Differentials - is this valid or just sloppy but right?

  • Thread starter pellman
  • Start date
666
2

Main Question or Discussion Point

The proper time is defined by

[tex]d\tau^2=g_{\mu\nu}dx^\mu dx^\nu[/tex]

Suppose we have flat space time with one space dimension.

[tex]d\tau=\sqrt{dt^2-dx^2}[/tex]
[tex]=dt\sqrt{1-\frac{(dx^2)}{(dt^2)}}[/tex]
[tex]=dt\sqrt{1-\left(\frac{dx}{dt}\right)^2}[/tex]

Can this be rigorous?
 

Answers and Replies

robphy
Science Advisor
Homework Helper
Insights Author
Gold Member
5,399
676
Regard [itex]x[/itex] as [itex]x(t)[/itex], then [tex]d(\ x(t)\ )=\left(\frac{dx}{dt}\right)dt[/tex]
 
979
1
Nothing wrong with juggling differentials --- just realise that they don't always obey the same algebraic rules as reals or complex numbers, which is sensible, since they're not. See non-standard analysis: http://en.wikipedia.org/wiki/Non-standard_analysis
 
666
2
And something I only just noticed. From [tex]d\tau=\sqrt{dt^2-dx^2}[/tex] we see that one cannot express [tex]d\tau[/tex] in terms of first order changes in t and x. That is, there are no numbers A and B such that [tex]d\tau=Adt+Bdx[/tex]. The slope of the graph of [tex]\sqrt{x^2}[/tex] is singular at x=0. There is probably something significant for the proper time concept here.
 
410
0
The proper time is defined by

[tex]d\tau^2=g_{\mu\nu}dx^\mu dx^\nu[/tex]

Suppose we have flat space time with one space dimension.

[tex]d\tau=\sqrt{dt^2-dx^2}[/tex]
[tex]=dt\sqrt{1-\frac{(dx^2)}{(dt^2)}}[/tex]
[tex]=dt\sqrt{1-\left(\frac{dx}{dt}\right)^2}[/tex]

Can this be rigorous?
Let's just say that [tex]d\tau=\sqrt{dt^2-dx^2}[/tex] is short hand for the volume form on the worldline, i.e., it has no meaning until you integrate it. Let us write

[tex]{\omega} = \sqrt{ \left( \frac{dt}{d\lambda}\right) ^2- \left(\frac{dx}{d\lambda}\right)^2} d\lambda[/tex]

Here [tex]d\lambda[/tex] is a oneform. If you integrate this quantity over the parameter [tex]\lambda[/tex] you will get the volume of the wordline, or, more informally it's arclength.
 
2,945
0
The proper time is defined by

[tex]d\tau^2=g_{\mu\nu}dx^\mu dx^\nu[/tex]

Suppose we have flat space time with one space dimension.

[tex]d\tau=\sqrt{dt^2-dx^2}[/tex]
[tex]=dt\sqrt{1-\frac{(dx^2)}{(dt^2)}}[/tex]
[tex]=dt\sqrt{1-\left(\frac{dx}{dt}\right)^2}[/tex]

Can this be rigorous?
Do you mean to ask if it is rigorous? If so then yes, it is rigorous.

Pete
 
410
0
Do you mean to ask if it is rigorous? If so then yes, it is rigorous.

Pete
Well... besides not being well defined =) I don't think you'll find any mathematician that would put their name to it =). It's just shorthand.
 

Related Threads for: Differentials - is this valid or just sloppy but right?

  • Last Post
Replies
4
Views
2K
Replies
10
Views
2K
  • Last Post
Replies
6
Views
2K
Replies
5
Views
682
Replies
8
Views
596
  • Last Post
Replies
5
Views
1K
Replies
20
Views
695
Replies
24
Views
2K
Top