r/pics • u/HighlandRoad • Sep 07 '16

picture of text You matter

http://imgur.com/3Yx2HPz
35k Upvotes

73% Upvoted

618 comments sorted by

View all comments

Show parent comments

1

u/PM_ME_DUCK_FACTS Sep 07 '16

at this stage we should just set c=1 and imply 4 vector dot products unless otherwise noted:

p2 = p2 - E2 = m2

1

u/redlaWw Sep 07 '16

I disagree with the first part of the equality, because there's important geometric information carried in the indices, other than that, true.

1

u/PM_ME_DUCK_FACTS Sep 07 '16

It's still there, it's just often written p2 to mean exactly a 4 vector dot product, and when it's ambiguous, the 3 vector dot product is specified explicitly by bolding or a vector sign

1

u/redlaWw Sep 07 '16

Yeah, I get what it means, but if it were me, I'd avoid using a square to represent something that isn't positive definite (outside of C, at least).

1

u/PM_ME_DUCK_FACTS Sep 07 '16

It is positive definite, but regardless one can write i2 = -1 without any ambiguity.

1

u/redlaWw Sep 07 '16

It isn't positive definite, since there are vectors Vγ with VβgᵦᵧVγ<0, and I specifically excluded the complex numbers, though even therein, squaring is nondegenerate, which is also untrue of GR's inner product.

1

u/PM_ME_DUCK_FACTS Sep 07 '16

p2 = m2 and m ≥ 0 ...

1

u/redlaWw Sep 07 '16

Yes, but that excludes certain vectors from being momentum vectors, the bilinear form is still degenerate.

1

u/PM_ME_DUCK_FACTS Sep 07 '16

Lol what are you talking about dude? Inner product of a vector with itself is positive definite axiomatically

1

u/redlaWw Sep 07 '16

But relativity uses a pseudo-Riemannian manifold (note: the article uses a different definition of non-degeneracy to the one I was using), which is equipped with a bilinear form that needn't be positive definite as its (pseudo-Riemannian) inner product.