Something incredible I worked out!:

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!

There's been a number of occasions where I've been trying to prove something and ended up with x = x...

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!

i did some calculations and i came up with

E/2=(mc^2)/2

It only took me like 8 seconds, 10 tops.
YouTube&t=40

You made a joke as well, right?

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?

Jeez, if you have to explain it....😉

Originally posted by @sonhouse
Jeez, if you have to explain it....😉

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.

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.

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 @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?

Wouldn't it be cubed given the relationship between radius and volume?

Originally posted by @deepthought
Wouldn't it be cubed given the relationship between radius and volume?

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!

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!

c^6 maybe.

Originally posted by @deepthought
c^6 maybe.

Exponents add so if it starts at (c^2) ^3 that would be c^5.

nope

a power 'to a power' you multiply

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

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

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

Entirely correct, but a little care is needed as the operation is not associative.

2^(3^4) = 2^81 = huge
(2^3)^4 = 8^4 = 64*64 = (2^6)^2 = 2^12 = 4096 = considerably less huge

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

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

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.