Differential equation for affine parameter

[stevendaryl]

Suppose you have a smooth parametrized path through spacetime $x^\mu(s)$. If the path is always spacelike or always timelike (meaning that $g_{\mu \nu} \dfrac{dx^\mu}{ds} \dfrac{dx^\nu}{ds}$ always has the same sign, and is never zero), then you can define a smooth function of $s$, $\tau(s) = \int \sqrt{|g_{\mu \nu} \dfrac{dx^\mu}{ds} \dfrac{dx^\nu}{ds}|}$. Then in terms of $\tau$, you can say that the parameter $s$ is affine if it satisfies:

$s = A \tau + B$

for constants $A$ and $B$. Or you can write it as a differential equation:

$\dfrac{d^2 \tau}{ds^2} = 0$

This implies a differential equation for $x^m$, via $d \tau = \sqrt{|g_{\mu \nu} dx^\mu dx^\nu|}$:

$\frac{1}{2} \partial_\lambda g_{\mu \nu} \dfrac{dx^\mu}{ds} \dfrac{dx^\nu}{ds} \dfrac{dx^\lambda}{ds} + g_{\mu \nu} \dfrac{d^2 x^\mu}{ds^2} \dfrac{dx^\nu}{ds} = 0$

This equation doesn't assume that $x^\mu(s)$ is a geodesic, it only assumes that it is either always timelike or always spacelike.

Now, my question is: Can we formulate a similar differential equation for deciding whether $s$ is affine when we relax that constraint on $x^\mu(s)$? That is, is there a differential equation describing an affine parameter for a path $x^\mu(s)$ that allows the path to be lightlike at points?

 

[stevendaryl]

Following up my own post, I see that there is a definition of an affine parameterization of a geodesic, particularly. If $x^\mu(s)$ is a geodesic, then for any parameter $s$,

$\dfrac{dx^\mu}{ds} \nabla_\mu \dfrac{dx^\nu}{ds} = f(s) \dfrac{dx^\nu}{ds}$

In the particular case where $s$ is an affine parameter, then $f(s) = 0$

This doesn't define an affine parameter, in general, but only for geodesics, but maybe that's the only case in which we care about parameterizations?

 

[PeterDonis]

This doesn't define an affine parameter, in general, but only for geodesics

Yes, and geodesics cannot change type: a geodesic that starts out spacelike (timelike, null) will always remain spacelike (timelike, null). So a geodesic can't be used to study what it looks like you want to study in the OP, i.e., curves that can change from spacelike or timelike to null, and back again.

maybe that's the only case in which we care about parameterizations?

I don't think so for parameterizations in general, since the definition of path curvature, which will be nonzero for any non-geodesic curve, requires a parameterization.

However, it might be that geodesics are the only curves for which the concept of affine parameterization is useful.

 

[vanhees71]

I think an affine parameter for a curve in a (pseudo-Riemannian) spacetime is one, for which
$$g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}=C=\text{const}.$$
In relativistic physics for a time-like curve you can use $C=c^2$. Then the affine parameter is the proper time (or better "Eigenzeit") of a particle along its world line.

For a light-like curve $C=0$, and there is no physically distinguished affine worldline parameter. Any such affine parameter is as good as any other. Time like curves usually describe "light rays" (or "naive photons", but I'd prefer to say "light rays") in the sense of geometrical optics.

For a space-like curve $C<0$. Also here, I'm not so sure, whether there is a generally valid physical interpretation of the affine parameter, for which $C=-1$ in that case. Maybe in some cases one can define it as a measure for the path length along a spatial curve (I guess it's most probably possible to interpret it in this way within a static non-rotating coordinate chart).

In relativistic mechanics the trick to always automatically describe the worldlines of massive particles with an affine paramter is to write the Lagrangian in the form (signature (+---))
$$L=-\frac{m}{2} g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu} + L_{\text{int}}(x,\dot{x}),$$
where $L_{\text{int}}$ is a 1st-rank function of $\dot{x}$, i.e.,
$$\dot{x}^{\mu} \frac{\partial L_{\text{int}}}{\partial \dot{x}^{\mu}}=L_{\text{int}}.$$
Then since $L$ is not explicitly dependent on the world-line parameter, $\lambda$, the "Hamiltonian"
$$H=\dot{x}^{\mu} \frac{\partial{L}}{\partial \dot{x}^{\mu}}-L=\frac{m}{2} g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}=\text{const}.$$
Setting $H=m c^2/2$ you choose the affine parameter as the proper time of the particle along its worldline, i.e., $\lambda=\tau$.

A geodesic in GR is simply described by the special case of $L_{\text{int}}=0$ (no other interactions than gravitation). Then you describe the motion of a test particle in the gravitational field (aka curved spacetime) of other bodies.

 

[PeterDonis]

In relativistic physics for a time-like curve you can use $C=c^2$. Then the affine parameter is the proper time (or better "Eigenzeit") of a particle along its world line.

Or you can use $C = 1$. Or any other constant. This is just a change in your choice of units for time.

For a space-like curve $C<0$. Also here, I'm not so sure, whether there is a generally valid physical interpretation of the affine parameter, for which $C=-1$ in that case.

Sure there is. Spacelike curves work like timelike curves: the physically relevant affine parameter is arc length along the curve (measured with a ruler instead of a clock). Different constant choices of $C$ correspond to different choices of units for distance.

 

[stevendaryl]

In relativistic mechanics the trick to always automatically describe the worldlines of massive particles with an affine paramter is to write the Lagrangian in the form (signature (+---))
$$L=-\frac{m}{2} g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu} + L_{\text{int}}(x,\dot{x}),$$
where $L_{\text{int}}$ is a 1st-rank function of $\dot{x}$, i.e.,
$$\dot{x}^{\mu} \frac{\partial L_{\text{int}}}{\partial \dot{x}^{\mu}}=L_{\text{int}}.$$
Then since $L$ is not explicitly dependent on the world-line parameter, $\lambda$, the "Hamiltonian"
$$H=\dot{x}^{\mu} \frac{\partial{L}}{\partial \dot{x}^{\mu}}-L=\frac{m}{2} g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}=\text{const}.$$
Setting $H=m c^2/2$ you choose the affine parameter as the proper time of the particle along its worldline, i.e., $\lambda=\tau$.

My issue is how to define the affine parameter for null paths.

A geodesic described using an affine parameter $s$ will satisfy

Affine parameter: $\frac{d^2 x^\mu}{ds^2} + \Gamma^\mu_{\nu \lambda} \frac{dx^\nu}{ds} \frac{dx^\mu}{ds} = 0$

Using a non-affine parameter $\lambda$ gives an extra term:

Nonaffine parameter: $\frac{d^2 x^\mu}{d\lambda^2} + \Gamma^\mu_{\nu \lambda} \frac{dx^\nu}{d\lambda} \frac{dx^\mu}{d\lambda} = f(\lambda) \frac{dx^\mu}{d\lambda}$

Oh, wait a minute... Maybe it works this way? You use the form for nonaffine parameters to define a geodesic, and then you scale $\lambda \rightarrow s$ to eliminate the right-hand side to get an affine parameter?

 

[romsofia]

Do you have access to "A relativists toolkit" by Eric Poisson? In 1.3, he talks about affine parameterization in the null case.

However, if you don't (and i'd suggest watching from the start if you have time) here is a video:

 

[vanhees71]

Or you can use $C = 1$. Or any other constant. This is just a change in your choice of units for time.Sure there is. Spacelike curves work like timelike curves: the physically relevant affine parameter is arc length along the curve (measured with a ruler instead of a clock). Different constant choices of $C$ correspond to different choices of units for distance.

Mathematically that's clear. I was only a bit uncertain, whether there is a proper physical interpretation of such spacelike curves.

 

[vanhees71]

My issue is how to define the affine parameter for null paths.

A geodesic described using an affine parameter $s$ will satisfy

Affine parameter: $\frac{d^2 x^\mu}{ds^2} + \Gamma^\mu_{\nu \lambda} \frac{dx^\nu}{ds} \frac{dx^\mu}{ds} = 0$

Using a non-affine parameter $\lambda$ gives an extra term:

Nonaffine parameter: $\frac{d^2 x^\mu}{d\lambda^2} + \Gamma^\mu_{\nu \lambda} \frac{dx^\nu}{d\lambda} \frac{dx^\mu}{d\lambda} = f(\lambda) \frac{dx^\mu}{d\lambda}$
Oh, wait a minute... Maybe it works this way? You use the form for nonaffine parameters to define a geodesic, and then tyou scale $\lambda \rightarrow s$ to eliminate the right-hand side to get an affine parameter?

There's no problem with null geodesics. Just take
$$L=-\dot{x}^{\mu} \dot{x}^{\nu} g_{\mu \nu}$$
as the Lagrangian. Since $L=-H=\text{const}$ for a null geodesic you must set this constant to 0 and solve for the equations under this constraint, which of course read
$$\ddot{x}^{\mu}+{\Gamma^{\mu}}_{\alpha \beta} \dot{x}^{\alpha} \dot{x}^{\beta}=0.$$
The affine parameter is of course arbitrary here. There's no physically distinguished parameter as proper time in the time-like case.

 

[PeterDonis]

I was only a bit uncertain, whether there is a proper physical interpretation of such spacelike curves.

The physical interpretation is that they represent a "constant time" slice of an object like a ruler; if the object's motion is Born rigid, any such slice represents the spatial length of the ruler.

 

[stevendaryl]

$$\ddot{x}^{\mu}+{\Gamma^{\mu}}_{\alpha \beta} \dot{x}^{\alpha} \dot{x}^{\beta}=0.$$
The affine parameter is of course arbitrary here.

Once again, if you know that the path is a geodesic, then it satisfying the above equation implies that the parameter is affine. But is there a notion of an affine parametrization for a non-geodesic? There is for timelike paths that are not geodesics.

 

[romsofia]

Once again, if you know that the path is a geodesic, then it satisfying the above equation implies that the parameter is affine. But is there a notion of an affine parametrization for a non-geodesic? There is for timelike paths that are not geodesics.

Although I personally don't like to talk about this idea (I never truly "got" it), wouldn't this concept just be parallel transporting on a path that isn't a geodesic? I don't think it covers ALL the cases of this popping up, but maybe that would be a start: Looking at non-geodesic parallel transport stuff?

 

[ergospherical]

Once again, if you know that the path is a geodesic, then it satisfying the above equation implies that the parameter is affine. But is there a notion of an affine parametrization for a non-geodesic? There is for timelike paths that are not geodesics.

There's no such notion because affine parameterisation pertains to geodesics only. It's named as such because the geodesic equation is invariant under affine reparameterisation $\lambda \rightarrow a\lambda + b$.

 

[stevendaryl]

Okay, I now have a characterization of the affine parameter for a geodesic. I'm now unenthusiastic about whether it's worth doing, but I think I've done it...

For any smooth parameter $\lambda$, a geodesic is characterized by:

$\dfrac{d^2 x^\mu}{d\lambda^2} + \Gamma^\mu_{\nu \sigma} \dfrac{dx^\nu}{d\lambda} \dfrac{dx^\sigma}{d\lambda} = f(\lambda) \dfrac{dx^\mu}{d\lambda}$

So here's a way to choose $\lambda$: Pick a coordinate system with three spacelike coordinates and one timelike coordinate and let $\lambda = t = x^0$

$\dfrac{d^2 x^\mu}{dt^2} + \Gamma^\mu_{\nu \sigma} \dfrac{dx^\nu}{dt} \dfrac{dx^\sigma}{dt} = f(t) \dfrac{dx^\mu}{dt}$

Choosing $\mu = 0$ gives:
$\dfrac{d^2 x^0}{dt^2} + \Gamma^0_{\nu \sigma} \dfrac{dx^\nu}{dt} \dfrac{dx^\sigma}{dt} = f(t) \dfrac{dx^0}{dt}$

Now, since $x^0 = t$, we have $\dfrac{dx^0}{dt} = 1$ and $\dfrac{d^2 x^0}{dt^2} = 0$. So we have the simplification:

$\Gamma^0_{\nu \sigma} \dfrac{dx^\nu}{dt} \dfrac{dx^\sigma}{dt} = f(t)$

So we have a characterization of $f(t)$ in terms of $\Gamma$ and the 4-velocities.

For an affine parameter $s$ we have:

$\dfrac{d^2 x^0}{ds^2} + \Gamma^0_{\nu \sigma} \dfrac{dx^\nu}{ds} \dfrac{dx^\sigma}{ds} = 0$

We can write: $\dfrac{dx^\mu}{ds} = (\dfrac{dx^\mu}{dt})/(\dfrac{ds}{dt})$. Using our function $f(t)$, we can rewrite the geodesic equation for $x^0$ as:

$\dfrac{d^2 x^0}{ds^2} + f(t) \dfrac{1}{(\frac{ds}{dt})^2}= 0$

Or using $t = x^0$

$\dfrac{d^2 t}{ds^2} + f(t) \dfrac{1}{(\frac{ds}{dt})^2}= 0$

Calculus gives us: $\dfrac{d^2 t}{ds^2} = - \dfrac{d^2 s}{dt^2} / (\frac{ds}{dt})^3$. So we have:

$- \dfrac{d^2 s}{dt^2} / (\frac{ds}{dt})^3+ f(t) \dfrac{1}{(\frac{ds}{dt})^2}= 0$

Which simplifies to

$- \dfrac{d^2 s}{dt^2} / \frac{ds}{dt}+ f(t) = 0$

That can be rewritten as

$\frac{d}{dt} log(\frac{ds}{dt}) = f(t)$

So we have a characterization of $\frac{ds}{dt}$:

$\frac{ds}{dt} = e^{\int f(t) dt}$

Integrating again gives:

$s(t) = \int e^{\int f(t') dt'} dt$

Because of the two indefinite integrals, we can see that if $s$ is a solution, then so is:

$s' = a s' + b$

 

[vanhees71]

Once again, if you know that the path is a geodesic, then it satisfying the above equation implies that the parameter is affine. But is there a notion of an affine parametrization for a non-geodesic? There is for timelike paths that are not geodesics.

I was talking about the light-like case, if I remember right. There you don't have a physically preferred affine parameter. For time-like curves you can choose "path length", $s$, for which $g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}=1$ or "proper time", $\tau$, for which $g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}=c^2$.

At least for time-like curves following as the equation of motion from the action principle. The most convenient Lagrangian for such a case is
$$L=-\frac{m}{2} g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}+L_{\text{int}},$$
where $L_{\text{int}}(x,\dot{x})$ is a rank-1 homogeneous function of $\dot{x}^{\mu}$, i.e.,
$$\dot{x}^{\mu} \frac{\partial L_{\text{int}}}{\partial \dot{x}^{\mu}}=L_{\text{int}}.$$
From this combination of the "square form" of the free-particle Lagrangian and the parameter-independent formulation of the action you have
$$g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}=\text{const}$$
as a conservation law, i.e., the world-line parameter is affine.

For a derivation (in the SRT case for a particle in an em. field, but there's no additional problem for the GRT case and other than em. interactions), see

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf
Sect. 2.4.3.

 

[cianfa72]

Spacelike curves work like timelike curves: the physically relevant affine parameter is arc length along the curve (measured with a ruler instead of a clock). Different constant choices of $C$ correspond to different choices of units for distance.

As we discussed so far in some PF thread, the 'process' to physically measure the length of a spacelike curve is not so obvious. For a timelike curve it is simple as read the elapsed time shown from a clock carried along the given timelike curve.

For a spacelike curve we cannot directly measure its length juxtaposing a ruler one after the other.

 

[PeterDonis]

For a spacelike curve we cannot directly measure its length juxtaposing a ruler one after the other.

Not directly, no. In the idealized case we imagine a ruler that already has exactly the right length and whose motion is such that the spacelike curve in question lies in one of the ruler's local "rest frames", i.e., that it is orthogonal at every point to the worldline in the congruence of worldlines that describes the ruler and passes through that point.

Of course it is impossible in practice to always ensure that we have a single ruler that meets the above criterion. However, we can still imagine a congruence of curves I described, which is orthogonal at every point to the spacelike curve in question, and if that congruence is Born rigid, then we can make measurements with rulers of segments of spacelike curves at different times along the congruence, and transport each segment along the congruence until it is coincident with a segment of the spacelike curve whose length we want to measure. In other words, we measure different segments of the "spacelike curve" at different times, and rely on the static nature of the congruence to ensure that the lengths of the segments we measure don't change with time, so we can treat measurements made at different times as though they were made all at once, as in the idealized case above.

 

[cianfa72]

Not directly, no. In the idealized case we imagine a ruler that already has exactly the right length and whose motion is such that the spacelike curve in question lies in one of the ruler's local "rest frames", i.e., that it is orthogonal at every point to the worldline in the congruence of worldlines that describes the ruler and passes through that point.

You mean the ruler's worldtube (the congruence of worldlines) has a spacetime extension such that its 'boundary' worldlines intersect the two endpoints of the given spacelike curve.

In other words, we measure different segments of the "spacelike curve" at different times, and rely on the static nature of the congruence to ensure that the lengths of the segments we measure don't change with time, so we can treat measurements made at different times as though they were made all at once, as in the idealized case above.

Why we can assume the congruence of worldlines to be static ?

 

[PeterDonis]

You mean the ruler's worldtube (the congruence of worldlines) has a spacetime extension such that its 'boundary' worldlines intersect the two endpoints of the given spacelike curve.

Yes.

Why we can assume the congruence of worldlines to be static ?

We can't assume it; I said if the congruence is Born rigid. Physically, you would try to arrange that by placing the rulers appropriately and limiting what can happen to them while the measurements are taking place.

 

[cianfa72]

We can't assume it; I said if the congruence is Born rigid. Physically, you would try to arrange that by placing the rulers appropriately and limiting what can happen to them while the measurements are taking place.

The worldlines of Born rigid static congruence in your post #17 are orbits of a timelike Killing vector field. On tangent space at each point of them the orthogonal complement to timelike Killing vector field evaluated at that point defines a spacelike hyperplane distribution.

The 'static' property of the congruence does actually mean such spacelike hyperplane distribution is integrable, i.e. by Frobenius theorem there exist a smooth function $f$ such that its 'level sets' are spacelike hypersufaces orthogonal to the congruence foliating the spacetime (at least the spacetime region of interest).

If the above is correct, the value the function $f$ take on each level set can be taken as the 'timelike coordinate' w.r.t. 'time translate' the segments (rulers) we can employ to indirectly measure the length of the given spacelike curve -- as you explained above.

 

[PeterDonis]

The worldlines of Born rigid static congruence in your post #17 are orbits of a timelike Killing vector field.

Not necessarily, although that is certainly one good way to obtain such a congruence.

On tangent space at each point of them the orthogonal complement to timelike Killing vector field evaluated at that point defines a spacelike hyperplane distribution.

I don't know why you include the word "distribution" here. With that word removed, locally, any Born rigid congruence has an orthogonal complement that defines a local spacelike hyperplane. However, unless the congruence has zero vorticity, this local spacelike hyperplane cannot be extended globally and still maintain the orthogonality property.

The 'static' property of the congruence does actually mean such spacelike hyperplane distribution is integrable

Yes, if the congruence is static, it has zero vorticity and is hypersurface orthogonal.

If the above is correct, the value the function $f$ take on each level set can be taken as the 'timelike coordinate'

Yes.

w.r.t. 'time translate' the segments (rulers) we can employ to indirectly measure the length of the given spacelike curve -- as you explained above.

Yes.

 

[cianfa72]

Not necessarily, although that is certainly one good way to obtain such a congruence.

So the definitory property of any static congruence is zero vorticity. The conditions of Frobenius theorem hold hence the congruence is actually hypersurface orthogonal even though the worldlines of the static congruence are not necessarily orbits of a timelike KVF.

 

[PeterDonis]

So the definitory property of any static congruence is zero vorticity.

The term "static", strictly speaking, would mean a timelike Killing congruence (i.e., integral curves of a timelike Killing vector field) with zero vorticity (if the vorticity is nonzero it would be called "stationary"). Applying the term "static" to a congruence that is Born rigid with zero vorticity, but not necessarily a Killing congruence, seems to me to be a reasonable extension of the term "static", but it is not standard. That's why I specified "Born rigid" in an earlier post. (Note that a timelike Killing congruence must be Born rigid, although the converse is not true.)

The conditions of Frobenius theorem hold hence the congruence is actually hypersurface orthogonal even though the worldlines of the static congruence are not necessarily orbits of a timelike KVF.

That's correct, the Frobenius theorem can be applied to any congruence. The congruence does not even have to be Born rigid; for example, you can apply the Frobenius theorem to the congruence of "comoving" worldlines in FRW spacetime to show that this congruence is hypersurface orthogonal, even though it has nonzero expansion and is therefore not Born rigid.

 

[cianfa72]

That's correct, the Frobenius theorem can be applied to any congruence. The congruence does not even have to be Born rigid; for example, you can apply the Frobenius theorem to the congruence of "comoving" worldlines in FRW spacetime to show that this congruence is hypersurface orthogonal, even though it has nonzero expansion and is therefore not Born rigid.

So the condition to apply the Frobenius theorem for congruence hypersurface orthogonality is actually zero vorticity, I believe.

 

[PeterDonis]

the condition to apply the Frobenius theorem for congruence hypersurface orthogonality is actually zero vorticity

Yes. More precisely, the theorem says that any congruence has zero vorticity if and only if it is hypersurface orthogonal; the two conditions are logically equivalent.

 

[cianfa72]

Ok, my doubt is that if the congruence discussed in post#17 and later was not, strictly speaking, static then the time translation of the segments (rulers) along the congruence would result in a change of their measures since they are not carried along integral curves of a timelike KVF.

Or, said in other words, is the 'Born rigid' condition that actually allows their measures to stay unchanged when they are carried along the congruence ?

 

[PeterDonis]

is the 'Born rigid' condition that actually allows their measures to stay unchanged when they are carried along the congruence ?

Go back and read my post #23 again (my first paragraph in particular). It answers this.

 

[cianfa72]

Go back and read my post #23 again (my first paragraph in particular). It answers this.

Sorry, I read it again however I didn't get the point

In post #23 you talked about Born rigid + zero vorticity as a reasonable definition of 'static' congruence. Are those two conditions sufficient to get the rulers's length unchanged when carried along the 'static' congruence ?

 

[PeterDonis]

In post #23 you talked about Born rigid + zero vorticity as a reasonable definition of 'static' congruence.

Yes.

Are those two conditions sufficient to get the rulers's length unchanged when carried along the 'static' congruence ?

Yes. Actually even the zero vorticity condition is not necessary if one is willing to adopt a more abstract definition of "length unchanged" than the one we have been using.