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

ergospherical
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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##?
 
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ergospherical said:
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.

ergospherical said:
##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?

ergospherical said:
what is even the physical interpretation of the component ##S^0##?
Try reconsidering this question after you have computed ##u^\alpha S_\alpha##.
 
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##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})##.
 
ergospherical said:
##S \cdot u = 0##
Yes, exactly.

ergospherical said:
The local inertial frame is chosen only for convenience.
You don't need to choose a frame at all.

ergospherical said:
##\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##.)
 
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PeterDonis said:
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.

PeterDonis said:
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##.
 
ergospherical said:
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. :wink:

ergospherical said:
##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##.
 
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PeterDonis said:
##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.
 
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.
 
ergospherical said:
##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.
 
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ergospherical said:
the book
Which book?

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