5
$\begingroup$

I know this question has been asked in other forms, generally regarding the balance of forces. This time I want to focus on motion. I've got a laser accelerometer on my desk. It tells me that I'm accelerating at $9.8~\rm m/s^2$. For the first experiment I'm travelling in space. I pick a nearby star and discover that I move about $490$ meters in $10$ seconds from that star. For the next experiment I'm on the surface of Earth. I measure the same acceleration with my laser accelerometer. I pick a spot (the center of the Earth) and discover I don't move at all in $10$ seconds. How is acceleration without motion possible?

$\endgroup$
  • $\begingroup$ You didn't move because you were standing on the floor and the floor was preventing you from moving. Take the floor away and you would be moving just fine. $\endgroup$ – CuriousOne Jul 3 '16 at 0:15
  • $\begingroup$ I just tried that. My accelerometer went to zero, so we're not dealing with the same situation. $\endgroup$ – Donald Airey Jul 3 '16 at 0:17
  • 1
    $\begingroup$ You were asking why you weren't moving. The answer to that is "floor". :-) $\endgroup$ – CuriousOne Jul 3 '16 at 0:18
  • $\begingroup$ I asked "how can I be accelerating without moving?" When I tried your suggestion, I stopped accelerating. $\endgroup$ – Donald Airey Jul 3 '16 at 0:19
  • 2
    $\begingroup$ You totally forgot the first and most important question: Am I moving and/or accelerating relative to what? $\endgroup$ – Diracology Jul 3 '16 at 1:01
11
$\begingroup$

In relativity (both flavours) we consider trajectories in four dimensional spacetime, and acceleration is a four-vector not a three-vector as in Newtonian mechanics. We call this four-acceleration while the Newtonian acceleration is normally referred to as coordinate acceleration.

Supoose we pick some coordinate system $(t,x,y,z)$ and measure the trajectory of some observer in these coordinates. The way we usually do this is to express the value of the coordinates as a function of the proper time of the observer, $\tau$. That is the position is given by the functions $\left(t(\tau), x(\tau), y(\tau), z(\tau)\right)$. The proper time $\tau$ is just the time recorded by a clock travelling with the observer, so we are describing the trajectory by how the position in our coordinates changes with the observer's time.

If we start by considering special relativity, i.e. flat spacetime, then the four-velocity and four-acceleration are calculated by differentiating once and twice respectively wrt time, just like in Newtonian mechanics. However we differentiate wrt the proper time $\tau$. So the four-velocity $U$ and four-acceleration $A$ are:

$$ \mathbf U = \left( \frac{dt}{d\tau}, \frac{dx}{d\tau}, \frac{dy}{d\tau}, \frac{dz}{d\tau} \right) $$

$$ \mathbf A = \left( \frac{d^2t}{d\tau^2}, \frac{d^2x}{d\tau^2}, \frac{d^2y}{d\tau^2}, \frac{d^2z}{d\tau^2} \right) $$

The four acceleration defined in this way is coordinate independent, and it behaves in a very similar way to Newtonian acceleration. For example we can (though we usually don't) write a relativistic equivalent of Newton's second law:

$$ \mathbf F = m \mathbf A $$

where $\mathbf F$ is the four-force.

To complete the comparison with Newtonian mechanics we can choose our $(t,x,y,z)$ to be the coordinates in which the accelerating observer is momntarily at rest, and in these coordinates the four-acceleration becomes the proper acceleration, which is just the acceleration felt by the observer. Let me emphasise this because we'll use it later:

the four-acceleration is equal to the acceleration felt by the observer in their rest frame.

Anyhow, this is all in flat spacetime, and in flat spacetime a non-zero four-acceleration means that in every inertial frame the position of the observer is changing with time. This ties up with the first part of your paragraph where you're talking about your position relative to a star changing with time. However in general relativity the expression for the four-acceleration has to include effects due to the curvature, and it becomes:

$$ A^\alpha = \frac{d^2x^\alpha}{d\tau^2} + \Gamma^\alpha_{\,\,\mu\nu}U^\mu U^\nu \tag{1} $$

I've written this using Einstein notation as it's rather long to write out otherwise. The index $\alpha$ is zero for $t$, one for $x$, two for $y$ and three for $z$. The new parameters $\Gamma^\alpha_{\,\,\mu\nu}$ in the equation are the Christoffel symbols that describe how the spacetime is curved.

The difference from flat spacetime is that now we can have a (locally) inertial frame, where the spatial coordinates are not changing with time, and we can still have a non-zero four-acceleration. That is even if $x$, $y$ and $z$ are constant, so $d^2x/d\tau^2$ etc are zero, the contribution from the Christoffel symbols means the four-acceleration $\mathbf A$ can still be non-zero.

And in general relativity it's still true that the four acceleration is the same as the acceleration felt by the observer in their rest frame, and this is the link to the second part of your question. Because of the curvature you can be (spatially) at rest on the surface of the Earth with respect to the distant star but still have a non-zero four-acceleration. But remember that above we said:

the four-acceleration is equal to the acceleration felt by the observer in their rest frame.

That means even though you are at rest in your coordinates your non-zero four-acceleration means you still feel an acceleration. That acceleration is of course just what we call gravity.

Response to comment: Moving in a straight line

The obvious way to define motion in a straight line is to say that the acceleration is zero. In Newtonian mechanics this is just Newton's first law, where the acceleration is the coordinate acceleration $\mathbf a$. Likewise in relativity (both flavours) a straight line means the four-acceleration $\mathbf A$, defined by equation (1) above, is zero. Looking at equation (1), the only way for $\mathbf A$ is if the $dx^\alpha/d\tau^2$ term exactly balances out the Christoffel symbol i.e.

$$ \frac{d^2x^\alpha}{d\tau^2} = -\Gamma^\alpha_{\,\,\mu\nu}U^\mu U^\nu \tag{2} $$

This equation is called the geodesic equation, and it describes the trajectory of a freely falling particle in a curved spacetime. That is, it is the equation for a straight line in curved spacetime or more formally a geodesic.

Actually solving the geodesic equation is usually hard (like most things in GR) but for an overview of how this equation describes things falling in Earth's gravity see How does "curved space" explain gravitational attraction?.

Footnote: The elevator, the rocket, and gravity: the equivalence principle

The above discussion provides a nice way to understand the elevator/rocket description of the equivalence principle. See this article for a full discussion, but in brief suppose you are inside a lift with the doors closed so you can't see out. You can feel a force pulling you down with an acceleration of $1$g, but you can't tell if the lift is stationary on the Earth and you're feeling gravity, or if you're in outer space and the lift has been attached to a rocket accelerating at $1$g.

To see why this is we take equation (1) and rewrite it as:

$$ \mathbf A = \mathbf A_\text{SR} + \mathbf A_\text{GR} \tag{3} $$

where $\mathbf A_\text{SR}$ is the term we get from special relativity, $d^2x^\alpha/d\tau^2$, and $\mathbf A_\text{GR}$ is the term we get from general relativity, $\Gamma^\alpha_{\,\,\mu\nu}U^\mu U^\nu$.

But all you can measure is $\mathbf A$. Remember that $\mathbf A$ is equal to the acceleration in your rest frame, so if you have a set of scales in the lift you can measure your weight, divide by your mass, and you get your proper acceleration $\mathbf A$.

The point is that although you can experimentally measure the left side of equation (3) the equivalence principle tells us that you can't tell what is on the right hand side. If the elevator is blasting through space on a rocket $\mathbf A_\text{GR}$ is zero and all your acceleration comes from $\mathbf A_\text{SR}$. Alternatively if the elevator is stationary on Earth $\mathbf A_\text{SR}$ is zero and your acceleration comes from the $\mathbf A_\text{GR}$ term. The equivalence principle tells us that there is no way for you to tell the difference.

$\endgroup$
  • $\begingroup$ Space is curved in the presence of mass. I understand this much. We look at the Earth and, from a Newtonian perspective, say that there must be a constant acceleration pulling the Earth towards the sun, therefore there's a force. However, when we look at the same problem as a projection of a four dimensional manifold onto three dimensional space, we see that the Earth is travelling in a straight line and there is no 'force' of gravity. Is this a proper way to try and understand how the Christoffel symbol in the equation above provides something that looks like acceleration? $\endgroup$ – Donald Airey Jul 4 '16 at 1:12
  • $\begingroup$ @DonaldRoyAirey: I've extended my answer to try and address your comment. Basically, yes, the Christoffel symbol provides something that looks like acceleration. $\endgroup$ – John Rennie Jul 4 '16 at 5:20
  • $\begingroup$ BTW, I tried the experiment you suggested with a very wide and very tall elevator cab. It was several miles wide and my instruments were very sensitive. When I dropped two balls in the first experiment, they landed on the floor fractionally closer to each other than when I dropped them. In the second experiment, they landed on the floor exactly the same distance apart as when I dropped them. What does that say about your Equivalence Principle? $\endgroup$ – Donald Airey Jul 5 '16 at 19:51
  • $\begingroup$ @DonaldRoyAirey: the equivalence principle says that acceleration and gravity are locally indistinguishable. The equation for the four-acceleration calculates the value at a point in spacetime, and the EP tells us that the SR and GR contributions cannot be distinguished in this calculation. If you integrate the four-acceleration over a large enough distance then you will indeed find tidal variations in a gravitational field, but this is not a violation of the EP. $\endgroup$ – John Rennie Jul 6 '16 at 5:01
  • $\begingroup$ Brilliant answer, I'm learning so much! If I may ask two naive questions: (i) I'm always confused by what is meant by the word proper in these kinds of discussions, but here you neatly describe it as in the observer's frame. So I wonder, in case of proper time $\tau$ that we use to describe trjactory of observer, how do have access to $\tau$, i.e. the observer's clock, to begin with? Do we not only have a measure of time in our own frame? (ii) in discussing Eq. 1, you said one can be spatially at rest on Earth but still being accelerated. Is this then a temporal acceleration? Thanks $\endgroup$ – user929304 Mar 18 '18 at 17:10
7
$\begingroup$

The accelerometer doesn't measure 'movement' it measures acceleration. (I know that sounds tautological, but bear with me.) That is, it should read zero when in inertial (unaccelerated) motion, and when it reads non-zero a frame of reference at rest relative the instrument is non-inertial.

In other words the accelerometer is a tool for identifying inertial frames.

So here's the kicker: despite the treatment of physics you'll get in a Physics 101 classroom, standing still on on the floor of the lab is not inertial movement and a coordinate system that is at rest relative the floor is not an inertial frame (even neglecting the rotation of the planet). But one falling freely in that lab is an inertial frame.

Gravity, which we treat as a force in Physics 101 is more properly conceived of as a inertia pseudo-force like the centrifugal and Coriolis (pseudo-)forces. (And like those other pseudo forces it exhibits the feature of being precisely proportional to the mass of the affected object, so that all objects exhibit the same acceleration in the same circumstances.)

This is a core feature of general relativity.

For day-to-day purposes it remains easier to work in the non-inertial frame where the floor is at rest and simply add the inertial pseudo-force of magnitude $mg$ directed downward to your calculation.

$\endgroup$
  • $\begingroup$ I'm not sure that I want to live in your reference frame. ;-) $\endgroup$ – David White Jul 3 '16 at 1:11
  • 2
    $\begingroup$ It's the changing back to the planet-relative frame that hurts. ;-) Well, that and the bone-mass loss and muscular atrophy if you avoid switching back to the planet-relative frame. $\endgroup$ – dmckee Jul 3 '16 at 1:14
  • $\begingroup$ The experiment as conceived is not concerned with Special Relativity inertial reference frames. We accept that both experiments are non-inertial. Given that a foundation of GR is the Equivalence of Motion Principle, both experiments generate the same net force and the same acceleration. How come one results in a change in space and the other doesn't? $\endgroup$ – Donald Airey Jul 3 '16 at 2:08
  • 2
    $\begingroup$ In the lab on the planet you diverge from you putative position if you hadn't been accelerated by the floor by the same distance that you cover in the spacecraft. The correct basis of measurement is the geodesic associated with your kinematic conditions at the start of the experiment. $\endgroup$ – dmckee Jul 3 '16 at 2:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.