Differentials - is this valid or just sloppy but right?

[pellman]

The proper time is defined by

$$d\tau^2=g_{\mu\nu}dx^\mu dx^\nu$$

Suppose we have flat space time with one space dimension.

$$d\tau=\sqrt{dt^2-dx^2}$$
$$=dt\sqrt{1-\frac{(dx^2)}{(dt^2)}}$$
$$=dt\sqrt{1-\left(\frac{dx}{dt}\right)^2}$$

Can this be rigorous?

 

[robphy]

Regard $x$ as $x(t)$, then $$d(\ x(t)\ )=\left(\frac{dx}{dt}\right)dt$$

 

[genneth]

Nothing wrong with juggling differentials --- just realize 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

 

[pellman]

And something I only just noticed. From $$d\tau=\sqrt{dt^2-dx^2}$$ we see that one cannot express $$d\tau$$ in terms of first order changes in t and x. That is, there are no numbers A and B such that $$d\tau=Adt+Bdx$$. The slope of the graph of $$\sqrt{x^2}$$ is singular at x=0. There is probably something significant for the proper time concept here.

 

[lbrits]

The proper time is defined by

$$d\tau^2=g_{\mu\nu}dx^\mu dx^\nu$$

Suppose we have flat space time with one space dimension.

$$d\tau=\sqrt{dt^2-dx^2}$$
$$=dt\sqrt{1-\frac{(dx^2)}{(dt^2)}}$$
$$=dt\sqrt{1-\left(\frac{dx}{dt}\right)^2}$$

Can this be rigorous?

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

$${\omega} = \sqrt{ \left( \frac{dt}{d\lambda}\right) ^2- \left(\frac{dx}{d\lambda}\right)^2} d\lambda$$

Here $$d\lambda$$ is a oneform. If you integrate this quantity over the parameter $$\lambda$$ you will get the volume of the wordline, or, more informally it's arclength.

 

[pmb_phy]

The proper time is defined by

$$d\tau^2=g_{\mu\nu}dx^\mu dx^\nu$$

Suppose we have flat space time with one space dimension.

$$d\tau=\sqrt{dt^2-dx^2}$$
$$=dt\sqrt{1-\frac{(dx^2)}{(dt^2)}}$$
$$=dt\sqrt{1-\left(\frac{dx}{dt}\right)^2}$$

Can this be rigorous?

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

Pete

 

[lbrits]

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.