# Why doesn't Diffeomorphism Invariance lead to Scale Invariance?

1. Jan 30, 2012

### 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.

2. Jan 30, 2012

### Matterwave

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

3. Jan 30, 2012

### lugita15

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

4. Jan 30, 2012

### Ben Niehoff

Scale transformations are not diffeomorphisms.

5. Jan 30, 2012

### lugita15

Why not? They're clearly smooth.

6. Jan 30, 2012

### 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.

7. Feb 1, 2012

### lugita15

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?

8. Feb 1, 2012

### TrickyDicky

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) wich is more related to local gauge transformations (internal symmetries).

9. Feb 1, 2012

### PAllen

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.

10. Feb 1, 2012

### lugita15

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?

Last edited: Feb 1, 2012
11. Feb 1, 2012

### Mentz114

I have a question, if I may.

Is this an example of diffeomorphism invariance ? Transforming coordinates xA -> xa

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

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

Last edited: Feb 1, 2012
12. Feb 2, 2012

### PAllen

Yes, this exemplifies diffeomorphism invariance.

13. Feb 2, 2012

### PAllen

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).

14. Feb 2, 2012

### lugita15

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?

15. Feb 2, 2012

### 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".

16. Feb 2, 2012

### 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.

17. Feb 2, 2012

### 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.

Last edited: Feb 2, 2012
18. Feb 2, 2012

### 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.

19. Feb 2, 2012

### 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.

20. Feb 2, 2012

### PAllen

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

21. Feb 3, 2012

### PAllen

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.

22. Feb 3, 2012

### 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.

Last edited: Feb 3, 2012
23. Feb 3, 2012

### Ben Niehoff

Good point, I agree.

24. Feb 4, 2012

### Finbar

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.

25. Feb 4, 2012

### 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.