Insights Rindler Motion in Special Relativity: Hyperbolic Trajectories - Comments

[stevendaryl]

Greg Bernhardt submitted a new PF Insights post

Rindler Motion in Special Relativity: Hyperbolic Trajectories

Continue reading the Original PF Insights Post.

 

[Histspec]

Rindler Motion in Special Relativity: Hyperbolic Trajectories

Some historical remarks:
This sort of motion is also known as hyperbolic motion, which was implicitly used by Minkowski in his famous talk on space and time (1908), search for "acceleration-vector" and "hyperbola of curvature" in
http://web.mit.edu/redingtn/www/netadv/SP20130320.html

The name was given by Max Born 1909, see
Born's remarks on hyperbolic motion
where he used the formula
$$\begin{cases}
x=-q\xi,\\
t=\frac{p}{c^{2}}\xi.\end{cases}$$
where $q=\sqrt{1+p^{2}/c^{2}}$. The modern formula follows with $\xi=c^{2}/g$ and $p=c\sinh(g\tau/c)$.

A nice summary was given by Sommerfeld in 1910, see
Sommerfeld's remarks on hyperbolic motion
where he used an imaginary time coordinate and imaginary rapidity,
$$x=r\cos\varphi,\ y=y,\ z=z,\ l=r\sin\varphi$$

 

[Orodruin]

Great Insight, but a few things I would like to underline.

The relations $V\cdot V = 1$ (in units where $c = 1$) and $V\cdot A = 0$ are generally true, not only for hyperbolic motion. The first by definition and the second as a direct consequence of that definition (just differentiating 1 wrt the proper time). This also gives a very straight-forward way of integrating the equation of motion for constant proper acceleration in terms of the proper time. We would have $A\cdot A = -g^2$ and $V\cdot V = 1$ generally gives the possibility of parametrising $V$ as $V^0 = \cosh(\theta)$ and $V^1 = \sinh(\theta)$ for motion in one spatial dimension. Differentiating wrt proper time then directly leads to
$$
-\dot\theta^2 = -g^2 \quad \Longrightarrow \quad \dot \theta = \pm g \quad \Longrightarrow \quad \theta = \pm g\tau \mp \theta_0,
$$
where $\theta_0$ is an integration constant. Directly integrating the 4-velocity then leads to
$$
t = \frac{1}{g} \sinh(g\tau - \theta_0) + t_0, \quad x = \pm \frac{1}{g} [\cosh(g\tau - \theta_0) - 1] + x_0,
$$
where $t_0$ and $x_0$ are integration constants chosen such that $t(\theta_0/g) = t_0$ and $x(\theta_0/g) = x_0$. Clearly, $t_0$ and $x_0$ are just translations of the solution in time and space, whereas $\theta_0$ represents a shift in the proper time. I have always found this direct integration a more direct way of deriving the hyperbolic motion than that you would typically find in textbooks, which typically is based on using coordinate time, solving coupled differential equations, and/or reference to the instantaneous rest frame.

Edit: A small inconsistency, you also introduce the 4-velocity as $(V^0, V^1)$, but later you place the spatial component first.

 

[Greg Bernhardt]

Looking forward to Part 2: Rindler Coordinates!

 

[ZapperZ]

Coincidentally, the Rindler motion is covered in this just-released video when discussing the Unruh effect in accelerating frames:



Zz.

 

[Greg Bernhardt]

Part 2 is up!
https://www.physicsforums.com/threa...y-part-2-rindler-coordinates-comments.944475/

 

[Houdini]

The equations of relativity simply confirm the elementary fact that the speed of light is a universal physical
constant and nothing more.
Where is the hyperbolic movement when we fall radially to a black hole?

 

[Ibix]

The equations of relativity simply confirm the elementary fact that the speed of light is a universal physical
constant and nothing more.

I tell chemists that we covered their whole subject in two pages of notes on atomic and molecular orbitals in my introductory quantum mechanics course. This claim is similar.

Where is the hyperbolic movement when we fall radially to a black hole?

A hovering object follows a hyperbolic worldline in the local inertial frame of a free falling observer.

 

[cianfa72]

Btw, I would like to ask for clarification on the following statement

Note: the spacetime “position” $X$ is only a spacetime vector in Special Relativity. In the curved spacetime that is considered for General Relativity, it is no longer a vector, because it’s not possible to add positions that you add vectors. However, the proper velocity and proper acceleration continue to be spacetime vectors.

Reference: https://www.physicsforums.com/insights/rindler-motion-in-special-relativity-hyperbolic-trajectories/

I take it as only in case of SR flat spacetime there is an affine structure that turns it into an affine spacetime. Hence, starting from a given arbitrary event, it makes sense to define a "position vector" $X$ as an element of the 'translation vector space' that enters in the definition of affine space itself.

In the context of GR, however, where we cannot endow the underlying spacetime of an affine structure it does not make any sense to talk about "position vectors". From the viewpoint of velocity and acceleration, instead, they actually "live" in tangent space at each point (event) so they are well-defined as 4-vectors in tangent space also in GR.

 

[Orodruin]

Btw, I would like to ask for clarification on the following statement

I take it as only in case of SR flat spacetime there is an affine structure that turns it into an affine spacetime. Hence, starting from a given arbitrary event, it makes sense to define a "position vector" $X$ as an element of the 'translation vector space' that enters in the definition of affine space itself.

In the context of GR, however, where we cannot endow the underlying spacetime of an affine structure it does not make any sense to talk about "position vectors". From the viewpoint of velocity and acceleration, instead, they actually "live" in tangent space at each point (event) so they are well-defined as 4-vectors in tangent space also in GR.

Correct. Much like you cannot add positions on a sphere.

 

[cianfa72]

Correct. Much like you cannot add positions on a sphere.

Basically you're saying that on a 2D sphere --starting from an arbitrary point-- there is no way to define arcs on it (for example geodesic segments) such that they become elements in a set complying with the axioms of vector space structure.

 

[Orodruin]

Basically you're saying that on a 2D sphere --starting from an arbitrary point-- there is no way to define arcs on it (for example geodesic segments) such that they become elements in a set complying with the axioms of vector space structure.

Indeed you cannot endow the sphere with an affine structure. If you attempt to do it using geodesic segments, then translations will generally not commute and they will not be 1-to-1 maps from the tangent space to the sphere itself (e.g., the antipode of the point you pick as your origin is going to be mapped to by several different tangents).

 

[cianfa72]

If you attempt to do it using geodesic segments, then translations will generally not commute and they will not be 1-to-1 maps from the tangent space to the sphere itself (e.g., the antipode of the point you pick as your origin is going to be mapped to by several different tangents).

Take a generic 2D surface and suppose that, starting from an arbitrary point, there exist always a way to define an 1-to-1 map from vectors in the tangent space at that point to the points on the surface.

From what you said, however, it is not enough. What can go wrong is that translations do not commute in the following sense: from point A go to the point B you get by 1-to-1 mapping vector $V$ in the tangent space at A. Next do the same starting from B using the vector $W$ in the tangent space at B. At the end of these two operations you will get a point C on the surface.

By definition translations do not commute if what you get flipping the above two operations is generally a different point on the surface (call it D).

 

[pervect]

Basically you're saying that on a 2D sphere --starting from an arbitrary point-- there is no way to define arcs on it (for example geodesic segments) such that they become elements in a set complying with the axioms of vector space structure.

I've always liked a remarks from Misner on this topic in "Precis of General Relativity", https://arxiv.org/abs/gr-qc/9508043

Think of a Caesarian general hoping to locate an outpost. Would he understand that 600 miles
North of Rome and 600 miles West could be a different spot depending on
whether one measured North before West or visa versa?

The reason this remark is applicable to the discussion is that vector addition, by defintion, must commute. However, the operation of "going north" and "going west" does not commute, because it matters which one you do first. Thus displacements cannot form a vector space, because their addition does not commute.

A very literal interpretation of Misner's remark uses a connection different than the Levi-Civita connection. In the Levi-Civita connection, geodesics on the sphere are great circles, but interpreting "600 miles west" as motion along a geodesic requires abandoning the Levi-Civita connection for some other connection where circles of constant lattitude are geodesics. This connection is a bit ugly mathematically in that it requires torsion, something that we don't bother with in GR. One imagines that the very notion that "motion in a straight line" is called geodesic motion is also foreign to the Roman general, but it will hopefully be familiar to many readers of this forum.

A further note, which is a bit of a digression, but sadly I cannot resist. The general name for the curve of motion followed by maintaining a constant compass heading on a sphere is a loxodrome, and in general a loxodrome is different than a great circle. See for instance https://en.wikipedia.org/wiki/Rhumb_line.

Let's move onto the point I want to make, how to revise Misner's scenario into a form that does use the Levi-Civita connection.

Formally, in this revised formulation, "North" and "West" are tangent vectors, which exist at every point on the sphere, including Rome. Also, in this formulation, we note that geodesics can be specified by giving a starting point, and a starting tangent vector.

We can now make a different, more verbose equivalent to Misner's remark that DOES use the Levi-Civita connection, demonstrating that the connection is not the fundamental issue here with the lack of commutation. In this revised statement, one army starts out at a point (Rome), and moves along a great circle geodesic whose initial tangent vector points "north", proceeds along this geodesic for the stated distance (600 miles), makes a left turn, then proceeds around this second great circle geodesic, and makes a camp.

A second army starting out from Rome, moves along a great circle whose initial tangent vector points "west", then makes a right turn, and marches 600 miles in a straight line (great circle). The point here is that the two armies still do not generally arrive at the same point. It may be convenient to relocate Rome so that it is on the equator if one wants to preform a more detailed analysis to convince oneself of the truth of this remark.

 

[PeterDonis]

interpreting "600 miles west" as motion along a geodesic requires abandoning the Levi-Civita connection for some other connection where circles of constant lattitude are geodesics.

Either that, or interpreting "600 miles west" as "follow the great circle whose tangent vector points due west at the point where you are now".

 

[cianfa72]

The reason this remark is applicable to the discussion is that vector addition, by defintion, must commute. However, the operation of "going north" and "going west" does not commute, because it matters which one you do first. Thus displacements cannot form a vector space, because their addition does not commute.

In other words there is no way to define an affine connection (not necessarily metric compatible -- i.e. Levi-Civita) on the sphere as manifold to turn it in an affine space.

 

[pervect]

In other words there is no way to define an affine connection (not necessarily metric compatible -- i.e. Levi-Civita) on the sphere as manifold to turn it in an affine space.

I might be being overcautious, but I'd stick with metric compatible connections unless I saw or came up with a more rigorous proof.

 

[Orodruin]

not necessarily metric compatible -- i.e. Levi-Civita

Levi-Civita is not the only metric compatible connection. In order to make it unique you also need to require that the connection is torsion free. The connection described in

requires abandoning the Levi-Civita connection for some other connection where circles of constant lattitude are geodesics. This connection is a bit ugly mathematically in that it requires torsion

is indeed metric compatible. It can be found as (I believe) the first example of a connection in my book even before introducing metrics and while having torsion may be a bit ugly, the form of the connection coefficients is very simple. It should also be mentioned that this connection does not include the poles, but it is a flat connection on the spheres minus the poles.

Either that, or interpreting "600 miles west" as "follow the great circle whose tangent vector points due west at the point where you are now".

… or doesn’t require geodesic motion and interpret 600 miles west as simply going 600 miles along the loxodrome.

The most solid reason that the sphere cannot be turned into an affine space - regardless of connection - is that it has the wrong topology. While an affine space is homeomorphic to $\mathbb R^n$, with all homotopy groups being trivial, the sphere has a non-trivial second homotopy group $\pi_2(\mathbb S^2)$.

 

[cianfa72]

The connection where circles of constant lattitude are geodesics is indeed metric compatible. It can be found as (I believe) the first example of a connection in my book even before introducing metrics and while having torsion may be a bit ugly, the form of the connection coefficients is very simple. It should also be mentioned that this connection does not include the poles, but it is a flat connection on the spheres minus the poles.

As you said on the sphere as manifold the connection where circles of constant lattitude are geodesics is metric compatible and it is flat. From my understanding it turns the sphere into an affine manifold (at least on all points other than poles ).

Does it mean that displacements (defined as geodesic segments of the above connection) actually commute ?

It can be found as (I believe) the first example of a connection in my book even before introducing metrics

Btw, which is the book you were talking about ?

 

[Houdini]

I tell chemists that we covered their whole subject in two pages of notes on atomic and molecular orbitals in my introductory quantum mechanics course. This claim is similar.

A hovering object follows a hyperbolic worldline in the local inertial frame of a free falling observer.

There is no existing "local inertia framework of a free falling observer". The local inertia framework of a free falling observer is an oxymoron of no physical importance.

 

[pervect]

Let's adopt a specific coordinate system on the sphere, lattitude and longitude. There are an infinite number of alternatives that we could adopt, but I'm trying to stick with something familiar to make some examples that I hope will be helpful.

Note that this coordinate choice above implies that we are excluding (or removing) a couple of points from the manifold, the north and south pole. This affects some of the topological arguments greatly. The topological arguments are of course valid when there assumptions are met, we're changing the rules by excluding some points that are coordinate singularities from the manifold.

Along with our coordinates, we also choose a connection, the same one I mentioned previously. This connection is metric compatible, but has non-zero torsion. It's not the Levi-Civita connection we usually use in GR. With this connection, circles of constant lattitude are geodesics, as are circles of constant longitude.

If we measure displacements in arc-minutes, they commute. In to Misner's example, if we go 500 arc minutes north, and 500 arc minutes west, in that order, we wind up at the same place as if we reversed the order, going west first. One arc minute is roughly a mile at the equator, unless I've messed up the conversion. But the length of an arc minute changes with lattitude, it's not a constant distance.

So, we've defined commuting displacements on the sphere, but they don't represent physical distances.

On a more abstract level, we often represent vectors by differential operators. In our example, the partial derivative with respect to lattitude and the partial derivative with respect to longitude are differential operators which represent vectors. And these vectors commute. As I recall, the basis vectors in any coordinate basis commute.

However, while these vectors commute, they don't represent a constant physical distances. In our example, 500 arc minutes west represents a different distance at the equator than it does at some other lattitude.

 

[PeterDonis]

As I recall, the basis vectors in any coordinate basis commute.

Yes, I believe that is a necessary and sufficient condition for a basis to be a coordinate basis.

 

[Orodruin]

Yes, I believe that is a necessary and sufficient condition for a basis to be a coordinate basis.

Let us remember what that statement means though. It refers to the commutation of vector fields as differential operators on a function and may be viewed also as the commutation of the flows of the fields. This is fundamentally different from what was discussed earlier here, which was the commutation of consecutive geodesic displacements.

The two coincidence when all connection coefficients are zero in the coordinate basis.

Does it mean that displacements (defined as geodesic segments of the above connection) actually commute ?

 

[cianfa72]

It refers to the commutation of vector fields as differential operators on a function and may be viewed also as the commutation of the flows of the fields. This is fundamentally different from what was discussed earlier here, which was the commutation of consecutive geodesic displacements.

For the commutation of any geodesic displacements, it is required vanishing connection coefficients in the coordinate basis. Otherwise it is true that geodesic displacements along integral curves of basis vectors commute, however it does not extend to any geodesic displacement.

 

[Orodruin]

For the commutation of any geodesic displacements, it is required vanishing connection coefficients in the coordinate basis.

No. Look at polar coordinates. Again, you need to differentiate between commuting flows of vector fields and commuting geodesics.

 

[cianfa72]

No. Look at polar coordinates. Again, you need to differentiate between commuting flows of vector fields and commuting geodesics.

Sorry, if we move along the flows of polar basis vector fields (i.e. geodesic segments of the associated connection) for fixed given amounts of the associated parameters (say $\Delta t$ along the former and $\Delta s$ along the latter) we will end up in a spot.

Then if we move from the same starting point in the reversed order (according to the same given amounts as above), we will end up in the same spot !

In this sense geodesic segments of the above connection actually commute, don't you ?

 

[Orodruin]

i.e. geodesic segments of the associated connection

This would require you to define a connection that is not defining an affine structure on $\mathbb R^2$.

Edit: Either way, the comment was in reply to

For the commutation of any geodesic displacements, it is required vanishing connection coefficients in the coordinate basis.

The point is that, in polar coordinates, consecutive geodesic displacements commute, but the connection coefficients are not zero in the coordinate basis. (Using the Levi-Civita connection)

 

[cianfa72]

The point is that, in polar coordinates, consecutive geodesic displacements commute, but the connection coefficients are not zero in the coordinate basis. (Using the Levi-Civita connection)

Sorry my bad...I was confusing polar coordinate on the plane with spherical coordinates $\theta, \phi$ on the 2D sphere.

Btw, spherical coordinates $(\theta, \phi)$ on the 2D sphere are basically the same as lattitude and longitude ?

 

[Orodruin]

Sorry my bad...I was confusing polar coordinate on the plane with spherical coordinates $\theta, \phi$ on the 2D sphere.

Btw, spherical coordinates $(\theta, \phi)$ on the 2D sphere are basically the same as lattitude and longitude ?

Basically, but not exactly. See https://en.wikipedia.org/wiki/Spherical_coordinate_system

 

[cianfa72]

Basically, but not exactly. See https://en.wikipedia.org/wiki/Spherical_coordinate_system

Yes there are some differences (a fixed offset between polar angle $\varphi$ and lattitude), however it seems that coordinate curves of constant longitude and constant lattitude on the sphere are actually the same as the coordinate curves of constant spherical coordinates $\theta$ and $\varphi$ respectively.