The period of a simple pendulum

[guv]

Homework Statement

How to calculate the period of a simple pendulum in lab reference frame and the pendulum frame if special relativistic effect is considered? This is a very silly question, it is just for the purpose of understanding. Imagine we have a simple pendulum with a watch as pendulum bob. There is another watch initially synchronized with the watch on the pendulum. The rest mass of the watch is $m$ and the rest length of the pendulum is $l$. The question is in which frame (lab versus pendulum) the reading is $2 \pi \sqrt{ \frac{l}{g}}$ for one oscillation, or neither frame will have that reading? Can we find the reading on each watch?

Relevant Equations

$$2 \pi \sqrt{ \frac{l}{g}}$$

While not exactly correct, we will continue to use Newtonian gravitational force and tension force in the lab frame. We will not concern ourselves with GR, besides the approximation is reasonable for low velocity and small mass.

In the lab frame, the forces acting on the pendulum is weight and tension, in an instantaneous tangential/radial frame,
$$\gamma m g \sin \theta = \gamma^3 m a_t $$
where $a_t$ is the tangential acceleration in the rest frame.

It's not trivial to find the speed of the pendulum in the lab frame because the mass of the pendulum keeps changing as it falls, note that the pendulum lengths remains $l$,
$$ \gamma mg l \cos \theta - (\gamma - d (\gamma) ) m g l \cos (\theta + d \theta) = d \gamma m c^2 $$

It can be seen that neither frame can have the $2 \pi \sqrt{ \frac{l}{g}}$ result. Let me know if I made any mistake in the arguments or equations. Thanks,

 

[anuttarasammyak]

To be more realistic in non relativistic mechanics, we need elliptic integrals. How do you incorporate it in your survey ?

 

[guv]

I am afraid the integral is more difficult than the elliptical integral used for non relativistic version of the period calculation. I don't think we can actually integrate it analytically. I am soliciting comments on the equations assembled for now. Thanks!

 

[anuttarasammyak]

It is Just a general observation wrt the problem
Lab frame is not an IFR due to Earth gravity. Say three clocs are
C1: at rest in lab frame at the height of the maximum height of the pendulum
C2: at pendulum. proper time of pendulum
C3: at rest in lab frame at the height of the lowest reach of the pendulum.

In a period of sequence, e.g. the pendulum starts and comes back to the maximum height, the time of the clocks show are
$$C2<C3<C1$$
or
$$C3<C2<C1$$
I have not sure which is right.

 

[bob012345]

It's not trivial to find the speed of the pendulum in the lab frame because the mass of the pendulum keeps changing as it falls, note that the pendulum lengths remains $l$,
$$ \gamma mg l \cos \theta - (\gamma - d (\gamma) ) m g l \cos (\theta + d \theta) = d \gamma m c^2 $$

In your notation, what is $d$ and $d(\gamma)$ ?

 

[bob012345]

I am afraid the integral is more difficult than the elliptical integral used for non relativistic version of the period calculation. I don't think we can actually integrate it analytically. I am soliciting comments on the equations assembled for now. Thanks!

My suggestion is reduce the $\gamma$ factor to first order in $\beta$ and notice $\beta$ is related to $\dot{\theta}$. Then treat the problem as a classical perturbation problem.

 

[bob012345]

@guv , Are you still interested in this problem? Any progress? How did you get this:

$$\gamma m g \sin \theta = \gamma^3 m a_t$$

 

[guv]

@bob012345 Yes I am still interested. $d \gamma$ and $d(\gamma)$ are the same, sorry I am still struggling with the forum latex syntax. I should have fixed that after fixing the other smaller issues during editing. The approximation of $\gamma$ with $\beta$ is an excellent suggestion, thanks.

Double checking the notes, I got the equation wrong! It should be $\gamma mg \sin \theta = m a_t$ where $a_t$ is the proper acceleration. The rest can be done with perturbation theory.

 

[bob012345]

@bob012345 Yes I am still interested. $d \gamma$ and $d(\gamma)$ are the same, sorry I am still struggling with the forum latex syntax. I should have fixed that after fixing the other smaller issues during editing. The approximation of $\gamma$ with $\beta$ is an excellent suggestion, thanks.

Double checking the notes, I got the equation wrong! It should be $\gamma mg \sin \theta = m a_t$ where $a_t$ is the proper acceleration. The rest can be done with perturbation theory.

I think there should be a minus sign and also you can still use the small angle approximation.

 

[bob012345]

@bob012345 Yes I am still interested. $d \gamma$ and $d(\gamma)$ are the same, sorry I am still struggling with the forum latex syntax. I should have fixed that after fixing the other smaller issues during editing. The approximation of $\gamma$ with $\beta$ is an excellent suggestion, thanks.

Double checking the notes, I got the equation wrong! It should be $\gamma mg \sin \theta = m a_t$ where $a_t$ is the proper acceleration. The rest can be done with perturbation theory.

One thing that bothers me about that equation is that it makes the period shorter instead of longer in the lab frame. Since moving clocks run slower and the pendulum is a moving clock, I think it should have a longer period as measured in the lab frame due to relativity than when not counting relativity. Since $\gamma$ is always larger than one, the equation; $-\gamma mg \sin \theta = m a_t$ reduces to $$\theta''=-\frac{g\gamma}{l}\theta$$

Gamma is not fixed but if one used an average value $\bar{\gamma}$ the period would be on the order of $$T=2\pi\sqrt{\frac{l}{g \bar{\gamma}}}= \frac{T_0}{\sqrt{\bar{\gamma}}}$$ which is smaller that $T_0$.

 

[guv]

Okay, so that's related to the equation I wrote initially in the first post. Notice that $a_t$ is the proper acceleration in the rest frame of the pendulum. In the lab frame, the tangential acceleration is $\gamma^3 a_t$ which is equal to $\theta''$. My initial equation would be correct if I said $a_t$ is tangential acceleration in the lab frame. Therefore, to use $\theta''$, we would have $$\theta'' = -\frac{g}{\gamma^2 l} \theta$$

I wrote the initial equation in magnitude sense, yes to properly solve the equation, the negative sign would be critical.

 

[bob012345]

Since the goal is just the change in the period, perhaps a more direct approach might work which is to just compute the time dilation due to a clock on the pendulum bob swinging along the arc path with a non-constant velocity. That would amount to integrating $dt'=\gamma dt$ with appropriate bounds and treating $\gamma$ as a function of time.

 

[bob012345]

@guv , did you get an answer yet for the period? I found it's a lot easier to do what I mentioned above than actually trying to compute $\theta$. I found that to first order;

$$T=T_0\left[1 + \frac{ l g \theta_0^2}{4c^2}\right]$$

 

[guv]

@bob012345 No, I haven't gotten around to derive the period but I think we can do that now with a few approximations as you mentioned. It concerns me that in your solution there is no mass. Since the kinematics and dynamics are coupled, I had assumed the mass $m$ would also show up in the solution. But perhaps the effect is so minimal, the mass is not part of the solution.

 

[bob012345]

@bob012345 No, I haven't gotten around to derive the period but I think we can do that now with a few approximations as you mentioned. It concerns me that in your solution there is no mass. Since the kinematics and dynamics are coupled, I had assumed the mass $m$ would also show up in the solution. But perhaps the effect is so minimal, the mass is not part of the solution.

Since in a normal pendulum the mass drops out and since the gamma factor in your equation is non-dimensional I am not surprised there is no mass involved in the solution (assuming I did it correctly). I worked three different approaches and they all gave the same formula to first order. Also, as I said I did not see the need to actually solve for $\theta$ and then figure the period. I forgot to mention that $T_0=2\pi\sqrt{\frac{l}{g }}$ in the formula above.

 

[guv]

You are correct, the mass term won't show up in the final solution. The dynamics and kinematics are coupled by they are coupled in such a way the mass would cancel exactly. Do you mind showing one way how you obtained the answer. I also agree with your solution up to an approximation, i.e. the difference between your solution and mine is on an order less than the solution you have. Thanks,

 

[bob012345]

You are correct, the mass term won't show up in the final solution. The dynamics and kinematics are coupled by they are coupled in such a way the mass would cancel exactly. Do you mind showing one way how you obtained the answer. I also agree with your solution up to an approximation, i.e. the difference between your solution and mine is on an order less than the solution you have. Thanks,

Sure. I started with the basic fact of time dilation. Let the lab frame be $t'$ and the moving clock on the pendulum bob be $t$. Since the speed is not constant, we have for each differential of time $$
dt'=\gamma dt= \frac{dt}{\sqrt{1-\beta^2}}$$ I assumed that to first order $\theta(t)=\theta_0cos(\omega t)$ thus $v(t)=l \dot{\theta(t)}=-l \omega \theta_0sin(\omega t)$ then I integrate over a quarter cycle;
$$\int_0^{\frac{T'}{4}}dt'= \int_0^{\frac{\pi}{2\omega}}\frac{dt}{\sqrt{1-\beta^2}}= \int_0^{\frac{\pi}{2\omega}}\frac{dt}{\sqrt{1-k^2 sin^2(wt)}}$$ where $k^2=\frac{l^2 \omega^2 \theta_o^2}{c^2}$,$\omega^2=\frac{g}{l}$ and $\frac{\pi}{2\omega}$ is the time when $t$ is a quarter cycle. Changing the variable of integration to $\tau=\omega t$ this can be put into the form of a Complete Elliptic Integral of the First Kind
$$\frac{1}{\omega}\int_0^{\frac{\pi}{2}}\frac{d\tau}{\sqrt{1-k^2 sin^2(\tau)}}$$ which has a series solution we can use the first couple of terms. Putting it together we have;
$$\int_0^{\frac{T'}{4}}dt'=\frac{1}{\omega}\left[\frac{\pi}{2}\left(1+\frac{k^2}{4} \right)\right]$$ or
$$T'=\frac{2\pi}{\omega}\left[1+\frac{l^2 \omega^2\theta_0^2}{4c^2}\right]=2\pi{\sqrt{\frac{l}{g}}}\left[1+\frac{l g \theta_0^2}{4c^2}\right]$$

My solution assumes if you were riding on the pendulum bob you would calculate the ideal period but I'm not really sure. It might be better to interpret it as the effect relativity adds to an ideal non-relativistic pendulum. I would be interested in seeing your approach.

 

[guv]

Thanks, yes, I ended up with a second order non-linear differential equation using the equations we assembled and solved directly. It's in the form of $\theta'' = - \frac{g}{l} [ 1 - \sigma \theta_0^2 + \sigma \theta^2] \theta$ where $\sigma = \frac{gl}{c^2}$. This clearly approximates to the Newtonian solution. With the argument $\overline{\theta} \approx \frac{\theta_0}{2}$, the two solutions have the same order. But the more exact solution for period will be a little less than the solution you show above. Thanks for your help!

 

[bob012345]

Thanks, yes, I ended up with a second order non-linear differential equation using the equations we assembled and solved directly. It's in the form of $\theta'' = - \frac{g}{l} [ 1 - \sigma \theta_0^2 + \sigma \theta^2] \theta$ where $\sigma = \frac{gl}{c^2}$. This clearly approximates to the Newtonian solution. With the argument $\overline{\theta} \approx \frac{\theta_0}{2}$, the two solutions have the same order. But the more exact solution for period will be a little less than the solution you show above. Thanks for your help!

At one point I got the same exact equation $\theta'' = - \frac{g}{l} [ 1 - \sigma \theta_0^2 + \sigma \theta^2] \theta$ and tried to solve it using perturbation theory without success. I see you took an average value for $\theta^2$ to estimate the period. If you use the rms value $\theta_{rms} = \frac{\theta_0}{\sqrt{2}}$ you will get what I got. One of my alternate methods was to take the rms value of $\beta$ in your original equation $\theta'' = -\frac{g}{\gamma^2 l} \theta=-\frac{g}{ l}\left(1-\beta^2\right) \theta=-\frac{g}{ l}\left(1- \frac{1}{2}\sigma \theta_0^2\right) \theta$ which gives the same period. My argument for using the rms value is that it seems more physical in that the power of a sinusoidal current through a resistor is related to the rms value not the average. Also, what makes you say the more exact solution for period will be slightly smaller? My series expansion predicts the higher order terms all add to the period.

 

[pbuk]

Homework Statement:This is a very silly question

No, it is not clever enough to be silly.

The question is in which frame (lab versus pendulum) the reading is $2 \pi \sqrt{ \frac{l}{g}}$ for one oscillation, or neither frame will have that reading?

The period of a simple pendulum is not $2 \pi \sqrt{ \frac{l}{g}}$ in Newtonian mechanics: this is an approximation based on $\sin \theta \approx \theta$ for small $\theta$ where the error of this approximation is many orders of magnitude greater than any relativistic effect.

 

[bob012345]

No, it is not clever enough to be silly.The period of a simple pendulum is not $2 \pi \sqrt{ \frac{l}{g}}$ in Newtonian mechanics: this is an approximation based on $\sin \theta \approx \theta$ for small $\theta$ where the error of this approximation is many orders of magnitude greater than any relativistic effect.

This thread assumes the small angle approximation. It is not about $sin(\theta)$ vs. $\theta$.