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Dividing two differentials gives a total derivative how?

  1. Sep 22, 2011 #1
    Hello
    I am studying some differential geometry. I think I have understood the meaning of "differential" of a function:

    [itex] \text{d}f (V) = V(f) [/itex]

    It is a 1-form, an operator that takes a vector and outputs a real number.
    But how is it related to the operation of "total derivative" ?
    For example, in special relativity we can perform a boost with velocity [itex]v[/itex] along the [itex]x[/itex] axis and transform the differentials like this:
    [tex]
    \text{d}x' = \gamma (\text{d}x-v \, \text{d}t) \qquad \text{d}y'=\text{d}y
    [/tex]
    [tex]
    \text{d}t' = \gamma\left(\text{d}t-v\, \text{d}x\right) \qquad \text{d}z'=\text{d}z
    [/tex]
    then the [itex]x[/itex] component of the velocity
    [tex]
    u^1=\frac{dx}{dt}
    [/tex]
    in the new frame will have the expression
    [tex]
    u'^1=\frac{dx'}{dt'} = ? = \frac{\text{d}x'}{\text{d}t'}= \frac{u^1 -v}{1-{u^1 v}}
    [/tex]
    but how is it mathematically possible? a differential is a tensor, not a scalar, how can I divide a tensor by another tensor and obtain a scalar?

    P.S. notice the difference between [itex]\text{d}[/itex] and [itex]d[/itex]
     
    Last edited: Sep 22, 2011
  2. jcsd
  3. Sep 22, 2011 #2

    Fredrik

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    Note that if you just drop the d's from this calculation, all the calculation does is to determine the velocity associated with the (straight) world line from (0,0) to (t,x) in another inertial coordinate system. (A Lorentz transformation leaves (0,0) unchanged and takes (t,x) to (t',x'), so x/t becomes x'/t'). I would say that the calculation with the d's in place is exactly the same thing, with a weird notation for the variables. It doesn't have anything to do with 1-forms.
     
  4. Sep 22, 2011 #3
    It's conceptually the same thing only if [itex]u[/itex] is constant I think. If the world line of the particle is not straight I don't think it's correct to use the transformations without the d's.
    And those must be 1-forms, since Lorentz transformations leave the metric invariant:
    [tex]
    \text{d}t^2 - \text{d}x^2 - \text{d}y^2 - \text{d}z^2 =
    \text{d}t'^2 - \text{d}x'^2 - \text{d}y'^2 - \text{d}z'^2
    [/tex]
    in which the notation means, for example
    [tex]
    \text{d}y^2 = \text{d}y \otimes \text{d}y
    [/tex]
    that is a tensor product of two 1-forms.
     
  5. Sep 22, 2011 #4

    Hurkyl

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    You could observe that one tensor is a scalar multiple of another tensor. This happens somewhat frequently when you have a one-dimensional space of tensors....
     
  6. Sep 22, 2011 #5

    Fredrik

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    Right, but you can take your dx, dy, etc to be coordinate differences between two points on a tangent to the world line. In that case, you're just dealing with numbers, not differentials.

    Sure, the metric can be expressed as [itex]g=g_{\mu\nu}\text{d}x^\mu\otimes\text{d}y^\nu[/itex], but that doesn't mean that the dx, dy, etc in the velocity calculation must be considered differentials.


    Since [itex]\text{d}x^\mu/\text{d}x^0[/itex] doesn't make sense, I guess the real question is if velocity components can be calculated as [tex]\frac{\text{d}x^\mu(v)}{\text{d}x^0(v)}=\frac{v(x^\mu)}{v(x^0)}=\frac{v^\mu}{v^0},[/tex] where v is some tangent vector. There's obviously a choice of v involved. I suggest [itex]v=\dot C(s)[/itex], the tangent vector at C(s) of the world line C, defined by [tex]\dot C(s)f=(f\circ C)'(s).[/tex] Its components in the coordinate system x are [tex]\dot C^\mu(s)=\dot C(s)x^\mu=(x^\mu\circ C)'(s)[/tex] (The first equality defines the notation on the left). So [tex]\frac{\text{d}x^\mu(\dot C(s))}{\text{d}x^0(\dot C(s))} =\frac{(x^\mu\circ C)'(s)}{(x^0\circ C)'(s)}.[/tex] The numerator is a velocity component. The whole thing isn't.
     
  7. Sep 22, 2011 #6

    mathwonk

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    as others have said, v/w = t means that tw = v, so if it is possible for a tensor to be a scalar multiple of another tensor, then it is equally possible for the quotient of those tensors to be a scalar.
     
  8. Sep 23, 2011 #7
    This seems reasonable, but as you know tensors can be represented by matrices, so are you saying that if a matrix equals a scalar multiplied by another matrix then it implies that the quotient of those matrices is a scalar?
    [tex]
    M = a N \Rightarrow M/N = a ?
    [/tex]
    My only problem is that the operator "/" usually takes two scalars as input and then outputs another scalar, right? How can you feed it matrices?
    I'm not saying it's wrong, just that probably we can't call it "quotient", and that the operation we are doing is not "dividing", it must be something else, but what?
     
  9. Sep 23, 2011 #8

    kith

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    Just like 1/x or x-1 is the inverse of a number x with respect to multiplication, A-1 is the inverse of a matrix A with respect to matrix multiplication.

    Wether you use A-1 or 1/A is a matter of convention (and like for the numbers, A/B is just a shorthand notation for A * 1/B).
     
  10. Sep 23, 2011 #9

    Fredrik

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    In this case, we're talking about cotangent vectors, so the matrices would be 1×4. I think what mathwonk is saying is just that the notation v/w=t can be defined to mean v=wt. But I don't understand what Hurkyl has in mind. There's a theorem that says that two (bounded) linear functionals on the same (normed) vector space have the same kernel if and only if one of them is a number times the other. So for [itex]\text{d}x^3[/itex] to be a number times [itex]\text{d}x^0[/itex], the "coordinate system" would have to be such the 3 component of a tangent vector is 0 if and only if the 0 component is 0. Wouldn't the 3 axis have to coincide with the 0 axis for that to happen? (That would mean that x doesn't qualify as a coordinate system at all, let alone as an inertial coordinate system in SR).
     
  11. Sep 23, 2011 #10
    I hope what I'm saying is not too trivial, but it took me a while to figure this out myself:

    but the differential of f is a linear map

    that describes the local change of the values of f , e.g., the differential of f(x)=x^2

    is 2xdx , so that the line with slope 2xo is the line that best approximates (in a delta-eps.

    sense) the change of f(x) in a 'hood (neighborhood) of xo. This is the simplest case, i.e.,

    of maps from Reals to Reals. In higher-dimensional cases, you have many more directions

    to take care of, so you deal with a full tangent space . So your differential gives you the

    best approximation in a given direction , and this direction is described by a vector in

    the tangent space. And, a technical point about M/N=a : the quotient of objects in

    a structure should be an object in that structure , so M/N =a would mean the scalar

    matrix a, not the number a, since M,N are matrices.

    So, formally, as a 1-form, the

    map assigns to each point
     
  12. Sep 24, 2011 #11
    I think he simply said that in a 1-dimensional vector space every vector is a multiple of the other ones. Right. But aren't we reasoning in 4 dimensions?
    About the whole "differential-derivative" stuff I think I'm figuring it out, thanks everyone for the help.
     
  13. Sep 24, 2011 #12

    Fredrik

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    I actually didn't quite understand what it was you wanted to know there. Was it that the formula [tex]df=\frac{\partial f}{\partial x^i}dx^i[/tex] can be interpreted as [tex]\text{d}f|_p(v)=v(f)=v^i\frac{\partial}{\partial x^i}\bigg|_p f=\frac{\partial f(p)}{\partial x^i}v^i\ ?[/tex]
     
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