Capiert

Why don't we use mv universally (instead), for non_relativistic objects too?(E.g. You physicists have a dark big hole, in your understanding about dark energy & dark matter, e.g. a mass defect. Wouldn't it be easier to drop the (energy as) problem? E.g. Does momentum have the same dark (=unknown) mass? How do we know light is massless if it has energy E?

Can we say any energy form has a rest mass? Ok, so does that mean Einstein's (E=m*(c^2)) formula is wrong (as an equality, equation)? & should be written so E=~m*(c^2), as an approximation instead, or E~m*(c^2)? Yes, but if we start with the rest mass speed c, & remove (=subtract) the translational KE speed v, do we not decrease the rest mass's energy? (A mass in motion can surely not be declared completely at rest.) Let v'=c-v. Ok, but isn't momentummom=m*v "mass" m multiplied by speed v? Isn't there a (rest) mass term? (to assume?). (You did say momentum, instead of impulse, or momentum impulse, or impulse momentum.)

+1 (=Thanks)

Thanks for your tip. Try green coffee extract (400 mg/day before your meal), it (=the chlorogenic acid, .. helps, &) burns fat for more energy (Joules, but that's biochem). (You'll have to toss a few (=many) coins (away) for it's price. )

I disagree (but I'm only saying that to prevent confusion). It's made of 2 decisions: a toss, & then specifying which. I can't say I agree (but that's nothing new). I agree.That is the 1st decision. It is biased (so to speak) to find "any" stander. If we collect data that's what we get. (Let's say we've tossed for multi_billion times, as rediculous as it sounds, so we have enough stander counts.) However if we become more biased (please remember we have (biasedly) selected for (any) edge_standers; because heads & tails have been excluded (by our bias=decision(s) rules), we can take that "same data" & ask how many standers happened only on the 1st tosses. (=That is more bias, &) we will find that new number is much less. How do you explain that difference in those 2 numbers? It's quite significant. Why?If you toss 10 times & have no selection rules (=preferences of head, tails, edge, i.e biases) then you only know you tossed, but you do not know what received the larger counts. The bias (results) is which data "you" want (=preferred to see). If "you" want "only" the 1st tosses (count, wrt to the total=all) then you have (biasedly) selected (only) that (set of data) to plot your statistics. The percentages will show that in the plot. It's ok, (maybe my wrong (informal) vocabulary?) you just need a bit of time (to let things settle & clear) to understand because you are logical you will find the answer. But as you see it is not because I have a different perspective than you. But I'm confident you will get it. I don't think so. But I'm sorry, because I'm not sure what (=how) you mean with that context. "Up to" is a limiting expression meaning "not more than". If you mean: if I tossed only once, versus if I tossed ten times, then naturally 10 tosses has better results. But those probability fractions (e.g. 1/N=1/6000) can be used as multiplying factors with the total number of tosses (e.g. 1 or 10) to give you the expected results. E.g. 1*1/6000, versus 10*1/6000. In that sense, you can see it wasn't why we have 1/6000 for any, versus (1/6000)*(1/6000) for "only" the 1st toss. 1/6000 can be the 1st toss, but does NOT have to be (the 1st toss). (1/6000)*(1/6000) is for a selected toss of "your" choice; & you can choose from all possibilities within the 6000 (virtually speaking). Have you at least understood that the results are from a selection (=choice, bias) (of (the total) data)? I.e. a specific percent (that had selection rules). No selection rules means all the data (E.g. multi_billion 100%; but (uselessly) NOT sorted into H, T, E.) Completely (=100%) neutral means you weren't looking for anything, & found everything that told you nothing but the tossing's grand total. Doesn't that ring a bell?How do you show that mathematically? Answer: with the extra *(1/N). I agree.Good. It looks like you have got it. ?

"one of any" is the key there.(It should become obvious to you.) If "any"=N possibilities, you forgot you are (also) excluding N-1 other possibilities (from any) when you want either (but) "only" the 1st or 7th or 216th,.. (See #15 again please.) E=(1/N)*(1/N). The extra (1/N) is needed for the further restriction (of only 1st (instead of any)).

Yes, looking at Lord Antares 1/6000 0.0002=1/5000 comes into that range! Please proceed Studiot.

Please reread #15 again, I've added some math & explaination, & then tell me what you think. I agree. Yes (I agree). Yes, I'm with you. Buckshot: In an attempt to persuade you above. Yes, it does not matter which ball we have chosen; the important thing is whether we have selected "only 1" (or NOT). Good, then maybe now you can see a parallel similarity, in example. No I haven't. You are right. But there is something different about excluding all other tosses. Yes, but there is a difference between those (any) versus specifying that it must be only 1 specific toss from all possibilities (of any).

As soon as you limit to a single number (of sequence) you have increased the complexity. Let us say the edge_stander probability for any toss is E=1/6000, =1/N, N=6000. Now if you want E for only the 1st toss, then that is 1/(N^2) but it does not matter, as you said, it could be 1 of any other than that, e.g. 9th, 22nd, 741st, .. or 6000th, etc. But let's say it did not matter whether we received a stander on either the 1st or 2nd toss, instead: then E=2/(N^2). How about a toss on either the 1st or 2nd or 3rd toss, then E=3/(N^2). I think you get the idea how it's going. If it didn't matter which toss, then any is E=N/(N^2), =1/N. Above, I don't have to use 1st, 2nd, 3rd; instead other numbers could be used e.g. 8th, 21st, 103rd, for E=3/(N^3). Am I wrong? I'd say real comment, because I experienced it so; but that's only my choice of expression. Fine. I used the word possibility (instead, in the header) maybe there is a difference (even in the math?) Does that mean they had something like at least 6000^2=36000000 tosses to display that each toss order (of sequence) was ruffly distributed? (e.g. stander on 1st toss: 1, 2nd toss: 1, 3rd toss: 1 etc for the ideal case, which doesn't really happen so perfect, but I can't show it too well, forgive me please.) Caution: special is distinguished, from the group. Special is only 1 of the group('s number). But "1st" & "any" are NOT the same. It is more difficult to get a stander only on the 1st toss. If you color 1 ball (of many) in buckshot & shoot at a fly expecting the coloured ball to hit it (=the fly); & repeat that experiment over & over, you will find much less color balls hits than the other balls, because there are more of them (other non_coloured balls). Please, don't forget, I was asking for something more practical than the typical probability. E.g. what we really find. 1/6000 sounds good. (I guess I was lucky, I didn't need so many attempts.) I still can't imagine an exclusive 1st toss is just as easy to obtain (but you guys & gals have your methods).

Yes. Experience.(I didn't get 1 on the 1st try (in my life) & I don't know anybody else, either.) You've put an extra limitation on the math, so if you collected the data (of the whole earth, let's say a million to keep the math simple although it's wrong) & found 1000 people had an edge_stander toss; & asked how many had it (=the edge_stander) on their (very) 1st toss & only 1 person reported (=1/1000) that they did, then you would know what I mean. E.g. ~(1/1000)^2=1/1000000. The statistics of being 1st versus not 1st (=simply being any) depend on the square of being (at all).

Ok let's start counting to infinity.The sooner we start, the sooner we'll be finished. A 1st toss (stander) is rarer, than any other toss stander. (I'm (talking about) looking at world statistics, historically.)

Phrases like that indicate the fraction is smaller than expected; because the tosses needed to get success are (really) (much) larger, which the math should reveal (for me) but did NOT. When we add an opposing opinion it seems to me (that) the math is more complicated than what we have done (=formulated).

The possibility of getting an edge_stander on the very 1st toss (in life (=a lifetime)) is even more rarer than to get 1 (=edge_stander) at all. That (1st toss edge_stander) is a (completely) different statistic & probabability. I do NOT say it is impossible, but I've NEVER seen it (happen). I think we agree (there). ?(#5 is edited.)

It won't happen on my 1st toss; I've already experienced that! (=History for my statistic. Anything that happens after can NOT be my 1st toss; because it has a different number of toss.) Sorry if that sounds harsh; but that is what I mean by more practical. E.g. How many times must a coin be tossed till an edge_stander happens, guaranteed (=practically)? Generally, (maybe) once in a lifetime for (only) some (people). The rest never experience it (= an edge_stander), no matter how long they toss. The odds are (for me) useless if I don't see practical results that can be verified experimentally. That's science: experimentally reproduceable, independently.

Thanks. Fascinating stuff.But I'm interested in something more practical. Although Janus (#11, #16, #18) is doing a good job at getting us started, (~5% does NOT seem real or correct to me); I would expect (with this probabilty stuff) after flipping the coin 20 times (that) I would get an edge_stander e.g. 5%=0.05=1/20. But in fact the statistics (of everybody trying in the whole world) say an edge stander is much rarer than 1/20. The bar's case (Acme #15) is the 2nd example I (now) know; but I suspect there are many more examples I have not heard of. Simply put, if I flipped the coin 100 times, I still would NOT get an edge_stander i.e. NOT even 1/100.

I once flipped a coin & it finally landed on edge. I was just wondering how you would calculate the probability of that (happening, because it never happened again). That (edge_stand) would be the (undecided) maybe answer, between yes & no. Assuming we know the diameter D thickness T density (rho) & leaving you free to choose your own flipping method (calculation, e.g. height h, twist rate, etc what(ever) you need, for the normal way people flip coins & coin contour): is there a way to calculate that probability (for edge standing)? E.g. (100%=) 1=H+E+T heads, edge, tails (H=1-T-E, E=1-H-T, T=-H-E), E=1/10000000. Signed SheerLuck Homes.

Yes but Einstein gave us the "general" formula of rest mass energy, as E=m*(c^2). Is it reversable? Can we say any energy form has a rest mass? (Matter is a wave, with (repelling) particle properties.) (Is not matter simply a property also?)

Yes, I noticed that (it was that formula) later (again), & also commented (about that formula) in #90, (due to the ambiguous uncertainty on my side, just to be sure). That formula is also a mixture of scalars & vectors; instead of only scalars (as you said). But thank you, for clearing my uncertainty. Both of your derived (simplified) formulas 2*v3=v1+v2 thus receives vectors; NOT scalars. That's what perhaps confused me into ambiguity because I did not expect you (might) make a wrong assumption. (I thought maybe you meant another formula because your assumption did NOT make sense to me.) The speeds v1 & v2 are vectors (NOT scalars). Thus, in both cases (mine & yours) we have the same value for v3 (=0.05 m/s), so there is no difference between them. So against what you wrote, 0.95 m/s does NOT happen for v3 in #88 (example 1). Or have I missed something, & made a mistake? (If so) Please explain.

From your 1st Example, you forgot something (made a false assumption, & over_simplified (if I'm not mistake). Sorry.). The equation KE3=(m1*KE1+m2*KE2+mom1*mom2)/(m1+m2) is NOT purely scalars, (but instead) it is a mixture of KE (scalars) & momentum (vectors). You did NOT include the momentum term (mom1*mom2=-0.9 kg*J) in your simplification which should read (before simplifing) m3*(v3^2)/2=(m1*m1*(v1^2)/2+m2*m2*(v2^2)/2+m1*m2*v1*v2)/(m1+m2) simplifing m=m1=m2=1 kg, (m3=m1+m2), & cancelling mass, we get v3^2=((v1^2)+(v2^2)+2*v1*v2)/(2^2) v3^2=(((1 m/s)^2)+((0.9 m/s)^2)+2*(1 m/s)*(-0.9 m/s))/4 v3^2=(1 ((m/s)^2)+0.81 ((m/s)^2)-1.8 ((m/s)^2))/4 v3^2=(0.01/4) ((m/s)^2) v3^2=0.0025 ((m/s)^2 v3=(+/-) 0.05 m/s. That is what you should have gotten from my formula even by simplifying, but (it looks like) you forgot the speeds were squared terms (if I'm not mistaken). (Too much simplification is not healthy, too much complexity neither.) The equation KE3/v3=KE1/v1+KE2/v2 is also NOT only scalars but (instead, it is) a mixture of KE (scalars) & v speed (vectors). Using KE1~m1*((v1^2)/2=1 kg*((1 m/s)^2)/2=0.5 J KE2~m2*((v2^2)/2=1 kg*((-0.9 m/s)^2)/2=0.405 J KE3~m3*((v3^2)/2=2 kg*((0.05 m/s)^2)/2=0.0025 J we get 0.0025 J/(0.05 m/s)=0.5 J/(1 m/s)+0.405 J/(-0.9 m/s) 0.05 J*s/m=0.5 J*s/m-0.45 J*s/m. A scalar*vector is a vector (which you know). The division operation can be seen as multiplying the inverse. Your 2nd example is more interesting, & thanks for pointing it out to me. m3*KE3=m1*KE1+m2*KE2+mom1*mom2 0.5 kg*J=0.5 kg*J+0 kg*J+0 kg*J But I think you are confusing things (a bit?) because I said (to you) energy does NOT (always) add correctly. That is my complaint & the classic (=best) example to show that it (=energy) doesn't add correctly! How can you expect to get the same energy output as input? The KE before a collision will NOT give the same energy as after a non_elastic collision. That is exactly what I am complaining about. You are (unintentionally) turning the table around & trying to (help) tell me that I have made a mistake there; but I am trying to (help) tell you the error lies in energy: when you try to add it (=energy) it (=energy) will not do that (addition) correctly. At least not always. I have given you a formula which will equate energy; & it is NOT simple addition as you can see. I have known your 2nd example for many years (in my excel sheet as m1=4 kg, & m2=4 kg v1=1 m/s, & v2=0 m/s; the total KE is 1 J; NOT 2 J, like the input (energy) of the 1st mass, m1 !) Your 2nd example is not new to me, (instead) it is my major proof_complaint (arguement, evidence, clue, hint). Knowing that (fault in energy) I have tried to make the numbers balance. An equation (now) exists. & you have seen it.

Elastically: There I would propose (introducing) a 4th term KE4(/v4) because the (formula's) structure is already established. For the syntax, we may let even postscript numbers be for the same mass; & the same for odd (postscript numbers). E.g. before vs after (collision), let m1=m3 m2=m4. At that point (in time, instance=circumstance, (=in that) case) I would then propose that the common mutual speed be called u for united (speed vector) of a non_elastic collision, to reduce confusion; instead of v3. (& perhaps (something like) mt=m1+m2 for the (common=) total mass (syntax, instead of m3.) In that way I could avoid prime (') notation for (other things; instead of for) after collision; (but (unfortunately) my notes & files are cluttered with it (=primes)). COM (before & after): mom1+mom2=mom3+mom4. Let m4*KE4=-mom1*mom2. For me an inelastic collision is the continuation of (what happened after) a non_elastic collision. Said differently, an elastic collision has an intermediate (state, actually several (states, or stages); each=all) called a non_elastic collision. I hope you understand what I mean. ? I see a continuum.

If we stateKE3/v3=(KE1/v1)+((KE2/v2) as KE3=((KE1/v1)+(KE2/v2))*v3 we can see vector_like attributes (=character) in the use of the KE(s in brackets). E.g. Let m1=1 kg m1=4 kg v1=1 m/s v2=-1 m/s ( where KE1~m1*(v1^2)/2=1 kg*((1 m/s)^2)/2=0.5 J KE2~m2*(v2^2)/2=4 kg*((-1 m/s)^2)/2=2 J m3=m1+m2=1 kg+4 kg=5 kg v3=(mom1+mom2)/m3=((1 kg*1 m/s)+(4 kg*(-1 m/s))/5 kg=-0.6 m/s KE3~m3*(v3^2)/2=5 kg*((-0.6 m/s)^2)/2=0.9 J so ) 0.9 J=((0.5 J/(1 m/s))+(2 J/(-1 m/s)))*(-0.6 m/s) particularly so as 0.9 J=(0.5 J*(s/m) - 2 J*(s/m))*(-0.6 m/s) i.e. the 0.5 J & -2 J part (of the equation); continued, that gives 0.9 J=-1.5 J*(s/m)*(-0.6 m/s). The -1.5 J (part) is the vector_like part we are searching for; but must be multiplied by 0.6 to give the correct value. However, polarity is lost because minus is squared. I have done nothing exceptional in that demonstration, only restating com conservation of momentum. But by now it should be obvious, that energy does NOT add correctly. That is how 2 J plus 0.5 J gives 0.9 J. If I am not mistaken: The person that declared the conservation of energy must have been out of his mind! Or do I see things wrong? If so please explain.

That's what I originally intended to do; but with your help I changed my mind a bit trying to clean things up (& figure things out). (I thought #81 made it clear, we're not dealing with energy vectors, anymore.) I would still like to get things to vector energy but the math tells me that's not completely possible (at least not yet). I'm still open for the idea, but I've given up a bit & have settled with that normal standard KE will do the job, so why should I make things more difficult than that, when I can't reap any advantage (that I intended). The math is set up so I can research the hows & whys it fails or works. It's not that I didn't warn you occational errors had happened & things I couldn't explain. I came here for help to straighten things out & try to figure these things out. I must start with some sort of assumption (as the truth, even if it turns out not to be true), to put it to the test; & I use it (intensely) until it fails, & convinces me otherwise. That has always been my basic strategy. Some hypothesises are obvious, while others are much more subtle.

You've marked me with a -1. Would you please help show me what I've done wrong? Maybe I've made a typo? If I'm not mistaken, your 3 sentences (#82) were discussing KEs as vectors: meaning if KE1=0.5 J & KE2=-0.5 J cancel (to zero), then you would expect mom1*mom2=-1 kg*J would still remain. But I tried to tell you energy vectors don't work (the way I want (them to), KE2 needs to be positive, instead) so I don't persist in maintaining them (KE as vectors). That means (instead), that the (negative) mom1*mom2 term will cancel some of the (non_vector, =positive) KE instead. I was only being curtious (for your benefit) in presenting a redundant vector format (that doesn't really work), because you persisted in maintaining (commenting on) vector energy syntax. If you want me to develop a vector energy formula that works (=fits), then I'm sorry but it's going to take me a lot of time (if it's possible, at all?). Your -1 evaluation was moot because you've mixed up formulas & thus were not on track. Yes, (but why do you want to do such a wrong thing with that formula?) I said, "yes" (but with what? Answer: mom1*mom2.) That's why I could not follow you when you said: Because mom1*mom2 had already been cancelled with the KEs. Are we getting any (bit) closer to an agreement?

Yes. (But it's no problem to devectorize: simply square, then root.At present I don't see a need to maintain that superfluous clutter, yet. So I didn't bother with it (=my vectorized energy).) Yes. I'm not sure what you mean there. ?Please clarify. What do you mean, still have mom1*mom2? That (KE3) equation does not work for energy vectors, (as stated above), it works for (non_polar) energy. As you've seen they (=mom1*mom2) cancel the mass*energy terms, if needed; & they (all, together) become the remaining energy term (KE3) when divided by the (total) mass. As far as I am concerned, (of course) they (=mom1*mom2) must be in the equation (now) otherwise you will not get the correct energy (value) out. (All that is only algebra.) If you want a formula that uses my KE vectors (instead) it is (something like) KE3~((m1*((KE1^2)^0.5)+m2*((KE2^2)^0.5)+mom1*mom2)/(m1+m2))*(mom1+mom2)/(((mom1+mom2)^2)^0.5) where to produce polarity (or vectorize) KE1~m1*((v1^2)/2)*v1/((v1^2)^0.5) KE2~m2*((v2^2)/2)*v2/((v2^2)^0.5). But (divide by) zero (e.g. speed) does not work in excel, so an if statement has to be used instead for the zero case. The last set of terms (as a factor) (mom1+mom2)/(((mom1+mom2)^2)^0.5) is only used to derive the (total) KE3's "polarity" from the total momentum. I.e. to vectorize the KE3 answer.

Thanks, both of you. Excuse me please, this time I mean your non_polar KE (not my vectorized KE). Let m1=1 kg m2=1 kg v1=1 m/s v2=-1 m/s mom1=m1*v1=1 kg*1 m/s=1 kg*m/s mom2=m2*v2=1 kg*(-1 m/s)=-1 kg*m/s KE1~m1*(v1^2)/2=1 kg*((1 m/s)^2)/2=0.5 J KE2~m2*(v2^2)/2=1 kg*((-1 m/s)^2)/2=0.5 J (not my negative value) KE3~(m1*KE1+m2*KE2+mom1*mom2)/(m1+m2) KE3~(1 kg*0.5 J+1 kg*0.5 J+1 [kg*m/s]*(-1 [kg*m/s]))/(1 kg + 1 kg) KE3~(1 kg*J - 1 kg*kg*m*m/(s*s))/(2 kg), J=kg*m*m/(s*s) KE3~0 kg*J/(2 kg) KE3~(0/2) J KE3~0 J Sorry, your arguement does not hold. Can we agree?