4-vector dot A = invariant => A is a 4-vector?
in [Relativity]
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From: blackhead on 16 Feb 2010 17:32 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? ..
From: Androcles on 16 Feb 2010 17:40 "blackhead" <larryharson(a)softhome.net> wrote in message news:c464c9f2-44cd-46b1-a169-aeaeb4b0c757(a)z19g2000yqk.googlegroups.com... > 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? No.
From: mpalenik on 16 Feb 2010 17:57 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.
From: blackhead on 17 Feb 2010 06:10 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?
From: Androcles on 17 Feb 2010 06:20
"blackhead" <larryharson(a)softhome.net> wrote in message news:ece445ab-a1be-4cef-97a2-d10b94fe3e58(a)d27g2000yqf.googlegroups.com... 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? ============================================== Every component of a vector is itself a vector. http://mathworld.wolfram.com/VectorSpace.html |