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

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The discussion centers on the adequacy of basic mathematical structures, such as those presented in Griffiths' "Introduction to Electrodynamics," for explaining vectors and pseudovectors without delving into advanced mathematics like group theory. Participants express confusion over the definitions and behaviors of vectors and pseudovectors, particularly regarding the cross product and coordinate transformations. There is debate about whether the Cartesian basis vectors should be classified as vectors or pseudovectors, with some arguing that the treatment of these concepts in introductory texts may be oversimplified or misleading. The conversation highlights the importance of understanding how different authors define terms and the implications of coordinate system transformations on vector properties. Ultimately, the need for a consistent and clear framework for understanding these concepts is emphasized.
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I'm an enginner and not a physics, however is quite a long time (years) that I try to study by myself math and physics to better undestand them deeply.. I'm quite disappointedd today as being very finichy, I alwasy end up at a point where things are not clear.. FOR EXAMPLE IN RECENT DAYS: PSEUDOVECTORS
Is it really possible to explain the concepts of vectors and pseudovectors (differencies) only using the simple mathematical structure used by authors like Griffiths (Introduction to electrodynamics), as the claim to do, without using more deep mathematics like group theory, etc?
Reading the above mentioned reference I ended up in even more confusion and disappoiting.
Please do not try to explain the topics using theory that I'm not ready as I just want to know if the approch of basics books like the quoted one is rigorous or not enough consistent to the proof.


more details:
Following their approach I 'm indeed tempted to say that the "k" base vector, being the result of "i x j" vectors is a pseudovector, and this sounds to me a contraddiction.

Furthermore the same author on applying coordinate trasformations, trasforms only vectors components and conclude that "a vector change sign" on a coordinate inversion ( ie x1'= -x1, x2'=-x2, x3'=x3). This to mee is very misleading as on a change of cordinate system you need to specify also how you like to change the vector base and, based on this choice, it will derive how components of a vector will consequently change (or not).. in any case for any choice you have had made on the new vector base, the new components of a vector will change in order that the geometrical object vector ( linear combination of base elements waighted by the proper components) will be the same.

if I apply this approach, two vectors are always the same geometrical objects regardless the coordinates system (usefull only to identify points) and vector base trasformation (and consequently the trasformation of components of the vectors) chosen and therefore their cross product do not change sign under any trasformation. And so will do their cross product. There is no place in this last way of argumenting for pseudovectors.

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. 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".

This last authors, later on in the text ( p949), then, however, mention about passive and active trasformation where indeed vectors behave differently.. but they are on their turn now not very clear on what that definition and assesments mean.

What I would like eventually to know is if concerning the matter of understanding many physical questions like the case of pseudovectors, the simple physics and math approach of the above mentioned references (and similar ones) is a bit naif and bluffing or if it can have a rigorous based on precise definitions and operations. In case of negative answer I would like to ask what level of mathemathical theory/methods is really consistent on consistently treating such concepts.
At the time being I do not need demonstrations based on superior mathematics ( group theory, etc) as I'm not ready for them.

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
 
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Words have more than one meaning. You need to be careful about comparing what one author says about a word with what another author says about the same word. Although they are using the same word, they may be talking about different things. I would try to avoid getting frustrated by different authors describing things differently. It is an inevitable consequence of the fact that there are more concepts than there are words for the concepts.
 
The mathematics used by physicists is generally not rigorous. If you look hard enough you can generally find minor wrinkles that would take another level of mathematical formalism to straighten out.

Take ##\vec F = m\vec a##, for example. That doesn't quite fit the theory of vector spaces, where ##\vec F## and ##\vec a## would be vectors in the same vector space and ##m## would be a scalar. If you really wanted to, you could sort all that out with some abstract mathematical formalism. But, why bother?

We study physics to understand and model physical phenomena. The theory itself doesn't need to be mathematically watertight. In fact, there is a lot of powerful mathematics used in physics that is very difficult or impossible to justify fully rigorously. If you ask for too much rigour, then you may lose a lot of powerful and useful mathematics.

If rigour is your primary concern, you should study pure mathematics or logic and set theory.
 
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Hi PeroK,
thank you for your answer. I would like not to be too rigorous but at least having a consistent idea and to be rigorous enough to answer questions on the matters and solve exercises or situations for example. If I learn from an Author that the cross product of two vector is a pseudovector than at the question wheather the base vector "k" of a cartesian base (id est the commonly used in cartesian corrdinate system) is a vector or a pseudovector I'm not able to answer and I got in confusion.
I do not think that is to be too rigorous.
Sometimes students are pointed to be too few critics and learn lesson by memory.. then they are not able to independently use the concept to pratical situation ( different from the examples the teacher has shown them). I tried to deep in a bit to really learn and to be ready to use the lesson in general different cases / or for replying to questions.
Is "k" indeed a vector.. why to it the argument to demostrate the pseudovector natur like in the case of othe cross vector is not leading to the same conclusion?
Thaks again
 
The Cartesian basis vectors are not vectors in the physical sense. They are associated with the Cartesian coordinate system.

Griffiths is talking about physical qualities. Like displacement (of a particle), momentum (of a particle) and angular momentum (of a particle).
 
Aleberto69 said:
... question wheather the base vector "k" of a cartesian base (id est the commonly used in cartesian corrdinate system) is a vector or a pseudovector I'm not able to answer and I got in confusion....
It depends how you want your coordinate system to behave under certain transformations like mirroring: Do you want it to preserve "handedness" or not?
https://en.wikipedia.org/wiki/Orientation_(vector_space)
 
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Dale said:
Words have more than one meaning. You need to be careful about comparing what one author says about a word with what another author says about the same word. Although they are using the same word, they may be talking about different things. I would try to avoid getting frustrated by different authors describing things differently. It is an inevitable consequence of the fact that there are more concepts than there are words for the concepts.
Hi Dale,
Thank you for your reply and suggestions.
When I started my life adventure, at 6 as a child student I loved math (numbeers) and llater physics too ( the one you study when you are at secondary) as the had only one right solution.. I was not in confort with poetry or with football talking as, like we say in Italy " in Italy we have 60 milion master coach of the national team and each of them exaclty knows how to win the wolrd cup".. so I decided to love math and physic.. However if now they are a metter of many concept and too word, then it will a bit disappoint me.

I want to go haead in the belive that there is a common language which should be clear and consistent.

Thanks to help telling me your opinion on the questions whether for understand pseudovector simple vector algebra is structured enough consistently or if some higher level math( the lower of the higher level the better) whoulkd be necessary to consider.
TY
 
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PeroK said:
The Cartesian basis vectors are not vectors in the physical sense. They are associated with the Cartesian coordinate system.

Griffiths is talking about physical qualities. Like displacement (of a particle), momentum (of a particle) and angular momentum (of a particle).
Hi PeroK, it sounds a good point...So the vectors of the base are a special cases. fair enough. Are they the only one or are there other example where the corss product of two vectors is still a vector?
But vieeing again from the point of view of the example that generate pseudovector ( eg: angular momentum) I still do not get the point of pseudovectors. Griffiths says that a vector change sign if the cordinate system is inverted. while cross products of them not;
I have many comments: A: vectors themselvse do not change on changing coordinate system; B: at least it is true taht the components of a vector change on changing coordianate system but in any case it depend on how the base of the new coordinate system is chosen; C: regardelss the choice of the new base the new components ( I would say the trasformed components) change in the way that with the trasformation of the vector base keep the vector ( phisical or geometricl opbject) the initial one.. so:
a'=(ax' i'+ay' j'+az' k') = (ax i+ay j+az k)=a the vector a is always itself the rappresentation changes.
and
c'=(ax' i'+ay' j'+az' k') x (bx' i'+by' j'+bz' k') = (ax i+ay j+az k) x (bx i+by j+bz k)= c

I do not see where appear a - sign and why the psedovector need to be introduced

In the above ax' etc are the new component and i' etc the new base vectors..
I furthermore assumed** above that the choice of the new base ( base trasformation) was to change of sign to any initial vector base too ( which is one choice) namely i'=-i, j'=-j etc
the result above is cordinateless and touch the geometrical nature of vectors however if we want to obtain the result also by calculation, I believe that if we use the rule of the detreminant in calculating cross product it needs to add a - sign when dealing with the inverted base as it is left handed.
(the usual determinant rule actually is valid only for righ handed base)

**By the way I have a question on vector base associated to a coordiante system : in any coordinate system there are many all valid choice of fixing a vector base ? some of them are natural but not the only one possible. is this correct?
 
Aleberto69 said:
I still do not get the point of pseudovectors. Griffiths says that a vector change sign if the cordinate system is inverted. while cross products of them not;
Sounds confused. Vectors in 3D don't have a sign, their components have. Maybe a translation error.

This is true tough:
M(a) x M(b) = -M(a x b)
where a, b are vectors in 3D, while M is a mirror transformation.

So if you have something defined via the cross product, like torque, and you mirror your scenario, then you also have to negate the torque.
 
  • #10
A.T. said:
It depends how you want your coordinate system to behave under certain transformations like mirroring: Do you want it to preserve "handedness" or not?
https://en.wikipedia.org/wiki/Orientation_(vector_space)
Thank you A.T. you raised a good point .. I think that handness "belongs" to base vactors and not to coordinates.. that because ( it is a doubt I have and a question that I asled in another reply to PeroK) with a change of coordiantes but even before with coordinate system in general .. coordinates needs to identify points , then for vectors you need to fix a vector base.. now tough there are more natural vector base rellay ther is a degrre of freedom on chosing the base .. once the base is set, then given a vector you can calculate the coeeficent of the linear combination of that base whch result in you vector.. those number are the components.. clarified that.. I think that the handness of the base matters but just in the way you calculate the the cross product with the determinat rule ( we are assuming here orthonormal cartesian coordinate wher at leaast the missing information is the handness) for left hand bases the rule needs a - sign at the and ( the determinat rule as we know it is indeed valid only with right handed bases) .
that is in accordance with the fact that in a left handed base i_lh x j_lh= -k_lh
regardless of the handness, by my calculation the cross product of two vectors behave like a vector and pseudovector are non necessary ( so far it is my opinion.. I would like to understand why I'm wrong as all the rest of the world web tells that is a pseudovector)
 
  • #11
The issue of pseudo vectors is subtle. The idea is that if we do a transformation such that every vector is mapped to the negative, then the cross product of two vectors is not mapped to the cross product in the transformed system.

The I,j,k basis vectors are defined according to a right-hand rule. If we defined one of them as the cross product of the other two (note that there is nothing special about k), then we would get a lefthand system under the above transformation.

Whereas, we would take our new Cartesian basis according to the right hand rule. That's how they are defined - explicitly as vectors with a certain mutual relationship.

Note that there is nothing special about the I,j k vectors. If a, b, c are vectors and it just happens that c = a x b, then after the above transformation ##c' \ne a' \times b'##.

Whereas, if we define the physical quantity represented by c as the cross product of the quantities represented by a and b, then in the transformed system we have ## c' = a' \times b'## by definition.

It's all about how things are defined. I, j k are defined as the Cartesian basis vectors.
 
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  • #12
PS more fundamentally, if we take a transformation that maps the ##i,j,k## basis vectors to something that is not a right-hand Cartesian basis, then the transformed vectors are no longer identified as ##i, j, k## vectors.
 
  • #13
A.T. said:
Sounds confused. Vectors in 3D don't have a sign, their components have. Maybe a translation error.

This is true tough:
M(a) x M(b) = -M(a x b)
where a, b are vectors in 3D, while M is a mirror transformation.

So if you have something defined via the cross product, like torque, and you mirror your scenario, then you also have to negate the torque.
Hi A.T. thanks for rising That point.. I do not understand example whith mirrors.. as it is not completely clear what, when using it, it does rellay means.. I look on a mirror an experiment and I use the "real world" base vectors or I look inside the mirror .. see a different pointing vector and a different base vector... By my side it is somenthing that doesn't help speacking the same language... then I would say .. is the world seen on a mirrorr useful for some undestanding on mecchanics or electromagnetics? I would say no.. Then: is what I see on a mirror thru? My dog doesn't recognioce itself on a mirror ( on Tv do not recognize dogs too).. My heart appear on the left in a mirror , but I'm sure it is on the rigth.. So I do not understand.. I like the language of math if we want to use math to do physics. If we want to consider mirror ( transformation) is ok, however I would like to better understand what you mean with it.. indeed the M(.) operator you used.. could you better specify it? what does it mean mirroring a vector? is that a coordinate/ base-vector trasformation.. so is it a "passive" trasformation ( trasformation of the observer) or an "active" one ( trasformation of the system/ object) or no of the privious one.. My concern agaisnt Griffiths text is on passive coordiante trasformation ( namely the observer change coordinate sistem to reppresent vectors). so what mathematical operation I have to intend when you say M(a)..what happen to coordiantes, ( real or mirrored) to base vectors ( real /mirrored) to components of a vector ( real / mirrores)..?
 
  • #14
PeroK said:
The issue of pseudo vectors is subtle. The idea is that if we do a transformation such that every vector is mapped to the negative, then the cross product of two vectors is not mapped to the cross product in the transformed system.

The I,j,k basis vectors are defined according to a right-hand rule. If we defined one of them as the cross product of the other two (note that there is nothing special about k), then we would get a lefthand system under the above transformation.

Whereas, we would take our new Cartesian basis according to the right hand rule. That's how they are defined - explicitly as vectors with a certain mutual relationship.

Note that there is nothing special about the I,j k vectors. If a, b, c are vectors and it just happens that c = a x b, then after the above transformation ##c' \ne a' \times b'##.

Whereas, if we define the physical quantity represented by c as the cross product of the quantities represented by a and b, then in the transformed system we have ## c' = a' \times b'## by definition.

It's all about how things are defined. I, j k are defined as the Cartesian basis vectors.
Hi PeroK,
I have an issue with your text .. I do not se properly the latex formulas ( eg I see ##c' \ne a' \times b'## ??????).
However I do not uderstand even at your beginning when you say "...we do a transformation such that every vector is mapped to the negative" ... I do not understand what map does mean.. in my question and diisagreement with Griffiths I refer to coordinate system trasformation ( that some author refer as passsive trasformation , namely of the observer).. This kind of operation left the object vector unchanged... what change is its rappresentation... if toghether with the coordinates you change the choice of the basae vactors than the components of the vector for that new base change them selfs accordingly ( I do no want now to better call covariant and controvariant words) .. hovever the two trasformation ( of base vectors and components works toghether for having the same geometrical object.

When you say that a vector is mapped on the negative I do not clearly understand wht you mean... it sounds more that I identify the vector with its negative... but this is not, in my opinion, a passive trasormation ( trasformation of the observer namely of the coordiante system). .may be it is an active one.. " you force changing the sign of the vector" may be... please ...kindly better clarify "map in to the negative"
 
  • #15
I meant the transformation that maps the components ##(x,y,z)## to components ##(-x,-y,-z)##.
 
  • #16
Aleberto69 said:
Hi A.T. thanks for rising That point.. I do not understand example whith mirrors.. as it is not completely clear what, when using it, it does rellay means..
Imagine you compute the forces and lever arms in the right leg, and from them the torques acting at the joints by using the cross product.

Now you want to transfer all this to the left leg. So you mirror all vectors at the sagittal plane, but now the torques in the left leg look as if they were defined using the left-hand rule, instead of right hand-rule, So if you want to have a consistent torque definition (e.g. right hand-rule) for both legs, you have to take this into account when mirroring your mechanics, and flip all pseudo-vectors.
 
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  • #17
Aleberto69 said:
Hi PeroK,
I have an issue with your text .. I do not se properly the latex formulas ( eg I see ##c' \ne a' \times b'## ??????).
However I do not uderstand even at your beginning when you say "...we do a transformation such that every vector is mapped to the negative" ... I do not understand what map does mean.. in my question and diisagreement with Griffiths I refer to coordinate system trasformation ( that some author refer as passsive trasformation , namely of the observer).. This kind of operation left the object vector unchanged... what change is its rappresentation... if toghether with the coordinates you change the choice of the basae vactors than the components of the vector for that new base change them selfs accordingly ( I do no want now to better call covariant and controvariant words) .. hovever the two trasformation ( of base vectors and components works toghether for having the same geometrical object.

When you say that a vector is mapped on the negative I do not clearly understand wht you mean... it sounds more that I identify the vector with its negative... but this is not, in my opinion, a passive trasormation ( trasformation of the observer namely of the coordiante system). .may be it is an active one.. " you force changing the sign of the vector" may be... please ...kindly better clarify "map in to the negative"
About the ijk coomment.. A: I know that k is not special... I could have pointed on j x k= i as well.... then I do not understand what is the difficulty with handness of the base.... when with start with vectors.. and base vectors ... nobody care of saying nothing about handness... the problem arised with cross product... if you define it geometriclally with the right hand rule it is a coordinatelless definition.. so it always works and is not depending on the trasformation ( becouse it does not use anything of it).. if in the context of trasformations ( let here restrict them to rotation and inversion) and their numerical dealing we want to use the determinant rule we have just to bear in mind that ( for numerical correct calculation) we neeed to use -i_lh -j_lh , -k_lh ( using a - sign) in the first row of the matrix for "x" product calculation when the base vector is left handed "- sign", while the "- sign" is not requested for right handed base... that in order to get consistency with:
i_lh x j_lh = -k_lh
however this - sign is just added to be able to use the trick of the determinant rule and it is not related to pseudovector....



By the way .. in a post before ...you said that k ( i o j) are special vector.. and so the proof that k is not a pseudovector is not applicable... well I think better to this question... and I tought the following: any vector regardless of its naure is a linear combination of i j k and some coeeficents ( components) . than the cross vector products has the associative property and the linear property namely the linear sounds like this: (Na) x (Mb= NM (axb) with N, M numbers and a,b vectors... so that any cross product vector can be reduced to a sum of terms involving i x j as i x i , i x k.... ect
if those terms are vectors and not pseudovector.. as they are a special cases.. also their sum is a vector and not a pseudovectors... the conclusion is therefore that any vector cross product should be considered as being reported to this argument/calculation and that therefore is a vector and not a pseudov ... what is wrong with this argument?
 
  • #18
Aleberto69 said:
About the ijk coomment.. A: I know that k is not special... I could have pointed on j x k= i as well.... then I do not understand what is the difficulty with handness of the base.... when with start with vectors.. and base vectors ... nobody care of saying nothing about handness... the problem arised with cross product... if you define it geometriclally with the right hand rule it is a coordinatelless definition.. so it always works and is not depending on the trasformation ( becouse it does not use anything of it).. if in the context of trasformations ( let here restrict them to rotation and inversion) and their numerical dealing we want to use the determinant rule we have just to bear in mind that ( for numerical correct calculation) we neeed to use -i_lh -j_lh , -k_lh ( using a - sign) in the first row of the matrix for "x" product calculation when the base vector is left handed "- sign", while the "- sign" is not requested for right handed base... that in order to get consistency with:
i_lh x j_lh = -k_lh
however this - sign is just added to be able to use the trick of the determinant rule and it is not related to pseudovector....



By the way .. in a post before ...you said that k ( i o j) are special vector.. and so the proof that k is not a pseudovector is not applicable... well I think better to this question... and I tought the following: any vector regardless of its naure is a linear combination of i j k and some coeeficents ( components) . than the cross vector products has the associative property and the linear property namely the linear sounds like this: (Na) x (Mb= NM (axb) with N, M numbers and a,b vectors... so that any cross product vector can be reduced to a sum of terms involving i x j as i x i , i x k.... ect
if those terms are vectors and not pseudovector.. as they are a special cases.. also their sum is a vector and not a pseudovectors... the conclusion is therefore that any vector cross product should be considered as being reported to this argument/calculation and that therefore is a vector and not a pseudov ... what is wrong with this argument?
Basis vectors are defined as vectors relating to a specific coordinate system. They are not defined as the cross product of two other vectors.
 
  • #19
PeroK said:
I meant the transformation that maps the components ##(x,y,z)## to components ##(-x,-y,-z)##.
ok.. x y z are coordinates or components? in cartesian system it is easy to confuse the two concepts..
However let assume you confirm "components", this means that your trasformation of coordinate implies also this trasformation of components of a vector.. I would also add that it implies also a trasformation of vector base id est i'=-i' etc... so that the vector that is a geometrical object is represented in two way but is always itself.
namely

v= x * i + y * j +z * k and x' * i' + y' * j' +z' * k' is the same object ( since it is the observer who changed.. not the object)
proof :
now let substitute x' * i' + y' * j' +z' * k' indeed (-x) * (-i) + (-y) * (-j) + (-z) * (-k)

this lead to x * i + y * j +z * k

let now having two vectors with components a1,a2,a3 and b1,b2,b3 as for the reppresntation in the initial system

they have components -a1,-a2,-a3 and -b1,-b2,-b3 in the trasformed ( inverted) coordinate system

the vector "a" (geometrical object independent on the coordinates system) is

a= -a1*i' - a2 * j' - a3 * k'

and similarly for b ( as reppresented in the trasformed system) we have

b= -b1*i' - b2 * j' - b3 * k'

to calculate the cross product I can follow two way..
A
1: first replace -i where i read i' and -j when I spot j' etc
2: calculate the cross product and it will apper as it is calculated in the not trasformed system

B:
1: perform directly the cross product of (-a1*i' - a2 * j' - a3 * k' ) x ( -b1*i' - b2 * j' - b3 * k')
Doing this I will find terms like (-a1)*(-b2)* (i'x j') = a1*b2* (i'x j') = a1*b2* (-k')
the last calculation is because the trasformed base is left handed and so the right espression is i' x j' =-k'
( to convince yourself of this try to draw the trasformed base and use the rigth and rule on i' x j')

eventually as -k' = k
you simplify the terms in the terms a1*b2* k that summed to the others ( non zero ones) give the same results you got with method A
I do not see any pseudovector need here
 
  • #20
A.T. said:
Imagine you compute the forces and lever arms in the right leg, and from them the torques acting at the joints by using the cross product.

Now you want to transfer all this to the left leg. So you mirror all vectors at the sagittal plane, but now the torques in the left leg look as if they were defined using the left-hand rule, instead of right hand-rule, So if you want to have a consistent torque definition (e.g. right hand-rule) for both legs, you have to take this into account when mirroring your mechanics, and flip all pseudo-vectors.
Hi A.T. apologize for not completely understend... what do you mean for the right leg and left leg?
could you traslate the example using coordinate trasformations...? I try to start with it..

I have a force and a lever arm ... let
F is a vector its components are (0,1,0) ( tey are not coordinates.. tehy are components) ore equivalently
Fx=0 Fy=1; Fz=0

this means that I have fixed a base for vectors and that

F = Fx * i + Fy * j +Fz * k = 1 *j = j

let assume that this is the reppresentation av the vector in the chosen coorddiante / base vector system

let call X the vector position of the point of application of the force F
similalry X is a vector ( position vectro) that is reffeerd as "level arm"

let say the position ( lever arm ) has components (1,0,0)
id est that the position vector is
X = 1 * i + 0 * j +0 * k = 1*i *i

( apologize if a repeat and go step by step but I'm rusty with meccanich as I'm an electronic engineer)

the torque is

X x F = i x j = k ( module 1 directoion of k)


Now I want to see the picture having set a different set of coordinates and base vectors....
that is what i mean for a coordinate trasformation... i do not know if you mean the same when you talk about " Now you want to transfer all this to the left leg"... please confirm or clarify....

keeping on my mening of the trasformation

F is represented in the new system by the components ( 0,-1,0) but they are the cooefficent (components) to be applied different base too ( the base is now no more i, j, k but i' j' and k' and the relationship
i'=-i , j=- j' k' = -k)


it is important to not that with the trasformation ( said to be "passive") I did not change anything of the sistem namele the force and the level arms.. they are still the original one.. I just chouse another base vector for reppresenting them..


with this in mind
the vectors are represented in this way

F= -j'
X=-i'

the torque as calculated in this new refernece is

X x F = (-i') x (-j') = i' x j'

bearing in mind that the new basis is left hand I need to apply the right result to the corssproduct of vectors of the base and in particular that

i' x j'= -k' ( that because the base is left hand in the trasformed system of coordinate/ base vector)



in conclusion

X x F = -k'

but as we stated in chosing the new base vector

k'= -k


and so

X x F keep being the same ( 1 in modulus pointing towords k namely in the opposite direction ok k')


I do not see any change of direction and any need of pseudovector definition



probably the error or misunderstnding is in differnet meaning of trasformations.

I 'm not able to get yours .. I hope you got mine
 
  • #21
PeroK said:
Basis vectors are defined as vectors relating to a specific coordinate system. They are not defined as the cross product of two other vectors.
agree!! but it is commonly accepted that if the base vector is RHanded i x j = k. That is not a definition of k but somethig that is commonly accepted to be true.. any cross product can be reduced to a sum of pieces .. all of them including a pair of vector from the base (and a coefficent that came from the components )... so the sum of a lot of cross products of base vectors ( scalared by coefficent)... all of them are vectors (I belive) and therefore a vector is their sum ( sum of vector is a vector).. conlusion is that any cross product is a vector as well.. why not?
 
  • #22
Aleberto69 said:
what do you mean for the right leg and left leg?
1000_F_197216121_eTIoqeth04sJPTwemAU9CAEC1e2WCsoh.jpg

Aleberto69 said:
i do not know if you mean the same when you talk about " Now you want to transfer all this to the left leg".
You know the torque vectors in the right leg, and now you to want to know the torque vectors in the left leg, assuming symmetry at the sagittal plane. All in the same coordinate system.
 
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  • #23
Aleberto69 said:
agree!! but it is commonly accepted that if the base vector is RHanded i x j = k. That is not a definition of k but somethig that is commonly accepted to be true.. any cross product can be reduced to a sum of pieces .. all of them including a pair of vector from the base (and a coefficent that came from the components )... so the sum of a lot of cross products of base vectors ( scalared by coefficent)... all of them are vectors (I belive) and therefore a vector is their sum ( sum of vector is a vector).. conlusion is that any cross product is a vector as well.. why not?
That's right. So, the question is how we define the process of taking the cross product. I found the following discussion

https://math.stackexchange.com/questions/264771/cross-product-and-pseudovector-confusion
 
  • #24
A.T. said:
View attachment 358520

You know the torque vectors in the right leg, and now you to want to know the torque vectors in the left leg, assuming symmetry at the sagittal plane. All in the same coordinate system.
I thank you for your kindness... why don 't we discuss in italian .. can you?
anyway ....did you read my attempt to explain my doubts in my last post to you? have you some comments on that insted of being hilarious?.. is there something that I didn't explain well ? or that I didn't calculate well..? I would rather prefer answer on the metter...but I do not see a word on your reply demonstrating you get the point of my questions. did you really understend what I was asking for about coordinate trasformations, cross products and eventually the needs of pseudotensor in relation to trasformation ?
 
  • #25
PeroK said:
That's right. So, the question is how we define the process of taking the cross product. I found the following discussion

https://math.stackexchange.com/questions/264771/cross-product-and-pseudovector-confusion
Hi PeroK Thank you for the reply. I read the link and in the first replying posts it is a bit more complex than simple linear algebra. I ' have not the knoledge to completely get that comment.. I do not kow what you want to enphasize about the link, however the last post of it talk about left end and right hand tuple.. It is a bit late here now in the nigth and I ll try to read it more carefullly again tomorrow.
did you understand ( more than me) what the posts conclude ?.. could you summarize and simplify them for me if yes?
One things that I get is that it sounds that things are more complicated than expected and probably that the simple vector algebra and its definitions is not the right framework to fully and consitently get an answer to the concept of pseudovector.
 
  • #26
Hi PeroK again ..I read posts on the link you gave me again...
I spotted two interesting things that come to my side.. one is on the second to last post..
It treats the matter with tensor using the levi-cita one. I started studing tensor and I understnd them a bit..
What I noticed is that they say "on active trasformation only components change" .. I agree with that....
as we force differnet vectors but we keep the observer the same.... so in this sense it make sanse that a cross product behave differently and do not change sign.. but this is another story respect to passive trasformation and indeed the coordinate sysstem trasformation... I think then That Griffiths mixed up things...
I found the same conclusion on the last post as well , where the author talk about a simple mathematical trick to get the same results on calculus using a last step of trasformation more than considering the changed handness of the trasformed base vector.. so psudovectors would be in his opinion just a mathematical artefact to get things right while doing a conceptual mistake... what is your point of view on that post?
 
  • #27
The only way I can make sense of pseudovectors is using the physical argument that @A.T. made. I.e. consider a physical system reflected in a mirror.

If we restrict out attention to simple physical transformations - rotations and translations - then there are only vectors. It's only when we consider improper transformations that the issue of pseudovectors arises.

I can check what Griffiths says when I get home next week.

I agree with you that some of the arguments in that discussion are not convincing.
 
  • #28
Aleberto69 said:
One things that I get is that it sounds that things are more complicated than expected and probably that the simple vector algebra and its definitions is not the right framework to fully and consitently get an answer to the concept of pseudovector.

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.
 
  • #29
Aleberto69 said:
I would rather prefer answer on the metter...
You asked what the point of pseudovectors is. I gave you a practical example from physics, where it matters if something is a vector or pseudovector.

You can find more examples here:
https://en.wikipedia.org/wiki/Pseudovector

320px-BIsAPseudovector.svg.png

Here, when you mirror ##\vec{I}##, then ##\vec{B}## is not just mirrored, but mirrored and negated. So ##\vec{I}## and ##\vec{B}## behave differently under a mirror transformation.

This anti-symmetric behavior might seem somewhat weird. But it becomes clearer, when you consider that in physics we (mis)use the result of a cross-product (a vector) to represent something which might be better understood as the result of an exterior-product (a bi-vector):
https://en.wikipedia.org/wiki/Pseudovector#Geometric_algebra

It's also important to note, that the actual observable physics is perfectly symmetric between the two mirrored cases above. If you shoot electrons with symmetric initial velocities through each ##\vec{B}## field , their deflections and thus paths will be symmetric. It's only that not directly observable quantity ##\vec{B}##, whose mathematical representation was chosen by a non-symmetric convention, that unsurprisingly breaks the symmetry. But only in our diagrams and calculations.

Why did physics go with cross-product (vector) instead of exterior-product (a bi-vector)? I guess it's simpler to visualize/draw arrows, than parallelograms with orientation arrows around their boundary (see diagram below). You can also apply the math you already know from other vector quantities, you just have to remember the exceptions for mirror transformation. But you have a similar issue with transforming direction-vectors vs. position vectors, so you have to consider what kind of "vector" you are transforming anyway.

See also:
https://en.wikipedia.org/wiki/Exterior_algebra#Cross_and_triple_products

231px-Exterior_calc_cross_product.svg.png

Note that the result of the external-product, the bi-vector represented by the blue parallelogram with blue orientation arrows, behaves perfectly symmetrically under mirror transformations, because it exists completely within the plane spannend by the input vectors a and b.

It's only when you get out of this plane to get the cross product direction (by applying the right-hand-rule to both, the original and to the mirrored variant) that the symmetry is broken.



See also this (here the term 'axial vector' is used for 'pseudovector'):
https://marctenbosch.com/quaternions/#h_8

 
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  • #30
Hi A.T.
thank you for your reply..
I agree with you that in your example "Here, when you mirror �→, then �→ is not just mirrored, but mirrored and negated. So �→ and �→ behave differently under a mirror transformation."

However I want to ask you a question about that example : are the the two skeched "I" in your picture the same phisical object (made by the same moving electrons) or are they two different objects, one being the mirrored version of the other?
If your replay will confirm that they are " two different object ... each the mirrored of the other" then I think to understand what is the misunderstanding. The metter is explained by at least two text "Mathematical Methods for Physics and Engineering" ( Riley, Bence etc) at page 948 and also in " Classical Electrodynamics" ( John David Jackson) at page 267 and following.

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"

I try to skech the question in the attached jpeg.

In the active trasformation you have two "I" currrents loop and two "B" magnetic fields.
For each element of the "I" on the left, let call it "dI", positioned at (x1, y1,z1) you have a correspondent element mirrored dI_mirrored at (-x1,y1,z1) .. and in a similar way for the B fields.
You are rigth when say that I and B behave differently when considering their mirrored "alter ego" and this is remarcable... and probably is for this reason that the concept of axial vector ( pseudovector) and polar vector (vector) is usefull. However in this scenario there is only one observer ( system of coordinate x,y,z and only one base vectors) and a specifc system situation of a current and another current which is the mirrored of the first one..

In the passive trasformation instead you consider only one current I and the generated B field.. there are no other entities mirrored.. However now you condisder two different observer each of them have decided of fixing a different coordinate system. In particular observer two decided for the same y and z axis but liked to have the x axse fliped.. Also in this case there is an invertion of coordinates and of vector components. indeed the generic infinitesimal element of "I" . let call it "dI" is positioned at coordinate (x1,y1,z1) as seen from the first observer and at coordinate (-x1,y1,z1) as seen from the second observer..

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)
 

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  • #31
Aleberto69 said:
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.
 
  • #32
PeroK said:
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.
 
  • #33
PeroK said:
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.
 
  • #34
Ibix said:
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.
 
  • #36
PeroK said:
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"
 
  • #37
weirdoguy said:
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?
 
  • #38
PeroK said:
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.
 
  • #39
Aleberto69 said:
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.
 
  • #41
Aleberto69 said:
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:
A.T. said:
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.


active-vs-passive-jpg.jpg
 
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  • #42
Aleberto69 said:
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.

Aleberto69 said:
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.
Aleberto69 said:
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.
 
  • #43
" ..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
 
  • #44
Aleberto69 said:
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.
 
  • #45
Aleberto69 said:
" ..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.
 
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  • #46
Honestly
A.T. said:
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?
 
  • #47
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.
 
  • #48
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.
 
  • #49
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.
 
  • #50
Aleberto69 said:
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.
 
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