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The metric components in a two-dimensional spacetime are given in terms of the coordinates $(t, x)$ by$$ds^2 = -\cosh x\,dt^2 + dx^2.$$Consider a particle that is "held in position" at $x = 1$. What is the acceleration of this particle, i.e., if the particle has unit mass, how much force must be exerted to hold it in place?

Qmechanic
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user265817
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1 Answers1

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Parametrize the particle's worldline w.r.t. $t$: $$~x^\mu (t) = (t,1)$$

Its four-velocity is: $$u^\mu =\frac{d x^\mu}{d \tau}$$

To evaluate this we use the fact that:

$$d \tau^2 = coshx~dt^2-dx^2$$

Also use $x=1$ and $dx=0$:

$$d \tau^2 = cosh(1) dt^2$$

Therefore:

$$u^\mu = \frac{d x^\mu}{d \tau} = \frac{d t}{d \tau} \frac{d x^\mu}{d t} = \frac{1}{\sqrt{cosh(1)}} (1,0)$$

Its four-acceleration is given by:

$$a^\mu = \frac{du^\mu}{d \tau}+ \Gamma^{\mu}_{\alpha \beta} u^\alpha u^\beta$$

The first term we can immediately see is zero. Since the x-component of velocity is zero we can also discard a number of the Christoffel terms so that we are left with:

$$a^\mu = \Gamma^{\mu}_{00} u^0 u^0$$

If you work out the Christoffel symbols (I'll leave that to you) to get an expression for $a^\mu$, you can find the proper acceleration experienced by the particle by taking the magnitude of the four-acceleration:

$$|a| = \sqrt{g_{\mu \nu} a^\mu a^\nu}$$

Jold
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  • Please do not post complete solutions to homework-like questions. Our policy on this can be found here which includes: “If someone posts an answer to a homework-type question that gives away a complete or near-complete solution, in most cases it will be temporarily deleted.” Please consider deleting this answer yourself. – garyp Apr 06 '16 at 22:54
  • @garyp: "Please do not post complete solutions to homework-like questions. [...]" -- Please note that jld's answer (in the present version above) is quite deliberately left incomplete by including "If you work out the Christoffel symbols (I'll leave that to you) ...". Even more importantly, jld's answer seems to be (only) deriving a value of coordinate acceleration, and therefore appears not to be addressing the OP by user265817 at all (which is asking instead to determine acceleration, and force). – user12262 Apr 07 '16 at 05:24
  • @garyp Apologies. I've removed the final answer as well as the expression for four-acceleration. – Jold Apr 07 '16 at 14:14
  • @user12262 I derived both the x-coordinate acceleration AND the physical (proper) acceleration (magnitude of 4-accel). They just happened to be equal due to the form of the metric ($g_{11}=1$). – Jold Apr 07 '16 at 14:18
  • jld: "I derived [...] the physical (proper) acceleration (magnitude of 4-accel)." -- I'm not convinced that's equivalent. You seem to start out with a manifestly coordinate-dependent quantity: $$u^{\mu} = \frac{dx^{\mu}}{d\tau}.$$I don't see what this "coordinate four-velocity" might have to do e.g. with $$\gamma_{\mathbf v}(c, \mathbf v);$$and consequently I don't see what the quantity $|a|$ you calculate from $u^{\mu}$ might have to do e.g. with $$| \mathbf a | := |~\frac{d}{d\tau}[~\mathbf v~]\mid_{\mathbf v = \vec 0}~|.$$ p.s. +1 for your way of addressing a homework-like question. – user12262 Apr 07 '16 at 16:33
  • @user12262 The components $u^\mu = \frac{d x^\mu}{d \tau}$ are certainly coordinate dependent. The quantities $dx_\mu dx^\mu$ and $u_\mu u^\mu$ are coordinate invariant. On curvilinear/non-inertial coordinates the definition of the acceleration four-vector makes use of the connection via a covariant derivative: $$a^{\mu} := u^{\nu} \nabla_\nu u^{\mu}.$$ This is done to ensure that $a_\mu a^\mu$ is also invariant, and corresponds to the square of the magnitude of the acceleration the particle would actually experience. Thanks for the +1 :) – Jold Apr 07 '16 at 18:12
  • jld: "[...] $a{\mu}a^{\mu}$ is also invariant_" -- Ok, so my "coordinate acceleration" was unjustified. (My apologies, and thanks for having helped me to this insight). It (the "four-acceleration magnitude squared") is indeed an unambiguous and (regarding the OP stipulations) specific quantity. Or (recalling the earlier version of your answer): is it perhaps just a plain ("dimension"-less) real number?? I still wonder how/whether it relates to the (dimensional) quantity $$|~\frac{d}{d\tau}[~\mathbf v~]\mid_{\mathbf v = \vec 0}~|$$ (but this may not necessarily have been the OP question). – user12262 Apr 07 '16 at 18:51
  • @user12262 In general: $$\mathbf a = (\mathbf v \cdot \nabla) \mathbf v$$ where $\nabla$ is the covariant derivative. When the metric is Minkowskian the operator $\mathbf v \cdot \nabla$ reduces to $d/d \tau$, however this is not true in general. The quantities appear dimensionless because in GR we tend to work in geometric units where c=G=1. – Jold Apr 08 '16 at 17:47
  • jld: "In general: $\mathbf{a = (v \cdot \nabla) v}$" -- Above we had (coordinate four-velocity) $u^{\mu}$ with coordinate dependent components; while I had introduced $\mathbf v$ in contrast. Let me spell out its coordinate independent norm for the flat case: $$| \mathbf v_{PQ}[~A~] | := \text{lim}{{s^2[~\varepsilon{AP},\varepsilon_{AQ}~]\rightarrow 0}}[ \sqrt{(\text{Max}{{J,K}}[~s^2[~\varepsilon{PJ},\varepsilon_{QK}~]~])~/~s^2[~\varepsilon_{AP},\varepsilon_{AQ}~]}],$$ for participants $P,Q$ at rest wrt. each other (i.e. in particular with constant and equal mutual ping durations). – user12262 Apr 12 '16 at 19:47
  • jld: "When the metric is Minkowskian the operator $v \cdot \nabla$ reduces to $d/d\tau$" -- Well, that's not quite the same as applying $d/d\tau$ outright in any case. "The quantities appear dimensionless because in GR we tend to work in geometric units where $c=G=1$." -- But, notably, re-introducing "$c$" as a (non-zero) dimensional symbol in the above dimensionless formulas doesn't seem to restore the appropriate dimension of $\mathbf a$. (And surely similarly for $G$, too.) So what's missing? – user12262 Apr 12 '16 at 19:48
  • @user12262 I'm not familiar with the notation you're using in the first comment. "Well, that's not quite the same as applying $d/d \tau$ outright in any case." You're right, it's not. That definition of acceleration is chosen because the operator $d/d \tau$ does not map tensors onto tensors while the operator $\mathbf v \cdot \nabla$ does. "But, notably, re-introducing "c"... So what's missing?" The function cosh(x) implies dimensionless x. If you're working in more common units the function would need to be cosh(Ax) for constant A with dim. $[L]^{-1}$. – Jold Apr 12 '16 at 20:40
  • jld: "I'm not familiar with the notation you're using [...]" -- Does this present a difficulty in matching my notation to notation which you might be using for the same purpose? Or perhaps instead your being not familiar with the purpose itself? This question and these preliminaries may be helpful. "If you're working in more common units [then ...]" -- Surely a particular value of a dimensionful quantity is independent of particular choices of units; and is not just any coordinate value either. – user12262 Apr 13 '16 at 21:10
  • @user12262 Those questions didn't help. Did you invent this notation yourself or did you get it from some other source? If the latter please forward it :). "Surely a particular value of a dimensionful quantity is independent of particular choices of units; and is not just any coordinate value either." What I'm saying is, the question as posed assumes that lengths are dimensionless. The function cosh(x) doesn't make sense otherwise. So you could say the question was expressed badly. If the question were posed better it would use cosh(Ax). – Jold Apr 13 '16 at 23:13
  • jld: "Did you invent this notation yourself or did you get it from some other source? If the latter please forward it" -- Gladly; see my recent comments there. "the question as posed assumes that lengths are dimensionless. [...] So you could say the question was expressed badly." -- That's what I was trying to point out. Of course, my complaint is therefore not only to OP writer @user265817 but also to those responding without being critical; and, to be fair, surely to certain people who inspired such a badly expressed question being asked. – user12262 Apr 14 '16 at 06:16