Prove Fermi-Walker Transport of Gyroscope's Spin Vector w/No Applied Moment

[ergospherical]

With no applied moments, it is asked to prove that a gyroscope Fermi-Walker transports its spin vector $S_{\alpha} = - \dfrac{1}{2} \epsilon_{\alpha \beta \gamma \delta} J^{\beta \gamma} u^{\delta}$. In a local inertial frame $u^{\alpha} = (1, \mathbf{0}) = \delta^{\alpha}_0$ and $\dfrac{d\mathbf{S}}{dt} = 0$ since a co-moving observer sees no precession of the spin axis. Put this condition in the form $u \cdot \partial S^i = 0$. If one further assumes that the time derivative $u \cdot \partial S^0$ of the time component $S^0$ is a constant, say $k$, then one can put $u \cdot \partial S^{\alpha} = k \delta^{\alpha}_0 = ku^{\alpha}$, or\begin{align*}
u \cdot \partial S = ku
\end{align*}and then since $S \cdot u = - \dfrac{1}{2} \epsilon_{\alpha \beta \gamma \delta} J^{\beta \gamma} u^{\delta} u^{\alpha} = 0$,\begin{align*}
0 = u \cdot \partial(S \cdot u) = -k + S \cdot a
\end{align*}and finally $u \cdot \partial S = (S \cdot a)u$, which is the equation of Fermi-Walker transport ($a = u \cdot \partial u$ is the four-acceleration of the gyroscope). What is the justification for putting the time derivative of $S^0$ to a constant? As a guess, it's the zero applied moment condition - but what is even the physical interpretation of the component $S^0$?

 

[PeterDonis]

In a local inertial frame

I don't think you should need to assume any frame for this; Fermi-Walker transport can be defined in terms of tensor equations valid in any frame.

$u^{\alpha} = (1, \mathbf{0}) = \delta^{\alpha}_0$ and $\dfrac{d\mathbf{S}}{dt} = 0$ since a co-moving observer sees no precession of the spin axis.

Here you are considering $\mathbf{S}$ to be a 3-vector; but in your original equation the spin $S_\alpha$ is a 4-vector (or covector, as you've written it). Try computing the 4-vector dot product $u^\alpha S_\alpha$. What do you get?

what is even the physical interpretation of the component $S^0$?

Try reconsidering this question after you have computed $u^\alpha S_\alpha$.

 

[ergospherical]

$S \cdot u = 0$ since $\epsilon_{\alpha \beta \gamma \delta}$ is anti-symmetric in $\alpha, \delta$ whilst $u^{\alpha} u^{\delta}$ is symmetric in $\alpha, \delta$, as in the OP. The local inertial frame is chosen only for convenience. Also, $\mathbf{S}$ is the spatial part of $S^{\alpha} = (S^0, \mathbf{S})$.

 

[PeterDonis]

$S \cdot u = 0$

Yes, exactly.

The local inertial frame is chosen only for convenience.

You don't need to choose a frame at all.

$\mathbf{S}$ is the spatial part of $S^{\alpha} = (S^0, \mathbf{S})$.

Yes, I know that. But since you have shown that $S \cdot u = 0$, in any local inertial frame in which $u = \left( 1, \mathbf{0} \right)$, what will $S^0$ be?

More importantly, now that you have shown that $S \cdot u = 0$, what does that imply about Fermi-Walker transport? (Hint: take the derivative of $S \cdot u$.)

 

[ergospherical]

But since you have shown that $S \cdot u = 0$, in any local inertial frame in which $u = \left( 1, \mathbf{0} \right)$, what will $S^0$ be?

This means that $S^0 = 0$ in the local inertial frame. It makes sense because in a general frame $S \cdot u = -S^0 u^0 + \mathbf{S} \cdot \mathbf{u}$ which implies that $S^0 = \gamma^{-1} \mathbf{S} \cdot (\gamma \mathbf{v}) = \mathbf{S} \cdot \mathbf{v}$, and $\mathbf{v} = \mathbf{0}$ in a local inertial frame.

More importantly, now that you have shown that $S \cdot u = 0$, what does that imply about Fermi-Walker transport? (Hint: take the derivative of $S \cdot u$.)

$u \cdot \partial (S \cdot u) = -k + S \cdot a$, where $k$ is the time derivative of $S^0$ evaluated at the instant the gyroscope is co-moving in this particular local inertial frame, so $k = S \cdot a$.

 

[PeterDonis]

This means that $S^0 = 0$ in the local inertial frame.

Yes. Which removes the need to try to find a physical interpretation for it.

$u \cdot \partial (S \cdot u) = -k + S \cdot a$, where $k$ is the time derivative of $S^0$

$k$ must be zero because $S^0$ is zero. More generally, $\partial(S \cdot u)$ must be zero because $S \cdot u$ is zero. But expanding out $\partial(S \cdot u)$ and setting it equal to zero should give an equation that is relevant to Fermi-Walker transport of $S$.

 

[ergospherical]

$k$ must be zero because $S^0$ is zero... But expanding out $\partial(S \cdot u)$ and setting it equal to zero should give an equation that is relevant to Fermi-Walker transport of $S$.

First part doesn't follow: e.g. if $f(x_0) = 0$, then $f'(x_0)$ is not necessarily zero...
Likewise, $S^0 = 0$ does not imply $k\equiv dS^0/dt = 0$.

It has been derived that $k = S\cdot a \, (\neq 0)$ and therefore $u \cdot \partial S = (S \cdot a)u$, which is the condition for Fermi-Walker transport.

But I was wondering what was the motivation to assume that the time-derivative of $S^0$ is constant, in the first few lines of the solution.

 

[ergospherical]

In fact, I wonder whether the book made a little linguistic slip. They wrote that $k$ is a constant of proportionality, but I reckon $k$ is not constrained to be constant. Rather it is some time-dependent factor $S \cdot a$. Awkward misnomer, probably.

 

[PeterDonis]

$S^0 = 0$ does not imply $k\equiv dS^0/dt = 0$.

If we are talking about $S^0$ in a particular local inertial frame, yes, you're right. I misspoke.

If we look at $S \cdot u$, we know that vanishes everywhere on the worldline, not just at a single point, so its derivative (along the worldline) must be zero. Looking again at your OP, though, it appears you are already taking that into account.

 

[PeterDonis]

the book

Which book?

$k$ is not constrained to be constant. Rather it is some time-dependent factor $S \cdot a$

Your OP shows that $k = S \cdot a$, so the question is whether $S \cdot a$ can vary along the worldline, i.e., whether $u \cdot \nabla \left( S \cdot a \right)$ must vanish or not.

 

[ergospherical]

It’s Lightman’s problem book on relativity, about 3 pages into chapter 11 on angular momentum (I can check exactly later).