E=mc2 passes tough test

[origin]

Here is a link to Ohio state astronomy site. It looks like any articles written on the subject by scientific journals are so old that they are not on line anymore. This is such a well documented effect that it is found in lower level physics in and astronomy courses.<br /><br /> <div class="Discussion_UserSignature"> </div>

 

[vandivx]

<blockquote><font class="small">In reply to:</font><hr /><p>Einstein used the classic formula E=(mc^2)/2 , for v=c.<br />Somehow he concluded that the total energy is twice the kinetic energy for any mass. <p><hr /></p></p></blockquote><br /><br />the problem is that in classical physics as they are tought we typically only look at half of the total picture most of the time, for example momentum of a body is p=mv and sum of the momentum by integration to obtain the body's kinetic energy is KE=1/2mv(sq)<br /><br />now the problem is that all this takes into account only half of the relevant picture, I mean that for every body going this way with momentum p there has to be another body going opposite way also with momentum p, same for kinetic energy<br /><br />total momentum is then 2p=2mv and total kinetic energy (2KE)=2*1/2mv(sq)=mv(sq) and when v=c we then have total kinetic energy KEtotal=mc(sq)<br /><br />imagine the anihilation of particle/antiparticle pair with all the energy (ideal situation) that it liberates being channeled into pushing appart two bodies (all initially at rest in lab) - the total KE of both bodies is then as above, that is mv(sq) which equals the total rest mass energy mc(sq) of the particle/antiparticle pair that went into production of that kinetic energy<br /><br />it never happens that the energy of anihilation would somehow produce motion of body going off one way only, the standard way the momentum and kinetic energy subjects are being tought in schools are not physicist's way but more like ingeneer's way (making it look falsely as if there is only one side to these things and quite naturaly a locomotive engineer for example is only ever concerned with half of the equation) and that's a problem when physics is being tought that way to budding physicists IMO<br /><br />btw have a look at the ingeneer vs engineer... for a good laugh, at the moment I was confused on the matter and so looked it up and this is what I found (had to host it as the location didn't all <div class="Discussion_UserSignature"> </div>

 

[vidar]

There is no proof that: E=1.0mc^2 or E=42mc^2<br />http://en.wikipedia.org/wiki/The_Answer_to_Life,_the_Universe,_and_Everything<br /><br />That E=mx riddle can’t be settled.<br />

 

[vidar]

It seems to me that the E=mx riddle cause a Theocrat vs Technocrat rivalery.<br />Einstein used God when cornered.<br />http://en.wikipedia.org/wiki/Einstein#Bohr_versus_Einstein<br />

 

[origin]

Umm, So your first post is a joke, right?<br /><br />Your second post has absolutely nothing to do with E=mc^2.<br /><br />PS: I think we can pretty confindently say that Einstein was wrong about quantum mechanics and God is a gambling man, er diety.<br /><br /> <div class="Discussion_UserSignature"> </div>

 

[vandivx]

err, deity<br /><br />many a time people make of initially nonsense or nonseriously meant threads good threads, everything is what you make of it<br /><br />it certainly appears that Einstein was wrong, still I can't see how that which we see can be true and I strongly suspect we don't understand QM enough as yet and that once we do the probabilistic interpretation will take on different look, reason is that since I don't believe in any god that would be throwing the dice I am left with uncaused chance happenings, that is happenings that have no physical or causal base, that just happen... perhaps this last is actually the base to be accepted as starting position in our knowledge?<br /><br />most damning is that all that primary uncaused behaviour results in perfect causal behaviour on larger scales which would seem to betray its inital uncaused behaviour, true riddle if I've seen one<br /><br />riddles always tempted me and they make physics interesting no end, already I sense a way out of this one<br /><br />ps I am greatly dissapointed at the general non response to my last post which I thought was ingenious if I may say so myself<br /><br />vanDivX <div class="Discussion_UserSignature"> </div>

 

[vandivx]

ok, but you make big HOP direct to life and intelligent no less, I am sure it could use more expansion getting there<br /><br />I would be satisfied to work out for starters the chance and apparently uncaused statistical random happening with perfectly causal results on macro scale when it comes to ordinary matter<br /><br />I suppose the e=mc(sq) discussion stopped being interesting for people (I don't know what do I do that I so often kill threads) and so it doesn't matter if we go a bit off topic<br /><br />vanDivX <div class="Discussion_UserSignature"> </div>

 

[vidar]

Yes, the E= 42 mc^2 is a joke, <br />(- It is a good one too, - if I may add.)<br /><br />However, science is so far from being able to prove the big c^2 factor (ca 10^12) too even seriously consider whether Einstein’s “1.0†factor is correct.<br /><br />God is not a dice-throwing gambler. That would be a contradiction to claim. The fact that Bohr and Einstein didn’t understand QM, doesn’t change the nature of things, - only their imaginations. But I ,at any rate, am convinced He didn’t throw bombs either, - not even BigBang ones.<br /><br />I think Einstein is taken far too seriously. The equation and his person have become icons for scientific genius. To reveal such a sci-fi joke can’t be done without making whole cold-war science the same. Now it seems that the protection of the “1.0†factor has rather become emotional and religious, than truly scientific. That’s dangerous, - and that’s not a joke.<br /><br />Einstein’s factor is of cause not “42â€, (the symbol for the “Answer to Life, the Universe, and Everythingâ€).<br />- The Einstein factor is “y2†(Gamma2).<br />http://en.wikipedia.org/wiki/Special_relativity#Reference_frames.2C_coordinates_and_the_Lorentz_transformation<br />

 

[alokmohan]

Einstein is hardly matter of joke.,No good sense of humour.

 

[emperor_of_localgroup]

It may sound laughable, but VIDAR may be right. I rechecked the derivation of E=mc² in a physics book. I don't know if anyone else has already pointed this out, but the derivation clearly use Binomial Expansion, which is generally approximated by dropping all terms except the first 2 terms. So, no one really knows what is the exact multiplier with mc². It has to be larger than 1, but smaller than 2.<br /><br /><br />Unless someone can direct me to a different derivation without any approximation.<br /> <div class="Discussion_UserSignature"> <font size="2" color="#ff0000"><strong>Earth is Boring</strong></font> </div>

 

[origin]

It is truly arrogant that you guys think you have found something so simple yet fundemantally wrong with a bedrock of physics, that has been missed over the past 80 years.<br /><br />Yes, it is laughable.<br /><br /> <div class="Discussion_UserSignature"> </div>

 

[emperor_of_localgroup]

I can't believe you people take everything so seriously. No one is saying Eienstein is wrong, I'm saying if the derivation technique I have seen in a physics text is exactly what what is used in original derivation (which I doubt 100%), then the factor '1' with mc² is not exact but approximation. <br /><br />Any one familiar with Binomial expansion is aware that the higher terms are usually dropped because they are 'negligible', but not 'zero'. <br /><br />In case of E=mc², the binomial expansion is carried out on [1-(v/c)²]^-1/2. As you can see object speed v is generally much much smaller than c. In such case higher terms in the expansion will have very very small effect and the factor can be approximated as '1'. But I don't know what the factor would be if v is comparable with c. <br /><br />That's why I asked if there's any method without approximation. This binomial expansion method may have been introduced in textbooks for its simplicity so that stupids like me can understand where E=mc² has come from. <br /><br /><br /><br />Ok, I get to think for a while and I'll correct myself. Rest energy is m_oc², with a factor of exactly '1'. But kinetic energy, when in motion, gets screwed up because of extra terms of binomial expansion, it is not <br /> m_ov²/2 .<br /><br /> <div class="Discussion_UserSignature"> <font size="2" color="#ff0000"><strong>Earth is Boring</strong></font> </div>

 

[vandivx]

"Unless someone can direct me to a different derivation without any approximation."<br />---<br /><br />as I understand it, the full binomial series is the exact thing and if you take only subset you get approximation<br /><br />taking just the first term is ok when velocity is non relativistic<br /><br />vanDivX <div class="Discussion_UserSignature"> </div>

 

[vidar]

emperor_of_localgroup:<br />"Unless someone can direct me to a different derivation without any approximation." <br /><br />vanDivX<br />as I understand it, the full binomial series is the exact thing and if you take only subset you get approximation <br />taking just the first term is ok when velocity is non relativistic <br />--------------------<br /><br /><br />Here is the real approximation: <br />½ = 0.5 ≈ 1<br /><br />That makes:<br />E = ½ mc^2 ≈ 1 mc^2 = /> E = mc^2

 

[vidar]

I was thinking that the classic Newton energy equations could be used in a modern way to approximate Einstein’s total energy; Et. Newton said that Et = Ek + Ep => ½ mc^2 + mgh. I thought I maybe could make Ep about as large as Ek at c.<br /><br />Ep must be relative too, because it depends on the gravity it’s related to. Classical Newton relates to Earth, which gives his apple a potential energy of Ep ≈ 30*m (at 3 meters high).<br />However; what if an object is to be related to the Sun, rather than the Earth? Then the gravity would be 28 times g.<br />Furthermore, what if the object is related to our galaxy, the Milky Way? Then the gravity would be about 28*10^10 as great, - supposed its core has a gravity equal to the galaxy’s stars altogether.<br />Finally; the distance to the galaxy’s core is about 2.5×10^20 m.<br />(I stopped trying to imagine a final collapse of the universe to another bigbangboom. I find no reason to believe in that theory anyway.)<br /><br />Anyway; that makes the potential energy related to our galaxy about Ep ≈ m 10^31<br />That far exceeds Einstein’s E ≈ m 10^12 and his mysterious *2 factor.<br />Best of all, E = Ek + Ep annul the mathematical ban of real speed of light.<br />We might break the light barrier in the future, like the sound barrier was finally broken in 1961.<br />That’s the good news, not being stuck here forever.<br />

 

[origin]

<font color="yellow">Et = Ek + Ep => ½ mc^2 + mgh</font><br />No. Ek = 1/2 mV^2<br /><br /><font color="yellow">Ep must be relative too, because it depends on the gravity it’s related to. Classical Newton relates to Earth</font><br />No. It is not related to any place - it is the acceleration due to gravity. <br /><br /><font color="yellow">Furthermore, what if the object is related to our galaxy, the Milky Way? Then the gravity would be about 28*10^10 as great, - supposed its core has a gravity equal to the galaxy’s stars altogether. <br />Finally; the distance to the galaxy’s core is about 2.5×10^20 m. Anyway; that makes the potential energy related to our galaxy about Ep ≈ m 10^31</font><br />No, this is not correct. I will let you figure out why. Here is a hint - based on your logic if Newton's apple was on the other side of the universe it would have an astronomically high PE.<br /><br /><br /> <div class="Discussion_UserSignature"> </div>

 

[vidar]

Origin wrote:<br />No, this is not correct. I will let you figure out why. Here is a hint - based on your logic if Newton's apple was on the other side of the universe it would have an astronomically high PE.<br />--------------------------------------------------<br /><br />My point exactly, any object has a astronomical potential energy relative to other astronomical objects, not merely Earth. Newton’s apple doesn’t have to be ‘on the other side of the universe’ (whatever that is). Its here and has potential energy related to our galaxy's core as well. (Unless he ate it).<br /><br />There is no centre of the universe that pulls objects. Also bigbangers believe so, even though they believe there once was for the shortest thinkable moment of time. (That genius, by the way, to answer both ‘yes and no’, rather than answer ‘haven’t the faintest idea’, to the question about the beginning of time.)<br />

 

[origin]

<font color="yellow">My point exactly, any object has a astronomical potential energy relative to other astronomical objects, not merely Earth.</font><br /><br />OK, let's try again...<br /><br />The PE of a mass on earth increases as it's distance from earth increases. But only to a point, then the farther from the earth you get the lower the PE becomes. This is because the acceleration from gravity decreases by the square of the distance but the distance is of course linear. Do you see why your calculation of the PE realtive to the galactic center is wrong now?<br /><br /> <div class="Discussion_UserSignature"> </div>

 

[vidar]

I’ll try to explain again.<br /><br />A satelite in orbit over Earth has a potential energy at Ep=mgh. When it finally fall, the energy will be release by the impact with Earth, - no doubt.<br />Likewise, an asteriode orbiting the sun has a potential energy related to the distance and the gravity to the Sun.<br />Likewise, our Sun has a potential energy related to the gravity of the the Milky way’s Galactic Center and the distance to it. Any star that finally fall towards the Galactic Center will release that potential energy in the impact.<br /><br />Interesting enough, scientists now believe the Galactic Centers are massive black holes.<br />http://en.wikipedia.org/wiki/Galactic_Center<br />Consequently, Newton's apple had a potential energy related to the gravity and distance to our galaxy’s black hole (before he ate it).<br />