# Does acceleration change according to inertial frame of reference?

1. Sep 30, 2014

### David Carroll

I apologize if this has been discussed before. I am no physicist, I am just trying to see if I am understanding special relativity correctly. Please be advised I am dumb, so be gentle.

If a spaceship (sorry for the cliché, but it's a little bit cumbersome to keep using the phrase "inertial frame of reference" or the like) is moving at such a speed that time dilates by a factor of 2, this means that the length of the spaceship is also contracted by a factor of 2, yes? (All, of course, from the perspective of me, the "stationary" observer).

If an object within the spaceship is accelerating at some given rate according to the passengers, would that object also be accelerating at the same rate from MY perspective?

She gotta tv eye on me, she gotta tv eye on me.......

2. Sep 30, 2014

### Orodruin

Staff Emeritus
No, the passengers will see the object accelerating at a different acceleration than the one you would measure. Just as you have to use a relativistic transformation law for the velocities, you have to use a relativistic transformation law for the accelerations.

3. Sep 30, 2014

### A.T.

It's crucial that the space ship is inertial, so always do mention that.

Yes. The proper acceleration of the object is frame invariant, and for every inertial frame the proper acceleration of the object equals its coordinate acceleration in that frame.

4. Sep 30, 2014

### jartsa

Let's consider a standing person that starts walking inside that spaceship.

Let's say the person is some kind of standard person doing everything normally in his own frame, so that we can say the final walking pace according to you is half the normal, while the step length is half the normal.

The person is increasing his distance from the start position at one fourth the normal rate according to you.

Of course it took him twice the normal time to get going, according to you. One fourth of the normal velocity change happened in twice the normal time, so the acceleration observed by you is 1/8 of the normal.

EDIT: That was a special case where acceleration is parallel to motion.

Last edited: Sep 30, 2014
5. Sep 30, 2014

### Orodruin

Staff Emeritus
Sorry, but this is simply not true. For example, for rectilinear motion, the proper acceleration $\alpha$ is given by
$$\alpha = \frac{d}{dt}[\gamma v] = \gamma^3 a,$$
where $a$ is the coordinate acceleration $d^2x/dt^2$ and $v$ is the coordinate velocity $dx/dt$. That acceleration is equal to the proper acceleration only holds in the object's own instantaneous rest frame.

More generally
$$\alpha^2 = \gamma^6 [a^2 - (\vec v \times \vec a)^2].$$

6. Sep 30, 2014

### Fredrik

Staff Emeritus
This will probably confuse the OP, so I will try to unconfuse him. There are two kinds of "acceleration": coordinate acceleration and proper acceleration. If x(t) is the position at time t (according to some coordinate system), then the coordinate acceleration at t (in the same coordinate system) is x''(t) (the second derivative of x at t).The coordinate acceleration is different in different coordinate systems.

The proper acceleration is (regardless of what coordinate system you're using) equal to the coordinate acceleration in the momentarily comoving inertial coordinate system.

They're equal in the (momentarily) comoving inertial coordinate system, but not in an arbitrary inertial coordinate system.

7. Sep 30, 2014

### David Carroll

Okay, trying to make sense of this. I forgot to add that in my scenario, the direction of the acceleration is parallel to the direction of motion of the spaceship (acceleration is due only to speed increase, not vector change in this get-up).

Here's what I'm thinking. We have an object accelerating within the spaceship at 2 meters per second squared for the passengers on board. At second 1, mark 2 meters. Second 2, mark 6 total meters traveled. Second 3, mark 12 total meters traveled. But 2 meters, 6 meters, and 12 meters are 1 meter, 3 meters, 6 meters respectively for myself, the "stationary" observer. Furthermore, the 1 second, the 2 second, and the 3 second are for me 2 seconds, 4 seconds, and 6 seconds respectively. Mathematically, this would be .5 meters per second squared acceleration for me, the outside observer, no? So, this kind of acceleration is for me the inverse of the square of the time dilation factor?

My next question is: wouldn't this mean that the gravitational field (and maybe some other fields as well) gets weaker within the spaceship from my frame of reference? After all, if time dilates, length contracts, and mass dilates, wouldn't this mean that two massive objects within the spaceship that are accelerating toward each other due to their gravitational fields are for me: 1) becoming more massive, 2) closer to each other, but 3) paradoxically accelerating MORE SLOWLY than they should be from my frame of reference?

8. Sep 30, 2014

### A.T.

As others noted, the above is only true in classical physics, but not in Relativity which is the context here. Sorry for the confusion.

9. Oct 1, 2014

### jartsa

You made some error, because it should be inverse of the cube of the time dilation factor. (See posts 4 and 5)

Fields become shorter in the direction of the motion, but force is unchanged.

Fields become weaker in the direction transverse to the motion, but distances are unchanged.

If you want references, google "force relativity".

10. Oct 1, 2014

### David Carroll

Okay. Understood. Thank you.

11. Oct 2, 2014

### Staff: Mentor

I think you mean distances become shorter in the direction of motion, correct?

Also, "force" does not include gravity in this context.

12. Oct 2, 2014

### Staff: Mentor

For non-gravitational forces (say oppositely charged objects attracting each other), yes, the accelerations would appear slower to you if the spaceship was moving at relativistic velocity.

For gravity, things get a lot more complex, because if you have two objects inside the spaceship whose gravity you can measure, then spacetime is no longer flat, and all the discussion in this thread has assumed flat spacetime.

13. Oct 3, 2014

### jartsa

I meant everything is Lorentz contracted, including fields.

If you are speeding through the universe, measuring the curvature of the space, you can do a curvature map of a large area of the universe in short time. That proves that the areas of curvature are contracted.

14. Oct 5, 2014

### pervect

Staff Emeritus
There is a way to do something similar like this that actually works. The Riemann curvature tensor descries the curvature of space-time, and one can perform a local Lorentz boost of this tensor using the usual tensor rules for a rank 4 tensor. Basically one creates a local frame field using local basis vectors, then boosts the basis vectors appropriately.

To go much further one need a physical interpretation of the Riemann curvature tensor. The most important parts of this tensor correspond in normal circumstances to to tidal forces, so one can get a somewhat intuitive idea of how gravity transforms by imagining you boost, not "gravity", but rather "tidal gravity". There are limits to how well this works, as will be discussed below.

There are other components of the Riemann than the so called "electrogravitic" parts that are responsible for tidal forces. These are the parts responsible for frame dragging effects (magnetogravitic parts) and purely spatial distortions (topogravitic parts). So there is already a bit of a disconnect between viewing the Riemann as "just" tidal gravity.

An additional issue is that the "tidal force" interpretation has issues for observers with sufficiently high accelerations, though for reasonable accelerations the errors are minor, for infinite proper accelerations the errors become infinite . This is why the tidal forces for an observer falling into a black hole are finite while the tidal forces for static observers hovering near the event horizon become infinite as one approaches the horizon closer and closer. The Riemann is always finite, to be clear, the issue is that the notion of "tidal forces" is not quite a perfect match for the actual Riemann.

An third problem that raises its head is accounting for gravitational time dilation. Gravitational time dilation does not arise naturally when interpreting gravity as a force - there is no other "forces" other than gravity that causes time dilation, and in fact the existence of gravitational time dilation is one of the observed features of gravity that does not easily fit into the idea that gravity is a force. Gravitational time dilation suggests, instead, that gravity is something else - curved spacetime. The data to show the existence of gravitational time dilation was not there before Einstein, but it is now. Time dilation is now so routinely observed that we have to account for it in precision timekeeping here on the Earth, but its very existence was not suspected until GR predicted it.

Gravitational time dilation does NOT transform as a tensor, even defining it requires defining a coordinate system. (One might regard g_00 as defining time dilation, in which case it does transform as one component of a tensor - but one needs to specify the whole coordinate system to be able to calcuIate that single component). Given a particular coordinate system for the moving observer (such as a Fermi Normal coordinate system), then one can compute the time dilation in that coordinate system, but the process of computation generally turns out to be very difficult in practice, with no closed form solutions, and require a fairly advanced knowledge of the mathematical formalisms of curved space-time. The minimum requirements would be to know the metric tensor, the Riemann tensor, and the geodesic equations, plus knowledge of how to go from the metric tensor to the Christoffel symbols needed for the geodesic equations , and how to go from the meric tensor to the Riemann tensor.

Last edited: Oct 5, 2014