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Mathematically a vector is defined as an element of vector space which obeys certain properties. While reading about the special theory of relativity, I came to know about another definition of vectors which was stated as follows:

Formally, then, a vector is any set of three components that transform in the same manner as a displacement when subjected to a transformation (say rotation).

My question is, can it be proved that the vector definition by the physicists is equivalent or is at least a special case of the mathematical definition of vectors?

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    $\begingroup$ The proof is trivial, because the physicist's definition is the same as the mathematician's definition with some additional conditions. $\endgroup$ – Ben Crowell May 17 '18 at 15:01
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    $\begingroup$ One difference is that physics vectors can be moved, math vectors can't. In elementary physics we don't usually have language that distinguishes between a vector and a vector field. Also, physics usually adds extra structure that allows for the definition of a scalar product, but we don't bother to point this out in the language. And furthermore, we don't bother to distinguish between vector, co-vector, and pseudo-vector. $\endgroup$ – garyp May 17 '18 at 15:48
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It is a discrepancy in terminology. So the mathematical term for what physicist use would be a Lorentz(or other symmetry group)-invariant vector field. (a covariant version namely a 1-form can also be constructed). They are geometrical objects deep down. So technically speaking a vector field or a 1-form are vectors at every point of the manifold where they are defined.

The usual case where this happens is in the context of special relativity when one first encounters "4-vectors". These are then mathematical vector fields, that is $\mathbf{x} = x^\mu\partial_\mu$ (physicist just care about the components $x^\mu$). Then if you have a Lorentz transformation $\Lambda$ which you can express as a matrix in components $\Lambda^\mu_{\;\nu}$, transforms our vector $\Lambda \mathbf{x} = \Lambda^\mu_{\;\nu}x^\nu \partial_\nu$. If a certain object doesn't transform like this then physicist say it is not a (Lorentz)-vector. One can check the formal construction of such objects in any book on differential geometry.

Physics are usually interested in the symmetries of a given scenario, and since usually vectors that do not follow the symmetry are not "physical", one refers as vectors to the interesting ones.

EDIT: If you are asking for simpler scenarios, just consider in linear algebra where you can see the same issue at work when you rotate your basis vectors but the "arrow", that is what a physicist kind of has in mind, stays pointing in the same direction after the basis change (although the linear combination coefficients, the component, did change).

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    $\begingroup$ It's not an abuse of terminology, it's simply a difference in usage between two fields. Both fields have a clearly defined mathematical definition of what a vector is. The definitions just don't happen to agree. By the way, the vector-scalar system, and the term "vector," were invented by physicists (Gibbs and Heaviside), so if anyone gets to claim ownership of the term, it's physicists. $\endgroup$ – Ben Crowell May 17 '18 at 14:59
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    $\begingroup$ I don't really mind who claims the name I take no sides, that discussion is beyond the question i think. However vague notions of vectors already existed, check (en.wikipedia.org/wiki/Euclidean_vector). $\endgroup$ – ohneVal May 17 '18 at 16:49
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Yes, vectors in physics obey the vector space axioms, but they also have specific behaviour under rotations so not all mathematicians' vectors are physicists' vectors..

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  • $\begingroup$ Can you cite a proof or at least show the way how to proceed with the proof? $\endgroup$ – Icchyamoy May 17 '18 at 10:17
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    $\begingroup$ Physics vectors add: if A and B are vectors so is A+B. They can be multiplied by members of a field (i.e. the real numbers): if A is a vector so is 2A. So physics vectors obey the vector space axioms. On the converse side, a shopping list is a perfectly valid mathematical vector (price lists form the dual space) but there's no way to rotate them. $\endgroup$ – RogerJBarlow May 17 '18 at 11:36
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    $\begingroup$ @Icchyamoy "how to proceed with the proof": First show that "displacement" obeys all the properties. So displacement is a mathematical-vector. Then "a (physical)-vector is any set of three components" that behaves like that, hence has the same properties, hence is a mathematical-vector. Adding the transformation property on top to define a physics-vector doesn't change that. $\endgroup$ – Bob Jacobsen May 20 '18 at 19:37
  • $\begingroup$ Can you please specify which behavior or recommend some resource? $\endgroup$ – Apoorv Potnis May 28 at 11:02
  • $\begingroup$ A rotation is a square (3x3 or whatever) matrix which transforms (physics) vectors such that scalar products are unchanged. Hence they are unitary: their transverse (or Hermitian Conjugate) is equal to their inverse. Under a rotation R all physics vectors transform as v'=Rv. For a mathematical vector this behaviour is optional (you can't rotate a shopping list), for physics vectors it is essential. (There is the option of covariant or contravariant variation but let's leave that aside.) $\endgroup$ – RogerJBarlow May 29 at 16:48

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