I was trying to see what results I would get if I were to incorporate relativistic corrections into the case of a harmonic oscillator in one dimension. I thought that if the maximum velocity of the oscillating body were to approach relativistic velocities, the measurement of the time period of motion and related measurements should change. I assumed a lab frame $(t,x)$.

The equation of motion should be \begin{equation}\frac{dp}{dt}+kx=0 \end{equation} where $t$ is the time measured in the lab frame. $\frac{dp}{dt}$ is calculated to be $\gamma^3m_0\ddot{x}$(since only $a_{\mid\mid}$ exists for this motion). Naively, I took this as my solution in the following way: \begin{equation}\gamma^3m_0\ddot{x}=-kx \end{equation} \begin{equation}\implies\ddot{x}=-\frac{k}{\gamma^3m_0}x \end{equation} \begin{equation}\implies\ddot{x}=-\omega^2x \end{equation} So I took the angular frequency $\omega$ simply like that. However, I forgot that $\gamma$ is a function of $\dot{x}$ and this obviously complicates things... for instance the period becomes a function of velocity. I'm not sure if my approach is correct and I don't have the tools to solve the differential equation I would get if i were to open up $\gamma$, so my questions are the following:

  1. Am I doing this right?
  2. Am I missing something conceptually?
  3. Is there a better way to approach the problem?
  4. And if, by some miracle, my last expression for $\ddot{x}$ is correct, could I have some help solving the differential equation?

EDIT: If you are posting the relativistic expression, please post the corresponding evaluation of the Classical limit of your expression.

  • $\begingroup$ (4.) I don't think you can find an analytical solution for that. Also, How did you get $\gamma^3$? Can you post the calculation? Because I think you have missed something essential, when taking the derivative of $\mathbf{p}$ since the mass $m$ is not a constant. $\endgroup$
    – iharob
    Jul 3, 2015 at 13:06
  • $\begingroup$ @iharob You can find the derivation here: en.wikipedia.org/wiki/Relativistic_mechanics#Force $\endgroup$ Jul 3, 2015 at 13:09

2 Answers 2


You can get an exact solution for $t(p)$, although it involves a rather nasty integral that I'm not sure can be written in closed form. Here's how:

The equations of motion are $$ \frac{dp}{dt} = -kx \qquad \frac{dx}{dt} = \frac{1}{m} \frac{p}{\sqrt{1 + p^2/m^2 c^2}}. $$ This second equation can be obtained by taking the equation $p = m v /\sqrt{1 - v^2/c^2}$ and solving for $v$. Differentiating the first equation and plugging in the second then yields $$ \ddot{p} = - \omega^2 \frac{p}{\sqrt{1 + p^2/m^2 c^2}}. $$ where $\omega^2 = k/m$ as usual. This is a second-order ODE of the form $\ddot{p} = f(p)$, and so can be solved (at least in principle) via the method of quadratures: \begin{align} \dot{p} \ddot{p} &= - \omega^2 \dot{p} \frac{p}{\sqrt{1 + p^2/m^2 c^2}} \\ \frac{d}{dt} \left( \frac{1}{2} \dot{p}^2 \right) &= \frac{d}{dt} \left( - \omega^2 m^2 c^2 \sqrt{ 1 + p^2/m^2 c^2 }\right) \\ \frac{1}{2} \dot{p}^2 &= - \omega^2 m^2 c^2 \sqrt{ 1 + p^2/m^2 c^2 } + C, \end{align} where $C$ is a constant determined by the initial conditions. In particular, if we take the case where the particle is released from rest a distance $A$ from the origin, then we have $p_0 = 0$ and $\dot{p}_0 = - kA$, implying that $$ C = \frac{1}{2} k^2 A^2 + \omega^2 m^2 c^2; $$
and so we have $$ \frac{dp}{dt} = \pm \sqrt{ k^2 A^2 + 2 \omega^2 m^2 c^2 - 2 \omega^2 m^2 c^2 \sqrt{ 1 + p^2/m^2 c^2 } }. $$

This is a separable equation, so in principle, then, we have an implicit solution for $t(p)$: $$ t(p) = \int_0^p dp \left[ 2 C - 2 \omega^2 m^2 c^2 \sqrt{ 1 + p^2/m^2 c^2 } \right]^{-1/2}. $$ Making everything as dimensionless as possible, we redefine $\tilde{p} = p/mc$ and $\tilde{C} = C/\omega^2 m^2 c^2$; the integral then becomes $$ t(\tilde{p}) = \frac{1}{\omega} \int_0^\tilde{p} d\tilde{p} \left[ 2 \tilde{C} - 2 \sqrt{ 1 + \tilde{p}^2 } \right]^{-1/2}. $$ I don't know how to solve this integral, personally. Mathematica tells me that there's a solution in terms of incomplete elliptic integrals, but it's not terribly illuminating; and it seems pretty hopeless to try to invert this to get $p(t)$ and thereby get $x(t)$.

However, this expression can still be used to find the period of oscillation. If we think about the oscillation, it will take one-quarter of the object's period to go from $p = 0$ to a maximum of $p$ (where $\dot{p} = 0$.) This will be precisely where quantity in square brackets in the above integral vanishes. In other words, the period $\tau$ will be given by $$ \tau = \frac{4}{\omega} \int_0^{\sqrt{\tilde{C}^2 - 1}} d\tilde{p} \left[ 2 \tilde{C} - 2 \sqrt{ 1 + \tilde{p}^2 } \right]^{-1/2} $$ Note that the dependence on amplitude of the period is encoded in the constant $\tilde{C}$. That said, I'm still not sure how to solve this integral, and Mathematica doesn't much care for it either. If nothing else, it's amenable to numerical integration now.

EDIT: In the Newtonian limit, we will have $\tilde{p} \ll 1$ at all times, and $$ \tilde{C} = 1 + \frac{1}{2} \frac{k^2 A^2}{\omega^2 m^2 c^2} = 1 + \frac{\omega^2 A^2}{2 c^2} \quad \Rightarrow \quad \sqrt{\tilde{C}^2 - 1} \approx \frac{\omega A}{c}. $$ So the period in this limit becomes \begin{align*} \tau &\approx \frac{4}{\omega} \int_0^{\omega A/c} d\tilde{p} \left[ 2 + \frac{\omega^2 A^2}{c^2} - 2 - \tilde{p}^2 \right]^{-1/2} \\ &= \frac{4}{\omega} \int_0^{\omega A/c} d\tilde{p} \left[ \frac{\omega^2 A^2}{c^2} - \tilde{p}^2 \right]^{-1/2} \\ &= \frac{4}{\omega} \left[ \arcsin \left( \frac{c \tilde{p}}{\omega A} \right) \right]_0^{\omega A/c} \\ &= \frac{4}{\omega} \left[\frac{\pi}{2} \right]= \frac{2\pi}{\omega}. \end{align*} So the Newtonian result for the period is recovered. One could do a similar expansion to solve for $p(t)$ in this case, since we would get something like $\omega t = \arcsin (c \tilde{p}/\omega A).$

  • 1
    $\begingroup$ Thanks a lot! The approach seems legit, but I'm curious to know whether this boils down to the more familiar expression when we take the classical limit... $\endgroup$ Jul 3, 2015 at 15:12
  • $\begingroup$ @PrishChakraborty: I've added the Newtonian limit to my derivation above. $\endgroup$ Jul 3, 2015 at 15:29
  • $\begingroup$ Thank you very much for your help! However, are you certain that a cleaner solution doesn't exist? $\endgroup$ Jul 3, 2015 at 15:48
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    $\begingroup$ @PrishChakraborty: Nope! There could well be a cleaner way to do this; my method of solution could be just one of many ways to get at the result. That said, I'm not optimistic that a closed-form solution for $x(t)$ and $p(t)$ actually exists, simply because closed-form solutions rarely exist for non-linear ODE systems. $\endgroup$ Jul 3, 2015 at 15:55
  • $\begingroup$ I hoped for an expression that would perhaps make the relativistic effects a little more obvious, but maybe I'm just being blind, or greedy. $\endgroup$ Jul 4, 2015 at 5:38

I was trying to see what results I would get if I were to incorporate relativistic corrections into the case of a harmonic oscillator in one dimension. [...] $ \gamma^3~m_0~\ddot{x} = -k x $

Since your question is specificly about harmonic motion, we might (instead) insist on this solution

$$x[~t~] := x_{\text{max}}~\text{Sin}[~\omega~(t - t_0)~],$$

and ask about the expression, as a function of $x$, of a corresponding suitable conservative force (or eventually, the "shape" of a corresponding potential).

Consequently, we'd insert the "harmonic solution" into the left-hand side of your equation:

$$ \gamma^3~m_0~\ddot{x} \mapsto -m_0~\omega^2~x_{\text{max}}~\text{Sin}[~\omega~(t - t_0)~]~/~\sqrt{ 1 - \left(\frac{x_{\text{max}}~\omega~\text{Cos}[~\omega~(t - t_0)~]}{c}\right)^2 }$$

and ask how to express it as a function of the variable $x$ (instead of variable $t$). That's of course straightforward by inserting the harmonic solution:

$$ \begin{align*} \gamma^3~m_0~\ddot{x} ~~~ & \mapsto -m_0~\omega^2~x~/~\sqrt{ 1 - \left(\frac{x_{\text{max}}~\omega~\text{Cos}[~\text{ArcSin}[~x/x_{\text{max}}~]~]}{c}\right)^2 } \\ ~ & = -m_0~\omega^2~x~/~\sqrt{ 1 - \left(\frac{x_{\text{max}}~\omega~\sqrt{1 - (x/x_{\text{max}})^2}}{c}\right)^2 } \\ ~ & = -m_0~\omega^2~x~/~\sqrt{ 1 - \left(\frac{x_{\text{max}}~\omega}{c}\right)^2 + \left(\frac{x~\omega}{c}\right)^2 }. \end{align*} $$

Of course we should require $0 \lt x_{\text{max}}~\omega \lt c$, whereby the square root is always defined, and real.
In the ("non-relativistic") limit that $x_{\text{max}}~\omega \ll c$ (and since $0 \le |x| \le x_{\text{max}}$) this force expression approaches the right-hand side of your equation: $-m_0~\omega^2~x \equiv -k~x$.

It is also easily integrated to obtain the corresponding "relativistic harmonic potential form"

$$V[~x~] := m_0~c^2 ~\sqrt{ 1 - \left(\frac{x_{\text{max}}~\omega}{c}\right)^2 + \left(\frac{x~\omega}{c}\right)^2 } - V_{\text{ref}};$$

where the term with explicit $x$ dependence in the limit $x_{\text{max}}~\omega \ll c$ approaches $m_0~c^2~\frac{1}{2}~\left(\frac{x~\omega}{c}\right)^2 = \frac{1}{2}~m_0~\omega^2~x^2 \equiv \frac{k}{2}~x^2$.

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    $\begingroup$ Prish Chakraborty: "Thanks for your contribution." -- You're welcome. "Please read the edit!" -- I did; and added some additional evaluations to my answer. $\endgroup$
    – user12262
    Jul 4, 2015 at 17:32

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