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

• I
Gold Member

## Summary:

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)

## Main Question or Discussion Point

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?

## Answers and Replies

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mathman
Science Advisor
Let u and v be vectors then $u\cdot v=|u||v|cosA$. Your calculation neglected }u|}v|.

Gold Member
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.

mathman
Science Advisor
|u||v| is the product of the norms, not the inner product.

Gold Member
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

mathman
Science Advisor
I can't comment any further, since I am not familiar with this subject.

WWGD
Science Advisor
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Does the identity $u.v=|u||v|cos\theta$
hold in Minkowski space?

Gold Member
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.

mathman
Science Advisor
If $u\cdot v=|u||v|cos(\theta)$, how is $cos(\theta) \gt 1$? It requires a weird definition for dot product.

WWGD
Science Advisor
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2019 Award
I guess it is a variant of the Cauchy-Schwarz inequality.

Gold Member
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.

WWGD
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Gold Member
2019 Award
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.

Gold Member
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.

WWGD
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Gold Member
2019 Award
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.