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Quick tensor question

  1. Sep 21, 2011 #1
    Greetings,

    I have been going through Bernard Schutz’s “A First Course in General Relativity”, and I have a quick question on tensors. On page 73, the book defines an (M,N) tensor as “a linear function of M one-forms and N vectors into the real numbers”, which I’m fine with – I had no trouble with the problems at the end of the chapter. However, on page 57, it says “Notice that an ordinary function of position, f(t,x,y,z), is a real-valued function of no vectors at all. It is therefore classified as a (0,0) tensor”.

    Why couldn’t (t,x,y,z) be thought of as a vector, since it is an ordered 4-tuple? This would make the position function f a one-form, so it would output a scalar after receiving the (t,x,y,z) vector as input - ie., would do exactly what the familiar f(t,x,y,z) does.

    Basically, my question isn’t really about SR or GR at all, but more about the distinction between a vector and an ordered n-tuple in general – it previously seemed like just an arbitrary difference in ways of thinking about it to me, but now it seems like it and its relation to the difference between (0,0) tensors and (0,1) tensors, or (1,0) tensors for that matter, is important for distinguishing between these different types of tensors.

    Thank you for any help you can give.

    -HJ Farnsworth
     
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  3. Sep 21, 2011 #2

    Fredrik

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    A tensor at a point p in a manifold M is a multilinear function [itex]T:V^*\times\cdots\times V^*\times V\times\cdots\times V\rightarrow\mathbb R[/itex], where V is the tangent space at p and V* is the cotangent space at p. A tensor field is a function that assigns a tensor at p to each p.

    Your 4-tuple (t,x,y,z) (the coordinates of the point p) is not a member of V. More importantly, if you change to an arbitrary coordinate system the change in the coordinate 4-tuple won't (in general) be linear, so the numbers in your 4-tuple won't (in general) transform in the way required by the tensor transformation law. If the coordinate change is a linear function, like a Lorentz transformation, it will transform as required, and you can think of your 4-tuple as a tensor if you want to. (Strictly speaking, it's a member of the wrong vector space, but if the components transform the right way, who cares).
     
    Last edited: Sep 22, 2011
  4. Sep 21, 2011 #3

    WannabeNewton

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    Well, for a manifold M, a scalar function [itex]f[/itex] and a one form [itex]\tilde{\omega } [/itex] are not exactly the same in terms of mappings. For one, [itex]f:M \mapsto \mathbb{R} [/itex] whereas [itex]\tilde{\omega }:T(M) \mapsto \mathbb{R} [/itex]. [itex]f[/itex] is not a function of a vector in the sense that [itex]f[/itex] takes [itex]p \in M[/itex] as its argument not [itex]V \in T_{p}(M)[/itex]. Remember that another condition for something to be a one - form is that, when mapping from one chart to another, [itex]\omega _{\mu '} = \frac{\partial x^{\mu }}{\partial x^{\mu '}}\omega _{\mu }[/itex]. A scalar function [itex]f[/itex] does not transform like this; in fact scalar quantities remain invariant under such conditions. Also note that for [itex]\tilde{\omega }\in T^{*}_{p}(M), V\in T_{p}(M)[/itex] we have that [itex]\tilde{\omega }(V)[/itex], which gives you a real number, can be written in component form as [itex]\omega _{\nu }V^{\nu } [/itex]. Could you do the same with a scalar field [itex]f[/itex]?

    EDIT: I completely missed Fredrik's post when I typed this up so I apologize for the redundancy.
     
  5. Sep 21, 2011 #4
    Thank you for your responses.

    I sort of get it, but not as well as I'd like yet. You both say that one-forms and scalar functions take arguments from different sets. The trouble I was having was differentiating between those sets, and I think that you are both saying that a key difference between them is that their elements transform differently under coordinate transformations - not necessarily linearly in the case of the scalar function arguments, ie., points on a manifold.

    First of all, did I read what you told me right? Second of all, would you be able to provide me with a simple example of a coordinate change where treating a scalar function of a point and a one-form function of a vector with the same components as that point doesn't work out - I think that would make the difference between n-tuples and vectors a lot more clear to me.

    Again, many thanks for your quick responses and help.

    -HJ Farnsworth
     
  6. Sep 21, 2011 #5

    atyy

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    (t,x,y,z) is just 4 numbers that index a point on the manifold - it is not an "arrow" that a vector is. It does not make sense to multiply (t1,x1,y1,z1) with a number k and say that the resulting (t2,x2,y2,z2)=k(t1,x1,y1,z1) is an object pointing in the same direction - (t2,x2,y2,z2) is just a different point on the floppy manifold.

    If you have a curve on the manifold, the tangent vector at a point (t,x,y,z) on that curve is a vector. It is represented by 4 numbers at that point (v1(t,x,y,z),v2(t,x,y,z),v3(t,x,y,z),v4(t,x,y,x)). The 4-tuple (v1,v2,v3,v4) represents an arrow, and is a vector (and can be multipled by a scalar to get another vector pointing in the same direction, and can be added to a vector pointing in another direction to get a resultant pointing in a direction between both vectors).

    If you represent tangent vectors to curves as column vectors, then one forms are row vectors. You can multiply a row and a column vector to get a number - that's how one forms act on vectors.
     
  7. Sep 22, 2011 #6
    Thanks again for the replies.

    I'm pretty sure I get it. A scalar function need not be linear, so treating an n-tuple meant to represent a point as a vector and a scalar function f as a one-form does work out if f happens to be linear, but otherwise doesn't satisfy the definition of a tensor. For example, if I have f = tx + y, point P1 = (1,1,1,1) and point P2 = (2,2,2,2), then if I treat the two points as vectors so that v2 = 2v1, I do not get f(v2) = 2f(v1), so f cannot be a 1-form with P1 and P2 as vectors. Thinking about it this way makes it a lot easier to understand the difference between the mappings that Fredrik and WannabeNewton were talking about. Still, could someone please confirm that I am thinking about it right?

    Random follow-up question that occurred to me this very moment: In the book I am reading, f is still considered a tensor - a (0,0) tensor - and it takes scalar quantities as arguments. Can an (M,N) tensor also take scalar quantities as arguments, and not necessarily be linear in its scalar arguments? It seems like it could, so long as it was set up so that it was still linear in its vector and one-form arguments. For example, a (1,1) tensor f(a,b,v,p), where a and b are scalars, v is a vector, and p is a one-form, and f(a,b,k*v,p) = f(a,b,v,k*p) = k*f(a,b,v,p), regardless to the values of a, b, and k?

    Thanks once more.

    -HJ Farnsworth
     
  8. Sep 22, 2011 #7

    atyy

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    Try thinking more geometrically. A vector is an arrow. A point on a manifold is not an arrow. The 4-numbers are just labels like latitude and longitude - you would never think of latitude and longitude as forming an "arrow" on the surface of the earth, although it is a 2-tuple.

    Vectors come about as tangents to curves - like velocity is a vector because it is a tangent to a curve. Generally position (like latitude and longitude) is not a vector. In special relativity and Newtonian mechanics, you can treat position as a vector, because space is "rigid and square", but that doesn't carry over to GR. Position is not and never a vector. Vectors are tangents, like veclocity, acceleration etc.
     
  9. Sep 22, 2011 #8

    Fredrik

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    The f you picked isn't the best example of a non-linear function. :smile: (It's linear).

    There are several different ways to define "spacetime" in SR. We can choose to define it as the vector space [itex]\mathbb R^4[/itex] if we want to, but we can also choose to define it as the smooth manifold [itex]\mathbb R^4[/itex]. If we make it a manifold, then I would call a real-valued linear function on the tangent space at p a cotangent vector at p, not a 1-form. I would define a 1-form as a cotangent vector field, i.e. as a function that assigns a cotangent vector at p to each p.

    If we instead define spacetime as a vector space, then things get pretty different. A vector space doesn't have tangent spaces. If the only coordinate transformations we ever consider are Lorentz transformations, we can let spacetime (the vector space [itex]\mathbb R^4[/itex]) take over the role of all those tangent spaces. (In other words, we set [itex]V=\mathbb R^4[/itex] in my first post).

    You need to figure out which one of these definitions of spacetime you're trying to understand. If you want to understand both, focus on one now, and leave the other for later. Since you're reading a GR book, you should probably focus on the manifold version of spacetime. I sometimes recommend this post and the posts I linked to in it, to people who would like to learn the basics of differential geometry. For example, click the second link and read post #5 to see how a coordinate change induces a change in the basis vectors of the tangent space.

    The convention is to not consider these things tensors.
     
  10. Sep 22, 2011 #9
    Thanks for your reply, atyy,

    I understand what you're saying about the geometrical viewpoint. I think part of the cause of my original confusion was that I had read sources that basically state that you can think of points as vectors - but those sources were mostly for vector calculus taught specifically for 3D Euclidean geometry, and were specifically referring to point vectors, so any vector (a,b) used to describe a point in fact implictly meant the difference of the points (a,b) and (0,0).

    Going to points on a sphere, subtracting points like this would be weird, since the vectors involved would be on planes tangent to the sphere at those points, so subtracting the points themselves would give vectors through the sphere. If the sphere itself is the manifold, I think a point subtraction would yield a kind of "curved" vector along the sphere (though straight in the spherical geometry), but I am not positive about that - since I'm just starting to learn about different manifolds, I don't want to get ahead of myself. I definitely think you are right in that I should learn to understand the concepts from a geometrical point of view in addition to a more "function-based" point of view.

    Thanks once more.

    -HJ Farnsworth

    P.S. Any thoughts on the scalar arguments in (M,N) tensors question at the end of my previous post?
     
  11. Sep 22, 2011 #10
    Thanks, Fredrik, just saw your post. I'll probably be busy with some of those links for a little while...

    Regarding the question I had at the end, why are they not conventionally considered tensors? - it seems like they would fit the definition well enough. Also, if those aren't considered tensors, why is a scalar function considered a (0,0) tensor at all? I'm probably just quibbling over a minor technicality in the definition I have, but I figure I may as well ask anyway.

    Thanks again.

    -HJ Farnsworth
     
  12. Sep 22, 2011 #11

    Fredrik

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    It's always hard to answer why a definition looks exactly the way it does. The answer is usually along the lines of "people have found it useful and easy to work with". I don't think this is an exception, but it would take a lot of work to make a convincing argument, so I'll just say that it seems to me that including those things would make the definition weird and complicated.

    In the case of scalars, it might have something to do with the fact that scalar fields and tensor fields are all sections of vector bundles over the manifold.

    A question like this is easier to answer. They either satisfy it or they don't, and they don't satisfy the one I posted. Did you have another definition in mind?
     
  13. Sep 22, 2011 #12

    atyy

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    A (0,0) tensor does not take a scalar as its argument (not in the sense that a (1,0) tensor takes a vector as its argument). A (0,0) tensor is a scalar (ie. a number).

    A (1,0) tensor is a row vector that takes a column vector as its argument and outputs a scalar. A (0,1) tensor is a column vector that takes a row vector as its argument and outputs a scalar.

    f(t,x,y,z) as a (0,0) tensor field is a scalar field - at every point in spacetime there is a number. Similarly, a (1,0) tensor field means that at every point in spacetime there is a row vector. In f(t,x,y,z), don't read the (t,x,y,z) as the argument, since a one form field will be (w1(t,x,y,z),w2(t,x,y,z),w3(t,x,y,z),w4(t,x,y,z)), but (t,x,y,z) is not the "vector" that is the input for the one form - it is simply the point in spacetime at which the one form exists.
     
    Last edited: Sep 22, 2011
  14. Sep 22, 2011 #13
    Thank you for the replies.

    The definition I had in mind was from Schutz: An (M,N) tensor is a function of M one-forms and N vectors into the real numbers that is linear in each of its M and N arguments.

    I took that to mean it could also be a function of scalars, for two reasons - 1. they weren't specifically forbidden by that definition (like if I say f is a function of a variable x, that does not negate that it is also a function of a variable t), and 2. the book I am using said that "a normal function of position, f(t,x,y,z), is a real-valued function of no vectors at all. It is classified as a (0,0) tensor", which I took as implying that a tensor could be a function of scalars as well as vectors and one-forms.

    The (0,0) tensor as a scalar seems a lot neater - by the Schutz definition above, it is a function of 0 one-forms and 0 vectors into the real numbers, so it simply is just a real number.

    The stuff I was saying at the end of that one post does clearly contradict Fredrik's definition of a tensor, though I'm still not sure whether it contradicted my book's definition, which could easily be modified to exclude arguments that aren't vectors and one-forms. Maybe some quick math would show that it inherently already does, but I don't think that's necessary, as I understand the answer to my questions well enough at this point, and am now on a track to learn a bunch of differential geometry.

    Thanks for being so helpful.

    -HJ Farnsworth
     
  15. Sep 22, 2011 #14

    atyy

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    Perfect.
     
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