Wikipedia:Reference desk/Archives/Mathematics/2016 August 12

= August 12 =

Proof of mass - energy equivalence
What is the detailed proof of mass-energy equivalence (starting from certain axioms of motion as indicated on article talk page?) Thanks.--213.233.84.12 (talk) 12:08, 12 August 2016 (UTC)
 * It depends on whether the conservation of mass is given in advance.
 * If the conservation of mass is not given in advance, then what can only be proved (a-priori) is the equivalence between the kinetic energy and the relativistic mass being added to the rest mass because of the kinetic energy (this relativistic mass being added is therefore the difference between the whole relativistic mass and the rest mass).
 * As for how one can prove this a-priori (even if the conservation of mass is not given in advance), you can simply calculate the kinetic energy, while bearing in mind four well-known Newtonian equations: v = dx/dt, p = mv, F = dp/dt, E$k$ = &int;Fdx: All of that leads us to the first conclusion: E$k$ = &int;vd(mv). Now you should also bear in mind the Lorentz transformation of mass: m = &gamma;m$0$: This, along with the first conclusion, leads us to the following conclusion: E$k$ = (m-m$0$)c$2$. QED.
 * As for the rest mass itself: One cannot prove (a-priori) it's equivalent to energy. What can only be proved (a-priori) - regarding the rest mass - is, that if the system conserves the mass - and the rest mass "vanishes" (somehow), then it becomes kinetic energy - whose quantity is m$0$c$2$, m$0$ being the quantity of the rest mass before it vanished. The proof for this result becomes quite easy (actually it becomes totally logical rather than mathematical), once you've received the previous conclusion mathematically proved in the previous paragraph. Additionally, when assuming the conservation of energy - after receiving the previous result, one logically concludes that every rest mass (whose quantity is m$0$) which can vanish - is potential energy (whose quantity is therefore m$0$c$2$). However, this does not mean (even if one does assume the conservation of mass and the conservation of energy) that one can prove (a-priori) that any rest mass can really vanish. Only experiments can show that (as they did); In other words, when not leaning on experiments, not only isn't it a-priori provable that mass is equivalent to energy, it's also a-priori unprovable that mass can become energy at all. HOTmag (talk) 13:16, 12 August 2016 (UTC)

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 * I think you can get pretty far using basic assumptions (energy conservation, that SR limits to Newtonian physics at small velocities and how to Lorentz transform EM waves). In the original paper, Einstein considers a body at rest with an energy $$E_1$$ emitting two photons back-to-back, each with energy $$\epsilon/2$$. After this decay, we have two photons and a new body (still at rest) with an energy $$E_2$$. Energy conservation tell us
 * $$E_1 = \epsilon + E_2$$
 * Einstein then wants to view this process in a new inertial frame, and crucially he knows how to change an electromagnetic wave's energy as he changes frame. In a new frame, boosted at velocity $$v$$, the original body has energy $$E_1^\prime$$ and the final body $$E_2^\prime$$. From his established rules for how to boost the electromagnetic waves, and requiring energy conservation in the new frame, he knows now that
 * $$E_1^\prime = \epsilon\gamma + E_2^\prime$$
 * where $$\gamma$$ is the usual relativistic one. He defines the kinetic energy of body $$i$$ to be $$K_i(v)=E_i^\prime - E_i$$. That is: the difference between the energies in the boosted frame and the rest frame. By subtracting the two energy conservation equations he finds that
 * $$K_1(v) - K_2(v) = \epsilon(\gamma - 1).$$
 * Now the smart bit is for Einstein to say that the velocity $$v$$ was entirely arbitrary. He requires that this relation holds in all frames regardless of $$v$$. He makes one additional assumption, that for very small velocities his kinetic energy expression should recover the standard one
 * $$K_i(v) = \frac{1}{2}m_iv^2 + \mathcal{O}(v^3)$$
 * So what happens for very small velocities? We see that
 * $$\frac{1}{2}(m_1-m_2)v^2 = \epsilon(\gamma -1) + \mathcal{O}(v^3) = \frac{1}{2}\epsilon v^2  + \mathcal{O}(v^3)$$
 * The only way for this to hold for every small velocity is if the mass difference between body 1 and body 2 is equal to the energy emitted in the decay. He has shown that for energy conservation to be compatible with his relativistic Doppler formula for EM waves in different inertial frames, while assuming that everything correctly limits to Newtonian physics at small velocities, mass must be lost as a body emits EM energy. In this sense, he has shown that mass and energy are equivalent under the assumptions of SR. 193.60.182.106 (talk) 10:59, 14 August 2016 (UTC)
 * I think Einstein's proof assumes two many assumptions (few of which you've mentioned in the paragraph beginninn with the word "Now"), but he receives too little, i.e. what he receives - only refers to bodies emitting photons: How about the potential energy inside a mass which emits no photons? Can he prove every object can emit photons? If it can't, then is its mass still equivalent to energy? Einstein doesn't answer these questions. Further, by adding assumptions as Einstein did, I can formulate a much shorter proof - even without Calculus (which I've used in my previous response) - but with three assumptions only, and without leaning on Dopler formula - but rather on Newtonian mechanics only. The first assumption is as follows: The velocity of a body having no rest mass, does not depend on any inertial frame. Now Let's look at such a body (e.g. a photon or a gluon or a gravition or whatever), which really has no rest mass (although it may have some relativistic mass, perhaps). According to the first assumption mentioned above, the body's velocity - let's call it c - must be constant (i.e. it does not depend on any inertial frame, and consequently neither the body's velocity nor its relativistic mass depends on each other). Now, according to the well known Newtonian equations: v = &Delta;x/&Delta;t, p = mv, F = &Delta;p/&Delta;t, E$k$ = &Sigma;F&Delta;x, which entail the conclusion: E$k$ = &Sigma;v&Delta;(mv), one concludes that the kinteic energy of that body (whether it's a photon or a gluon or a gravition or whatever) - whose relativistic mass is m, is: E$k$ = &Sigma;c&Delta;(mc). However, since c is constant (as I've proved above), then the body's kinetic energy must be: E$k$ = &Sigma;c&Delta;(mc) = c$2$&Sigma;&Delta;m = mc$2$. Now let's look at a "bigger" object (e.g. a pair of electorn and positron), whose mass is M, and which vanishes while emitting photons (or gluons or gravitons or any particles having no rest mass). According to my second assumption - that is the conservation of mass, the total (relativistic) mass m of those photons (or gluons or gravitons or whatever) must be equal to the original mass M of the bigger object that has just vanished while emitting them, so their total kinetic energy - which is mc$2$ (as I've proved above) - must be equal to Mc$2$. Adding my third assumption - that is the conservation of energy, one can logically infer, that before this bigger object - whose mass was M - vanished, its total energy was Mc$2$. QED. HOTmag (talk) 13:56, 14 August 2016 (UTC)

Another derivation given here. This is based on the fact that the center of mass of an isolated box that is initially at rest, will not change if there are no external forces acting on the box. If we then consider a box containing two masses that exchange a photon, and consider the movement of the mass emitting the photon during the the time that the photon is on its way to be intercepted by the other mass, you end up with what looks like a change in the center of mass of the system (we need to be agnostic about how we account for the center of mass during the time of flight of the photon, but after it has been absorbed, the center of mass is unambiguously defined according to non-relativistic physics). If we then demand that there has been no change in the center of mass and also impose conservation of mass, then that implies a mass transfer that is given by the energy transfer. The equivalence between energy and mass must then hold in general, because any process that would allow the transfer of energy without the equivalent mass transfer could be coupled to an electromagnetic transfer of energy and its equivalent mass, which would then allow you to violate conservation of mass. Count Iblis (talk) 21:38, 15 August 2016 (UTC)
 * Is somehow electromagnetic mass involved in the proof? How about the proof involving plastic collision? Does conservation of mass imply a sum between ordinary mass and electromagnetic mass due to electromagnetic momentum?--213.233.84.70 (talk) 21:59, 18 August 2016 (UTC)

Z/P-value for skewed data
I want to validate my assumption. To do so, I need to explain what I'm reading in more detail that most want to read. This paper describes multiple sets of cars. The goal is to get MPG above 33. The problem is that MPG for a single car is variable, depending on the driver, conditions, etc... So, each car has a set of MPG values taken at multiple times. Then, they go into the probability that any single MPG value will be above 33 MPG. So, this is NOT about getting an average MPG for a single car above 33. This is about a single MPG measure being above 33. So, they used statistics. They got the population of MPG measurements and assumed that they have a normal distribution. They calculated the mean and standard deviation of the population of MPG measurements. They used that to calculate a Z value and then a P value. They came up with the claim that the probability of a single MPG value being above 33 is 72.4%. I then looked at data for MPG and found that the data has a long left tail and a very short right tail. So, there is a larger population on the left side of the "lump" than the right side. The median is well to the left of the mean as well. Therefore, my assumption is that because there is a left-skewed population, their claim that 72.4% of the MPG measurements being above 33 is inflated. The actual percent will be lower. Is that correct? Would it actually be higher? 209.149.113.4 (talk) 14:09, 12 August 2016 (UTC)


 * Did you find that the median is less than 33? If so, let the median be m<33. Then no more than 50% of the cases are greater than m, so fewer than 50% of the cases are greater than 33, contrary to what they report. Your observation that the data are skewed would invalidate their assumption of normality. Loraof (talk) 15:54, 12 August 2016 (UTC)
 * If they have enough data points, you could use a normality test on the null hypothesis that the data are normally distributed. But if there are not many data points, that test might have low power to reject the null when the null is false. Loraof (talk) 16:36, 12 August 2016 (UTC)


 * I don't have their data. I'm reviewing a paper that is about computer programs for cars. They are making the argument that there is no need to log a continuous MPG value because over 72.4% of the time the MPG will be above 33 (a lot of regulation issues involved here that is creating the argument, but that isn't important for this topic). If they are underestimating and the skewed values make it actually higher than 72.4%, that is better. They strengthen their argument. I think that the long low-end tail makes the percent less than 72.4%. If it is below 70%, it completely busts their argument (again, regulation issues). Because statistics is outside my field of expertise, I didn't want to make an invalid criticism. My criticism is currently: "Your claim that 72.4% of the MPG readings will be above 33 is based on the assumption of a normal distribution, which you state is a false claim. Checking similar data, I have found that MPG values tend to be strongly skewed to the low end because the left tail is much longer than the right tail. There will be less MPG readings above any specific P-value than would be found in a normal distribution because the population is skewed to the left side. Therefore, your value of 72.4% is an inflated estimate. If it is off by just 2.4%, it will be at the 70% cutoff and your claim about continuous MPG readings will be invalidated." If I have it backwards, I obviously don't want to make the criticism. That is why I've asked here if my criticism appears valid. 209.149.113.4 (talk) 16:54, 12 August 2016 (UTC)


 * It seems to me that may not be possible to say anything if you don't have access to their data. As our article skewness says, for left-skewed data the median could be less than or greater than the mean. Do you know the mean and median? Are their daya on a particular model, or on a fleet? You say you've looked at data and found it left-skewed—was that also data on a particular model, or on a fleet? Loraof (talk) 17:55, 12 August 2016 (UTC)


 * Thanks. I will reread earlier sections to see if they mention the median. They only mention the mean in that section. Because the entire paper isn't about MPG reading, it may be omitted. If so, I will change my statement from a criticism to a request for information. 209.149.113.4 (talk) 18:24, 12 August 2016 (UTC)

Sufficient Condition for Relaxation Method for Integer Linear Programming
Hi, I am looking for a condition that when holds for some linear program, then we can use the relaxation method for solving its integer programming variation. Strictly speaking, I am looking for a sufficient condition such that $$Ax\leq b$$ implies $$Ax'\leq b$$ where $$x_i^'=[x_i]$$ (that is, it equals the nearest integer of $$x_i$$). 31.154.81.54 (talk) 16:10, 12 August 2016 (UTC)


 * From Integer linear programming:
 * While in general the solution to LP relaxation will not be guaranteed to be optimal, if the ILP has the form $$\max\mathbf{c}^\mathrm{T} \mathbf{x}$$ such that $$A\mathbf{x} = \mathbf{b}$$ where $$A, \mathbf{b},$$ and $$\mathbf{c}$$ have all integer entries and $$A$$ is totally unimodular, then every basic feasible solution is integral.


 * I really doubt you're going to find anything better than that. Loraof (talk) 16:29, 12 August 2016 (UTC)


 * The reason I don't think there's anything useful for you is this. Suppose you have a constraint set and a point x such that $$Ax\le b, $$ and in one of the binding constraints there is a positive coefficient on an $$x_i$$ where $$x_i$$ is greater than 1/2 and less than 1 (and all other $$x_i$$ values are integers). Then that constraint will be violated by the system with revised variable values, since $$x_i$$ will go up to 1 and the rest of the variables will be unaffected, so the value of the left side of the constraint will increase, violating the constraint. Loraof (talk) 16:49, 12 August 2016 (UTC)


 * From my experience with IP's I'd have to agree. Basically if the solution to the relaxed problem isn't already integral then rounding to the nearest integer rarely has any benefit. The integral points in a polytope may lie quite far away from its surface, so the nearest integer point to a vertex is rarely going to be feasible. This seems a bit counterintuitive if you're used to thinking about polytopes in 2 or 3 dimensions, but when you get into 20 or 30 dimensions things look a bit different and your intuition is extremely unreliable. To be more specific, you might have a relaxed solution with a variable y =.7, but where the maximum value of y in the polytope is .8. So really the only integral solutions have y=0 and rounding y to 1 won't work. This sounds a bit contrived but in my experience it's the most common situation. You're much better off using integer cuts of some kind than to use rounding. Maybe the condition should be that no further integer cuts exist, but this would be the same as saying all vertices are integral anyway. --RDBury (talk) 22:41, 12 August 2016 (UTC)

Thank you for the answers. The total unimodularity condition sounds rather powerful to me. However, for my very specific set of equations (that is relatively hard to explain) does not satisfy this condition, as A contains real zeros rather than only {0,1,-1}. So, I came up with a weaker version of the question: What condition on A guarantees that $$Ax\leq b$$ implies $$A[x]\leq \lambda b$$ where $$\lambda\in\R$$ is some scalar, and b is a vector as before?80.246.137.183 (talk) 08:06, 13 August 2016 (UTC)

as the first digit
In California the auto license plates are sequential. The formula is a seven digit number letter combination. The first digit is always a single number then three letters in alphabetical order then three numbers is sequence. The first license was 1AAA001  then 1AAA002 etc until  they reached 1ZZZ999 then they start all over but with a "2" as the first number. My question is ...how many licenses would have been created with the number 1 as the first digit? This id definitely not homework. Also I would love to know the formula used to solve the problem. w. s.carter — Preceding unsigned comment added by 72.132.36.185 (talk) 22:23, 12 August 2016 (UTC)


 * In the question, there's no variation possible in the first digit, so we can just ignore it. The only part of the problem that's not immediately obvious is the number of combinations of 3 letters in alphabetical order. To count them, we can represent each valid 3-letter sequence as a string "ABC...Z" with markers added before the letters that occur in the sequence. With this notation, the sequence "AFP" is represented as
 * And "AAD" as
 * The number of valid 3-letter sequences is the number of valid ways you can intersperse 3 $s among the letters. Since the last symbol in the 29-symbol sequence is always a Z, we can ignore it in our counting. In the remaining 28 symbols, exactly 3 positions are $s. If we modify our notation so that each letter is represented simply as an "@", "AFP" is now
 * And "AAD" is now
 * There are $${28 \choose 3} = 3276$$ such sequences, and therefore 3276 valid 3-letter combinations. Each of these can be followed by 1000 possible 4-digit endings. There are therefore $$3276 \times 1000 = 3276000$$ possible plate numbers beginning with 1. --72.78.149.18 (talk) 02:13, 13 August 2016 (UTC)
 * And "AAD" is now
 * There are $${28 \choose 3} = 3276$$ such sequences, and therefore 3276 valid 3-letter combinations. Each of these can be followed by 1000 possible 4-digit endings. There are therefore $$3276 \times 1000 = 3276000$$ possible plate numbers beginning with 1. --72.78.149.18 (talk) 02:13, 13 August 2016 (UTC)
 * There are $${28 \choose 3} = 3276$$ such sequences, and therefore 3276 valid 3-letter combinations. Each of these can be followed by 1000 possible 4-digit endings. There are therefore $$3276 \times 1000 = 3276000$$ possible plate numbers beginning with 1. --72.78.149.18 (talk) 02:13, 13 August 2016 (UTC)
 * There are $${28 \choose 3} = 3276$$ such sequences, and therefore 3276 valid 3-letter combinations. Each of these can be followed by 1000 possible 4-digit endings. There are therefore $$3276 \times 1000 = 3276000$$ possible plate numbers beginning with 1. --72.78.149.18 (talk) 02:13, 13 August 2016 (UTC)


 * However, I suspect that the original poster did not mean to say that the letters must be in alphabetical order: there would be no strong reason to do that. Rather, I suspect they meant that the plates are given out in alphabetical order of the 3-letter sequences.  If so, there would be 26&sup3; = 17,576 possible letter combinations&mdash;except that the authorities would surely screen out a number of possibilities such as FUK and GOD.
 * Looking through 329 digital photos that I took when I was in San Francisco in 2014 to find ones where ordinary California license plates on cars are legible, the letter combinations I found were: ZHK, FQY, FHY, XSA, SGZ, GIY, ALA, YHR, FCP, UVX, DCC, VIC, XBZ, EUH, EZH, DHX, JTD, BND, DEA, WRP, HGX, SCV, DYU, and the one I rented, FSY. 6 out of the 24 have the letters in alphabetical order, or just 2 more than would be expected by chance alone.
 * Incidentally, while doing this I was amused to see that while walking around Russian Hill I had photographed the same station wagon twice: once turning from Lombard St. (the non-wiggly part) onto Hyde St., and then 22 minutes later on Filbert St. approaching Leavenworth St. Must've been another tourist! --69.159.9.219 (talk) 05:29, 13 August 2016 (UTC)
 * Vehicle registration plates of California has more info on the CA license plates. According to it, the 1 as a starting number hasn't been used for a few decades now, at least for normal plates; there are exceptions and exemptions as in all things governmental. The numbers aren't assigned sequentially, for example there is usually a gap when the plate is restyled every so often. Also according to the article, they're currently on 7 and according to my estimate they start on a new starting digit every 5 years or so. Which means they will be running out sometime in the next decade. Perhaps they will continue on with AAAA001 after that, or perhaps some mega-fuel shortage will force everyone to ride bicycles by then. --RDBury (talk) 08:07, 13 August 2016 (UTC)