# Can there be acceleration without velocity?

If I was in a spaceship continually accelerating at $9.81m/s^2 =1g$ in a straight line, I would reach near light speed within a year.

On Earth, we are in a gravitational field of $1g$, which according to the equivalence principle means we are effectively accelerating at $9.81m/s^2$.

What does this mean? How can we be accelerating without increasing our velocity? Is there some way of looking at this, from some other inertial frame, that means we are travelling near light speed?

• notice that $g$ is the acceleration and object would have if there were no other forces, that is, if it was on free fall. On Earth's surface objects are not on constant free fall, they are eventually stopped by other forces, so they are not under a constant acceleration. – user83548 Dec 27 '15 at 16:19
• – Qmechanic Dec 27 '15 at 16:26
• How can we be accelerating... probably because we're not accelerating? There are two forces acting on us: gravity and normal. – Kyle Kanos Dec 27 '15 at 17:58
• Have a look at my answer to Does the pilot of a rocket ship experience an asymptotic approach to the speed of light?. At a constant $1g$ acceleration a rocket ship approaches the speed of light asymptotically and never reaches it. – John Rennie Dec 27 '15 at 18:05
• I don't think this should be closed. The question makes a good point - the equivalence principle tells us that gravity and acceleration are (locally) the same thing, but if so why isn't the $1g$ acceleration I feel right now changing my velocity? The answer is that it is changing my velocity relative to a freely falling observer. The second point about reaching the speed of light highlights the fact that while the (Rindler) metric for an accelerating observer is uniform no gravitational field is uniform. I think expanding these statements would maake an interesting answer. – John Rennie Dec 27 '15 at 19:01

Let me paraphrase the question to make it clear what I'm answering. The equivalence principle tells us that acceleration and gravity are (locally) equivalent. But everyday experience tells us that an accelerating object speeds off into the distance at an ever increasing velocity, while the $1g$ gravitational acceleration I'm feeling right now hasn't increased my velocity at all in the 54 years I've been experiencing it. So how can acceleration and gravity be equivalent?

The answer is that acceleration is measured relative to a freely falling observer i.e. an observer who isn't accelerating. If you are aboard the rocket accelerating at $1g$ then in your own coordinate frame you are at rest and your own velocity isn't changing. However if you throw one of your crew mates out of the air lock you'll see them accelerate away from you at $1g$. Your velocity is changing relative to your freely falling crew mate.

Likewise here I am on the surface of Earth experiencing an acceleration of $1g$, and in my own coordinate frame I am at rest and my velocity isn't changing. However if I throw you down a mine shaft I will see you accelerate away at $1g$. The gravitational acceleration is changing my velocity relative to you.

So the two scenarios are indeed equivalent. Sitting here at my laptop typing this I really am accelerating at $1g$, and as a result of this acceleration my velocity really is increasing relative to a freely falling observer.

But there remains a big difference. If my rocket has enough fuel I can carry on accelerating indefinitely and the unfortunate soul I threw out of the air lock will keep accelerating (relative to me) indefinitely. Hence as the question says:

If I was in a spaceship continually accelerating at $9.81m/s^2 =1g$ in a straight line, I would reach near light speed within a year.

By contrast if I throw you down a mineshaft you'll accelerate away at first, but only for a limited time. As you reach the centre of the Earth (it's a deep mineshaft) your acceleration relative to me will decrease to zero then change direction. If I wait long enough you'll start moving back towards me again. This is very different to the behaviour we see with an accelerating rocket.

The reason for the difference is that the equivalence principle tells us that acceleration and gravity are only equivalent locally. There are no completely uniform gravitational fields in nature - all gravitational fields change with distance. The equivalence principle applies only over a distance small enough that the gravitational field is approximately uniform. A fair comparison with the rocket would require the acceleration of the rocket to change in the same way the gravitational acceleration changed.

Can there be acceleration without velocity?

No.

If I was in a spaceship continually accelerating at 9.81m/s² =1g in a straight line, I would reach near light speed within a year.

No problem with that, we've doubtless all read about the relativistic rocket.

On Earth, we are in a gravitational field of 1g, which according to the equivalence principle means we are effectively accelerating at 9.81m/s².

Actually, the principle of equivalence applies only to an infinitesimal region. See the second paragraph here. It is "nowhere precisely realized in the real world". And we aren't effectively accelerating. Your velocity just isn't changing. It's like you're accelerating. You feel a force on your feet, but you aren't actually accelerating.

What does this mean? How can we be accelerating without increasing our velocity?

It means we aren't accelerating. See this where Einstein referred to a gravitational field as inhomogeneous space, which people typically describe as curved spacetime. Standing on the ground in inhomogeneous space / curved spacetime is like accelerating in homogeneous space / flat spacetime , but it isn't exactly the same. If it was, if you really were accelerating, your electrons would be emitting radiation. You'd be glowing like the Ready-Brek kid, and you're not.

Is there some way of looking at this, from some other inertial frame, that means we are travelling near light speed?

No. We just aren't travelling at near light speed. However the wave nature of matter does involve something of this ilk, which is the underlying reason why the principle of equivalence "works". To appreciate this it's best to start with a single light beam, and work up from there.

Can there be acceleration without velocity?

Yes. The equivalence principle shows that.

Einstein:

We arrive at a very satisfactory interpretation of this law of experience, if we assume that the systems K and K' are physically exactly equivalent, that is, if we assume that we may just as well regard the system K as being in a space free from gravitational fields, if we then regard K as uniformly accelerated. This assumption of exact physical equivalence makes it impossible for us to speak of the absolute acceleration of the system of reference, just as the usual theory of relativity forbids us to talk of the absolute velocity of a system; and it makes the equal falling of all bodies in a gravitational field seem a matter of course.

There are some pretty good arguments here http://www.mathpages.com/home/kmath528/kmath528.htm that show how electromagnetic radiation cannot be emitted from uniformly accelerating observers. See also Does a constantly accelerating charged particle emit EM radiation or not?

The very slight inhomogeneity of the gravitational field on earth does not come into the picture, as we can always choose a smaller region of space to get a region with arbitrarily small curvature.

First, you have to sort out what velocity and acceleration means here. The acceleration you're talking about is what one would measure with an accelerometer, which essentially measures the 'bendiness' of your worldline in spacetime. And spacetime has $3+1$ dimensions.

Thus, if by velocity, you mean either a spatial $3$-vector or its magnitude (speed), then it becomes unsurprising that you can have an acceleration without a velocity: you're ignoring one component of the four-velocity in spacetime.

For example, in a static spacetime in coordinates in which the metric the metric takes the form $$\mathrm{d}s^2 = -e^{2U}\mathrm{d}t^2 + e^{-2U}h_{ij}\mathrm{d}x^i\mathrm{d}x^j\text{,}$$ where $e^U$ is the gravitational redshift factor, a function of the spatial coordinates $x^i$ only, a stationary observer would have a four-velocity $u = e^{-U}\partial_t$, i.e. with vanishing components in the spatial directions. Meanwhile, the four-acceleration vector is given by the covariant derivative of gravitational redshift factor: $$a^\mu = U^{;\mu} = g^{\mu\nu}U_{,\nu}\text{.}$$ For a spherically symmetric gravitational field (Schwarzschild spacetime in Schwarzschild coordinates), you will have a nonzero radial component $a^r$ essentially because the four-velocity has a temporal component $$u^t = e^{-U} = \left(1-\frac{2M}{r}\right)^{1/2}$$ that is not constant with respect to $r$, giving you a non-vanishing derivative.