3 ms·

But why mv^2 ? Is the universe 2D? And it's E=mc^2 because the only 'consensus' value in the universe is c? Why is that? And so the 'mass' (whatever that is)

But why mv^2 ? Is the universe 2D? And it's E=mc^2 because the only 'consensus' value in the universe is c? Why is that? And so the 'mass' (whatever that is) must be moving at the speed of light for the equation to make sense, even though it's stationary? The blogs demonstrate great factual knowledge and 'mastery', but don't really explain anything IMHO

prerok 2y ago
Well, depends on your viewpoint, I guess. When you really get into the details, it turns out we don't "really understand" anything. We are just making more and more detailed observations and then creating mathematical models of these behaviors. For example, we observe that space is curved around mass. We can model that and it helps us understand what's going on, so it's useful. We don't, however, understand what exactly is curved and what is this empty space that curves.
af3d 2y ago
If you break up "mv^2" into its constituent dimensions you get m(d/t)^2 = m(d^2)(t^-2). Now the so-called kinetic energy of the object only "manifests itself" whenever there is a change in velocity of the object in question. Well, the derivative of velocity is an acceleration, so the object in acceleration would be represented as mdt^-2, aka "a force". Hence the energy of the system is simply that force acting over some distance d. As to the internal/intrinsic energy of a given object, think of it as "hidden potential energy". It is essentially the energy that was required to turn photons into the matter that you, and I, and everything else are made of! The equation itself is mc^2 simply because that is what you get when you rearrange and simplify the experimentally-verified equations which it was drawn from. Likewise, for c is nothing more than the measured value of the speed of light in vacuum for any observer. Of course the choice of units is completely arbitrary. Whether you state it in miles per hour, kilometers per second, or whatever, the ratio remains constant.
pharrington 2y ago
The mv^2 isn't a geometrical property - it's because the kinetic energy can also be thought of as the integral of an object's momentum with respect to time.

prerok 2y ago

Well, depends on your viewpoint, I guess. When you really get into the details, it turns out we don't "really understand" anything. We are just making more and more detailed observations and then creating mathematical models of these behaviors. For example, we observe that space is curved around mass. We can model that and it helps us understand what's going on, so it's useful. We don't, however, understand what exactly is curved and what is this empty space that curves.

af3d 2y ago

If you break up "mv^2" into its constituent dimensions you get m(d/t)^2 = m(d^2)(t^-2). Now the so-called kinetic energy of the object only "manifests itself" whenever there is a change in velocity of the object in question. Well, the derivative of velocity is an acceleration, so the object in acceleration would be represented as mdt^-2, aka "a force". Hence the energy of the system is simply that force acting over some distance d. As to the internal/intrinsic energy of a given object, think of it as "hidden potential energy". It is essentially the energy that was required to turn photons into the matter that you, and I, and everything else are made of! The equation itself is mc^2 simply because that is what you get when you rearrange and simplify the experimentally-verified equations which it was drawn from. Likewise, for c is nothing more than the measured value of the speed of light in vacuum for any observer. Of course the choice of units is completely arbitrary. Whether you state it in miles per hour, kilometers per second, or whatever, the ratio remains constant.

pharrington 2y ago

The mv^2 isn't a geometrical property - it's because the kinetic energy can also be thought of as the integral of an object's momentum with respect to time.