0
$\begingroup$

I was playing a bit with the equations of relativity, and just wanted to know how much time it would take to accelerate a mass to a certain (relativistic) velocity. Now this problem turns out to be a lot harder than I expected, as an integral doesn't seem to arise naturally from the problem.

The problem is basically this, we have an object of mass m and apply a constant force to it. In the classical case it just has a constant acceleration, but here that clearly is not the case a gamma factor comes in between. My line of attack has been using the fact that in the restframe of the mass it is actually really constantly accelerating at a certain acceleration. I then used the relativistic velocity addition formula to define a new kind of algebra:

$$v\oplus u= \frac{u+v}{1+uv} $$

Expressing velocities as a fraction of the lightspeed. Then I define multiplication of a velocity with a natural number in the following way:

$$ 0 \otimes v = 0 $$ $$ (n+1) \otimes v = v \oplus(n\otimes v) $$

This is what I used to get a nice numerical approximiation, although it get's very very messy after only a few iterations(but that's no problem for mathematica). If a constant force is applied that would lead to an acceleration of a in a classical context, and that force is applied to the system for T seconds in the restframe that leads to a change in speed for the external frame of:

$$\Delta v = \lim_{n->\infty} n\otimes a \frac{T}{n}$$

This get's more and more accurate the bigger I make n, and converges to the 'real' value when I let n approach infinity(as the classical approximiation get's more and more accurate over smaller and smaller time intervals). I don't know if there exist a nicer approach that actually leads to an explicit formula or at least some integral containing an explicit formula or something.

In mathematica I used n=T=3000 and assumed an acceleration of 0.001c/s. This gave me the following graphs, for speed and time dilation as function of (proper) time respectively:

enter image description here

The first looks a lot like the inverse tangent, although I couldn't find any formula of that form to actually fit. Is there actually a way to get a nice formula for either speed or time dilation as function of time?

And just to test the above approach: Imagine travelling in a spaceship accelerating at 10 m/s^2 constantly(a bit more than 1 g), then these graphs represent steps of time of about 300 seconds(time needed to reach 0.001 c classically), so that means that in 5000*300 seconds or about 17 days you should reach that speed as measured on a clock in the spaceship and thus reach a time dilation of about 74, integrating that second graph would then give the time for an outside observer, and would be a lot greater than 17 days. Does this sound reasonable at all?

It would be nice if an easier approach exists, if not I at least hope this one is somewhat correct.

$\endgroup$
  • $\begingroup$ Change velocities to rapidities and then the problem becomes linear. You're seeking to construct infinitessimal boosts and the one parameter boost group in $SO(1,3)$. Have a look at my anser here $\endgroup$ – WetSavannaAnimal Dec 1 '17 at 3:30
1
$\begingroup$

LINK : My answer as user82794 (former diracpaul) here : Inertia on relativistic mass when particle is near speed of light.

A Non-Relativistic Interpretation of Relativistic Results

(warning : all stuff here is old-fashioned, but gives you the answers)

When self-studying special relativity, an exercise was created in order to understand non-relativistically the relativistic result that a force could not accelerate a particle to speed greater than $\:c\:$. The exercise, the Figure and the solution are already in LaTeX and are copy-paste here. May be is useful.

The solution to the following exercise yields a non-relativistic interpretation of the relativistic result that energy $\:E\:$ has inertia, that is resistance to be accelerated, as material objects. The inertia of material objects is expressed by their inertial mass $\:m\:$ while the inertia of energy $\:E\:$ is expressed by its mass equivalent$\: m_{E}=E/c^{2}\:$. Also, based on this interpretation, an explanation of the impossibility of a force to accelerate an object to velocity greater than that of light ($c$) is given.

enter image description here

A body of mass $\:m_{o}\:$ initially at rest is moving on a surface without friction under the influence of a constant force $\:\mathbf{f}\:$, as in Figure above. As it is moving, it is carrying away material from a straight line above it. The straight line has a constant linear mass density $\:\rho_{\ell}=f/c^{2}\:$, where $\:\rm{f}\:$ the magnitude of the constant force $\:\mathbf{f}\:$ and $\:c\:$ a quantity with dimensions of speed.

The exercise concerns the determination of the following :

(a) the position $\:x(t)\:$, the velocity $\:\upsilon(t)\:$ and the mass $\:m(t)\:$ as functions of time.

(b) any explicit relation, if there exists, between above quantities.

Answers :

(for a proof see in the link above)

Position $\:x(\upsilon)\:$ as function of speed $\:\upsilon$ :

\begin{equation} x\left(\upsilon\right)=\dfrac{m_{o}c^{2}}{f}\left( \dfrac{1}{\sqrt{1-\dfrac{\upsilon^{2}}{c^{2}}}}-1\right) \tag{01} \end{equation}

Speed(1)$\:\upsilon(x)$ as function of position $\:x$ : \begin{equation} \upsilon\left(x\right) = c \sqrt{1-\dfrac{1}{\left(1+\dfrac{f}{m_{o}c^{2}}x\right)^{2}}} \tag{02} \end{equation}

Position(2)$\:x(t)\:$ as function of time $\:t$ : \begin{equation} x\left(t\right)=c \left[\sqrt{t^2+\left(\dfrac{m_{o}c}{f}\right)^{2}}-\left(\dfrac{m_{o}c}{f}\right)\right] \tag{03} \end{equation}

Speed $\:\upsilon(t)$ as function of time $\:t$ :

\begin{equation} \upsilon\left(t\right)=c \sqrt{1-\dfrac{1}{\left(\dfrac{f}{m_{o}c}t\right)^2+1}} \tag{04} \end{equation}

From (04) the time needed to accelerate the body from rest to speed $\:\upsilon$ is :

\begin{equation} \boxed{\:\:t=\dfrac{m_{o}}{f}\dfrac{\upsilon}{\sqrt{1-\dfrac{\upsilon^{2}}{c^{2}}}}=\dfrac{\gamma\,m_{o}\,\upsilon}{f} \:\: \vphantom{\dfrac{\frac12}{\sqrt{1-\dfrac{\upsilon^{2}}{c^{2}}}}}} \tag{05} \end{equation} Equation (05) is expressed as : \begin{equation} \dfrac{f\,t}{m_{o}\,c}=\dfrac{\upsilon}{c\sqrt{1-\dfrac{\upsilon^{2}}{c^{2}}}}=\dfrac{\gamma\,\upsilon}{c}=\sqrt{\gamma^{2}-1} \tag{06} \end{equation} From (06), or directly from (04) \begin{equation} \boxed{\:\:\gamma\left(t\right)=\sqrt{\left(\dfrac{f}{m_{o}c}\right)^{\!2}\!t^{2}+1}\:\: \vphantom{\dfrac{\dfrac12}{\dfrac12}}} \tag{07} \end{equation}

For the proper time $\:\tau\left(t\right)\:$ as function of the time $\:t\:$ we have \begin{equation} \mathrm d \tau=\dfrac{1}{\gamma\left(t\right)}\,\mathrm d t =\dfrac{1}{\sqrt{\left(\dfrac{f}{m_{o}c}\right)^{\!2}\!t^{2}+1}}\,\mathrm d t \tag{09} \end{equation} and under the condition $\:\tau\left(0\right)=0\:$ \begin{equation} \tau\left(t\right) =\int\limits_{0}^{t}\mathrm d \tau=\int\limits_{0}^{t}\dfrac{1}{\gamma\left(t\right)}\,\mathrm d t =\int\limits_{0}^{t}\dfrac{1}{\sqrt{\left(\dfrac{f}{m_{o}c}\right)^{\!2}\!t^{2}+1}}\,\mathrm d t =\dfrac{m_{0}c}{f}\sinh^{-1}\left(\dfrac{f}{m_{o}c}\,t\right) \tag{10} \end{equation} that is \begin{equation} \boxed{\:\:\tau\left(t\right) =\dfrac{m_{0}c}{f}\sinh^{-1}\left(\dfrac{f}{m_{o}c}\,t\right)\:\: \vphantom{\dfrac{\dfrac12}{\dfrac12}}} \tag{11} \end{equation} and inversely \begin{equation} \boxed{\:\:t\left(\tau\right) =\dfrac{m_{0}c}{f}\sinh\left(\dfrac{f}{m_{o}c}\,\tau\right)\:\: \vphantom{\dfrac{\dfrac12}{\dfrac12}}} \tag{12} \end{equation} Inserting this expression of $\:t\:$ in (04) we have the following function $\:\upsilon \left(\tau\right)\:$, that is the speed as function of the proper time $\:\tau$ \begin{equation} \boxed{\:\:\upsilon \left(\tau\right) =c\tanh\left(\dfrac{f}{m_{o}c}\,\tau\right)\:\: \vphantom{\dfrac{\dfrac12}{\dfrac12}}} \tag{13} \end{equation}

Mass(3) $\:m(t)$ as function of time $\:t\:$ and mass $\:m(\upsilon)\:$ as function of $\:\upsilon$ :

\begin{equation} m\left(t\right)=\sqrt{m_{o}^{2}+\left(\dfrac{f t}{c}\right)^{2}}=\dfrac{m_{o}}{\sqrt{1-\dfrac{\upsilon^{2}}{c^{2}}}}=m(\upsilon) \tag{14} \end{equation}

From above equations we have :

\begin{equation} \lim_{t\rightarrow +\infty}x\left(t\right) = +\infty \tag{15} \end{equation}

\begin{equation} \lim_{t\rightarrow +\infty}\upsilon\left(t\right) = c \;,\quad \upsilon\left(t\right) < c \tag{16} \end{equation}

\begin{equation} \lim_{t\rightarrow +\infty}m\left(t\right) = +\infty \tag{17} \end{equation}

The force could not accelerate the particle to speed greater than $\:c\:$ because its work is feeding continuously back as mass (inertia) to this particle.


enter image description here

enter image description here


(1) Equation (02) is identical to
\begin{equation} v=c\sqrt{1-\left(\dfrac{m_0 c^2}{F\cdot x + m_0 c^2}\right)^{2}} \end{equation} given without proof by @jaromrax here : Is there a formula that gives the position of an object depending on the time, but which doesn't allow the object to surpass the speed of light?.


(2) Equation (03) is identical to
\begin{equation} x(t) = \frac{c^2}{\alpha}\left[\sqrt{1 + \frac{\alpha^2 t^2}{c^2}}-1 \right]\;,\; t \ge 0 \end{equation} given without proof by @Alfred Centauri in the same link here : Is there a formula that gives the position of an object depending on the time, but which doesn't allow the object to surpass the speed of light?. Simply replace \begin{equation} \alpha = \dfrac{f}{m_{0}} \end{equation}


(3) In Modern Physics there exists only the mass at rest $\:m_{o}\:$, while the $^{\prime\prime}$motional mass$^{\prime\prime}$ $\:m\:$ is not used any more. In above answer think of $\:m(t)$ and $\:m(\upsilon)\:$ as auxiliary functions that make elaboration easier.

$\endgroup$
-3
$\begingroup$

Every velocity has an energy associated with it. Even when relativity comes into play.

Pick the velocity and therefore the energy (however you choose the rest mass) for start and finish. Calculate the difference.

Then using whatever force you wish and the mass chosen, calculate the rate of energy addition to the body.

Finally, divide.

It seems like that would lead fairly straightforwardly to the answer as relativity would only come into play in the final energy's calculation. (Well, real world and all that, but clearly real world is not being considered so HOW the force is applied and maintained is not important.)

$\endgroup$
  • 1
    $\begingroup$ So you are saying, calculate the energy at the end and beginning of the journey, then subtract to find the energy difference and then divide to find the time needed? That doesn't even work in a classical context though, if I have an end velocity in mind and want to calculate the time needed to reach that velocity under a certain force and use your energy approach, I will overestimate the time needed. The amount of energy I add to the system by applying a constant force doesn't vary lineairly, as energy is proportional to the square of velocity. $\endgroup$ – Dirkboss Nov 29 '17 at 0:07
  • 1
    $\begingroup$ -1. Even classically, a constant force does not give a constant rate of energy addition, so this method doesn't work. $\endgroup$ – Chris Nov 29 '17 at 0:46

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.