I Is a Manifold with a Boundary Considered a True Manifold?

[PeterDonis]

See what I added above, before seeing your response.

Ah, yes, I see. Yes, I agree the "one parameter group" part is important.

 

[jbergman]

Your link here just goes to the stackexchange home page.

Thanks I fixed it.

https://physics.stackexchange.com/q...folds-physically-equivalent?noredirect=1&lq=1

 

[PeterDonis]

- A vector field $v$ induces a flow $\phi_t$ which shifts points a parameter distance $t$ along the integral curves of $v$.

To rephrase this in terms of the key item from Wald that @PAllen pointed out: a vector field induces a one-parameter group of transformations from the manifold to itself. The group parameter is $t$.

 

[PAllen]

Thanks I fixed it.

https://physics.stackexchange.com/q...folds-physically-equivalent?noredirect=1&lq=1

Fine, and this does not disagree with anything I have stated, nor answer the question I raised. It does clarify than since GR is theory of psuedo-Riemannian manifolds, unqualified use of diffeomorphism in the context of GR, implicitly includes that the metrics are related by a pullback of the diffeomorphism. Wald appendix C.3 does answer my question, showing not any equivalence between isometry, in general, and kvf, but instead, an equivalence between a one parameter group of isometries from a manifold to itself, and a kvf.

 

[jbergman]

Well, if you want, I can provide dozens of quotes calling GR a diffeomorphism invariant theory. Further, the use of pulllbacks or pushforwards does not in any way restrict the mapping between manifolds, per se. It simply adds a rule that metric on each is determined by the mapping combined with the metric on the other. I have a whole shelf of GR textbooks, and not one ever considers a diffeomorphism without also using a pullback or pushforward.

This link doesn't go to any discussion, just the home page.

Here is the fixed link. https://physics.stackexchange.com/q...folds-physically-equivalent?noredirect=1&lq=1

In the mathematics literature a diffeomorphism is a smooth invertible mapping between manifolds or more precisely an isomorphism in the category of smooth manifolds. It says nothing about what happens to metrics.

An isometry is an isomorphism in the category of Riemannian or Pseudo-Riemannian manifolds, smooth manifolds equipped with metrics.

Precisely, as I stated before,

$f : M \rightarrow M'$ such that $g = f^{*}g'$. So every diffeomorphism with the above property by definition is an isometry. If we have a map smooth invertible map $f' : M \rightarrow M'$ such that $g \ne f^{*}g'$, $f'$ is still a diffeomorphism but not an isometry.

What am confused about is not addressed in this post. That is, what is the connection between diffeomorphims and kvfs?? The latter are equivalently, particular congruences generating a vector field, or particular vector fields (with associate integral curves), that meet certain properties. None of these properties specify a diffeomorphism at all.

Not sure what more I can say here. Every vector field on a smooth manifold generates a flow. Each flow is a one-parameter group $\theta_t : M \rightarrow M$ and induces a family of diffeomorphisms from the manifold to itself for each value of the parameter, $t$. We say that the vector field generates the flow because the derivative with respect to the group parameter at each point gives a tangent vector or that the flow contains all the integral curves of the vector field.

Now in general the flows induced by a vector field do not have the property that
$g = \theta_t^{*}g$ and are not isometries. Flows induced by Killing Vector fields do have the property that $g = \theta_t^{*}g$, hence are not *just* a family of diffeomorphisms but are also a continuous family of isometries.

One subtle point is that the difference between a diffeomorphism induced by a flow versus a general mapping between manifolds is that you aren't free to change the metric since it is a map to the same underlying manifold which is why I don't use $g'$ here and we should think of these as automorphisms in the appropriate category.

Also, as I stated earlier, not every Isometry is generated by KVF, only the continuous ones.

There is a passing comment in the wikipedia entry on isometries to kvf's as generators. Unfortunately, there is no further explanation or reference associated, and I don't understand it. I would guess there is something to it, but I remain search of any explanation. A related confusion, is that a pseudo-Riemannian manifold may have no kvfs at all (e.g. I think this is true of some of Misner's mixmaster universes). Yet one may readily define isometric mappings between manifolds, one of which is mixmaster (which would, of course, make the other one mixmaster as well). If I am right that a pseudo-Riemannian manifold may have no kvfs, I genuinely don't understand the comment about kvfs as generators of isometries.

Are you familiar with Lie Groups and Left Invariant vector fields? It's essentially the same idea.

 

[PeterDonis]

I don't think that GR is a diffeomorphism invariant theory.

Often when physicists say that GR is diffeomorphism invariant, they mean that the laws of GR are diffeomorphism invariant. Or, to put it another way, applying a diffeomorphism to any solution of the Einstein Field Equation will give another solution, but that other solution is not necessarily physically equivalent to the first one.

Another thing physicists often mean when they say that GR is diffeomorphism invariant is in the passive sense of "diffeomorphism", i.e., a coordinate transformation. As long as we transform all tensor quantities properly when we do the coordinate transformation, we end up with the same physical solution, just described in different coordinates.

Mathematicians might describe these operations in different ways, which, as I said before, might mean we just have a difference here between mathematicians' and physicists' terminology.

 

[PeterDonis]

Now in general the flows induced by a vector field do not have the property that $g = \theta_t^{*}g$ and are not isometries.

This might require a little more detail, since if we are talking about transformations from a manifold to itself, the metric is the same as before.

Consider, for example, the Milne congruence that @PAllen described earlier. In standard Minkowski coordinates on the upper "wedge" of Minkowski spacetime, i.e., the interior of the future light cone of the origin, this congruence is generated by the vector field $U = \gamma(v) \left( \partial_t + v \partial_x \right)$, where $v = x / t$ and $\gamma(v) = 1 / \sqrt{1 - v^2}$.

Computing the Lie derivative $\mathscr{L}_U g$ is straightforward and shows that it is nonzero. The question is, what does this Lie derivative actually represent? The manifold is still Minkowski spacetime both before and after the transformation. So what "change" is $\mathscr{L}_U g$ describing?

 

[PAllen]

Often when physicists say that GR is diffeomorphism invariant, they mean that the laws of GR are diffeomorphism invariant. Or, to put it another way, applying a diffeomorphism to any solution of the Einstein Field Equation will give another solution, but that other solution is not necessarily physically equivalent to the first one.

Another thing physicists often mean when they say that GR is diffeomorphism invariant is in the passive sense of "diffeomorphism", i.e., a coordinate transformation. As long as we transform all tensor quantities properly when we do the coordinate transformation, we end up with the same physical solution, just described in different coordinates.

Mathematicians might describe these operations in different ways, which, as I said before, might mean we just have a difference here between mathematicians' and physicists' terminology.

Actually, Wald takes a pure active approach, but assumes one uses the pullback along with the diffeomorphism itself. At the end of Appendix C, he discusses the active/passive difference, and (per his usage) shows they are completely equivalent as to physics, differing only 'philosophically'. Walds' usage is the one I am familiar with - by using the pullback, any active or passive diffeomorphism becomes an isometry, and this is what is meant by diffeomorphism invariance in GR.

 

[vanhees71]

To rephrase this in terms of the key item from Wald that @PAllen pointed out: a vector field induces a one-parameter group of transformations from the manifold to itself. The group parameter is $t$.

I think, it's important that it's a transformation $M \rightarrow M$. That's also the meaning of a symmetry as defined by a Killing vector field: transporting line-elements along the corresponding curves doesn't change their metrics, i.e., the Riemannian space is symmetric under such a mapping.

 

[jbergman]

This might require a little more detail, since if we are talking about transformations from a manifold to itself, the metric is the same as before.

Consider, for example, the Milne congruence that @PAllen described earlier. In standard Minkowski coordinates on the upper "wedge" of Minkowski spacetime, i.e., the interior of the future light cone of the origin, this congruence is generated by the vector field $U = \gamma(v) \left( \partial_t + v \partial_x \right)$, where $v = x / t$ and $\gamma(v) = 1 / \sqrt{1 - v^2}$.

Computing the Lie derivative $\mathscr{L}_U g$ is straightforward and shows that it is nonzero. The question is, what does this Lie derivative actually represent? The manifold is still Minkowski spacetime both before and after the transformation. So what "change" is $\mathscr{L}_U g$ describing?

So, I wasn't familiar with the Milne congruence but I did take the Lie Derivative of your vector field and it wasn't zero as you stated.

The problem with your statement is that the Milne congruence is not an isometry unless you pullback the metric from Minkowski space onto your new coordinates. If you don't do that you only have a diffeomorphism. $\mathscr{L}_U g$ tells us how the metric changes when we take a small step on our manifold in the direction of the integral curves of $U$. If it is not zero then we are moving in a direction that the metric changes, i.e. the distance between points changes.

Really, a Milne congruence is just a coordinate transformation which is a slightly different thing, IMO. The problems with viewing it as a congruence are two-fold.

1. The congruence is only onto a subset of Minkowski space. Typically when I can only parameterize a small part of the overall manifold, I consider that a coordinate chart.
2. We have to change the metric. Again, I think it's more helpful to think that there is an intrinsic manifold $M$ and the consider the auto-isomorphisms of $M$. Things like KVFs if we are interested in the smooth symmetries. Then something like the Milne coordinates is just a different coordinate representation.

 

[jbergman]

$\mathscr{L}_U g$ tells us how the metric changes when we take a small step on our manifold in the direction of the integral curves of $U$. If it is not zero then we are moving in a direction that the metric changes, i.e. the distance between points changes.

This statement was imprecise. Probably a better way to put it is that $\mathscr{L}_U g$ tells us how the associated flow deforms the metric. If it is 0, then there is no deformation. If it is non-zero then infinitesimal steps along the flow lead to a deformation of the space such that if one pulls back the metric along the flow it will change. In short distances are changing between points locally if the are all moved along the flow.

 

[cianfa72]

$\mathscr{L}_U g$ tells us how the metric changes when we take a small step on our manifold in the direction of the integral curves of $U$. If it is not zero then we are moving in a direction that the metric changes, i.e. the distance between points changes.

As previously pointed out, I would say that $\mathscr{L}_U g=0$ is not a condition on the "values" the metric tensor $g$ takes pointwise along the integral curves of $U$. $\mathscr{L}_U g=0$ condition does mean a null change of distance between points as we 'flow' them along the integral curves of $U$.

 

[jbergman]

As previously pointed out, I would say that $\mathscr{L}_U g=0$ is not a condition on the "values" the metric tensor $g$ takes pointwise along the integral curves of $U$. $\mathscr{L}_U g=0$ condition does mean a null change of distance between points as we 'flow' them along the integral curves of $U$.

Agree. It was an imprecise statement. What we are doing is actually changing distance between points. Now, if you were to pullback the unchanged metric under that map it would change.

The easiest example I can think of is a stretching say on the x direction mapping every point (x, y) to (2x, y). If you pullback the standard euclidean metric under that map the metric change. Anyways, I like the way you phrase it.

 

[PeterDonis]

The problem with your statement is that the Milne congruence is not an isometry

That's not a problem, it's the point: I was showing that the Milne congruence is indeed not an isometry, because it doesn't satisfy Killing's equation--yet it's a transformation from Minkowski spacetime to Minkowski spacetime, so the metric of the manifold itself remains the same.

$\mathscr{L}_U g$ tells us how the metric changes when we take a small step on our manifold in the direction of the integral curves of $U$.

But that's the point: the metric doesn't change. The metric of Minkowski spacetime is the same everywhere. So the fact that $\mathscr{L}_U g$ is nonzero cannot be telling us that the metric changes, because it doesn't change. So what is it telling us? What is changing along integral curves of U? That's the question.

If it is not zero then we are moving in a direction that the metric changes, i.e. the distance between points changes.

"The distance between points changes" makes sense, yes. But "the metric changes" does not make sense, because, as above, it doesn't change.

Really, a Milne congruence is just a coordinate transformation

No, it isn't. It's a perfectly good congruence. The definition I gave is obviously a vector field; it's obviously not any kind of coordinate transformation.

1. The congruence is only onto a subset of Minkowski space.

Many congruences only cover a portion of a particular spacetime. That's not a problem, it's just a limitation that needs to be recognized when the congruence is defined. That's why I specified that the Milne congruence I gave is only defined on the "upper wedge" of Minkowski spacetime.

1. Typically when I can only parameterize a small part of the overall manifold, I consider that a coordinate chart.

This makes no sense; the vector field I gave, as noted above, is not a coordinate chart, it's a vector field. I don't see how your terminology here makes sense.

You can of course define a Milne coordinate chart on the same "wedge" of Minkowski spacetime, in which the vector field U I gave gives the timelike basis vector at each point. But to define a coordinate chart you have to do more than just give the timelike basis vector. A vector field is not the same as a coordinate chart.

1. We have to change the metric.

I did no such thing when I defined U. The Minkowski metric remains the same, as I have already noted several times.

You can of course, as noted above, define a Milne coordinate chart, in which the metric takes a different form--but it's still the metric of flat Minkowski spacetime, as you can easily verify by computing its Riemann tensor and seeing that it vanishes.

1. Again, I think it's more helpful to think that there is an intrinsic manifold $M$ and the consider the auto-isomorphisms of $M$. Things like KVFs if we are interested in the smooth symmetries. Then something like the Milne coordinates is just a different coordinate representation.

I see what you mean here, but I still think your viewpoint has issues, as I noted above.

 

[jbergman]

That's not a problem, it's the point: I was showing that the Milne congruence is indeed not an isometry, because it doesn't satisfy Killing's equation--yet it's a transformation from Minkowski spacetime to Minkowski spacetime, so the metric of the manifold itself remains the same.

Right. It is a diffeomrophism but not an isometry. We have our Minkowski space call it $M$ with metric $g$. And let's call the flow generated by your vector field, $U$ $\theta$. Each $\theta_t$ are diffeomorphisms of Minkowski space. But they are not isometries. What does that mean? It means that it isn't a symmetry transformation that respects the Minkowski metric. It only respects the smooth structure. In other words it won't preserve the distances between points. That is why it isn't one of the symmetry transformations of the Lorentz Group.

But that's the point: the metric doesn't change. The metric of Minkowski spacetime is the same everywhere. So the fact that $\mathscr{L}_U g$ is nonzero cannot be telling us that the metric changes, because it doesn't change. So what is it telling us? What is changing along integral curves of U? That's the question.

It's telling us that the mapping isn't a symmetry transformation with respect to the metric and corresponds to a different physical situation.

"The distance between points changes" makes sense, yes. But "the metric changes" does not make sense, because, as above, it doesn't change.

To be precise, I am saying that $g \ne \theta_t^{*}g$. What does that mean? Our mapping has moved points in a way that they are no longer the distance from them because if you evaluate their distances using the pullback of the metric you will get different distances. With symmetry transformations that isn't the case.

Concretely given two tangent vectors $u$ and $v$ at point $p$, we can calculate the metric as $g_p(u, v)$ or we could compute the metric using a pullback by pushing the tangent vectors forward to $\theta_t(p)$ as $g_{\theta_t(p)}(\theta_t^{*}(u), \theta_t^{*}(v))$. This is exactly what the Lie Derivative tells us as it just is the limit of these pullbacks.

No, it isn't. It's a perfectly good congruence. The definition I gave is obviously a vector field; it's obviously not any kind of coordinate transformation.

I am not sure what is meant by the term congruence. As I think of it, a vector field induces a family diffeomorphisms of a manifold or submanifold in this case of Minkowski space via its flow. You are correct that it isn't a coordinate transformation unless we pick a specific diffeomorphism and define a coordinate chart with it.

 

[PeterDonis]

It means that it isn't a symmetry transformation that respects the Minkowski metric.

Yes, it does. That's the point I've been trying to make.

In other words it won't preserve the distances between points.

More precisely: the distance between pairs of points varies with the curve parameter $t$. But that's because of the way the curves "sit" in the Minkowski geometry. It has nothing to do with the Minkowski geometry itself. That stays the same.

To be precise, I am saying that $g \ne \theta_t^{*}g$.

But this seems obviously false since $g$ is the Minkowski metric in both cases.

Concretely given two tangent vectors $u$ and $v$ at point $p$

But that's not what we're talking about. We're talking about two different curves in the Milne congruence, i.e., two different integral curves of the vector field U that I gave. You can't have two different integral curves of U at the same point.

If we take two points $p$, $q$ that both have the same curve parameter $t_0$, one on each curve, and then we flow those points along their respective curves to points $p'$, $q'$ at curve parameter $t_1$, we find that the Minkowski distance from $p'$ to $q'$ is different from the Minkowski distance from $p$ to $q$. The Lie derivative being nonzero is basically telling us this for two neighboring curves.

Whereas, if we take two neighboring integral curves of a Killing vector field in Minkowski spacetime, the Minkowski distance between pairs of points with the same curve parameter stays the same as we flow the points along the curves. That is what the Lie derivative being zero tells us.

If this is what you were trying to say, then we're in agreement, but I don't see how what you have been saying is the same as the above. At the very least, your choice of terminology seems misleading to me.

I am not sure what is meant by the term congruence.

It means a set of curves that cover an open region of a manifold, i.e., every point in the open region lies on exactly one curve in the set. Equivalently, it is the set of integral curves of a non-vanishing vector field on an open region of a manifold. (If the manifold itself is open, the open region may be the entire manifold, but it doesn't have to be.)

 

[jbergman]

More precisely: the distance between pairs of points varies with the curve parameter $t$. But that's because of the way the curves "sit" in the Minkowski geometry. It has nothing to do with the Minkowski geometry itself. That stays the same.

Correct, but a symmetry transformation is one that the state of the manifold is unchanged. By changing distances between points on the curves you have changed the configuration of things. The simplest example is of a map on a globe. If you have a smooth mapping of the globe to itself which stretches some parts and compresses others but still covers the globe, the shapes of the countries will change after those transformations. That's all this is saying and that is what the pullback of the metric tells us. I can't state it any clearer without writing out a concrete example in some detail which shows how the pullback of the metric behaves under a transformation that is not an isometry. Again the simplest example is a stretching of a single coordinate like the x-axis. I will try and write it up later.

 

[PeterDonis]

a symmetry transformation is one that the state of the manifold is unchanged. By changing distances between points on the curves you have changed the configuration of things.

This doesn't make sense either. The manifold doesn't change. The curves don't change. They're geometric objects that are just there; they don't change from one thing to another. All that changes is which particular pairs of points on two neighboring curves you are looking at.

a smooth mapping of the globe to itself which stretches some parts and compresses others but still covers the globe

But, again, this is not what we are talking about. Minkowski spacetime itself is not being stretched or compressed at all.

I can't state it any clearer without writing out a concrete example in some detail which shows how the pullback of the metric behaves under a transformation that is not an isometry.

If you can write out explicitly exactly what "metric" is changing as we flow along the integral curves of the vector field U that I gave, that would help, yes. But if you are just going to write down the Minkowski distance between points on two different curves with the same curve parameter, and show that that changes as we flow along the curves, don't bother, because I have already agreed that that changes. What I don't see is how that change is to be described as a change in "the metric". Indeed, we have to use the same Minkowski metric to evaluate the distances between pairs of points with the same curve parameter as we flow along the curves; otherwise we would have no basis for saying that that distance changed.

Another possibility might be that you are going to write the "distance" in Milne coordinates, using that form of the metric, between two points with different spatial coordinates, at the same Milne coordinate time. But that "metric"--the Milne metric at constant Milne coordinate time--does not give a Minkowski distance, because it is not measured along a spacelike geodesic of Minkowski spacetime. It's measured along a spacelike hyperbola. (Those spacelike hyperbolas, interestingly, are integral curves of a KVF of Minkowski spacetime.)

 

[PeterDonis]

$\mathscr{L}_U g$ tells us how the associated flow deforms the metric.

It might be worth noting that the reason an issue arose in this thread regarding this particular kind of ordinary language description is that it doesn't seem to work for manifolds with constant curvature. In a manifold that doesn't have constant curvature, if you move along any flow that is not a Killing flow, the local metric itself will change. For example, in Schwarzschild spacetime, the metric is a function of $r$ (by which I mean here the "areal radius", which is an invariant, even though it is also used as a coordinate in some charts), so any flow that has $r$ changing along its integral curves will not be a Killing flow and the local metric itself--the underlying geometry of the manifold itself, not just "distances between points" on neighboring curves--will change. So for this case, the ordinary language description of $\mathscr{L}_U g$ as telling us how "the metric changes" along the flow works fine. It just breaks down in the edge case of a manifold with constant curvature, because in such a manifold the metric doesn't change at all; it's the same everywhere. So we have to be more careful in describing what a nonzero $\mathscr{L}_U g$ actually means.

 

[jbergman]

This doesn't make sense either. The manifold doesn't change. The curves don't change. They're geometric objects that are just there; they don't change from one thing to another. All that changes is which particular pairs of points on two neighboring curves you are looking at.But, again, this is not what we are talking about. Minkowski spacetime itself is not being stretched or compressed at all.

In my example the sphere was the manifold. The sphere did not change shape or distances did not change. But what did change were that if I had two points $p$ and $q$, then the distance between $f(p)$ and $f(q)$ changes. The maps of the countries were just to highlight the distances that changed between two pairs of points under the mapping.

If you can write out explicitly exactly what "metric" is changing as we flow along the integral curves of the vector field U that I gave, that would help, yes. But if you are just going to write down the Minkowski distance between points on two different curves with the same curve parameter, and show that that changes as we flow along the curves, don't bother, because I have already agreed that that changes. What I don't see is how that change is to be described as a change in "the metric". Indeed, we have to use the same Minkowski metric to evaluate the distances between pairs of points with the same curve parameter as we flow along the curves; otherwise we would have no basis for saying that that distance changed.

Here is an example for the 2-d euclidean plane. We have a stretching in the x-direction.
$$V = 2x\partial_x$$
which induces a flow,
$$\theta_t(x,y) = (x+2xt, y)$$
Our metric is just
$$ds^2 = dx^2 + dy^2$$

Now for concreteness we will pick $t = 2$ and $p = (1,0)$
With that,
$$\theta_{t=2}(x, y) = (x+4x, y) = (5x, y)$$.
and $q = \theta_{t=2}(1,0) = (5, 0)$.
So now the question is if we pull the metric back from point $q$ to the point $p$ what does it look like? Since the metric is the same everywhere this is relatively easy. For tangent vectors $u_p$ and $v_p$

$$(\theta_{t=2}^{*}g)_p(u_p, v_p) = g((\theta_{t=2}^{*})_p(u_p), (\theta_{t=2}^{*})_p(v_p))$$

So how does $(\theta_{t=2}^{*})_p$ push forward tangent vectors? It is just the jacobian of the map at that point. In this case it's just $(\theta_{t=2}^{*})_p = diag(5,1)$.

So we get that

$$(\theta_{t=2}^{*}g)_p(u_p, v_p) = g((5u^0, u^1), (5v^0, v^1)) = 25u^0v^0 + u^1v^1$$
which is clearly not equal to $g(u_p,v_p) = u^0v^0 + u^1v^1$.

 

[PeterDonis]

Here is an example for the 2-d euclidean plane.

And in this example, as you explicitly state:

the metric is the same everywhere

So whatever is "changing" along the flow of the vector field you describe, it isn't the metric.

Note that this is not saying any of the math you wrote down is wrong. Of course it's not. It's entirely about how you describe what the math is saying in ordinary language. In ordinary language, it doesn't seem like "the metric is changing along the flow of the vector field V" is a good description, since you explicitly say the metric isn't changing.

Also, your example actually doesn't even illustrate what we were discussing before, that the distance between pairs of points on neighboring integral curves of the vector field changes. If we take any two neighboring integral curves, i.e., curves $y = k$ and $y = k + \delta$ for any real $k$ and infinitesimal $\delta$, the Euclidean distance between pairs of points with the same parameter obviously stays the same (i.e., $\delta$, at least if we adopt the obvious parameterization) as we flow along V.

A better example would be $W = 2 y \partial_x$, since for this flow, even though it has the same integral curves as your V, the parameterization along them varies with $y$, so the Euclidean distance between pairs of points on neighboring curves with the same parameter must vary along the flow.

 

[jbergman]

Also, your example actually doesn't even illustrate what we were discussing before, that the distance between pairs of points on neighboring integral curves of the vector field changes. If we take any two neighboring integral curves, i.e., curves $y = k$ and $y = k + \delta$ for any real $k$ and infinitesimal $\delta$, the Euclidean distance between pairs of points with the same parameter obviously stays the same (i.e., $\delta$, at least if we adopt the obvious parameterization) as we flow along V.

A better example would be $W = 2 y \partial_x$, since for this flow, even though it has the same integral curves as your V, the parameterization along them varies with $y$, so the Euclidean distance between pairs of points on neighboring curves with the same parameter must vary along the flow.

I don't see that as being an essential feature, here. What we are interested in is how the distance of *any* neighboring points change under the flow not necessarily on different integral curves. The differential of $d\theta_t$ which is used to push forward the tangent vectors contains information about how the map distorts space in each tangent direction.

 

[PeterDonis]

how the map distorts space

But, as I've been saying repeatedly now, the map doesn't distort "space", because, as you said explicitly (I quoted you), the metric stays the same. So whatever the map is distorting, it isn't "space".

Again, this is not saying any of the math you are describing is wrong; obviously it's not. It's just about how to describe the math in ordinary language.

 

[cianfa72]

So we get that
$$(\theta_{t=2}^{*}g)_p(u_p, v_p) = g((5u^0, u^1), (5v^0, v^1)) = 25u^0v^0 + u^1v^1$$
which is clearly not equal to $g(u_p,v_p) = u^0v^0 + u^1v^1$.

Sorry, does it mean $\mathscr{L}_V g$ actually vanishes or not ?

I believe it doesn't vanish for both vector fields $V = 2x\partial_x$ and $V = 2y\partial_x$ since neither are actually isometries of Euclidean plane (yet their integral curves are both straight lines parallel to the $x$ axis).

If we take any two neighboring integral curves, i.e., curves $y = k$ and $y = k + \delta$ for any real $k$ and infinitesimal $\delta$, the Euclidean distance between pairs of points with the same parameter obviously stays the same (i.e., $\delta$, at least if we adopt the obvious parameterization) as we flow along V.

As obvious parameterization do you mean the euclidean length along each integral curve in the congruence ?

 

[jbergman]

Sorry, does it mean $\mathscr{L}_V g$ actually vanishes or not ?

I believe it doesn't vanish for both vector fields $V = 2x\partial_x$ and $V = 2y\partial_x$ since neither are actually isometries of Euclidean plane (yet their integral curves are both straight lines parallel to the $x$ axis).

$$(\mathscr{L}_V g)_p = \frac{d}{dt}\Bigr|_{t=0} (\theta_t^{*}g)_p = \lim_{t \rightarrow 0} \frac{(\theta_t^{*})_p(g_{\theta_t(p)}) - g_p}{t}$$
So we see it is just the limit of the difference of the pullback as we move along the flow towards p. In the previous post, I calculated the pullback for a specific $t$. Here we, do it for an arbitrary $t$.

First we need to recalculate the pushforward for tangent vectors which is again just the jacobian.
$$\theta_t^{*} = d\theta_t = diag(1+2t, 1)$$
So, now plugging this into the limit we get,
$$(\mathscr{L}_V g)_p = \lim_{t \rightarrow 0} \frac{(\theta_t^{*})_p(g_{\theta_t(p)}) - g_p}{t} = \lim_{t \rightarrow 0} \frac{(1+2t)^2dx^2 +dy^2 - dx^2 - dy^2}{t}$$
$$(\mathscr{L}_V g)_p = \lim_{t \rightarrow 0} \frac{4t^2dx^2 + 4tdx^2 }{t} = 4 dx^2$$
So, we see that the Lie Derivative is non-zero as expected. As a sanity check, a more conventional way to calculate the Lie Derivative is as follows,
$$\mathscr{L}_V g_{\mu\nu} = V^{\alpha}\partial_{\alpha}g_{\mu\nu} + (\partial_{\mu}V^{\alpha})g_{\alpha\nu} + (\partial_{\nu}V^{\alpha})g_{\mu\alpha}$$

$$\mathscr{L}_V g_{xx} = 2(\partial_{x}V^{x})g_{xx} =4 $$

 

[cianfa72]

Some comments about your calculation.

First one: I believe the pushforward operator should have a different symbol from the pullback, something like $\theta_{*t}$ for instance.

Second one: $\mathscr{L}_V g_{\mu\nu}$ should be the $(\mu,\nu)$ component of the tensor $\mathscr{L}_V g$ in coordinate basis -- i.e. $(\mathscr{L}_V g)_{\mu\nu}$, right ?

 

[jbergman]

Some comments about your calculation.

First one: I believe the pushforward operator should have a different symbol from the pullback, something like $\theta_{*t}$ for instance.

Yes. However, conventions vary on the notation. I believe Wald uses the upper star for both and some authors just use $dF$ for the push forward of $F$ when applied to tangent vectors.

Second one: $\mathscr{L}_V g_{\mu\nu}$ should be the $(\mu,\nu)$ component of the tensor $\mathscr{L}_V g$ in coordinate basis -- i.e. $(\mathscr{L}_V g)_{\mu\nu}$, right ?

Yes.

 

[cianfa72]

Even for $V = 2y\partial_x$ the Lie derivative $(\mathscr{L}_V g)_{\mu\nu}$ does not vanish, indeed:
$$(\mathscr{L}_V g)_{xy} = (\mathscr{L}_V g)_{yx} = (\partial_{y}V^{x})g_{xx} =2$$

A better example would be $W = 2 y \partial_x$, since for this flow, even though it has the same integral curves as your V, the parameterization along them varies with $y$, so the Euclidean distance between pairs of points on neighboring curves with the same parameter must vary along the flow.

Maybe I misinterpreted your claim: which is the parameterization such that the Euclidean distance between pairs of points on neighboring curves with the same parameter changes along the flow ?

 

[PeterDonis]

which is the parameterization such that the Euclidean distance between pairs of points on neighboring curves with the same parameter changes along the flow ?

The components of $W$ are $(2y, 0)$. So an integral curve of $W$ will have a parameterization $(x, y) = (2y_0 t, y_0)$, where $y_0$ is a constant (the unchanging $y$ coordinate of the curve) and $t$ is the parameter.

Now consider two adjacent curves with $y_0 = k$ and $y_0 = k + \delta$. At parameter value $t = 0$, corresponding points on the two curves are $(0, k)$ and $(0, k + \delta)$, so the Euclidean distance between them is $\delta$. At parameter value $t = 1$, corresponding points on the two curves are $(2 k, k)$ and $(2 k + 2 \delta, k + \delta)$, so the Euclidean distance between them is now $\sqrt{4 \delta^2 + \delta^2} = \sqrt{5} \delta$, which is larger.

 

[cianfa72]

ok, which is the obvious parameterization to be used for the case $V=2x\partial_x$ ?

 

[PeterDonis]

ok, which is the obvious parameterization to be used for the case $V=2x\partial_x$ ?

You should be able to work that out by analogy with what I said in post #179. I'll give you the first step: the components of $W$ for this case are $(2x, 0)$.

 

[cianfa72]

You should be able to work that out by analogy with what I said in post #179. I'll give you the first step: the components of $W$ for this case are $(2x, 0)$.

$(x,y)=(t^2,y_0)$ ?

 

[PeterDonis]

$(x,y)=(t^2,y_0)$ ?

No. Try writing out explicitly the equation for the $x$ component of $W$ that you need to integrate.

 

[cianfa72]

No. Try writing out explicitly the equation for the $x$ component of $W$ that you need to integrate.

$\dot x=2x \Rightarrow x=e^{2t} \Rightarrow (e^{2t},y_0)$

 

[PeterDonis]

$\dot x=2x \Rightarrow x=e^{2t} \Rightarrow (e^{2t},y_0)$

Yes.

 

[jbergman]

The components of $W$ are $(2y, 0)$. So an integral curve of $W$ will have a parameterization $(x, y) = (2y_0 t, y_0)$, where $y_0$ is a constant (the unchanging $y$ coordinate of the curve) and $t$ is the parameter.

This parameterization doesn't look correctly since for $t=0$ you want the mapping to be the identity mapping. I believe it should be $\theta_t(x,y) = (x+ 2yt, y)$.

 

[jbergman]

$\dot x=2x \Rightarrow x=e^{2t} \Rightarrow (e^{2t},y_0)$

This also doesn't look correct to me. I already gave the parameterization in the post where I computed the pullback.

We have two first order ODEs of the following form.
$$ \frac{d}{dt}\Bigr|_{t=0}\theta_t(x,y) = (2x, 0)$$
with initial conditions
$$\theta_t(x,y) = (x,y)$$
The solution to this is just
$$\theta_t(x,y) = (x + 2xt, y)$$

 

[DrGreg]

This also doesn't look correct to me. I already gave the parameterization in the post where I computed the pullback.

We have two first order ODEs of the following form.
$$ \frac{d}{dt}\Bigr|_{t=0}\theta_t(x,y) = (2x, 0)$$
with initial conditions
$$\theta_t(x,y) = (x,y)$$
The solution to this is just
$$\theta_t(x,y) = (x + 2xt, y)$$

Your solution isn't a one-parameter group i.e. $\theta_{s+t} \neq \theta_s \circ \theta_t$.

I'd like to change the notation slightly and write$$(x_t, y_t) = \theta_t(x_0, y_0).$$Then the differential equation to solve is$$
\frac{d}{dt}\theta_t(x_0,y_0) = V(\theta_t(x_0,y_0)) = V(x_t, y_t) = (2x_t, 0)
$$for all $t$, not just $t=0$, i.e.$$
\frac{dx_t}{dt} = 2x_t \quad ; \quad \frac{dy_t}{dt} = 0
$$which has solutions$$
x_t = x_0 \, e^{2t} \quad ; \quad y_t = y_0
$$i.e.$$
\theta_t(x_0, y_0) = (x_0 \, e^{2t}, y_0)
$$

 

[jbergman]

Your solution isn't a one-parameter group i.e. $\theta_{s+t} \neq \theta_s \circ \theta_t$.

I'd like to change the notation slightly and write$$(x_t, y_t) = \theta_t(x_0, y_0).$$Then the differential equation to solve is$$
\frac{d}{dt}\theta_t(x_0,y_0) = V(\theta_t(x_0,y_0)) = V(x_t, y_t) = (2x_t, 0)
$$for all $t$, not just $t=0$, i.e.$$
\frac{dx_t}{dt} = 2x_t \quad ; \quad \frac{dy_t}{dt} = 0
$$which has solutions$$
x_t = x_0 \, e^{2t} \quad ; \quad y_t = y_0
$$i.e.$$
\theta_t(x_0, y_0) = (x_0 \, e^{2t}, y_0)
$$

Yikes. Bumbled the differential equation. Thanks for the correction!

 

[PeterDonis]

This parameterization doesn't look correctly since for $t=0$ you want the mapping to be the identity mapping. I believe it should be $\theta_t(x,y) = (x+ 2yt, y)$.

Yes, you're right, there should be a constant term in $x$ to make the transformation general; what I wrote was really for the special case of $x = 0$ for $t = 0$.

 

[jbergman]

Yes, you're right, there should be a constant term in $x$ to make the transformation general; what I wrote was really for the special case of $x = 0$ for $t = 0$.

My solution was wrong as pointed out by Dr. Greg.

 

[PeterDonis]

My solution was wrong as pointed out by Dr. Greg.

@DrGreg was talking about the vector field $V = 2 x \partial_x$ that you originally brought up, not the vector field $W = 2 y \partial_x$ that I originally brought up. My post that you just responded to was about $W$. The differential equation for $W$ is $\dot{x} = 2 y$, which integrates to $x = 2yt + C$, where $C$ is a constant that obviously must be the starting value of $x$ so that for $t = 0$ the mapping is the identity.

 

[cianfa72]

So the explicit calculation in #175 becomes:
$$\theta_t^{*} = d\theta_t = diag(e^{2t}, 1)$$
Hence as expected:
$$(\mathscr{L}_V g)_p = \lim_{t \rightarrow 0} \frac{(e^{2t})^2dx^2 +dy^2 - dx^2 - dy^2}{t}= \lim_{t \rightarrow 0} \frac{(e^{4t} - 1)}{t} dx^2=4 dx^2$$
At the end of the day, both vector fields $V=2x\partial_x$ and $V=2y\partial_x$ are not KVFs for the Euclidean metric.

A point looking weird to me is that in the former case if we take two points on two neighboring integral curves (i.e. straight lines parallel to the $x$ axis) with the same curve parameter $t_0$ and flow them along their respective integral curves up to the same parameter $t_1$, the Euclidean distance between them does not change. Yet the Lie Derivative of the euclidean metric tensor $g$ along $V=2x\partial_x$ is not null.

 

[DrGreg]

A point looking weird to me is that in the former case if we take two points on two neighboring integral curves (i.e. straight lines parallel to the $x$ axis) with the same curve parameter $t_0$ and flow them along their respective integral curves up to the same parameter $t_1$, the Euclidean distance between them does not change. Yet the Lie Derivative of the euclidean metric tensor $g$ along $V=2x\partial_x$ is not null.

But take two neighbouring points on the same curve and "flow" them both by the same increase in parameter...

 

[cianfa72]

But take two neighbouring points on the same curve and "flow" them both by the same increase in parameter...

Ah ok, so what the Lie derivative of the tensor metric tells us applies even for the 'distance' of neighboring points on the same integral curve (both "flowed" by the same increase of the parameter), not just for points on neighboring integral curves in the congruence.

 

[jbergman]

Ah ok, so what the Lie derivative of the tensor metric tells us applies even for the 'distance' of neighboring points on the same integral curve (both "flowed" by the same increase of the parameter), not just for points on neighboring integral curves in the congruence.

Yes. That's the point. I think we are getting two hung up on the integral curves, here. A vector field induces a smooth maps of the manifold to itself via the flows.

The Lie derivative is the limit of the pullback of the metric along the integral curve that the point is on. But it uses the differential of the smooth maps associated with the flow to compute the pullback by pushing forward tangent vectors.

This differential encodes information about distance changing in all directions.

So, in effect the Lie Derivative is like a directional derivative. It gives the first order change in the metric when moving along the integral curve at a point. But that change can encode changes in distance in any direction.

It's like an ordinary directional derivative of a vector quantity. The direction just specifies the direction of the step to take. It says nothing of the direction in which the vector will vary along that step.

 

[jbergman]

To beat a dead horse, there are ways the Lie Derivative of the metric can be non-zero.

1. The induced flow of the vector field moves points in a direction that the metric varies on a manifold
2. The induced flow moves points in the same neighborhood at different speeds there by distorting the distance between them. This is reflected in the pushforward of the induced flow and leads to tangent vectors being stretched or squished.

 

[sysprog]

This is reflected in the pushforward of the induced flow and leads to tangent vectors being stretched or squished.

I'm not sure that the 'squished' negative vector properly emerges when we take the first derivative (velocity) of position wrt time.

 

[jbergman]

I'm not sure that the 'squished' negative vector properly emerges when we take the first derivative (velocity) of position wrt time.

I'm not sure I understand this comment. Are you talking about Minkowski space?

 

[cianfa72]

To beat a dead horse, there are ways the Lie Derivative of the metric can be non-zero.

1. The induced flow of the vector field moves points in a direction that the metric varies on a manifold
2. The induced flow moves points in the same neighborhood at different speeds there by distorting the distance between them. This is reflected in the pushforward of the induced flow and leads to tangent vectors being stretched or squished.

About 2. I take it as, even though the metric tensor did not change along the induced flow of the vector field, a non-zero Lie Derivative would imply that the induced flow moves points in a neighborhood such that the 'distance' between them changes when 'flowed' by the same increase of the integral curves parameter.