Originally posted by turin
I HATE ITEX!
I will type the corresponding html next to the itex.
well, feel free to use html. i actually preferred the html, since it seems to fit more nicely with the text, however it doesn t display on some peoples browsers, and of course it can t do as much stuff as tex.
I have this appreciation. My major prof. adamantly declares that a tensor is defined by its transformation properties. If this is true, then it is obvious that the rule of thumb you mention here is nonsense. If it is a bad way to think of it in terms of the transformation properties (I seem to vaguely remember you discouraging this way of thinking), then please remind me.
physicists definition of a tensor:
an object with (r,s) raised, lowered indices (and therefore a coordinate dependent object) that transforms in such and such a way when you change coordinates
mathematicians definition of a tensor:
a tensor product of r,s vectors, covectors. the mathematicians definition of a vector and covector is such that it makes no reference to coordinates, and thus neither does the definition of a tensor. it is an exercise for the reader in most math books to check that when you look at the coordinate components of a tensor, they transform in the physicists way when you change your choice of coordinates.
you take your pick as to which definition. it is nice to understand both definitions, and then one doesn t have to adamantly adhere to one or the other. but certainly one can have a preference, mine is the mathematicians definition.
I wasn't thinking that the index free quality indicated scalar-ness. I was more concerned with the apparent contraction of two apparent vectors. According to my major prof:
- the contraction of two rank 1 tensors (vectors) is a rank 0 tensor (scalar)
you cannot contract two (1,0) rank tensors. you can only contract a (1,0) tensor with a (0,1) tensor. of course, since the metric (if you are doing Riemannian geometry, known to physicists as relativity) and the symplectic form (if you are doing symplectic geometry, known to physicists as classical mechanics) are both nondegenerate, you can always convert one of your (1,0) rank tensors into a (0,1) rank tensor and then contract them (known to physicists as raising and lowering indices). but in the absence of a metric or symplectic form, there is no canonical isomorphism between the vector space of tangent vectors ((1,0) rank tensors) and the vector space of covectors/dual vectors ((0,1) rank tensors), and therefore you cannot contract them.
if you would like any of those terms explained further, please ask.
- \partial_\mu (that is, ∂μ) and dx^\mu (that is, dxμ) are tensors (at least in Minkowski space-time)
therefore:
yes indeed, those are both tensors (they are basis tensors, and therefore coordinate dependent). also, none of this is particular to Minkowski space.
- the object in question still looks like a scalar to me, unless I radically change my understanding of tensors.
yes indeed, the contraction of dx^\mu and \mathbf{\partial_\mu} is indeed a scalar. in fact, it is 1.
What about the proper time interval:
d\tau^2 = dx_\mu dx^\mu (that is, dτ2 = dxμdxμ)?
It seems like there are two inconsistent ways of looking at it:
- either this is a contraction and therefore a scalar
contraction makes a scalar if the contracted objects are a (1,0) tensor and a (0,1) tensor. that is not the case here, so that thing is not a scalar. the fact that it doesn t carry any indices indicates that it is a geometric object, independent of coordinates.
- or this is a 1-form with components dx_\mu (that is, dxμ).
it is also not a 1-form. it is a tensor product of 2 1-forms. if it were also antisymmetric, i would call it a 2-form, but it is not antisymmetric, so i will call it a (0,2) rank tensor.
but you could have figured out that it was a (0,2) rank tensor just by looking at its coordinate components g_{\mu\nu}. 2 lowered indices on the coordinate components = (0,2) rank tensor.
Do the components of any 1-form form a vector basis, and do the components of any vector form a covector basis?
components do not form a basis, since they are not vectors. the components of a 1-form happen to transform like the basis vectors of tangent space, and components of the tangent vector happen to transform like the basis vectors of the cotangent vector space, but this does not mean that the components are themselves vectors
in fact, this point of confusion is exactly the reason that i dislike the physicists definition of a tensor. you become confused about what is a vector, what is a covector, and what are just components.
i said above "happen to transform", but it is no coincidence. recall that any vector (for any vector space. i m thinking linear algebra here) can be written like this:
\mathbf{x}=x_1\mathbf{e}_1+x_2\mathbf{e}_2+...x_n\mathbf{e}_n
here, \mathbf{x} is a vector, which exists in any basis, but has different components. the components live in some field (sometimes these guys are called scalars in math class, but i won t use that word here. for physicists, scalar means something that is invariant under coordinate transformations). thus the components are not vectors. the basis vectors are vectors, but they also depend on your choice of basis (obviously).
if you make a change of basis, you can achieve this by multiplying the basis vectors by some matrix to get a new basis. then you multiply the components by the inverse of that matrix to get the components of the vector in the new basis. the vector itself has matrix times matrix^-1, and thus doesn t change. it is independent of your choice of basis. it is only the coordinates that depend on your choice of basis, and they change in the opposite way that the basis vectors themselves change.
this is why the components of a tangent vector transform like the basis vectors of the cotangent space.
My major prof. would say that it is OK to call v^\mu (that is, vμ) a vector because it implies all of the components (and I guess because it implies the basis?). What say you? I don't want to be picky, just trying to get a handle on the different notational formalisms.
yeah, all physicists do this. it is fine to call v^\mu a vector. in fact, i do it myself whenever i am doing physics. but just keep in the back of your head that v^\mu are really the
components of a vector, strictly speaking, they are not the vector itself. since, in physics, we only ever deal with components, we can replace them in our minds. but be aware that doing so will lead to confusion when you try to do this in a math class. and when it comes time to start doing non-Abelian gauge theory, you will wish you were in the math camp, instead of the physics camp.
Ya. If you meant for that wedge to be in there, then I get it. Again, I'm not trying to be picky, but I have been given the impression lately that these kinds of notational issues are important.
indeed they are (in my opinion)
I thought the basis vectors were \partial_\mu (that is, ∂μ) and that the 1-forms were dx^\mu (that is, dxμ). Did you mean \partial_\mu (that is, ∂μ) here?
no.
\mathbf{\partial}_\mu are the basis vectors for the tangent space, and dx^\mu are the basis vectors for the cotangent space. since df and dg are 1-forms (by definition, a 1-form is a member of the cotangent space), they can be written in terms of the basis of that space. of course, since the basis vectors of any vector space are themselves members of that vector space, dx^\mu is itself a 1-form, it is a basis 1-form. but this 1-form depends on your coordinates. and likewise for \mathbf{\partial}_\mu
