2
$\begingroup$

[To put this question in context, I was reading "How much time will it take to move an object whose length is equal to one light year?" and don't really feel satisfied by the answer that has been upvoted and accepted. This question builds off of the question and comments I left on that answer.]

Consider the following scenario: You float in an otherwise empty universe near the end of a very long uniform, solid rod. The rod is long enough that light takes a measurable amount of time to traverse its length - so, say, it's 1 light-year in length. The material of the rod is not important other than that it is fairly stiff and strong and (for simplicity) that it is non-dissipative and non-dispersive. The rod has a total length $L$, a total mass $M$, a density $\rho$, a speed of sound $v_s$, and (if it matters) a cross-sectional area $A$. All of the values (except for the length) should be taken to be reasonably-sized given the constraints of the problem (specifically, the mass $M$ should be very large).

If you push on the rod, how long would it take for the rod to move some finite distance $d$?

If you were to strike the end of the rod with a hammer, a compression wave would form in the rod and propagate down its length at the speed of sound $v_s$. But even if the impulse is very large, the naively calculated velocity of the center of mass ($v_{CM}$) of the rod would be very small (since the rod has such a huge mass $M$).

Now, I'm totally on board with the idea that the wave packet (and therefore the information concerning the push) will take a certain, constant amount of time to traverse the length of the rod, namely $t_1=\frac{L}{v_s}$. However, it is not clear to me that (as the answer-giver suggests in the other thread) this means that $t_1$ is the same amount of time it will take for the far end of the rod to move the distance $d$. Instead, I would think that it should take an amount of time $t_2$, where $t_2$ must be no smaller than $t_1$ but is asymptotically related to the ratio $\frac{d}{v_{CM}}$.

Furthermore, for a sustained force $F$ being applied to the rod, I would expect a classical acceleration of $a=\frac{F}{M}$. That means that my expected asymptotic relation should be $\sqrt{\frac{2d}{a}}$.

The only case where I could see the original logic being correct is if the rod were very easily compressible and the the distance being compressed were very small. In that case, one could compress the rod a distance $d$ quickly and then hold the compression in place to prevent rebounding as the wave packet propagates away.

Is my intuition about this problem correct? If not, why not? If so, how can one calculate the correct form of $t_2$ which takes into account the speed of sound in the rod and its immense length? Considering the scale of the rod, does the problem change if gravitational or non-classical effects are taken into account?

$\endgroup$
0
$\begingroup$

enter image description here

Suppose the near end of the rod is moved a distance $d$ with uniform speed $v$ away from us, after which it is brought to rest. This takes time $t_{rigid}=d/v$; this is the time that far end of a perfectly rigid rod would take to move a distance $d$.

But in an elastic rod, the compression wave has finite speed, call it $v_s$. This speed is as seen in the reference frame of the rod. Since during initial time duration $t_{rigid}$ the rod is moving with speed $v$, the compression wave speed as observed by us during time $0 <t< t_{rigid}$ is $v_s'=(v_s+v)/(1+v_sv)$. We have adopted $c=1$ units here. For time $t\geq t_{rigid}$, the compression wave speed as observed by us returns to $v_s$.

During time duration $t_{rigid}$ the compression wave travels a distance $v_s't_{rigid}$. Rest of the distance, $L-v_s't_{rigid}$, is traveled by the compression wave in time equal to $(L-v_s't_{rigid})/v_s$. Therefore the total time for the compression wave to reach the far end is equal to $t_{rigid}+(L-v_s't_{rigid})/v_s$. As the compression wave reaches the far end, the far end stretches out by a length $d$ at the speed of the wave, which takes time equal to $d/v_s$. Assuming that the wave does not bounce back (say due to presence of some absorber fitted at this end) we have \begin{align} t_{elastic} & =t_{rigid}+\frac{L-v_s't_{rigid}}{v_s}+\frac{d}{v_s}\\ & = \frac{d}{v}+\frac{L+d}{v_s}-\frac{1+v/v_s}{1+v_sv} \end{align}

$\endgroup$

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.