## Lengthwise Accelerated Rod

Discussions on classical and modern physics, quantum mechanics, particle physics, thermodynamics, general and special relativity, etc.

### Lengthwise Accelerated Rod

In the locked thread SSDZ's Vixra Papers, the issue of an accelerated rod with clocks at both ends came up. It was unfortunately left hanging in the air due to the nature of that thread, so here is another 'bite at the cherry'.

To keeps thing simpler, we ignore any mechanical compression or stretch that a 'rigid' rod may suffer when pushed or pulled from one end.* On a Minkowski spacetime diagram, the end where the rod is being pushed with a constant force will follow a hyperbolic worldline, like hyperbola A in this diagram.**

The distance of the vertex of the hyperbola from the origin will be c2/a, where a is the acceleration measured with an accelerometer at the point where the force is applied. The origin is where the two blue lines cross. If the rod has a rest length L, the front end of the rod will follow the hyperbola B, with vertex a distance a+L from the origin. Hence, the acceleration that an accelerometer fixed to the front of the rod will register is c2/(a+L). This is algebraically smaller than the acceleration of the rear end.

Now, we must not be confused by the fact that the rod is temporarily compressed when the force is applied (as described in **); the reason for the smaller proper acceleration is that relative to the Minkowski coordinates, the rod Lorentz contracts as it picks up speeds. This means that during the acceleration the front end moves marginally slower than the rear end, as observed in the coordinate system; hence the front end suffers less time dilation than the rear end and clocks that were synchronized before the acceleration will gradually lose sync. Relative to coordinate clocks, both clocks will record less elapsed time, but the (faster) rear clock (A) will lose more time than the (slower) front clock (B).

As observed by persons riding on the rod, the accelerating rod does not Lorentz contract, but keeps its proper length L (ignoring mechanical compression as in *). In the original coordinate system the rod will however gradually Lorentz contract. If the acceleration is stopped after some coordinate time t, the whole rod will move uniformly at speed v = a t and the rod will have a constant Lorentz contracted length: L' = $\gamma$L, where $\gamma$ is the Lorentz factor.***

By Einstein's equivalence principle, the above scenario is qualitatively the same as if the rod has been sitting vertically on its rear end on Earth's surface. The rod will experience marginally less gravity at the top than at the bottom and apart from a minor constant compression, the length of the rod will be L, as measured by us on Earth. But to an inertial observer free-falling at a speed v past the rod, the rod will appear to be length contracted to the same extent as above.

Since the distance a for an acceleration of 1g is about one light year, the difference in acceleration for a practical length rod is obviously pretty small, but if it is kept up for a long time, the difference in time can be significant. This has been practically demonstrated by clocks on Earth and in space.

I hope this clears up any confusion that might have arisen out of the aborted thread. If not, please shout.

--
Regards
Jorrie

* If we keep the acceleration mild and the rod reasonably short, it makes little difference to the situation. Forces propagate through the rod at the speed of sound in the rod, so when abruptly accelerated by pushing it, the rod will compress/expand, i.e its length will oscillate for a short time and quickly settle into a very slightly compressed length state. The reverse will happen when the acceleration abruptly stops.

** This slide is from prof John Mallinckrodt's talk: Simple, Interesting, and Unappreciated Facts about Relativistic Acceleration. If the link does not work, Google the title and you should get it. The "Identical Asymptotic Light Cones" are explained in the slides, but essentially means that both ends of the rod will eventually approach the speed of light and the rod will appear to have zero length in the original inertial frame.

*** The Lorentz factor $\gamma = \frac{1}{\sqrt{1-v^2/c^2}}$
Last edited by BurtJordaan on October 5th, 2014, 3:42 pm, edited 1 time in total.
Reason: typo

BurtJordaan
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### Re: Lengthwise Accelerated Rod

Neat! Really good slide and explanation. I had been wondering, so it was definitely helpful.

I realized that the front clock runs faster, because it is AS IF it is higher in a gravity well (as you say, equivalence principle). But I couldn't square that with the gravity-less SR picture.

This is the kind of thing that is so helpful it is tempting to make a sticky of it just so it won't disappear and get lost. Or put a link to it in some annotated physics bibliography thread, to keep tabs on it.

Marshall
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### Re: Lengthwise Accelerated Rod

Marshall » 06 Oct 2014, 00:00 wrote:I realized that the front clock runs faster, because it is AS IF it is higher in a gravity well (as you say, equivalence principle). But I couldn't square that with the gravity-less SR picture.

One other thing that I failed to mention and which is not easily understood, is that if the acceleration at the rear end is stopped, the front clock accelerates for a longer time than the rear clock to reach the final speed. This is valid not only in the inertial reference frame, but also as measured by the clocks on the rod. I repeat prof. Mallinckrodt's slide here for ease of reference.

Suppose we stop the acceleration when the rod is in the position indicated by 3 and 4 in the diagram. The front and rear have identical speeds at that stage, as measured in either the reference frame or the relatively moving rod frame. It is easy to see on the diagram that in the reference frame the front takes longer to reach that speed than the rear, but what about the front and rear clocks themselves?

As we have seen, the acceleration of the front is less than that of the rear, so in order to reach the same speed, the front has to accelerate for a longer time than the rear, as measured on their own clocks. And again, this has nothing to do with compression/decompression of the rod. Even if the rod could have had infinite stiffness (impossible), this difference in acceleration time would have been there.

I have been asked this question before: how do we know that an accelerated rod would work like this? True, nobody has ever accelerated a rod exactly like this AND measured this tiny relativistic effect. The best answer that I have is that if things did not work like this, the whole of relativity theory must be wrong. And by 'wrong' I do not mean the expected breakdown at quantum mechanical level, but that it must then be broken at macro scales as well - something that physicists have tried to do since its inception and have never succeeded in.

--
Regards
Jorrie

BurtJordaan
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### Re: Lengthwise Accelerated Rod

I guess this thread would not be complete, unless something is said about the so called "Bell's Spaceship Paradox" in relation to the accelerated rod. We know that this is only an apparent paradox when the analysis is done incorrectly.*

If instead of pushing the rear end of the rod and allowing the front end to follow its own spacetime path, we force the two ends to always accelerate identically, as measured by their respective accelerometers. It boils down to this: the rear end is pushed with an acceleration a and the front end is pulled with an acceleration a, obviously in the same direction. Now the coordinate length of the rod will not Lorentz contract; instead, the proper length of the rod will have to increase, placing it under increasing stress. It will eventually break when the stress exceeds its tensile strength, despite a relatively moderate, but constant acceleration being applied to both ends.

Since both front and rear clocks are now moving at the same speed as measured by inertial system observers, both clocks will lose an equal amount of time relative to the inertial reference clock. Hence, if we stop the accelerations simultaneously, the two rod clocks will still be in sync, but they will no longer be in sync with the reference clock (both behind by the same amount).

This then concludes the accelerated rod scenario, unless anyone wants to add or ask something. I will later include a reference in the SCF Physics Library 'sticky', in order to make it reasonably easy to reference, without cluttering the forum with another sticky.

--
Regards
Jorrie

Last edited by BurtJordaan on October 7th, 2014, 1:07 pm, edited 1 time in total.
Reason: minor 'word correction'

BurtJordaan
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### Re: Lengthwise Accelerated Rod

This whole discussion is fascinating, Jorrie!
This is a good idea: "a reference in the SCF Physics Library 'sticky', in order to make it reasonably easy to reference,..."

The SCF Physics Library becomes an annotated bibliography of links to (among other things) useful threads or individual posts. Even a small-to-moderate step in that direction could be a help to people. Including me, I lose track of good stuff and waste time hunting for it.

For some reason this accelerated rod business (which is a visual intuition-training exercise) reminds me of the "tethered galaxy" thought experiment, which is also a visual mind-trainer.

As I was reading your post the word "stickypedia" came to mind unasked. Never heard it before. doesn't exactly fit this particular situation with the SCF Physics Library, just on the outskirts.

Marshall
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