Inertial and non inertial frames

[TrickyDicky]

No, look more carefully. It is given as a rank-1 tensor, A^\lambda.

It is obvious to me you don't distinguish the difference anyway. Maybe someone can help you with that, I don't have time right now.

 

[Dale]

It is obvious to me you don't distinguish the difference anyway. Maybe someone can help you with that, I don't have time right now.

That is funny, your claiming that I don't distinguish the difference between a rank-1 and a rank-2 tensor. Luckily, I do have the time to help you with that.

A rank-1 tensor has 1 index. It can either be up (A^{\mu}) or down (A_{\mu}).

A rank-2 tensor has 2 indices. They can either be up (A^{\mu\nu}) or down (A_{\mu\nu}) or mixed ({A^{\mu}}_{\nu} or {A_{\mu}}^{\nu})

 

[Mentz114]

I have to add to the confusion by citing the case of the hovering static observer in the Schwarzschild vacuum. The acceleration 4-thing has one non-zero component in the r-direction as one would expect. But by boosting the frame field by just the right amount β(r) it's possible to make all four components vanish ( I can show the calculation). I have been under the impression that all tensorial quantites transform properly under this kind of transformation. From this it looks like the acceleration 4-object does not transform like a tensor. Am I making some unwarranted assumption ?

It is confusing and I'd like to get it untangled.

 

[PAllen]

I have to add to the confusion by citing the case of the hovering static observer in the Schwarzschild vacuum. The acceleration 4-thing has one non-zero component in the r-direction as one would expect. But by boosting the frame field by just the right amount β(r) it's possible to make all four components vanish ( I can show the calculation). I have been under the impression that all tensorial quantites transform properly under this kind of transformation. From this it looks like the acceleration 4-object does not transform like a tensor. Am I making some unwarrented assumption ?

It is confusing and I'd like to get it untangled.

It would help to show the calculation. That would be a complete surprise to me. Limit (tangent vector - tangent vector [ || transported] )/<scalar invariant> seems like it must define a true vector to me.

[And to me, vector = rank 1 tensor; though I would usually call it a vector].

 

[Nugatory]

Who says it is not?

You, I thought...
At least that's how I read "You guys are mixing the 4-acceleration in SR with its definition in GR, here we were AFAIK discussing it within SR, where its definition is not a tensor, but a 4-vector".

 

[Nugatory]

I have to add to the confusion by citing the case of the hovering static observer in the Schwarzschild vacuum. The acceleration 4-thing has one non-zero component in the r-direction as one would expect. But by boosting the frame field by just the right amount β(r) it's possible to make all four components vanish ( I can show the calculation). I have been under the impression that all tensorial quantites transform properly under this kind of transformation. From this it looks like the acceleration 4-object does not transform like a tensor. Am I making some unwarrented assumption ?

It is confusing and I'd like to get it untangled.

You might want to start another thread for this question... This one is getting a bit polluted... But yes, I would be very interested in seeing that calculation.

 

[Mentz114]

You might want to start another thread for this question... This one is getting a bit polluted... But yes, I would be very interested in seeing that calculation.

OK, I'll start another thread. I'm tex'ing it up.

(I corrected the spelling of 'unwarranted' in my previous reply.)

 

[TrickyDicky]

You, I thought...
At least that's how I read "You guys are mixing the 4-acceleration in SR with its definition in GR, here we were AFAIK discussing it within SR, where its definition is not a tensor, but a 4-vector".

When the rank is not specified it is universally understood that the word tensor refers to rank >1 tensors.
Did you really not know this?

 

[DrGreg]

For the record, if the worldline of a particle is parametrised by proper time as x^\alpha(\tau), in any coordinate system in GR or SR, inertial or non-inertial, then the 4-velocity U and the 4-acceleration A are defined by
\begin{align}<br /> U^\alpha &amp;= \frac{dx^\alpha}{d\tau} \\<br /> A^\alpha &amp;= \frac{dU^\alpha}{d\tau} + \Gamma^\alpha_{\beta \gamma} U^\beta U^\gamma<br /> \end{align}
Both U and A are 4-vectors, i.e. rank-1 tensors.

All of the above is true in both SR and GR, in inertial or non-inertial coordinates. Of course in Minkowski coordinates in SR, all the \Gamma^\alpha_{\beta \gamma}s are zero so the expression for 4-acceleration can be simplified.

I thought this was well-known stuff covered in every GR textbook but apparently not from the evidence of this thread.

 

[Dale]

When the rank is not specified it is universally understood that the word tensor refers to rank >1 tensors.
Did you really not know this?

When the rank isn't specified then it should be understood that the word "tensor" refers to a tensor of an arbitrary unspecified rank, including rank 0 and rank 1.

 

[Nugatory]

When the rank is not specified it is universally understood that the word tensor refers to rank >1 tensors.

I'm sorry, but I that answer still isn't helping me make any sense of your post #85. In what way is the GR definition of proper acceleration different than the SR definition, so that "mixing the 4-acceleration in SR with its definition in GR" will lead to problems/errors?

 

[Jeronimus]

Accelerometers do not measure acceleration. They only measure different parts of the accelerometer being accelerated differently. Or otherwise said, different parts of the accelerometer being in different inertial frames of references. They measure the resulting space shift of parts of the accelerometer.

But what about a point on a disc rotating? The point(object) on the disc keeps switching the inertial reference frame it is at rest in constantly.
edit: An infinitely small accelerometer would measure nothing even on a switch of the inertial frame of reference, but such accelerometers unfortunately do not exist. The accelerometer, even the smallest, could never have all it's parts within the same inertial reference frame at all times when on a rotating disc.

An accelerometer in free fall in the right force field measures nothing, even though it is accelerating. This is because all of it's parts are in the same inertial frame at all times(ideally)

Acceleration is relative.

 

[Mentz114]

Accelerometers do not measure acceleration. They only measure different parts of the accelerometer being accelerated differently. Or otherwise said, different parts of the accelerometer being in different inertial frames of references. They measure the resulting space shift of parts of the accelerometer.

But what about a point on a disc rotating? The point(object) on the disc keeps switching the inertial reference frame it is at rest in constantly.

An accelerometer in free fall in the right force field measures nothing, even though it is accelerating. This is because all of it's parts are in the same inertial frame at all times(ideally)

Acceleration is relative.

Relative to what ? Have you read the other posts in this thread ?

 

[Dale]

Accelerometers do not measure acceleration.

The unqualified word "acceleration" is ambiguous in this context. Accelerometers do measure proper acceleration, they do not measure coordinate acceleration.

They only measure different parts of the accelerometer being accelerated differently.

This is simply false.

But what about a point on a disc rotating? The point(object) on the disc keeps switching the inertial reference frame it is at rest in constantly.
edit: An infinitely small accelerometer would measure nothing even on a switch of the inertial frame of reference, but such accelerometers unfortunately do not exist. The accelerometer, even the smallest, could never have all it's parts within the same inertial reference frame at all times when on a rotating disc.

Usually when we refer to accelerometers in theoretical discussions or thought experiments we are referring to the 6 degree of freedom kind that measures linear accelerations on three axes and rotations on three axes. Such an accelerometer would measure both the rotation and the centripetal acceleration on a rotating disk.

 

[PAllen]

Accelerometers do not measure acceleration. They only measure different parts of the accelerometer being accelerated differently. Or otherwise said, different parts of the accelerometer being in different inertial frames of references. They measure the resulting space shift of parts of the accelerometer.

It is not helpful to define your own physics. We define what an accelerometer measures as acceleration of the device as whole, given that the acceleration is sufficiently uniform (over space) and slowly changing over time.

But what about a point on a disc rotating? The point(object) on the disc keeps switching the inertial reference frame it is at rest in constantly.
edit: An infinitely small accelerometer would measure nothing even on a switch of the inertial frame of reference, but such accelerometers unfortunately do not exist. The accelerometer, even the smallest, could never have all it's parts within the same inertial reference frame at all times when on a rotating disc.

An accelerometer on the edge of rotating disk would measure that the acceleration is radial and constant.

An accelerometer in free fall in the right force field measures nothing, even though it is accelerating. This is because all of it's parts are in the same inertial frame at all times(ideally)

Free fall is defined as inertial motion in (general) relativity. Doing anything with gravity in SR is an abuse of SR. If you mean some other force, an accelerometer must be built out of matter that is non-responsive all known forces (except gravity, if you consider that a force - because nothing be neutral to gravity); else you can't call it an accelerometer.

Acceleration is relative.

Acceleration as measured by an accelerometer is absolute. All inertial frames will compute the response of a given accelerometer (in some state of motion) to be the same - even though the coordinate description of the motion will be different for each frame. If you are willing to use laws expressed in terms of tensors, with connection coefficients, then an accelerating observer's coordinate calculations will also yield the same result.

I must warn that this is not the forum to propose your own version of physics. You are welcome to pursue such an endeavor - but not on physicsforums.

 

[Jeronimus]

Maybe someone could define acceleration. To me, acceleration is the context of SR, is that which is responsible for changing which inertial frame of reference an object becomes at rest in.

I could not possibly define it to be the change in velocity over time (seen from the observer's point of view who is supposed to be accelerating), since in SR an observer considers himself at rest.

So here i am, waiting for how you define acceleration in the context of SR seen from an accelerating observer's point of view.

 

[Mentz114]

Maybe someone could define acceleration. To me, acceleration is the context of SR, is that which is responsible for changing which inertial frame of reference an object becomes at rest in.

I could not possibly define it to be the change in velocity over time (seen from the observer's point of view who is supposed to be accelerating), since in SR an observer considers himself at rest. So here i am, waiting for how you define acceleration in the context of SR seen from an accelerating observer's point of view.

Try the Wiki article. There's a definition there.

http://en.wikipedia.org/wiki/Four-vector

Imagine you're sitting in a small boat on a calm lake so the boat is at rest wrt the water. If you then throw a brick towards aft so it leaves the boat, what happens ? The brick was accelerated by your arm and the reaction ( Newton's third) pushes the boat and you. You are still at rest in the boat even if it accelerates, as long as you go with it.

 

[Jeronimus]

Imagine you're sitting in a small boat on a calm lake so the boat is at rest wrt the water. If you then throw a brick towards aft so it leaves the boat, what happens ? The brick was accelerated by your arm and the reaction ( Newton's third) pushes the boat and you. You are still at rest in the boat even if it accelerates, as long as you go with it.

Nice example, except that you do not feel acceleration when you push the brick away of you. What you feel/sense is different parts of your body being accelerated differently, and the resulting change of your body structure. This is what you sense, and not acceleration itself.
At the point your fingers touch the brick, your fingers get compressed because the brick resists being accelerated (why it resists and why inertial mass is connected to gravitational mass i do not know yet exactly). This compression moves through your body, and changes the structure which then can be sensed by you. But that is not acceleration you sense.

 

[Mentz114]

Nice example, except that you do not feel acceleration when you push the brick away of you. What you feel/sense is different parts of your body being accelerated differently, and the resulting change of your body structure. This is what you sense, and not acceleration itself.
At the point your fingers touch the brick, your fingers get compressed because the brick resists being accelerated (why it resists and why inertial mass is connected to gravitational mass i do not know yet exactly). This compression moves through your body, and changes the structure which then can be sensed by you. But that is not acceleration you sense.

Suppose you had a friend on the boat and she threw the brick ? Would you feel anything ?
Whether you did or not, an accelerometer would register the push on the boat.

You can make an accelerometer by standing a brick an its small face on a skateboard. Give the skateboard a good push and the brick falls over !

 

[Jeronimus]

Suppose you had a friend on the boat and she threw the brick ? Would you feel anything ?
Whether you did or not, an accelerometer would register the push on the boat.

You can make an accelerometer by standing a brick an its small face on a skateboard. Give the skateboard a good push and the brick falls over !

Of course it falls over, since you accelerated the bottom part of the brick touching the skateboard first. You accelerated different parts of the brick differently with the resulting change in position you measured. But you did not measure acceleration itself.

 

[PAllen]

Maybe someone could define acceleration. To me, acceleration is the context of SR, is that which is responsible for changing which inertial frame of reference an object becomes at rest in.

Where did you get this idea from? Acceleration of a body in an inertial frame in SR is taken as derivative of 4-velocity by proper time.

I could not possibly define it to be the change in velocity over time (seen from the observer's point of view who is supposed to be accelerating), since in SR an observer considers himself at rest.

Again, where do you get this from. SR says only inertial observers consider themselves at rest. The standard SR equations only apply to inertial frames; observers at rest in these frames detect no acceleration (e.g. using an accelerometer).

So here i am, waiting for how you define acceleration in the context of SR seen from an accelerating observer's point of view.

SR typically does not deal with accelerated observers; certainly Einstein never defined accelerated observers for SR. It readily deals with accelerated motion in inertial frames.

It is possible to define coordinates representing an accelerated observer, in SR. If you do, acceleration is now absolute derivative of 4-velocity by proper time. This type of derivative involves what are called connection coefficients (which are zero in an inertial frame in SR). In these coordinates, the coordinate 3-velocity of the observer will be zero (its 4-velocity will not be zero - it will have a time component). However, because of the nature of the absolute derivative, the 4-acceleration will be nonzero, and the proper acceleration (given by the norm of the 4-acceleration) will be what is measured by an accelerometer.

 

[Jeronimus]

Where did you get this idea from?

It's a pretty simple idea. You accelerate(instantaneous to keep it simple) and as a result, someone who observes you accelerating will conclude that you are no longer at rest in your former inertial frame of reference, but you are at rest in another frame.

Switching the frame an object is at rest in, is a direct result of acceleration. (seen by an observer in another inertial frame of reference)

I could continue with what acceleration is a result of, but i'd rather not.

Again, where do you get this from. SR says only inertial observers consider themselves at rest. The standard SR equations only apply to inertial frames; observers at rest in these frames detect no acceleration (e.g. using an accelerometer).

But isn't accelerating an object, seen from an observer in another IFR at rest, a constant switch of IFRs the object is at rest in? I think it is.

 

[TrickyDicky]

For the record, if the worldline of a particle is parametrised by proper time as x^\alpha(\tau), in any coordinate system in GR or SR, inertial or non-inertial, then the 4-velocity U and the 4-acceleration A are defined by
\begin{align}<br /> U^\alpha &amp;= \frac{dx^\alpha}{d\tau} \\<br /> A^\alpha &amp;= \frac{dU^\alpha}{d\tau} + \Gamma^\alpha_{\beta \gamma} U^\beta U^\gamma<br /> \end{align}
Both U and A are 4-vectors, i.e. rank-1 tensors.

All of the above is true in both SR and GR, in inertial or non-inertial coordinates. Of course in Minkowski coordinates in SR, all the \Gamma^\alpha_{\beta \gamma}s are zero so the expression for 4-acceleration can be simplified.

I thought this was well-known stuff covered in every GR textbook but apparently not from the evidence of this thread.

Nope, you are wrong, I hoped someone would come up to correct it but apparently, and incredibly, no one did.
Have you tried reading Carroll notes for instance? Then you should realize that in GR the expression with christoffel corrections transforms as a tensor, not as a contravariant vector.

 

[Dale]

Maybe someone could define acceleration.

Again, your question is ambiguous in this context. You need to specify if you want a definition for coordinate acceleration or proper acceleration. Both are easy to define. Coordinate acceleration is the second time derivative of an object's position. Proper acceleration is the first covariant derivative of the object's tangent vector.

So here i am, waiting for how you define acceleration in the context of SR seen from an accelerating observer's point of view.

The above definitions hold for any point of view, inertial or not.

 

[TrickyDicky]

I'm sorry, but I that answer still isn't helping me make any sense of your post #85. In what way is the GR definition of proper acceleration different than the SR definition, so that "mixing the 4-acceleration in SR with its definition in GR" will lead to problems/errors?

See my answer to DrGreg.

 

[TrickyDicky]

When the rank isn't specified then it should be understood that the word "tensor" refers to a tensor of an arbitrary unspecified rank, including rank 0 and rank 1.

Then I guess for you a book entitled "Tensor calculus" should deal with real functions differential calculus and with vector calculus, right?

 

[stevendaryl]

Nope, you are wrong, I hoped someone would come up to correct it but apparently, and incredibly, no one did.
Have you tried reading Carroll notes for instance? Then you should realize that in GR the expression with christoffel corrections transforms as a tensor, not as a contravariant vector.

What are you talking about? Are you saying that Dr. Greg wrote down the wrong expressions for 4-velocity and 4-acceleration? Are you saying that he's wrong to call them 4-vectors? Are you saying that he's wrong to say that a 4-vector is a rank-1 tensor? Are you saying that he's wrong that this is "well-known stuff"? What do you mean wrong?

 

[PAllen]

It's a pretty simple idea. You accelerate(instantaneous to keep it simple) and as a result, someone who observes you accelerating will conclude that you are no longer at rest in your former inertial frame of reference, but you are at rest in another frame.

Switching the frame an object is at rest in, is a direct result of acceleration. (seen by an observer in another inertial frame of reference)

I could continue with what acceleration is a result of, but i'd rather not.

You made a statement about SR. As has been clarified multiple times, SR distinguishes inertial and non-inertial frames, and, as formulated by Einstein, dealt only with inertial frames. Also, FYI, pre-relativistic physics made the same distinction. Newton in Principia had extensive discussions of the absolute character of rotation and acceleration. SR did not change this (and GR, not really either).

But isn't accelerating an object, seen from an observer in another IFR at rest, a constant switch of IFRs the object is at rest in? I think it is.

There is a limited sense in which you can treat it this way. However, a derivative is has a value at a single point. If the comoving inertial frame changes, there is accelerated motion in each and every co-moving inertial frame involved.

 

[TrickyDicky]

What are you talking about? Are you saying that Dr. Greg wrote down the wrong expressions for 4-velocity and 4-acceleration? Are you saying that he's wrong to call them 4-vectors? Are you saying that he's wrong to say that a 4-vector is a rank-1 tensor? Are you saying that he's wrong that this is "well-known stuff"? What do you mean wrong?

Try reading what you quote.

 

[stevendaryl]

Try reading what you quote.

What you said made no sense, which is why I asked for clarification. Everything that Dr. Greg wrote was correct, so I have no idea what you are talking about when you call it wrong.