Inertial and non inertial frames

[Dale]

Anyway, are you saying that since SR postulates don't include arbitrary motion, proper acceleration should not be included in SR?

No, I didn't say that. I don't think that I said anything remotely close to that.

 

[haael]

My three euro-cents to this topic.

Inertial and non-inertial coordinate frames in SR are different in the thing that the latter are curvilinear! That's all.

Transformation between two inertial frames in SR always maps straight lines into straight lines. Transformation between inertial and non-inertial frame always maps straight time-like line into a curved one.

Acceleration of a frame is its internal property and one does not need any other frame to check it. Acceleration is a change of speed with time. In SR time is one of the dimensions. "Speed" of a frame is an angle between time and space base vectors of the frame. Acceleration is change of that angle with time. That means, the angle between the base vectors is not constant. Back in the terms of speed, a frame is accelerating if it has different speed at some moment compared to its own speed a moment before. The speed of a frame is compared to the speed of the very same frame at a different moment. We don't need any other frame to check it.

Inertial frames of reference are the frames that use straight lines as the time axis. The transformations of frames with different speed (boosts) change the straight lines into other straight lines of different direction. There is no preferred direction in SR, so this is a symmetry. On the other hand, transformations with frames of a different acceleration map straight time-like lines into curved ones. There is no symmetry in SR that requires straight and curved lines to be equivalent, so there is an absolute acceleration.

In GR, the concept of "straight" is replaced with a concept of "geodesic". A geodesic may be straight or may be not and that is an objective fact (one of the axioms in fact). If a particle is moving along a geodesic, it feels no acceleration. If a geodesic happens to be a straight line, a particle is not affected by any force at all. If a geodesic is not a straight line and a particle is moving along it, then it is free-falling due to gravitational force, but can not feel it. If a particle is moving along a non-straight non-geodesic line, then it feels non-gravitational acceleration. Finally, if a particle is moving along a straight but non-geodesic line, then it means that it is opposing gravity. This is often called hovering, but it need not be - for example humans standing on Earth surface are a case of that situation. We are not free-falling, but instead we are a subject to a force that exactly balances gravity.

So, in GR:
1. straight geodesic lines - no acceleration at all
2. non-straight geodesic lines - gravitational acceleration (free fall)
3. straight non-geodesic lines - force cancelling gravity
4. non-straight non-geodesic lines - unbalanced non-gravitational acceleration

In SR every geodesic is straight, so we have to consider only cases 1 and 4.

 

[TrickyDicky]

No, I didn't say that. I don't think that I said anything remotely close to that.

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.

 

[TrickyDicky]

Inertial and non-inertial coordinate frames in SR are different in the thing that the latter are curvilinear! That's all.

That is coordinate acceleration, but here it is proper acceleration that is the issue.

 

[haael]

That is coordinate acceleration, but here it is proper acceleration that is the issue.

A particle is properly accelerating if it is moving along a curved path. The curvature of a line is an objective fact, so there is an absolute acceleration.

 

[Mentz114]

A particle is properly accelerating if it is moving along a curved path. The curvature of a line is an objective fact, so there is an absolute acceleration.

Good point. You don't need to compare a curve to a straight line to tell it is curved.

 

[TrickyDicky]

A particle is properly accelerating if it is moving along a curved path. The curvature of a line is an objective fact, so there is an absolute acceleration.

Sure, but you were talking about SR and curvilinear coordinates, and that is what I responded to. Don't confuse coordinates with physical observables, that is a very common thing around here.

 

[Mentz114]

Sure, but you were talking about SR and curvilinear coordinates, and that is what I responded to. Don't confuse coordinates with physical observables, that is a very common thing around here.

Proper acceleration is a tensor and cannot be transformed away.

 

[A.T.]

AFAIK, we were discussing "absolute" acceleration, which people here has identified with proper acceleration explaining it away as a "local observable". But that suggests we should call local observables like proper time, "absolute time" which would be both wrong and confusing in a theory that bans absolute time and space.

It is neither wrong nor confusing. In this context "absolute vs. relative" just means "frame invariant vs. frame dependent". Proper time intervals are "frame invariant" or "absolute". The notion that "everything is relative in Relativity" is a very common misconception. In fact Einstein originally wanted to call it "theory of invariants", to put the emphasis on the "absolute" quantities.

 

[bgq]

Thank you all for your replies, they are very helpful.

 

[TrickyDicky]

It is neither wrong nor confusing. In this context "absolute vs. relative" just means "frame invariant vs. frame dependent". Proper time intervals are "frame invariant" or "absolute".

It might have been not the best example to make my point, that's true, but what I considered wrong was saying time is absolute in SR and I maintain it. That is not the same as saying that the proper time interval is frame invariant which is of course right.

The notion that "everything is relative in Relativity" is a very common misconception. In fact Einstein originally wanted to call it "theory of invariants", to put the emphasis on the "absolute" quantities.

Nonody here has claimed that "everything is relative".

 

[harrylin]

[..] when we say accelerating (and so non inertial), we mean accelerating with respect to a certain frame (According to both classical and special relativity). What is this frame?
According to the classical theory, the answer is very clear: It is the absolute rest frame, but in SR it seems (to me) that there is something missing. It is not the issue how to test whether a frame is inertial or not, I know the whole story of this, but the issue is what initially makes some frames inertial and the others not? If we say that non inertial frames are those that are accelerating, this has no meaning unless we specify with respect to what frame.

I think that you are using the argument of Newton ("rotating bucket"), enhanced by arguments of Langevin (radiation, "twin paradox")¹. Indeed, quite some early relativists took the stationary ether model for granted despite the fact that it plays no direct role in the equations - the issue being that it can interpreted as playing an indirect role. Later many relativists replaced this by the block universe interpretation². Welcome to the "family of the knowing".

1. https://www.physicsforums.com/showthread.php?p=4111027
2. https://www.physicsforums.com/showthread.php?t=583606

 

[TrickyDicky]

Proper acceleration is a tensor and cannot be transformed away.

It is actually a vector but I don't know what you mean.

 

[Mentz114]

It is actually a vector but I don't know what you mean.

For example, we can find the acceleration of a comoving frame field by calculating U^β∇_αU_β. This is manifestly a tensor, so its components can't all be reduced to zero by a coordinate transformation.

But you know this so why the question ?

 

[bgq]

Thank you very much Harrylin, the links are interesting.

 

[TrickyDicky]

For example, we can find the acceleration of a comoving frame field by calculating U^β∇_αU_β. This is manifestly a tensor, so its components can't all be reduced to zero by a coordinate transformation.

But you know this so why the question ?

I can't see the relation of what you are posting with what is being discussed in the thread.

 

[PAllen]

For example, we can find the acceleration of a comoving frame field by calculating U^β∇_αU_β. This is manifestly a tensor, so its components can't all be reduced to zero by a coordinate transformation.

But you know this so why the question ?

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.

 

[Mentz114]

I can't see the relation of what you are posting with what is being discussed in the thread.

You started it. Your interventions are generally cryptic and unintelligible to me so I'll leave you to it.

 

[Mentz114]

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.

Yes this is so. But I did state clearly I was talking about a vector field, a congruence. I have never found a case where one is satisfied and the other not. I'm not smart enough to work out when it will happen.

 

[PAllen]

Yes this is so. But I did state clearly I was talking about a vector field, a congruence. I have never found a case where one is satisfied and the other not. I'm not smart enough to work out when it will happen.

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.

 

[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}<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.