4-vector dot A = invariant => A is a 4-vector?
in [Relativity]
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From: Dono. on 19 Feb 2010 03:16 On Feb 18, 11:20 pm, mpalenik <markpale...(a)gmail.com> wrote: > 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'. Thank you, got it.
From: Schoenfeld on 24 Feb 2010 00:18 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.
From: mpalenik on 24 Feb 2010 18:45 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.
From: Schoenfeld on 24 Feb 2010 23:40 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". You need to express B as a field equation, which requires a theory (postulates). So generally speaking, let B be a frame-dependent field B(x0,x1,x2,x3) = b0(x0,x1,x2,x3) e0 + b1(x0,x1,x2,x3) e1 + b2(x0,x1,x2,x3) e2 + b3(x0,x1,x2,x3) e4 where b0,b1,b2,b3 are frame-dependent scalar fields x0,x1,x2,x3 are frame-dependent coordinates e0,e1,e2,e3 are the frames basis vectors To prove B is a CONTRAVARIANT VECTOR field (and thus frame- independent), then you need to show that b'0(x'0, x'1, x'2, x'3) = b0(x0,x1,x2,x3) @x'0/@x0 + b0(x0,x1,x2,x3) @x'0/@x1 + b0(x0,x1,x2,x3) @x'0/@x2 + b0(x0,x1,x2,x3) @x'0/@x3 + AND b'1(x'0, x'1, x'2, x'3) = b1(x0,x1,x2,x3) @x'1/@x0 + b1(x0,x1,x2,x3) @x'1/@x1 + b1(x0,x1,x2,x3) @x'1/@x2 + b1(x0,x1,x2,x3) @x'1/@x3 + .. .. .. WHERE B'(x'0,x'1,x'2,x'3) = b'0(x'0,x'1,x'2,x'3) e'0 + b'1(x'0,x'1,x'2,x'3) e'1 + b'2(x'0,x'1,x'2,x'3) e'2 + b'3(x'0,x'1,x'2,x'3) e'4 b'0,b'1,b'2,b'3 are b0,b1.. in the transformed cordinate system x'0,x'1,x'2,x'3 are x0,x1,.. in the transformed coordinate system e'0,e'1,e'2,e'3 is the transformed frames basis vectors Note that x' describes THE COORDINATE TRANSFORMATION and is expressed GENERALLY as a set bijective-functions: x'0 = x0'(x0, x1, x2, x3) x'1 = x1'(x0, x1, x2, x3) x'2 = x2'(x0, x1, x2, x3) x'3 = x3'(x0, x1, x2, x3) Note that (@x'_n/@x_m) are the first-order partial derivatives of the coordinate transformation (or elements of the Jacobian matrix). All of the above cumbersome equations are neatly compacted in what is called Tensor Calculus, which deals with geometric objects at this level - without knowing the actual geometry of the coordinate systems, or the actual coordinate transformation equation itself. It only requires that THEY EXIST. Tensor Calculus was created as a notational system to "wrap up" the above systems of equations into a tidy (but delicate) notational system. All of the above can be simply stated as: B is a contravariant vector if B'^i = B^r (@x'^i/@x^r ) B is a covariant vector if B'_i = B_r (@x^r / @x'^i) using Einstein summation convention (i.e. sum over repeated indices) NOTE: Modern Tensor Calculus tends to reinterpret this "notational system" as "abstract geometric objects" deriving from topology analysis (i.e. manifolds, tangent spaces, etc). For me, it is much simpler to treat tensors as Jacobi and Ricci did, and that is the Classical Tensor Calculus approach which I highly recommend. Now finally, to answer your interpretations of the OP's question, then he after making empirical measurements of B in some frame, he needs to pick another frame and make the same measurements, and if the measurements transform according to the above equations, then he can say whether or not those measurements transformed as a vector. Of course, you cannot prove B' is a vector emprically, only falsify it by repeating the experiment until B' does not transform as a vector.
From: mpalenik on 25 Feb 2010 00:26
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. Let's take the energy momentum 4-vector. You can construct it by measuring energy and momentum in any given frame, or you can construct it by measuring energy and momentum in one frame and applying a coordinate transformation. So, the OP's question was "given that these measurements have the property that the inner product with a 4 vector is invariant, does that mean that they transform as a 4-vector." We had established that A was a 4 vector. B is a set of frame dependent measurements but we are GIVEN the information that B*A is invariant. He was asking does this imply that B transforms as a 4 vector (or "is a 4-vector" to use his words) or is it possible that it is something else. Or perhaps to put it a bit more concisely, does this mean that we can acheive the frame dependant, coordinate representation of B' by applying a Lorentz transformation to B. Maybe my language was a little sloppy, but geeze. I think it's pretty clear that's what his question was. |