Dividing two differentials gives a total derivative how?

In summary, the differentials of a function are a 1-form related to the operation of total derivative, and it is possible to divide a tensor by another tensor to obtain a scalar.
  • #1
ArthurB
17
0
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]
 
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  • #2
ArthurB said:
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?
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.
 
  • #3
Fredrik said:
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.

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.
 
  • #4
ArthurB said:
how can I divide a tensor by another tensor and obtain a scalar?
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...
 
  • #5
ArthurB said:
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.
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.

ArthurB said:
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.
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.
 
  • #6
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.
 
  • #7
mathwonk said:
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.

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?
 
  • #8
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).
 
  • #9
kith said:
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).
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).
 
  • #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
 
  • #11
Fredrik said:
But I don't understand what Hurkyl has in mind.

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.
 
  • #12
ArthurB said:
About the whole "differential-derivative" stuff I think I'm figuring it out,
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]
 

FAQ: Dividing two differentials gives a total derivative how?

1. What is a total derivative?

A total derivative is a mathematical concept that represents the overall change in a function with respect to its input variables. It takes into account all possible changes in the function, including changes in multiple variables simultaneously.

2. How is the total derivative calculated?

The total derivative is calculated using the chain rule, which involves taking the partial derivatives of the function with respect to each input variable and multiplying them by the corresponding changes in those variables.

3. What is the relationship between dividing two differentials and finding the total derivative?

Dividing two differentials is a way to find the total derivative of a function. It involves dividing the change in the function by the change in the input variables, and results in the total derivative.

4. Can the total derivative be used to approximate changes in a function?

Yes, the total derivative can be used to approximate the change in a function for small changes in the input variables. This is known as the linear approximation or the tangent line approximation.

5. How is the total derivative used in real-world applications?

The total derivative is used in many fields of science and engineering, including physics, economics, and biology. It is used to model and predict changes in complex systems, and is essential for understanding and solving differential equations.

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