Is it that hard to do Relativity in accelerating frames?

[davidge]

According to the Lorentz transformations, in the absence of gravity, the relations between coordinates of a primed system to those of a unprimed system are $$x'^{\ \alpha} = \Lambda^{\alpha}{}_{\beta}x^{\beta}$$ For the Lorentz invariance to be satisfied we must have $\Lambda^{\alpha}{}_{\beta}$ constant, because then $$dx'^{\ \alpha} = \Lambda^{\alpha}{}_{\beta}dx^{\beta}$$
My questions are:
- is it always possible to "adjuste" a accelerating frame so that it becomes unaccelerated?
- the equations get much more complicated if we assume the accelerating frames without making any adjustment or do they become just a little more complicated?

 

[Orodruin]

Using accelerated frames in special relativity is very much like using curvilinear (and not necessarily orthonormal) coordinates in regular calculus. Depending on whether on not you find that difficult (along with seeing time as just another coordinate), you may or may not find relativity in accelerated frames difficult.

 

[Dale]

It also depends strongly on the problem. For certain problems picking a good non inertial coordinate system dramatically simplifies the problem. Picking a bad coordinate system makes things even worse, of course.

 

[Orodruin]

It also depends strongly on the problem. For certain problems picking a good non inertial coordinate system dramatically simplifies the problem. Picking a bad coordinate system makes things even worse, of course.

Just like when using curvilinear coordinates in regular calculus!

 

[vanhees71]

It's in fact pretty simple. Just take arbitrary coordinates $q^{\mu}$ to parametrize the components of the spacetime four-vector in Minkowski coordinates. Then you have for the invariant increment
$$\mathrm{d} s^2=\eta{\mu \nu} \mathrm{d} x^{\mu} \mathrm{d} x^{\nu} = \eta_{\mu \nu} \frac{\partial x^{\mu}}{\partial q^{\rho}} \frac{\partial x^{\nu}}{\partial q^{\sigma}} \mathrm{d} q^{\rho} \mathrm{d} q^{\sigma},$$
i.e., the pseudometric components with respect to the holonomous (!) curvilinear coordinates are given by
$$g_{\rho \sigma} = \eta_{\mu \nu} \frac{\partial x^{\mu}}{\partial q^{\rho}} \frac{\partial x^{\nu}}{\partial q^{\sigma}}.$$
BTW: You are now already very close to the general tensor framework needed for general relativity. The only thing one must learn is that there is no spacetime four-vector anymore but only a differential manifold with tangent vectors and tangent co-vectors and a pseudo-Riemannian (Lorentzian) fundamental form, which physicists like to simply call "the metric" although it's not positive definite ;-).

 

[davidge]

Thank you all.

BTW: You are now already very close to the general tensor framework needed for general relativity. The only thing one must learn is that there is no spacetime four-vector anymore but only a differential manifold with tangent vectors and tangent co-vectors and a pseudo-Riemannian (Lorentzian) fundamental form, which physicists like to simply call "the metric" although it's not positive definite ;-).

It's interesting to know this.

It's in fact pretty simple. Just take arbitrary coordinates $q^{\mu}$ to parametrize the components of the spacetime four-vector in Minkowski coordinates. Then you have for the invariant increment

$$\mathrm{d} s^2=\eta_{\mu \nu} \mathrm{d} x^{\mu} \mathrm{d} x^{\nu} = \eta_{\mu \nu} \frac{\partial x^{\mu}}{\partial q^{\rho}} \frac{\partial x^{\nu}}{\partial q^{\sigma}} \mathrm{d} q^{\rho} \mathrm{d} q^{\sigma}$$

Is this the same as working on the problem in the intertial frame and after changing to the accelerating frame?

 

[robphy]

One aspect that [to me] isn't so clear, however, is making and interpreting measurements by an accelerating observer (e.g. Piecewise inertial, uniformly accelerating, piecewise uniformly accelerating, rotating, arbitrary, etc) ... they don't just carry over from the inertial case. For example, as I currently understand it, radar measurements yield different results from those associated with clocks and "rods" (somehow defined).

 

[pervect]

According to the Lorentz transformations, in the absence of gravity, the relations between coordinates of a primed system to those of a unprimed system are $$x'^{\ \alpha} = \Lambda^{\alpha}{}_{\beta}x^{\beta}$$ For the Lorentz invariance to be satisfied we must have $\Lambda^{\alpha}{}_{\beta}$ constant, because then $$dx'^{\ \alpha} = \Lambda^{\alpha}{}_{\beta}dx^{\beta}$$
My questions are:
- is it always possible to "adjuste" a accelerating frame so that it becomes unaccelerated?
- the equations get much more complicated if we assume the accelerating frames without making any adjustment or do they become just a little more complicated?

I'm not sure of the context of the question. If one is in the flat space-time of special relativity, one can always find a global inertial frame of reference. If one is in the curved space-time of General Relativity, in general there isn't any such thing, nor is there any general global Lorentz transform.

The key point I think is defining what one means by "frame". Is it a global coordinate system? Is it a purely local thing (a tangent space, and/or a set of basis vectors for the tangent space near some particular point). Or perhaps it's a congruence of time-like worldlines, each worldline representing an "observer" in the frame.

Once one has an agreed on definition of what a "frame" really is, it becomes much easier to answer questions like the one you posed.

 

[davidge]

The key point I think is defining what one means by "frame". Is it a global coordinate system? Is it a purely local thing (a tangent space, and/or a set of basis vectors for the tangent space near some particular point). Or perhaps it's a congruence of time-like worldlines, each worldline representing an "observer" in the frame.

What would you personally define to be a frame from these that you mentioned?

 

[pervect]

What would you personally define to be a frame from these that you mentioned?

I'm not totally consistent, but usually I regard a frame as being defined by a set of basis vectors, one set at every point along the worldline of some particular "observer". One thinks of the observer as "carrying along" a set of basis vectors, or transporting them.

If the worldline is a geodesic, the frame is "not accelerating". If the basis vectors are fermi-walker transported, the frame is "not rotating". (I believe that in the case where the observer is not accelerating, i.e. is a geodesic observer, fermi-walker transport is equilvalent to parallel transport. But don't shoot me if I'm wrong.)

So by this definition, a non-accelerated frame of reference exists if geodesics exist. This is nearly always, though singularities (where geodesics are incomplete, i.e. they just stop) would be an exception, I suppose.

 

[PeterDonis]

I believe that in the case where the observer is not accelerating, i.e. is a geodesic observer, fermi-walker transport is equilvalent to parallel transport.

This is correct.

 

[davidge]

Just to know if I'm getting things right, I have constructed an example. The coordinates of one system are labeled $y$, while those of the other system are $x$.

$$y^1 = x^1 \text{Cos}(wt') \\ y^2 = x^2 \text{Sin}(wt') \\ y^3 = x^3 \\ y^0 = t' = x^0 = t$$ where $w$ is the angular velocity.
Because $dy^{\alpha} = (\partial y^{\alpha} / \partial x^{\beta}) dx^{\beta}$ and $\eta_{\sigma \kappa} = (\partial y^{\alpha} / \partial x^{\sigma}) (\partial y^{\beta} / \partial x^{\kappa}) \eta_{\alpha \beta}$, we have for the space time interval $$d \tau '{\ }^2 = dt^2 - (\text {Cos}^2(wt)(dx^1)^2 + \text{Sin}^2(wt)(dx^2)^2 + (dx^3)^2)$$ Does it make sense?

At a constant time $t = n \pi / 2$, $n \in \mathbb{N}$ the proper time in the primed ($y$) system is equal to $dx^3$!

 

[DrGreg]

1. You have forgotten to include $\partial y^1 / \partial x^0$ and $\partial y^2 / \partial x^0$.
2. I assume you intended this to be a "rigidly rotating" coordinate system, so $$y^1 = x^1 \cos(\omega x^0) - x^2\sin(\omega x^0) \\y^2 = x^1 \sin(\omega x^0) +x^2\cos(\omega x^0)$$

 

[davidge]

Thanks Dr Greg. So would the proper time become $$d \tau '{\ }^2 = dt^2 - (\text {Cos}^2(wt)(dx^1)^2 + \text{Sin}^2(wt)(dx^2)^2 + w( \text{Cos}^2(wt) - \text{Sin}^2(wt))(dt)^2 + (dx^3)^2)$$

 

[haushofer]

It is similar to doing Newtonian mechanics in accelerating frames. Only the coordinate transformations differ.

 

[vanhees71]

This is correct.

Nearly correct! Fermi-Walker transport is parallel transport if the (timelike) worldline along which you transport is a geodesic. In Minkowski space it's only the same, if the worldline is given by
$$x^{\mu}(s)=u^{\mu} s+x_0^{\mu},$$
where $u^{\mu}=\text{const}$ with $u^{\mu} u_{\mu}=1$ and $x_0^{\mu}=\text{const}$, where the $x^{\mu}$ are the components of the spacetime vector wrt. a Minkowski basis. Then indeed $a^{\mu}=\ddot{x}^{\mu}=0$, and Fermi-Walker transport is just parallel transport (which is trivial in a flat space).

Fermi-Walker transport is more general. It works along any time-like worldline and transports some Minkowski basis (tetrad) from one point to another such that it is locally non-rotating. For details, see the new Section on Fermi-Walker transport in my SRT FAQ, which is unfortunately down right now :-(. I hope our sysadmins bring the server up soon (although we have holidays till including Monday of course):

http://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf

 

[DrGreg]

Thanks Dr Greg. So would the proper time become $$d \tau '{\ }^2 = dt^2 - (\text {Cos}^2(wt)(dx^1)^2 + \text{Sin}^2(wt)(dx^2)^2 + w( \text{Cos}^2(wt) - \text{Sin}^2(wt))(dt)^2 + (dx^3)^2)$$

You've made a mistake somewhere in your working. Think about it. Do you really expect the answer to depend on the angle you have rotated through? You should find all the angles cancel out, either via $\cos^2 \omega t + \sin^2 \omega t = 1$ in some places, or $2\cos \omega t \sin \omega t - 2\cos \omega t \sin \omega t = 0$ in other places.

 

[martinbn]

Nearly correct!

Why nearly? You are saying exactly the same thing.

 

[PeterDonis]

Fermi-Walker transport is parallel transport if the (timelike) worldline along which you transport is a geodesic.

Yes, that's what pervect said and I said was correct.

 

[vanhees71]

Yes, that's what pervect said and I said was correct.

Argh! Sorry, I misread the statement. So it's indeed fully correct. Sorry again!

 

[pervect]

So, from the GR point of view (which may not be what the OP is interested in), we can construct a local frame of the sort I mentioned by considering an event in space-time, and creating four vectors at that point. Usually we pick four vectors of unit length that are orthogonal to each other, a tetrad or vierbein <<wiki link>>. One vector, formally called the four-velocity, points essentially in the time direction. We have a lot of freedom to pick these vectors at any given event, corresponding to spatial rotations and velocity changes (local Lorentz boosts).

Then, given that we know how and what parallel transport is, we can create a non-accelerated frame by finding the curves, called geodesics, that parallel transport the four-velocity along itself. Geodesics should exist as long as there are no space-time singularities. Once we have the geodesics, we use parallel transport again to transport the spatial vectors along the geodesic we just found. This process constructs a non-accelerating, non-rotating "frame of reference" from the initial event and the initial choice of the tetrad at that event.

 

[davidge]

we can create a non-accelerated frame by finding the curves, called geodesics, that parallel transport the four-velocity along itself

This motivates me to ask another question:

In the OP post I was thinking of a acceleration caused by "human" hand, that is a acceleration not caused by gravity. Would such acceleration cause space-time to curve? (I thought only gravity caused space-time to curve.)

 

[pervect]

Only gravity curves space-time. In the absence of gravity, space-time is flat, and one can replace "parallel transport" with the ordinary notion of parallel from Euclidean geometry ( at least for the spatial axes, one might take a detour into Lorentzian geometry rather than Euclidean geometry when talking about space-time geometry. I'll stick to the Euclidean viewpoint for this post.)

The non-rotating condition on the space axes is simply that each axis remains parallel to itself as time progresses.

Geodesic motion for the GR curved space-time case can be replaced with constant velocity motion for the flat space-time of SR. Constant velocity motion is natural motion, the motion the observer makes in the absence of any external forces. So if one has an observer at some event and some specific state of motion (velocity), the path that observer naturally takes with no forces acting on it defines the origin of a frame that's non-accelerating. In the four-vector treatment, the four-velocity of the observer is is regarded as a vector and part of the tetrad that defines a frame. In a three-vector treatment, the frame is regarded as being defined by only the three spatial vectors.

The condition that all spatial axes remain parallel to themselves defines non-rotating.

 

[vanhees71]

This motivates me to ask another question:

In the OP post I was thinking of a acceleration caused by "human" hand, that is a acceleration not caused by gravity. Would such acceleration cause space-time to curve? (I thought only gravity caused space-time to curve.)

Note that within General Relativity gravity doesn't accelerate test particles, but their proper acceleration vanishes, i.e., they move along geodesics in 4D spacetime.

 

[davidge]

Note that within General Relativity gravity doesn't accelerate test particles, but their proper acceleration vanishes, i.e., they move along geodesics in 4D spacetime.

Or, to put it in another way, in the reference frame of the particle, is it never accelerating (nor moving in space) and thus is following a geodesic (moving only through time in space time), while in other frames it is actually moving in a curved path?

 

[Dale]

Or, to put it in another way, in the reference frame of the particle, is it never accelerating (nor moving in space) and thus is following a geodesic

No, definitely not. Whether or not a given worldline is a geodesic is invariant. If a particle is non inertial then its worldline will not be a geodesic even in its own reference frame!

 

[davidge]

No, definitely not. Whether or not a given worldline is a geodesic is invariant. If a particle is non inertial then its worldline will not be a geodesic even in its own reference frame!

Oh, I should expect it, because in the proper reference frame $ds^2 = d \tau^2$ and $ds^2$ is invariant. But how can a particle be non inertial in its own frame?

 

[Dale]

Oh, I should expect it, because in the proper reference frame $ds^2 = d \tau^2$ and $ds^2$ is invariant. But how can a particle be non inertial in its own frame?

You can attach an accelerometer to a particle and determine if it is non inertial. That accelerometer reading does not change simply because you switch the math from one frame to another.

 

[davidge]

You can attach an accelerometer to a particle and determine if it is non inertial. That accelerometer reading does not change simply because you switch the math from one frame to another.

I got it

 

[Dale]

I got it

excellent!

For a more geometrical approach, consider a flat piece of paper. I can draw a straight line and I can draw a curved line. You can tell which is straight and which is curved without reference to any coordinate system. It is an invariant geometrical fact.

Now you can add coordinate systems by drawing grids on the paper. A grid which goes parallel to a given line is called that line's frame. So in the curved line's frame it is at a constant coordinate position, but that does not change the invariant fact that it is curved.