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What does it mean: transforms like a vector

  1. Jun 30, 2013 #1
    What does it mean: "transforms like a vector"

    Reading Zee's new "Einstein Gravity in a Nutshell" I just don't get it what exactly is meant by "transforms like a ...". He explicitly writes that not any odd column of numbers is a vector, so I am trying to figure out what exactly is meant. Here is what I think I did understand so far:

    - "transforms" refers to the transformation from one coordinate systems to another.
    - Specifically a Vector [itex]V^\nu[/itex] shall transform like [itex]dx^\nu[/itex] in [itex]dx'^\mu = \frac{∂x'^\mu}{∂x^\nu}dx^\nu[/itex], in particular
    [tex]V'^\mu(\vec{x}') = \frac{∂x'^\mu}{∂x^\nu}V^\nu(\vec{x})[/tex]
    If I take the equality above as the definition of [itex]V'^\nu[/itex], the statement "transforms like ..." does not have any deeper meaning. The "transforms like ..." only has a deeper semantic meaning, if [itex]V'^\nu[/itex] is defined independently of the equality above and we can then say, "Hey, this equality does not hold for any odd column of numbers but only for those which we want to call a vector".

    Now I ask: what is this independent definition of [itex]V'^\nu[/itex]?

    To be more specific: suppose [itex]V^\mu(\vec{x})\in R^2[/itex] is some physical quantity with a length and a direction and let our unprimed coordinate system be polar coordinates. For the sake of argument let [itex]V^\mu(r,\theta) = (l,\xi)[/itex], where [itex](l,\xi)[/itex] is the physical quantity written in polar coordinates.

    Let the primed coordinate system be cartesian coordiinates, meaning [itex]x'=r \cos(\theta)[/itex] and [itex]y'=r \sin(\theta)[/itex].

    Assume that [itex](l,\xi)[/itex] is indeed a vector, then I can tranform it by [itex]\frac{∂x'^\mu}{∂x^\nu}[/itex] into a [itex]V'(r\cos(\theta),r\sin(\theta))=(x_0',y_0')[/itex] of the primed coordinate system. (Still correct?)

    My hunch would be that "transforms like a vector" means that [itex]x_0' = l\cos(\xi)[/itex] and [itex]y_0'=l\sin(\xi)[/itex], where the right hand sides are the "independent definition" I was looking for. Is that true?
     
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  3. Jun 30, 2013 #2

    WannabeNewton

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    You are correct that defining a geometric object in terms of how it transforms under a change of coordinates is absolutely ridiculous but this is how vectors, tensors etc. were defined classically. You can find the geometric, coordinate independent definition of tensors in any differential topology/smooth manifolds text and in good GR texts like Carroll and Wald. See here as well: https://www.physicsforums.com/showpost.php?p=4348374&postcount=5

    The reason we say transforms like a vector is because each covector in the the dual coordinate basis ##\{dx^{1},...,dx^{n}\}## transforms in exactly that way but we aren't talking about the components of a vector transforming under the usual vector transformation law; we are talking about each covector in the dual basis itself transforming to a new one under the change of coordinates. Similarly each vector in the coordinate basis for the tangent space ##\{\partial_1,...,\partial_n\}## transforms like a covector ##\partial_{\mu'} = \frac{\partial x^{\nu}}{\partial x^{\mu'}}\partial_{\nu}## however we are talking about the basis vectors themselves transforming from one coordinate system to another and not the components of some covector.
     
    Last edited: Jun 30, 2013
  4. Jun 30, 2013 #3

    WannabeNewton

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    If you've ever used Zee's QFT text, this one has a similar flavor i.e. very warm, witty, conversational, and focuses more on physical insights than on mathematical formalism. On the other hand if you've not used Zee's QFT text before, that would be something to do as well :)! The only thing I don't like is that Zee uses ##D_{\mu}## for the covariant derivative as opposed to the standard ##\nabla_{\mu}##, since ##D_{\mu}## usually shows up in QFT in the form of the gauge covariant derivative.
     
  5. Jul 1, 2013 #4
    Thanks for the reply, but it is only a small help, as it leaves the question still open. So I try again, from a slightly different angle. I'll check your link to the other thread in due time, but meanwhile:

    Well then, lets get this more concrete: Zee defines [itex]S^\mu_\nu = \frac{∂
    x'^\mu}{∂x^\nu}[/itex] and says that [itex]V^\nu[/itex] is a vector if it transforms as [itex]V'^\mu=S^\mu_\nu V^\nu[/itex]. For every [itex]\mu,\nu\in\{1,\dots\}[/itex] the [itex]V^\mu[/itex] etc is nothing fancy but just a number.

    What number is it, if not the components of the vector? And what components could it be other than the components with regard to the unit vectors of the coordinate system?:confused:
     
  6. Jul 1, 2013 #5

    WannabeNewton

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    The components ##V^{\mu}## of a vector ##V## as represented in some coordinate system ##x^{\mu}## will transform as ##V^{\mu'} = \frac{\partial x^{\mu'}}{\partial x^{\nu}}V^{\nu}##. For this we can clearly state that these are the components of vectors. But if we are given a dual coordinate basis ##\{dx^{1},...,dx^{n}\}## then these will also transform as ##dx^{\mu'} = \frac{\partial x^{\mu'}}{\partial x^{\nu}}dx^{\nu}##. Notice the difference in the second case: we aren't talking about the components of things transforming but rather the elements of the dual basis themselves transforming under the change of coordinates. For example ##dx^{1'} = \frac{\partial x^{1'}}{\partial x^{\nu}}dx^{\nu}## gives us the new dual basis vector ##dx^{1'}## itself under the coordinate transformation. On the other hand ##V^{1'} = \frac{\partial x^{1'}}{\partial x^{\nu}}V^{\nu}## simply tells us what the specific component in the slot labeled ##1## transforms to under the coordinate transformation. This is why we say ##dx^{\mu}## transforms like a vector because the individual dual basis vectors transform in a similar way to the components of vectors but of course they aren't the same thing as geometric objects. In other words the ##V^{\mu}## are numbers with respect to a basis but the ##dx^{\mu}## are dual basis vectors.
     
  7. Jul 1, 2013 #6

    D H

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    You are perhaps missing the point by thinking in terms of cartesian coordinates. There is no distinction between covariance and contravariance with Cartesian tensors (which is perhaps a misnomer). There's a big difference with generalized coordinates. For example the gradient of a scalar function is a covariant tensor. It doesn't "transform like a vector" in general. It transforms like a covector.
     
  8. Jul 2, 2013 #7
    :cry: I am not getting my point across. I am not after the difference between vector and covector. Nor am I after the statement "transforms like a vector" for the dx. The one and only thing I want to understand pertains to the equality
    [tex]V'^{\mu} = \frac{\partial x'^{\mu}}{\partial x^{\nu}}V^{\nu}[/tex]

    I can take this as the definition of [itex]V'^{\mu}[/itex]. That would be math, not physics. That would be trivial. There would be nothing that I would not understand.

    But in Physics this does not seem to be a definition. It is rather a test saying: if a [itex]V^\nu[/itex] fulfills this equality we call it a vector. But for this to be non-trivial, I need an independent derivation or definition of [itex]V'^\mu[/itex], maybe like:

    1. Let the unprimed coordinate system be polar coordinates. Let the primed coordinate system be cartesian. We restrict ourselves to two dimensions for now.
    2. In the unprimed coordinate system I make a measurement at [itex](r,\theta)[/itex]. My assumption is, that I can describe my measurement as components with regard to the unit vectors of the unprimed coordinate system. (Is this a valid assumption?) My measurement results in [itex]V(r,\theta)=(l,\xi)[/itex].
    3. In the primed coordinate system I make a measurement at the same physical point in space, namely at [itex](x',y')=(r\cos(\theta),r\sin(\theta))[/itex], and get [itex]V'(x',y')=(a,b)[/itex]. Again I assume that I can describe my measurement by means of the primed coordinate system.
    Note that I did not transform the measured values from one coordinate system to the other. I just did my measurement twice, once with the measurement devices valid in the unprimed coordinate system, then with the measurement devices of the primed coordinate system.

    This would give me an independent physical derivation/definition of [itex]V'^\mu[/itex]. And now it would make sense to say that [itex](l,\xi)[/itex] is a vector, if it fulfills [itex](a,b)=S\cdot(l,\xi)[/itex], where [itex]S^\mu_\nu=\frac{\partial x'^{\mu}}{\partial x^{\nu}}[/itex].

    Is this the story?
     
    Last edited: Jul 2, 2013
  9. Jul 2, 2013 #8

    WannabeNewton

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    Coordinate components do not have independent geometric meaning; they only have meaning in a specific coordinate system. It makes no sense to say coordinate components of a vector have independent physical/geometric meaning as they are inextricably linked to a coordinate chart. That is not how a vector is defined in differential geometry. We never define vectors in terms of anything related to coordinates; a vector has an independent geometric definition as I said in post #2 and D H stated in post #6. Take a look at the link I gave and this as well: https://www.physicsforums.com/showthread.php?t=689904&page=2
     
  10. Jul 2, 2013 #9

    D H

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    You're getting closer.

    Suppose there exists some multidimensional attribute V of some physical event, and suppose you have two different coordinate systems in which you can express V. Furthermore, suppose that those coordinate systems are such that you can distinguish covariant from contravariant tensors. (In other words, those coordinate systems something other than two sets of cartesian coordinates that are related via a proper rotation.)

    In general, there are three possible outcomes for the relation between V and V', the representations of that attribute in the two coordinate systems:
    1. [itex]V^{'i} = \frac {\partial x^{'i}}{\partial x^j} V^j[/itex]
    2. [itex]V_i^{'} = \frac {\partial x^j}{\partial x^{'i}} V_j[/itex]
    3. Neither of the above.

    If case 1 is true, the quantity in question "transforms like a vector" (better: is contravariant). If case 2 is true, the quantity in question "transforms like a covector" (better: is covariant). If neither is true (i.e., it's case 3), the quantity in question does not have a tensorial quality. Just because the quantity has n attributes does not mean it is a vector or a rank 1 tensor.
     
  11. Jul 7, 2013 #10
    I hope so:approve:

    Just to be sure. When you say [itex]V[/itex] is a multidimensional attribute of an event that can be expressed in the the different coordinate systems, then "expressing" means that [itex]V[/itex] can be written as a bunch of numbers [itex]V^j[/itex], and that these numbers are components with regard to the basis vectors of the unprimed coordinate system. (And similar for the [itex]V'^j[/itex] in the primed system.)

    The reason I am so unsure about all this is that I tried, as suggested in the original post, to see which kind of number pairs transform like a contravariant vector between cartesian and polar coordinates. And there seem to be surprisingly few.
     
    Last edited: Jul 7, 2013
  12. Jul 7, 2013 #11

    Dale

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    I don't understand the distinction you are trying to draw between a test and a definition. If X is defined by some definition then to test whether some Y is an X all you need to do is see if the definition of X is true for Y. I.e. a definition is a test.

    EDIT: I suppose the reverse is not always true. So you could have a test which is different from the definition, a sufficient condition.
     
    Last edited: Jul 7, 2013
  13. Jul 7, 2013 #12
    I can measure the length of a stick in England and in Germany, so there are two ways to get hold of the length of the stick. In England it could be v=1 inch, in Germany v'=25.4mm. Mathematically it is trivial that I can multiply any number of inch by 25.4mm/inch to get the corresponding mm. Physically it is, however, a not completely trivial statement that the size of the stick is the same in England and in Germany except for the necessary transformation of coordinates.

    In the same way I can multiply any n-tupel v with the partial derivatives between two coordinate systems to get another n-tupel v'. I can call the computed v' or actually the pair (v,v') anything I like, for example a bombombang. But only if a given physical objects exhibits v as measurement values in one coordinate system and v' in the other, it is called a vector or covector, depending on the partial derivatives used. (So much I learned in this thread.-)
     
  14. Jul 7, 2013 #13

    WannabeNewton

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    I'm not sure what you mean by "multidimensional event" but yes you have the right idea.
     
  15. Jul 7, 2013 #14

    Dale

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    OK, but what is the distinction you are making between a definition and a test?
     
  16. Jul 7, 2013 #15

    wle

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    You can derive this rule from the way the basis vectors transform when you change from one coordinate system to another.

    A very common convention from differential geometry, which turns out to be particularly elegant and convenient once you're used to it, is to identify the basis vectors in some coordinate system [itex]x^{\mu}[/itex] with the partial derivatives [itex]\frac{\partial}{\partial x^{\mu}}[/itex]. So, using this convention, you'd write a general vector (at a point) as something like
    [tex]V = V^{\mu} \frac{\partial}{\partial x^{\mu}} \,.[/tex]
    The partial derivatives change between coordinate systems according to the chain rule:
    [tex]\frac{\partial}{\partial x'^{\mu}} = \frac{\partial x^{\nu}}{\partial x'^{\mu}} \frac{\partial}{\partial x^{\nu}} \,.[/tex]
    This is where the component transformation rule you're quoting above comes from.
     
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