Is acceleration relative in the same way as the speed?

[Luigi Fortunati]

The two spaceships A and B approach each other at constant speed v:
A----------B

There is no way of knowing if B is standing still while A is moving at speed v (to the right), or if A is standing still while B is moving at speed -v (to the left).

This shows that speed is relative.

But if the two spaceships A and B on approaching at increasing speed (that is, if there is an acceleration), they can both legitimately support the other to accelerate, even when only one of them has its own accelerometer measuring a non-zero acceleration?

It seems to me not.

Is this last case sufficient to show that acceleration * is not * relative to the reference?

 

[Dale]

Is this last case sufficient to show that acceleration * is not * relative to the reference?

Yes, proper acceleration (the acceleration measured by an accelerometer) is invariant, for the reason you showed.

 

[Luigi Fortunati]

Yes, proper acceleration (the acceleration measured by an accelerometer) is invariant, for the reason you showed.

Ok.

And beyond this proper invariant acceleration, is there also another (real) acceleration that varies from one reference to another?

 

[Ibix]

There is also coordinate acceleration, which is the second derivative of your coordinate with respect to time. Since you are free to use any coordinate system you like (including ones with non-linear relationships to standard Cartesian coordinates) this can take any value depending on your choice of coordinates.

For example consider polar coordinates centred on A. A and B both have their engines off. B is moving towards A, passes it and then moves away. Somewhere along the way dr/dt switched signs, implying $d^2r/dt^2\neq 0$ even though neither rocket was under power at any time. So coordinate acceleration, although frequently mathematically useful, is not physically meaningful.

The distinction is useful for relativity and the study of rotational motion.

 

[vanhees71]

It's not so clear, which quantity you are discussing. The most simple acceleration is the four-vector
$$a^{\mu}=\frac{\mathrm{d}^2 x^{\mu}}{\mathrm{d} \tau^2},$$
which is a four-vector and transforms as such under Lorentz transformations,
$$\tilde{a}^{\mu} = {\Lambda^{\mu}}_{\nu} a^{\nu}.$$

 

[Luigi Fortunati]

There is also coordinate acceleration, which is the second derivative of your coordinate with respect to time. Since you are free to use any coordinate system you like (including ones with non-linear relationships to standard Cartesian coordinates) this can take any value depending on your choice of coordinates.

For example consider polar coordinates centred on A. A and B both have their engines off. B is moving towards A, passes it and then moves away. Somewhere along the way dr/dt switched signs, implying $d^2r/dt^2\neq 0$ even though neither rocket was under power at any time. So coordinate acceleration, although frequently mathematically useful, is not physically meaningful.

The distinction is useful for relativity and the study of rotational motion.

Is this one *real* acceleration?

 

[Ibix]

Is this one *real* acceleration?

What do you mean by "real"? It's not something you can measure with an accelerometer, if that's what you mean.

 

[Luigi Fortunati]

What do you mean by "real"? It's not something you can measure with an accelerometer, if that's what you mean.

By real I mean something that can be measured on the spot, for example it is real (precisely) the acceleration that is measured with the accelerometer.

It is "imaginary" (apparent) everything that is evaluated from afar, for example it is "apparent" (not "real"), the acceleration of the world that seems to rotate around us when we look at it from a carousel that turns (the world rotates but not around the carousel!).

For me it is "real" the rotation of the carousel, but not that of the people around.

 

[Ibix]

In that case you are interested in the proper acceleration. Coordinate acceleration is what you call apparent.

In general I'd recommend using "proper acceleration", rather than "real", since that's standard terminology.

 

[Dale]

In general I'd recommend using "proper acceleration", rather than "real", since that's standard terminology.

@Luigi Fortunati I would strongly second this recommendation. There are two cocncepts of acceleration: proper acceleration and coordinate acceleration. Labeling either one as “real” just invites a philosophical argument about what “real” means. But according to what you described you think of proper acceleration as “real”.

 

[vanhees71]

In that case you are interested in the proper acceleration. Coordinate acceleration is what you call apparent.

In general I'd recommend using "proper acceleration", rather than "real", since that's standard terminology.

Please, define clearly, what you mean by "proper acceleration". The literature is not unique using this term. I prefer to call manifestly covariant quantities "proper" (as do most modern textbooks), i.e., I'd call (in Sepcial Relativity)
$$a^{\mu} = \frac{\mathrm{d}^2}{\mathrm{d} \tau^2} x^{\mu} = c \frac{\mathrm{d} u^{\mu}}{\mathrm{d} \tau}$$
the "proper acceleration", and that are the components of a four-vector.

 

[Ibix]

Please, define clearly, what you mean by "proper acceleration". The literature is not unique using this term.

I was meaning "the reading on your accelerometer", which I think is the spatial components of the four-acceleration (assuming you actually have three mutually orthogonal accelerometers). Given the constraint that four-acceleration is orthogonal to four-velocity, are the two meanings not equivalent? Or am I being daft?

 

[Dale]

Please, define clearly, what you mean by "proper acceleration". The literature is not unique using this term. I prefer to call manifestly covariant quantities "proper" (as do most modern textbooks), i.e., I'd call (in Sepcial Relativity)
$$a^{\mu} = \frac{\mathrm{d}^2}{\mathrm{d} \tau^2} x^{\mu} = c \frac{\mathrm{d} u^{\mu}}{\mathrm{d} \tau}$$
the "proper acceleration", and that are the components of a four-vector.

Hmm, I wasn’t aware there was another usage. I thought that was what was always meant by proper acceleration. In any case, here we usually default to the usage in most modern textbooks.

 

[vanhees71]

E.g., Wikipedia claims that proper acceleration is
$$\vec{\alpha}=\frac{\mathrm{d \vec{w}}}{\mathrm{d} t},$$
where
$$\vec{w}=c \vec{u}=\frac{\mathrm{d} \vec{x}}{\mathrm{d} \tau}.$$
That quantity, however doesn't make any sense to me, if you ask me.

 

[vanhees71]

I was meaning "the reading on your accelerometer", which I think is the spatial components of the four-acceleration (assuming you actually have three mutually orthogonal accelerometers). Given the constraint that four-acceleration is orthogonal to four-velocity, are the two meanings not equivalent? Or am I being daft?

It's not clear to me, what you mean by "accelerometer". You need to give a concrete device and in which frame it's used.

Of curse the proper acceleration in both my sense and in Wikipedia's sense is Lorentz orthogonal to the four-velocity since
$$w_{\mu} w^{\mu}=c^2 u_{\mu} u^{\mu}=c^2=\text{const},$$
from which both
$$w_{\mu} a^{\mu} =\frac{1}{2} \frac{\mathrm{d} (w^{\mu} w_{\mu}) }{\mathrm{d} \tau}=0$$
and
$$w_{\mu} \alpha^{\mu} = \frac{1}{2} \frac{\mathrm{d} (w^{\mu} w_{\mu}) }{\mathrm{d} t}=0.$$

 

[Dale]

That quantity, however doesn't make any sense to me, if you ask me.

Hmm, Same here. I have not seen that before, but I don’t see how that quantity would be useful. It is not covariant, it is not equal to the reading on the accelerometer, and it is also not the acceleration in the reference frame.

Seems like a wiki error

 

[vanhees71]

Hmm, Same here. I have not seen that before, but I don’t see how that quantity would be useful. It is not covariant, it is not equal to the reading on the accelerometer, and it is also not the acceleration in the reference frame.

Seems like a wiki error

Yes, me too. Also they are schizophrenic, because within GR they use the proper definition of proper acceleration:

https://en.wikipedia.org/wiki/Proper_acceleration#In_curved_spacetime

 

[PeterDonis]

Wikipedia claims

This seems like a good time to reiterate that Wikipedia is not an acceptable source. That's not to say we never reference it here, just that when we do reference it, it's because we've already determined by other means that it happens to be reasonably accurate in that particular case. In cases where it isn't, we don't reference it; and we don't expect it to be accurate all the time.

The literature is not unique using this term.

I'm not aware of any modern literature that uses "proper acceleration" in any other way than the way we've been describing here. It's possible that there are older sources that do.

 

[Mark44]

E.g., Wikipedia claims that proper acceleration is
$$\vec{\alpha}=\mathrm{d \vec{w}}{\mathrm{d} t},$$

Shouldn't this be $\vec{\alpha}=\frac{\mathrm{d \vec{w}}}{{\mathrm{d} t}}$?

 

[vanhees71]

Indeed, I've corrected it in the original posting. It's of course not very clever to use this definition, because it's not covariant and leads to misunderstandings easily. In modern textbooks and in the scientific literature one uses covariant quantities, and proper acceleration is defined by (in my notation)
$$a^{\mu}=c \frac{\mathrm{d} u^{\mu}}{\mathrm{d} \tau},$$
where $\tau$ is the proper time of the particle, and
$$u^{\mu}=\frac{\mathrm{d} x^{\mu}}{\mathrm{d} s}=\frac{1}{c} \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}.$$
As in Newtonion physics, which is momentarily valid in the restframe of the particle at the instant of time in question, thus the modern definition of (proper) acceleration is
$$a^{\mu}=\frac{\mathrm{d}^2 x^{\mu}}{\mathrm{d} \tau^2}.$$
Of course, there might be still textbooks around, which treat the issue otherwise, because they haven't read the literature for 110 years!