18 Jun '18 08:06

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=m^2c^4 !!!!

I thought, WOW, Eureka!

- Joined
- 28 Dec '04
- Moves
- 52681

slatington, pa, usa- Joined
- 27 Oct '04
- Moves
- 78904

Cosmopolis- Joined
- 04 Feb '05
- Moves
- 29132

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?- Joined
- 28 Dec '04
- Moves
- 52681

slatington, pa, usa21 Jun '18 01:051 edit

Jeez, if you have to explain it....ðŸ˜‰*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?- Joined
- 04 Feb '05
- Moves
- 29132

21 Jun '18 06:33

You never know here. ðŸ™‚*Originally posted by @sonhouse***Jeez, if you have to explain it....ðŸ˜‰**

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.- Joined
- 28 Dec '04
- Moves
- 52681

slatington, pa, usa21 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. ðŸ™‚**

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.- Joined
- 27 Oct '04
- Moves
- 78904

Cosmopolis22 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?**- Joined
- 28 Dec '04
- Moves
- 52681

slatington, pa, usa22 Jun '18 23:001 edit

Oh yeah, 4/3 PI R cubed. Dang. Screwed up my own jokeðŸ™‚*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!- Joined
- 27 Oct '04
- Moves
- 78904

Cosmopolis23 Jun '18 02:13

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

So it would have to be E^3=M^3 C^5. Shoot!- Joined
- 28 Dec '04
- Moves
- 52681

slatington, pa, usa23 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.**- Joined
- 11 Nov '14
- Moves
- 16719

- Joined
- 28 Dec '04
- Moves
- 52681

slatington, pa, usa23 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- Joined
- 27 Oct '04
- Moves
- 78904

Cosmopolis23 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- Joined
- 11 Nov '14
- Moves
- 16719

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- Joined
- 27 Oct '04
- Moves
- 78904

Cosmopolis23 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