# Differentials - is this valid or just sloppy but right?

## Main Question or Discussion Point

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?

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robphy
Homework Helper
Gold Member
Regard $x$ as $x(t)$, then $$d(\ x(t)\ )=\left(\frac{dx}{dt}\right)dt$$

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

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.

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.

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

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.