25

I have learned about general relativity and how gravity arises from spacetime curvature. And I have always been taught that gravity is not a real force in the sense that

$$\frac{dp}{dt} = 0$$

And from this, gravity does not accelerate objects while they are in freefall. They are only accelerated when they are on the ground at rest.

On the other hand, when a spacecraft needs to reach a destination more quickly, they can use planets as velocity boosters. They use a gravitational assist from the planet to accelerate them to a greater velocity.

How can this be if gravity does not accelerate objects in freefall since it is not a force? I am seeing a contradiction here and it is confusing me. What am I missing in my conceptual understanding of gravity?

Qmechanic
  • 201,751
Tachyon
  • 1,866
  • 1
    in GR gravity (I learned this recently) isn't described as a force, though is it ? and Related (is gravity a force?) Good question btw; +1 for the last sentence about conceptual understanding of gravity – William Martens Jan 08 '23 at 19:57
  • 14
    Can you elaborate on how this differs from the fact that a dropped object picks up speed as it travels toward the floor? Or is that the essence of your question? – J. Murray Jan 08 '23 at 19:57
  • 2
    @J.Murray Well in the case of freefall, the object isn't accelerating or evening moving toward the ground, it is the ground that is accelerating toward the object. This is one interpretation I have heard about. Comparing what I have just said seems to contradict the idea of a gravitational assist where the object picks up speed to reach a planet faster and can be measured because the spacecraft reaches the planet sooner. This is what I am trying to understand. – Tachyon Jan 08 '23 at 20:04
  • 1
    It picks up speed relative to the Sun, with the help of the planet. – J.G. Jan 08 '23 at 22:29
  • 3
    @Tachyon do you have a good grasp of elastic collisions, as per an introductory mechancics class? Gravitational deflection of one body around another is almost exactly the same as an elastic collision between those two bodies if you zoom out far enough to consider only the parts of the trajectory that approximate straight lines. Momentum and kinetic energy, in particular, are exchanged according to the same formulae. – g s Jan 08 '23 at 23:28
  • 3
    We use various mathematical models to fit and predict measurements and observations. These models have their region of validity in coordinates and other variables used to measure and predict. There exists an overall consistency where the coordinate phase space overlap, and that is where confusion of terms used happens. Force is a Newtonian variable , and our senses exist a Nwetonian kinematics phase space, The words we use to define variables carry over to the more esoteric theories we discovered to be necessary , and that is what is confusing you. – anna v Jan 09 '23 at 05:46
  • 1
    It perhaps helps to consider a gravitational assist as an elastic collision between 2 bodies with one of them being very massive compared to the other one. – ratchet freak Jan 09 '23 at 12:56
  • 3
    If there's no special force assisting an airplane near the pole, why is it that a polar route from Boston to Tokyo is quicker than flying due west? – John Doty Jan 09 '23 at 17:35
  • 5
    Whether or not gravity is a force, where did you get the idea that gravity does not accelerate objects in free-fall; only on the ground at rest? That seems wholly counter-intuitive, at best… – Robbie Goodwin Jan 09 '23 at 23:07
  • Re, "in the case of freefall, the object isn't accelerating...toward the ground, it is the ground that is accelerating toward the object." No coordinate system is more special than any other. Either way you look at it, general relativity explains the relative acceleration between the COM of a spacecraft and the COM of a planet toward which the spacecraft is falling by showing that each of their world lines is a geodesic in curved spacetime. – Solomon Slow Jan 11 '23 at 00:08
  • @SolomonSlow Acceleration is what an accelerometer measures. An accelerometer sitting on the ground insists that it's accelerating upward. If you want to do the Newtonian thing, and say it isn't accelerating, you have to "correct" it. – John Doty Jan 11 '23 at 01:41
  • @JohnDoty, You're right. I was mixing metaphors there. I was saying that what Issac Newton would call "relative acceleration of the two bodies" can be explained by the geometry of their geodesics in GR. – Solomon Slow Jan 11 '23 at 01:47
  • 1
    @JohnDoty But if you experiment with an accelerometer and let it fall down to Earth, then it will not measure acceleration. You can think of an accelerometer as a spring with a mass attached, it measures acceleration by detecting whether or not the spring stretches or compresses. In the case of freefall, the spring will not stretch or compress, and therefore, the object is not accelerating. This is why some people say that the ground is accelerating toward the object. And I have heard others say that the accelerometer accelerates uniformly, which is why there is no detection of acceleration. – Tachyon Jan 13 '23 at 02:43
  • @Tachyon Yes. The view that the falling object accelerates is the Newtonian model. The view that the ground accelerates is the GR model. Your question is about the GR model. The Newtonian model is handier, with simpler math, but it needs conceptual crutches like absolute space and instantaneous action at a distance. We don't find these when we look for them: the phenomena don't always perfectly agree with Newton. But in most cases the imperfections are small, so we use the Newtonian model for its simplicity of application. – John Doty Jan 13 '23 at 15:16
  • @Tachyon I just need to say, your choice of username is flawless! totally marvelous 8) "Tachyon" – William Martens Jan 13 '23 at 15:30

3 Answers3

41

Well, gravity is a force and it isn't. What is a force anyway? It's what makes you accelerate, which is already a statement about a second-order derivative of one variable with respect to another, and now all of a sudden your coordinate system is important.

The point being made when someone says "gravity isn't a force" is that, if you express a body's location in spacetime, not space as a function of proper, not "ordinary" time along its path, gravity doesn't appear in the resulting generalization of Newton's second law in the same way as other forces do. In that coordinate system, the equation can be written as $\color{blue}{\ddot{x}^\mu}+\color{red}{\Gamma^\mu_{\nu\rho}\dot{x}^\nu\dot{x}^\rho}-\color{limegreen}{a^\mu}=0$, where the red (green) part is gravity (other forces). But this red/green distinction looks different, or disappears, if you look at things another, mathematically equivalent way. In particular:

  • Putting on Newton's hat This is the less elegant of two options I'll mention, one that uses pre-relativistic coordinates. If you look at the body's location in space, not spacetime as a function of ordinary, not proper time, the red term looks like the green term, hence like the stuff you learned from Newton. In particular, $\frac{dp^i}{dt}\ne0$.
  • Putting on Einstein's hat Even more elegantly, we don't need to leave behind the coordinates I suggested first to change our perspective. As @jawheele notes in a comment, we unlock the real power of GR if we use a covariant derivative as per the no-red formulation $\color{blue}{\dot{x}^\nu\nabla_\nu\dot{x}^\mu}-\color{limegreen}{a^\mu}=0$. This time, the equation's terms manifestly transform as a tensor, making the blue term the unique simplest coordinate-invariant notion of acceleration.

The main advantage of the $\Gamma$-based version is doing calculations we can relate back to familiar coordinates. This not only recovers Newtonian gravity in a suitable limit, it computes a correction to it.

Regarding the first bullet point above, have you ever spun on a big wheel? There's a similar perspective-changing procedure that says the dizziness you're feeling is due to something that's "not a force". You're still dizzy, though. This isn't a contradiction; they're just two different ways of deciding what counts as a force.

The good news is we don't need to "forget" GR to understand a gravity assist. How does it work? It exploits the fact that, if a planet's in the right place at the right time for you, the red term is very different from what the Sun alone would normally give you there. This has implications for the blue part even without wasting fuel on the green part. Or you can explain it without GR; your choice.

J.G.
  • 24,837
7

For $\dfrac{d}{dt}\vec P = \dfrac{d}{dt} (m\dfrac{d}{dt}\vec s)$, where $\vec s$ is the position of a gravitationally-affected body as measured by a distant observer, $t$ is the time measured by a distant observer, $\dfrac{d}{dt}\vec P \ne 0$. This is the difference between coordinate acceleration (the second derivative of position with respect to time) and proper acceleration (what a local accelerometer measures).

g s
  • 13,563
  • 1
    So it is a matter of reference frames then? I have seen from the links provided by commenters that locally, gravity is not a force, while a distant observer does measure a force? – Tachyon Jan 08 '23 at 20:19
  • 3
    @tachyon yes, but in this case I would prefer to use the similar term coordinate system rather than reference frame – g s Jan 08 '23 at 22:23
  • 1
    I see your point, thanks. – Tachyon Jan 08 '23 at 23:45
  • @gs So, as per your maths, the definition of a term in your equation is observer-dependent. A person right next to a falling object will see the object falling, so gravity sure is a force (force is defined by the thing that makes stuff move, isn't it?). A distant observer will see, well what? Wouldn't he also see the object falling? Or would he see it handing there in mid-air forever or moving upwards or dancing or what? He will obviously see object falling. The object start from zero and accelerate, where do the force comes from? – Atif Jan 10 '23 at 05:14
  • 1
    @Atif If the person next to the falling body is unsupported, the person will not see it falling, they will both just be in free-fall together. The person will, of course, be able to look down and see the distant planet coordinate-accelerating towards him, just as an observer distant from both will see the two accelerating towards their center of mass per Newton's laws. [1/2] – g s Jan 10 '23 at 08:49
  • 1
    @Atif [2/2] If the person next to the falling body is supported, the person will see the falling body coordinate-accelerate away from him. However, looking at his accelerometer, he will measure a proper acceleration equal and opposite the coordinate acceleration of the falling body, and will be able to conclude that he is the one who's accelerating. – g s Jan 10 '23 at 08:49
  • @gs So, as you just said, all observers see movement, right? As long as force is necessary for movement (how else will you define force?) there has to be force for the movement to happen. Whats the issue in this? – Atif Jan 10 '23 at 08:51
  • @Atif Two astronauts are on the international space station in a closed room looking at each other. What do they measure? – g s Jan 10 '23 at 09:06
  • @gs Look you can make your observers Helen Keller if you want, that wouldnt change the fact that you would have to explain observations of observers who can observe. Playing with definition of observers (your "observers" cannot observe now because that fit your "explanation" as you don't need any explanation if there is no observation) would do nobody any good. Even in a closed room as long as there is acceleration there are effects that can be observed. For example both in a free falling lift and space station in orbit there would be weightlessness which is because of gravity acceleration. – Atif Jan 10 '23 at 09:13
  • 2
    @Atif If you have further questions, I suggest searching the SE and posing a new question if you can't find an answer. If you are confident that this answer is wrong, feel free to downvote. I do not intend to debate on SE. – g s Jan 10 '23 at 09:18
  • 1
    @gs The closed-room way out of explaining wouldn't work here because there is still acceleration therefore force. Your way out of explaining would work only if there is no force but the question and your answer are only for situations when there is a force. If there is a force there is a movement. Without force no movement begin. Gravity start a movement in a direction, the falling object is not already in motion. Without force nothing start moving, thats in definition of force. – Atif Jan 10 '23 at 09:20
  • @gs The question I intend to ask is already asked. This is that question. Your answer has fundamental flaws making it meaningless. Comments are the right place to discuss answer and ask for explanation. You cannot say "I do not intend to debate on SE" after already debating (see your comments). We are all here to learn and seeking explanation is all this site is about. Debating may lead to truth. – Atif Jan 10 '23 at 09:25
3

An intutive description, no mathematics and not totally 100% true.

Take the example of the international space station. It goes round the earth in a (more or less) circular path. Inside the station you feel no gravity. This is despite that the station is constantly on a circular path - if this was a carousel on earth you would feel a "force" outwards. You would expect things in zero gravity to move towards the "outside" of the path - but they dont.

In the Newton equations this is explained by saying that gravity is a force. Remember that an object will continue with fixed speed on a straight path unless influenced by a force.

In general relativity it instead described by saying that space-time is curved. Here as well an object will continue with fixed speed on a straight path unless influenced by a force. The difference is that a straight path in curved space time might not look straight when seen from the "outside". In this description the space station moves in a straight path through space time - there is no gravition force involved.

Regardless of which set of rules you use (added: in doing calculations), gravity as a force or curved space time, the result will be almost the same. The very slight difference will be seen only in some cases but mostly with things moving at speeds at a significant fraction of the speed of light.

So what happens in a gravity assist in the GR description is that the space craft goes in a straight line in a curved space. You might imagine it as similar to a skate boarder going into a "bowl": down, round, up. The difference is that the "bowl" moves along with the planet. Gaining speed can be done by entering the curved space behind the moving planet and "steal" a bit of the planets forward motion. There is no force involved in this, the space craft simply follow the straight line in the curved space. But as the planet moves the curved space will change over time.

ghellquist
  • 834
  • 1
    "Inside the station you feel no gravity." You do. The weightlessness is due to gravity. If you are in a free-falling lift you will feel same weightlessness. Also, as you said, the space station is in movement. How do that movement start? There has to be a force to start it. That force is gravity. Not just that but force is a vector so it has a direction too, and that direction is always straight. So gravity force is continuously affecting the space station, otherwise the space station would just fly in a straight direction not keep going round and round. – Atif Jan 10 '23 at 09:39
  • "Regardless of which set of rules you use" The rules are fixed. You cannot pick which rules apply. The rules are hard, physical rules and don't care about your choices. Your choices are limited to explanation of facts. Your theories must always match the data. Moreover, you cannot change definitions of terms. otherwise you will be meaningless. However you explain it movement happen and movement cannot happen in absence of force when starting from stationary. So there is a force. Any theory or equation that say there is no force is wrong. – Atif Jan 10 '23 at 09:43
  • @Atif: I agree on one part of your post, and has added that I mean what rules you use in doing calculations. The physical world does what it does regardless of our calculations. – ghellquist Jan 10 '23 at 10:13