Can somebody explain this algebra equation please?

Basically, in special relativity, everyone agrees that the speed of light is c -- every reference frame will see a supernova (a sort of star explosion) as an expanding bubble of light, moving at speed c. This is a /very/ strange statement -- that you on a train don't see the bubble "flatten" but see it become a new bubble. The fact that everyone agrees on the speed of light allows us to describe their time t with a spatial coordinate s = c t.

Now, we can work out the exact coordinate transformation which must be occurring between coordinates (s, x, y, z)? (s', x', y', z') to allow this to happen. This is called the /Lorentz transformation/, and it goes like this, assuming a relative speed v along the x-axis: Let?

= v / c Let? = 1/?(1? ²) s' =?

· (s? X) x' =? · (x?

S) y' = y z' = z We thus have a way that the "position 4-vector" (s, x, y, z) transforms across different coordinates. We define a 4-vector as any vector which transforms like the position 4-vector. Thus, everyone agrees on the 4-vectors, even though they might *disagree* on the exact coordinates.

This is the same way normal vectors work: I will point an arrow in the downwards direction, but not all coordinate systems will agree with me that my "down" is their coordinates' "down". Still, we all agree on the /vector/, we just disagree on its /direction/. Now, you see, we want a "momentum 4-vector" -- something which every coordinate system will agree is the 4D momentum of a particle, no matter how it is traveling.

To do this, we explicitly specify the time coordinate? Of a particle at rest in the primed frame: R = (s', x') = (c? , 0) in the primed frame.

This time coordinate? Is called the "proper time" and is something which every other reference frame can calculate, by using a Lorentz transform to relate s to s'. Thus, a derivative with respect to it is also objective, and our traditional definition of momentum "m v" gives us: P = m dR/d?

= (m c, 0) in the primed frame. Transforming with a Lorentz transform to the unprimed frame, we can find that: P = (? m c,? M v) unprimed.

Mass-energy equivalence. Mass is evidence of energy...all forms of mass, all forms of energy. Consider a nuclear reaction in a water bath, all well insulated.

The nuclear reaction progresses and heats up the water bath. Yes particles are destroyed, but no mass is destroyed. If no energy escapes the system, then no mass escapes the system.

Yes the nuclear vessel dropped in mass, but the heated water bath gained mass due to gaining energy (but not particles). It isn't obvious for our common observation. You'd need a balance accurate to the nanogram to detect the mass difference when 1 kg of ice melts to water.

I cant really gove you an answer,but what I can give you is a way to a solution, that is you have to find the anglde that you relate to or peaks your interest. A good paper is one that people get drawn into because it reaches them ln some way.As for me WW11 to me, I think of the holocaust and the effect it had on the survivors, their families and those who stood by and did nothing until it was too late.