GR and acceleration

1. Nov 17, 2007

daandezwart

Hi all!

I'm a first year physics student and I've got some questions I'm hoping someone is able to enlighten me on.

Since I'm only in my first year and I haven't had a full course on GR, I'm most likely to have made some errors, so please point out the flaws in my reasoning and gaps in my knowledge :)

A few days ago I was travelling by train to the university. When it pulled out of the station, I began thinking about its acceleration.

Consider an isolated particle in the following situations:
a) constant velocity in a straight line
b) constant acceleration in a straight line

Let's say, for simplicity, they're both in the x-direction. Then, when you look at their x-t diagrams you'll notice that (a) is a straight line and (b) is curved. So my conclusion from those diagrams is that acceleration could be described as a "deviation from a straight line in space time" or "change of direction in space time". I'm not entirely satisfied with either definition (as I'll get to in a moment), but I'm unsure how else to call it.

Then I started thinking about what I know concerning GR. As I understand it, matter causes space time to curve and what we perceive as acceleration is actually following a straight line on a curved space (here is where I feel my previous definitions start to break down..).

Next thought experiment. Say there's a sufficiently isolated star with a certain mass M which curves space time in a certain way. If I were to fly at it in my, well, space ship with constant velocity, it'd be experiencing an acceleration towards the star when I near it. If I were to counteract the acceleration (I want to pass the star, not hit it), I'd have to accelerate in the opposite direction and thus take a different path through space time. In this case, to stay on course, acceleration would mean I'm (to rephrase my previous defintion) "changing from a 'predetermined' path through space time".

Although I feel this should be enough to ask my question, I'd like to include a more general approach. As I understand it, from a given mass configuration, we're able to define the gravitational potential. Drawing a V-x diagram of such a potential, we'd be able to draw an x-t diagram displaying the path of a particle when it should enter such a potential. I've tried to draw several x-t diagrams resulting from V-x diagrams and all the x-t diagram seems to follow a similar pattern to the 'ordinary accelerating' particle diagram I mentioned earlier. Therefore, I might as well conjure up a gravitational potential resulting in the exact acceleration pattern of the particle as if it were accelerating towards a particular mass. I'm sure in the context of GR something similar is possible, am I right?

There seems to be a connection between 'ordinary acceleration' and 'gravitational acceleration' (obviously, since there's no reason to have two types of acceleration). But, gravitational acceleration is the result of the curving of space time, does that mean the converse is also true? And, if so, how does it work? As I recall an increase in the velocity of a particle means an increase in the energy of that particle and therefore the particle itself has an influence on the curving of space time, but it just doesn't feel 'enough' and I certainly lack the appropriate knowledge to calculate it (for some reason I'm inclined to think curving space time requires a lot of effort <grin>).

So, when the train pulls out of the station, does it actually curve space time? And, getting back to my previous definition, does it then follow that 'changing direction in space time' causes the actual curving? And this is where it gets seriously strange. I've tried to imagine that principle and it seems to lead to bizarre situations and questions, so I'm inclined to say 'no'. But then again, doesn't that mean we'd have to have two types of acceleration? Surely, a particle doesn't need to know the cause of its acceleration to be able to accelerate...

There's obviously something wrong with my line of reasoning. Could someone please point it out and clarify?

Best regards,
Daan

2. Nov 17, 2007

cesiumfrog

That's correct.

Do you realise that falling objects follow geodesics ("straight lines") in GR, whilst objects supported against gravity (like yourself sitting on the Earth's surface) do not (and hence feel acceleration)?

Hence the path of the station through space-time is curved, and the path of the train is more so (and it's mass-energy also, albeit negligibly, can be said to "curve" the surrounding "volume" of space-time, but that's a different issue).

3. Nov 17, 2007

daandezwart

Ouch, my post must have sounded like complete nonsense. I apologize :)

I think I now see where I went wrong and you seem to have nailed the exact spot (I think I mixed up straight and curved lines in space with straight and curved lines in space time in my star v.s. space ship example, while they're not necessarily the same). I'll think about it some more (I've run into some trouble trying to figure out some other examples), consult some books and to try to grasp all I can. I'm really looking forward to a full course on GR! (for which, unfortunately, I'll have to wait atleast two years ...) .

Thank you again!

Best regards,
Daan.

4. Nov 17, 2007

Chris Hillman

Path curvature =/= spacetime curvature

Hi, Daan, welcome to PF! You already got an answer from Cesium but I'll chime in anyway with a more sophisticated response

(EDIT: that is to say, while I am certainly not trying to go over your head, I probably will, in which case you can print out this post, put it in a drawer, and look at it again in a year or two.)

But you know about Minkowski geometry, right?

Indeed, it appears as a hyperbola in Minkowski spacetime, the analog of a circle, i.e. a curve with constant Minkowskian path curvature, namely the magnitude of the constant acceleration.

That's exactly correct; that's what path curvature is all about!

You might not have seen either euclidean or Minkowskian path curvature, but this corresponds in Riemannian/Lorentzian geometry respectively to the covariant derivative of the unit tangent vectors to a curve (timelike curve, in the Minkowski case). For simplicity let's just consider plane curves.

Consider first a parameterized curve in the euclidean plane $(x(\lambda),y(\lambda))$, whose unit tangent vector is
$$\vec{t} = \frac{1}{\sqrt{\dot{x}^2+\dot{y}^2}} \; \left[ \begin{array}{c} \dot{x}\\ \dot{y} \end{array} \right]$$
where dots denote differentiation wrt the parameter. We have two orthonormal frames, the obvious one
$$\vec{e}_1 = \left[ \begin{array}{c} 1 \\ 0 \end{array} \right], \; \vec{e}_2 = \left[ \begin{array}{c} 0 \\ 1 \end{array} \right]$$
and the frame consisting of the unit tangent vector and its normal $\vec{f}_1 = \vec{t}, \; \vec{f}_2 = \vec{n}$. The element of the circle group SO(2) which rotates the first frame into the second is (using the left action)
$$P = \frac{1}{\sqrt{\dot{x}^2+\dot{y}^2}} \; \left[ \begin{array}{cc} \dot{x} & \dot{y} \\ -\dot{y} & \dot{x} \end{array} \right] = \left[ \begin{array}{cc} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{array} \right]$$
This orthogonal linear operator is simply a rotation by angle $\theta$. The rate of rotation wrt the given parameter $\lambda$ is given by the anti-symmetric operator
$$P^{-1} \, \dot{P} = \left[ \begin{array}{cc} 0 & \mu \\ -\mu & 0 \end{array} \right]$$
where
$$\mu = \frac{d\theta}{d\lambda} = \frac{\dot{x} \, \ddot{y} - \dot{y} \, \ddot{x} }{\dot{x}^2 + \dot{y}^2}$$.

(Something to tuck away in your memory for retrieval when you study Lie groups: the operator-valued one-form
$$P^{-1} \, dP = \left[ \begin{array}{cc} 0 & \mu \, d\lambda \\ -\mu \, d\lambda & 0 \end{array} \right] = \left[ \begin{array}{cc} 0 & d\theta \\ -d\theta & 0 \end{array} \right]$$
is known as the (left) Maurer-Cartan form and in general, it allows us to obtain a left-invariant frame on a Lie group from any parameterization. In this case, it is pretty trivial, but notice that our angle does indeed give a notion of length on the unit circle which is invariant under rotations! More generally, a (left and right!) invariant volume form on the rotation group SO(n) is essentially Haar measure, which is a very important notion in probability, analysis, statistics, Lie theory, and physics.)

An arc length parameter s satisfies $\dot{x}^2+\dot{y}^2=1$ (which defines it uniquely, up to an additive constant). This parameterization is particularly important because the distance, measured along our curve, between two points on the curve is simply the difference of the arc parameter values at the two points. Note that this distance along the curve is a geometrical quantity; it does not depend upon our parameterization of the curve, and we can compute it even if we parameterized the curve by some other parameter. Similarly, while the coordinate axes we choose to write down our vectors and matrices are arbitary, the angle made between the second orthonormal frame taken at two points on the curve is not dependent upon our representation. Thus, the rate of turning with respect to arc length is an invariant of a curve, with respect to euclidean geometry. This is the (euclidean) path curvature
$$\kappa = \frac{d\theta}{ds} = \frac{d\theta}{d\lambda} \; \frac{d\lambda}{ds} = \frac{\dot{x} \, \ddot{y} - \dot{y} \, \ddot{x} }{ \left( \dot{x}^2 + \dot{y}^2 \right)^{3/2}}$$
I stress that this formula is valid for a parameterization of our curve by an arbitrary parameter, not just an arc parameter!

Nonetheless, for special choices of parameter, this formula takes on other forms which may look more familiar to some readers. If we are using an arc length parameterization, the path curvature becomes simply $\kappa = \dot{x} \, \ddot{y} - \dot{y} \, \ddot{x}$. For the Monge parameter $\lambda=x$, the path curvature is
$$\kappa = \frac{y^\prime}{ \left(1+{y^\prime}^2 \right)^{3/2} }$$
where we write $dy/dx = y^\prime$.

In the same way we can define Minkowski path curvature, using SO(1,1) instead of SO(2), and then we find
$$\kappa = \frac{\dot{x} \, \ddot{y} - \dot{y} \, \ddot{x} }{ \left( \dot{x}^2 - \dot{y}^2 \right)^{3/2}}$$
which is an invariant of our curve wrt Minkowski geometry, in which the usual squared distance $ds^2 = dx^2+dy^2$ is not invariant but the squared spacetime interval $d\tau^2 = dx^2-dy^2$ is invariant. (Here, we usually take x to be a time coordinate and y to be a spatial coordinate).

The procedure sketched above works for some other Kleinian geometries. See for example Guggenheimer, Differential Geometry, Dover reprint, for a nice discussion of a notion of affine curvature, which is "less precise" than either euclidean or Minkowki path curvature since affine geometry is a common generalization of the two, in which the notions of area, straight line, and convex combination, are all preserved, but the notions of length and angle are not.

The first chapter of MTW, Gravitation, Freeman, 1973, discusses this.

This is the same issue which has been raised here by a half dozen presumably independent posters suffering from a common misconception in just the last few weeks! Namely: the notion of the path curvature of a curve can be generalized to higher dimensions and to curves in curved manifolds. In gtr, the path curvature of a world line is precisely its acceleration vector (the covariant derivative of the unit tangent vector taken wrt itself). This is nonzero iff there are physical forces acting on the observer whose world line is the given curve. In particular, a freely falling observer experiences no "body force" and thus no acceleration, so his world line is a (timelike) geodesic, the analog of a straight line in our two plane geometries above. In gtr, gravitational phenomena are represented entirely via the Riemann curvature tensor of spacetime itself, which is treated as a Lorentzian manifold. This is a different type of curvature, motivated by the Gaussian curvature of a surface. Path curvature always has units of reciprocal length while Gaussian curvature has units of reciprocal area, so there is no possibility of confusing spacetime curvature with path curvature.

Right, you'd be bending your world line away from the geodesic arc it would otherwise follow!

No, not in gtr! You are probably thinking of the Newtonian potential, but that is not how the gravitational field is represented in gtr.

I think your question is: can a suitable isolated configuration of mass-energy cause any desired test particle motion outside the source of the gravitational field? This is difficult to answer briefly because it involves comparing motions occuring in different spacetime manifolds!

Right, except that in gtr, "gravitational acceleration" is not a property of the field but of the motion of an observer who is bending his world line. For example, the world line of a person standing on the surface of the Earth has constant path curvature vector, pointing radially outward and with magnitude equal to the usual Newtonian value (modulo some quibbles which need not concern us here).

Better to say the result of an observer attempting to bend his world line from a geodesic path, i.e. attempting to avoid falling freely. It is true that the geometry of spacetime puts constraints on the possible motions which various observers can exhibit, whether they accelerate or not.

In practice, its mass is neglible, but in principle, yes. Wheeler's slogan says "the density and momentum of mass-energy tells spacetime how to curve; the curvature of spacetime (plus additional factors under an observer's control, like whether to turn on his rocket engine) tell an observer how to move".

Clearly, the problem was that you confused path curvature and spacetime curvature.

BTW, when you study the Bianchi classification of the three-dimensional real Lie groups, these arise as the symmetry groups of certain differential operators such as path curvature. For example,
$$L(y) = \frac{y^{\prime \prime}}{{y^\prime}^3}$$
is invariant under Bianchi II geometry, aka the group of 3x3 upper triangular matrices with ones on the diagonal, UT(3), and solving L(y) = constant gives certain parabolas which are carried to others of their kind by the group UT(3) in its natural action on the plane. Similarly, the euclidean path curvature
$$K(y) = \frac{y^{\prime \prime}}{\left( 1+ {y^\prime}^2 \right)^{3/2}}$$
is invariant under Bianchi VII_0 geometry, i.e. the euclidean group E(2), and solving K(y)=constant gives circles of a particular radius, which are carried to others of their kind by E(2) in its natural action on the plane. And the Minkowski path curvature
$$M(y) = \frac{y^{\prime \prime}}{\left( 1- {y^\prime}^2 \right)^{3/2}}$$
is invariant under Bianch VI_0 geometry, i.e. E(1,1), and solving M(y) = constant gives certain hyperbolas (curves of constant path curvature) which are carried to others of their kind by E(1,1) in its natural action on the plane. And
$$S(y) = \frac{1 + {y^\prime}^2 + y \, y^{\prime\prime}} {\left( 1+ {y^\prime}^2 \right)^{3/2}}$$
is invariant under SL(2,R) in its natural action on the plane.

Last edited: Nov 17, 2007
5. Nov 17, 2007

pervect

Staff Emeritus
Some other posters have responded at great length, but I'll try and give a very quick overview in the interest of an easily digestible post.

The concept you are probably reaching for when you say "straight line" is that of a geodesic, which is roughly equivalent to the closest thing to a straight line possible on a curved surface. For instance, on the curved surface of the Earth, a geodesic would be a "great circle".

Particles that accelerate to remain stationary with respect to some massive object (or are held up by the ground) do not follow geodesic paths, while particles in free fall do follow geodesic paths.

There's a lot of mathematics needed to learn the concept of a geodesic rigorously. One way of viewing a geodesic in GR is that it is a path that extremizes proper time. See for instance Taylor & Wheeler A call to action. This is more specific to GR than other defintions of geodesic, which apply to general curved surfaces (like the earth), but it may involve the least number of new concepts.

The more general way of describing a geodesic path (one that applies to curved surfaces like the Earth) is that a geodesic parallel transports a tangent vector along a curve so that it remains parallel to the curve. If you think of your nose as being a vector, then you could say that a geodesic path is to follow your nose, wherever you go. (And, don't turn your head! :-)).

However, it will probably take a fair number of specialized courses for you to really appreciate what "parallel transport" is. This is one of the introductory topics in differential geometry. "Schild's ladder" is a fairly intuitive way of grasping this, but I'm not going to get into it in detail here. Unfortunately, the only reference I have for this simple and intuitive idea of "parallel transport" is in MTW's textbook gravitation, I haven't found an online source.

6. Nov 18, 2007

daandezwart

Thank you for the warm welcome

I must admit, a lot did (especially the last part), but it certainly gives me something to work towards. I've often been accused of trying to run before I'm able to walk and in this case it seems I'm trying to participate in a 3000m steeplechase

Nevertheless, I don't think it'll hurt to try to take a look at it and I certainly will (I must have read your post about ten times already). I've got a million questions both conceptually as well as mathematically, but I'll limit myself in asking those questions here as I don't think this is the appropriate place for them: after all, this is a forum, not a university

I very much appreciate the comments the three of you made and each contributed in their own way to make me see the errors I made. And, yes, there were plenty, so thanks for not ridiculing me

I've got one more question: I've gotten hold of two books which I hope will introduce me to both GR as well as give me a more solid foundation in physics related mathematics in general. Do any of you know either of them and, if so, are they any good?

"A first course in general relativity" (Bernard F. Schutz, Cambridge U. Press, 1985)
"Mathematic of classical and quantum physics" (F.W. Byron Jr. and R.W. Fuller, Dover publications, 1970).

Best regards,
Daan

7. Nov 18, 2007

Chris Hillman

Hi, Daan,

Contrary to your fears, I thought your initial post showed more insight than many I've seen And you definitely shouldn't expect to understand what I wrote write away, certainly not without struggling with Guggenheimer's excellent textbook.

One difficulty to bear in mind is that the innovation of Minkowski (introducing a geometric interpretation of str) is REALLY useful in thinking about str, but pictorial discussions are unfortunately a bit awkward at PF, hence our preference to send inquirers to good textbooks until such time as they have mastered the geometric picture sufficiently to correctly understand a verbal description of some spacetime diagram. Sometimes a picture really is worth several thousand words, and often the only short answer is "draw a diagram". Which of course won't help if the inquirer doesn't know how to do that!

I don't know the second but the first is one of a dozen gtr textbooks to which I give my highest recommendation. In fact, the textbooks by Schutz and D'Inverno are probably the best introductory undergraduate textbooks I've seen for ambitious young spacetime adventurers. Slightly more demanding but also superb is the textbook by Carroll. The textbook by Ohanian and Ruffini is also very good.

We're getting OT, but here's a thought which possibly should go in a sticky somewhere on "Advice for New Posters at PF, Especially Young Ones" (and I don't mean you here!):

IMO, many newbies are much too sensitive. IMO, high school (which tends to provide the youngest posters at PF, as far as I can make out) is a good place for students to start coming to grips with a basic difference between childhood and adulthood: the world of childhood is nurturing and supportive while the world of adulthood is competitive and demanding (at least in the U.S., at least in scholarly pursuits). I've taught at the college level, and I've seen many college freshman suffer from suddenly being dumped into a highly competitive environment where real demands are being placed upon them for the first time in their lives. And real criticism being leveled at them. For example, college papers are not, or should not be, the same thing at all as high school essays, and as a former college writing tutor I've seen many a shocked reaction on the part of some distraught freshman to the grade given to his first college paper. IMO it's never too early to start trying to learn to write well. I urge all posters to try hard not only to use proper spelling/grammar but to get their basic facts straight, order their thoughts carefully, express their questions well, and generally make a strong initial impression. Of course, I and others here recognize that when one is confused about topic T, it may be impossible to formulate a clear question about topic T, but very often when a bright student tries to formulate a clear question, he will be able to answer it himself! Even if he is not able to do so, his care in proposing his conundrum will be visible to and appreciated by all.

Last edited: Nov 18, 2007