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Qmechanic
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# Is the Invariant interval S$S$ between two points independent of the path taken?

Due to a misunderstanding of what I was asking, I'm re-asking this question (Is the Invariant interval S between the singularity and the present, the same for any point in space in an FLRW universe?Is the Invariant interval S between the singularity and the present, the same for any point in space in an FLRW universe?) in a much more general sense.

The invariant interval is defined as:

$$ds=\sqrt{g_{\mu\nu}dx^{\mu}dx^{\nu}}$$

or rather (as per Hamilton's General Relativity, Black holes, and cosmology equation 2.13):

$$ds^{2}=g_{\mu\nu}dx^{\mu}dx^{\nu}$$

Because it is a scalar $$ds$$ may be written as an exact differential form (As per Hamilton equation 2.11 referenced above):

$$ds=\frac{\partial s}{\partial x_{\mu}}dx_{\mu}$$

Where summation over \mu is implied. Note that (for geometric consistency) $$\frac{\partial s}{\partial x_{\mu}}$$ can be identified with the metric:

$$ds^{2}=\left(\frac{\partial s}{\partial x^{\mu}}dx^{\mu}\right)\left(\frac{\partial s}{\partial x^{\nu}}dx^{\nu}\right)$$

$$=\left(\frac{\partial s}{\partial x^{\mu}}\frac{\partial s}{\partial x^{\nu}}+\frac{\partial s}{\partial x^{\nu}}\frac{\partial s}{\partial x^{\mu}}\right)dx^{\mu}dx^{\nu}$$

Which implies that:

$$\left(\frac{\partial s}{\partial x^{\mu}}\frac{\partial s}{\partial x^{\nu}}+\frac{\partial s}{\partial x^{\nu}}\frac{\partial s}{\partial x^{\mu}}\right)=\left\{ \frac{\partial s}{\partial x^{\mu}},\frac{\partial s}{\partial x^{\nu}}\right\} =g_{\mu\nu}$$

But this expression is familiar, the generalized gamma matrices $$\gamma^{\mu}$$ are defined by:

$$\left\{ \gamma_{\mu},\gamma_{\nu}\right\} =\gamma_{\mu}\gamma_{\nu}+\gamma_{\nu}\gamma_{\mu}=2g_{\mu\nu}$$

Which means that $$\frac{\partial s}{\partial x_{\mu}}$$ can be identified with the generalized gamma matrices:

$$\frac{\partial s}{\partial x^{\mu}}=\frac{1}{\sqrt{2}}\gamma_{\mu}$$

The second equation can also be written as:

$$ds=\frac{\partial s}{\partial x_{\mu}}dx_{\mu}=\overrightarrow{\nabla}s\cdot d\overrightarrow{r}$$

Integrating now over some arbitrary interval $$\{a,b\}$$:

$$S=\intop_{a}^{b}\overrightarrow{\nabla}s\cdot d\overrightarrow{r}$$

Via the fundamental theorem of calculus, it is clear that the interval is independent of the path taken between the points $$\{a,b\}$$. Did I mess this up somewhere?

EDIT: Here's an approach not assuming s is an exact differential (as per the the objections voiced below)

The invariant interval is defined as:

$$ds=\sqrt{g_{\mu\nu}dx^{\mu}dx^{\nu}}$$

or rather:

If we wish, this can be rearranged as:

$$0=dx\cdot g\cdot dx-ds^{2}=dx^{\mu}g_{\mu\nu}dx^{\nu}$$

One can write the metric tensor in terms of local basis:

$$g_{\mu\nu}=e_{\mu}\bullet e_{\nu}$$

Where $$\cdot$$ denotes the standard dot product and $$\bullet$$ the tensor product. Used in the preceding equation, the above yields:

$$0=\left(e_{\mu}dx^{\mu}\right)\bullet\left(e_{\nu}dx^{\nu}\right)-ds^{2}$$

(note $$dx\cdot e=dx^{\mu}e_{\mu}$$) This can be simply factored to obtain:

$$0=\left(e_{\mu}dx^{\mu}-ds\right)\bullet\left(e_{\nu}dx^{\nu}+ds\right)$$ Which, for a given metric, gives two different solutions to $$ds$$.

$$0=\left(e_{\mu}dx^{\mu}-ds\right)\qquad0=\left(e_{\nu}dx^{\nu}+ds\right)$$

Algebraically, this corresponds to the Clifford algebra as we have the relationship:

$$\left\{ e_{\mu},e_{\nu}\right\} =e_{\mu}e_{\nu}+e_{\nu}e_{\mu}=g_{\mu\nu}$$

Which means that $$e_{\mu}$$ can be identified with the generalized gamma matrice $$\gamma_{\mu}$$:

$$e_{\mu}=\frac{1}{\sqrt{2}}\gamma_{\mu}$$

Taking either solution for ds individually, the integral for s between two nearby points appears to be independent of the path taken. Note that either solution to $$s$$ individually could not be considered as proper time between events, but is simply a geometrical invariant.

The relationship between our starting equation and our two solutions now is entirely analogous to that between the Klein gordon equation and the Dirac equation. Solutions to the former are not necessarily solutions to the latter. Apparently no-one liked my first question using differential forms, so I wrote it up this way.

Also, if it eases concerns of undefined intervals, one can simply consider a flat space, since this argument itself is general.

replaced http://physics.stackexchange.com/ with https://physics.stackexchange.com/

Due to a misunderstanding of what I was asking, I'm re-asking this question (Is the Invariant interval S between the singularity and the present, the same for any point in space in an FLRW universe?Is the Invariant interval S between the singularity and the present, the same for any point in space in an FLRW universe?) in a much more general sense

The invariant interval is defined as:

$$ds=\sqrt{g_{\mu\nu}dx^{\mu}dx^{\nu}}$$

or rather (as per Hamilton's General Relativity, Black holes, and cosmology equation 2.13):

$$ds^{2}=g_{\mu\nu}dx^{\mu}dx^{\nu}$$

Because it is a scalar $$ds$$ may be written as an exact differential form (As per Hamilton equation 2.11 referenced above):

$$ds=\frac{\partial s}{\partial x_{\mu}}dx_{\mu}$$

Where summation over \mu is implied. Note that (for geometric consistency) $$\frac{\partial s}{\partial x_{\mu}}$$ can be identified with the metric:

$$ds^{2}=\left(\frac{\partial s}{\partial x^{\mu}}dx^{\mu}\right)\left(\frac{\partial s}{\partial x^{\nu}}dx^{\nu}\right)$$

$$=\left(\frac{\partial s}{\partial x^{\mu}}\frac{\partial s}{\partial x^{\nu}}+\frac{\partial s}{\partial x^{\nu}}\frac{\partial s}{\partial x^{\mu}}\right)dx^{\mu}dx^{\nu}$$

Which implies that:

$$\left(\frac{\partial s}{\partial x^{\mu}}\frac{\partial s}{\partial x^{\nu}}+\frac{\partial s}{\partial x^{\nu}}\frac{\partial s}{\partial x^{\mu}}\right)=\left\{ \frac{\partial s}{\partial x^{\mu}},\frac{\partial s}{\partial x^{\nu}}\right\} =g_{\mu\nu}$$

But this expression is familiar, the generalized gamma matrices $$\gamma^{\mu}$$ are defined by:

$$\left\{ \gamma_{\mu},\gamma_{\nu}\right\} =\gamma_{\mu}\gamma_{\nu}+\gamma_{\nu}\gamma_{\mu}=2g_{\mu\nu}$$

Which means that $$\frac{\partial s}{\partial x_{\mu}}$$ can be identified with the generalized gamma matrices:

$$\frac{\partial s}{\partial x^{\mu}}=\frac{1}{\sqrt{2}}\gamma_{\mu}$$

The second equation can also be written as:

$$ds=\frac{\partial s}{\partial x_{\mu}}dx_{\mu}=\overrightarrow{\nabla}s\cdot d\overrightarrow{r}$$

Integrating now over some arbitrary interval $$\{a,b\}$$:

$$S=\intop_{a}^{b}\overrightarrow{\nabla}s\cdot d\overrightarrow{r}$$

Via the fundamental theorem of calculus, it is clear that the interval is independent of the path taken between the points $$\{a,b\}$$. Did I mess this up somewhere?

EDIT: Here's an approach not assuming s is an exact differential (as per the the objections voiced below)

The invariant interval is defined as:

$$ds=\sqrt{g_{\mu\nu}dx^{\mu}dx^{\nu}}$$

or rather:

If we wish, this can be rearranged as:

$$0=dx\cdot g\cdot dx-ds^{2}=dx^{\mu}g_{\mu\nu}dx^{\nu}$$

One can write the metric tensor in terms of local basis:

$$g_{\mu\nu}=e_{\mu}\bullet e_{\nu}$$

Where $$\cdot$$ denotes the standard dot product and $$\bullet$$ the tensor product. Used in the preceding equation, the above yields:

$$0=\left(e_{\mu}dx^{\mu}\right)\bullet\left(e_{\nu}dx^{\nu}\right)-ds^{2}$$

(note $$dx\cdot e=dx^{\mu}e_{\mu}$$) This can be simply factored to obtain:

$$0=\left(e_{\mu}dx^{\mu}-ds\right)\bullet\left(e_{\nu}dx^{\nu}+ds\right)$$ Which, for a given metric, gives two different solutions to $$ds$$.

$$0=\left(e_{\mu}dx^{\mu}-ds\right)\qquad0=\left(e_{\nu}dx^{\nu}+ds\right)$$

Algebraically, this corresponds to the Clifford algebra as we have the relationship:

$$\left\{ e_{\mu},e_{\nu}\right\} =e_{\mu}e_{\nu}+e_{\nu}e_{\mu}=g_{\mu\nu}$$

Which means that $$e_{\mu}$$ can be identified with the generalized gamma matrice $$\gamma_{\mu}$$:

$$e_{\mu}=\frac{1}{\sqrt{2}}\gamma_{\mu}$$

Taking either solution for ds individually, the integral for s between two nearby points appears to be independent of the path taken. Note that either solution to $$s$$ individually could not be considered as proper time between events, but is simply a geometrical invariant.

The relationship between our starting equation and our two solutions now is entirely analogous to that between the Klein gordon equation and the Dirac equation. Solutions to the former are not necessarily solutions to the latter. Apparently no-one liked my first question using differential forms, so I wrote it up this way.

Also, if it eases concerns of undefined intervals, one can simply consider a flat space, since this argument itself is general.

Rollback to Revision 7
ACuriousMind
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I'll explain it this time with minimal math. ForDue to a chosen setmisunderstanding of coordinateswhat I was asking, the interval between two points appears to be independent of the path taken. The twin paradox is a perfect exampleI'm re-asking this question (for flat spaceIs the Invariant interval S between the singularity and the present, the same for any point in space in an FLRW universe?). in a much more general sense

Consider one observer at “rest”, whilst the other one speeds off. Considering just the inertialThe invariant interval is defined as:

$$ds=\sqrt{g_{\mu\nu}dx^{\mu}dx^{\nu}}$$

or rather (stationary in free space) coordinate systemas per Hamilton's General Relativity, the (stationaryBlack holes, and cosmology equation 2.13) observer witnesses:

$$ds^{2}=g_{\mu\nu}dx^{\mu}dx^{\nu}$$

Because it is a scalar (proper) time pass$$ds$$ may be written as an exact differential form $$\Delta\tau$$(As per Hamilton equation 2.11 referenced above):

Let us denote the interval$$ds=\frac{\partial s}{\partial x_{\mu}}dx_{\mu}$$

Where summation over \mu is implied. Note that ($$S$$for geometric consistency) witnessed by the observer to have units of distance. $$S^{2}=c^{2}\Delta\tau$$$$\frac{\partial s}{\partial x_{\mu}}$$ can be identified with the metric:

For simplicity let us assume$$ds^{2}=\left(\frac{\partial s}{\partial x^{\mu}}dx^{\mu}\right)\left(\frac{\partial s}{\partial x^{\nu}}dx^{\nu}\right)$$

$$=\left(\frac{\partial s}{\partial x^{\mu}}\frac{\partial s}{\partial x^{\nu}}+\frac{\partial s}{\partial x^{\nu}}\frac{\partial s}{\partial x^{\mu}}\right)dx^{\mu}dx^{\nu}$$

Which implies that:

$$\left(\frac{\partial s}{\partial x^{\mu}}\frac{\partial s}{\partial x^{\nu}}+\frac{\partial s}{\partial x^{\nu}}\frac{\partial s}{\partial x^{\mu}}\right)=\left\{ \frac{\partial s}{\partial x^{\mu}},\frac{\partial s}{\partial x^{\nu}}\right\} =g_{\mu\nu}$$

But this expression is familiar, the other observer (twin if you will) to travel some arbitrarygeneralized gamma matrices (obviously not traveling faster than light) distance$$\gamma^{\mu}$$ are defined by:

$$\left\{ \gamma_{\mu},\gamma_{\nu}\right\} =\gamma_{\mu}\gamma_{\nu}+\gamma_{\nu}\gamma_{\mu}=2g_{\mu\nu}$$

Which means that $$\Delta x$$ before the two observers position again coincides. Then clearly in$$\frac{\partial s}{\partial x_{\mu}}$$ can be identified with the chosen coordinate systemgeneralized gamma matrices:

$$S^{2}=c^{2}\Delta\tau^{2}=c^{2}\varDelta t^{2}-\triangle x^{2}$$$$\frac{\partial s}{\partial x^{\mu}}=\frac{1}{\sqrt{2}}\gamma_{\mu}$$

WhereThe second equation can also be written as:

$$ds=\frac{\partial s}{\partial x_{\mu}}dx_{\mu}=\overrightarrow{\nabla}s\cdot d\overrightarrow{r}$$

Integrating now over some arbitrary interval $$\Delta t$$ is the time measured by$$\{a,b\}$$:

$$S=\intop_{a}^{b}\overrightarrow{\nabla}s\cdot d\overrightarrow{r}$$

Via the traveling observer. Obviously, for a given coordinate choicefundamental theorem of calculus, it is clear that the invariant interval is independent of the path taken between two points since the motion of the traveling observer was entirely arbitrarypoints $$\{a,b\}$$. Did I mess this up somewhere?

Generalizing toEDIT: Here's an FLRWapproach not assuming s is an exact differential (Friedmann-Lemaitre-Robertson-Walkeras per the the objections voiced below) universe,

The invariant interval is defined as:

$$ds=\sqrt{g_{\mu\nu}dx^{\mu}dx^{\nu}}$$

or rather:

If we know that all comoving frames will measurewish, this can be rearranged as:

$$0=dx\cdot g\cdot dx-ds^{2}=dx^{\mu}g_{\mu\nu}dx^{\nu}$$

One can write the samemetric tensor in terms of local basis:

$$g_{\mu\nu}=e_{\mu}\bullet e_{\nu}$$

Where (proper) time$$\cdot$$ denotes the standard dot product and $$\bullet$$ the tensor product. Used in the preceding equation, the above yields:

$$0=\left(e_{\mu}dx^{\mu}\right)\bullet\left(e_{\nu}dx^{\nu}\right)-ds^{2}$$

(intervalnote $$dx\cdot e=dx^{\mu}e_{\mu}$$) regardless of position.This can be simply factored to obtain:

It follows that$$0=\left(e_{\mu}dx^{\mu}-ds\right)\bullet\left(e_{\nu}dx^{\nu}+ds\right)$$ Which, for any particular choice of coordinate framea given metric, an observer will witness all othergives two different solutions to (observable) bodies in$$ds$$.

$$0=\left(e_{\mu}dx^{\mu}-ds\right)\qquad0=\left(e_{\nu}dx^{\nu}+ds\right)$$

Algebraically, this corresponds to the universeClifford algebra as having traversedwe have the same interval.relationship:

I got a lot of flack$$\left\{ e_{\mu},e_{\nu}\right\} =e_{\mu}e_{\nu}+e_{\nu}e_{\mu}=g_{\mu\nu}$$

Which means that $$e_{\mu}$$ can be identified with the generalized gamma matrice $$\gamma_{\mu}$$:

$$e_{\mu}=\frac{1}{\sqrt{2}}\gamma_{\mu}$$

Taking either solution for this questionds individually, but it seems pretty elementarythe integral for s between two nearby points appears to be independent of the path taken. Note that either solution to $$s$$ individually could not be considered as proper time between events, apparently I hadbut is simply a geometrical invariant.

The relationship between our starting equation and our two solutions now is entirely analogous to specify that I'm onlybetween the Klein gordon equation and the Dirac equation. Solutions to the former are not necessarily solutions to the latter. Apparently no-one liked my first question using one setdifferential forms, so I wrote it up this way.

Also, if it eases concerns of coordinates (though isn't that typical?) Isundefined intervals, one can simply consider a flat space, since this right guys? maybe I'm missing somethingargument itself is general.

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