0
$\begingroup$

For anholonomic system, i.e. Gravitation Eq. 9.22

$$[e_\mu, e_\nu] =c_{\mu\nu}^\alpha e_\alpha$$ where $$[e_\mu, e_\nu]\neq 0$$ for some $\mu,\nu$, the states of the system dependent on its path.

Question 1.a: However, is it still possible to construct some sort of geodesic equation for the anholonomic system? i.e. $$\nabla_u v=0$$

Question 1.b: If it is possible, how would it be different from the usual geodesic equation, ie. Gravitation Eq. 10.28 $$\frac{dv^\alpha}{d\lambda}+ \Gamma ^\alpha _{\gamma \beta} v^\gamma \frac{dx^\beta}{d\lambda}=0$$

Question 2: Especially, since the solution would be path dependent, what physical meaning would it be carrying?

$\endgroup$

1 Answer 1

2
$\begingroup$

A non-holonomic system is different from a nonholonomic basis, which is what MTW refers to. A non-holonomic basis is simply one which cannot be induced by a coordinate chart (i.e. it cannot be written in the form $\big\{\frac{\partial}{\partial x^\alpha}\big\}$ for some coordinate system $x^\alpha$). The wiki article you link is talking about something completely different.


However, is it still possible to construct some sort of geodesic equation for the anholonomic system?

Sure. It looks basically the same - however, the connection coefficients in a non-coordinate basis are different:

$$\nabla_\mu \mathbf X = \left(\frac{\partial}{\partial x^\mu} X^a + \omega_\mu\ ^a_{\ \ b} X^b\right)\mathbf e_a$$

where the vector $\mathbf X$ has been expanded in a non-coordinate basis $\mathbf e_a$. Contrast this with the expression we get when we write $\mathbf X$ in a coordinate basis:

$$\nabla_\mu \mathbf X = \left(\frac{\partial}{\partial x^\mu} X^\nu + X^\sigma\Gamma^\nu_{\mu \sigma}\right) \frac{\partial}{\partial x^\nu}$$

Whereas the $\Gamma$'s are the components of (usually) the Levi-Civita connection, the $\omega$'s are the components of the spin connection (at least, they are when the $\mathbf e$'s are chosen to be orthonormal). One can show that

$$\omega_\mu\ ^a_{\ \ b} = e^a_{\ \ \nu} e_b^{\ \ \lambda} \Gamma^\nu_{\mu \lambda} +e_b^{\ \ \lambda} \frac{\partial}{\partial x^\mu} e^a_{\ \ \lambda}$$


If it is possible, how would it be different from the usual geodesic equation?

Not much different. Consider a worldline with components $x^\mu(\lambda)$. The tangent vector to the worldline can be expressed as

$$\dot{\mathbf x} = \dot x^\mu \frac{\partial}{\partial x^\mu} = \dot x^\mu e^a_{\ \ \mu} \mathbf e_a \equiv \dot x^a \mathbf e_a$$

where $\dot x^a \equiv e^a_{\ \ \mu} \frac{dx^\mu}{d\lambda}$ is not the time derivative of the coordinate $x^a$ (there is no such coordinate!), but rather a linear combination of such time derivatives. From there, the geodesic equation becomes

$$\ddot x^\mu + \dot x^a \dot x^b \omega^\mu_{\ \ ab} = 0$$

where $\omega^\mu_{\ \ ab} \equiv g^{\mu \nu} g_{ac} \omega_\nu \ ^c_{\ \ b}$


Especially, since the solution would be path dependent, what physical meaning would it be carrying?

In light of the fact that a non-holonomic system and a nonholonomic basis are completely different things, this question doesn't make sense. I will say, however, that non-coordinate bases are necessary if you want to consider spinor fields on spacetime, rather than just tensor fields.


Example: Consider $\mathbb R^2$, equipped with the Euclidean metric. In Cartesian coordinates $\{x,y\}$, the coordinate-induced basis vectors are $\left\{\frac{\partial}{\partial x}, \frac{\partial}{\partial y}\right\}$. The metric components in this basis are simply $g_{\mu\nu}=\pmatrix{1 & 0 \\ 0 & 1}$, and the connection coefficients $\Gamma^\mu_{\ \ \nu\sigma}$ all vanish.

If we remove the origin from consideration, we can switch to polar coordinates $\{r,\theta\}$ and the corresponding induced basis $\left\{\frac{\partial}{\partial r},\frac{\partial}{\partial \theta}\right\}$. Since $x=r\cos(\theta)$ and $y=r\sin(\theta)$, we can write

$$\frac{\partial}{\partial r} = \frac{\partial x}{\partial r} \frac{\partial}{\partial x} + \frac{\partial y}{\partial r} \frac{\partial}{\partial y} = \cos(\theta)\frac{\partial}{\partial x} + \sin(\theta)\frac{\partial}{\partial y} = \left(\frac{x}{\sqrt{x^2+y^2}}\right)\frac{\partial}{\partial x} + \left(\frac{y}{\sqrt{x^2+y^2}}\right)\frac{\partial}{\partial y}$$ and $$\frac{\partial}{\partial \theta} = -r\sin(\theta)\frac{\partial}{\partial x} + r\cos(\theta)\frac{\partial}{\partial y} = -y\frac{\partial}{\partial x} + x \frac{\partial}{\partial y}$$

We have expressed the new basis vectors as linear combinations of the old ones. We can find the new metric components: $$g_{rr} = \frac{\partial x}{\partial r} \frac{\partial x}{\partial r} g_{xx}+\frac{\partial x}{\partial r} \frac{\partial y}{\partial r} g_{xy}+\frac{\partial y}{\partial r} \frac{\partial x}{\partial r} g_{yx}+\frac{\partial y}{\partial r} \frac{\partial y}{\partial r} g_{yy}$$ $$=\cos^2(\theta) (1) + \cos(\theta)\sin(\theta) (0) + \sin(\theta)\cos(\theta)(0)+ \sin^2(\theta) (1) =1$$ and similarly, $$g_{r\theta}=g_{\theta r} =0$$ $$g_{\theta\theta} = r$$

We could also find the new connection coefficients; you can find the relatively simple calculation here.

The key thing to understand is the interplay between the change of coordinates and the change of basis. Our procedure from changing the metric components and the connection coefficients from a basis $\left\{\frac{\partial}{\partial x},\frac{\partial}{\partial y}\right\}$ to a basis $\left\{\frac{\partial}{\partial r},\frac{\partial}{\partial \theta}\right\}$ is derived from the corresponding coordinate shift from $\{x,y\}$ to $\{r,\theta\}$. What if we want to use a basis which cannot be induced by such a coordinate change?

For instance, consider the orthonormal polar basis $\{\mathbf e_r, \mathbf e_\theta\}$, where $$\mathbf e_r = \left(\frac{x}{\sqrt{x^2+y^2}}\right)\frac{\partial}{\partial x} + \left(\frac{y}{\sqrt{x^2+y^2}}\right)\frac{\partial}{\partial y} \equiv e_r^{\ \ x}\frac{\partial}{\partial x} + e_r^{\ \ y} \frac{\partial}{\partial y}$$ $$\mathbf e_\theta = \left(\frac{-y}{\sqrt{x^2+y^2}}\right)\frac{\partial}{\partial x} + \left(\frac{x}{\sqrt{x^2+y^2}}\right)\frac{\partial}{\partial y} \equiv e_\theta^{\ \ x}\frac{\partial}{\partial x} + e_\theta^{\ \ y} \frac{\partial}{\partial y}$$

It is straightforward to show that given some generic smooth function $f$, $[\mathbf e_r,\mathbf e_\theta]f \neq 0$, which means that $\mathbf e_r$ and $\mathbf e_\theta$ do not commute. This is in contrast to $\left\{\frac{\partial}{\partial r},\frac{\partial}{\partial \theta}\right\}$ (or indeed, $\left\{\frac{\partial}{\partial \alpha},\frac{\partial}{\partial \beta}\right\}$ for any $\alpha,\beta$) because mixed partial derivatives always commute. It follows that this basis simply does not correspond to any choice of coordinates; by using it, we have decoupled the coordinate system from the tangent vector basis.

$\endgroup$
3
  • $\begingroup$ Thank you! However, could you provide some hints over "nonholomonic basis" could not be expressed as in basis $\{\partial_{x^\alpha}\}$? The book didn't mentioned anything about it. Does it mean the basis had to contain factor such as $\delta_{ab}$ or $\partial^2_{x^\alpha}$? $\endgroup$ Commented May 16, 2020 at 21:03
  • 1
    $\begingroup$ @ShoutOutAndCalculate I have expanded my answer considerably to give you an example of such a basis. Does this answer your question? $\endgroup$
    – J. Murray
    Commented May 16, 2020 at 21:54
  • $\begingroup$ Yes. Thank you. That was an excellent explanation. $\endgroup$ Commented May 17, 2020 at 5:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.