Undergrad Tensor calculation, giving|cos A|>1: how to interpret

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The discussion centers on the interpretation of a tensor calculation involving vectors in Minkowski space, specifically regarding the angle between two vectors, Ai and Bi, where the cosine of the angle results in a value greater than one. The author of the referenced book concludes that the angle is "not real," which raises questions about the meaning of angles and orthogonality in this context. Participants explore the definitions of inner products and norms under the Minkowski metric, noting that the standard identity for dot products may not hold as expected. The conversation also touches on the nature of time-like and space-like vectors, as well as the concept of null vectors being self-orthogonal. Overall, the discussion highlights the complexities of geometric interpretations in relativistic frameworks.
nomadreid
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TL;DR
In the Minkowski metric, using the usual inner product relationship, finding the angle A between (1,0,0,0) and (2^0.5, 0, 0, (3^0.5)/c gives cos A = -2^0.5, which indicates that A is not real, but what is it? (source given)
On pages 42-43 of the book "Tensors: Mathematics of Differential Geometry and Relativity" by Zafar Ahsan (Delhi, 2018), the calculation for the angle between Ai=(1,0,0,0) (the superscript being tensor, not exponent, notation) and Bi=(√2,0,0,(√3)/c), where c is the speed of light, in the Minkowski metric with the (- - - +) convention, i.e.,
ds2=-dx2-dy2-dz2+c2⋅dt2,
is carried out with the calculation cos A = gijAiBi/√((AiAi)(BjBj)) (Einstein summation convention). The result, -√2, then tells us that |cos A| = √2, which the author uses to conclude: "the angle between Ai and Bi is not real." I do not understand this: does he mean that the concept of angle has no meaning here, or that the angle is complex (whatever that means), or what?
 
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Let u and v be vectors then ##u\cdot v=|u||v|cosA##. Your calculation neglected }u|}v|.
 
By }u|}v| I presume you mean the inner product. But the calculation is less straightforward because we are in curved space. The author defines the inner product between the vectors as AiBi=gijAiBj, and the square of the norm (length)of vector Ai is thus AiAi= gijAiAj. With the Minkowski metric, g11=g22=g33= -1, , g44=c2, and gij=0 for i≠j. This makes the two vectors unit vectors, and the calculation for the inner product is -√2.
 
|u||v| is the product of the norms, not the inner product.
 
Ah, OK. I did not forget the product of the norms in the calculation; as I pointed out, these are (in this metric) unit vectors, so the product of the norms is 1:
A2=gijAiAj=g11A1A1=1
B2=gijBiBj=g11B1B1+g44B4B4=1
 
I can't comment any further, since I am not familiar with this subject.
 
Does the identity ##u.v=|u||v|cos\theta ##
hold in Minkowski space?
 
WWGD, the author of the book I am citing assumes that the identity holds in Minkowski space (and other metric spaces), with the definitions of inner product and norm taking into account the metric tensor in the manner I cited in my other posts in this thread. But he also allows for the angle to be "not real", giving a cosine value outside of that usually defined for cosine, which is what puzzles me, and which prompted my original question.
 
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If ##u\cdot v=|u||v|cos(\theta)##, how is ##cos(\theta) \gt 1##? It requires a weird definition for dot product.
 
  • #10
I guess it is a variant of the Cauchy-Schwarz inequality.
 
  • #11
IMHO it is not such an unusual definition (defined in post #3): after all, usually ds2=<dr, dr>= <eidxi,ejdxj> =<ei,ej>dxidxj=gijdxidxj, where <.,.> is the inner product, and ei is a coordinate basis vector. So, the definition of the inner product as given seems to be natural, and fulfills all the requirements of an inner product. I am glad to be corrected, of course.
 
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  • #12
nomadreid said:
IMHO it is not such an unusual definition (defined in post #3): after all, usually ds2=<dr, dr>= <eidxi,ejdxj> =<ei,ej>dxidxj=gijdxidxj, where <.,.> is the inner product, and ei is a coordinate basis vector. So, the definition of the inner product as given seems to be natural, and fulfills all the requirements of an inner product. I am glad to be corrected, of course.
I would be interested in understanding the meaning of orthogonality regarding the time coordinate.
 
  • #13
WWGD said:
I would be interested in understanding the meaning of orthogonality regarding the time coordinate.
The key point to go on would be, I suppose, the definition of two vectors being orthogonal if their inner product (as defined in my previous post) is equal to zero. One remark out of this source (which I need to work out an example for) is that a null vector, that is, one with magnitude zero, need not be the zero vector (which means that null vectors are self-orthogonal, which sounds rather paradoxical). To complete the connection with relativity, the author defines vectors with real magnitudes as "time-like" and those with imaginary magnitudes as "space-like." Intuition flew out the window some time ago.
 
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  • #14
There are quadratic forms Q ( obviously not standard inner products) for which there are nonzero a with
Q(a,a)=0. I will look for examples.
 
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