How does imploding light relate to Einsteins Theory of Relativity (in laymans terms)?

Question

I'm researching hyper-luminary space travel, black holes, gravitational collapse, imploding light, E=mc2, etc., and I don't quite understand most of it, as all refers to the Special Theory of Relativity (E=mc2) which I don't fully comprehend -- mostly because When i look it up in dictionary or encyclopedia (on-line or otherwise) it uses os many mis-understood physics terms and descriptions that I cannott grasp the simplicity of it -- I get confused and need a simple explanation so I grasp it conceptually in my mind.

Can you help me to understand this?

Answer 1

When you don't understand a subject, it's easy to get confused by terms that really don't belong in the subject, but have been misused by others that don't understand it either. "Imploding light" is such a term that is non-existent in serious papers on relativity. Light does not "implode". A light bulb that has a vacuum "implode" because, when shattered, the bulb collapses catastrophically because of the vacuum. Likewise, failed stars can simliarly "implode" under tremendous and sudden gravitational pull, and even superdense neutron stars can "implode" into a black hole. Once a black hole forms, light that is passing by too closely can be pulled into it and never come out again. This might be what you're thinking about. Even though light photons have no mass, in relativity theory, gravity can nevertheless exert a pull on it, and bend light rays.

There's actually 2 levels of relativity theory, the first called "Special Relativity", which is the simpler one, and the 2nd called "General Relativity", which is by far the harder, more complex one. While the equation E=mc² was derived in Special Relativity, things like "hyper-luminal space travel", "black holes", "gravitational collapse" are all things that can only be understood and treated in General Relativity, as all of them involve curved spacetime, a subject that Special Relativity does not cover. Einstein first developed and published Special Relativity after just a few years, but took him far longer to develop and publish General Relativity afterwards, which is really a generalization of Special Relativity. Here's a quick summary of the assumptions made for each as a starting point:

Special Relativity: 1) All the laws of physics are the same for all uniformly moving frames of reference 2) The speed of light is always the same for all observers in all moving frames of reference

General Relativity 1) It's not possible to tell if you're at rest or in a gravitational free fall, if you can't see outside your spacecraft 2) All inertial motion is a spacetime geodestic

Unfortunately, all relativity theory is mind-bending stuff, and defies intuitive understanding, as it seems to lead to many paradoxical situations, which is one of the reasons why it was much resisted throughout the 20th century, particularly by laymen. I wish I could offer you a "simple explanation" so that you can "grasp it conceptually", but I can't. What's important is understanding the mathematical implications of the basic premises that I have outlined above for relativity theory, both special and general.

Answer 2

In physics, E = mc2 is the equation that expresses an equivalence between energy (E) and mass (m), in direct proportion to the square of the speed of light in a vacuum (c2). This formula proposes that when a body has a mass, it has a certain energy equivalence, even "at rest". This is opposed to the Newtonian mechanics, in which a massive body at rest has no kinetic energy, and may or may not have other (relatively small) amounts of internal stored energy (such as chemical energy or thermal energy), in addition to any potential energy it may have from its position in a field of force. That is why a body's rest mass, in Einstein's theory, is often called the rest energy of the body. The E of the formula can be seen as the total energy of the body, which is proportional to the mass of the body. Conversely, a single photon travelling in empty space cannot be considered to have an effective mass, m, according to the above equation. The reason is that such a photon cannot be measured in any way to be at "rest" and the formula above applies only to single particles when they are at rest, and also systems at rest (i.e., systems when seen from their center of mass frame). Individual photons are generally considered to be "massless," (that is, they have no rest mass or invariant mass) even though they have varying amounts of energy and relativistic mass. Systems of two or more photons moving in different directions (as for example from an electron-positron annihilation) will have an invariant mass, and the above equation will then apply to them, as a system, if the invariant mass is used. This formula also gives the quantitative relation of the quantity of mass lost from a resting body or a resting system (a system with no net momentum, where invariant mass and relativistic mass are equal), when energy is removed from it, such as in a chemical or a nuclear reaction where heat and light are removed. Then this E could be seen as the energy released or removed, corresponding with a certain amount of relativistic or invariant mass m which is lost, and which corresponds with the removed heat or light. In those cases, the energy released and removed is equal in quantity to the mass lost, times the speed of light squared. Similarly, when energy of any kind is added to a resting body, the increase in the resting mass of the body will be the energy added, divided by the speed of light squared.

Answer 3

Energy = Matter (speed of light in a vacume)^2
// so the number will be very very large, which is the energy. This means, take a little bit of matter and you will get an aweful lot of energy from it.

Answer 4

Read "The Elegant Universe" by Brian Greene.