Falling object dispute

1. Aug 27, 2004

Tojen

Hello to the board. I don't know a lot about physics or relativiity, but I'm here to try to settle a friendly dispute I'm having with a friend. He was taught that the mathematics of General Relativity state that an object doesn't fall to Earth, but Earth rises up to meet it. That flies in the face of common sense (if I drop a pencil and the Earth rises to meet it, why don't I rise up with the Earth too?), and seems to contradict the Principle of Equivalence as well, though I can't quite explain why right now. Can anyone here set one of us straight?

2. Aug 27, 2004

Staff Emeritus
That is just wrong. People get taught all sorts of crazy things about modern physics, and that's just a sad example.

What happens according to general relativity is that the earth's mass warps spacetime near it, and the easiest paths theough the warped spacetime are the world-lines of falling objects. So when the object falls it is taking in a sense the shortest path through spacetime, given the curvature of spacetime near the earth. It's important to note that time is there in all of this; the object's world line is a set of points (x,t) giving the successive positions, and times, of its passage.

3. Aug 27, 2004

Artorius

Would it be correct to say that the acceleration of the ball is caused by an increase in the curvature of spacetime as the ball approaches the surface of the Earth?

4. Aug 27, 2004

NateTG

There is no 'preffered' frame of reference. For an ant standing on the pencil, it certainly seems like the earth is coming up to hit the pencil, and that is a completely valid frame of reference, even in Newtonian mechanics (where it might be considered an accelerated reference frame). In practice people generally select a frame of reference that is convenient for whatever reason.

What your friend might be trying to describe is that in GR the reference frame of the pencil is not considered to be accelerated, while the surface of the earth is. That is, imagine, for a moment that you're inside a box without windows. If the box is in free-fall you can't tell whether it's floating in space, or falling into a planet. Similarly, it's impossible to distinguish between a constant acceleration (like some wierd kind of rocket drive) or the box standing on the surface of the earth. In Newtonian Mechanics, the surface of a planet is considered to be unaccelerated, in GR it is considered to be accelerated. Conversely, in GR, the falling pencil is considered to be unaccelerated, and in Newtonian Mechanics it is considered to be accelerated.

5. Aug 27, 2004

Staff Emeritus
Actually I think the spacetime curvature gets less as you approach the center of mass. The acceleration is due to the curvature of the worldline, which I repeat passes through time as well as space.. its curvature makes it cover more space in equal units of time.

6. Aug 27, 2004

Integral

Staff Emeritus
It is not a result of Relativity which says that the earth raises to meet the pencil, it is basic Newton. Any 2 bodies with mass are effected by the gravitational force between them the pencil attracts the earth as much as the earth attracts the pencil. Newton also gives us the tools to compute how much the earth moves due to the force of the pensil....Not much.

So while your friend is sort of correct the fact is the distance which the earths moves under the influence of the falling pencil is computable but not measurable.

7. Aug 28, 2004

aekanshchumber

understand it, in most simple language without using relativity. let a pencil is droped from some height, earth exerts force on it accordin to newton's third law pencil will exert same force to earth in the opposite direction. but the mass of earth is so large that its acceleration towards the pencil is zero, but not so for the pencil its acceleration toward the earth is 'g' hence it will move towards the earth not earth towards the pencil.

8. Aug 28, 2004

Tojen

Thanks for the replies everyone. I was starting to think I had won until I read this and now I'm not sure:

That sounds to me like the Earth, being the accelerated object, is actually rising to meet the pencil, the unaccelerated object. Is that just from the pencil's perspective (it doens'nt "know" it's falling)? Because from my perspective, and I would think from Earth's as well, the pencil looks like the accelerated object.

9. Aug 28, 2004

pmb_phy

Tha'ts crazy talk. If someone says that then what they have in mind is that the frame of reference which is attached to the surface of the Earth is a non-inertial frame of reference and its accelerating relative to an inertial frame - Perdiod. It doesn't mean that the surface of the Earth is accelerating up to the particle since to speak of such an acceleration you have to reference what it is its accelerating with respect to. Consider two particles which are dropped from the opposite sides of the Earth at the same time. If the surface of the Earth is accelerating up to each particle and the size of the Earth is not chaning then how exactly does he explain that?

Pete

10. Aug 28, 2004

mijoon

This seems to be a rather popular piece of psuedo-science, as I've heard it several times, in several different places. I believe that it is the pop-sci-jibberish version of something which Hawkings once wrote.

11. Aug 28, 2004

pmb_phy

Its more misunderstanding than it is psuedo-science. An object on the surface of the Earth is considered to be accelerating with respect to an inertial frame of reference. That is mistaken to mean that the object is accelerating with respect to something which is at rest at a great distance from the Earth (e.g. "an object doesn't fall to Earth, but Earth rises up to meet it"). People are not focusing on what the object is accelerating with respect to and are focusing on the fact that its accelerating.

Pete

12. Aug 29, 2004

Tojen

Okay, I think I have it right. To be fair to my friend, he says it's the mathematics which state that Earth rises to meet an object, and I don't know the first thing about the math of GR.

But that would seem to violate the Principle of Equivalence. If it's impossible to tell, under the right circumstances, if I'm moving towards something or it's moving towards me, the math is stating emphatically that it's one and not the other. (He was also told by the same professor that it's impossible to fully understand GR without understanding its mathematics.)

13. Aug 30, 2004

Chronos

I kind of like the idea that the pencil simply falls to the floor. The earth is so much more massive than the pencil, it does not even move a planck length towards it. Since the planck length is the minimum unit of motion permitted, the earth does not move at all.

Last edited: Aug 30, 2004
14. Aug 30, 2004

aekanshchumber

thanks, if some one like my explanation .

15. Aug 31, 2004

Fredrik

Staff Emeritus
Even a point that isn't moving at all through space is "moving" through time. Every single physical object traces out a path in spacetime, called its "world line". In special relativity as well as in non-relativistic (Newtonian) mechanics, the world lines that are straight lines (technically "geodesics") define a special class of coordinate systems called inertial systems. Objects whose world lines aren't geodesics are said to be accelerating.

The reason I'm using terminology from differential geometry to define inertial systems and acceleration in special relativity and non-relativistic mechanics is that it isn't too difficult to see that these definitions make sense in general relativity too. In fact, there's no other way to generalize the concepts "inertial system" and "acceleration" to general relativity that makes sense.

What I'm saying is that in all of these theories, non-relativistic, special relativistic and general relativistic, the definition of an inertial ("not accelerating") system is the same. So, in all those theories, we can say that if the world line of an object isn't a geodesic, the object is accelerating.

In the non-relativistic and special relativistic theories, the world line of a pencil that falls to the ground isn't a geodesic, but the world line of a person standing on the ground is a geodesic. This means that the pencil is accelerating, and the person isn't.

But in general relativity, a heavy object like the earth changes the geometry of spacetime. The change is such that the world line of a pencil that falls to the ground is a geodesic (in a curved spacetime), and the world line of a person standing on the ground isn't a geodesic. So in general relativity, the pencil isn't accelerating, but the person (and the ground) is.

16. Aug 31, 2004

Olias

Next time you go by train from say..London to Glasgow, if Glasgow is moving towards the train, then you would NOT feel any fatigue, as you have not travelled to Glasgow from London, Glasgow has travelled to London!

I believe it was Eddington, or one of his students who stated the above.

Glasgow Time remains static, London time travels with the observer as that is the source of his journey, thus you get tired andfeel travel/motion sickness.

17. Aug 31, 2004

HallsofIvy

Whoever told you that has a very poor understanding of both General and Special Relativity- it looks like he pinning everything on the word "relativity". There was, after all, "relativity" in Newtonian Mechanics- it is sometimes referred to as "Gallilean relativity"- If you are in a completely closed room moving at constant velocity, you cannot do any experiment that does not refer to the outside, to determine how fast you are moving or even that you are moving at all- speed is always relative to some other object.
Of course, Newton and Gallileo didn't know about electricty and magnetism. When it was shown that the magnetic force of a moving charge depended on the speed of the object, it raised the possibility that you could do some sort of experiment with an electromagnetic field (say, light- for example, the Michaelson-Morley experiment) to determine the speed of your lab without reference to anything outside (an "absolute" rather than "relative" speed). When all such experiments gave a null result, a new "relativity" had to be developed to explain why.

In any case, Newtonian Mechanics does NOT have to "assume the surface of a planet is unaccelerated" nor does GR assume it is. I think your informant had in mind Einstein's observation that, locally, standing on the surface of a planet and experiencing gravity is indistinguishable from being in an accelerating "elevator" in empty space. That is NOT saying that the surface of the earth is accelerating.

In both Newtonian and Einsteinian mechanics, you are free to choose whatever frame of reference you want. Since you are standing on the surface of the earth, it is natural to choose that as your frame and see the ball as "falling" toward the earth. On the other hand, if you we the ball, you would see the earth moving very rapidly toward you! No GR required for that. If you were look at the moon orbiting around the earth, in order to be very accurate, you might need to use the fact that they both rotate around the common center of mass. Again, that's basic Newtonian mechanics.

By the way, in your original post you said "if I drop a pencil and the Earth rises to meet it, why don't I rise up with the Earth too?" Well, you do of course- you don't stay in the same place realtive to the pencil! That's why it appears that the earth is not moving- you choose to the frame of reference in which you are at rest.

18. Sep 4, 2004

jdstokes

Actually, an object undergoing free fall is locally inertial (geodesic) in the flat Minkowski space of special relativity.

19. Sep 6, 2004

Fredrik

Staff Emeritus
The world line of a falling object is certainly not a geodesic. I'm not sure what you mean by "locally inertial (geodesic)". Are you saying that the tangent of the world line is a geodesic? That's true for any world line, not just the ones describing objects in free fall. If a pencil that falls to the ground is "locally inertial", then so is a train that slows to a halt at a station.

20. Sep 6, 2004

pervect

Staff Emeritus
The world line of a falling object is definitely a geodesic, as long as you neglect air resistance and other such minor effects. (The emission of gravitational radiation is one of the minor effects which can cause deviation of a falling object from a geodesic, however, this is *extremely* small for a falling pencil, *much* much smaller thain air resistance).

For a worldline to be a geodesic means that the tangent vector of the worldline, when parallel transported along the worldline, remains tangent to the curve.

Given a metric, there is a unique notion of parallel transport which does not change the vector product gab vavb. A corollary of this result is that the length (i.e in GR the Lorentz interval) of a vector does not change when parallel transported, nor does the angle between any two vectors change as a result of parallel transport.

The most intuitive way of defining parallel transport, IMO, is "Schild's ladder". Alas, I don't find any online treatment's of "Schild's ladder" on the WWW, even on Greg Egan's site, but it's in "Gravitation" by Misner, Thorne & Wheeler.

21. Sep 7, 2004

da_willem

I looked it up for a Schwarzschild geometry and in this case it is indeed true that the curvature (as in the componnents of the Riemann tensor) gets less as you approach the center of mass (decreasing r).

About your second remark; I've heard before that the curvature makes you cover more space in equal units of time. and that's where the acceleration comes from. Can you (or somebody else) elaborate on this, maybe a little more mathematical, or provide me a link to a relevant webpage?

22. Sep 7, 2004

Fredrik

Staff Emeritus
Yes, in general relativity, this is certainly true. But I was answering someone who was talking about special relativity.

23. Sep 7, 2004

pervect

Staff Emeritus
Where did you look this up?

For the Schwarzschild metric, which is a point mass, the components of the Riemann tensor increase as you go towards the center. In terms of the unit basis vectors of an observer the "r" components go as

$$\large R_{\hat t \hat r \hat t \hat r} = -2GM/r^3$$

MTW, pg 820, "The nonsingularity at the gravitational radius"

If you were moving towards the center of a distributed, non-pointlike body, I would expect different behavior, so I don't necessarily disagree with the remark you were responding to, I haven't worked that case out (but it sounds right). The Schwarzschild case, though, definitely has the Riemann components increasing as you decrease r.

as I will explain, this tensor component represents the tidal "stretching force" in the radial direction. There are other components of the tensor, but they behave in a similar manner.

The link between space-time curvature and acceleration is the geodesic deviation equation

Two nearby _freely moving_ obsevers separated by spacelike vector Ek will accelerate away from each other with an acceleration given by

$$\large \frac{D^2 E^k}{d \tau^2} = -R^{\hat j}{}_{\hat \tau \hat k \hat \tau} E^k$$

Here $$\large \tau$$ is the proper time of the observer(s). Since they are intially co-moving and close together, their clocks run at the same rate.

The symmetries of the Riemann make this equivalent to

$$\large -R_{\hat \tau \hat j \hat \tau \hat k} E^k$$

When we translate this remark out of tensor language,when we look at an observer freely falling near a black hole, we say that two observers separated by a distance delta-r in the radial direction will accelerate away from each other. The magitude of the acceleration is -2GM/r^3 * delta-r, the value of the apprpriate component of the Riemann tensor * the separation, and it can be interpreted as the tidal stretching force / unit mass, or the tidal acceleration.

For a web reference, I can't come up with anything better than John Baez's

http://math.ucr.edu/home/baez/gr/geodesic.deviation.html

I hope the tensor notation isn't too confusing, but I'm afraid it may be. A few more examples might help.

EDIT for readibility, I'm going to replace $$\tau$$ with t from now on, $$\tau$$ looks too much like $$r$$.

Our observer has coordinates

$$r, \theta, \phi, t$$ and unit vectors of $$\hat r, \hat \theta, \hat \phi, \hat t$$

Suppose there were a "tidal torque" that caused the end of a rod pointed in the $$\hat r$$ direction to accelerate in the $$\hat \theta$$ direction? The associated component of the Riemann would be

$$\large R_{\hat t \hat \theta \hat t \hat r}.$$

There isn't any such component in actually, but there is a compresive tidal force, for a dispacement in the $$\hat \theta$$ direction, that points in the same direction, which is given by

$$\large R_{\hat t \hat \theta \hat t \hat \theta}$$ and has a value of GM/r^3.

You can basically think of the 16 componets $$R_{\hat t \hat i \hat t \hat j}$$ as being a collection of tidal stretching/compressive forces, when i=j, and tidal torques, when i is not equal to j.

Last edited: Sep 7, 2004
24. Sep 8, 2004

da_willem

My apologies for my incorrect statement. I looked it up in "A first course in GR" but the Riemann tensor was written down in terms of a few functions wich I had to put in and I made a little mistake. Sorry.

Thank you for your clear and thorough answer to my second question. I now have a clear understanding of the Riemann tensor components as representing tidal forces. This is in agreement with the fact that the Riemann tensor vanishes in a homogeneous gravitational field, in wich there are no tidal forces. The r components you give are in agreement with what you classically would expect from tidal forces:

The tidalforce causing an acceleration between two falling objects a distance $\vec{d}$ away in a gravitational field $\vec{G}$ is $m(\Delta \vec{G} )$. When the objects are an infinitesimal distance ($\vec{d}$) away from eachother aligned radially: $F_r= m(\Delta G_r} = m( \nabla G_r) \cdot \vec{d} = m dG_r/dr$. With $G_r=GM/r^2$ differentiating to r yields $F= (-2GM/r^3) d$ the same as the GR result.

25. Sep 8, 2004

pervect

Staff Emeritus
I'm glad you could read all the little subscripts in my last post - you've got the right idea. Note that the agreement of the Einsteinian and classical results for the tidal force is found only when the observer measuring the tidal forces uses his own local clocks and rods. You'll often see the Riemann not normalized in this manner. This is not a major conceptual issue, but if you actually try to carry out calculations, esp using the values of Riemann from sources you find on the net, it may become important. The value of the Riemann you see with the "hats" are measured in a coordinate system where |gii| = 1. This includes coordinates in the angular directions such as the $$\theta$$ direction.