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2

If only the cyclic coordinate $q(t)$ varies with time (if it doesn't, $q$ is superfluous), the Lagrangian, or the essential physical situation, doesn't vary. Hence the initial value of $q$ doesn't determine the path, which is only possible if the path is closed.


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It appears that you are confusing the force you contribute (F1), with the force necessary to overcome friction (F2). The force necessary to overcome friction, is fixed by the mass of the object and the surface friction. As an example let F2 be 100N, lets assume you can only provide 80N, then you will not be able to move the object. If you get a friend and ...


0

If your friend's energy+ yours= F1, then you would see your own energy expenditure halved, which we know cannot be the case. If your friend helps you push the object, then you are no longer applying the same force, or (lazy answer) the force is no longer localised and motivates the part of the object most subject to friction. So, once the object is first ...


3

All the definitions you've posted are correct, and they aren't in conflict with each other, although they are a bit imprecise. I'll try to explain in more detail what these concepts are. Probability I'll assume we don't need to attempt to define what probability is. :) I'll just note that it's formalized in mathematics under the aegis of measure theory. ...


3

To understand the difference between probability and probability density consider the difference between mass and density. Density is the mass per unit volume, so to find the mass you multiply the density by the volume: $$ mass = density \times volume $$ In some cases the density will be a function of position and we have to write it as a function of the ...


-1

It has been many years since I have used this information (so please forgive my inaccuracies, but I wanted to get you PART of the way there with an answer (since no one has responded yet). The probability density is the integral (area under the curve) of the probability. I think the reason for the funky discrepancy between the two definitions is because.. ...


3

I'm sure everyone has had that concern when we encountered the definition for the first time, in school. There is a valid reason why this definition is still persisted with, despite the deficiency that you hit on. The most popular (and simple) forces in physics (also the ones with which we begin learning physics) are conservative forces, implying that the ...


35

If you're pushing a 10-ton truck and it's not moving, you are not doing any work on the truck because the distance $ds=0$ and the nonzero force $F$ isn't enough for the product $F\cdot ds$ to be nonzero. Your muscles may get tired so you feel that you're "doing something" and "spending energy" but it's not the work done on truck. You're just burning the ...


1

I'm not familiar with ISO 5725 (a 1994 revision of a 1986 document, apparently "reviewed and confirmed" in 2012), and it seems that I have to buy it to read it. A 2008 vocabulary of metrology put out by the BIPM and also cited by Wikipedia has definitions much closer to my intuition, and to common usage among folks I know who specialize in precision ...


7

The "shift in the meaning" refers to some attempts to reinterpret the terminology that were made by a metrological document, ISO 5725, in 2008. That may be described as a bureaucratic effort by a few officials – really bureaucrats of a sort – and as far as I know, the "shift in the meaning" hasn't penetrated to the community of professionals. The people ...


3

We have a perfectly unambiguous definition of temperature for canonical ensembles, and this temperature may be negative in bounded-energy systems. This kind of negative temperature is indisputable, and some would argue it has been realized in spin-inversion experiments. The problem is that there are two decent but imperfect definitions for the entropy of a ...


3

What I am asking, then, is whether someone on StackExchange might be able to shed some light on the matter as to how there can be a disagreement about something that seems should be a mathematical fact. The main disagreement seems to be about which definition of the word "entropy" in the context of statistical physics is "correct". Definition is an ...


2

Simply the thermodynamical quantities used in the original paper were not suitable for that problem. They in particular calculated $T=\frac{\partial U}{\partial S}$ where $ U$ is the internal energy and $S$ the entropy. However a wrong definition of entropy has been used. Mathematicians has proved that the use of that specific entropy was wrong and that ...


4

It's the differential relationship between internal energy and entropy: \begin{align} dU &= T\,dS + \cdots \\ \frac{\partial S}{\partial U} &= \frac 1T \end{align} As energy is added to a system, its internal entropy changes. Remember that the (total) entropy is $$ S = k \ln\Omega, $$ where $\Omega$ is the number of available microscopic states that ...


0

As a metrologist, I am glad of this interest in correct notation, often not enough pondered also among metrologists but essential for understanding with each others. I would say first that any numerical value of an experimental result is always expressed as a rational, not irrational, number, because the number of digits is always bounded by the position of ...



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