A simple pendulum moving at a relativistic speed - how does the period change?

Question

I've been pondering the precise mechanism of time dilation for the example of a simple pendulum in two different situations:

1. The observer and ground are at rest in one frame of reference; the pendulum is moving at high speed with respect to that frame.

2. The observer is at rest in one frame of reference; the pendulum and the ground together move at high speed with respect to that frame.

user8260 has pointed out that in situation 1, in the pendulum's frame $g$ is greater by $\gamma^2$ compared to $g$ in the observer's frame. Thus in the pendulum's frame the period is less than it is in the observer's frame by a factor of $1/\gamma$, just as one would expect from time dilation.

But what about situation 2? Here, compared to the pendulum frame, the observer sees the pendulum with the same length in a stronger gravitational field, yet observes a longer period. Does the inertial mass of the pendulum change differently than its gravitational mass? Also, does the analysis depend on whether the plane of swing is perpendicular or parallel the direction of motion?

Answer 1

As in physics in general, a suitable choice of coordinates makes our life so much better. Time dilation in this problem is somewhat a more trivial effect, and the transformation of gravitational field is somewhat a more complicated phenomenon. With this in mind, let me reformulate slightly the two situations:

Case 2. Pendulum is at rest with respect to the Earth (and some observer moves with respect to them, observes time dilation etc etc)

Case 1. Pendulum is set above the Earth, which moves relativistically below it (and some observer moves with the Earth, observes time dilation etc)

So, let us settle the physics first, and the observer effects last.

Case 2: Classical physics problem, nothing to settle.

Case 1: From the pendulum's point of view, the gravitational field is generated by a moving body (=>the field is unknown). From the Earth frame, a relativistic body moves in a gravity field (=>the equations of motion are unknown).

One might transform the energy-momentum tensor of the Earth from the Earth rest frame to the pendulum frame, but special care should be taken about the fact that the Earth ceases to be spherical in the new frame (though its density does increase as $\gamma^2$). Additionally, it is not clear appriori that the motion of the Earth doesn't cause any additional forces.

I propose to use a straightforward yet more secure method of transforming the metric tensor from the Earth frame to the pendulum frame, and hence obtain the gravity, acting on the pendulum.

In the Earth rest frame the metric tensor is known to be $$g_{\mu\nu}=\left(\begin{array}{cccc} &1-2U & 0 & 0 & 0 &\\ &0 & 1-2U & 0 & 0 &\\ &0 & 0 & 1-2U & 0 &\\ &0 & 0 & 0 & -1-2U &\\ \end{array} \right),$$

where $U$ is the Newtonian potential of the Earth. This expression corresponds to the so called weak field limit, when the metric tensor is nearly flat. We use the standard notation of MTW ($c=1$, signature $(+++ -)$, Einstein's summation rule etc) and refer to this book for further details on linearized gravity.

Transformation of the field to the pendulum frame:

Lorentz tranformation matrix is given by:

$$ \Lambda_{\mu'}^{~\mu}=\left(\begin{array}{cccc} &\gamma & 0 & 0 & \beta \gamma &\\ &0 & 1 & 0 & 0 &\\ &0 & 0 & 1 & 0 &\\ &\beta\gamma & 0 & 0 & \gamma &\\ \end{array} \right), $$ with $\beta=\dfrac{v}{c}, \gamma=(1-\beta^2)^{-1/2}$ and $v$ being the relative velocity of the pendulum with respect to the Earth rest frame.

The transformed metric tensor is obtained by: $$g_{\mu'\nu'}=\Lambda_{\mu'}^{~\mu}\Lambda_{\nu'}^{~\nu} g_{\mu\nu}=\left(\begin{array}{cccc} &1-2U\dfrac{1+\beta^2}{1-\beta^2} & 0 & 0 & -\dfrac{4 U \beta}{1-\beta^2} &\\ &0 & 1-2U & 0 & 0 &\\ &0 & 0 & 1-2U & 0 &\\ &-\dfrac{4 U \beta}{1-\beta^2} & 0 & 0 & -1-2U\dfrac{1+\beta^2}{1-\beta^2} &\\ \end{array} \right)$$

In the pendulum frame (further primes in the indices are omitted!):

It is known that only the term $g_{44}$ determines the newtonian potential. One can see that by writing out the lagrangian for the pendulum:

$$ \mathcal{L}=\dfrac{1}{2}g_{\mu\nu} u^\mu u^\nu=\\ =\dfrac{1}{2}((u^1)^2+(u^2)^2+(u^3)^2-(u^4)^2)-\\ -U((u^2)^2+(u^3)^2+4 u^1 u^4 \beta \gamma^2+((u^1)^2+(u^4)^2)\dfrac{1+\beta^2}{1-\beta^2}) $$

Here $u^\mu$ is the 4-velocity of the pendulum. As the latter moves non-relativistically (in its own frame), we may consider $u^4\gg u^1,u^2,u^3$ and $u^4\approx \mathrm{const}$, which leaves:

$$ \mathcal{L}=\dfrac{1}{2}((u^1)^2+(u^2)^2+(u^3)^2)-U(u^4)^2\dfrac{1+\beta^2}{1-\beta^2} $$

If the pendulum as a whole didn't move with respect to the Earth, we would have $\beta = 0$ and

$$ \mathcal{L}_0=\dfrac{1}{2}((u^1)^2+(u^2)^2+(u^3)^2)-U(u^4)^2 $$

Effectively, therefore, the pendulum in its rest frame experiences the gravitational field magnified by the factor of $\dfrac{1+\beta^2}{1-\beta^2}$. The pendulum frequency is thus magnified by $\dfrac{(1+\beta^2)^{1/2}}{(1-\beta^2)^{1/2}}$.

Remarks: the neglected terms in the lagrangian are either $\dfrac{v}{c}$ or $(\dfrac{v}{c})^2$ smaller than the kept leading terms. Hence, up to $\dfrac{v}{c}$ accuracy the direction of motion doesn't affect the pendulum frequency.

Finally, lets add time dilations to get the final answers. Let the period of the pendulum in the case when observer, the Earth, and the pendulum do not move with respect to each other be $T_0$. Then:

Case 1: In the pendulum frame, as we have seen it has the period of $\dfrac{(1-\beta^2)^{1/2}}{(1+\beta^2)^{1/2}} T_0$. Then in the observer frame, due to time dilation, the period is $\dfrac{1}{(1+\beta^2)^{1/2}}T_0$.

Case 2: In the pendulum frame the period is $T_0$. In the observer frame the period is $\dfrac{T_0}{(1-\beta^2)^{1/2}}$.

To conclude, the two cases are quite different due to the different physics happening. In one case the observed period changes due to the change of the reference frame, whereas in the other there is an additional factor due to the fact that the gravity of a moving source is not the same as that of a still source.

Answer 2

When g increases in the moving frame the period should decrease, not increase. Recall, the period goes like $\sqrt\frac{l}{g}$, so a large gravitational acceleration translates to fast oscillations, or a short period. In any case, the stationary reference frame should always measure a period that is larger by a factor of $\gamma$ than what the moving frame sees. This time dilation shouldn't be correlated with the changing of the effective density in the moving frame - that is something of a red herring. Really, this isn't a lorentz invariant system (because of the large stationary mass defining the gravitational field) so this might cause some of the confusion. Yes, the gravitational field looks different in a moving frame, and yes, this should cause a change in the period, but this isn't the same thing as the usual time dilation effect one expects in special relativity. Both effects are at play here. Finally, the density of the earth that the moving observer sees will change by $\gamma^2$, not simply $\gamma$. However, as detailed in the answer above, one cannot simply deduce the the moving pendulum's frequency from the density alone because this is a non-inertial frame and one should do a full calculation.

The situation in question 2 is analogous, but now, from the moving frame's perspective, we have a moving pendulum in an altered gravitational field. We should really formulate this problem properly in the language of tensors. From the moving frame's viewpoint the equation of motion of the mass at the end of the pendulum is (very roughly)$(dv/dt+v\Gamma'v)=F'$, where $\gamma'$ is the christoffel symbol in the moving frame and F' is the force. F' transforms as a vector, while $\gamma$ almost transforms as a vector, but with an extra additive piece. In any case, it is the christoffel symbol built from the metric g'_uv in the moving frame, which does transform as a tensor. We may thus obtain g'_uv by acting on g_uv with the usual transformation matrix, and we know g_uv since it's zero-zero component is simply related to the usual Newtonian potential.

Now, one could do all the algebra here, but the main point is that since both sides are tensors that transform the same way, the end result is guaranteed to be invariant. Thus, if we just wanted to get some answer we would naturally use the stationary frame where the calculation is easier. Thus, we may conclude that the stationary frame gets $\sqrt(l/g)$ and the moving frame sees $\gamma$ times this.

One further note. This question is somewhat analogous to the following problem. Let observer A be stationary and let B be moving. Observer A measures the time, $t_a$, between two events and observer B likewise measures $t_b$. Now, observer A thinks $t_a < t_b$, while observer B thinks $t_b < t_a$. These are just real numbers, so $t_a=t_b$! Obviously something went wrong and this is a standard "paradox" in special relativity. The answer is the same: we have to consider the transformation of the non-time components of the four-vector connecting the two events in order to see how one observer interprets the measurements of another. As an exercise, try to calculate B's coordinates for the events and t_b given A's coordinates, and then use this to go back and get A's coordinates from B's coordinates. I.e., transform from A to B to A. One might think that the time is getting "slower" with each boost but the two transforms cancel each other (obviously). I think that this is essentially a simplified version of the problem here.