4-vector dot A = invariant => A is a 4-vector?
in [Relativity]
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From: mpalenik on 25 Feb 2010 08:39 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 (i.e. (E', x', y', m) is not (E, x, y, m) under a coordinate transformation). 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.
From: mpalenik on 25 Feb 2010 11:36 On Feb 25, 8:28 am, mpalenik <markpale...(a)gmail.com> wrote: > 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. And just so we're clear, before you tell me that's not the ONLY way to get a scalar, I'm not discounting contracting a tensor, i.e. T^mu_mu as a method of obtaining an invariant, or other such tensor operations. The question is whether or not there are frame dependent quantities that *don't* transform as a 4-vector but will give an invariant when contracted with a 4-vector. And once again, I AGREE with you that the answer is no, there are not. And you can prove this, but your first answer didn't really address this.
From: Schoenfeld on 25 Feb 2010 20:26 On Feb 25, 11:39 pm, mpalenik <markpale...(a)gmail.com> wrote: > 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 (i.e. (E', x', y', m) is not (E, x, y, m) > under a coordinate transformation). 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. Okay. So reading all that, I look at question as: QUESTION "If A.B = INVARIANT SCALAR and A is an invariant vector, prove B MUST be an invariant vector". By invariant vector, we mean covariant or contravariant which is acceptable usage of word invariant. PROOF We know that A.B = A'.B', so in tensor form [1] A'^r B'_r = A^r B_r We chose A as contravariant, so now we must show that given [1], B is covariant for all B. So we know that A' is contravariant a priori [2] A'^r = A^j (@x'^r/@x^j) Substitute [2] into LHS of [1] [3] A'^r B'_r = A^j (@x'^r/@x^j) B'_r Substituting covariant B'_r = B_j (@x^j/@x'r) in RHS of [3] [4] A'^r B'_r = A^j (@x'^r/@x^j) B_j (@x^j/@x'r) Simplifying [5] A'^r B'_r = A^j B_j KroneckaDelta^r_r Thus recover equation [1] [6] A'^r B'_r = A^j B_j [7] So in step 4 we assumed B was covariant we recovered equation [1] as the jacobian elements cancelled out to a contraction of the KroneckaDelta tensor. The question is now was the ONLY POSSIBLE way to recover [1] the subsitution made in [4]? And yes, it is based on the uniqueness of the inverse. For example A*B = A*B*10*C implies that C = 1/10 and only 1/10 So observe that A^j B_j = A^j B_j (@x'^r/@x^j) C^j_r Uniquely implies 'C' cancel out the previous term such that A'^r B'_r = A^j B_j (@x'^r/@x^j) (@x^j/@x'r) = A^j B_j But rather than cancelling it, just rewrite it as A'^r B'_r = A^j (@x'^r/@x^j) B_j (@x^j/@x'r) and you can clearly see that given [1], B' is UNIQUELY B'_r = B_j (@x^j/@x'r) which is covariant. QED If you chose A_r as covariant then the above would show B is uniquely contravariant.
From: mpalenik on 26 Feb 2010 01:55 On Feb 25, 8:26 pm, Schoenfeld <schoenfeld.fore...(a)gmail.com> wrote: > On Feb 25, 11:39 pm, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > > 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 (i.e. (E', x', y', m) is not (E, x, y, m) > > under a coordinate transformation). 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. > > Okay. So reading all that, I look at question as: > > QUESTION > "If A.B = INVARIANT SCALAR and A is an invariant vector, prove B MUST > be an invariant vector". By invariant vector, we mean covariant or > contravariant which is acceptable usage of word invariant. > > PROOF > > We know that A.B = A'.B', so in tensor form > > [1] A'^r B'_r = A^r B_r > > We chose A as contravariant, so now we must show that given [1], B is > covariant for all B. > > So we know that A' is contravariant a priori > > [2] A'^r = A^j (@x'^r/@x^j) > > Substitute [2] into LHS of [1] > > [3] A'^r B'_r = A^j (@x'^r/@x^j) B'_r > > Substituting covariant B'_r = B_j (@x^j/@x'r) in RHS of [3] > > [4] A'^r B'_r = A^j (@x'^r/@x^j) B_j (@x^j/@x'r) > > Simplifying > > [5] A'^r B'_r = A^j B_j KroneckaDelta^r_r > > Thus recover equation [1] > > [6] A'^r B'_r = A^j B_j > > [7] So in step 4 we assumed B was covariant we recovered equation [1] > as the jacobian elements cancelled out to a contraction of the > KroneckaDelta tensor. > > The question is now was the ONLY POSSIBLE way to recover [1] the > subsitution made in [4]? And yes, it is based on the uniqueness of the > inverse. > > For example > > A*B = A*B*10*C implies that C = 1/10 and only 1/10 > > So observe that > > A^j B_j = A^j B_j (@x'^r/@x^j) C^j_r > > Uniquely implies 'C' cancel out the previous term such that > > A'^r B'_r = A^j B_j (@x'^r/@x^j) (@x^j/@x'r) = A^j B_j > > But rather than cancelling it, just rewrite it as > > A'^r B'_r = A^j (@x'^r/@x^j) B_j (@x^j/@x'r) > > and you can clearly see that given [1], B' is UNIQUELY > > B'_r = B_j (@x^j/@x'r) > > which is covariant. > > QED > > If you chose A_r as covariant then the above would show B is uniquely > contravariant.- Hide quoted text - > > - Show quoted text - And hence we have a more general version of the exact same thing I wrote several posts ago, where I used the Lorentz matrices (written as L) instead of general Jacobian matrices, since the problem dealt specifically with flat, Minkowski spacetime.
From: mpalenik on 26 Feb 2010 01:57
On Feb 25, 8:26 pm, Schoenfeld <schoenfeld.fore...(a)gmail.com> wrote: > On Feb 25, 11:39 pm, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > > 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 (i.e. (E', x', y', m) is not (E, x, y, m) > > under a coordinate transformation). 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. > > Okay. So reading all that, I look at question as: > > QUESTION > "If A.B = INVARIANT SCALAR and A is an invariant vector, prove B MUST > be an invariant vector". By invariant vector, we mean covariant or > contravariant which is acceptable usage of word invariant. > > PROOF > > We know that A.B = A'.B', so in tensor form > > [1] A'^r B'_r = A^r B_r > > We chose A as contravariant, so now we must show that given [1], B is > covariant for all B. > > So we know that A' is contravariant a priori > > [2] A'^r = A^j (@x'^r/@x^j) > > Substitute [2] into LHS of [1] > > [3] A'^r B'_r = A^j (@x'^r/@x^j) B'_r > > Substituting covariant B'_r = B_j (@x^j/@x'r) in RHS of [3] > > [4] A'^r B'_r = A^j (@x'^r/@x^j) B_j (@x^j/@x'r) > > Simplifying > > [5] A'^r B'_r = A^j B_j KroneckaDelta^r_r > > Thus recover equation [1] > > [6] A'^r B'_r = A^j B_j > > [7] So in step 4 we assumed B was covariant we recovered equation [1] > as the jacobian elements cancelled out to a contraction of the > KroneckaDelta tensor. > > The question is now was the ONLY POSSIBLE way to recover [1] the > subsitution made in [4]? And yes, it is based on the uniqueness of the > inverse. > > For example > > A*B = A*B*10*C implies that C = 1/10 and only 1/10 > > So observe that > > A^j B_j = A^j B_j (@x'^r/@x^j) C^j_r > > Uniquely implies 'C' cancel out the previous term such that > > A'^r B'_r = A^j B_j (@x'^r/@x^j) (@x^j/@x'r) = A^j B_j > > But rather than cancelling it, just rewrite it as > > A'^r B'_r = A^j (@x'^r/@x^j) B_j (@x^j/@x'r) > > and you can clearly see that given [1], B' is UNIQUELY > > B'_r = B_j (@x^j/@x'r) > > which is covariant. > > QED > > If you chose A_r as covariant then the above would show B is uniquely > contravariant.- Hide quoted text - > > - Show quoted text - And hence we have a more general version of the exact same thing I posted a little while back in the thread, where I used the Lorentz matrices (written as L) instead of general Jacobian matrices. I hope this doesn't appear twice. My last reply didn't show up for some reason. |