I How deep does it need to go for a physics theory to be consistent?

[PeroK]

We need therefore to agree on the meaning of "trasformation". The authors talk about "active" and "passive" trasformation. and depending on what of the two concept you apply then you end up with different assesment.
Considering the above classification you are refferring to "active trasformation" while my questino was about "passive trasformation"

The two circumstances are complitely a different situation.....

In the active trasformation you have only one observer ( cordinate system) and two surces (currents i and I mirrored).
the passive trasformation is simply the different point of view of two observer looking only one surce ( current I). In the passive trasformation you ave mirrored coordinate and mirrored base vectors associated to the two cordinate systems.. then you see only one source "I" and the only effect being the "B" fields..

The unique B fields is always hte same for the two observer..


My questioning, from the beginning is that some authors define the concept of pseudovector based on the trasformation of coordinate system ( indeed a passive trasformation... indeed when a different observer look at the experiment having fixed another reference system of coordinates).. I think that this assesment of that authors was not correct.

Your point is correct as it is referred to another meaning of the concept of "trasformation" ( the active one)

Here's what I think.

The whole issue depends on whether the laws of physics are the same in a mirror system. The situation with the active transformation is clear. The cross product produces a pseudovector that transforms differently from a vector.

In the passive transformation, there is no change to the physical system. But, for reasons that I don't fully understand, the cross product changes to follow the left-hand rule in the case of a reflection (improper transformation). I guess if it didn't, then the results for active and passive transformations would be different.

 

[PeroK]

But, for reasons that I don't fully understand, the cross product changes to follow the left-hand rule in the case of a reflection (improper transformation).

PS The definition of the cross product that I use is the right-hand rule (determinant rule) for a right-hand orthonormal basis. And, it's not clear what should happen to the cross-product operation under a passive reflection.

However, there must be a more general way of defining the cross product - which would tell you what to do in terms of a reflection. In any case, whatever the mathematical basis, it is in line with the case of an active reflection and the requirement that the magnetic field, for example, be reflected. That requires the cross product to transform under a passive reflection.

 

[Ibix]

And, it's not clear what should happen to the cross-product operation under a passive reflection.

Isn't it just that you want scalars like $(\vec a\times\vec b)\cdot\vec c$ to be invariant? So you want $\vec a\times\vec b$ to point in the same direction, which requires a handedness flip in the cross product when you flip the handedness of your coordinate system.

 

[PeroK]

Isn't it just that you want scalars like $(\vec a\times\vec b)\cdot\vec c$ to be invariant? So you want $\vec a\times\vec b$ to point in the same direction, which requires a handedness flip in the cross product when you flip the handedness of your coordinate system.

Perhaps, but what's the mathematical justification for that being a scalar? Under improper transformations.

 

[A.T.]

However, there must be a more general way of defining the cross product - which would tell you what to do in terms of a reflection.

Isn't that the external product, that I mention in post #29?
https://en.wikipedia.org/wiki/Cross_product#As_an_external_product

 

[Aleberto69]

PS The definition of the cross product that I use is the right-hand rule (determinant rule) for a right-hand orthonormal basis. And, it's not clear what should happen to the cross-product operation under a passive reflection.

However, there must be a more general way of defining the cross product - which would tell you what to do in terms of a reflection. In any case, whatever the mathematical basis, it is in line with the case of an active reflection and the requirement that the magnetic field, for example, be reflected. That requires the cross product to transform under a passive reflection.

Hi PeroK,

I have no doubts about the rules for the cross product and theri differences.

Let consider a cartesian orthonormal system of cordinates and its natural base vector i', j' and k' ( natural meaning that the unit vector of the base point to the same direction towards the relative coordinate increases)
Assume then that that system of cordinates is the inverted of the standard one ( i , j ,k)..
Easy to note that i,j,k is right handed and that i', j', k' is left handed.
What I'm talking about is just a passive trasformation in the sense that the two system of coordinates and related natural bases are two different systems that two different observers decided to adopt to study the system.
Well, what I believe is that the right hand rules (which is not the determinant rule) holds for both systems of reference. The right and rule applies indeed for vectors regardless the coordinate system adopted. The rigth hand rule applies to arrows which are geometrical object existing regrdeless the coordinate system adopted for describing them.
So if you sketch the 2 system of references and in particular the versors i,j,k and i',j',k' you will verify that for both ijk the right hand rule works well. In particular for i',j',k', it happens that i' x j' = -k' .That result is an easy geometrical verification....
What doesn't work for the left hand basis instead is the rule of the determinant.
Many authors indeed tell that the determinant rule ( as we know it) is valid only for orthonormal right handed basis.
For left handed orthonormal basis however ther is a modified determinant rule which use in the first row ( the row of the vector base) -i' , -j', -z' instead of i' , j', z'...
If you use this rules you get for example i' x j' = det ( -i' , -j' , -k'
1, 0 , 0
0, 1 , 0)

which gives -k' that is the right result ( the same you get by the right hand rule applied to the vectors i' and j' geometrically calculating the cross product i' x j').

In conclusion ... the right hand rule applies always as it is a geometrical operation on the arrows ( vectors).

The determinant rule is more a mathematical tool for using components of vector and for having a easy rule to remember for the cross product. However it is indeed coordinate system dependent and indeed involves the components of the vectors ( respect to a specific coordinate system and base vectors) and involve the unit vectors of the base relative to the specific coordinate system. For these dependece the rule as we usual know it applies only to orthonormal right hand vector base... for orthonormal left hand vector base the determinant rule can be still used provided that you change the sign to the first row of the matrix..

For not orthonormal base vector the determinat rule is wrong ( both with the + and the - sign

Let demostrate this last statement...

Assume the observer base vector is orthogonal, righ handed but not normal.

let assume that the base vectors are i' = 2 i and j'= 2 j and k'= 2 k ( so the length of the each base vector is 2)

than it is easy to see ( grafically) that i' x j' = 4 k = 2 k' ( correct result)

However if we try to use the determinant rule written as usual I got i' x j' = k' which is wrong


Conclusion:
1) Right hand rule .... "always valid ... apply it to arrows ( geometric object)"
2) Determinat rule: involving components and base vectors... be carfull: "ask yourself before appling it what kind of base is referred to and ask yourself what correction need to apply to the rule to get the right answer"

 

[Aleberto69]

This. For me, all those things got resolved with the notion of orientation/orientability/volume forms/orientation bundles/other things in the context of differential geometry. I always wanted as much mathematical rigour as possible, and that ended with me specialising in mathematical physics/differential geometry. But you know, sometimes it's better to just give up on some things, 'cause most of people have limited time resources, especially if you're learning physics just for fun.

TY weirdoguy.... you are probably rigth... and probably also other people in this thread menitoned bivectors which probaly more easily clarify the point.. I 'll get an insight to this more sophisicated tools later on soon.
However, restricting the matematics to the simple vector algebra.. do you think that the seed of the problem could be explainded in term of the misunderstanding about "active" and "passive" trasformation as defined by Jakson for example?

 

[Aleberto69]

PS The definition of the cross product that I use is the right-hand rule (determinant rule) for a right-hand orthonormal basis. And, it's not clear what should happen to the cross-product operation under a passive reflection.

However, there must be a more general way of defining the cross product - which would tell you what to do in terms of a reflection. In any case, whatever the mathematical basis, it is in line with the case of an active reflection and the requirement that the magnetic field, for example, be reflected. That requires the cross product to transform under a passive reflection.

right-hand rule and determinant rule are not the same thing in my opinion..
determinant rule give the same result of right ahand rule only for orthonormal rigth hand base vector.

 

[PeroK]

right-hand rule and determinant rule are not the same thing in my opinion..
determinant rule give the same result of right ahand rule only for orthonormal rigth hand base vector.

You're missing the point that these formulations of the cross product are special cases of the exterior product that @A.T. has provided a link to.

 

[PeroK]

Isn't that the external product, that I mention in post #29?
https://en.wikipedia.org/wiki/Cross_product#As_an_external_product

Yes, that's what I was missing.

 

[A.T.]

My questioning, from the beginning is that some authors define the concept of pseudovector based on the trasformation of coordinate system ( indeed a passive trasformation...

Your point is correct as it is referred to another meaning of the concept of "trasformation" ( the active one)

What I wrote here:

M(a) x M(b) = -M(a x b)

where a, b are vectors in 3D, while M is a mirror transformation.

Applies regardless if M represents an active or passive mirror transformation. A transformation matrix doesn't encode the type of transformation. So regardless if you interpret it as active or passive mirror transformation, you will get some inconsistency, because:

M(a) x M(b) ≠ M(a x b)

In your diagram below, the passively transformed I-vectors and loop-positions in the primed frame (bottom diagram), will have the same numerical components, as the actively transformed I-vectors and loop-positions on the left in the top diagram. So if you apply the same cross-product-definition to both, you get a B-vector pointing up (positve Z) for both. But for the passive case (bottom diagram), that contradicts the passively transformed B-vector which still points down (negative Z).

So regardless if you do an active or passive mirror transformation, you have to flip the pseudo-vectors, if you want to preserve the cross-product-definition.

 

[PeroK]

Many authors say ( Griffiths is one of them) that magnetic filed and angular momentum are pseudovectors and not vectors ( as they are the result of a crss product of vectors) while other authors say the contrary.

Griffiths (EM) introduces the cross product using the determinant form (eq 1.13 and 1.14). Then, exercise 1.10 asks you to calculate how the cross product of two vectors transforms under an inversion ($\bar x = -x, \bar y = -y, \bar z = -z$).

Griffiths does not discuss the subtleties here. For example:
$$\hat x \times \hat y = \hat z$$The components of $\hat z$ are $(0, 0, 1)$ and $(0, 0, -1)$ in the two coordinates systems. That's if we transform $\hat z$ as a vector. However, if we define $\vec c = \vec a \times \vec b$, then we want the components of $\vec c$ to transform differently under inversion. (Griffiths assumes we get this through calculation, but looking more closely, that calculation is not justified. In order to justify it mathematically, we need the broader scope of the exterior product, as discussed above.)

Note that if $\vec c$ is a vector that just happens to equal $\vec a \times \vec b$ component-wise, that is very different from the physical quantity represented by $\vec c$ being so defined. The Cartesian unit vectors are dimensionless basis vectors that have the given vectorial relationship involving the cross product. Whereas, the cross product of a displacement and a linear momentum; or, the cross product of displacement and electric current, are very different physically and hence mathematically.

I admit I hadn't noticed these subtleties before. It's very interesting.

Eg: Riley-Hobson-Hence in "Mathematical methods for Physics and Engineering".. pag 948 say that "It is worth nothing that, although pseudotensor can be usefull mathematical object , the description of real phisical world must usually be in term of tensors( i.e. scalars , vectors, etc) .......velocity, magnetic field strength or angular momentum can only be described by a vector and not by a pseudovector".

I've no idea what Riley and Hobson mean here. It's physics that demands pseudovectors, as far as I can see.

I would appreciete just possible rigorous demonstrations based on simple vectors algebra definition or the comment that such a structure is not enough to describe this kind of complexity
Thank you in advance for your replies

You don't need to dive into the full rigour. You only need to be able to see the subtleties and where the simple treatement in Griffiths lacks detail.

 

[Aleberto69]

" ..So if you apply the same cross-product-definition to both ".... do you understand the determinant rule?

what I'm saying is ( only appling to the case of passive trasformation) that the second observer knows that his base vector is left handed and therefore he should be aware that the determinant rule is not the rigth formula for calculating the cross product . For that reference system the rigth formula is the determinant rule and a minus sign after that.. if in the primed system of reference you do not do that and you use " the same rule for dot product" you would get also that i' x j' = k' while i' x j' = -k'
I do not see inconsistency and need of pseudovector.. only I need to say that determinant rule do not applies if the observer adopt a
left handed vector base.

Different thing is the atctive trasformation.. in this case the components of the involved vectors are mirrored ( x changed of sign in the example) however the base vector is still the same for the two dfferent set (current / displacement) of object.

otherwise i have the question:
Your example understand the Biot- Savart rule... which is an integral... looking to just the integrand it is a cross product... so... if an ionconsistecy there is ( as you say) it shoulod be for any cross product of any vectors.
I arrive to the conclusion that any vector of the base vector is a pseudovector... that becouse alltought they are not defined by the cross product however the cross product relation exist...
indeed i x j = k and j x k =i and k x i = j
are therefore i j k pseudovector too

 

[PeroK]

what I'm saying is ( only appling to the case of passive trasformation) that the second observer knows that his base vector is left handed and therefore he should be aware that the determinant rule is not the rigth formula for calculating the cross product .

How do you know your basis vectors are right or left handed? How do you know which is your right hand and which is your left? There is nothing mathematically to tell you. There would be nothing mathematically wrong with the determinant rule. The other observer would have no way to tell he was using a "wrong-handed" system of coordinates. He would simply be working with a different sign convention for angular momentum and magnetic field.

It's only when you compare the sign convention of the two observers you see that the differences.

 

[A.T.]

" ..So if you apply the same cross-product-definition to both ".... do you understand the determinant rule?

... the second observer knows that his base vector is left handed and therefore he should be aware that the determinant rule is not the rigth formula for calculating the cross product .´´

It is the "right" formula, because that is how the cross product is defined, regardles of the handedness of the system.

Having two different formulas for cross product depending on the handedness of the system, would be an alternative, but it's not the current convention. Likely beacuse it would be inparctical to always keep the handedness of the system in mind, when doing vector physics. With the current convention you only have to consider it when you transfrom between two systems of different handedness. But you don't even have to know which system has which handedness, just that they differ as you can deduce from the transformation matrix, so you flip the pseudo-vectors.

 

[Aleberto69]

Honestly

It is the "right" formula, because that is how the cross product is defined, regardles of the handedness of the system.

Having two different formulas for cross product depending on the handedness of the system, would be an alternative, but it's not the current convention. Likely beacuse it would be inparctical to always keep the handedness of the system in mind, when doing vector physics. With the current convention you only have to consider it when you transfrom between two systems of different handedness. But you don't even have to know which system has which handedness, just that they differ as you can deduce from the transformation matrix, so you flip the pseudo-vectors.

I reply here to PeroK too.

Honestly I do not have an answer to the question How the observer knows the handenss of the coordinate system that he is adopting.. However the following:

1) I can reference many Authors and text that says " the determinant rule applies only to rigth handed system of coordinates / base vectors" . I can give reference if you need

2) There are other operations that are constrained by the nture of the base vector and coordinate system.. for example the scalar product calculated as a1*b1+a2*b2+ a3*b3 ( ai and bi being the components of two vectors respect to the adopted base ) is valid only if the base is orthonormal... How can the observer knows if the base vector is orthonormal then? has he mathematical instrument to say that is or is not orthonormal.. the question is the same for handness.. I belived that the observer in some way need to have this information.. eventually base vector and coordinates need to perform calculation and computing.. however you need to know something about the reference system you are using ( I think)

3) you didn't answer to the observation and question wheras k is a pseudovector or not .. Indeed k is defined as a vector of the reference base but at the same time is also true the relation k= i x j
and for this cross produc the same arguments used for the Biot-Savart expression for calculating B would lead to the conclusion that k is a pseudovector. ( the same applies for i and for j)


In any case . I'm a bit busy at work rigth now so I have limited time and I can answer only in the breaks, however I started and want to read better the contents of the link A.T. provided on cross product and also keep reading other authors explanation too.
for example : In post #23 from PeroK I noticed a comment :

• 2
  Yes, there is more to the story. If you want to fully formalize those things in a convenient mathematical framework you need to move into the realm of the differential forms. Try looking for the keyword "pseudovector" in the book "The Geometry of Physics" by Theodore Frankel.
  – Giuseppe Negro
  CommentedDec 24, 2012
• 
  
• 
  
• I would be curious to follows the suggestion but unfortunately I do not have that book.
• have some of you the book and can report what that author says?

 

[PeroK]

A basis is orthonormal if the basis vectors obey the rule $e_i \cdot e_j = \delta_{ij}$. Whether a basis is orthonormal depends, therefore, on the definition of the inner product.

If you have an orthonormal basis in three dimensions, then you can define a cross product by defining $e_1\times e_2 = e_3$ and the others in cyclical fashion.

Imagine this were an abstract space. How do you tell right handed from left handed? The question for 3D Euclidean space is quite subtle. How would you communicate right-handedness mathematically? It's a serious point.

 

[PeroK]

PS the usual Euclidean inner product is defined so that the usual basis vectors are orthonormal. Or, alternatively, by assuming that the angle between them is a right angle.

 

[Aleberto69]

Apologize Perok but your explanation is not convincing me...
1)
If I have an abctract space how can I verify that ei dot ej = dij?

In my opinion is a similar problem of verifying that the three vectors are left handed..
I could say if e1 x e2 = e3 (and ciclically) then the base is right handed otherwise il left..

Furthermore
If I have only the coordinate instead you do not have any mathematical information if the base is orthonormal and if it is right handed or not.

2) Still waiting arumenting if the result of i x j is a pseudovector

3) have you verified in some text the assesment that "the determinant rule for cross product is valid also for rigth hand base vector"? let me know and state if that author are then wrong.

 

[PeroK]

Apologize Perok but your explanation is not convincing me...
1)
If I have an abctract space how can I verify that ei dot ej = dij?

The inner product must be defined somehow. You calculate the inner product on each pair of basis vectors using the definition.

One way to define an inner product is to define it explicitly on the basic vectors. Note that a basis is an algebraic concept and the inner product is an analytic concept- in particular, it gives you the concept of the length of each vector.

In my opinion, you'd be better to learn more about these introductory concepts than to dive down into differential geometry and external products.

Pseudovectors are a side issue, in my opinion.

 

[PeroK]

2) Still waiting arumenting if the result of i x j is a pseudovector

That's been fully explained. The basis vectors are dimensionless. They are vectors by definition.