# Why does adding kinematical conditions between the coordinates prevent the Riemannian line element to “preserve the Euclidean structure”?

I was reading Kinetic Energy and Riemannian Geometry in The Variational Principles of Mechanics by Cornelius Lanczos; here is the concerned excerpt:

Let us define the line-element of a $$3N$$- dimensional space by the equation: $$\overline{\mathrm ds}^2 = \sum_{i\,=\,1}^Nm_i~(\mathrm dx_i^2 + \mathrm dy_i^2 + \mathrm dz_i^2)\tag{15.11}$$

[...] The form of the line-element $$(15.11)$$ of $$3N$$-dimensional configuration space of $$N$$ free particles has a Euclidean structure, and that the quantities $$\sqrt{m_i}x_i~~~ \sqrt{m_i}y_i~~~\sqrt{m_i}z_i$$ have to considered rectangular coordinates of that space.

If the rectangular coordinates are changed to arbitrary curvilinear coordinates according to the transformation equations $$x_1= f_1(q_1,q_2,\ldots, q_n),\\ .......................\\ .......................\\ z_N= f_{3N}(q_1,q_2,\ldots, q_n),\tag{12.8}$$ the geometry remains Euclidean, although the line-element is given by the more general Riemannian form: $$\overline{\mathrm ds}^2 = \sum_{i, \, k\,= \,1}^n g_{ik}~\mathrm dx_i\mathrm dx_k$$ with $$n= 3N\,.$$

Let us now consider a system with given kinematical conditions between the coordinates. We can handle such a system in two different ways. We may consider once more the previous configuration space of $$3N$$ dimensions but restrict the free-movability of the C-point by the given kinematical conditions which take the form $$f_1(x_1,\ldots, z_n) = 0,\\ ..................\\ f_m(z_1,\ldots, z_n)= 0\,.\tag{15.15}$$ Geometrically, each one of these restricting equations signifies a curved hyper-surface of $$3N$$-dimensional space. The intersections of these hyper-spaces determines a subspace of $$3N-m= n$$ dimensions, in which the C-point is forced to stay. This subspace is no longer a flat Euclidean but a curved Riemannian space.

Another way of attacking the same problem is to express from the very beginning the rectangular coordinates of the particles in terms of $$n$$ parameters $$q_1\ldots,q_n\,.$$ These parameters are now the curvilinear coordinates of an $$n$$-dimensional space whose line element can be obtained by differentiating each side of $$(12.8)$$ and substituting them in the expression $$(15.11)\,.$$ The line-element takes the form: $$\overline{\mathrm ds}^2 = \sum_{i, \, k\,= \,1}^n a_{ik}~\mathrm dq_i\mathrm dq_k\,.\tag{15.16}$$ The $$a_{ik}$$ are here given the functions of the $$q_i\,.$$ The line element is now truly Riemannian not only because the $$q_i$$ are curvilinear coordinates, but because the geometry of the configuration space does not preserve the Euclidean structure of the original $$3N$$-dimensional space, but in infinitesimal regions.

I couldn't comprehend the above marked statements:

From $$(15.11),$$ the author concludes that the $$3N$$-dimensional space is Euclidean. Then transforming the coordinates to generalised coordinates $$q_1,\ldots, q_n,$$ he told the line-element $$\overline{\mathrm ds}^2 = \displaystyle\sum_{i, \, k\,= \,1}^n g_{ik}~\mathrm dx_i\mathrm dx_k$$ develops again the Euclidean Geometry but the line-element being in the more general Riemannian form.

$$\bullet$$ How does the line-element $$\overline{\mathrm ds}^2 = \displaystyle\sum_{i, \, k\,= \,1}^n g_{ik}~\mathrm dx_i\mathrm dx_k$$ develop Euclidean Geometry as stated by the author? For the line-element to be developing Euclidean geometry, the $$\mathrm dx_i\mathrm dx_k$$ terms must not be there, isn't it?

In the very next para he worked on a system with given kinematical conditions between the coordinates and got the form of the line-element as:

$$\overline{\mathrm ds}^2 = \sum_{i, \, k\,= \,1}^n a_{ik}~\mathrm dq_i\mathrm dq_k\,.$$

He asserted the line-element $$(15,16),$$ above, is purely Riemannian and it does not develop the Euclidean Geometry since "the geometry of the space doesn't preserve the Euclidean structure".

$$\bullet$$ Notice both the line-elements viz. $$\overline{\mathrm ds}^2 = \displaystyle\sum_{i, \, k\,= \,1}^n g_{ik}~\mathrm dx_i\mathrm dx_k$$ and $$\overline{\mathrm ds}^2 = \displaystyle\sum_{i, \, k\,= \,1}^n a_{ik}~\mathrm dq_i\mathrm dq_k$$ are in the general Riemannian form and looking almost alike.

Still, the former preserves the Euclidean Geometry while the latter develops geometry which does not "preserve the Euclidean structure of the $$3N$$-dimensional space".

Why is it so? Why does, by introducing kinematical conditions between the coordinates, avert the line-element $$\overline{\mathrm ds}^2 = \displaystyle\sum_{i, \, k\,= \,1}^n a_{ik}~\mathrm dq_i\mathrm dq_k$$ unlike $$\overline{\mathrm ds}^2 = \displaystyle\sum_{i, \, k\,= \,1}^n g_{ik}~\mathrm dx_i\mathrm dx_k$$ to preserve the Euclidean Structure?

Of course, I wouldn't expect same behaviour from both cases as the first one deals with free-particles while the second one deals with a system with given kinematical conditions i.e., constraints.

But, I'm not getting, in spite of appearing to be exactly same, why the first line-element can develop the Euclidean Geometry while the second one not.

Could anyone shed light on this?

I have discussed about this in the chat where John Rennie put forth:

The curvature is coordinate independent, so it does not depend on the choice of coordinates. $$^\S$$

Yes. Starting from a flat space it's possible to choose curved coordinates in which the metric doesn't look like a Euclidean metric. However the apparent curvature is due to your choice of coordinates and not a property of the space. $$^{\S\S}$$

I thought if that is so, then it seems pretty okay to me for the first case as the $$3N$$ dimension was Euclidean while after transforming the coordinates to generalised one, the line-element $$\overline{\mathrm ds}^2 = \displaystyle\sum_{i, \, k\,= \,1}^n g_{ik}~\mathrm dx_i\mathrm dx_k$$ also develops Euclidean Geometry.

But for the second case, the author explicitly mentioned the line-element $$\overline{\mathrm ds}^2 = \displaystyle\sum_{i, \, k\,= \,1}^n a_{ik}~\mathrm dq_i\mathrm dq_k$$ didn't develop Euclidean Geometry not due to just the curvilinear coordinates:

The line element is now truly Riemannian not only because the $$q_i$$ are curvilinear coordinates, but because the geometry of the configuration space does not preserve the Euclidean structure of the original $$3N$$-dimensional space.

That's the thing I didn't get why it didn't preserve the Euclidean structure "even though curvature is coordinate dependent".

Where am I misinterpreting and mistaking?

$$^\S$$ Link I

$$^{\S\S}$$ Link II

• Comment to the post (v1): Consider to double check for $x\leftrightarrow q$ typos, and $3N \leftrightarrow n$ summation limit typos. – Qmechanic Sep 19 '16 at 14:38
• @Qmech, Are you talking about $(12.8)\,?$ I would read between the lines then..... – user36790 Sep 19 '16 at 19:50
• No, eq. (12.8) is OK. – Qmechanic Sep 19 '16 at 20:04
• Exactly, @Qmech, where did I mess up? I know $n$ is not same in the two cases. Hmmm. – user36790 Sep 19 '16 at 20:42

## 1 Answer

Lanczos is essentially saying that if one imposes constraints on a $3N$-dimensional Euclidean space (which is an affine space with an Euclidean metric of Euclidean signature), one gets (under certain regularity assumptions) an embedded $n$-dimensional submanifold, with the generalized coordinates as local coordinates. The submanifold inherits a Riemannian metric by pullback from the ambient Euclidean space. Be aware that the word Euclidean has different meanings in mathematics and physics, cf. my Phys.SE answer here.