2

I recently learned that a vector in mathematics (an element of vector space) is not necessarily a vector in physics. In physics, we also need that the components of the vector on a coordinate transformation as the components of the displacement vector change. So, if my understanding is correct, if $|\mathbf{c}_1|, |\mathbf{c}_2|, |\mathbf{c}_3|,\, \ldots \,,|\mathbf{c}_n|$ are the components of a vector $\mathbf{A}$ and $f$ is the function of transforming coordinates (change of basis), then $$f(\mathbf{A}) = \sum_{i=1}^n{f(\mathbf{c}_i)}$$ where $\mathbf{A} = \sum_{i=1}^n\mathbf{c}_i$.

That is to say, the transformed vector by applying $f$ to it should be equal to the vector formed by the vector components which have been transformed by applying $f$ to them.

Am I correct?

Apoorv Potnis
  • 1,274
  • Your equation defines what it means for $f$ to be linear. The physicist's insistence that vector components transform properly is only really relevant to vector fields, where the basis can be different at different points in space, and the coordinate transformation changes each basis in a different way. – jacob1729 May 31 '19 at 08:54
  • @jacob1729 So if I define a function which changes the basis, is linear and acts on a vector space in which the elements are shopping lists, are shopping lists vectors in the physicists' sense? – Apoorv Potnis May 31 '19 at 08:59
  • @jacob1729 Or I need that all kinds of coordinate transformations be linear? – Apoorv Potnis May 31 '19 at 09:05
  • 1
    What you define components (${\bf c_i}$) are the basis vectors in linear algebra. I do not see the reason you should use the modulus. A part this observation, vectors of linear algebra are exactly the same as vectors in physics. Quite often physicists use freely the term vector also for vector fields, including the simplest case of an affine euclidean space. That is usually source of confusion. – GiorgioP-DoomsdayClockIsAt-90 May 31 '19 at 10:36
  • @ApoorvPotnis you can't have a vector field of shopping lists is the point I think. Even if you chose a shopping list for each point in space, the list entries would have nothing to do with the physical space. What is interesting about vectors is that you can attach a $n$ dimensional vector to each point in an $n$ dimensional space, and each component relates to one of the dimensions. Thus when you transform coordinates they need to change appropriately. – jacob1729 May 31 '19 at 12:19
  • @GiorgioP I know that I am being pedantic but according to Physics, 5th ed. by Halliday, Resnick and Krane, components are the magnitudes of the vectors along the basis vectors. For example, for a vector $\mathbf{A}$ in $\mathbb{R}$, $|\mathbf{a}| \cos{\theta}$ is the component along the $x$-axis and $|\mathbf{a}| \cos{\theta} \hat{\mathbf{i}}$ is the vector component. – Apoorv Potnis May 31 '19 at 13:18
  • Also, if the definitions are the same in maths and physics, why don't physics books define vectors simply as elements of a vector space. – Apoorv Potnis May 31 '19 at 13:21
  • @jacob1729 Yes, shopping lists are not vectors because they do not represent the physical space. If we transform the coordinates, then one object does not change into another as one component changes into another (apples do not change into oranges when we change coordinates). So, is what I have written in the question correct or not? – Apoorv Potnis May 31 '19 at 13:27
  • @ApoorvPotnis I'm not sure. You haven't written it the way I think about it, so I'll have to think. Could you give an example of a transformation and the $f$ that it produces? – jacob1729 May 31 '19 at 14:52
  • Meanwhile, is this a duplicate: https://physics.stackexchange.com/questions/241610/physicists-definition-of-vectors-based-on-transformation-laws?rq=1? – jacob1729 May 31 '19 at 14:59
  • @ApoorvPotnis $|{\bf a}|$ is not the same thing as $|{\bf a}| cos \theta$ . The reason the definition in the elementary Physics textbooks vectors are not defined as elements of a vector space is because they try to provide some intuition about them. But it is a very simple exercise for everybody who has understood the concept of vector space to show that the two concept are equivalent. – GiorgioP-DoomsdayClockIsAt-90 May 31 '19 at 21:25
  • @GiorgioP Correction in my previous comment. "$\mathbf{A}$ in $\mathbb{R}$" should be "$\mathbf{a}$ in $\mathbb{R}^2$". I never said that $|\mathbf{a}|$ is the same thing as $|\mathbf{a}| \cos \theta$. – Apoorv Potnis Jun 01 '19 at 16:47
  • @GiorgioP It may be very trivial to show that the definitions in maths and physics are equivalent, but I'm unable to prove it. Can you show a short proof as an answer or suggest any book which proves the same? – Apoorv Potnis Jun 01 '19 at 16:50

4 Answers4

4

I recently learned that a vector in mathematics (an element of vector space) is not necessarily a vector in physics

A vector in "physics" is exactly the same thing as you have defined it in "mathematics".

Any vector space contains a basis $e_i$ upon which each element can be expanded as $$ v = \sum_k v^k e_k. $$ By definition of basis as tangent vectors to a set of curves, one can show that they must transform in a certain way, say given a transformation matrix $\Lambda$. Since the vector $v$ must be independent of the representation, if the basis transform using $\Lambda$ then the components must transform using the inverse matrix $\Lambda^{-1}$.

$\Lambda$ (respectively $\Lambda^{-1}$) are what physicists refer to as covariant (respectively contravariant) transformation laws for the basis (respectively vector components).

Same holds for dual forms and tensors mutatis mutandis.

gented
  • 5,940
  • 14
  • 25
  • Can you give some further readings for your answer? I haven't studied differential geometry yet. I also haven't studied tensors and dual forms. I've completed one course on introductory linear algebra. – Apoorv Potnis May 31 '19 at 09:21
  • Also, the definitions are equal, shouldn't objects such as a "barrel of fruit containing pears, apples and bananas" be vectors in physics as well? – Apoorv Potnis May 31 '19 at 09:23
  • 1
    @ApoorvPotnis As further readings you can take any text in geometry of analytical mechanics really, these are just the standard definitions of vectors and vector spaces. Coming to your second point about "barrel of fruit" - if you define operations on such objects such as addition and multiplication by a scalar then yes, they will be elements of a vector space and can be expanded onto some basis thereof. – gented May 31 '19 at 19:38
  • But what physical sense can we make of 'rotations' which transform apples into bananas? – Apoorv Potnis Jun 02 '19 at 03:52
  • 1
    @ApoorvPotnis This is a different question and has nothing to do with something being a vector or not. Then you should go back and ask what physical sense you can make of your objects in the first place - also it is a common misconception that in physics everything is "interpretable": it isn't. Physics creates models to make predictions, as long as they work with the experiments we consider them fine. – gented Jun 02 '19 at 10:18
1

Yes, your statement is correct, however... I haven't seen your use of "component" in a long time. Your use is the strictly correct meaning of component. That is, components are vectors. But the term is often used to mean the "coordinates" of a vector. That is in $\vec{v}=v_{x}\hat{i}+v_{y}\hat{j}+v_{z}\hat{k}$ the actual components are $v_{x}\hat{i}, v_{y}\hat{j}$ and $v_{z}\hat{k}$. But people almost always mean $v_{x}, v_{y}$ and $v_{z}$ when they say "components." And $f\left(v_{y}\hat{j}\right)$ (vector argument) and $f\left(v_{y}\right)$ (scalar argument) are not the same thing. In most situations a function (transformation) taking a vector argument will not be defined for a scalar argument.

Steven Thomas Hatton
  • 1,965
  • Can you please clarify what statement in particular is "correct"? There is no "displacement vector" in transformation laws and no function $f$ is required in any vector definition. – gented May 31 '19 at 08:07
  • That is one definition of vector. That is, a thing that transforms in the same way a displacement transforms. More generally, this is applied to tensors. I believe that's how Einstein defined things in The Meaning of Relativity. I believe Feynman also used that definition. Some physicists don't like the "component dependent" definition. – Steven Thomas Hatton May 31 '19 at 08:14
  • That's really not how vectors are defined. Vectors, forms and tensors are defined exclusively as linear forms and what you call "displacement transform" is a mere consequence of the change of basis for tangents to a curve (which is how basis may be defined). – gented May 31 '19 at 08:26
  • Yes. I actually used component as vector but then edited it to be precise. I assume in my question that component is a scalar and vector component to mean as a vector (as per my physics book). – Apoorv Potnis May 31 '19 at 08:37
  • @gented I don't even know what you mean by "linear form". I'm confident that it means something different for me. And I can assure you that yours is not the only accepted definition of vector. – Steven Thomas Hatton May 31 '19 at 08:54
  • @StevenThomasHatton All the properties that you claim to be "different definition of a vector" automatically follow from the one definition that you have in mathematics. You shouldn't call it "alternative" definitions because it gives OP the understanding that there are indeed different ways to define vectors, which isn't true. – gented May 31 '19 at 09:17
  • @gented decades ago, as a freshman, I was invited to attend a graduate course on the foundational problems in modern physics. I asked the professor what the best definition of "tensor" is. His response was: "A tensor is a geometric object, independent of coordinates." A vector is a tensor. You may believe you have the One True Definition, but there are multiple ways of defining what a vector is. I checked to book I am currently reading, and they give two different definitions in the same section, and claim them to be equally fundamental. https://mitpress.mit.edu/contributors/h-behnke – Steven Thomas Hatton May 31 '19 at 09:49
  • @gented That professor's name was John Archibald Wheeler. – Steven Thomas Hatton May 31 '19 at 09:52
1

You are correct in saying they are different. Physics vectors are mathematical vectors, but not necessarily vice versa.

For example, Birkhoff and Maclane "A Survey of Modern Algebra", p162 of the 1953 edition:

A vector space $V$ over a field $F$ is a set of elements, called vectors, such that any two elements $\alpha$ and $\beta$ of $V$ determine a (unique) vector $\alpha+\beta$ as sum, and that any vector $\alpha$ from V and any scalar $c$ from $F$ determine a scalar product $c.\alpha$ in $V$, with the properties

$V$ is an Abelian group under addition

$c.(\alpha+\beta)=c.\alpha+c.\beta , \qquad (c+c').\alpha=c.\alpha+c'.\alpha$ (Distributive laws)

$(cc').\alpha=c.(c'.\alpha),\qquad 1.\alpha=\alpha$

Hence sets of functions form a vector space. So do simple shopping lists. This brings in the dual space, dimensionality and the basis but there is nothing about physical space, tangents, pointing arrows or all the familiar properties us physicists ascribe to a vector.

Physics vectors have the additional property that they can be transformed (by a rotation). Vector equations must remain valid if they are transformed. So if ${\bf A}={\bf B} + {\bf C}$ then ${\bf f(A)=f(B)+f(C)}$. The transformation must be linear. Your result follows, and shows that, if you have a basis $\{{\bf e}_i\}$ and write ${\bf A}=\sum_i c_i {\bf e_i}$ then the function can be written as a matrix multiplication.

RogerJBarlow
  • 10,063
  • So if I define a function which changes the basis, is linear and acts on a vector space in which the elements are shopping lists, are shopping lists vectors in the physicists' sense? – Apoorv Potnis May 31 '19 at 08:59
  • Or I need that all kinds of coordinate transformations be linear? – Apoorv Potnis May 31 '19 at 09:01
  • that is not a correct answer. Rotations are available also for "mathematical vectors". Just the language looks less familiar. However, an automorphism of a 3D vector space implemented through one orthogonal matrix of determinant +1 represents a rotation. – GiorgioP-DoomsdayClockIsAt-90 May 31 '19 at 10:42
  • The set of all functions defined on the range [0,1] forms a (mathematical) vector space. But such a function has no 'direction' and although you can apply an automorphism it is not a 'rotation' in any real physical; sense. It is far more general than the definition of a (physics) vector given in, say, https://en.wikipedia.org/wiki/Euclidean_vector – RogerJBarlow May 31 '19 at 10:55
  • Your shopping list might be 3 tins of beans, 2 cans of coke, and a pizza. If you rotate it in a meaningful physical sense that just rotates the cans and the pizza and doesn't affect the vector. You can change the basis (3 beans, 1 coke, 1 pizza+coke special) which is a 'rotation' but not in the physicosts' sense. – RogerJBarlow May 31 '19 at 11:10
  • I actually want the mathematically rigorous meaning of "rotate it in a meaningful physical sense". And does the transformation being linear capture the notion of the same? – Apoorv Potnis May 31 '19 at 13:30
  • In 2d: rotation through an angle $\theta$. In 3D, rotation though the 3 Euler angles, or equivalent. Rotations are a subset of linear transformations in which scalar products are unchanged. – RogerJBarlow May 31 '19 at 13:41
  • @RogerJBarlow A function in the range [0,1] does have a direction and you can rotate it. Of course, since it does not lives in the same space as space displacement, you cannot "see" in the same way. It requires some training. but a similar training is required also for the "usual' displacements. Vectors, in the form as we know, here called "vectors of physicists" did not appear in the physics textbooks until the end of the nineteenth century. – GiorgioP-DoomsdayClockIsAt-90 May 31 '19 at 21:18
  • @RogerJBarlow I can also define functions which kind of 'rotate' the vector space of beans, coke and pizza. But coke transforming into pizza doesn't make physical sense, at least in the usual sense. I want some mathematical way which distinguishes between the transformations which make physical sense and which not. – Apoorv Potnis Jun 01 '19 at 16:53
0

I found out that this answer by user joshphysics answers my question satisfactorily.

Apoorv Potnis
  • 1,274