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c2 = E/mWhere E = energy, m = mass, c = speed of light, d = distance, t = timeWe can see from this that as m goes toward 0 E goes towards ∞Then the speed of light goes toward ∞
Mike I got this farQuotec2 = E/mWhere E = energy, m = mass, c = speed of light, d = distance, t = timeWe can see from this that as m goes toward 0 E goes towards ∞Then the speed of light goes toward ∞Do the simple sums of E=mc^2 for a mass of 1kg, .1kg and .01kg and you will quickly see that as mass gets lower then Energy gets lower.Simplistically we might say that in the frame of the centre of mass an object of 1kg has an energy of 9*10^16 joules, an object of .1 kg --> 9*10^15 joules, .01kg --> 9*10^14 joules. You will note that sqrt[(9*10^16)/1] = +/- 3*10^8m/ssqrt[(9*10^15)/.1] = +/- 3*10^8m/ssqrt[(9*10^14)/.01] = +/- 3*10^8m/sFeel free to continue this making your mass smaller and smaller - the energy will get smaller and smaller and the +ve square root of the ratio will continue to be the speed of light.For your guidance for your claim to be true you need an equations as follows xy=k, if k is a constant then as x tends to get very small then y tends to get very big and vice versa. BTW handling infinity, dividing through by infinities and presuming that division by zero is allowable is very dangerous in both maths and physics and if not handled with great care will often lead to erroneous results. As the first mathematically claim is simply wrong I am not going to continue reading the post.
But Mike - the energy of the entire universe is nothing like E=mc^2 - that's mass energy equivalence of a particle in the frame of reference of the centre of mass. If you start adding kinetic energy, potential energy then perhaps you might get a whole different kettle of fish. You cannot even account for any radiation in that formula. Wikipedia has a very nice page that will start to guide you through what the equation E=mc^2 encompasses - your interpretation is entirely incorrect.