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Modelling of gas molecule interacting with the wall of a chamber faithful usage of expression in state comparison: states involved are state before elastic collision with wall and state after collision with wall. \begin{eqnarray*} p_{wall}+m*v_{x} & = & p_{wall}'+m*v_{x}' \end{eqnarray*} The above resolves as: \begin{eqnarray*} m*v_{x}-(m*v_{x}') & = & p_{wall}'+p_{wall} \end{eqnarray*} where $p_{wall}=0$ so we have \begin{eqnarray*} m*v_{x}-(-(m*v_{x})) & = & p_{wall}'+0 \end{eqnarray*} so we end up with \begin{eqnarray*} m*v_{x}+m*v_{x} & = & p_{wall}' \end{eqnarray*} in the end we got: \begin{eqnarray*} p_{wall}' & = & 2*m*v_{x} \end{eqnarray*} \begin{eqnarray*} p_{wall}' & = & 2*p_{x} \end{eqnarray*} so the wall gets double the momentum as the initial momentum of the colliding particle. But this is wrong. Equality is only applicable when the two states are equal literally but not when they are just isomorphic and we know that equal upto isomorphism means equality. But if that notion is not captured in the outset then things are doomed to make no sense. The gist of the problem at hand is this: the two states are the reverse of each other, one is in a momentum gaining state and the other is in a momentum loosing state, but the above outset implies that the wall after the collision is also in a momentum gaining state. A simple Sum does not do justice to the description of the situation. Weather we look from beyon the paper plane or from above it does not change that the two states are different (one is still gaining whilst the other is loosing) either. Furthermore the use of $v_{x}$and $v_{x}'$ implies a difference, but when we express the one in terms of the other that difference is lost, tho we know the two are two different velocities. A better symbology would use indices. So if we capture the above into a new formulation we can arrive at two approaches: 1.) \begin{eqnarray*} p_{wall}+m*v_{x} & = & -(p_{wall}'+m*v_{x}') \end{eqnarray*} \begin{eqnarray*} 0+m*v_{x} & = & -p_{wall}'-m*v_{x}' \end{eqnarray*} \begin{eqnarray*} m*v_{x}+m*v_{x}' & = & -p_{wall}' \end{eqnarray*} \begin{eqnarray*} m*v_{x,1}+(-m*v_{x,2}) & = & -p_{wall}' \end{eqnarray*} \begin{eqnarray*} m*(v_{x,1}-v_{x,2}) & = & -p_{wall}' \end{eqnarray*} \begin{eqnarray*} p_{wall}' & = & -m*\Delta v_{x} \end{eqnarray*} \begin{eqnarray*} p_{wall}' & = & -\Delta m*v_{x} \end{eqnarray*} \begin{eqnarray*} p_{wall}' & = & -\Delta p_{x} \end{eqnarray*} or 2.) \begin{eqnarray*} p_{wall}+m*v_{x} & = & p_{wall}'-m*v_{x}' \end{eqnarray*} \begin{eqnarray*} 0+m*v_{x} & = & p_{wall}'-m*v_{x}' \end{eqnarray*} \begin{eqnarray*} m*v_{x}+m*v_{x}' & = & p_{wall}' \end{eqnarray*} \begin{eqnarray*} m*v_{x,1}+(-m*v_{x,2}) & = & p_{wall}' \end{eqnarray*} \begin{eqnarray*} m*(v_{x,1}-v_{x,2}) & = & p_{wall}' \end{eqnarray*} \begin{eqnarray*} p_{wall}' & = & m*\varDelta v_{x} \end{eqnarray*} \begin{eqnarray*} p_{wall}' & = & \Delta m*v_{x} \end{eqnarray*} \begin{eqnarray*} p_{wall}' & = & \Delta p_{x} \end{eqnarray*} The minus sign in the first captures the momentum as a net gain for the wall after the collision, whereas the second one misses to to tell us if it is a gain or loss. As we see we have to be careful using equations faithfuly. When we look at the first approach what it tells to us is that the second state is the reverse of the first as a whole while the second one does not claim so. If one thinks about it the former part in the beginning of this section is an attempt of using math for physics without using physical laws or bypassing it. As it turns out newtons law already captures that notion as in: $actio=reactio$ One just has to turn that into the correct relation: $actio=-(actio')$ to see what is missing, even tho we have to admit the above mathematical attempt without explicitly referring to physical laws could still be turned into the right thing when one frames the notions right in the mathematical language. This way the error is shown to be twofold, one by bypassing physical laws, the second in the mathematical flaw of describing the situation. A third one would be to dismiss this in the knowledgebase for the public. That said we have to admit we don't know the full extent of the consequences for other formulae stated/held in the public knowledgebase. Whatever it may be (how much of it creeps into other stuff, how much of difference it causes etc.) it has to be adressed. So the conclusion is either we are doing math without physics but badly or we are doing badly physical maths. To depict some of the aspects of work involved in elucidating the thought process pertaining to the situation at hand we have to make some analogies to known math as follows. One of those aspects is object identity! We would be appalled to take it true that $1=-(1)$. Tho the object to the left of the equation and the object within parenthesis on the right hand are the same, we know this equality fails. So did we assert something wrong above ? Since we might be inclined to believe based on the similarity that $p_{wall}+m*v_{x}=-(p_{wall}'+m*v_{x}')$ that this expression in its outline corresponds to $1=-(1)$. The confusion arises from the view that object identity has to hold in the case of an expression, to further what we mean look at here: $1=-(-1)$. We know this to be a true expression, tho the objects left and right side of the equation 1 and -1 are obviosly not the same. The otherness here is expressed by a negation which itself is part of or sticks to the number. In our case the otherness is expressed via $'$. It also sticks to the operation $+$ itself as is shown here, i.e. $-'$ corresponds to a $+$: \begin{eqnarray*} p_{wall}+m*v_{x} & = & -(p_{wall}'+m*v_{x}') \end{eqnarray*} \begin{eqnarray*} p_{wall}+m*v_{x} & = & -(p_{wall}'\:-'\:m*v_{x}') \end{eqnarray*} Here it is on us to take the apostrophes outside the parens or do some substitution for some elements of the paren expression. i.e. \begin{eqnarray*} p_{wall}+m*v_{x} & = & -'(p_{wall}-m*v_{x}) \end{eqnarray*} \begin{eqnarray*} p_{wall}+m*v_{x} & = & -(p_{wall}'+m*v_{x}') \end{eqnarray*} \begin{eqnarray*} p_{wall}+m*v_{x} & = & -p_{wall}'-m*v_{x}' \end{eqnarray*} But you see we wrote the shorthand version with $+$ instead of the quoted $-$, not to invoke confusion, but because it's an expansion step before applying the minus surrounding the expression as a whole (which comes from the mirror symmetry of the system). So in actuallity when we compare to $1=-(-1)$ that $1$ corresponds to $p_{wall}+m*v_{x}$ and $-1$ corresponds to $'(p_{wall}-m*v_{x})$. There is no object identity as $p_{wall}+m*v_{x}\neq'(p_{wall}-m*v_{x})$ as is not the case that $1=-(1)$. The resemblence of the left side with the right $(p_{wall}'+m*v_{x}')$ is an intended one because it looks similar, but nevertheless the left side is not equal to the right side as we have $(p_{wall}'+m*v_{x}')='(p_{wall}-m*v_{x})$. So we are not concluding logically ill. Our description is of the form $a+b=-'(a-b)=-(a'\:-'\:b')=-(a'+b')$ where the distribution of the apostrophe over the expression elements happens first and the ``otherness'' of $-$ corresponds to a negation i.e. $-'=-(-)=+$. Another word about object Identity, what do we think when we consider a raw egg to omelette, is it the same egg? Ofcourse it is the same egg, it is just in another state, so there is the notion of sameness and otherness involved here as is in our case.One more thing to notice in the distribution of the apostrophe over the elements in parentheses is that it seems to apply uniquely, i.e. the otherness of an operator is it's negation, the otherness of the momentum of the wall is it's difference to it's prior state, the otherness of the gas molecule is its state after the collision. So it applies over a range of different objects, falling into the context required for “otherness” for each object from different categories at hand. Or said another way otherness of objects are state related (state change), whereas operators retain their negation sense from ordinary math. We said that there are some more aspects to our overall statements. These are more on the perceptional/cognitional level of mathematical symbology itself. One has to admit some cognitional concoctions of the equals sign. It bears similarity with a pictorial mirror plane. a=b. Since it's a mirror b when moved over to the other side changes sign. This gives one the false impression that b is already the opposite of a, when in actuality what the equals sign means is that a is just b. So when we set out to compare two states which are the opposites of each other we may be lead to the elusive idea to state it like a=b when in fact what we mean is a=-b. But by posing just that we are only at the global level. For the description of the states a and b to relative to each other we have to go another level down to the block level and state in that block (or frame) level how a and b relate to each other again. So now we know where the first minus comes from and also why there is double negation involved and that the formula at the beginnig of this elucidation is only a negation because at least one of these is forgotten to be stated properly. Hence we risk either doubling p_{x} or nulling it, but never getting the right answer. Last but not least we should say that a transformation is not a statement or rather a transformation is not the same thing as a de-novo statement. What about mathematical symbols like equals. Are they easy ? For example it is deceiving you into left and right handedness, but it's function is about $"isness"$ . you can't say your left arm is your right arm, because your left arm isn't your right arm. So the equals sign has no metalogical property of expressing intrinsically something about right and leftness. At best your left and right arm are isomorphic you can say, but that's it. the one is the opposite of the other in some sense. So what happens here is we forget on the language level to tuck a minus sign on to the expression. As a result we are negating in a logical sense the sentence and using the conclusion derived from such a negetad A ( $\lnot A$ ). In our approach we do however tuck a minus sign onto the expression. What then happens is that it gives a double negation in a logical sense. And we know double negations don't change the truth value of the sentence. So we are logically on firm grounds with regards to our conclusion.
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Wed, 02 Jun 2021 19:21 GMT