Quadrupole Moment Time Variation: Does Coordinate Choice Matter?

[AdirianSoan]

[Moderator's note: Thread spun off from previous discussion due to topic change.]

Does the observed quadrapole moment change over time when considering a relatively moving object, for certain choices of observer coordinates?

My suspicion is that it does (Terrell-Penrose rotation implies gravity isn't spherically symmetric, as it has to "travel further" for some choices of coordinate), but the effect is negligibly weak.

 

[PeterDonis]

Does the observed quadrapole moment change over time when considering a relatively moving object, for certain choices of observer coordinates?

Choice of coordinates cannot affect any direct observable.

Terrell-Penrose rotation implies gravity isn't spherically symmetric

I don't know where you are getting this from, but it's wrong. Terrell-Penrose rotation is an optical effect that has nothing to do with any symmetry or lack thereof of the spacetime geometry. Whether or not "gravity is spherically symmetric" depends on the spacetime geometry, which in turn depends on the distribution of stress-energy.

it has to "travel further" for some choices of coordinate

Terrell-Penrose rotation is not due to "choices of coordinate". It's an observable effect, and as above, choice of coordinates cannot affect any direct observable.

 

[AdirianSoan]

For a more straightforward way of asking this question, is gravity spherically symmetric from the perspective of a reference frame from which the source of gravity is moving?

My intuition says no, for a variety of reasons, including that the total distance involved is greater for points that are not directly in the line of motion.

 

[PeterDonis]

is gravity spherically symmetric from the perspective of a reference frame from which the source of gravity is moving?

It depends on what you mean by "spherically symmetric". If you are talking about invariant symmetries of the spacetime geometry, those are there regardless of your choice of reference frame. So, for example, the Schwarzschild spacetime geometry is spherically symmetric, and that is true regardless of any choice of reference frame.

I suggest that you think up a concrete scenario that illustrates the issue you are asking about, instead of just asking about it in the abstract. Having a specific concrete scenario will make it much easier, I think, to discuss the issues involved.

the total distance involved is greater for points that are not directly in the line of motion

What "total distance involved" are you talking about?

Again, I think a specific scenario will be very helpful in clarifying and discussing the issues involved.

 

[AdirianSoan]

Two clocks are synchronized with each other, some distance apart. A large moving object is moving towards one clock we'll call A at some velocity V, on a vector which intersects a point equidistant to both clocks.

When the object reaches the point of equidistance, B should be, if I have defined the situation correctly, on a vector orthogonal to the velocity vector of the object. At this point in time, I expect A and B to experience difference gravitational accelerations; in particular I expect A to experience greater acceleration.

Or, taking the speed of the object to just below C, I expect the gravitational acceleration at B to be approximately zero.

 

[PeterDonis]

A large moving object is moving towards one clock we'll call A at some velocity V, on a vector which intersects a point equidistant to both clocks.

Is the object moving on a line perpendicular to the line connecting the two clocks? (All of this description is, I take it, in the frame in which the two clocks are at rest and synchronized.) If this is the case, I don't understand how you are drawing any of the inferences you describe. But if it is not the case, I don't understand how the large object is supposed to be moving relative to the two clocks.

 

[AdirianSoan]

The object is moving in a line, a line perpendicular to which forms a right triangle with the two clocks.

If we call the axis of motion X, when it is equidistant, one of the clocks can be directly along an orthogonal axis, say, Y.

 

[pervect]

[Moderator's note: Thread spun off from previous discussion due to topic change.]

Does the observed quadrapole moment change over time when considering a relatively moving object, for certain choices of observer coordinates?

My suspicion is that it does (Terrell-Penrose rotation implies gravity isn't spherically symmetric, as it has to "travel further" for some choices of coordinate), but the effect is negligibly weak.

I believe the quadrupole moment is a 3-tensor, see for instance MTW page 975, where (in its trace-reduced form, the one used for gravitational wave calculations) it's expressed as an integral over dx^3, i.e. 3 dimensional space, as opposed to dx^4, an integral over space-time.

While normally we say that tensors are independent of frame of reference, that argument is usually applied to 4-tensors. I haven't seen any 4-tensor version of the quadrupole moment in my text, but I assume it'd be a normal geometric invariant object, which could be regarded as coordinate independent. To get the 3-tensor version, we need to project it. Loosely speaking, that's a mathematical operation that splits space-time 4-tensor into 3-space+time, then throws away the "time" part as not being interested. This projection process, though, in the usual interpretation, makes the quadrupole moment or any other 3-tensor a "different tensor" when one choses a different projection. The process of going from a "stationary frame" to a "moving frame" is an example of making a "different projection".

Note that when we say that tensors are "independent of coordinates" or "independent of the frame of reference", we don't mean that the components of the tensor don't change when we change frames. Rather, the components change in a well-defined manner. It's the entity as whole that is regarded as being "the same entity", even though it has different components, when we change frames.

 

[PeterDonis]

The object is moving in a line, a line perpendicular to which forms a right triangle with the two clocks.

This seems to be saying that the object will hit one of the clocks. But your previous post said the object would pass a point equidistant between the two clocks. So I still don't understand what you are describing.

Can you give coordinates for the clocks and the equation of the line along which the object moves, in the frame you are using?

 

[AdirianSoan]

It will hit one of the clocks, yes.

So, making up numbers, if we treat the point at which it is equidistant as 0, its velocity is 55,0,0 (let's say light seconds per minute), and the clocks are at .01,0,0 and 0,.01,0

 

[PeterDonis]

if we treat the point at which it is equidistant as 0, its velocity is 55,0,0 (let's say light seconds per minute), and the clocks are at .01,0,0 and 0,.01,0

You can't have all of these be true in the same scenario.

First, since the two clocks will respond differently to the gravity of the large object (because their placement when the scenario starts is not symmetric with respect to the large object), they won't stay the same distance apart, so it is impossible to pick any reference frame in which both remain at rest. The best you can do is to pick a frame in which one clock remains at rest: the obvious one to pick is clock A, since the line along which the large object moves (and along which it will hit clock A) can be, say, the X axis of that frame.

Second, if we pick a frame, as above, in which clock A is at rest, the velocity of the large object in that frame won't be constant; the large object will be accelerating towards clock A in this frame (which is just the appropriate coordinate transformation of the fact that, in a frame in which the large object is at rest, clock A will be accelerating towards the large object). So you can't just specify the velocity of the large object as a single number.

In short, while the numbers you give above could characterize the initial conditions of the scenario, the only thing that will stay the same as the scenario progresses is the coordinates of one of the clocks (as above, I would suggest picking clock A as that clock). The other numbers will change.

 

[AdirianSoan]

Also, simultaenity means the two clocks won't agree on when the object reaches that point, hence the initial specification of different coordinate observers. Which means when the object reaches 0,0,0 from A's perspective, it hasn't from B's.

So let's ignore the perspectives of both A and B. Let's consider the perspective of a third clock C at -1,-1,0 and have the other coordinates be the coordinates it observes for A, B, and the object. Edit: The coordinates it observes at the point in time it observes the object at* 0,0,0

Am I correct in guessing that it will observe B accelerating towards a region slightly behind the object and A accelerating towards the object itself? And does that imply that C's perspective is that the quadrapole moment of the object's gravitational field isn't spherically symmetric? Or do we ignore simultaenity for these purposes?

 

[AdirianSoan]

Hrm. Hang on.

If there is a light on the back of a relativistic ship we are viewing from the side (er, so it passes next to us, rather than through us), Terrell-Penrose rotation implies that, although the light is older (for some parts of the trip) and therefore should be dimmer (longer light cone), the increased perceived rate of time passage of the light precisely offsets this (a clock next to the light will tick faster for our observer than a clock at the front of the ship, since as it is approaching the clock at the back has a longer light path, and as it is retreating after it passes it has a shorter light path)

If I think of gravity in terms of gravitons, the same phenomenon should arise, preserving spherical symmetry. More gravitons are emitted, relative to B.

There is probably an equivalent way of modeling it, treating motion as a type of curvature, which doesn't require thinking in terms of gravitons, but I am unfamiliar with it if so.

 

[Nugatory]

If I think of gravity in terms of gravitons...

Don't. Gravitons are a hypothetical particle that may or may not be found in an as-yet-undiscovered successful theory of quantum gravity (so have no place in a discussion of general relativity), but even if such a theory is eventually found its gravitons won't act the way you're thinking.

 

[AdirianSoan]

Then, setting aside motion as curvature, I'm back to the question of why gravity should be spherically symmetrical considered from the perspective of an observer in relative motion. Considering it as emanating from the source, the path it travels is longer in one direction than another. Considering it as moving with the source, some of the gravity is rotated in time and the axis of motion (equal distance in space, longer distance in time), which I think amounts to the same thing.

With implied quotation marks around that entire paragraph because I am doing some damage to language to try to convey the question.

Hrm. Or to switch back to the light on the back of the ship, it is rotated towards all observers orthogonal to the direction of motion. In order to maintain the same number of photons for each observer (maintain constant observed intensity), it has to move faster in time. I assume we can't use time contraction to describe why gravity isn't distorted in the same fashion. Can we use the effect of time contraction on energy?

 

[Ibix]

Then, setting aside motion as curvature, I'm back to the question of why gravity should be spherically symmetrical considered from the perspective of an observer in relative motion.

You first need to define "from the perspective of an observer in motion".

It is certainly true that a pair of observers far apart moving rapidly towards a gravitating mass (e.g. a pair of ships approaching the solar system with a couple of AUs lateral separation) will see their courses abruptly change (i.e., the rate of change of their separation changes abruptly) as they pass it, where slower observers would see a more gradual change. This does not mean that the spacetime is not spherically symmetric - merely that these observers have picked a coordinate system that obscures the symmetry, however natural the system may be to them.

 

[Nugatory]

I'm back to the question of why gravity should be spherically symmetrical considered from the perspective of an observer in relative motion.

You've been tossing that phrase "spherically symmetrical" around without defining it... and a coordinate-free definition is a bit trickier than you might expect.
The spacetime outside of a Schwarzschild black hole (or any other spherical mass distribution) is spherically symmetrical. You can choose coordinate systems that make this symmetry obvious or that obscure it, but it's always there.

Considering it as emanating from the source...

It doesn't. This line of thought pretty much guarantees misunderstanding and confusion

Considering it as moving with the source...

It doesn't This is a four-dimensional spacetime that we're talking about, so nothing is moving. You can choose coordinates in which one or more of the position coordinates of the gravitating body are functions of the time coordinate, but that's just a way of constructing a coordinate system that obscures the symmetry of the spacetime.

 

[AdirianSoan]

Alright, let's switch to yet another question that is asking the same thing.

A very large mass is moving at relativistic speed, a significant distance away. I am observing from a point such that the closest point in it's motion is some distance away to my right.

I observe it slightly to the left of where it "really" is, because simultaenity. If I had absurdly precise detection accelerometers without any interference, would these accelerometers measure acceleration towards the point I observe it at, or at the point it "really" is?

 

[Nugatory]

I observe it slightly to the left of where it "really" is, because simultaenity.

The fact that you had to put the word "really" in quotation marks suggests that you still haven't completely defined the problem in your own mind. But you do agree that this situation is indistinguishable in all respects from the one in which the gravitating body is at rest and you are moving relative to it?

 

[Ibix]

If I had absurdly precise detection accelerometers without any interference,

...they would measure zero because you are in freefall.

If you simply release a cloud of particles inside your spacecraft you could, in principle, detect tidal gravity by comparing the slight differences in their trajectories and define "down" this way. Whether or not that definition of down points towards where the mass is "now" depends on your choice of simultaneity condition. Because that is what defines where the object is "now".

 

[AdirianSoan]

The fact that you had to put the word "really" in quotation marks suggests that you still haven't completely defined the problem in your own mind. But you do agree that this situation is indistinguishable in all respects from the one in which the gravitating body is at rest and you are moving relative to it?

In this case it means I think the object is really where you see it, and calculating out where it would be if not for simultaneity, and insisting it is "really" there, is misleading yourself, but am not 100% certain of this assertion. So my expectation is that the acceleration behaves as though the object is where you observe it to be.

And effect is symmetrical, in a sense; from the perspective of the moving object, your position is also shifted (closer to it's line of motion). The chirality is reversed, but I don't think this matters much.

What gets weird in treating the object as being where you see it is when you consider other observers in your inertial frame nearer the object. If they flash light at you indicating their observed acceleration, the only reasonable conclusion is that the curvature of the moving object is itself massively curved by its motion. (In particular, consider the case when an observer is situated such that they are on one side of the object from their perspective, and the other side of the object from yours.)

Mind, I am fine with this. It has some interesting implications.

 

[AdirianSoan]

they would measure zero because you are in freefall.

Ah. Fair. Let's suppose they are measuring acceleration compared to a set of distant objects which you begin at relative rest with. (There is a complication there, because your motion relative to them will change everything again. Can this be handwaved away?)

Edit:
Wait, nevermind, your tidal approach was more apt anyways. My brain wasn't processing it correctly.

 

[Ibix]

Ah. Fair. Let's suppose they are measuring acceleration compared to a set of distant objects which you begin at relative rest with.

This is just a form of coordinate acceleration. Exactly what it shows you will depend on the exact trajectories of the objects and how you define your simultaneity criterion - just as the tidal gravity approach does. This keeps coming back to your simultaneity definition, and always will.