4-vector dot A = invariant => A is a 4-vector?
in [Relativity]
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From: mpalenik on 17 Feb 2010 07:47 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. It only qualifies as a 4 vector if that scalar product is frame invariant. For example, P = (E, px, py, pz) is a 4 vector because P*P is invariant (E^2 - p^2 = m^2). No matter what frame you write (E, px, py, pz) in, you will always get P*P = m^2. X = (0, x,y,z) is not a 4 vector because X*X will not always come out to be the same number, depending on your choice of coordinates.
From: blackhead on 17 Feb 2010 08:52 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? > It only qualifies as a 4 vector if that scalar product is frame > invariant. > > For example, P = (E, px, py, pz) is a 4 vector because P*P is > invariant (E^2 - p^2 = m^2). No matter what frame you write (E, px, > py, pz) in, you will always get P*P = m^2. > > X = (0, x,y,z) is not a 4 vector because X*X will not always come out > to be the same number, depending on your choice of coordinates.- Hide quoted text - > > - Show quoted text -
From: mpalenik on 17 Feb 2010 10:02 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.
From: mpalenik on 17 Feb 2010 10:39 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 from this, we can see B' = L^-1B 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.
From: eric gisse on 17 Feb 2010 18:54
blackhead 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, tensorial quantities are defined through their transformative properties. [,,,] |