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.