# Relativistic particle under increasing force

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

• 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. – Robert Poenaru Jul 26 '17 at 13:19
• 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? – Derangedmuon Jul 26 '17 at 13:28
• Well...if you want to study a relativistic particle, I guess you have to drop the newton's law :D feelsbadman – Robert Poenaru Jul 26 '17 at 13:31
• 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. – Derangedmuon Jul 26 '17 at 13:35
• 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). – Frobenius Jul 26 '17 at 21:23

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!

• 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. – Ben Crowell Jul 26 '17 at 16:26
• Yes. I was just trying to preserve something of Derangedmuon's approach. – Philip Wood Jul 26 '17 at 16:55

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

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 $\,t\,$ $$\left(1\!-\!\left[\dfrac{m_{0} c^{2}}{m_{0}c^{2}+\mathrm{W}\left(x\right)}\right]^{2}\right)^{-\frac12}\mathrm{d}x=c\,\mathrm{d}t \tag{17}$$ 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) $$x\left(t\right)=\dfrac{m_{o}c^{2}}{f_{0}}\left[\sqrt{1+\left(\dfrac{f_{0}t}{m_{o}c}\right)^{2}}-1\right] \tag{18}$$

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