4-vector dot A = invariant => A is a 4-vector?
in [Relativity]
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From: mpalenik on 25 Feb 2010 00:32 BTW, I only skimmed your post--but thanks for the extensive, yet totally unnecessarry, description of Jacobian matrices. My guess is that Jackson E&M is meant to be used in the context of flat, Minkowski spacetime, and the Jacobian matrices that you deal with are the standard Lorentz boost matrices. And once again, to rehash, you don't need to repeatedly measure B' in every possible frame, as the problem *GIVES* you the information that A*B is invariant and it *GIVES* you the information that A is a 4- vector.
From: Schoenfeld on 25 Feb 2010 01:24 On Feb 25, 3:26 pm, mpalenik <markpale...(a)gmail.com> wrote: > On Feb 24, 11:40 pm, Schoenfeld <schoenfeld.fore...(a)gmail.com> wrote: > > > > > > > On Feb 25, 9:45 am, mpalenik <markpale...(a)gmail.com> wrote: > > > > On Feb 24, 12:18 am,Schoenfeld<schoenfeld.fore...(a)gmail.com> wrote: > > > > > On Feb 17, 8:32 am, 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? > > > > > It's true in generalized geometry as the inner product of two tensors > > > > results in a decrement of the covariant and contravariant rank. So the > > > > inner product of two rank-1 tensors (vector) results in a rank-0 > > > > tensor (invariant scalar). > > > > > The generalized inner product of two vectors U and V is > > > > > UV = g_ij U^i V^j > > > > = U^i V_i = U_i V^i = invariant > > > > > The angle between the two vectors is > > > > cos(theta) = UV/|U||V| > > > > > The existence of theta follows from the Cauchy-Schwarz inequality > > > > -1 <= UV/ (|U||V|) <= +1 > > > > > It's true in geometric algebra as well, as the the dot product on an N- > > > > vector with an M-vector is an |N-M| vector. Vectors are > > > > specializations of a multivectors (or blades). So generally speaking, > > > > the dot product of an R-blade A_R with an S-blade B_S is defined as > > > > > A_R . B_S = (AB)_(|R-S|) > > > > > where (AB) refers to the geometric product. > > > > Although, he was kind of asking the opposite question. If the inner > > > product of a 4 vector and another set of 4 numbers makes an invariant > > > scalar, does that mean that the second set of numbers is also a 4- > > > vector. The fact that that's true isn't obvious from what you've > > > written. > > > > The second set of 4 numbers isn't obtained by using a coordinate > > > transformation, it's obtained by making a series of frame dependent > > > measurements. Since that's the case, it's not obvious that A*B = > > > A'*B', where A is a 4-vector automatically implies that B is also a 4- > > > vector, although you can prove that this is the case.- Hide quoted text - > > > You can never "prove" that B is a "vector" as it is a "set of > > measurements". > > Maybe I'm not being clear. Not only are you not clear, your interpretation of the OPs question was unclear. I've responded comprehensively and conclusively. [...]
From: Schoenfeld on 25 Feb 2010 02:57 On Feb 25, 3:32 pm, mpalenik <markpale...(a)gmail.com> wrote: > BTW, I only skimmed your post--but thanks for the extensive, yet > totally unnecessarry, description of Jacobian matrices. My guess is > that Jackson E&M is meant to be used in the context of flat, Minkowski > spacetime, and the Jacobian matrices that you deal with are the > standard Lorentz boost matrices. I didn't reference any specific geometry or coordinate transformations at all. > And once again, to rehash, you don't need to repeatedly measure B' in > every possible frame, as the problem *GIVES* you the information that > A*B is invariant and it *GIVES* you the information that A is a 4- > vector. [1] If your question is: "If A . B = invariant and A is a vector, prove B a vector" Proof: The dot product is DEFINED for vectors so B is necessarily a vector. [2] If your question is: "If A_x B_x + A_y B_y + ... = invariant and A is a vector, prove the set of B forms a vector" Proof: - Overview: I define "invariant" as a "rank-0 tensor" so A and B are necessarily invariant vectors (i.e. either contravariant or covariant rank-1 tensors). [2.1] Grab any rank-1,1 tensor M^u_v (i.e. a mixed variance square matrix) [2.2] Decompose the M tensor (matrix) into a OUTER PRODUCT of two tensors A,B (vectors) M^u_v = A^u B_v Note: A^u is a CHOSEN contravariant vector. B_v is a CHOSEN covariant vector. [2.3] Contract the tensor M M^u_u = A^u B_u = invariant scalar which proves an invariant scalar can be expressed as either a contraction of a mixed variance square matrix or the tensor INNER product of a covariant/contravariant vector pair. [2.4] Observe that A^u B_u = A_u B^u A^u B_u = M^u_u (g_iu A^u)(g^iu B_u) = g_iu g^iu M^u_v A_i B^i = KroneckaDelta^i_i * M^i_i = M^i_i [3] THINK ABOUT WHAT YOU ARE SAYING The statement A.B = INVARIANT means A.B = A'.B' which necessarily means B is a vector as it can be transformed to B'. [4] If you WEAKENED your RHS to A . B = (SCALAR NOT NECESSARILY INVARIANT) then B is still a vector but not contravariant or covariant. [5] EVERY SCALAR CAN BE EXPRESSED AS THE DOT PRODUCT OF TWO VECTORS SCALAR = SCALAR = SCALAR1 + SCALAR2 + SCALAR3 + ... = SCALAR1*SCALARA + SCALAR2*SCALARB + SCALAR3*SCALARC + .... = SCALAR1*SCALARA e0.e0 + SCALAR2*SCALARB e1.e1 + ... = (SCALAR1*e0).(SCALARA*e0) + (SCALAR2*e1).(SCALARB*e1) + ... = (SCALAR1 e0 + SCALAR2 e1 + ...) . (SCALARA e0 + SCALARB e1 + ...) = VECTOR . VECTOR But this does not necessarily mean your LHS is invariant or your RHS vectors are invariant.
From: mpalenik on 25 Feb 2010 08:28 On Feb 25, 2:57 am, Schoenfeld <schoenfeld.fore...(a)gmail.com> wrote: > On Feb 25, 3:32 pm, mpalenik <markpale...(a)gmail.com> wrote: > > > BTW, I only skimmed your post--but thanks for the extensive, yet > > totally unnecessarry, description of Jacobian matrices. My guess is > > that Jackson E&M is meant to be used in the context of flat, Minkowski > > spacetime, and the Jacobian matrices that you deal with are the > > standard Lorentz boost matrices. > > I didn't reference any specific geometry or coordinate transformations > at all. > > > And once again, to rehash, you don't need to repeatedly measure B' in > > every possible frame, as the problem *GIVES* you the information that > > A*B is invariant and it *GIVES* you the information that A is a 4- > > vector. > > [1] If your question is: > "If A . B = invariant and A is a vector, prove B a vector" > > Proof: The dot product is DEFINED for vectors so B is necessarily a > vector. > > [2] If your question is: > > "If A_x B_x + A_y B_y + ... = invariant and A is a vector, prove the > set of B forms a vector" Yes, this is the OP's question, not my question. I wasn't asking a question. > > Proof: > - Overview: I define "invariant" as a "rank-0 tensor" so A and B are > necessarily invariant vectors (i.e. either contravariant or covariant > rank-1 tensors). Yes, I don't disagree with this. My point is, it's not obvious from this statement that the only way to get an invariant is by taking the dot product of two vectors. I agree that this is the definition of a scalar, but answering with the definition of a scalar as a rank zero tensor doesn't make it clear WHY B must necessarily be a vector. Listen to me--I do not doubt what you're saying--if you read my answer to the OP, you'll see that I agreed with this. I just disagree that you're explanation is very helpful to his original question. Jackson uses the invariance of A*B to prove that B is a 4-vector. He was asking if this is always valid or just in the one specific case. > > [2.1] Grab any rank-1,1 tensor M^u_v (i.e. a mixed variance square > matrix) > > [2.2] Decompose the M tensor (matrix) into a OUTER PRODUCT of two > tensors A,B (vectors) > > M^u_v = A^u B_v > > Note: > A^u is a CHOSEN contravariant vector. > B_v is a CHOSEN covariant vector. > > [2.3] Contract the tensor M > > M^u_u = A^u B_u = invariant scalar > > which proves an invariant scalar can be expressed as either a > contraction of a mixed variance square matrix or the tensor INNER > product of a covariant/contravariant vector pair. Yes, but it I don't believe that proves that it can't be expressed as something else. Again, I don't disagree with what you're saying. My point is that while it DOES clearly prove that A*B is invariant if A and B are 4-vectors, it doesn't clearly prove that A*B can't be invariant if B isn't a 4-vector. Once again, I am not saying and never have said that it could be invariant if B is not a 4-vector. I am saying you are not clearly proving this to the OP. BTW, I believe that Dono is the OP--he sent me a couple e-mails and he tends to change his username. > > [2.4] Observe that A^u B_u = A_u B^u > > A^u B_u = M^u_u > (g_iu A^u)(g^iu B_u) = g_iu g^iu M^u_v > A_i B^i = KroneckaDelta^i_i * M^i_i > = M^i_i > > [3] THINK ABOUT WHAT YOU ARE SAYING > > The statement A.B = INVARIANT means A.B = A'.B' which necessarily > means B is a vector as it can be transformed to B'. Again, I know this and understand this, my point is that your answer didn't make this part clear, which is important to the OP's question. *THIS* would have been a more useful answer (or part of 1), because in the original problem, we do not know that the measured quantity B' is the transformation of B except by the fact that A*B is invariant. For example, B could be (E, x, y, m), and B' could be (E', x', y', m'), which is not a transformation of a 4-vector, and the OP's question was "could something like this make an invariant with a 4- vector." I completely agree that the answer is "no." I just don't think providing the definition that you did clearly demonstrates why this is true. Anyway, I'm snipping the rest.
From: mpalenik on 25 Feb 2010 08:34
On Feb 25, 2:57 am, Schoenfeld <schoenfeld.fore...(a)gmail.com> wrote: > > > The statement A.B = INVARIANT means A.B = A'.B' which necessarily > means B is a vector as it can be transformed to B'. > I just want to emphasize this one more time, the way the problem in Jackson is phrased, the existence of B' does not on its own mean that B is a 4-vector. B' is constructed by frame dependent measurements, in this case (w, kx, ky, kz). In this case B *is* a 4-vector but it could easily have been something like (E', x', y', m) which does not transform as a 4-vector. We are using the invariance of A*B to PROVE that B transforms as a 4-vector. I agree that this is completely valid but the OP's question was "is the only possible way that A*B can be invariant if B is a 4-vector." The answer is yes--but although A*B DOES give an invariant when B is a 4-vector, it is not clear by this definition that the ONLY way for A*B to be invariant is for B to be a 4-vector. |