Why doesn't Diffeomorphism Invariance lead to Scale Invariance?

[lugita15]

One of the foundations of General Relativity is diffeomorphism invariance - the fact that the laws of physics are invariant under smooth coordinate transformations, and thus the laws must involve tensors. My question is, why doesn't this imply scale invariance; after all, isn't a change of scale about the smoothest transformation you can have? Yet the universe is manifestly not scale invariant. Where am I going wrong?

Any help would be greatly appreciated.

Thank You in Advance.

 

[Matterwave]

What exactly (in a mathematical sense) do you mean by a change in scale?

 

[lugita15]

What exactly (in a mathematical sense) do you mean by a change in scale?

r→λr, for example (it doesn't have to keep the origin fixed, it can be an affine transformation).

 

[Ben Niehoff]

Scale transformations are not diffeomorphisms.

 

[lugita15]

Scale transformations are not diffeomorphisms.

Why not? They're clearly smooth.

 

[PAllen]

In GR, the diffeomorphism is coupled to transformation of tensors (pull backs is, I guess, the modern word for this). As a result, such transform would be literally as clerical as a change from meters to centimeters.

 

[lugita15]

In GR, the diffeomorphism is coupled to transformation of tensors (pull backs is, I guess, the modern word for this). As a result, such transform would be literally as clerical as a change from meters to centimeters.

Are you saying that scale invariance is a trivial consequence of diffeomorphism invariance? But Feynman says that a house made of matchsticks can be made to have fantastic architectural structures which would collapse if the matchsticks were blown up to human scales and their mass was modified appropriately. He uses this argument, which he traces to Galileo, to demonstrate that he laws of physics are not scale invariant. Yet the only forces involved are gravity and electromagnetism, both of which can be described (at least classically) by tensor equations and thus respect diffeomorphism invariance. So what am I missing?

 

[TrickyDicky]

Are you saying that scale invariance is a trivial consequence of diffeomorphism invariance? But Feynman says that a house made of matchsticks can be made to have fantastic architectural structures which would collapse if the matchsticks were blown up to human scales and their mass was modified appropriately. He uses this argument, which he traces to Galileo, to demonstrate that he laws of physics are not scale invariant. Yet the only forces involved are gravity and electromagnetism, both of which can be described (at least classically) by tensor equations and thus respect diffeomorphism invariance. So what am I missing?

You seem to be conflating global and local invariance, a global change of size like in your example is not what is usually understood as a scale transformation (dilatation) which is more related to local gauge transformations (internal symmetries).

 

[PAllen]

Are you saying that scale invariance is a trivial consequence of diffeomorphism invariance? But Feynman says that a house made of matchsticks can be made to have fantastic architectural structures which would collapse if the matchsticks were blown up to human scales and their mass was modified appropriately. He uses this argument, which he traces to Galileo, to demonstrate that he laws of physics are not scale invariant. Yet the only forces involved are gravity and electromagnetism, both of which can be described (at least classically) by tensor equations and thus respect diffeomorphism invariance. So what am I missing?

No, I'm saying diffeomorphism invariance as used in GR makes a mapping that looks like re-scaling not behave that way, due to the way tensor are transformed. A physical re-scaling (trying to build a house with giant matchsticks) is not a diffeomorphism as used in GR because it would correspond to not transforming tensors. On the other hand, the same re-scaling mapping coupled with transforming tensors would correspond to measuring in centimeters rather than meters, and be of no physical consequence.

 

[lugita15]

No, I'm saying diffeomorphism invariance as used in GR makes a mapping that looks like re-scaling not behave that way, due to the way tensor are transformed. A physical re-scaling (trying to build a house with giant matchsticks) is not a diffeomorphism as used in GR because it would correspond to not transforming tensors. On the other hand, the same re-scaling mapping coupled with transforming tensors would correspond to measuring in centimeters rather than meters, and be of no physical consequence.

You seem to be making a distinction between "passive" transformations and "active" transformations. I thought the whole point of diffeomorphism invariance according to Einstein was to eliminate the physical significance of this distinction. For instance, it shouldn't matter whether you rotate your ruler by 5 degrees clockwise or you rotate the system by 5 degrees counterclockwise. Similarly it shouldn't matter whether you shrink your ruler or enlarge the system.

What would you need to do physically to change the tensors in order to make the system physically the same? Do you just need to change the stress-energy tensor and hence the mass of the matchsticks?

 

[Mentz114]

I have a question, if I may.

Is this an example of diffeomorphism invariance ? Transforming coordinates x^A -> x^a

$$T_{ab}u^a v^b = \Lambda^{A}_{a} \Lambda^{B}_{b} T_{AB} \Lambda^{a}_{A} u^A \Lambda^{b}_{B} v^B =T_{AB} u^A v^B ,\ \ \Lambda^{A}_{a} = \frac{\partial x^A}{\partial x^a}$$

This is an example of the tensorial property that scalar contractions of tensors are invariant, within certain constraints on $\Lambda^A_a$

 

[PAllen]

I have a question, if I may.

Is this an example of diffeomorphism invariance ? Transforming coordinates x^A -> x^a

$$T_{ab}u^a v^b = \Lambda^{A}_{a} \Lambda^{B}_{b} T_{AB} \Lambda^{a}_{A} u^A \Lambda^{b}_{B} v^B =T_{AB} u^A v^B ,\ \ \Lambda^{A}_{a} = \frac{\partial x^A}{\partial x^a}$$

This is an example of the tensorial property that scalar contractions of tensors are invariant, within certain constraints on $\Lambda^A_a$

Yes, this exemplifies diffeomorphism invariance.

 

[PAllen]

You seem to be making a distinction between "passive" transformations and "active" transformations. I thought the whole point of diffeomorphism invariance according to Einstein was to eliminate the physical significance of this distinction. For instance, it shouldn't matter whether you rotate your ruler by 5 degrees clockwise or you rotate the system by 5 degrees counterclockwise. Similarly it shouldn't matter whether you shrink your ruler or enlarge the system.

What would you need to do physically to change the tensors in order to make the system physically the same? Do you just need to change the stress-energy tensor and hence the mass of the matchsticks?

Imagine the tensor T associated with each point in a manifold. Imagine stretching the manifold so a ball a radius 1 becomes a ball of radius 2. The transformation of T doesn't just carry its value as a point 'moves', but is also adjusted by the Jacobian of the stretch transformation. The result is that the mass in the ball is the same as before, and that, by measurement and physics, you cannot distinguish that the stretch has taken place. The use of the tensor transformation ensures that there is no difference between 'moving points' and 'relabeling points'.

Meanwhile, as Ben Niehoff noted much earlier, a physical scaling is not diffeomorphism as used in GR (because a physical scaling does not correspond to use of the tensor transformation 'pullbacks' that preserve metric and physical properties).

 

[lugita15]

Meanwhile, as Ben Niehoff noted much earlier, a physical scaling is not diffeomorphism as used in GR (because a physical scaling does not correspond to use of the tensor transformation 'pullbacks' that preserve metric and physical properties).

Again, is the distinction you're making between active and passive transformations?

You're saying that a physical scaling is not a diffeomorphism because the tensors don't transform in the right way. What else is not a diffeomorphism even though we might naively expect it to be? A physical rotation? A physical boost? A physical translation?

 

[PAllen]

Diffeomorphism invariance (modern name for general covariance) is mathematical device with no physical significance. Any theory at all can be cast in such a way as to have this 'property'.

Physically significant principles that Einstein was really getting at are "no prior geometry", or "the principle of minimal coupling".

 

[Ben Niehoff]

In general, an overall scale has nothing to do with the smooth structure of a manifold. Remember that a manifold doesn't have to have a distance function defined on it at all, so an overall scale is meaningless from the perspective of diffeomorphisms. It can have meaning when you give the manifold a Riemannian metric.

There are some cases in which a scale transformation IS a diffeomorphism. This can only happen on Riemannian manifolds that do not have a characteristic scale. For example, you can do it on flat Euclidean space, because flat space looks the same at all scales.

However, a scale transformation on a sphere is NOT a diffeomorphism, because a sphere has a characteristic scale: the radius. Given the metrics of two spheres,

$$ds_a^2 = a^2 (d\theta^2 + \sin^2 \theta \; d\phi^2),$$
$$ds_b^2 = b^2 (d\psi^2 + \sin^2 \psi \; d\chi^2),$$
it shouldn't be too hard to convince yourself that there is no diffeomorphism (i.e., relation between $(\theta,\phi)$ and $(\psi, \chi)$) that can change a^2 into b^2.

 

[Ben Niehoff]

By the way, when discussing global properties like scale transformations, you should be careful not to be deceived by the local expression of the Riemannian metric.

For example: The metric on a circle of radius R is given by

$$ds^2 = R^2 \; d\theta^2,$$
together with the identification

$$\theta \sim \theta + 2 \pi.$$
The metric looks just like the metric on a straight line, but the identification is important. This is what tells us the circumference of this circle is $2 \pi R$.

Now let's try a (local) diffeomorphism of the form

$$\theta = a \varphi,$$
for some constant a. So now we have

$$ds^2 = a^2 R^2 \; d\varphi^2,$$
and, importantly,

$$\varphi \sim \varphi + \frac{2 \pi}{a}.$$
Therefore we find that the circumference is still $2 \pi R$. So the circle is exactly the same size! This shows that the diffeomorphism we wrote down is not a scale transformation at all, even though it locally rescales the metric!

The fact that a circle has a characteristic scale should have been a clue. In fact, I think infinite, flat space is the only Riemannian manifold on which scale transformations are diffeomorphisms.

 

[PAllen]

Yes, but even locally, let's say you have two points theta=0 and theta=1. Let a=2. Then, they map to phi=0 and phi = 1/2. Lo, and behold, the interval between them is unchanged (4 * 1/4).

That's what I'm getting at in saying that diffeomorphism invariance is victory by definition.

 

[Finbar]

Isn't the point just this...

I can make a transformation


$x^\mu \to a x^\mu$

Which would amount to a diffeomorphism and will have no physical effect provided we transform any tensors that live on the manifold appropriately. In particular the metric tensor should transform such that
$ds^2= g_{\mu \nu} dx^\mu dx^\nu$
is invariant. Physically this means that I don't change lengths and thus I'm not making a scale transformation. Remember it is the metric that really defines scales on a manifold. We could have a manifold with coordinates x on but no metric and preform diffeomorphisms just fine. So if we do have a metric the proper way to make a scale transformation is to re scale the metric

$g_{\mu \nu} \to a^2 g_{\mu \nu}$

This is not a diffeomorphism since we are not transforming the underlying manifold. We are instead rescaling the metric and hence changing the physical size of all objects that live on the manifold.

 

[PAllen]

Isn't the point just this...

I can make a transformation


$x^\mu \to a x^\mu$

Which would amount to a diffeomorphism and will have no physical effect provided we transform any tensors that live on the manifold appropriately. In particular the metric tensor should transform such that
$ds^2= g_{\mu \nu} dx^\mu dx^\nu$
is invariant. Physically this means that I don't change lengths and thus I'm not making a scale transformation. Remember it is the metric that really defines scales on a manifold. We could have a manifold with coordinates x on but no metric and preform diffeomorphisms just fine. So if we do have a metric the proper way to make a scale transformation is to re scale the metric

$g_{\mu \nu} \to a^2 g_{\mu \nu}$

This is not a diffeomorphism since we are not transforming the underlying manifold. We are instead rescaling the metric and hence changing the physical size of all objects that live on the manifold.

Yes, I think that's a reasonable way to put it.

 

[PAllen]

In general, an overall scale has nothing to do with the smooth structure of a manifold. Remember that a manifold doesn't have to have a distance function defined on it at all, so an overall scale is meaningless from the perspective of diffeomorphisms. It can have meaning when you give the manifold a Riemannian metric.

There are some cases in which a scale transformation IS a diffeomorphism. This can only happen on Riemannian manifolds that do not have a characteristic scale. For example, you can do it on flat Euclidean space, because flat space looks the same at all scales.

However, a scale transformation on a sphere is NOT a diffeomorphism, because a sphere has a characteristic scale: the radius. Given the metrics of two spheres,

$$ds_a^2 = a^2 (d\theta^2 + \sin^2 \theta \; d\phi^2),$$
$$ds_b^2 = b^2 (d\psi^2 + \sin^2 \psi \; d\chi^2),$$
it shouldn't be too hard to convince yourself that there is no diffeomorphism (i.e., relation between $(\theta,\phi)$ and $(\psi, \chi)$) that can change a^2 into b^2.

True, but you can pick any small region, and apply a transform that 'almost perfectly' scales any shape in the chosen region, on a coordinate basis. Yet, since the metric and all tensors transform, all measurements and physical observables will be unchanged.

 

[Sam Gralla]

The confusion here is one reason why general relativists have abandoned using coordinates to define scale transformations. (An even better reason is because a coordinate definition won't even work in general on a non-flat spacetime.) Instead, a scale transformation is defined as letting gab -> L^2 gab, i.e., multiplying the metric by some length L. Since the metric measures physical lengths, this corresponds to the intuitive notion of "resizing" as in Feynman's argument. Note that the vacuum field equations of GR are invariant, so that vacuum GR is scale invariant. If there were nothing but black holes and gravitational waves in the universe, you wouldn't be able to measure any scale.

Notice that the notion of diffeomorphisms or coordinates never came up. Relativists always try to make their definitions as coordinate-free as possible, so that potentially confusing issues like those raised in this thread don't come.

There is yet another definition of scale invariance which is adopted in the context of QFT. I tried to explain the differences in the appendix of http://arxiv.org/abs/1002.5045, but it may be a bit terse.

Edit: reading the thread again, I see Finbar already made this point.

 

[Ben Niehoff]

True, but you can pick any small region, and apply a transform that 'almost perfectly' scales any shape in the chosen region, on a coordinate basis. Yet, since the metric and all tensors transform, all measurements and physical observables will be unchanged.

Good point, I agree.

 

[Finbar]

The confusion here is one reason why general relativists have abandoned using coordinates to define scale transformations. (An even better reason is because a coordinate definition won't even work in general on a non-flat spacetime.) Instead, a scale transformation is defined as letting gab -> L^2 gab, i.e., multiplying the metric by some length L. Since the metric measures physical lengths, this corresponds to the intuitive notion of "resizing" as in Feynman's argument. Note that the vacuum field equations of GR are invariant, so that vacuum GR is scale invariant. If there were nothing but black holes and gravitational waves in the universe, you wouldn't be able to measure any scale.

Hi sam,

Won't black hole solutions break conformal in variance with the presence of the Schwarzschild radius providing a scale? It appears as a constant of integration when solving the vacuum equations. Granted the curvature is purely weyl and hence the solution should be invariant under a conformal transformation. I guess the logical contradiction is due to the singularity where the notion of a conformal transformation would break down.

 

[Sam Gralla]

Well the black hole is fine in terms of Feynman's idea of scale invariance (that somebody else posted about): if you scale its size you always still get a nice black hole. This is not true of stars, which come only in certain sizes (fixed by by scale non-invariant physics like nuclear fusion).

Another way to think about scale-invariance is the following. Suppose you were going to take a trip in your spaceship to visit some aliens you knew existed (but hadn't yet met). Being prudent, you want to call ahead and tell them you're coming, and ask them to build you a garage to fit your spaceship in it. How do you tell them how big to make the garage? If the theory of the universe is scale-invariant, you can't. (What are you going to say, make it twice the size of a black hole? Black holes come in all sizes.)

By the way I would be careful with the word "conformal". That often means something different.

 

[Ben Niehoff]

If there were nothing but black holes and gravitational waves in the universe, you wouldn't be able to measure any scale.

I am a little skeptical of this point. Just because the equations are scale-invariant does not mean that solutions to those equations have to be scale-invariant. As Finbar mentions, the Schwarzschild solution has a scale (and the Kerr solution has two scales).

Similarly, field equations for matter in flat space are Lorentz-invariant, but solutions that actually contain bits of matter are not Lorentz-invariant, because obviously those bits of matter have specific locations.

 

[Ben Niehoff]

Another way to think about scale-invariance is the following. Suppose you were going to take a trip in your spaceship to visit some aliens you knew existed (but hadn't yet met). Being prudent, you want to call ahead and tell them you're coming, and ask them to build you a garage to fit your spaceship in it. How do you tell them how big to make the garage? If the theory of the universe is scale-invariant, you can't. (What are you going to say, make it twice the size of a black hole? Black holes come in all sizes.)

Surely you can refer to a specific black hole. Or send them gravitational waves of a specific wavelength (assuming you have the means to create and detect them, and you know the aliens' relative motion to yours with sufficient accuracy).

 

[Finbar]

Surely you can refer to a specific black hole. Or send them gravitational waves of a specific wavelength (assuming you have the means to create and detect them, and you know the aliens' relative motion to yours with sufficient accuracy).

Agreed. To often people think physics is invariant under some transformation but are wrong because solutions break the invariance. translational invariance is the most obvious. The laws of physics may be the same in Paris and London but Paris is not London are different.

 

[lugita15]

Note that the vacuum field equations of GR are invariant, so that vacuum GR is scale invariant. If there were nothing but black holes and gravitational waves in the universe, you wouldn't be able to measure any scale.

So how does the presence of a stress-energy term break scale invariance? To use your language, how would we communicate a length to aliens using GR, without reference to common objects?

 

[TrickyDicky]

Diffeomorphism invariance (modern name for general covariance) is mathematical device with no physical significance. Any theory at all can be cast in such a way as to have this 'property'.

Physically significant principles that Einstein was really getting at are "no prior geometry", or "the principle of minimal coupling".

There seems to be some terminology discrepancies about what is meant by "diffeomorphism invariance", for some it includes what you call minimal coupling, because the diffeomorphism are understood as active diffeomorphisms(flows), not just invariance for passive diffeomorphisms (cordinate transformations); it is agreed that theories in general can be formulated in a coordinate invariant way, but only GR seems to be diffeomorphism invariant in the active way, demanded by the strong Equivalence principle and by identifying the universe with a (pseudo)Riemannian manifold.
As pointed out before a scale transformation is not a diffeomorphism in our spacetime since it is not flat, it is just a local diffeomorphism and therefore there is no global scale invariance.

 

[PAllen]

So how does the presence of a stress-energy term break scale invariance? To use your language, how would we communicate a length to aliens using GR, without reference to common objects?

The particular definition of scale invariance Sam was using in the appendix of the paper he referenced (and co-authored) was:

Multiply the metric by a constant. Is it still a solution of vacuum field equations (if original metric had G(tensor)=0, new metric will have G=0). This is trivially true.

With matter solution, multiply the metric, and the resultant G will be different. Then you can ask the nature of the consequent change to the tensor T (=G). This change will not be trivial, and may even violate an energy condition satisfied by the original T.

 

[Sam Gralla]

I am a little skeptical of this point. Just because the equations are scale-invariant does not mean that solutions to those equations have to be scale-invariant. As Finbar mentions, the Schwarzschild solution has a scale (and the Kerr solution has two scales).

I agree. What I meant (without saying it) was that if you considered some kind of "generic universe with black holes and gravitational radiation" (i.e., a sort of ensemble over all possible solutions), then you wouldn't be able to decide that "black holes are X big", or say anything about an overall scale. You could always take a specific solution, pick a black hole (maybe it only has one black hole anyway), and call that your "scale". But somehow in a generic universe it would be clear that there isn't any "absolute" scale.

This is different from our universe and its theory, where (e.g.) the Compton wavelength of the electron sets a real length scale.

 

[Sam Gralla]

Surely you can refer to a specific black hole. Or send them gravitational waves of a specific wavelength (assuming you have the means to create and detect them, and you know the aliens' relative motion to yours with sufficient accuracy).

Well, if you get to refer to a specific black hole, you might as well refer to your specific spaceship. Similarly, if you get to send the aliens gravitational waves, you might as well send them your spaceship.

I guess I wasn't clear about this, but the idea is not to use any shared knowledge of the particular state of the universe, but only to use the fact that you and the aliens both know the theory of the universe. With a gravity-only (or otherwise scale-invariant) theory, telling them the size of garage to make seems impossible. But with our real universe, you can always say "just measure the Compton wavelength of the electron and make the garage 10^15 times as big".

Of course, none of this is a physical example; it has in mind being able to talk to the aliens by some means that does not use the physical universe. Still, I find it a useful way to think about scale-invariance.

 

[lugita15]

I guess I wasn't clear about this, but the idea is not to use any shared knowledge of the particular state of the universe, but only to use the fact that you and the aliens both know the theory of the universe. With a gravity-only (or otherwise scale-invariant) theory, telling them the size of garage to make seems impossible.

So are you saying nothing in theory of general relativity (as opposed to a particular solution) allows you to communicate a length to aliens without referring to commonly known objects? Could you communicate a length using classical electromagnetic theory?

If you cannot communicate a length using either one of these theories, then how does Feynman's example of a matchstick house make sense? Surely the only forces at play here are gravity and electromagnetism.

 

[Sam Gralla]

So are you saying nothing in theory of general relativity (as opposed to a particular solution) allows you to communicate a length to aliens without referring to commonly known objects? Could you communicate a length using classical electromagnetic theory?

If you cannot communicate a length using either one of these theories, then how does Feynman's example of a matchstick house make sense? Surely the only forces at play here are gravity and electromagnetism.

If there are no sources, Classical electromagnetism is scale-invariant too. (I don't think this has been discussed yet, but in the definition of scale-invariance you get to rescale all fields by some fixed power of lambda. If you can find a power of lambda that preserves the equations of motion, then one says the theory is scale invariant. This corresponds to the physical interpretation we have been discussing.) So I claim you can't tell the aliens a length scale with just source-free electromagnetism and gravity (even if you let them interact; electrovacuum Einstein-Maxwell theory is scale-invariant too). But as soon as sources get involved scale-invariance breaks down. In principle you might be able to find equations for the sources that are scale-invariant, but in any case the types of sources in our real world (including Feynman's matchsticks) are not described by scale-invariant equations. To harp my my previous example, the Compton wavelength of the electron is definitely relevant for the description of wood. It's probably easiest to think about shrinking the matchsticks; at some point you reach the scale of the Compton wavelength of the electron and (whoops!) there aren't going to be matchstick-like solutions any more.