- 20 Jun '18 20:33

i did some calculations and i came up with*Originally posted by @sonhouse***After working out the Schwarzchild volume and many hours with pen and paper I discovered this amazing formula!**

E^2=m^2c^4 !!!!

I thought, WOW, Eureka!

E/2=(mc^2)/2

It only took me like 8 seconds, 10 tops.

YouTube&t=40

You made a joke as well, right? - 21 Jun '18 01:05 / 1 edit

Jeez, if you have to explain it....*Originally posted by @zahlanzi***i did some calculations and i came up with**

E/2=(mc^2)/2

It only took me like 8 seconds, 10 tops.

https://www.youtube.com/watch?v=L4aX0c6ffuI&t=40

You made a joke as well, right? - 21 Jun '18 06:33

You never know here.*Originally posted by @sonhouse***Jeez, if you have to explain it....**

Just now there are 100 posts in a thread arguing on the terence howard brain fart that sqrt(2) is a rational number.

You have to double-check who you're dealing with. - 21 Jun '18 16:01

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?*Originally posted by @zahlanzi***You never know here.**

Just now there are 100 posts in a thread arguing on the terence howard brain fart that sqrt(2) is a rational number.

You have to double-check who you're dealing with. - 22 Jun '18 14:13

Wouldn't it be cubed given the relationship between radius and volume?*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?** - 22 Jun '18 23:00 / 1 edit

Oh yeah, 4/3 PI R cubed. Dang. Screwed up my own joke*Originally posted by @deepthought***Wouldn't it be cubed given the relationship between radius and volume?**

So it would have to be E^3=M^3 C^5. Shoot! - 23 Jun '18 02:13

c^6 maybe.*Originally posted by @sonhouse***Oh yeah, 4/3 PI R cubed. Dang. Screwed up my own joke**

So it would have to be E^3=M^3 C^5. Shoot! - 23 Jun '18 16:32

Exponents add so if it starts at (c^2) ^3 that would be c^5.*Originally posted by @deepthought***c^6 maybe.** - 23 Jun '18 18:28

Yep, you right. May Blad.*Originally posted by @blood-on-the-tracks***nope**

a power 'to a power' you multiply

(c^2)^3 = c^2 x c^2 x c^2 (now add) = c^6 - 23 Jun '18 21:33

Entirely correct, but a little care is needed as the operation is not associative.*Originally posted by @blood-on-the-tracks***nope**

a power 'to a power' you multiply

(c^2)^3 = c^2 x c^2 x c^2 (now add) = c^6

2^(3^4) = 2^81 = huge

(2^3)^4 = 8^4 = 64*64 = (2^6)^2 = 2^12 = 4096 = considerably less huge - 23 Jun '18 21:44They only look as if they might be associative using the ^ symbol for 'to the power ' or, if you prefer, to indicate an index. With the notation we have to use here, they do look similar.

If they were written out using standard mathematical notation, it would be clear that they are quite different - 23 Jun '18 22:15

Well not really as 2^3^4 would be written as a superscript to a superscript. There is a convention for it, I think working from the right so 2^3^4 = 2^(3^4). But it's easy for me to have it the wrong way round. A similar problem exists for the vector product (AXB = |A||B|sin(angle)C where C is a unit vector perpendicular to the other two). I think one is meant to read left to right in that case, but it caused me no end of confusion as the non-associativity wasn't emphasized in my formal training.*Originally posted by @blood-on-the-tracks***They only look as if they might be associative using the ^ symbol for 'to the power ' or, if you prefer, to indicate an index. With the notation we have to use here, they do look similar.**

If they were written out using standard mathematical notation, it would be clear that they are quite different