How do accelerometers measure acceleration without using relative motion?

[Zula110100100]

So on another post somewhere it was mentioned that while velocity is relative, acceleration is absolute, this might be true for the magnitude of acceleration but how are we sure of the direction, am I accelerating or is the rest of the world.

Is there any design of accelerometer that doesn't work by measuring the relative motion, and therefore acceleration, of two components? Under the force of gravity the jar and cork are accelerated uniformly and so no measure of acceleration is given.

I understand that in most cases we can tell which of two bodies is accelerating by comparing measurements on their accelerometers, but if two people are are constant velocity in space, and one nears a black-hole(with no stars around for gravitational lensing) their relative velocity increases, but which one is near the black hole?

So if objects in relativity's gravity follows geodesics rather than accelerating how do their relative velocities increase?

 

[Vorde]

With regards to the acceleration question, I believe the direction of acceleration is relative simply because I dare anyone to define an object's direction without reference to something else.

To the black hole question, the answer goes back to the same point you made. While each sees the other recede at the same rate, only one observer feels the pull of the black hole; acceleration is absolute.

 

[Zula110100100]

Going in free-fall into a black hole one would not feel the accelration would they? Until the tidal effects spaghettied them

 

[Vorde]

I'm not quite sure, I know from experience that a person can detect acceleration, I once had a conversation where I was convinced of it. However I can readily believe that is due to a complex 'accelerometer' (I can think of a few mechanisms ) present in the brain.

I'm not sure whether or not ultimate acceleration can be discerned, however I am tempted to believe so as it was my understanding that is how the twin paradox is resolved.

 

[PAllen]

All you need to measure magnitude and direction of acceleration from your point of view (you being stationary relative to yourself) is a standard mass on a spring.

However, you are completely correct that direction of acceleration is observer dependent. In relativity, acceleration is a 4-vector, the proper time derivative of the the 4-velocity along some world line. The vector is a coordinate independent object, its magnitude is invariant, but its component expression is coordinate dependent, so 'natural' coordinates for different observers will lead to different component expression.

As to another of your questions, in curved spacetime, geodesics may converge. Thus, objects with no acceleration, initially at rest with respect to each other, can move together. This is simply a consequence of curvature: the parallel postulate fails on sphere - geodesics intersect. The same is true for geodesics in curved spacetime.

I think you may also have a confusion about what acceleration is in GR. An apple falling to Earth is not accelerating. An apple resting on the ground, is accelerating.

 

[Zula110100100]

I do understand the difference, or I am aware of it, I just have trouble accepting it. The way I look at it though is that these converging geodesics(if that is what I call the apple falling) does constitute acceleration. It has an effect on their relative velocities. In the case of to two people at rest and then one begins to be sucked into a black hole, not only would their relative velocity change, but the one entering the black hole would have time differences indicating they are in a different frame of reference? which i equate with a change in energy level and therefore acceleration...is this wrong?

 

[PAllen]

I do understand the difference, or I am aware of it, I just have trouble accepting it. The way I look at it though is that these converging geodesics(if that is what I call the apple falling) does constitute acceleration. It has an effect on their relative velocities. In the case of to two people at rest and then one begins to be sucked into a black hole, not only would their relative velocity change, but the one entering the black hole would have time differences indicating they are in a different frame of reference? which i equate with a change in energy level and therefore acceleration...is this wrong?

To understand relativity, you need to let go of concepts like 'at rest' without clarification of relative to what.

The apple falling is following an inertial (no force on it), geodesic path. Converging geodesics is not shown by comparison with an apple on the ground (which is accelerating), but with an apple falling on the other side of the world - the two free fall (geodesic) paths are approaching each other.

An equivalence principle analogy is that a person stepping off a cliff and beginning to fall is the same as stepping out of an accelerating rocket. The rocket and the person each see the other accelerating away (coordinate acceleration); only the rocket observer feels force (thus this is the non-inertial one with proper acceleration).

As for energy, I'm not sure what you are asking. Can you try to state a specific question about energy?

 

[D H]

With regards to the acceleration question, I believe the direction of acceleration is relative simply because I dare anyone to define an object's direction without reference to something else.

You are missing the point. How we represent something is not the same as the thing being represented. In this case, you are confusing the representation of the acceleration vector with the acceleration vector itself. Choose one particular reference frame to use as the basis for representing that acceleration vector and you will get some particular set of coordinates for vector. Choose a different frame and you will of course get a different set of coordinates. It is still the same vector, however. You say potato I say patahto ...

To the black hole question, the answer goes back to the same point you made. While each sees the other recede at the same rate, only one observer feels the pull of the black hole; acceleration is absolute.

Sans spaghettification, neither "feels" the pull of the black hole. An accelerometer doesn't sense acceleration due to gravity. An accelerometer senses acceleration relative to an inertial frame that is co-located and co-moving with the accelerometer. This is why some say that "acceleration is absolute". An accelerometer is a purely local experiment. No outside reference is needed. It is measuring acceleration relative to some attribute of local space-time itself -- and it does so without constructing that co-located, co-moving inertial frame.

 

[Zula110100100]

To clarify then, I mean, the two people in space are at rest(relative to each other) and I am not comparing the apple to the Earth and saying the apple is accelerating, I am comparing the apple falling near the Earth with an apple floating in space that was comoving at the beginning of the experiment. As the apple near the Earth falls nearer the Earth its trajectory would lead to an increase in relative velocity as compared to the apple far away from gravity.

The answer: "That is not acceleration in GR because it is not represented as such" is not a satisfactory answer to me. At least not with a reason WHY it isn't.

[EDIT] this is like saying all we did was redefine acceleration..

The most common reason I know of is the accelerometer deal, and the fact that one could not tell if he were accelerated by gravity or by a rocket.

An accelerometer is a purely local experiment. No outside reference is needed. It is measuring acceleration relative to some attribute of local space-time itself -- and it does so without constructing that co-located, co-moving inertial frame.
T 03:55 PM

Please explain to me the design of an accelerometer that works in such a way, I assert that a mechanical accelerometer can only measure a DIFFERENCE in acceleration of one part and another.

So if I had a weight on a spring, and I accelerate the spring and the weight, it measures no acceleration. In any case other than gravity it doesn't really work because you could put accelerometers on each component and see that each was accelerating, but in a uniform(or nearly uniform) acceleration field, every component of every accelerometer would accelerate evenly, giving the appearance of no acceleration.

Putting an accelerometer on a table and measuring upward acceleration requires you place it on that table, contacting it, measuring a push with my finger requires my finger contact it, so how is this local and without outside reference, in the case of a cork and a jar, with just the cork, we know nothing. it is only when measuring the acceleration required to move the cork with the jar that we read acceleration.

So consider a weight in a box, in freefall the box and weight accelerate evenly, so no measured acceleration, at the point that the box contacts the ground, the box is suddenly accelerated upward, causing the weight to continue moving down while force is built in the spring, at this point, we measure a relative acceleration. As the device reaches equilibrium it is left that the box is sitting on the ground, and the weight suspeneded by the spring that is pulled of equilibrium due to the fact that gravity is continuing to apply force that must be balanced by the spring. This extended spring shows us that upward acceleration is occurring but I attest that this is a flaw in the design of the accelerometer, not the universe.

 

[Zula110100100]

As for a specific question about energy, it would be:Is a change in inertial frame proportional to a change in relative kinetic energy? Is a change in relative kinetic energy proportional to a change in acceleration? and finally nearing gravity, in a free-fall, aren't the relativistic effects on time still occurring, like, falling into a black hole, time still slows down(relative to a previously co-moving observer removed from gravity), right? This to me indicates that the one "not accelerating" is changing inertial frame, and if the above questions are right...accelerating

 

[D H]

The answer: "That is not acceleration in GR because it is not represented as such" is not a satisfactory answer to me. At least not with a reason WHY it isn't.

[EDIT] this is like saying all we did was redefine acceleration..

What general relativity has done is to redefine the what constitutes an inertial frame.

First off, it is important to remember that different observers looking at the same object will see different accelerations. This isn't just a matter of redefining acceleration. It is reality. Even in Newtonian mechanics, an accelerating observer will obviously see something different from a non-accelerating observer. Inertial frames have a special place in Newtonian mechanics; all inertial observers in a Newtonian universe would see that object as undergoing the same acceleration. Even though inertial frames are the same in Newtonian mechanics and special relativity, that acceleration is invariant to all inertial observers no longer true in special relativity. To overcome this ambiguity, special relativity introduces the concept of proper acceleration. This concept of proper acceleration carries over to general relativity, but now under the guise of an inertial frame as defined in general relativity.

The most common reason I know of is the accelerometer deal, and the fact that one could not tell if he were accelerated by gravity or by a rocket.

The ideal accelerometers can serve as as a test for whether a frame is inertial is a very important concept in general relativity. The latter part of your statement is a bit off. Slap an accelerometer on a falling apple and it will register zero. Once it hits the ground and stops bouncing around, the accelerometer will register an acceleration of 1 g upward. That upward acceleration is caused by the ground, not gravity. It is this normal force that the accelerometer is measuring, not gravity. An apple on an a rocket in deep space accelerating at 1 g will also measure an acceleration of 1 g.

What you are missing is that the definition of an inertial frame is quite different in general relativity than Newtonian mechanics. In Newtonian mechanics, a frame that is inertial at the origin is inertial everywhere. In general relativity, that inertial characteristic is local rather than universal. Another big difference is that a free-falling frame is non-inertial in Newtonian mechanics but is inertial in general relativity.

Another thing you are missing is that gravity is a fictitious force in general relativity. An accelerometer cannot measure coriolis acceleration because it doesn't arise from a "real" force. Accelerometers cannot measure acceleration resulting from any fictitious force -- including gravitation.

Please explain to me the design of an accelerometer that works in such a way, I assert that a mechanical accelerometer can only measure a DIFFERENCE in acceleration of one part and another.

A completely different way to measure acceleration is to measure numerical differentiate the measured velocity with respect to some external object. This is a non-local experiment. The test masses in an accelerometer are internal to the accelerometer. No external measurements are needed: A local experiment.

So if I had a weight on a spring, and I accelerate the spring and the weight, it measures no acceleration. In any case other than gravity it doesn't really work because you could put accelerometers on each component and see that each was accelerating, but in a uniform(or nearly uniform) acceleration field, every component of every accelerometer would accelerate evenly, giving the appearance of no acceleration.

This extended spring shows us that upward acceleration is occurring but I attest that this is a flaw in the design of the accelerometer, not the universe.

Accelerometers cannot measure gravity. No local experiment can. This is not a flaw in the design of the accelerometer. It is the way the universe works.

 

[PAllen]

As for a specific question about energy, it would be:Is a change in inertial frame proportional to a change in relative kinetic energy? Is a change in relative kinetic energy proportional to a change in acceleration? and finally nearing gravity, in a free-fall, aren't the relativistic effects on time still occurring, like, falling into a black hole, time still slows down(relative to a previously co-moving observer removed from gravity), right? This to me indicates that the one "not accelerating" is changing inertial frame, and if the above questions are right...accelerating

This may be clear to you, but it is totally unclear to me. I'll try to answer some possible questions.

Kinetic energy is frame dependent in Galilean mechanics, SR, and GR. In none can you talk about 'proportionality' though. Changing frame changes kinetic energy according to 1/2 mv^2 in Galilean mechanics; by mc^2(1 - 1/(1-v^2/c)) in relativity (which works out almost the same when v is small compared to c).

Kinetic energy at a given moment for a given body is unrelated to acceleration. Of course, acceleration is related to change in kinetic energy. Integrate force*distance and you get change in kinetic energy.

Compare two free fall observers: one falling radially, the other orbiting. The orbiting one will see the radial clock slow down. Both are still inertial. If we posit a stationary observer where the orbital and radial observers separate, they will see the orbital clock slow and the radial free fall clock even slower. If we have a clock 'tossed' radially away from the planet and falling back to the stationary clock, this clock will be faster than the stationary clock (in fact, this radial out and back free fall path will be the path of longest elapsed time - fastest clock - between any two events).

I don't understand what conclusions you are trying to draw from all these different clock rates. Very little of what you say makes any sense to me.

 

[chingel]

Do I understand correctly that gravity is not really accelerating me, it just curves the space and the ground has to apply a force on me to make me move different than a "straight" line would, which is now actually bent? But why am I moving along the curved space if I am standing still? I can understand how a bullet flying past Earth would get pulled into the Earth due to the space curving, but why does it happen when I stand still?

Also, is gravitational time dilation and length contraction caused by the fact that acceleration and gravity is really the same thing? But how do I get the speed to calculate the time dilation, because if all I know is acceleration, I might have been accelerating for a million years or for one second?

To the OP, if you are in a spaceship and not accelerating, you will hover weightless. If you start accelerating, you will fly into the wall opposite to the direction of acceleration. To determine the rate of acceleration, place a scale underneath yourself. If you doubt that maybe you are actually accelerating in the other direction, you can test this by placing the scale on the opposite wall of which you are standing on and you would find that the scale falls back to you. You can determine acceleration this way without having to look anywhere else what other objects or spaceships are doing.

 

[D H]

Do I understand correctly that gravity is not really accelerating me, it just curves the space and the ground has to apply a force on me to make me move different than a "straight" line would, which is now actually bent?

You understand incorrectly.

There is a problem with how to represent acceleration even in Newtonian mechanics. Acceleration is in general frame-dependent even in Newtonian mechanics. There is one special class of frames where acceleration is invariant in Newtonian mechanics, and these are the Newtonian inertial frames. The acceleration vectors of some object in any two inertial frames are the same vector. The acceleration vector as expressed in and observed from an inertial frame is the canonical representation of that frame.

This problem gets even more involved in relativistic mechanics. In special relativity, the acceleration 3-vectors are not invariant across inertial frames. The solution is to use "proper acceleration," the acceleration with respect to a co-located, co-moving inertial frame. Note well:

• This proper acceleration is exactly what an ideal accelerometer would measure.
• The qualifier "proper" does not mean that any other representation of acceleration is somehow incorrect. To the contrary! All frames of reference are equally valid in relativity theory. Perhaps a better qualifier would be "canonical." But that isn't the label that was attached, and now we're stuck with "proper."

The problem gets yet hairier in general relativity. The concept of proper acceleration carries over rather nicely, with one caveat: What constitutes an an inertial frame is quite different in general relativity. The equivalence principle is a huge motivating factor for this change in meaning of "inertial frame." Another motivating factor is that the Newtonian laws of physics take on their simplest form in Newtonian inertial frames. Fictitious forces vanish in Newtonian inertial frames. The laws of physics per general relativity similarly take on their simplest form in general relativity. Just as fictitious forces vanish in Newtonian inertial frames, they also vanish in GR inertial frames. One side effect: Gravitational acceleration tautologically vanishes in a GR inertial frame. In GR, gravitation is a fictitious force!

 

[Zula110100100]

So in this picture the two components of the accelerometer are equally accelerated by the force, as would be with gravity, Yet it would measure an upward acceleration as the outer component(which every type of mechanical accelerometer we can make would have(as far as I know that is why I asked about types in use)) if the table were not there and the acceleration held constant it would not measure acceleration. Why in GR is gravity treated different than this? If I am being accelerated by the ground, not gravity, then what am I being accelerated relative to? Does acceleration in GR not add to my kinetic energy? If it does, my energy must be astronomical compared to whatever I am being accelerated relative to. If the answer to all this is just that "In GR we don't call it acceleration so...it must not be." then I am sort of disappointed with these forums.

And to PAllen what I mean is: two people in space at rest relative to each other: I would think that not counting gravity(in GR) the only way to "change inertial frames" and in so doing be no longer comoving is to accelerate. So if that is indeed the case, then my question is if one of the objects starts moving more quickly toward the surface of the Earth does it "change inertial frame relative to the one farther removed from gravity that it was previously comoving" no longer comoving with said other object?

 

[robphy]

[*]The qualifier "proper" does not mean that any other representation of acceleration is somehow incorrect. To the contrary! All frames of reference are equally valid in relativity theory. Perhaps a better qualifier would be "canonical." But that isn't the label that was attached, and now we're stuck with "proper."

I believe "proper" was used to suggest "property" or "ownership" (and, as you say, not something suggesting "correctness"). Minkowski introduced "Eigenzeit", which we call "Proper Time" (or what EF Taylor calls "wristwatch time" or what H.Bondi calls "private time"). Some authors have suggested that "Eigentime" might have been a better translation (similar to "Eigenvector" a.k.a. "Proper vector").

 

[PAllen]

So in this picture the two components of the accelerometer are equally accelerated by the force, as would be with gravity, Yet it would measure an upward acceleration as the outer component(which every type of mechanical accelerometer we can make would have(as far as I know that is why I asked about types in use)) if the table were not there and the acceleration held constant it would not measure acceleration. Why in GR is gravity treated different than this? If I am being accelerated by the ground, not gravity, then what am I being accelerated relative to? Does acceleration in GR not add to my kinetic energy? If it does, my energy must be astronomical compared to whatever I am being accelerated relative to. If the answer to all this is just that "In GR we don't call it acceleration so...it must not be." then I am sort of disappointed with these forums.

And to PAllen what I mean is: two people in space at rest relative to each other: I would think that not counting gravity(in GR) the only way to "change inertial frames" and in so doing be no longer comoving is to accelerate. So if that is indeed the case, then my question is if one of the objects starts moving more quickly toward the surface of the Earth does it "change inertial frame relative to the one farther removed from gravity that it was previously comoving" no longer comoving with said other object?

The essence of GR is that gravity is not a force. There is a feature of objects called inertia, which is their resistance to force. We call the numeric value of this inertia 'mass'. Instead of the 'law of inertia' and the 'law of gravity', GR only has the 'law of inertia': unless acted on by a force, a body moves in the straightest possible path - a geodesic of spacetime.

When the ground accelerates you, you are being accelerated relative to inertial motion, the same as the law of inertia states in Galilean mechanics. Note that if you are in a rocket, the floor is accelerating you; from your point of view, you feel force but (assuming no windows) you don't seem accelerated relative to anything, and your kinetic energy remains zero in your frame (because the force acts over zero distance in your frame). In GR, standing on Earth is simply identical to this situation. Relative to an inertial frame, when you stand on Earth you are being accelerated upward and your kinetic energy rapidly grows (in the free fall inertial frame). Note that the idea the KE is strictly frame dependent goes all the way back to Newton and Galileo - it is not new with relativity. There is no such thing as KE except in relation to a specified frame; it is always zero in an object's own frame

As to your final question, it still makes no sense to me. If two people in space are at rest, next to each other, and not under a force, they will remain at rest relative to each other. If one of them accelerates (measured by an accelerometer), by definition, they are not in an inertial frame any more. The rest of this paragraph comes across as gibberish to me. Sorry. Please try to state a precise scenario, and ask a specific question about it.

 

[Zula110100100]

Then I fear gibberish may be my only language. It seems I cannot find a way to describe the situation without using words which we cannot agree on a definition. I will try to describe a situation as clearly as I can, the specific question is: Does acceleration take place?

Suppose you have two objects at rest to each other, and each are approaching a gravitational body at the same rate. As the object nearer the gravitational body does whatever is done to cause it to move closer to that body at a greater rate than before, the objects would be no longer moving at the same rate. Did acceleration take place?

I am starting to believe I will know your answer. You will reply, "No, Since no acceleration is measured on an accelerometer, no acceleration has taken place"

Still you have not addressed the question about the method employed by the accelerometer, I am not sure if it posted correctly but in the previous post there was an attached image, could you please refer to the image and answer the specific question:

So in this picture the two components of the accelerometer are equally accelerated by the force, as would be with gravity, Yet it would measure an upward acceleration as the outer component(which every type of mechanical accelerometer we can make would have(as far as I know that is why I asked about types in use)) {is held by the table and the dampered mass is only held by the spring, applying a force on the spring in the downward direction} if the table were not there and the acceleration held constant it would not measure acceleration. Why in GR is gravity treated different than this?

To add extra description, the two components would be the inner weight and the outer shell. Unless you can describe an accelerometer that works without some form of two components that measure the force required to move a component not directly connected to the acceleration. From wikipedia:

Conceptually, an accelerometer behaves as a damped mass on a spring. When the accelerometer experiences an acceleration, the mass is displaced to the point that the spring is able to accelerate the mass at the same rate as the casing. The displacement is then measured to give the acceleration.

...

Under the influence of external accelerations the proof mass deflects from its neutral position.

So the gist of it is that it is made of two parts, a casing an a damped mass, in mentions that under external accelerations the mass is displaced, It does not explain that the device fails to work when under a uniform force, the mass and casing would be equally displaced, so the device is not flawless.

But, since that is not how it is defined, none of that makes sense to you either, and is certainly could not be what is happening

 

[PAllen]

Then I fear gibberish may be my only language. It seems I cannot find a way to describe the situation without using words which we cannot agree on a definition. I will try to describe a situation as clearly as I can, the specific question is: Does acceleration take place?

Just to clarify here, there is coordinate acceleration and proper acceleration. Consider a rocket in deep space passing a floating planetoid. The rocket sees the planetoid accelerate, the planetoid sees the rocket accelerate. They are both coordinate acceleration. But only the rocket locally measures acceleration with an accelerometer, with no reference to any other object; that is proper acceleration.

Suppose you have two objects at rest to each other, and each are approaching a gravitational body at the same rate. As the object nearer the gravitational body does whatever is done to cause it to move closer to that body at a greater rate than before, the objects would be no longer moving at the same rate. Did acceleration take place?

If the two objects are next to each other, they will not separate, they will remain at rest. I am going to guess what you might mean, but the problem isn't terminology, it unwillingness to be reasonably precise on your part. I am going to guess you have in mind two objects starting out at rest relative to each other, one e.g. 50 miles from earth, the other 100 miles (ignore air). Over time, the distance between them increases and they are no longer at rest relative to each other. Was that so hard? You can't expect people to guess all these things - you have to be specific. I am still not sure this is your scenario, but I will answer this one.

The main thing to realize is that in curved spacetime, inertial frames are local. There is no such thing as global inertial frame. Thus, each of these bodies has its own inertial frame for its 'immediate region', but extended as far as described, each of these bodies has its own separate local inertial frame. Another point, is that interpretation is needed about each being at rest relative to the other initially. One definition of 'at rest' is absence of red/blue shift. However, in this case, the 100 mile object will see redshift from the 50 mile object, while the 50 mile object will see blueshift from the 100 mile object. Thus, if each interprets things as if there were no gravity, they conclude they are not stationary. However, they could use light bouncing, and determine change in bounce time is change in distance (there are subtelties in this choice as well, but we can ignore that for simplicity). Then, indeed they see motion develop between them even though they are both inertial, and started out apparently at rest. What does this mean? It means there is space time curvature. A definition of curvature is that geodesics cease to remain parallel - they converge or diverge. Since each object is inertial, yet their distance grows, voila, you have demonstrated that there is detectible spacetime curvature over this distance.

What type of acceleration is this? Well, remember, each can only establish a local inertial frame. Beyond that, you just have coordinates. Thus each sees coordinate acceleration by the other, but neither experiences proper acceleration, which is a local measurement.

I am starting to believe I will know your answer. You will reply, "No, Since no acceleration is measured on an accelerometer, no acceleration has taken place"

Still you have not addressed the question about the method employed by the accelerometer, I am not sure if it posted correctly but in the previous post there was an attached image, could you please refer to the image and answer the specific question:

To add extra description, the two components would be the inner weight and the outer shell. Unless you can describe an accelerometer that works without some form of two components that measure the force required to move a component not directly connected to the acceleration. From wikipedia:
So the gist of it is that it is made of two parts, a casing an a damped mass, in mentions that under external accelerations the mass is displaced, It does not explain that the device fails to work when under a uniform force, the mass and casing would be equally displaced, so the device is not flawless.

But, since that is not how it is defined, none of that makes sense to you either, and is certainly could not be what is happening

As for accelerometers, I wasn't planning to answer this. This is a basic classical physics. You over complicate the whole picture. If I put a weight on a string that can measure tension, and release the object while holding the other end of the string, the direction and tension of the string tells me direction and magnitude of acceleration. The direction of acceleration points away from the weight along the line of the string. The magnitude of acceleration is given by the string tension. The 'thing' being tested for acceleration is whatever is attached to the other end of the string. This is the complete definition. With minimal thought, you can answer all of your scenarios in which you add unnecessary complications.

 

[Zula110100100]

Just to clarify here, there is coordinate acceleration and proper acceleration. Consider a rocket in deep space passing a floating planetoid. The rocket sees the planetoid accelerate, the planetoid sees the rocket accelerate. They are both coordinate acceleration. But only the rocket locally measures acceleration with an accelerometer, with no reference to any other object; that is proper acceleration.

This is really of no consequence to the discussion if you mean the rocket is firing because I never disagreed that mechanical acceleration was not acceleration, or that being in contact with the floor of that rocket let's you feel it

If the two objects are next to each other, they will not separate, they will remain at rest. I am going to guess what you might mean, but the problem isn't terminology, it unwillingness to be reasonably precise on your part. I am going to guess you have in mind two objects starting out at rest relative to each other, one e.g. 50 miles from earth, the other 100 miles (ignore air). Over time, the distance between them increases and they are no longer at rest relative to each other. Was that so hard? You can't expect people to guess all these things - you have to be specific. I am still not sure this is your scenario, but I will answer this one.

I never said they were right next to each other, I didn't give specific numbers but I did specify one is nearer, and that one is "the one farther removed from gravity". Either way the distance should not matter, if one is nearer at ALL then even if only a small amount it should accelerate more, if only a small amount.

One definition of 'at rest' is absence of red/blue shift. However, in this case, the 100 mile object will see redshift from the 50 mile object, while the 50 mile object will see blueshift from the 100 mile object. Thus, if each interprets things as if there were no gravity, they conclude they are not stationary.

I am trying very hard not to be rude and respond "You must put the idea of stationary away, in relativity, there is no way to know one is stationary, and could only be stationary relative to something else" But I know that you mean, *relative to each other* on the end of that, if you could please give me some of the same courtesy I would appreciate it.

But here's the thing, if gravity cannot accelerate, explain this situation:

My Accelerometer is my hand holding a string. My hand is placed 100miles away from the planet(ignoring air) and the string is a 50mi string connecting a weight 50miles above the surface to my hand, 100mi away from the surface. Do I measure a pull on the string or not?

This actually ties in with the complication of the accelerometer quite nicely, who decides when it is overcomplication and when it is just enough. You still choose an accelerometer made of two parts, the weight and your hand, and measure the force between them to get acceleration, if both are acted on by the same acceleration, there with be no force on the string.

As for accelerometers, I wasn't planning to answer this. This is a basic classical physics. You over complicate the whole picture. If I put a weight on a string that can measure tension, and release the object while holding the other end of the string, the direction and tension of the string tells me direction and magnitude of acceleration. The direction of acceleration points away from the weight along the line of the string. The magnitude of acceleration is given by the string tension. The 'thing' being tested for acceleration is whatever is attached to the other end of the string. This is the complete definition. With minimal thought, you can answer all of your scenarios in which you add unnecessary complications.

 

[Zula110100100]

Rereading you accelerometer remark:

The 'thing' being tested for acceleration is whatever is attached to the other end of the string. This is the complete definition.

If that is the case, is there no control? Not a very good experiment. The control is usually the casing, or hand in your simplified example. So only when the control is free of the acceleration to be tested is it valid. How can you set this up in gravity? There will be no "control"

[Correction:] The control is usually the weight? as the force is applied to the casing, and not the weight, so the force required to accelerate the weight with the casing is measured by the spring.

 

[PAllen]

This is really of no consequence to the discussion if you mean the rocket is firing because I never disagreed that mechanical acceleration was not acceleration, or that being in contact with the floor of that rocket let's you feel it

The purpose of this was to lay the groundwork for discussion of coordinate acceleration versus proper acceleration with non-controversial example. Of course, if you are only interested in controversy ...

I never said they were right next to each other, I didn't give specific numbers but I did specify one is nearer, and that one is "the one farther removed from gravity". Either way the distance should not matter, if one is nearer at ALL then even if only a small amount it should accelerate more, if only a small amount.

You were very ambiguous. You said: "Suppose you have two objects at rest to each other, and each are approaching a gravitational body at the same rate". How is someone to know the degree of distance or the orientation you have in mind from this. I had to guess. That should not be necessary.

I am trying very hard not to be rude and respond "You must put the idea of stationary away, in relativity, there is no way to know one is stationary, and could only be stationary relative to something else" But I know that you mean, *relative to each other* on the end of that, if you could please give me some of the same courtesy I would appreciate it.

Be rude if you want, but don't lie. What I said in the sentence you refer to was: "Another point, is that interpretation is needed about each being at rest relative to the other initially". Please note the wording and mis-representation.

But here's the thing, if gravity cannot accelerate, explain this situation:

My Accelerometer is my hand holding a string. My hand is placed 100miles away from the planet(ignoring air) and the string is a 50mi string connecting a weight 50miles above the surface to my hand, 100mi away from the surface. Do I measure a pull on the string or not?

This actually ties in with the complication of the accelerometer quite nicely, who decides when it is overcomplication and when it is just enough. You still choose an accelerometer made of two parts, the weight and your hand, and measure the force between them to get acceleration, if both are acted on by the same acceleration, there with be no force on the string.

An accelerometer making a local measurement is expected to be small. It can be one millimeter if desired. If it is large, it is no longer a local measurement, which raises additional issues in GR.

If you measure stress on an some extended object near another massive object, you measure tidal forces. GR in no way says tidal forces don't exist - they directly measure curvature over a distance, as distinct from local straightness, which GR classifies as inertial motion not acceleration. Tidal forces are irremovable. Local 'gravitational force' is trivially removable. GR distinguishes the former as a physical feature of spacetime, and the latter as a fictitious artifact of coordinate choice.

 

[PAllen]

Rereading you accelerometer remark:
If that is the case, is there no control? Not a very good experiment. The control is usually the casing, or hand in your simplified example. So only when the control is free of the acceleration to be tested is it valid. How can you set this up in gravity? There will be no "control"

[Correction:] The control is usually the weight? as the force is applied to the casing, and not the weight, so the force required to accelerate the weight with the casing is measured by the spring.

I cannot untangle what you mean here. The accelerometer measures the acceleration of what it is attached to, period. If it is your hand, then, big surprise, your hand can accelerate relative to your body, and the accelerometer is only sensitive to the objective, proper, acceleration of your hand. There is no 'relative to anything else'. A small accelerometer measures an invariant, local feature of the motion of whatever it is attached to. Other objects in the vicinity are irrelevant, casings are irrelevant. The accelerometer specifies a vector, magnitude and direction, irrespective of any complications you try to introduce.

 

[Zula110100100]

If you are unable to tell the difference between a force acting on only the casing and the force acting on both, and cannot understand what I mean in the picture, then I cannot make it any clearer to you.

Aside from that it doesn't really matter because what I get what what you are saying is acceleration or not, we don't call it acceleration in GR, we are able to cut it out since free-falling objects travel as if it rest to each other, EXCEPT in the case that tidal FORCES make a difference, then you get an irremovable force that is cause by NOT acceleration acting on the objects.

The problem I have really then is that if x were a variable representing the accuracy of measuring tidal force, and f(x) a function that gave you the largest possible distance at which tidal forces were not noticable with at that accuracy

Then as x approached 100%, f(x) would approach 0

There must be a better way to represent gravity, I know Newton's has problems too, and GR is an improvement.

But there should be an explanation that doesn't require such restrictions.

 

[pervect]

Did you just imply that you don't want the force of gravity to approach zero as a limit as the distance approaches infinity? ::scratches head::.

This may or may not help you very much, but your thinking seems to be entirely oriented around some concept of "force".

Forces used to be a tensor in special relativity, by virtue of requiring that all measurements be done in coordinate systems with special properties, these coordinate systems being known as "inertial frames".

However, when you look at more general transformations, for instance accelerating frames of reference, one findd that forces no longer transform as tensors.

For example, the Newtonian answer to "what is the force on an object at rest in an inertial reference frame" would be zero. When you switch to an accelerating reference frame, the Newtonian answer to what the force in the accelerating frame is is "you need to use an inertial refererence frame, you don't use accelerating frames in Newtonian Mechanics". You might find some discussion about how you can treat a non-inertial reference frame as inertial if you introduce "fictitious forces", but it will be stressed that these forces are fictitious.

The GR answer is to generally avoid forces whenever is possible. This tends to confuse people who have come to get used to them. The GR textbooks don't generally explain why they are avoiding forces, though you can note the avoidance by the general lack of disccusio of them. The fact that they no longer transform as tensors is, IMO, the fundamental reason.

GR does not rely anymore on defining special "inertial frames". So if you try to worry about "what is the GR equivalent of the Newtonian inertial frame", you're not going to get an answer. Except perhaps in the local sense - GR does have a clear idea of what a locally inertial frame is. It's just that these local frames are purely local, and no longer global - they only cover small regions of space-time.

So if you're used to thinking of forces as being tensors and having physical significance, they're no longer tensors. The only thing that made them tensors was defining special "inertial frames", and in GR defining inertial frames is no longer required, the formalism has been changed and modifed so that you don't really need them anymore. There are a few places where they might "sneak in", but if you exaine those places they sneak in along with "local inertial frames".

 

[Zula110100100]

Did you just imply that you don't want the force of gravity to approach zero as a limit as the distance approaches infinity? ::scratches head::.

What I meant was that as the sensitivity of the device measuring the tidal force gets better, the distance at which you can't tell it's there gets smaller, approaching 0 as the sensitivity increases.

The accelerometer question remains though, is there any way that a force that I apply can be represented as a fictitious force? I guess I am still "stuck on forces" but what does relativity call the equivalent of "force applied"?

 

[PatrickPowers]

My Accelerometer is my hand holding a string. My hand is placed 100miles away from the planet(ignoring air) and the string is a 50mi string connecting a weight 50miles above the surface to my hand, 100mi away from the surface. Do I measure a pull on the string or not?

You will measure a pull. Now have that same string and weight, but AWAY from the planet instead of toward. You will also measure a pull. So your device is telling you that there is acceleration in both opposite directions.

 

[Zula110100100]

GR does not rely anymore on defining special "inertial frames".

Does GR worry about proper acceleration? I know wikipedia is not a 100% reliable source, but it defines proper acceleration as:

In relativity theory, proper acceleration[1] is the physical acceleration (i.e., measurable acceleration as by an accelerometer) experienced by an object. It is acceleration relative to a free-fall, or inertial, observer who is momentarily at rest relative to the object being measured.

If that is the case, the GR does worry about inertial frames

What is your definition of local? One should not have to guess these things. Do you have a mathematical definition for the maximum distance that could be considered local? The only requirement I can see so far is that for two objects to be considered inertial they must be momentarily at rest.

So your device is telling you that there is acceleration in both opposite directions.

Does this verify my original postion that the magnitude is constant but the direction is not?

 

[PatrickPowers]

Does GR worry about proper acceleration?

Sure. It's general and worries about everything, even things like torsion.

What is your definition of local? One should not have to guess these things. Do you have a mathematical definition for the maximum distance that could be considered local? The only requirement I can see so far is that for two objects to be considered inertial they must be momentarily at rest.

The local frame is where Newtonian physics is a good enough approximation for your purposes. Since purposes vary greatly, one can't be more definite than that.

Does this verify my original postion that the magnitude is constant but the direction is not?

The magnitude isn't constant either. In a gravitational field With a weight and a string that long you would measure a pull in almost every direction, with the magnitude varying widely. That doesn't happen with an accelerometer.

 

[Zula110100100]

May I please have a mathematical definition of proper acceleration then?

If I calibrated my accelerometer to read 0 in minowski space and I was able to measure the force on the spring with infinite accuracy the moment I neared any gravitational body I would be able to measure the tidal effect on the accelerometer

The local frame is where Newtonian physics is a good enough approximation for your purposes.

In Newtonian physics gravity accelerates things, so the distance at which they work in a GR space containiny mass is null, because there is no distance in GR that gravity accelerates things?

If that is the case, how can one measure proper acceleration?

The magnitude isn't constant either. In a gravitational field With a weight and a string that long you would measure a pull in almost every direction, with the magnitude varying widely.

Do you mean as I change perspective? It seems that from a coordinate system fixed to my hand, it could only measure ONE vector of acceleration at a time

 

[PatrickPowers]

In Newtonian physics gravity accelerates things, so the distance at which they work in a GR space containiny mass is null, because there is no distance in GR that gravity accelerates things?

Whoops, you got me there. I was thinking of Newton's laws of motion, not his theory of gravitation.

Do you mean as I change perspective? It seems that from a coordinate system fixed to my hand, it could only measure ONE vector of acceleration at a time

Hmm. You can have a string and weight attached to your hand, but you can't have a coordinate system attached to your hand. You could have several strings and weights attached to your hand and notice that they are pulling in almost all different directions at the same time. The net force would depend on many things. It would be complicated and doesn't seem helpful.

The way I see it, Albert found out that once you establish a universal coordinate system then you have already lost. It can't be done. His theory helps you to do the best you can for your purposes.

As for proper acceleration, I don't know and others who do know have already answered, it seems to me.

 

[Zula110100100]

Hmm. You can have a string and weight attached to your hand, but you can't have a coordinate system attached to your hand. You could have several strings and weights attached to your hand and notice that they are pulling in almost all different directions at the same time. The net force would depend on many things. It would be complicated and doesn't seem helpful.

You can't have a coordinate system "attached" to you hand, but you can choose a coordinate system in which you hand is at rest, this is what I meant. The pull on all these string would cancel other than the net pull of gravity(wording issues, I know) These multiple string would complicate things, so to be simple I am using a unidirectional accelerometer with one weight and one string. It should measure the net acceleration I could be wrong here. To simplify further, let me add I am talking about a universe containing one large massed body and no other significant sources of gravity.

The way I see it, Albert found out that once you establish a universal coordinate system then you have already lost. It can't be done. His theory helps you to do the best you can for your purposes.

I am not establishing a universal coordinate system, but for anyone problem I must choose at least one coordinate system to work with(agree?), for this problem I am first choosing a cooridinate system in which my hand is always at the origin.

And I do appreciate that it helps, but to further out understanding of reality we must test each theory we have with every way we can think of, it's what Einstein did with Newtonian mechanics and to imply it need not be done is to say you are content with the level of understanding we currently have.

As for proper acceleration, I don't know and others who do know have already answered, it seems to me.

I will re-read the posts but I do not believe it has been satisfactorily answered in regards to this specific question.

 

[PAllen]

Just a few clarifying points:

1) Proper acceleration is a strictly local measurement. This follows immediately from its mathematical definition, which is second derivative with respect to proper time along a world line. You can't get more local than infinitesimal. (Note, in the formal definition, proper acceleration is a 4-vector; its norm (magnitude) is also frequently called proper acceleration when the context makes clear a scalar quantity is referenced. The vector is contravariant, the magnitude is a scalar invariant). Thus, an accelerometer to measure proper acceleration must be small.

2) Given the above, a weight on a long string is not considered an accelerometer (at least in the GR context under discussion). This type of experiment, instead, measures:

a) Newtonian terminology: in addition to how a small accelerometer would work, gravitational potential difference and tidal gravity, in a mixture depending on orientation.

b) GR terminology: a nonlinear combination of proper acceleration plus a measure of degree of curvature spanned by the long string.

3) The GR way of looking at things, I explained many posts ago, is motivated by replacing two laws with one (instead of a gravitational force law and a law of inertia, you have only a law of inertia). Given this foundation of GR, you cannot discuss or try to understand gravity in GR as a force. [Caveat: this is true in classical GR, classically interpreted. There are alternative viewpoints motivated in part by making GR look more like QFT forces. I prefer to avoid bringing these into discussion until the basics are understood.] Any time you want to say: "what about a force acting on everything the same way proportional to mass", that is exactly what caused Einstein to say: "What sort of force is that? It seems to affect all matter the same, acting on exactly the same quantity as inertia. Ah, it is actually inertia in disguise.".

 

[Zula110100100]

All you need to measure magnitude and direction of acceleration from your point of view (you being stationary relative to yourself) is a standard mass on a spring.

So you imply your method counts as an accelerometer, so long as the spring is infinitesimal in length, in a strong enough gravitational gradient all you would need is 1mm to measure a noticable pull on the spring... As your sensitivity to the measurement increases it becomes smaller and small that is needed.

Proper acceleration is a strictly local measurement. This follows immediately from its mathematical definition, which is second derivative with respect to proper time along a world line. You can't get more local than infinitesimal. (Note, in the formal definition, proper acceleration is a 4-vector; its norm (magnitude) is also frequently called proper acceleration when the context makes clear a scalar quantity is referenced. The vector is contravariant, the magnitude is a scalar invariant). Thus, an accelerometer to measure proper acceleration must be small.

I am neither a scientist or a mathematician so I apologize that my calculus terminology is lacking...

In a free-falling frame my acceleration I never change velocity, since A change of no velocity per proper time equals no acceleration, If I fire a rocket to escape(the not pull of) gravity, and use a frame that accelerates thus, then I still measure no change in velocity per proper time...so what is an example of proper acceleration please?

Now the object that is 50m from the surface does change its velocity(in my free-falling coordinate system 100m from surface) per proper time. So does an object 150m from surface, seems to slowly accelerate away? I am probably misusing proper acceleration here, but can you help clear that up?

 

[D H]

In a free-falling frame my acceleration I never change velocity

Velocity with respect to what? You are still appear to have the mistaken concept of absolute space, and you apparently do not understand the difference between proper acceleration and coordinate acceleration.

Keep in mind that the prefix "proper" does not mean "everything else is incorrect". A better prefix would be "eigen" or "characterestic".

 

[PAllen]

So you imply your method counts as an accelerometer, so long as the spring is infinitesimal in length, in a strong enough gravitational gradient all you would need is 1mm to measure a noticable pull on the spring... As your sensitivity to the measurement increases it becomes smaller and small that is needed.

Correct. The more extreme the curvature (GR equivalent of Newtonian gradient), the smaller the accelerometer must be to measure proper acceleration.

I am neither a scientist or a mathematician so I apologize that my calculus terminology is lacking...

In a free-falling frame my acceleration I never change velocity, since A change of no velocity per proper time equals no acceleration, If I fire a rocket to escape(the not pull of) gravity, and use a frame that accelerates thus, then I still measure no change in velocity per proper time...so what is an example of proper acceleration please?

Now the object that is 50m from the surface does change its velocity(in my free-falling coordinate system 100m from surface) per proper time. So does an object 150m from surface, seems to slowly accelerate away? I am probably misusing proper acceleration here, but can you help clear that up?

Here you are being limited by your math back ground, and I'm not sure I can help. I'll try.

A world line is a path through space and time. It represents, for example, the complete history of you, sitting down, running, flying in a plane or a rocket. Even if use coordinates where your world line is always of the form (t,0,0,0) (that is, you are always at the origin of your coordinate system, the only change in your coordinates being your wristwatch time), this does not make either your 4-velocity or 4-acceleration zero. The reason is that the derivative I referred to is really the covariant derivative (this involves a term called the connection as well as ordinary derivatives). One qualitative statement I can make is that at any point on your world line, there is a unique geodesic tangent to it. This geodesic would be the free fall path of an object you let go of at that point. To the extent that object moves away from you, you have proper acceleration.

 

[Zula110100100]

DH: With respect to my free-falling frame, at no point in this am I saying there is one correct coordinate system.

(that is, you are always at the origin of your coordinate system, the only change in your coordinates being your wristwatch time), this does not make either your 4-velocity or 4-acceleration zero

Right, so my 4-velocity would be a constant (ct, 0, 0, 0)?

 

[PAllen]

DH: With respect to my free-falling frame, at no point in this am I saying there is one correct coordinate system.



Right, so my 4-velocity would be a constant (ct, 0, 0, 0)?

Yes, you could construct such coordinates that your 4-velocity would be always be (1,0,0,0) in units of c=1. However, your proper acceleration vector would not, therefore, be zero unless an object you let go of stayed with you.

 

[Zula110100100]

What are the dr and dt in the swartzchild equation relative to, the center of the body M?