Misconceptions of E=MC^2

E=mc^2 has nothing to do with a mass travelling at the speed of light. Here, c is just a number, and c^2 forms the constant of proportionality between E and m. This number is also the speed of light, but that is (mostly) irrelevant to the equation.It is tested true every second of every day at every nuclear reactor in the world. None of the billion $ particle accelerators in the world would work at all if this equation was not true.There are very few equations in science that are better tested than this one...

No, it most certainly does not.Glad to clear up the misconception.

Absoluetly not. Pseudo-scientific bullshit perpetuated in popular science and all over the internet...

You know absolutely NOTHING about this subject, so why are you so intent on making a complete idiot of yourself?Take a look at this picture: {Source here}THAT is the creation of matter (electron/positron pair) from energy, perfectly obeying e=mc^2. This happens billions of times a day at the particle accelerators around the world. Why have we suddenly a forum full of arrogant idiots? You haven't a clue about this subject so why are you so sure that the problem lies with others and not with yourself?

No, mass is not a form of energy - mass is a measure of energy, irrespective of the form of the energy. The space-time curvature generated by a pair of photons is exactly the same after they have pair-created an eletcron and a positron. The energy will now be in the form of the rest-mass of the particles plus the interaction energy between them.The measured mass of 1kg of lead consists of almost entirely the binding energy (i.e. gluons) holding the quarks, and in turn, the nucleons together, plus the binding energy (i.e. photons) holding the electrons, and in turn, the atoms together, plus the vibrational energy (i.e. photons) of the temperature of the lead, and finally plus a vanishingly small amount of energy from the rest-mass of the quarks and eletrons making up the actual matter.If you put that lead into a container with 1kg of anti-lead, and managed perfect annihilation all the way down to just having photons left, AND managed to contain those photons within the container, then you would still have 2kg inside the container...Edited by cavediver, : No reason given.

The wisest words you have uttered all this thread. There is hope for you. Now apply that humility when you read my post above.

To quieten down and listen to what you are being told. For you to realise that you are very mistaken about all of this. Nearly everyone repsonding to you is reasonably well grounded in this subject. In my case, it was my profession for ten years. I'm not the one claiming insight in a field in which I have no experience...

So, just for the record Heinrik, you are contending that every particle factory in the world doesn't actually work, and every particle physicist in the world is lying, and hiding a huge conspiracy that we have never observed pair creation You are utterly deluded, and beyond help. Truly pathetic. I just hope that the hundreds of lurkers who have perused this thread have managed to see beyond your mad frothy ranting and raving and have actually perhaps learned something about particle physics.Bye bye

Arrogance is misplaced confidence. I can assure you that my confidence is very well placed

What you have to remember is that there is nothing special about nuclear reactions creating 'mass loss'. Chemical reactions do exactly the same. If we say Little Boy was 15kilotons, then a conventional explosion of 15,000 tons of TNT will also create a mass loss of around 0.6g.In both nuclear explosions and chemical explosuions, the energy released comes from the binding energy - binding energy of the nucleus and binding energy of the atoms/molecules. This binding energy has an assocaited mass via E=mc^2, and it is the loss of this energy that gives rise to the 'mass loss'. No actual matter was harmed in the course of either explosion

Yes, that is 'real' matter to non-matter. The rest mass of the electron and positron is released as energy which dictates the frequency of the two emitted photons.

If there's only two photons, and the the electron/positron were low energy (i.e low Kinetic Energy) then the photons will be 511KeV. E = h x v, where h is Plancks Constant and v is frequency (Greek nu actually), so a quick calc and you find that they are Gamma Ray photns with a freq of....

I think you've got your enegy units mixed up h is usually in Joule seconds.1 eV = 1.610^’19 J.

small v, so taylor expand so

Despite the appearance of that Wiki article, it's actually standard senior high school advanced calculus - though in these days of diminishing standards in the UK, it's out of single A-Level mathematics and only remains in one-and-a-half or double maths. I expect you did study it at one point... This wiki entry is better and provides the proof using basic integral calculus (work through it - it's not as bad as it looks)Essentially you are approximating a curve at a point by using a straight line with the same gradient at that point, plus a parabola with the same quadratic curvature at that point, plus a cubic, etcEdited by cavediver, : No reason given.Edited by cavediver, : No reason given.