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If I take any body in the shape of a rod and stretch that, after it reaches breaking stress it breaks at one point.

Even though we apply the same the stress on each and every part of the rod it broke at one point. If it's uniform it should break at all points because breaking stress is same for all the parts of body as it's uniform.

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    $\begingroup$ It depends on how you stretch it. Grasping each end and pulling certainly does not apply the same stress to each point of the rod. $\endgroup$ – chepner Oct 30 '18 at 13:17
  • $\begingroup$ There are several good answers which explain why the body brokes in one point. It even looks natural now. Funny story is that if you try to break a piece of spaghetti you will always get more that two pieces! (usually three). $\endgroup$ – lesnik Oct 30 '18 at 13:59
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Uniform bodies are idealizations like frictionless surfaces or no air resistance. They make the work easier. In reality there will be slight deviations in material properties (such as density, tensile strength, etc.) along various parts of the body. In your example these deviations become more and more important and extreme as the body is stretched (which the stretching method will also have some asymmetries as well) until we get a single breaking point.

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You could consider it as one more demonstration of the underlying quantum mechanical frame keeping atoms and molecules bonded together. Quantum mechanics is a probabilistic theory, and which bond will "break" depends on the square of the wavefunction describing the rod, with a probability which manifests in this one instance of breakage.

To get the probability curve for a hypothetical completely uniform rod would be a laborious process, as it consists of order of 10^23 atoms/molecules and the number of experiments needed to plot a probability of which atom or group of atoms "breaks" will take forever.

In reality, no rod can be completely uniform, as the other answer states. Even crystals have defects.

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    $\begingroup$ I'll take macroscopic, you take microscopic :) +1 $\endgroup$ – Aaron Stevens Oct 29 '18 at 4:16
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    $\begingroup$ @AaronStevens yes, the macroscopic defects would have to be added to the probability also, and would be more important as a break has to cover a plane replete with atoms and molecules. I just thought it fun to go back to the basic quantum mechanical nature. even if it were completely uniform, there would still be a "random" probability of breakage due to the probabilistic nature of quantum mechanics. $\endgroup$ – anna v Oct 29 '18 at 4:19
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You should consider the implicances of your question, which is mostly a philosophical issue regarding our human subjectivity. You are implying an ideal uniformity of states which wouldn't allow quantum behavior.

That would mean that atoms should have all the exact same state before the breaking, that is, considering them as particles, they should be all at the same angle, have the same rotational speed, same energy, etc. In such case, the momentum of the expansion process would be exactly the same for all couples, and the object would dissipate perfectly as smoke.

A very beautiful and utopic idea. But reality is not like that. Things might break even without stressing them. The probability of a body to remain intact without external action is not 100%.

Taking such logic to the extreme, it means that even if the material is perfect, all its couples of structural interactions have different degrees of resistance to stress... That means that absolutely all couples have different values, even if they are infinitesimally small!

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    $\begingroup$ +1 fantastic answer. However, assuming the perfect (impossible) state you describe (same state for all atoms) would @annav answer still apply, and therefore there should only be one breakage point, and never see the object dissipate like smoke? $\endgroup$ – user Oct 29 '18 at 10:45
  • $\begingroup$ @user right. That's precisely the point. Moreover, if we consider a thing as a set of interactions between its parts (e.g. a thing is a set of atomic bounding links), the "resistance" of each link would change permanently, according to our perception. That is, it changes each time it is measured. That means that microscopically, the same object would start to break always at a different point (microscopically, because the macroscopical factors are different and unrelated: the object is supposed to be "uniform"). $\endgroup$ – RAP - Reinstate Monica Cellio Oct 29 '18 at 11:54
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    $\begingroup$ This is the only answer that points out (indirectly) that actually the rod breaking in a single point is a false assertion; plenty of objects shatter when they break. $\endgroup$ – UKMonkey Oct 29 '18 at 15:07
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All it takes is one single tiny flaw in the microstructure of that rod of yours, anywhere along its length, to begin concentrating the stresses at that point, and all the stretch that happens thereafter will occur right at that point. That is where the rod begins to neck down, leaving progressively less and less cross-sectional area across which to distribute the imposed load, and that is where the rod will fail.

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TL;DR: No

One way of looking at this is to recognise that Physics is a series of constructed narratives involving modelling (or simulating) observations, mainly using (and developing) interpretations from mathematics (which is also a collection of powerful modelling languages).

The notion of idealised objects (including uniformity) comes from the simulation aspect of physics. When we model, we eliminate as much as possible from the problem, in order to be able to concentrate on part of that model that we wish to develop a narrative for (in most cases, to make a good prediction of). This is so that we can construct a useful prediction without having to record too many variables – we are looking for a model that allows us to learn something without making it too complex.

So uniformity is a notion, used in simulation only (so not found in the universe) that serves the purpose of reducing the domain to a workable model.

But, and this is relevant, when we are looking at breaking points and stress loads, we don't (and shouldn't) be modelling the situation with uniform objects, because the predictions will (as you point out in the question) not behave accordingly. Instead, we use alternative models that e.g. provide probabilities of a break to occur somewhere along the structure under stress.

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Does Prince Rupert's drop count?

It will break explosively and reasonably uniformly when damaged due to accumulated stress. This is of course not due to uniformity but rather to its peculiar composition. I wonder if you could redefine uniformity so that it will be covered.

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  • $\begingroup$ As you note, water-quenched molten glass drops are not an example of uniformity, but of extreme polarities of tension on the outer surface and compression on the inner. When the PRD tail is broken, the layers of defects associated with the compressive to tensile non-uniformity are exposed, and the there are so many fracture lines through which to bifurcate that the PRD explodes almost uniformly. Each increment of radius will have a different absolute tensile and compressive pressure, making it extremely defect-dense, hence subject to near-uniform explosive fracture/failure. $\endgroup$ – Thomas Lee Abshier ND Oct 30 '18 at 14:00
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    $\begingroup$ Such a material approaches having a uniformity of defects, rather than the defect-free uniformity of a perfect crystaline structure. $\endgroup$ – Thomas Lee Abshier ND Oct 30 '18 at 14:03
  • $\begingroup$ If we postulate that body can only break where it has defects, then an uniform body needs to have uniformity of defects if it is at all breakable. $\endgroup$ – alamar Oct 30 '18 at 15:07

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