I'm not sure but when a car is accelerating is the air in the car pushed backwards or not?

If so why did Einstein say that there is no way of finding out (besides tidal effects) the difference between being on the surface of the Earth, and being in a spaceship in deep space accelerating at 1g.

But if a rocket is accelerating and the air in it is moving like in the car pushed backwards than a balloon is going also backwards in this rocket.

But on the surface of the earth it is not moving (backwards) because the air and the ballon is under influence of the same gravity.

So in this experiment you can see the difference if you are in a rocket or on the surface of the earth.

Probably I miss something but what?

In this video is shown the ballon in the car: http://www.youtube.com/watch?v=y8mzDvpKzfY&feature=iv&src_vid=NblR01hHK6U&annotation_id=annotation_2968066189


  • $\begingroup$ The key point you are missing is that the acceleration is constant. If the rocket would suddenly start accelerating like your car, you'd see a difference, but this hypothetical rocket is always accelerating. $\endgroup$
    – Martin
    Feb 17 '16 at 14:18
  • $\begingroup$ Ok, but does the balloon keep standing a little skew ? $\endgroup$
    – Marijn
    Feb 17 '16 at 14:20
  • $\begingroup$ I don't fully understand what you mean by this. In a car, if you keep accelerating at the same rate, the balloon will stay in one position, which is a little skewed, yes. $\endgroup$
    – Martin
    Feb 17 '16 at 14:23
  • $\begingroup$ If the balloon is skewed in the car or rocket and not on the surface on the earth than it is possibel to see where you are in a rocket or on the surface $\endgroup$
    – Marijn
    Feb 17 '16 at 14:27
  • $\begingroup$ Ah, this is what you mean. It is only skewed in the car because you have horizontal acceleration (car) + vertical acceleration (gravity). In the rocket, it is not skewed (only vertical acceleration) as on earth. $\endgroup$
    – Martin
    Feb 17 '16 at 14:35

You are absolutely correct in thinking that the air inside an accelerating rocket would be pushed backwards. This pushing backwards of the air would create a pressure difference in the air inside the rocket, Making the air at the bottom more dense as compared to the air at the top, very similar to what we find on the earth. This leads to the development of buoyant forces in the rocket and lead to the following results.

As for the balloon, there may be two cases:

  1. When the balloon is denser than air:

When the balloon is denser than the air, it 'sinks' in the air and we see the balloon go in the direction opposite to the acceleration (also similar to what we see on earth when a normal air balloon falls to the ground).

Balloons sink!

  1. When the balloon is less dense as compared to the air.

When the balloon is less dense as compared to the air inside the rocket, it 'rises' in the air due to the buoyant forces acting on it (also similar to what we see on earth when a helium balloon rises in the air!) .

Helium balloon

Buoyant force also acts in the first case but it is not able to make the balloon float due to the balloon's density being greater than the air's (read about buoyancy).

So, we can easily see that being in a constantly accelerating rocket and being on earth is indistinguishable as long as the acceleration felt on the rocket is 9.8 m/s^2 (apart from tidal forces as you have mentioned).

To develop some perspective : ( i hope this helps clear your doubts)

enter image description here

enter image description here

  • $\begingroup$ But when the balloon is denser than the air wouldn't it has a skew backwards when it hangs on the ceiling while on earht is hangs straigt? $\endgroup$
    – Marijn
    Feb 17 '16 at 15:39
  • $\begingroup$ As i have already mentioned that we would see the balloon going against the direction of the acceleration of the rocket. In the case of the earth it would be towards the surface of the earth which is why the balloon hangs in the first place (it tries to go towards the earth's surface but is prevented by the tension in the string of the balloon.) $\endgroup$
    – Aniansh
    Feb 17 '16 at 15:42
  • $\begingroup$ So isn't that besides tidal effects a way of distinguishing where you are? $\endgroup$
    – Marijn
    Feb 17 '16 at 15:49
  • $\begingroup$ No it is not because if you hang the balloon in the same manner in the rocket as you did on earth, you would see that the balloon hangs from the ceiling of the rocket as well( assuming that the rocket is accelerating in the direction of its ceiling). The balloon would exhibit all the characteristics as it would on earth. $\endgroup$
    – Aniansh
    Feb 17 '16 at 15:51
  • $\begingroup$ And when it is not accelerating in the direction of its ceiling? $\endgroup$
    – Marijn
    Feb 17 '16 at 15:55

The behavior of a helium-filled balloon in an air-filled vehicle under the same constant acceleration is the same under all circumstances: Relative to the vehicle, the balloon will move in the same direction that the vehicle is accelerating in, due to bouyancy.

In a rocket, the balloon will migrate towards the front of the rocket, i.e., the direction that the rocket is accelerating in.

In your house, a helium-fill balloon will move toward the house's roof, which is the direction your house is accelerating in. From Einstein's perspective of gravity, a rock that you release in the house isn't accelerated toward the Earth. Instead the rock travels at a constant velocity, and just appears to be accelerating toward the Earth because your house is actually accelerating upward, because the surface of the earth is accelerating upward.

In the video, the vehicle is experiencing a combination of acceleration upward due to the upward acceleration of the Earth's surface, and acceleration forward due to the vehicle's engine. Using a vector addition of the two accelerations means that the vehicle's total acceleration is upward and toward the front, which is the same direction the balloon moves in as much as it can, given that it's tethered.

Technically, the Einstein perspective is that a rock you release doesn't experience any proper acceleration. The rock appearing to accelerate is due to gravity acting as a fictitious force, due to using an accelerating frame of reference. It's the surface of the Earth that's experiencing an upward proper acceleration.

Edit to address your follow-up question:

It would appear at first glance that it's merely a matter of perspective as to which is more correct, to say that the Earth's surface is moving at a constant speed, and a rock dropped from height of a couple meters accelerates towards the Earth, or to say that the rock is moving at a constant speed, and it's the Earth's surface that accelerates upward until it meets the rock. After all, what you can really measure is how the distance between the rock and the Earth varies with time, which would seemingly be the same equation from either perspective.

However, saying "it's all relative" like that turns out not to be precisely correct. It's possible to do an experiment to determine that the Earth's surface is indeed accelerating upward, and not just moving at a constant speed. According to special relativity (which is shown to be correct by lots of different experiments), two clocks that are moving at the same constant speed run at the same rate, as is expected. However, if two clocks a fixed distance apart are accelerating at a constant rate, the clock in back runs slightly slower than the clock in front, by a factor of approximately $$T_d \approx 1 + \frac{ah}{c^{2}} ,$$ where $a$ is the magnitude of the acceleration, $h$ is how much in front of the back clock the front clock is in the direction of acceleration, and $c$ is the speed of light. This effect is called gravitational time dilation, but it can be derived just using special relativity, without needing general relativity. The effect is very small, but it's large enough to be measurable with atomic clocks at two different heights. If you do the experiment, you find that gravitational time dilation shows that the Earth's surface is indeed accelerating upward at a rate $a=g$, where $g$ is the acceleration of gravity at the Earth's surface.

It may seem impossible that it's the Earth's surface that's accelerating instead of the rock, because the Earth's surface accelerating upward everywhere would mean that the Earth is exploding, which it clearly isn't. However, the acceleration of different parts of the Earth's surface is measured in different inertial frames of reference, such that the total picture is not one of the Earth exploding. And it isn't "cheating" to use different inertial frames of reference to conclude that the Earth's surface is accelerating, because there is no choice but to use different inertial frames of reference for different parts of the Earth's surface. Contrary to Newton's assumption, there is no inertial frame of reference that includes all of spacetime. The only possible way to measure proper acceleration precisely is to use inertial frames of reference that have an extent that's small compared to the size of gravitational features in that region of spacetime.

  • $\begingroup$ I don't understand why a rock which is released is not experiencing an acceleration while the earth does experience an acceleration. Why is that? $\endgroup$
    – Marijn
    Feb 17 '16 at 16:24
  • $\begingroup$ It's all relative. You can say that the stone is accelerating towards the earth and you may even say that both the earth and the stone are accelerating towards each other. It just doesn't matter which frame of reference you chose. $\endgroup$
    – Aniansh
    Feb 17 '16 at 18:04
  • $\begingroup$ I think that you should study regarding this topic. That is the only way you're going to understand this well. $\endgroup$
    – Aniansh
    Feb 17 '16 at 18:05
  • $\begingroup$ If the two clocks start at rest with respect to each other and start accelerating together, wouldn't the two clocks be at rest relative to each other? $\endgroup$
    – Aniansh
    Feb 18 '16 at 14:50
  • $\begingroup$ Are you refusing to accept that the frame of reference that we choose to observe the acceleration of the rock towards the earth doesn't matter? If yes then please explain to me why? $\endgroup$
    – Aniansh
    Feb 18 '16 at 14:57

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