f\left(\xi\right)\mathrm{d}\xi=\mathrm{d}\mathrm{W}\left(\xi\right) \tag{02} \end{equation} so integrating and considering the initial conditions \begin{equation} m\left(x\right)=m_{0}+\dfrac{1}{c^{2}}\int\limits_{\xi=0}^{\xi=x}f\left(\xi\right)\mathrm{d}\xi=m_{0}+\dfrac{1}{c^{2}}\mathrm{W}\left(x\right) \tag{03} \end{equation} where \begin{align} \!\!\!\!\!\!\!\!\!\!\!\!W(x) & =\int\limits_{\xi=0}^{\xi=x}f\left(\xi\right)\mathrm{d}\xi=\text{work done by force $f(x)$ from $0$ to $x$} \quad [W(x)> 0\:\: \forall x>0] \\ &=\left(m-m_{0}\right)c^{2}=\text{kinetic energy of the particle} \tag{04} \end{align} and on the other hand \begin{equation} f=\dfrac{\mathrm{d}p}{\mathrm{d}t}=\dfrac{\mathrm{d}\left(mv\right)}{\mathrm{d}t} =m\dfrac{\mathrm{d}v}{\mathrm{d}t}+v\dfrac{\mathrm{d}m}{\mathrm{d}t} \tag{05} \end{equation} so \begin{equation} f\left(x\right)\mathrm{d}x=\frac12 m\,\mathrm{d}v^{2}+v^{2}\mathrm{d}m \tag{06} \end{equation} Replacing in (06) the term $\,\mathrm{d}m\,$ by its expression from (02) \begin{equation} f\left(x\right)\mathrm{d}x=\frac12 m\,\mathrm{d}v^{2}+\dfrac{v^{2}}{c^{2}}f\left(x\right)\mathrm{d}x \tag{07} \end{equation} so \begin{equation} \dfrac{f\left(x\right)\mathrm{d}x}{m}=\frac12\dfrac{\mathrm{d}v^{2}}{\left(1-\dfrac{v^{2}}{c^{2}}\right)} \tag{08} \end{equation} Inserting the expression (03) for $\,m\,$ and replacing $\,f\left(x\right)\mathrm{d}x \rightarrow \mathrm{d}\mathrm{W}\left(x\right)\,$ we reach the following differential equation with respect to the separated variables $\,x\,$ and $\,v^{2}\,$ \begin{equation} \dfrac{\mathrm{d}\mathrm{W}\left(x\right)}{\left[m_{0}c^{2}+\mathrm{W}\left(x\right)\vphantom{\frac12}\right]}=\frac12\dfrac{\mathrm{d}\left(\dfrac{v^{2}}{c^{2}}\right)}{\left(1\!-\!\dfrac{v^{2}}{c^{2}}\right)} \tag{09} \end{equation} or \begin{equation} \dfrac{\mathrm{d}\left[m_{0}c^{2}+\mathrm{W}\left(x\right)\vphantom{\frac12}\right]}{\hphantom{\mathrm{d}}\left[m_{0}c^{2}+\mathrm{W}\left(x\right)\vphantom{\frac12}\right]}=-\frac12 \dfrac{\mathrm{d}\left(1\!-\!\dfrac{v^{2}}{c^{2}}\right)}{\hphantom{\mathrm{d}}\left(1\!-\!\dfrac{v^{2}}{c^{2}}\right)} \tag{10} \end{equation} that is \begin{equation} \mathrm{d}\left(\ln\left[m_{0}c^{2}+\mathrm{W}\left(x\right)\vphantom{\tfrac12}\right]\vphantom{\frac12}\!\!\right)= \mathrm{d}\left[\ln\left(1\!-\!\dfrac{v^{2}}{c^{2}}\right)^{-\frac12}\vphantom{\frac12}\!\!\right] \tag{11} \end{equation} Considering the initial conditions the integration of (11) yields \begin{equation} m_{0}c^{2}+\mathrm{W}\left(x\right)=m_{0}c^{2}\left(1\!-\!\dfrac{v^{2}}{c^{2}}\right)^{-\frac12} \tag{12} \end{equation} From equations (12) and (03) \begin{equation} m\left(\upsilon\right)=\dfrac{m_{o}}{\sqrt{1-\dfrac{\upsilon^{2}}{c^{2}}}} \tag{13} \end{equation} while solving (12) with respect to $\,v\,$ we have its following expression as function of $\,x\,$ \begin{equation} v\left(x\right)=c\sqrt{1\!-\!\left[\dfrac{m_{0} c^{2}}{m_{0}c^{2}+\mathrm{W}\left(x\right)}\right]^{2}} \tag{14} \end{equation} Since $\,\mathrm{W}\left(x\right)\,$ is a positive increasing function of $\,x\,$ and \begin{equation} \lim_{x\rightarrow +\infty}\mathrm{W}\left(x\right)=+\infty \tag{15} \end{equation} we have \begin{equation} \lim_{x\rightarrow +\infty}v\left(x\right)=c \:, \qquad 0<v\left(x\right)<c \quad \forall x \in \mathbb{R}^{+} \tag{16} \end{equation}

If we want to find the position as function of time, that is $\,x\left(t\right)\,$, then from equation (14) given that $\,v\left(x\right)=\mathrm{d}x/\mathrm{d}t\,$, we have after separation of the variables $\,x\,$ and $\,t\,$ \begin{equation} \left(1\!-\!\left[\dfrac{m_{0} c^{2}}{m_{0}c^{2}+\mathrm{W}\left(x\right)}\right]^{2}\right)^{-\frac12}\mathrm{d}x=c\,\mathrm{d}t \tag{17} \end{equation} To have analytic solution depends upon the possibility of analytic integration of the function of the lhs of (17).
For example in case of constant force, $\,f\left(x\right)=f_{0}>0 \quad \forall x \in \mathbb{R}^{+}\,$, the analytic solution is(1) \begin{equation} x\left(t\right)=\dfrac{m_{o}c^{2}}{f_{0}}\left[\sqrt{1+\left(\dfrac{f_{0}t}{m_{o}c}\right)^{2}}-1\right] \tag{18} \end{equation}


(1) See in my answer here :Inertia on relativistic mass when particle is near speed of light

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.