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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?

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  • $\begingroup$ 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. $\endgroup$ – jacob1729 May 31 at 8:54
  • $\begingroup$ @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? $\endgroup$ – Apoorv Potnis May 31 at 8:59
  • $\begingroup$ @jacob1729 Or I need that all kinds of coordinate transformations be linear? $\endgroup$ – Apoorv Potnis May 31 at 9:05
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    $\begingroup$ 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. $\endgroup$ – GiorgioP May 31 at 10:36
  • $\begingroup$ @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. $\endgroup$ – jacob1729 May 31 at 12:19
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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.

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  • $\begingroup$ 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. $\endgroup$ – Apoorv Potnis May 31 at 9:21
  • $\begingroup$ 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? $\endgroup$ – Apoorv Potnis May 31 at 9:23
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    $\begingroup$ @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. $\endgroup$ – gented May 31 at 19:38
  • $\begingroup$ But what physical sense can we make of 'rotations' which transform apples into bananas? $\endgroup$ – Apoorv Potnis Jun 2 at 3:52
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    $\begingroup$ @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. $\endgroup$ – gented Jun 2 at 10:18
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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.

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  • $\begingroup$ 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. $\endgroup$ – gented May 31 at 8:07
  • $\begingroup$ 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. $\endgroup$ – Steven Thomas Hatton May 31 at 8:14
  • $\begingroup$ 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). $\endgroup$ – gented May 31 at 8:26
  • $\begingroup$ 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). $\endgroup$ – Apoorv Potnis May 31 at 8:37
  • $\begingroup$ @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. $\endgroup$ – Steven Thomas Hatton May 31 at 8:54
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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.

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  • $\begingroup$ 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? $\endgroup$ – Apoorv Potnis May 31 at 8:59
  • $\begingroup$ Or I need that all kinds of coordinate transformations be linear? $\endgroup$ – Apoorv Potnis May 31 at 9:01
  • $\begingroup$ 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. $\endgroup$ – GiorgioP May 31 at 10:42
  • $\begingroup$ 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, en.wikipedia.org/wiki/Euclidean_vector $\endgroup$ – RogerJBarlow May 31 at 10:55
  • $\begingroup$ 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. $\endgroup$ – RogerJBarlow May 31 at 11:10
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I found out that this answer by user joshphysics answers my question satisfactorily.

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