Inertial and non inertial frames

[Mentz114]

There is a theorem that restricting a field (covariant) gradient to a path, versus the absolute derivative along a path, will always produce the same result. This was discussed in a recent thread here. I don't have a chance now to find it. Conceptually, if I am talking about particle, I think of absolute derivative on a path. If thinking of a congruence, then a gradient makes sense.

Thanks, for the info. I didn't know about the theorem. So there's no risk in using the sometimes much easier gradient method.

 

[A.T.]

That is not the same as saying that the proper time interval is frame invariant which is of course right

And the same is true for proper acceleration.

Nonody here has claimed that "everything is relative".

But you reject "absolute acceleration" just because "time and space are relative". This sounds like trying to make everything relative, just because something is.

 

[Dale]

So explain what you mean then, if you claim that relativity doesn't handle arbitrary motion, I take it you mean it is specific to inertial frames, but at the same type you claim you didn't mean by that that it can't deal with absolute acceleration, IOW arbitrary motion. Please clarify.

I never claimed that relativity doesn't handle arbitrary motion either.

I only made the obvious and uncontroversial statement that the principle of relativity is specific to inertial motion. I.e. it asserts the physical equivalence of different states of inertial motion. The principle of relativity's equivalence of inertial frames neither implies the equivalence of accelerated motion nor does it imply that accelerated motion cannot be treated or handled.

 

[FalseVaccum89]

A frame, on the other hand, is a local system of measuring rods and clocks. There is no apriori reason this should have anything to do with the underlying coordinate system. A frame is just a collection of vectors at a point, of known lengths and angles, against which you can make standard measurements of other vectors at that point.

This makes me wonder if the Einstein tensor in GR is a frame in and of itself. I know space and time aren't absolute, but what about their 4D manifold form, spacetime?

 

[TrickyDicky]

But you reject "absolute acceleration" just because "time and space are relative". This sounds like trying to make everything relative, just because something is.

I didn't reject anything, and certainly not "just because" of that, I'll refer you to PAllen posts in this thread, hopefully you'll understand them better than mine.

 

[TrickyDicky]

I always think of proper acceleration as the absolute derivative by proper time of the 4-velocity. This is clearly a 4-vector defined on a path.

Just to be precise, that would the 4-acceleration, proper acceleration is the 3-vector(according to wikipedia at least).

 

[Fredrik]

This makes me wonder if the Einstein tensor in GR is a frame in and of itself.

A tensor equal to a frame? That's like having a rational number equal to an irrational number.

I know space and time aren't absolute, but what about their 4D manifold form, spacetime?

The question doesn't make enough sense to have a yes or no answer. A coordinate system is a function that maps a subset of spacetime into $\mathbb R^4$. The set of all coordinate systems is part of the specification of a manifold, but no one coordinate system is in any way preferred over the others. So I guess you could choose to describe this situation as "spacetime is absolute", but it would be a different kind of "absoluteness" than when we're talking about space and time.

 

[harrylin]

[..] I know space and time aren't absolute, but what about their 4D manifold form, spacetime?

Basically yes: http://www.einstein-online.info/elementary/specialRT/spacetime

 

[TrickyDicky]

It might alleviate some of the confusion of the OP to remind that it is always a good thing not to mix concepts and objects "pre-SR" with SR ones, most people here are automatically thinking in SR terms, so even when talking about pre-SR concepts like classical acceleration or velocity which are 3-vectors they translate it to the spatial components of 4-vectors of SR, and mixing classical 3-vectors with SR 4-vectors might result in misunderstandings (not only for the OP).
Now proper acceleration is a 3-vector, but in SR terms is just the 4-acceleration omitting the null time component, and that is what most people here understands.
The problem comes in mixing the pre-SR concept of acceleration which didn't distinguish coordinate time from proper time since there was no time coordinate, and time was just a parameter:time and space were considered absolute in Newtonian physics-this created a big dispute with Leibniz that wanted to keep the strict Galilean relativity in which only time was absolute but I digress-, and thus acceleration was absolute, but then again at the time also velocity was absolute(not so in strict galilean relativity).
So to avoid useles discussion one should keep in mind that proper acceleration in SR is not the pre-SR absolute acceleration. However it is frame-invariant, as a local observable, locally measurable with an accelerometer, unlike proper velocity which relies on the calculation of the distant traveller proper time. The ultimate reason for this has no real answer as stated in #44.

 

[DrGreg]

I prefer to think of "proper acceleration" as a scalar rather than a vector and this removes some of the confusion.

Proper acceleration can then be defined either as the magnitude of the 4-acceleration, which makes it clear that it's a scalar invariant, or as the magnitude of the 3-acceleration as measured by a comoving locally-inertial frame.

It doesn't make sense to talk about an "invariant vector".

 

[haael]

4-speed and 4-acceleration are not covariant tensors!

4-speed is a formal vector obtained by dividing 4-momentum by the relativistic mass. The relativistic mass is a formal scalar, but it is not an invariant scalar, so the resulting object is not a tensor. It is mainly used in a proof that relativity reduces to Newtonian dynamics in a low speed limit. Its use in normal considerations is rare.

4-acceleration is a formal 4-vector correction one must add to transform one 4-speed into another. It is much less a tensor.

Both 4-speed and 4-acceleration have only 3 independent parameters. For 4-speed it can be seen from the definition and for 4-acceleration from the fact that there are 3 independent boost generators.

None of the lengths of 4-speed and 4-acceleration are invariant scalars.

The most useful thing one can do to describe acceleration in SR is to group 3 boost generators and 3 rotation generators into an antisymmetric 2-rank tensor. From this picture one gets clear impression that boosts are in fact rotations over the axis of the time dimension.

As for acceleration as an observable. 3 generators of boosts do not commute with themselves, nor with the rotation (spin) generators, nor with the momentum, nor with the Hamiltonian. The fact that they do not commute with Hamiltonian is obvious - acceleration changes the particle energy. So, you can use them as an alternative state parametrization, instead of Hamiltonian, momentum and spin, but you don't get any new information. However, this still proves that the acceleration is an observable and not a frame-dependent parameter like speed. It is just redundant, since energy, momentum and spin provide complete particle description.

All this can be read in the fist two chapters of Weinberg.

 

[Mentz114]

@haael, thanks for that. I admit to being confused about the 4-acceleration.

 

[Dale]

4-speed is a formal vector obtained by dividing 4-momentum by the relativistic mass. The relativistic mass is a formal scalar, but it is not an invariant scalar, so the resulting object is not a tensor. It is mainly used in a proof that relativity reduces to Newtonian dynamics in a low speed limit. Its use in normal considerations is rare.

I have never seen this definition before, but nobody here was talking about the four-speed, so the fact that it is not a tensor is not relevant to the conversation.

4-acceleration is a formal 4-vector correction one must add to transform one 4-speed into another. It is much less a tensor.
...
None of the lengths of 4-speed and 4-acceleration are invariant scalars.

I would like to see the definition that leads to this conclusion. Under the usual definition of four-acceleration it most definitely is a tensor and its norm is in fact an invariant scalar which is equal to the magnitude of the proper acceleration.

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

All this can be read in the fist two chapters of Weinberg.

Can you post or link Weinberg's definition?

 

[TrickyDicky]

I prefer to think of "proper acceleration" as a scalar rather than a vector and this removes some of the confusion.

Proper acceleration can then be defined either as the magnitude of the 4-acceleration, which makes it clear that it's a scalar invariant, or as the magnitude of the 3-acceleration as measured by a comoving locally-inertial frame.

It doesn't make sense to talk about an "invariant vector".

But the fact is that it isn't a scalar, it's a vector, and yes the component representation of vectors is frame-dependent so I assume we were all referring to its scalar magnitude when saying it is frame-invariant.

 

[TrickyDicky]

Under the usual definition of four-acceleration it most definitely is a tensor

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. In GR it is indeed the covariant derivative of proper velocity and equating it to 0 leads us to the geodesic equation.

 

[Nugatory]

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. In GR it is indeed the covariant derivative of proper velocity and equating it to 0 leads us to the geodesic equation.

And how is the SR acceleration 4-vector not a rank-1 tensor?

 

[Dale]

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. In GR it is indeed the covariant derivative of proper velocity and equating it to 0 leads us to the geodesic equation.

Four vectors are tensors. As Nugatory mentioned, they are specifically rank-1 tensors. The only confusion between GR and SR is whether they have upper or lower indices since there is no distinction using the Minkowski metric of SR.

 

[TrickyDicky]

And how is the SR acceleration 4-vector not a rank-1 tensor?

Who says it is not? And scalars are rank-0 tensors, nice eh? But people don't usually refer to scalars and vectors as tensors to avoid confusions.
Was that a rhetorical question, or did you want to make a serious point? ;)

 

[TrickyDicky]

Four vectors are tensors. As Nugatory mentioned, they are specifically rank-1 tensors. The only confusion between GR and SR is whether they have upper or lower indices since there is no distinction using the Minkowski metric of SR.

You were talking about the definition of tensor so don't pretend you were referring to vectors as 1-tensors, you specifically linked to the wikipage where it says that in GR 4-acceleration is the covariant derivative of 4-velocity wrt proper time, that is a tensor.

 

[Dale]

You were talking about the definition of tensor so don't pretend you were referring to vectors as 1-tensors, you specifically linked to the wikipage where it says that in GR 4-acceleration is the covariant derivative of 4-velocity wrt proper time, that is a rank 2- tensor.

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

I really don't know what you mean by "pretend". Four-vectors are rank 1 tensors, they are the same thing. It doesn't matter in the least if I referred to them as a four-vector in one place and a tensor in another. It is a little like complaining that I use the word "couch" in one place and "sofa" in another by saying that I was pretending to refer to a sofa as a couch.

 

[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
$$\begin{align} U^\alpha &= \frac{dx^\alpha}{d\tau} \\ A^\alpha &= \frac{dU^\alpha}{d\tau} + \Gamma^\alpha_{\beta \gamma} U^\beta U^\gamma \end{align}$$
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.