# Gravitational field strength

1. Aug 26, 2013

### muppet

"gravitational field strength"

Hi all,

I've been reading some lecture notes by G. t'Hooft, available from

http://www.staff.science.uu.nl/~hooft101/lectures/genrel_2010.pdf

On page 12, t'Hooft is discussing the Rindler space of an observer using co-ordinates $\xi^{\mu}$ with constant acceleration g in the $\xi^3$ direction when he says says: "The gravitational field strength is given by $\rho^{-2} \vec{g} (\xi)$",
where $\rho = 1+g\xi^3$ is the "local clock speed" (which, with a bit of help from MTW, I take to mean $\frac{d \tau}{d \xi^0}$).

I can't see that he defines "gravitational field strength", or $\vec{g}(\xi)$ anywhere (just a constant vector $\vec{g}=(0,0,g)$.)

Could someone help me understand what precisely is meant by the "gravitational field strength" in GR, as well as where the above formula comes from?

2. Aug 26, 2013

### Mentz114

There's no quick answer to this, because energy in GR is tricky. The nearest thing to the energy of the field can only be defined for spacetimes possessing a stationary Killing vector field. There is no tensor that fits the bill, so the proposed measures are at best coordinate dependent or require some other constraints. Examples are the Komar mass and the Landau-Lifshitz pseudo-tensor.

The local clock rate can be identified with the inverse of the $(d\xi^0)^2$ metric coefficient and I suspect that is what t'Hooft is using. The Rindler chart is a set of coordinates in Minkowski spacetime, so strictly it is SR, and one identifies the acceleration as similar to that of a 'uniform' gravitational field.

I'm not an expert and I think you'll get a better answer when some of the SA's log in.

Last edited: Aug 26, 2013
3. Aug 26, 2013

### robphy

In classical Newtonian gravity, "gravitational field strength" is also known as the "acceleration due to gravity".
"Gravitational field strength" (gravitational force per gravitational charge) is better because it is analogous to "electric field strength" (electrostatic force per unit electric charge).

4. Aug 26, 2013

### muppet

robphy: I know that he's talking about the effect that an observer in an accelerating lift/rocket ship/whatever will see something fall towards the floor. But I don't understand precisely what he means when he extends this to a local concept that depends on position $\xi^{\mu}$. There's only one accelerated observer, and if I understand the construction correctly, then he regards himself to be at constant $\xi^{i}$ for the spatial coordinates. Is he talking about something like the "acceleration" of an inertial observer in the Minkowski space (located at constant $x^i$) relative to the curves of constant $\xi^{i}$? If so, how do I see that this is given by the formula that t'Hooft claims it is?

5. Aug 26, 2013

### Mentz114

Yes, t'Hooft says this in 3.15 on the next page.

6. Aug 27, 2013

### PAllen

This section of t'Hooft is all SR, no GR yet.

A hint is to study the relation of equations 3.12 to the diagram. You should be able to justify the picture feature that lines of constant elevation from the floor of the elevator become more and more straight as represented in an inertial frame. This means that points above the floor of the elevator are undergoing less acceleration, thus less gravitational field strength. Try to use this hint plus 3.12 plus the time dilation expression 3.13 to derive observation ii) .

7. Aug 27, 2013

### muppet

PAllen: If I understand correctly, the basic physical point here is that the observer in the lift sees a distant object, say some star, to accelerate towards the floor of the lift more slowly than say a tennis ball that he's dropped from eye level -right?

Based on what you said about the curves of constant $\xi^3$, I tried computing the four-acceleration of a particle at constant $\xi^3$; in the frame of the accelerated observer, I find it to be
$$a^{\mu}=(0,0,g \rho^{-1},0)$$
- so is my calculation out by a factor of $\rho$ somehow, or am I calculating the wrong thing? (I'm not entirely sure if what t'Hooft means by the vector $\vec{g}(\xi)$ is different from the vector $\vec{g}=(0,0,g)$ describing the acceleration due to gravity within the elevator.)

Last edited: Aug 27, 2013
8. Aug 27, 2013

### Mentz114

I also calculated the 4-acceleration $a_\mu=\nabla_\nu \xi_\mu \xi^\nu$ and I agree with your result. But this is in the coordinate basis. I'll try the frame-field basis later, and I suspect the offending factor will go away.

9. Aug 27, 2013

### PAllen

The mystery should be resolved if you take the norm of the 4-acceleration using the Rindler metric. However, I see that metric norms have not been introduced at this point in the text. If you are not familiar with this, then I am not quite sure how t'Hooft expected this result to be derived. Note that he certainly hasn't introduced frame basis yet either (in fact, he never does, in this document).

Last edited: Aug 27, 2013
10. Aug 27, 2013

### muppet

Forgive me for being dense, but I don't see how this helps

The Rindler metric for this case is
$$ds^2 = \eta_{\mu nu} d x^{\mu} d x^{\nu} =-\rho^2 (d \xi^0)^2+(d \xi^1)^2+(d \xi^2)^2+(d \xi^3)^2$$
where $\eta$ is the ordinary Minkowski metric. The expression given for $a^{\mu}$ given above has zero time component, so the "non-standard" diagonal time entry shouldn't change anything.

Also, the way I actually worked out this vector was computing the proper acceleration of a particle following a trajectory of constant $\xi^3$ in locally inertial coordinates, using the expression 3.12 (together with the time dilation relation to relate the local proper time of this particle to that of the observer in the elevator) and transformed it to the frame of the accelerated observer using the chain rule ...

And having typed that out, I've remembered that in passing from the 4-velocity to the 4-acceleration, I need to take the covariant derivative, as it's a nonlinear coordinate transformation, even though we're in flat space. D'oh.

Last edited: Aug 27, 2013
11. Aug 27, 2013

### WannabeNewton

Well even if you start with the Rindler metric $g_{ab}$, take the covariant derivative operator $\nabla_a$ associated with $g_{ab}$ and compute $a^{b} = u^a\nabla_a u^b$ for a static observer ($u^b = (-\xi^a\xi_a)^{-1/2}\xi^b$ where $\xi^a$ is the time-like killing vector field) in the coordinate basis, you will still find that the components of the acceleration in this basis are $a^{\mu} = g\rho^{-1}\delta^{\mu}_{z}$ from which you get $a = g\rho^{-1}$ using $g_{ab}$. The "norm" $a$ of the acceleration is independent of any choice of basis so I don't know what exactly the author in that pdf is computing.

12. Aug 27, 2013

### Mentz114

I calculated the acceleration in the coordinate basis and the frame basis for this observer and the result is the same. I wondered why, now I know. Thanks.

13. Aug 27, 2013

### muppet

Have you done this calculation? I know that the norm is independent of the choice of basis; but I figured that as I hadn't taken the covariant derivative properly I was computing the norm of the wrong 4-vector! So I'm currently wading through a sea of Christoffel symbols...

14. Aug 27, 2013

### WannabeNewton

I did the calculation yes. $(-\xi^{a}\xi_{a})^{-1/2} = (1 + gz)^{-1} = \rho^{-1}$ so

$u^{\nu}\nabla_{\nu}u^{\mu}\\ = \rho^{-1}\nabla_{0}(\rho^{-1}\delta^{\mu}_{0})\\ = \rho^{-2}\Gamma ^{\mu}_{00}\\ = \frac{1}{2}\rho^{-2}g^{\mu\nu}(\partial_{0}g_{\nu0} + \partial_{0}g_{\nu0} - \partial_{\nu}g_{00})\\ = \frac{1}{2}\rho^{-2}g^{\mu z}\partial_{z}(1 + gz)^{2}\\ = g\rho^{-1}\delta^{\mu}_{z}$

15. Aug 27, 2013

### PAllen

I admit I didn't do the computation but simply assumed the very world famous author must be right (and the general statement therein "The gravitational field strength felt locally is ... is inversely proportional to the distance to the point" is obviously true. It is true that a norm is invariant, but if you compute 4-acceleration per some coordinates, you must take its norm per the metric in those coordinates, so I blithely assumed it would work out.

I agree with all the other calculations here, and now wonder whether the author's formula is simply a typo; it should be -1 exponent not -2.

16. Aug 28, 2013

### PAllen

Yet another way to confirm the results above is to compute in Fermi-Normal coordinates for a given uniformly accelerating reference observer. Here, the metric is:

d$\tau$^2 = (1+gx)^2 dt^2 - dx^2

The 4-velocity of a world line of constant x is then:

U = (1/(1+gx),0)

The only non-vanishing connection components are (first index meant as 'on top'):

[0,01] = g/(1+gx)
[1,00]=g(1+gx)

Using these the absolute derivative by proper time of U (other people call it covariant derivative) is:

A = (0,g/(1+gx)) // 4 acceleration

whose norm is, again, g/(1+gx).

Last edited: Aug 28, 2013
17. Aug 28, 2013

### muppet

Thank you all for your replies.

I have just one more question: why is it that we focus on curves of constant $\xi$? Surely, the physically relevant thing, in keeping with what I said above-

-is how this observer sees an object at rest in the global inertial frame accelerate in his coordinates?

I tried computing this by giving an object a 4-velocity (0,0,0,1) in the inertial frame and transforming this to a vector $u= (0,0,-sinh(g \xi^0), cosh(g \xi^0)$ in the accelerating frame, with 4- acceleration given by
$$a^{\mu}=\frac{d u^{\mu}}{d x^0}+\Gamma^{\mu}_{\rho \sigma}u^{\rho}u^{ \sigma}$$
and PAllen gave the non-zero Christoffel symbols before.

However, the answer I got was a terrible mess of hyperbolic functions, and didn't remotely resemble anything like t'Hooft's expression.

EDIT: Just to be clear, the accelerating observer is using coordinates $\xi^{\mu}$ locally, whilst $x^{\mu}$ are inertial coords in Minkowski space.

Last edited: Aug 28, 2013
18. Aug 28, 2013

### WannabeNewton

Well an observer at a constant $(x,y,z)$ in Rindler space-time is applying some force (say using a rocket) to hover in place against the gravitational field so the acceleration of the observer gives the force per unit mass needed to counter what would, in Newtonian mechanics, be the gravitational field strength at $(x,y,z)$. Keep in mind that we can really only say all this because the space-time is stationary i.e. there exists a time-like killing vector field, so the notion of being at rest in the gravitational field makes sense.

19. Aug 28, 2013

### muppet

The thing that gets me is that in Newtonian gravity, the observer who doesn't apply an "upwards" force $mg$ to remain at rest in a gravitational rest falls, with an acceleration given by $g$. Here, we've worked out what acceleration a particle must undergo in order to be perceived as being at rest by our accelerating observer; but it seems to me as if this isn't the same as the rate at which objects at rest in the inertial frame "fall" according to the accelerated observer, which is what I'd expected. (Unless I've made a mistake with my maths, which happens more often than not ).
I don't understand what you mean. The inertial observers (i.e. the free-floating observers of Minkowski space-time) follow geodesics of Rindler space-time so they don't have any proper acceleration. If you are talking about the coordinate acceleration of an inertial observer relative to a stationary observer in Rindler space-time, then you have to use the coordinates PAllen used earlier in the thread i.e. Fermi-Normal coordinates; these are local coordinates whose spatial origin is always centered on the stationary observer i.e. they are coordinates comoving with the stationary observer. If the inertial observer is coincident with the origin of the coordinates at some event, the coordinate acceleration of said observer in these coordinates at that event will be given by $\frac{\mathrm{d} ^2 x^{i}}{\mathrm{d} t^2} = -a^{i} + 2(a_jv^j) v^i$ where $a^i$ is the proper acceleration of the stationary observer and $v^i$ is the 3-velocity of the inertial observer at that event relative to the stationary observer. The first term is the usual Newtonian inertial acceleration whereas the second term is a relativistic addition to that.