# What happens when you push a rod?

• B

## Summary:

Does a rod keep its length when pushed, if so, how?
Note, I put this question here because I imagined it fit the best here, but mods feel free to move it if there's a better place.

So my girlfriend asked me a question today I couldn't answer. Imagine you have a crystal rod (for simplicity), and you push it a bit at one end, in the axial direction, and hold it there. As your finger hits the rod, presumably this starts a pressure wave traveling along the rod, gradually displacing the atoms in the crystal, resulting in the rod moving.

However, we learn at school that energy is always lost to heat, so if that is the case would the pressure wave "lose steam" and the rod end up a little shorter due to being poked? Or would the pressure wave make it to the other end of the rod with all the energy intact, ensuring the rod keeps its length?

She was also curious if it would it be different if the rod was in a vacuum.

We both had the feeling that the rod should keep its length, but I failed to come up with a good explanation why. My immediate thought was that I imagine the atoms in the crystal are located where they are due to the repulsive force of the neighboring atoms, and if one pokes atom A so it gets a little closer to its neighbor B, it will be more energetically favorable for B to move towards its neighbor C than back towards A. As B moves a bit closer to C, C will be in a similar position, and over time they will equalize to the same inter-atom distance that they had before A was poked.

Of course, I have no idea so that might be a bunch of rubbish, and she didn't quite buy this explanation because "what about the energy loss". I know energy isn't behaving quite the same once you get down to the level of atoms, but my knowledge is lacking so I couldn't explain it well.

Are there any decent B-level explanations for this, or?

Delta2

hutchphd
Homework Helper
Most of your analysis seems pretty good to me. I will try to point out the shortcomings. The compression wave moves very fast (maybe ##10^3m/s##) and so it bounces back and forth many times during this event. The rod may shorten slightly but will quickly find its uncompressed length as your finger is removed (lots of compression waves bouncing back and forth) There will be some energy expended continuously but that will come from your finger in the final analysis...the rod may heat slightly.
The difference between air and vacuum is of little consequence to the rod; it is much denser than either so it will not much care.

ChemAir, Delta2 and Lord Crc
Baluncore
As your finger hits the rod, presumably this starts a pressure wave traveling along the rod, gradually displacing the atoms in the crystal, resulting in the rod moving.
That is correct. You can think of the atoms being connected by springs. Another model would be to think of the rod as a cylindrical block of rubber that has a fixed volume.
As you apply a steady pressure to the rod, the point you touch will move away. That movement will propagate along the rod as a compression wave at the speed of sound. Where the rod is compressed axially it will also get slightly thicker. When that pressure wave reaches the far end of the rod, the far end will, (if it is unrestrained), extend away from the rod by twice the initial shortening, reflecting a tension wave, back up the rod to your hand where it will cancel with the applied pressure. So the rod is shortened during the first part, and then stretched during the second part. As it is being compressed, or extended, it is also being widened, or narrowed.
The result of the asymmetrical applied force will be movement of the rod away from the point where the pressure was applied.

Lord Crc and Delta2
etotheipi
Gold Member
2020 Award
I wonder how far it's possible to take Baluncore's idea. If you had ##N## particles connected by springs, and you exert a force ##F## on the first one, then you end up with ##N## coupled differential equations,\begin{align*}F - k_{12}(x_1 - x_2) &= m_1 \ddot{x}_1 \\ k_{12}(x_1 - x_2) - k_{23}(x_2 - x_3) &= m_2 \ddot{x}_2 \\ \dots \\ k_{N-1, N}(x_{N-1} - x_{N}) &= m_N \ddot{x}_N \end{align*}What happens if you take the limit as ##N \rightarrow \infty##? We can quite reasonable assert that ##k_{ij} = k = \text{constant}## and inversely proportional to the spring natural length, i.e. ##k = \lambda/l_0## in which case, if the total length of the rod is ##L##, we have ##k = (N \lambda)/L##. We can also put ##m_i = m = \text{constant}##.

Does anyone know how to take the continuum limit of this setup, in order to find ##\Delta x(t) = x_N(t) - x_1(t)##?

hutchphd
Homework Helper
This is the usual stuff of solid state physics. I'm not sure where you are trying to go here....the discrete problem is soluble and the solution is phonons + the center of mass moving. The exact nature of the phonons depends upon F(t). The nature of the lattice force is local and quantum mechanical.....but the usual model presumes small (and slow) excursions.
The continuum model is an elastic solid with various elastic moduli. Am I misunderstanding the question?

etotheipi
etotheipi
Gold Member
2020 Award
I'm not sure where I'm trying to go either, just playing around with stuff and wondering if anyone has better models. I'll have a look at phonons, thanks

hutchphd
Homework Helper
I guess you haven't looked much at solid state yet. There are some clever techniques (also much of the world is in a solid state !). Enjoy.

etotheipi
Baluncore
There are some clever techniques (also much of the world is in a solid state !). Enjoy.
We live in a universe of incompatible interfaces.

We do not see the internal isotropic solid, we must weigh it. We might ring it like a bell, but we are still only experiencing the bulk properties, through the shape of the external surface.

When you touch the end of a rod, you sense the acoustic impedance mismatch between the air and the rod. When you see the end of a rod, you are sensing the reflection of light from the step change in the electrical properties of the rod in air.

hutchphd
Thank you for the replies, lets see if I can manage to convey this in a sensible manner.

Summary:: Does a rod keep its length when pushed, if so, how?

Note, I put this question here because I imagined it fit the best here, but mods feel free to move it if there's a better place.

So my girlfriend asked me a question today I couldn't answer. Imagine you have a crystal rod (for simplicity), and you push it a bit at one end, in the axial direction, and hold it there. As your finger hits the rod, presumably this starts a pressure wave traveling along the rod, gradually displacing the atoms in the crystal, resulting in the rod moving.

However, we learn at school that energy is always lost to heat, so if that is the case would the pressure wave "lose steam" and the rod end up a little shorter due to being poked? Or would the pressure wave make it to the other end of the rod with all the energy intact, ensuring the rod keeps its length?

She was also curious if it would it be different if the rod was in a vacuum.

We both had the feeling that the rod should keep its length, but I failed to come up with a good explanation why. My immediate thought was that I imagine the atoms in the crystal are located where they are due to the repulsive force of the neighboring atoms, and if one pokes atom A so it gets a little closer to its neighbor B, it will be more energetically favorable for B to move towards its neighbor C than back towards A. As B moves a bit closer to C, C will be in a similar position, and over time they will equalize to the same inter-atom distance that they had before A was poked.

Of course, I have no idea so that might be a bunch of rubbish, and she didn't quite buy this explanation because "what about the energy loss". I know energy isn't behaving quite the same once you get down to the level of atoms, but my knowledge is lacking so I couldn't explain it well.
In general case, the length of the rod in medium term is likely to increase.

Consider a rod like a wind chime, hung horizontally so as to cause little damping, and you use not your finger but a hammer to ring it.
Your blow will set the rod ringing.
Assume your blow is not so strong as to cause brittle fracture or spallation of the chime.
The energy goes into sound waves that travel along the chime and reflect from ends.
Some of the sound energy is radiated into air (and would not be so radiated if instead of hanging the chime in air, you rang it in vacuum of space).
But the crystal of chime undergoes elastic hysteresis. Some of the energy is directed from collective oscillations into disordered thermal motion.
When the ringing dies down, due to the extra thermal motion, the chime is slightly warmer.
Most solids expand on heating. (Some, like invar, have complex structures that allow them to be exceptions). Therefore, a rung chime will be longer because of thermal expansion due to heat from the sound dissipated in the chime.
But after you have cooled the chime back to exact same temperature that it had before ringing?
Well, it was a crystal. The length of the chime consisted of bonds between specific atoms. When the chime was rung, the bonds were stretched, but assume that they were not broken. So when the chime was heated, the bonds were stretched but not broken... and after the chime was cooled back, the exact same bonds between same atoms had the same length, severally and combined.
The matter is different if the metal is soft and plastic, and your blow exceeds the plastic limit. Then the bonds inside the crystal do break, and the atoms form new bonds instead with new atoms. In which case the chime does get permanently shorter - and thicker. You may find the end of the chime mushrooming under your blow.

Lord Crc