Conservation of energy for stationary particle attached to string

[etotheipi]

I was going to put this in the homework forums, but on second thoughts it's more conceptual so perhaps here is better. It's about problem 4, chapter 6 of Wald. Part (a) is fine, $$u^a \nabla_a u^b = \frac{\xi^a}{(-\xi^c \xi_c)^{1/2}} \left( \frac{\nabla_a \xi^b}{(-\xi^c \xi_c)^{1/2}} + \frac{\xi^b \xi^c \nabla_a \xi_c}{(-\xi^d \xi_d)^{3/2}} \right) = \frac{\xi^a \nabla_a \xi^b}{(-\xi^c \xi_c)} + \frac{\xi^a \xi^b \xi^c \nabla_a \xi_c}{(-\xi^d \xi_d)^{5/2}}$$From the Killing equation we get $\nabla_a \xi_b + \nabla_b \xi_a = 0 \implies \xi^a \xi^c \nabla_a \xi_c = \xi^c \xi^a \nabla_c \xi_a = - \xi^a \xi^c \nabla_a \xi_c \implies \xi^a \xi^c \nabla_a \xi_c = 0$, so the above reduces to $u^a \nabla_a u^b = \frac{\xi^a \nabla_a \xi^b}{(-\xi^c \xi_c)}$.

Then consider $\begin{align*} \nabla^b \ln{V} = \frac{1}{2} \nabla^b \ln(-\xi^a \xi_a) =\frac{\nabla^b (-\xi^a \xi_a)}{2(-\xi^c \xi_c)} = \frac{-\xi^a \nabla^b \xi_a}{(-\xi^c \xi_c)} = \frac{\xi^a \nabla_a \xi^b}{(-\xi^c \xi_c)} \end{align*}$ so then the result $\boxed{a^b = \nabla^b \ln V}$.

Unfortunately, I am stuck on part (b). We already proved somewhere else in the book that there's a conserved quantity $u^a \xi_a$ along the worldline since $u^b \nabla_b (u^a \xi_a) = u^b u^a \nabla_b \xi_a + \xi_a u^b \nabla_b u^a = 0$ (because again the first term is a contraction of a symmetric and antisymmetric tensor, and the second term vanishes if $u^a$ is a tangent vector to a geodesic).

If this were a classical calculation, making use of the equilibrium condition by energy methods amounts to applying the principle of virtual work (virtual tension work + virtual gravitational work $\overset{!}{=} 0$) for some virtual displacement of the particle on the end of the string.

From (a) the force on the particle is $mV^{-1} [\nabla_a V \nabla^a V]^{1/2}$, but how is the force $F_{\infty}$ exerted on the observer at infinity related to $V$? I don't really understand this "energy at infinity" business very well

 

[pervect]

This looks like the problem about the force at infinity being exerted by a "massless string" , correct?

My take on this is a string is a stress energy tensor satisfying the continuity condition $\nabla_a T^{ab} = 0$ along the length of the string, with some boundary conditions at the end of the string. Wald I think refers to this condition as an "energy conservation" argument. The "massless" part is a coordinate dependent statement that $T^{00}=0$ for a static observer. The coordinate independent translation of this statement is that $T_{ab} \xi^a \xi^b = 0$, because a static observer's tangent vector is $\xi^a$, and $T_{ab} u^a u^b$ is the energy density for an observer with a 4-velocity $u^a$.

The tension along the length of the string at infinity is the force/unit normal area the string exerts in an orthonormal basis.

For this part, I suppose I'd recommend thinking about the stress-energy tensor of a string in tension in the Minkowskii metric first.

That's the general approach I remember using for this problem, but I don't recall all the fine details and my notation is probably a bit sloppy, in particular translating the stress-energy tensor of the string into a "force".

I don't recall anymore the details of how I solved for the tension in the string as a function of position.

 

[etotheipi]

Hmm, my guess would be that, at least in a flat spacetime, suppose the string is a region $\mathscr{S} \subseteq M$ where in some orthonormal chart $\phi$ you have $\phi(\mathscr{S}) = \{ (x,0,0,t) : 0 \leq x \leq L , t \in \mathbb{R}\}$, i.e. a timelike plane in $\mathbf{R}^4$. Because the string is under pure axial tension the stress energy tensor just looks like $T^{xx} = - \lambda$ with $\lambda > 0$, and $T^{\mu \nu} = 0$ for all the other components.

Must say I don't have a clue what to do with that, to generalise to the schwarzschild solution. Probably need to think about it a little more

 

[pervect]

That looks right for the stress-energy tensor of the string. I didn't capture the part about the tension being axial in my previous post, but it's a necessary condition. And somewhere along the line we need to include the idea that the string has a constant cross-section. It's not quite clear how to write this condition in coordinate independent notation to me at the moment. We might do something with congruences and the expansion scalar, perhaps. But let's stick to the Minkowskii case, and do a coordinate dependent treatment where we use cartesian coordinates (t,x,y,z).

As far as what to do with it - in Minkowskii space, the goal is to prove that the tension in the string is constant and that this is implied by the conservation of energy condition $\nabla_a T^{ab}=0$.

The only non-zero component of $T^{ab}$ is $\lambda$ which we can assume is a function of z. It can't be a function of x,y, or t. So $T^{zz} = \lambda(z)$

Writing out the non-zero components of $\nabla_a T^{ab} = \partial_a T^{ab}$ gives us $\frac{\partial}{\partial z} \lambda(z) = 0$. This is what we seek to prove, that that the tension in the string is constant in Minkowskii space.

In curved space-times, such as the Schwarzschild space-time, though, the tension in the string won't be constant along the length of the string.

If we were doing this in a coordinate-independent way, we'd have to make the answer work out in arbitrary coordinates in Minkowskii space-time. For instance, we might use spherical spatial coordinates, with the string being oriented radially. Or we might use Rindler coordinates which are still flat space-time. Both cases are interesting on their own in different ways, the Rindler case in particular has the fact that $g_{tt}$ isn't constant, while the spherical case gives us some insight into the spherical Schwarzschild metric. It'd be best if we could do this in a truly coordinate independent manner, but I'm afraid I don't quite see how to write that down :(.

 

[etotheipi]

Ah, okay! So the game-plan is to solve $\nabla_a T^{ab} = \partial_a T^{ab} + \Gamma^a_{ac} T^{cb} + \Gamma^{b}_{ac} T^{ac} = 0$, where the only non-vanishing equation for is for $b=z$, i.e. $\nabla_a T^{az} = \partial_z T^{zz} + \Gamma^{a}_{az} T^{zz} + \Gamma^{z}_{zz} T^{zz} = 0$. So for the given metric it's required to first compute $\Gamma^{a}_{az} = \frac{1}{2} g^{cd} \partial_{z} g_{dc}$ as well as $\Gamma^{z}_{zz}$? But maybe it would be better to keep everything in sphericals... I'll try playing around with it for a bit, thanks!

 

[PeterDonis]

I don't really understand this "energy at infinity" business very well

As Wald notes, the energy at infinity is just the conserved quantity $E = - m u^a \xi_a$, where $\xi_a$ is the timelike KVF of the spacetime. (The minus sign is due to the signature convention he is using, and $m$ is the rest mass of the object.) The name comes from the fact that, if the spacetime is asymptotically flat, then at infinity, $u^a \xi_a = -1$ and $E = m$, so we can think of $E$ more generally as the energy "left behind" in the object after it has been slowly lowered from infinity to some finite radius $r$ at which $| u^a \xi_a | < 1$, so $E < m$. The difference $m - E$ must then be extracted from the object and captured at infinity by the slow lowering process.

Note, btw, that the actual proof that $u^a \xi_a$ is a conserved quantity, as you note, requires that $u^a$ is the tangent vector to a geodesic worldline--but the worldline of an object "hovering" statically at some radius $r$ is not a geodesic! The wrinkle here is that $u^a \xi_a$ is also conserved along worldlines that are integral curves of the timelike KVF--the proof should be simple once you realize what the key property is of such worldlines with regard to the magnitude of $u^a$. So for any two worldlines that are both integral curves of the timelike KVF--including the "worldline at infinity"--there is a well-defined difference in energy at infinity between the two.

The fact that I just mentioned is key to answering part (b) of Wald's problem. If we imagine slowly lowering the object through some radius $r$, the radial gradient of energy at infinity, which can be expressed in terms of the gradient of $V$, will be equal to the force exerted at infinity, $F_\infty$.

So the game-plan is to solve $\nabla_a T^{ab} = \partial_a T^{ab} + \Gamma^a_{ac} T^{cb} + \Gamma^{b}_{ac} T^{ac} = 0$

I don't think the stress-energy tensor of the massless string has anything to do with answering Wald's problem. I think the point of the "massless" requirement is to make the SET of the string traceless, which means that the process of slowly lowering the object does not store any energy in the string or extract any energy from the string (heuristically, the increased tension in the string makes a negative contribution to "stored energy in the string" which exactly balances the increased positive contribution of the string's energy density as it is stretched).

 

[etotheipi]

As Wald notes, the energy at infinity is just the conserved quantity $E = - m u^a \xi_a$, where $\xi_a$ is the timelike KVF of the spacetime. (The minus sign is due to the signature convention he is using, and $m$ is the rest mass of the object.) The name comes from the fact that, if the spacetime is asymptotically flat, then at infinity, $u^a \xi_a = -1$ and $E = m$, so we can think of $E$ more generally as the energy "left behind" in the object after it has been slowly lowered from infinity to some finite radius $r$ at which $| u^a \xi_a | < 1$, so $E < m$. The difference $m - E$ must then be extracted from the object and captured at infinity by the slow lowering process.

Awesome, thanks for the explanation. I was struggling to figure out how you would realize "energy at infinity" physically, so this helps.

Note, btw, that the actual proof that $u^a \xi_a$ is a conserved quantity, as you note, requires that $u^a$ is the tangent vector to a geodesic worldline--but the worldline of an object "hovering" statically at some radius $r$ is not a geodesic! The wrinkle here is that $u^a \xi_a$ is also conserved along worldlines that are integral curves of the timelike KVF--the proof should be simple once you realize what the key property is of such worldlines with regard to the magnitude of $u^a$. So for any two worldlines that are both integral curves of the timelike KVF--including the "worldline at infinity"--there is a well-defined difference in energy at infinity between the two.

Yeah, I did actually wonder about this too! In this case, the $u^a$ are parallel to the Killing field and I think the slightly modified proof just again uses the antisymmetry, i.e. we can re-write $u^b \nabla_b (u^a \xi_a) = \cancel{u^b u^a \nabla_b \xi_a} + \xi_a u^b \nabla_b u^a = \frac{\xi_a \xi^b \nabla_b \xi^a}{-\xi^c \xi_c}$ and then use that $\xi_a \xi^b \nabla_b \xi^a = \xi_b \xi^a \nabla_a \xi^b = - \xi_b \xi^a \nabla^b \xi_a = - \xi^b \xi_a \nabla_b \xi^a \implies \xi_a \xi^b \nabla_b \xi^a = 0$ in order to show the numerator vanishes and deduce that $u^a \xi_a$ is conserved along this particular non-geodesic worldline.

The fact that I just mentioned is key to answering part (b) of Wald's problem. If we imagine slowly lowering the object through some radius $r$, the radial gradient of energy at infinity, which can be expressed in terms of the gradient of $V$, will be equal to the force exerted at infinity, $F_\infty$.

That piece of intuition makes the whole thing extremely easy (unless I screwed up...). In that case you just have $(F_{\infty})^a = -m \nabla^a V$, which is a vector of magnitude $\sqrt{(F_{\infty})_a (F_{\infty})^a} = m [\nabla_a V \nabla^a V]^{1/2}$, which is of course just $VF$.

Let me think for a little while about why exactly that argument works... I'll let you know if there's something I don't understand. Thanks!

 

[PeterDonis]

In this case, the $u^a$ are parallel to the Killing field and I think the slightly modified proof just again uses the antisymmetry, i.e. we can re-write $u^b \nabla_b (u^a \xi_a) = \cancel{u^b u^a \nabla_b \xi_a} + \xi_a u^b \nabla_b u^a = \frac{\xi_a \xi^b \nabla_b \xi^a}{-\xi^c \xi_c}$ and then use that $\xi_a \xi^b \nabla_b \xi^a = \xi_b \xi^a \nabla_a \xi^b = - \xi_b \xi^a \nabla^b \xi_a = - \xi^b \xi_a \nabla_b \xi^a \implies \xi_a \xi^b \nabla_b \xi^a = 0$ in order to show the numerator vanishes and deduce that $u^a \xi_a$ is conserved along this particular non-geodesic worldline.

Yes, exactly. (Physically, what you've done amounts to making use of the fact that the proper acceleration of any curve has to be orthogonal to the tangent vector to the curve.) Note that you had to make use of Killing's equation in the proof, so it only works for an integral curve of the KVF.

In that case you just have $(F_{\infty})^a = -m \nabla^a V$, which is a vector of magnitude $\sqrt{(F_{\infty})_a (F_{\infty})^a} = m [\nabla_a V \nabla^a V]^{1/2}$, which is of course just $VF$.

Yes, you've got it.