# Accelerated motion

1. Apr 22, 2006

### Ratzinger

I like to understand better how relativity treats accelerated motion.

1.How does SR handle it? I only have a formula in front of me that says how two inertial systems relate their observations of an accelerated object with each other. But what does SR say about accelerated observers?

2.How does GR deal with accelerated motion other than free fall (accelerating rockets)? Or does it only talk about free fall which would make sense to me because it is a theory of gravity?

thanks

2. Apr 22, 2006

### robphy

In SR, [4-]accelerated observers have non-geodesic worldlines.
In general, an accelerated observer cannot set up a coordinate system to cover all of Minkowski spacetime.

In GR, [4-]accelerated observers have non-geodesic worldlines.
An object in free fall has a geodesic worldline... it has zero [4-]acceleration.

3. Apr 22, 2006

### pervect

Staff Emeritus
Suppose you have an inertial coordinate system $x^i$. You have a formula in front of you that gives generalized coordinates $\chi^i$ in terms of the $x^i$.

I believe from past discussions that already know how to deal with transformations of $x^i$ via SR, and that you are familiar with the tensor notation being used here - i.e. you know that the $x^i$ transform according to the Lorentz transforms $x'^j = \Lambda^j{}_i x^i$, and you know that the square of the Lorentz interval is given by $\eta_{ij} x^i x_j$, where $\eta_ij$ is a Minkowskian metric. With a -+++ sign convention $\eta_{00}=-1$, $\eta_{11}=\eta_{22}=\eta_{33}=1$, and all other terms are zero.

If you have some questions on these points, please feel free to ask, I'm assuming you know all this already and need at most a reminder.

The question then arises - can you compute the Lorentz interval in terms of the new variables, $\chi^i$, which are nonlinear functions of the $x^i$?

The answer is yes - with a little work, you can write write ds^2 in terms of the new variables $\chi^i$ by using a general, non-Minkowskian metric $g_{ij}$

When one does this, is one doing SR or GR? I'm going to weasel out of this and not take a firm stand :-).

One is clearly using a non-Minkowskian metric, which looks like GR, but on the other hand conceptually all we have done is SR in generalized coordinates. One is not getting into the Einstein field equations, or curved space-time. So you'll find this topic discussed mainly in GR textbooks, so peadagogically that's where to look for enlightenment. On the other hand one isn't doing anything that involves different concepts or physics, just different notation and different mathematics.

I'm not quite sure how to answer your question about GR - it can handle accelrating rockets, it can also handle space-times which have a non-zero curvature tensor (unlike SR), it can handle gravity.

Probably the simplest thing to say is that in SR, we write

4-acceleration = $$a^i = \frac{d}{d\tau} u^i$$

where $u^i$ is the 4-velocity, and the 4-acceleration of a body following a geodesic is zero.

Because we are working with generalized coordinate systems, to express the 4-acceleration in generalized coordinates we need to write instead

4-acceleration = $\nabla_u u^i$, where u is a tangent to the curve. (I hope I've written that correctly).

In component language this can be expressed via the geodesic equation

$$a^i= \frac{\partial u^i}{\partial \tau} + \Gamma^i{}_{jk} u^j u^k$$

The $\Gamma^i{}_{jk}$ are the Christoffel symbols associated with the metric.

Note again that the $a^i$ here are all zero for an object following a geodesic, and give the 4-accleration for an object that is accelerating.

You can see from the above equations that an object at rest in an arbitrary coordinate system has a 4-accleration of $\Gamma^i{}_{00}$.

To understand the details of this, you'll need to learn how to take the covariant derivative, which is where Christoffel symbols are also explained.

Conceptually what we are doing is replacing partial differentiation which is valid only in an inertial coordinate system by a covariant derivative which we can use in any coordinate system - it's part of our move to go from inertial coordinates to generalized coordinates. We still haven't gotten to curved space-times yet, but we've taken a big step at this point to the mathematical formalisms needed to deal with GR.

Last edited: Apr 22, 2006
4. Apr 23, 2006

### Perturbation

$$a^i= \frac{du^i}{d\tau} + \Gamma^i{}_{jk} u^j u^k$$ to be picky.

5. Apr 23, 2006

### Ratzinger

First of all, thank you for answering. Very helpful and highly appreciated.

As for the GR part of my original post I meant this.

When forces act, bodies get kicked out of their geodesics. But these forces are no gravity, so we need more than the original geodesic equation to describe the motion, more than GR.

Curved spacetime is the background for the other three forces and the acceleration they cause, but it gives no equation of motion for non-gravitational forces. It gives the background on which the spaceship accelerates, but does not describe it like it does for a falling elevator.

So in this direction was my question: how does GR describe acceleration other than free fall?

6. Apr 23, 2006

### Stingray

I think that this sentence highlights your misunderstanding. GR does not say that objects must move on geodesics. That statement is derived from more general considerations once certain assumptions on the body have been made. GR does not have any problems handling non-geodesic motion. GR is a complete framework for classical mechanics; not just a theory of gravity.

Here is (one) standard way of doing mechanics in general relativity: Start with Einstein's equation. Take the divergence of both sides. With Bianchi's identity, this gives you

$$\nabla_{a} T^{ab} = 0[/itex] This is an equation of motion. To see this in a simple case, write down the stress-energy tensor for a point particle, and plug it in (there are technical problems with this, but ignore them for now). You'll get the geodesic equation for its trajectory. Now write down the stress-energy tensor for a pressureless fluid (dust). Plug it in, and you'll find that each particle moves on a geodesic. Then try adding pressure. You'll get the Navier-Stokes equations. These imply that particles are accelerated by interactions with their neighbors. To see what happens with a long-range force, let's add an electromagnetic field. This has a stress-energy tensor of its own, which has a known form in terms of the field $F^{ab}$. We can therefore write $T^{ab}=T^{ab}_{\mathrm{matter}} + T^{ab}_{\mathrm{em}}$. We are only interested in the motion of the matter (the fields are just governed by Maxwell's equations), so we want to find what happens to $T^{ab}_{\mathrm{matter}}$. After some manipulation, you can show that (I might have a sign error) [tex]\nabla_{a} T^{ab}_{\mathrm{matter}} = F^{ab} J_{a}$$,

where $J^{a}$ is the electromagnetic current vector (which is divergence-free).

Plug in the relations for a point charge, and you'll find the Lorentz force law. So again, acceleration is not a problem.

Last edited: Apr 23, 2006
7. Apr 23, 2006

### Stingray

I don't know what you're background is, but Synge's relativity book has a beautiful and very complete discussion of accelerated motion in GR that is not contained in any other textbooks that I've seen.

It turns out that it is possible to define a number of (curvature) scalars that characterize a particular accelerating worldline. One of these is the magnitude of the acceleration, but there are others as well. He goes on to explain how one can then measure each of these scalars by a number of fun experiments. For example, if an observer were to throw a ball "up" so that he could later catch it, the details of this process give some information about the curvatures of his own worldline. I think Synge called this process "ballistic suicide" . He had a great writing style.

8. Apr 23, 2006

### pervect

Staff Emeritus
$m a^i = F^i$ where F^i is the "4-force", and a^i is given by the geodesic equation I gave earlier - which is just the covariant derivative of the 4-velocity.

i.e.

$$a^i= \frac{du^i}{d\tau} + \Gamma^i{}_{jk} u^j u^k$$

the m in the first equation is just the invariant mass of the particle, i.e $$\sqrt{E^2-p^2}[/itex]. With MKS notation, for a charge q with a 4-velocity $u^j$ in an electromagnetic field with a Faraday tensor F, $\mathrm{Force}^i = q F^i{}_j u^j$. The general notion of a 4-force is $F^i = \frac{dP^i}{d\tau}$, where P is the energy-momentum 4-vector of a particle. The notion of a "particle" is mainly a SR notion, but it is used ocassionaly in GR for "test particles". Extended bodies in GR are modelled by a stress-energy tensor rather than treating them as particles. This means that we replace the energy momentum 4-vector with the energy-momentum 4-density, i.e. the stress-energy tensor $T^{ij}$. We replace charges and 4-velocities with a current density as well. Last edited: Apr 23, 2006 9. Apr 24, 2006 ### Ratzinger I’m still not fully getting it. I would like to show again how my thinking goes and maybe someone could point out where I went wrong. Assuming a big massive and electrically charged body free falling in gravity. As long as no electric field is present the Einstein field equation governs the motion of the body. But as soon as an electric field is switched on we need (tensorial) Maxwell equations ( Lorentz force law to be precise) to explain the accelerating non-geodesic motion. So of course we can describe acceleration of a charge by an electric field in curved spacetime, but we do need Maxwell equations, as much as we needed them in SR or in Galilean spacetime. So again here my statement: GR alone can not describe acceleration other than free fall. Suppose I’m right, I would like to know where to plug the Maxwell equations to describe non-geodesic motion in curved spacetime. Unfortunately my textbook only says there is the minimal-coupling law ( take laws which are valid in inertial coordinates in flat space and write them in coordinate-invariant form, law is true in curved spacetime) but then only treats free-falling particles. I see Stingray answered that somewhat, but again I'm not completly understanding it. 10. Apr 24, 2006 ### Stingray I disagree that Maxwell's equations are really separate from GR, but I think this is a semantic issue. Let me try a different route instead: even in pure gravity, an uncharged pointlike test particle doesn't necessarily move on a geodesic. The most clear example of this arises with spin. Spinning test bodies obey something known as Papapetrou's equation, which in general gives nonzero acceleration. On top of this, you can also think of an extended body in pure gravity. Point particles don't exist anyway. Even when neglecting the effects of spin and self-gravity, such an object's center-of-mass will not quite move on a geodesic in general. This is true even in Newtonian gravity, where the analogous statement is that a body's center-of-mass does not move as a point particle at that point. It's close in most astronomical situations, but you have higher multipole contributions to the equations of motion when objects get close to each other. Beyond these long-range interactions, there are also contact forces in almost every interesting problem. Right now, you have a nonzero acceleration because the floor is keeping you from falling through it. GR has no problems accounting for such things. If it did, it would have never been taken seriously! Pervect already described how to treat a point test charge in curved spacetime. A little more explicitly, you first have to have an electromagnetic field $F^{ab}$ that satisfies the curved spacetime Maxwell equations: [tex]\nabla_{a} F^{ab} = J^{b}$$

$$\nabla_{[a} F_{bc]} =0$$

where $J^{a}$ is the 4-current of the sources. It must satisfy $\nabla_{a} J^{a}=0$ in order for these equations to be consistent.

Now introduce a test charge with worldline $z(\tau)$. One can show that

$$\frac{ \delta^{2} z^{a}(\tau) }{ d\tau^{2} } = \frac{ d \dot{z}^{a} }{ d\tau } + \Gamma^{a}_{bc} \dot{z}^{b} \dot{z}^{c} = q F^{ab}(z) \dot{z}_{b}$$,

where $\dot{z}^{a} := dz^{a}/d\tau$.

Last edited: Apr 24, 2006
11. Apr 25, 2006

### Ratzinger

thanks Stingray and pervect

So plugging the non-gravitational force term (in tensorial form) in the geodesic equation on the side where the zero stands when only free fall. That gives non-geodesic motion.

So when I want to calculate my non-geodesic motion of me siting on my stool, I have insert the normal force on the RHS of the geodesic equation. Same for an accelerating rocket. Right?

Well, maybe only right in a rough way since there are problems as you pointed out, but not completly wrong either. Or not?

Could you expand on :"Extended bodies in GR are modelled by a stress-energy tensor rather than treating them as particles."

I thought the stress-energy tensor enters the Einstein equation as representation of the mass-energy distribution that determines the curvatur?

Last edited: Apr 25, 2006
12. Apr 25, 2006

### Stingray

Yep. That's pretty much correct. But I wouldn't use the words "RHS of the geodesic equation." The LHS is just the 4-acceleration. It's only called the geodesic equation when you set that equal to zero. But I'm just nitpicking.

Yes, the stress-energy tensor serves as the source in Einstein's equation. But it also contains most of the information that you'd want to know about a body. Standard things like density and pressure are just eigenvalues of the stress-energy tensor. The associated eigenvectors also have nice physical interpretations. So instead of talking about particle mechanics, we just worry about happens to the stress-energy tensor. More intuitive quantities can be derived from it later on if desired.

13. Apr 27, 2006

### pervect

Staff Emeritus
Basically, yes. If you take the covariant derivative of the 4-velocity and set it to zero, you have the geodesic equations. If you take the covariant derivative of the 4-velocity and set it equal to the 4-force, you have the equation of motion of a test particle with an external 4-force.

It does. But it is also represents nicely the properties of an extended body.

Suppose we have a small 3-volume dV. The stress energy tensor directly gives the energy-momentum 4-vector of this volume dV, via the following procedure.

Take a unit vector that's orthogonal (Minkowski-orthogonal) to all the spatial vectors that compose the 3-volume dV. Multiply this unit vector by the magnitude of the volume enclosed in the rest frame of the volume.

This is the vector representation of a volume. If you happen to be familiar with clifford algebras and/or hodge duals and wedge products, you may recognize this as the hodges dual of dx^dy^dz. The important thing is that it is a vector representation of a volume.

Call this vector $V^i$.

Then the amount of energy and momentum in the specified volume can be computed by

$$E^j= T^j{}_i V^i$$

You raise or lower indices in the usual way (with the metric) to get $T^{ij}$ or $T_{ij}$, which is a symmetric second rank tensor.

A very small volume element is similar in many ways to a point particle. (The analogy is not perfect, though.)

The volume elements composing an extended body could all be static in some particular frame (i.e. the momentum components are all zero), representing a rigid body. If they are not all zero in some frame, then the body is deforming.

Most pratical applications of GR involve massive objects like stars, which are essentially fluids.

You will see the 3-d versions of the stress-energy tensor in advanced treatments of classical (non-relativistic) fluid mechanics.

The Wikipedia article on the stress-energy tensor is currently fairly good, hopefully this goodness will not be edited away.

http://en.wikipedia.org/wiki/Stress-energy_tensor
http://www.efunda.com/formulae/solid_mechanics/mat_mechanics/stress.cfm

Last edited: Apr 27, 2006