Something incredible I worked out!:

Originally posted by @sonhouse
Didn't you see the oblique reference to the Shwartzchild radius? I upped it mathically to the Shwartzchild volume to account for the squaring of the Einstein formula?

But E^2=m^2c^4 is Einstein's formula.

The common E=mc^2 assumes that Energy must be positive.

Originally posted by @wolfgang59
But E^2=m^2c^4 is Einstein's formula.

The common E=mc^2 assumes that Energy must be positive.

Well, ignoring factors of the speed of light, the full equation is:

E^2 - p^2 = m^2

Which is just the genralization of Pythagoras' theorem to Minkowski Space applied to the energy momentum vector. It simply states that the length of the energy momentum vector is the mass. In the comoving frame the momentum is zero and we have E = m or E = -m. However, since the energy is just the length of a vector, it seems odd to assign it a negative value.

Consider a vector on the Euclidean plane, with components x and y. We have that:

L^2 = x^2 + y^2

Rotating our axes so that y=0 we get:

L = x or L = -x

Since negative length makes no sense it's normal to drop whichever one of these is negative.

Whether negative energy or mass exists is one of those empirical problems that can't really be addressed theoretically.

Originally posted by @deepthought
Well, ignoring factors of the speed of light, the full equation is:

E^2 - p^2 = m^2

Which is just the genralization of Pythagoras' theorem to Minkowski Space applied to the energy momentum vector. It simply states that the length of the energy momentum vector is the mass. In the comoving frame the momentum is zero and we have E = m or E = -m. H ...[text shortened]... or mass exists is one of those empirical problems that can't really be addressed theoretically.

What is 'p'? It is c in big Al's formula, is it the same here and if so why is it now 'p'? Ah, ignoring c. Ok, so what is p then?

Originally posted by @sonhouse
What is 'p'? It is c in big Al's formula, is it the same here and if so why is it now 'p'? Ah, ignoring c. Ok, so what is p then?

Sorry, I forgot people might not recognize that, p is the conventional symbol for momentum.

Originally posted by @deepthought
Sorry, I forgot people might not recognize that, p is the conventional symbol for momentum.

Ah, thanks. How do they eliminate velocity? I thought momentum and kinetic energy were dependent partially on velocity. Well, looking at the formula then, energy squared minus momentum squared equals mass squared.
What are the units involved? Mass, obviously, kilograms, energy? kinetic energy? How is that noted and also how is momentum notated?

Also, can't it be simplified to E-P=M? Why do all three factors have to be squared?

E^2 - P^2 = M^2. Implies E - P = M?

So 5 - 3 = 4, after all 25 - 9 DOES = 16

Originally posted by @sonhouse
Ah, thanks. How do they eliminate velocity? I thought momentum and kinetic energy were dependent partially on velocity. Well, looking at the formula then, energy squared minus momentum squared equals mass squared.
What are the units involved? Mass, obviously, kilograms, energy? kinetic energy? How is that noted and also how is momentum notated?

Also, can't it be simplified to E-P=M? Why do all three factors have to be squared?

In classical physics, momentum is equal to mass times velocity. In relativity, the 4-momentum (the 4-vector with time like component equal to the energy and space like components equal to the momentum) is the (rest) mass times the 4-velocity. Let y = 1/sqrt(1 - v^2/c^2) be the gamma factor then the 4-velocity is just (y, yv) where v is the classical 3-velocity. So it's still true that momentum is mass times velocity, it's just you need to be a little careful about what is meant by velocity.

The units are natural units, chosen so that the speed of light, Plank's constant, and Bolzmann's constant are equal to 1. These quantities act as constants of proportionality and it tidies up a lot of equations if one sets them equal to 1 and drops them.

As Blood noted, E^2 - p^2 = (E - p)(E + p) = E - p if and only if (E + p) = 1, which in general, isn't going to be the case.

Originally posted by @deepthought
In classical physics, momentum is equal to mass times velocity. In relativity, the 4-momentum (the 4-vector with time like component equal to the energy and space like components equal to the momentum) is the (rest) mass times the 4-velocity. Let y = 1/sqrt(1 - v^2/c^2) be the gamma factor then the 4-velocity is just (y, yv) where v is the classical 3 ...[text shortened]... (E - p)(E + p) = E - p if and only if (E + p) = 1, which in general, isn't going to be the case.

What do you mean by 'natural units'?

Originally posted by @sonhouse
What do you mean by 'natural units'?

Units where the speed of light is 1, Boltzmann's constant is 1, and Plank's constant is 1.

Originally posted by @deepthought
Units where the speed of light is 1, Boltzmann's constant is 1, and Plank's constant is 1.

I can see that simplifies analysis, are all those units 1 at the same time for the calcs?

Originally posted by @sonhouse
I can see that simplifies analysis, are all those units 1 at the same time for the calcs?

Yes. Physical laws consist of a collection of equations with some constants in them. Here we have:

E = mc^2
E = hf
E = kT

Given some unit of energy, choosing the units for mass, frequency and temperature fixes those constants of proportionality. We can turn this on its head and instead select values for the constants, which then forces a set of units on us. The natural units are just those where as many constants as possible have been set to 1.

Note that it's the reduced Plank's constant that is set to 1. Writing h slash as a single symbol is possible (there's a unicode character), but a monstrous pain (I have to find out what the code is). So we'd set Plank's constant as presented here to 1/2π.

Originally posted by @deepthought
Yes. Physical laws consist of a collection of equations with some constants in them. Here we have:

E = mc^2
E = hf
E = kT

Given some unit of energy, choosing the units for mass, frequency and temperature fixes those constants of proportionality. We can turn this on its head and instead select values for the constants, which then forces a set ...[text shortened]... (I have to find out what the code is). So we'd set Plank's constant as presented here to 1/2π.

Wow, that is frigging sneaky🙂 Plank just SET to half Pi!

Originally posted by @sonhouse
Wow, that is frigging sneaky🙂 Plank just SET to half Pi!

That's 1/( 2π ). But yes.

Originally posted by @deepthought
That's 1/( 2π ). But yes.

After all the mathy manipulations and you have gotten what you want, do you then have to un-unitize those constants to get real world results?

Originally posted by @sonhouse
After all the mathy manipulations and you have gotten what you want, do you then have to un-unitize those constants to get real world results?

Yes, but that's a problem for the experimentalists.