4-vector dot A = invariant => A is a 4-vector?
in [Relativity]
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From: Tom Roberts on 17 Feb 2010 20:04 blackhead wrote: > The scalar product of 2 4-vectors is an invariant. However, Page 530 > of Jackson's Electrodynamics makes the claim that because the phase of > a wave is an invariant and given by the scalar product of a 4 vector > with (w/c, K), then the latter is a 4 vector. > > Is this generally true? It depends on how "general" one wants to be. If one only considers 4-vectors in a given vector space with a dot product, then it is true. But if one considers 4-vector fields on a manifold, such as your example from Jackson, then considerably more care is required. I don't know the answer, but conjecture that the Poincar� lemma could be used to show this is true for any simply-connected manifold. Tom Roberts
From: blackhead on 18 Feb 2010 06:46 On 17 Feb, 15:39, mpalenik <markpale...(a)gmail.com> wrote: > On Feb 17, 10:02 am, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > > On Feb 17, 8:52 am, blackhead <larryhar...(a)softhome.net> wrote: > > > > On 17 Feb, 12:47, mpalenik <markpale...(a)gmail.com> wrote: > > > > > On Feb 17, 6:10 am, blackhead <larryhar...(a)softhome.net> wrote: > > > > > > On 16 Feb, 22:57, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > > On Feb 16, 5:32 pm, blackhead <larryhar...(a)softhome.net> wrote: > > > > > > > > The scalar product of 2 4-vectors is an invariant. However, Page 530 > > > > > > > of Jackson's Electrodynamics makes the claim that because the phase of > > > > > > > a wave is an invariant and given by the scalar product of a 4 vector > > > > > > > with (w/c, K), then the latter is a 4 vector. > > > > > > > > Is this generally true? > > > > > > > > . > > > > > > > Yes, it is. The inner product of two 4 vectors is a scalar, which > > > > > > should be invariant in any frame. Typically, you would show that A*A > > > > > > is invariant in any frame but it suffices to show that it's invariant > > > > > > when you take the product with another 4 vector. > > > > > > But if the scalar product of a 4 vector with 4 numbers is a scalar, > > > > > does that imply those 4 numbers are the components of a 4 vector? > > > > > You can arrange any 4 numbers into a vector and get a scalar when you > > > > take the scalar product with a 4 vector. That's why it's called a > > > > scalar product. > > > > Suppose these 4 numbers transform in a certain way under a coordinate > > > transformation, so that their scalar product with a 4-vector is a > > > scalar invariant. Must the 4 numbers transform as a 4-vector? > > > Yes, that's what I was trying to say in the first response. > > If you want a more complete answer, let's say we know A is a 4 vector > but we're unsure about B. > > in the first frame, we have A*B > > Now, let's transform into another frame, where we have A' and B' > > A transforms like a 4 vector, so we know that A' = LA > > So LA*B' = A*B = (L*L^-1)A*B Seems clear to me. > from this, we can see B' = L^-1B You've lost me. > In order to take a an inner product, if A is a vector, B must be a one- > form (or vector and co-vector, or covariant and contravariant vectors, > whatever terminology you use). The transformation rule for one-forms > is B' = (L^-1)B > > Therefore, B transforms like a one-form. So it is a 4-vector as well.- Hide quoted text - > > - Show quoted text -
From: mpalenik on 18 Feb 2010 16:08 On Feb 18, 6:46 am, blackhead <larryhar...(a)softhome.net> wrote: > On 17 Feb, 15:39, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > > On Feb 17, 10:02 am, mpalenik <markpale...(a)gmail.com> wrote: > > > > On Feb 17, 8:52 am, blackhead <larryhar...(a)softhome.net> wrote: > > > > > On 17 Feb, 12:47, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > On Feb 17, 6:10 am, blackhead <larryhar...(a)softhome.net> wrote: > > > > > > > On 16 Feb, 22:57, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > > > On Feb 16, 5:32 pm, blackhead <larryhar...(a)softhome.net> wrote: > > > > > > > > > The scalar product of 2 4-vectors is an invariant. However, Page 530 > > > > > > > > of Jackson's Electrodynamics makes the claim that because the phase of > > > > > > > > a wave is an invariant and given by the scalar product of a 4 vector > > > > > > > > with (w/c, K), then the latter is a 4 vector. > > > > > > > > > Is this generally true? > > > > > > > > > . > > > > > > > > Yes, it is. The inner product of two 4 vectors is a scalar, which > > > > > > > should be invariant in any frame. Typically, you would show that A*A > > > > > > > is invariant in any frame but it suffices to show that it's invariant > > > > > > > when you take the product with another 4 vector. > > > > > > > But if the scalar product of a 4 vector with 4 numbers is a scalar, > > > > > > does that imply those 4 numbers are the components of a 4 vector? > > > > > > You can arrange any 4 numbers into a vector and get a scalar when you > > > > > take the scalar product with a 4 vector. That's why it's called a > > > > > scalar product. > > > > > Suppose these 4 numbers transform in a certain way under a coordinate > > > > transformation, so that their scalar product with a 4-vector is a > > > > scalar invariant. Must the 4 numbers transform as a 4-vector? > > > > Yes, that's what I was trying to say in the first response. > > > If you want a more complete answer, let's say we know A is a 4 vector > > but we're unsure about B. > > > in the first frame, we have A*B > > > Now, let's transform into another frame, where we have A' and B' > > > A transforms like a 4 vector, so we know that A' = LA > > > So LA*B' = A*B = (L*L^-1)A*B > > Seems clear to me. > > > from this, we can see B' = L^-1B > > You've lost me. > > > > > In order to take a an inner product, if A is a vector, B must be a one- > > form (or vector and co-vector, or covariant and contravariant vectors, > > whatever terminology you use). The transformation rule for one-forms > > is B' = (L^-1)B > > > Therefore, B transforms like a one-form. So it is a 4-vector as well..- Hide quoted text - > > > - Show quoted text -- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text - Well, since A' = LA and A*B = LA*B', then LA*B' = (L*L^-1)*A*B = (L^-1)A'*B = A'*B'so B' = (L^-1)B
From: Dono. on 18 Feb 2010 18:23 On Feb 18, 1:08 pm, mpalenik <markpale...(a)gmail.com> wrote: > > Well, since A' = LA and A*B = LA*B', then LA*B' = (L*L^-1)*A*B =(L^-1)A'*B = A'*B'so B' = (L^-1)B- Hide quoted text - > How do you get (L^-1)A'*B = A'*B'? From your earlier: LA*B' = A*B = (L*L^-1)A*B the best that you can get is: A*B'=L^-1(L*L^-1)A*B =L^-1A*B so: B'=A^-1L^-1AB=(LA)^-1AB
From: mpalenik on 19 Feb 2010 02:20
On Feb 18, 6:23 pm, "Dono." <sa...(a)comcast.net> wrote: > On Feb 18, 1:08 pm, mpalenik <markpale...(a)gmail.com> wrote: > > > > > Well, since A' = LA and A*B = LA*B', then LA*B' = (L*L^-1)*A*B =(L^-1)A'*B = A'*B'so B' = (L^-1)B- Hide quoted text - > > How do you get (L^-1)A'*B = A'*B'? > Just by the definition we had before that A*B = A'*B' So, since we established that A*B = (L^-1)A'*B this also means that (L^-1)A'*B = A'*B'. |