Einstein says objects do not fall to the Earth?

[A.T.]

If the Sun is orbiting the Earth and Mars is orbitting the Sun, how do you explain that using classical physics?

In non-inertial frames there are inertial forces, additionally to Newtonian gravity. That is why the description from the inertial frame of the common center of mass is simpler, because it involves only Newtonian gravity.

 

[PeterDonis]

I was trying to argue that, experimentally, you could show that the Earth has a bigger influence on the "motion" of a ball than the ball has on the Earth.

"Bigger" here is coordinate-dependent (because "motion" is), so there will be no invariant way to show this. You will always be able to adopt a reference frame in which the ball is at rest and the Earth is moving. There's no measurement that can show that the ball is what is "really" moving.

It seems strange to say: it's equally valid to view the Sun as orbiting the Earth and Mars orbiting the Sun.

That's not what I said. I said it's equally valid (in principle) to adopt coordinates centered on the Sun, or coordinates centered on the Earth. Neither of those will result in the Sun orbiting the Earth but Mars orbiting the Sun. In coordinates centered on the Earth, the Sun orbits the Earth in a relatively simple trajectory, and Mars orbits the Earth in a much more complicated trajectory, with loops in it.

If the Sun is orbiting the Earth and Mars is orbitting the Sun, how do you explain that using classical physics?

You don't. Nobody ever proposed a model like this. The model that was used for many centuries before Copernicus and Kepler had everything orbiting the Earth. I would advise you to do some research into what that model actually said. It was nowhere near as naive and simplistic as you appear to believe.

 

[PeroK]

"Bigger" here is coordinate-dependent (because "motion" is), so there will be no invariant way to show this. You will always be able to adopt a reference frame in which the ball is at rest and the Earth is moving. There's no measurement that can show that the ball is what is "really" moving.

Some final questions.

In classical physics: I thought that Foucault's Pendulum and the Coriolis effect (for example) showed that the Earth is spinning (every 24 hours) and not that the Sun is orbiting the Earth every 24 hours. Is that not the case? Is there no experiment - in classical physics - that shows that the Earth is spinning?

In GR: is it not possible to conduct an experiment to show that the Earth orbits the Sun every year and not that the Sun orbits the Earth every day? It seems to me that the curvature of spacetime would be different in these two cases. Can the absolute curvature of spacetime not be measured?

I understand the idea about relativity of observations and motion. And I know that you can't detect absolute motion. But, I thought you could detect absolute acceleration? And an accelerating reference frame is equivalent to a gravitational field? But, is it impossible to measure the curvature of spacetime?

Final thought: a ball in local inertial motion starts a long way from a massive body and gradually gets closer to it. The ball gradually sees everything else "accelerate" until it collides with the massive body. Meanwhile, an observer on the massive body sees no significant changes to its motion.

This does not seem to me like a symmetric situation. The ball saw the rest of the universe "accelerate" and the massive object didn't. But, there is no experiment that shows that the ball entered a large gravitational field, while the massive object was the source of the gravitational field? I know they are both in "free fall", but it seems that their obseravtions are asymmetric.

I find it strange that you can't detect this asymmetry. The massive object is in massively curved spacetime all along, whereas the ball experiences increasingly curved spacetime.

 

[A.T.]

In classical physics: I thought that Foucault's Pendulum and the Coriolis effect (for example) showed that the Earth is spinning (every 24 hours) and not that the Sun is orbiting the Earth every 24 hours. Is that not the case? Is there no experiment - in classical physics - that shows that the Earth is spinning?

It shows that it's simpler to describe the Earth as spinning, because that eliminates the Coriolis and other effects related to a frame where the Earth doesn't spin.

In GR: is it not possible to conduct an experiment to show that the Earth orbits the Sun every year and not that the Sun orbits the Earth every day? It seems to me that the curvature of spacetime would be different in these two cases. Can the absolute curvature of spacetime not be measured?

The absolute curvature of space time is related to tidal effects, which are indeed frame invariant. And yes, of course it would be different.

I understand the idea about relativity of observations and motion. And I know that you can't detect absolute motion. But, I thought you could detect absolute acceleration?

You can detect proper acceleration, but gravity doesn't cause proper acceleration.

This does not seem to me like a symmetric situation. The ball saw the rest of the universe "accelerate" and the massive object didn't.

It's not symmetric because you introduced a third object: the rest of universe.

The massive object is in massively curved spacetime all along, whereas the ball experiences increasingly curved spacetime.

That is correct, if the ball is big enough it will be stretched by tidal forces, while the planet wont' be. But that doesn't imply anything about which of them is actually moving, beyond simplicity of calculations.

 

[PeroK]

It shows that it's simpler to describe the Earth as spinning, because that eliminates the Coriolis and other effects related to a frame where the Earth doesn't spin.

Thanks. I think I understand all this, but perhaps I'm looking for conclusions that modern physics, in general, does not like to draw.

 

[rajeshmarndi]

When we throw a ball up, it decelerates and ultimately comes to rest and then accelerates towards the surface. Is it the ball actually travel in curve path at constant velocity but since we see the ball always vertically above us, we don't see the ball path which is curve [Edited: and therefore we don't observe constant velocity of the ball either]. Just like in a half ellipse , we move ahead in time along the minor axis of the half ellipse from one end and the thrown ball(vertically above us) moving along the curve of the half ellipse and the ball meet with us at the other end of the minor axis.

 

[PeterDonis]

I thought that Foucault's Pendulum and the Coriolis effect (for example) showed that the Earth is spinning (every 24 hours)

The Foucault Pendulum shows that the Earth is spinning relative to the pendulum. The Coriolis effect is a coordinate effect, and can be eliminated, as A.T. pointed out, by choosing different coordinates.

and not that the Sun is orbiting the Earth every 24 hours

The Foucault pendulum and the Coriolis effect are separate sets of observations from the observation of the relative motion of the Sun and the Earth, so they have to be considered separately. (For example, the period of relative rotation of the Earth and the Foucault pendulum is not 24 hours; it's 23 hours, 56 minutes, and some number of seconds that I can't remember right now.)

is it not possible to conduct an experiment to show that the Earth orbits the Sun every year and not that the Sun orbits the Earth every day? It seems to me that the curvature of spacetime would be different in these two cases.

of course it would be different.

The situation that exists in our actual solar system can be described both ways, and the curvature of spacetime is the same. Of course if the relative masses of the Earth and the Sun were different, the curvature of spacetime would be different, but then experiments would show a lot of other differences as well. Given the experimental results we actually have, there is no experiment we can do that shows that the Earth "really" orbits the Sun; the best we can show is where the center of mass of the solar system is, and that the simplest description of the motion of the Sun and planets is given in a frame centered on that center of mass.

 

[PeterDonis]

When we throw a ball up, it decelerates and ultimately comes to rest and then accelerates towards the surface.

In a frame in which the surface is at rest, yes. But this is coordinate acceleration, not proper acceleration. An accelerometer attached to the ball will read zero, and an accelerometer attached to the surface will read nonzero, so the surface is what is accelerated in the sense of proper acceleration.

Is it the ball actually travel in curve path at constant velocity but since we see the ball always vertically above us, we don't see the ball path which is curve

No, it's that whether or not the ball's path is "curved" or not, in a coordinate sense, depends on the coordinates you pick; it's curved in coordinates in which the surface is at rest, but it's straight in coordinates in which the ball is at rest. If you want an invariant sense of "curved", then you have to define a path as "curved" if it has nonzero proper acceleration, and "straight" if it has zero proper acceleration; as above, that means the ball's path is straight and the surface's path is curved.

 

[A.T.]

Of course if the relative masses of the Earth and the Sun were different, the curvature of spacetime would be different

Yeah, that's how I meant it.

 

[inertiaforce]

Guys, take a look at this video. Brian Greene is also saying that Einstein's view was that the Earth rushes up and hits you. You do not fall to the earth. Watch this video from 9:30 to 15:00:

 

[A.T.]

Brian Greene is also saying that Einstein's view was that the Earth rushes up and hits you.

That is not just Einstein's view, but every free faller's view. Even in classical Newtonian mechanics you can view things from the free falling frame, where the Earth is moving, while the free falling object is still.

You do not fall to the earth.

See my earlier comments on the ambiguity of pop-sci language, aimed at a mass audience:
https://www.physicsforums.com/threa...do-not-fall-to-the-earth.781200/#post-4909705

 

[inertiaforce]

That is not just Einstein's view, but every free faller's view. Even in classical Newtonian mechanics you can view things from the free falling frame, where the Earth is moving, while the free falling object is still.See my earlier comments on the ambiguity of pop-sci language, aimed at a mass audience:
https://www.physicsforums.com/threads/einstein-says-objects-do-not-fall-to-the-earth.781200/#post-4909705[/QUOTE]

Thanks for your reference A.T. I agree with you. "Falling" appears to be an ambiguous term.

 

[Dale]

Note that while "falling" is ambiguous, "free-fall" and "free-falling" are not. They refer to the inertial object and the ground rushes up to meet the inertial object precisely because the ground is not free-falling and the inertial object is.

When Brian Greene uses the unambiguous scientific term he uses it correctly (at least in the brief clip I watched).

 

[A.T.]

Note that while "falling" is ambiguous, "free-fall" and "free-falling" are not. They refer to the inertial object and the ground rushes up ...

To me "rushing up" is ambiguous too. It could mean movement, which is relative. So even in classical mechanics the Earth can be "rushing up" in some frame.

 

[CKH]

But to say that the Earth orbits the Sun "and not vice versa" is to assert a preference for a certain system of coordinates (one centered on the Sun--or on the common center of mass), which is not, in principle, valid. Certain coordinates may be more useful than others, because the description of the solar system looks a lot simpler in terms of them, but that doesn't make other coordinates invalid; it just makes them less useful for certain purposes--but not for others.

But the choice is more than just a preference for a certain system of coordinates. The laws of motion are stated for an inertial frame. If you choose a non-inertial frame and incorrectly apply these laws of motion to that frame, fictitious force fields appear. These fields don't actually exist. They are the result of misapplication of physical law.

For example, if you contend that the Earth is stationary (that it is at rest in an inertial frame) and that the universe revolves around the earth, you need fictitious force fields (pseudo centrifugal and coriolis forces) to explain this. This is just a mistake in the application of physical law. The conclusion that the universe revolves around the Earth is untenable.

The choice of coordinates is not arbitrary when you apply physical laws that are expressed for an inertial frame. To apply the laws you must choose an inertial frame.

So long as you know how two coordinate systems are related, you can transform physical events from one system to another, no physics is involved in this, it is just math.

 

[A.T.]

But the choice is more than just a preference for a certain system of coordinates.
...
So long as you know how two coordinate systems are related, you can transform physical events from one system to another, no physics is involved in this, it is just math.

So, it is just a preference for a certain system of coordinates. For example a preference based on the simpler math in those coordinates.

 

[DrGreg]

But the choice is more than just a preference for a certain system of coordinates. The laws of motion are stated for an inertial frame. If you choose a non-inertial frame and incorrectly apply these laws of motion to that frame, fictitious force fields appear. These fields don't actually exist. They are the result of misapplication of physical law.

For example, if you contend that the Earth is stationary (that it is at rest in an inertial frame) and that the universe revolves around the earth, you need fictitious force fields (pseudo centrifugal and coriolis forces) to explain this. This is just a mistake in the application of physical law. The conclusion that the universe revolves around the Earth is untenable.

The choice of coordinates is not arbitrary when you apply physical laws that are expressed for an inertial frame. To apply the laws you must choose an inertial frame.

So long as you know how two coordinate systems are related, you can transform physical events from one system to another, no physics is involved in this, it is just math.

All this would be correct within Special Relativity.

In General Relativity there are no such things as truly inertial frames. However in a small enough region, where the tidal effects of gravity are negligible, you can find a locally-approximately-inertial frame.

 

[jerromyjon]

you can find a locally-approximately-inertial frame.

"Near Earth Objectivity"?

 

[Dale]

The choice of coordinates is not arbitrary when you apply physical laws that are expressed for an inertial frame. To apply the laws you must choose an inertial frame.

I mostly agree with that, but I would say "The choice of coordinates is not arbitrary when you apply THOSE physical laws that are expressed for an inertial frame. To apply THOSE laws you must choose an inertial frame."

Not all physical laws are expressed in terms of an inertial frame. For instance, the EFE is expressed in terms of tensors, and Lagrangian mechanics is expressed in terms of generalized coordinates. So you need to know for each specific physical law you are using whether or not the choice of coordinates is arbitrary. If the choice is not arbitrary for that specific law then, as you said, you cannot use a non-inertial frame with that law.

 

[DrGreg]

you can find a locally-approximately-inertial frame.

"Near Earth Objectivity"?

I don't understand what that means.

 

[jerromyjon]

Objective permanence for the agoraphobic. Look up at the sky and see through the manifolds... kind of like a joke I guess. I wasn't sure what to think it meant but in my endeavors I finally envisioned a manifold in spacetime so its all good.

 

[CKH]

Not all physical laws are expressed in terms of an inertial frame. For instance, the EFE is expressed in terms of tensors, and Lagrangian mechanics is expressed in terms of generalized coordinates. So you need to know for each specific physical law you are using whether or not the choice of coordinates is arbitrary. If the choice is not arbitrary for that specific law then, as you said, you cannot use a non-inertial frame with that law.

I rather expected that when I posted. Unfortunately I'm not up to speed on those subjects (tensors, Lagrangian mechanics and generalized coordinates) nor the physical law that uses them. So I cannot appreciate yet how physical law can be stated independently of any background conditions in which the law is said to apply.

.

 

[CKH]

So, it is just a preference for a certain system of coordinates. For example a preference based on the simpler math in those coordinates.

I don't see how you can say that. What I'm talking about is how a physical law is stated. If it's stated for an inertial frame (which is not unusual) then you cannot apply it directly to a non-inertial frame. That is not a matter of preference, it is a requirement. You may have some freedom to choose a particular inertial frame but you have no freedom to apply the law to non-inertial frames.

Take for example :

This law cannot be applied to non-inertial frames, it's not just a matter of convenience.

 

[CKH]

In General Relativity there are no such things as truly inertial frames. However in a small enough region, where the tidal effects of gravity are negligible, you can find a locally-approximately-inertial frame.

Is there a name for that beast? Apparently by definition an "Inertial frame" is global (over all time and space) rectilinear motion. Therefore we cannot say that the space station is at rest in any inertial frame. However, the motion of the space station is inertial in that it has nearly 0 proper acceleration.

How do we properly express this in physics? Do we say "local inertial frame" perhaps "local inertial space" or perhaps "momentarily comoving inertial rest frame". Momentary isn't particularly satisfying because whatever we name this frame it retains it's inertial motion (in the sense of 0 proper acceleration) over time. Can we call it's worldline an "inertial path"?

 

[A.T.]

...If it's stated for an inertial frame...

Here is where the preference comes in. We prefer to state the laws for inertial frames, because that is simpler.

 

[A.T.]

Apparently by definition an "Inertial frame" is global (over all time and space) rectilinear motion.

Not in curved space time. There it only applies to a small region approximately.

Therefore we cannot say that the space station is at rest in any inertial frame.

For one small object that creates only negligible gravity itself you can say this. Build the space-station bigger, and some parts will experience non-zero proper acceleration, so it its not entirely inertial. And when you have two massive celestial bodies, there is no way you can define global inertial coordinates, that have zero proper acceleration everywhere.

 

[CKH]

Here is where the preference comes in. We prefer to state the laws for inertial frames, because that is simpler.

OK. I'm curious about what the alternative might be. How would you express the physical law below for any arbitrary frame? Or does each frame have it's own different physical laws? The latter concept is not an attractive approach to defining physical law.

Whenever you mistreat a non-inertial frame as inertial and apply the above physical law, fields appear that are not real, that is they have no existence independently from that frame.

Re global inertial frame:

Not in curved space time. There it only applies to a small region approximately.

Right. So the strict definition of an "inertial frame" is something that only exists in principle. As I mentioned we need a term for "local inertial frame". Is that the term that is standard? For the space station there is an instantaneous local inertial frame, but there is also is also continuous inertial motion. If you consider the worldline of the space station, a small region of space around the worldline is "continuously inertial" in that proper acceleration approaches 0. Does such a worldline have a name? E.g. inertial path, geodesic path in spacetime?

 

[A.T.]

The latter concept is not an attractive approach

Hence we prefer not to use it

Does such a worldline have a name? E.g. inertial path, geodesic path in spacetime?

That’s how I would call it.

 

[Dale]

I rather expected that when I posted. Unfortunately I'm not up to speed on those subjects (tensors, Lagrangian mechanics and generalized coordinates) nor the physical law that uses them. So I cannot appreciate yet how physical law can be stated independently of any background conditions in which the law is said to apply.

.

No problem. This way at least you are aware of their existence, if not their mechanics.

Your point is correct, just not universally applicable. For laws expressed in terms of inertial frames you certainly do not have the option to use them in a non inertial frame as is.

For laws expressed in a covariant form or in terms of generalized coordinates the above doesn't apply. Luckily, most laws of physics have a known formulation or generalization in one of these types of structures.

 

[CKH]

Hence we prefer not to use it

You did not address my question directly. You claim that we just "prefer" to describe this physical law in an inertial frame. So what is an alternative expression of the physical law?

Perhaps you would say, "instead of using an inertial frame for this physical law, I will define a more complicated physical law for a frame fixed to the surface of the earth"?

How would you do that? I.e. how would you derive an equation for the corresponding physical law of inertia in this frame?

Do you believe that fictitious forces should be considered as having physical existence? (To me that's like believing "my shadow has an independent existence" or "when I turn my head it is equally valid to say my head is still, but my body twisted and caused the entire universe to accelerate, rotate about my head and then stop again".)

 

[A.T.]

Perhaps you would say, "instead of using an inertial frame for this physical law, I will define a more complicated physical law for a frame fixed to the surface of the earth"?

For example.

How would you do that? I.e. how would you derive an equation for the corresponding physical law of inertia in this frame?

Why derive? Laws can be simply postulated based on observation.

Do you believe that fictitious forces should be considered as having physical existence?

That's a vague philosophical question of little interest to me.

 

[stevendaryl]

OK. I'm curious about what the alternative might be. How would you express the physical law below for any arbitrary frame? Or does each frame have it's own different physical laws? The latter concept is not an attractive approach to defining physical law.

Whenever you mistreat a non-inertial frame as inertial and apply the above physical law, fields appear that are not real, that is they have no existence independently from that frame.

There is a covariant way of expressing $\vec{F} = \frac{d\vec{p}}{dt}$. First, introduce a scalar parameter, $s$, and a trivial equation of motion for $t$ itself:

Let $t(s)$ be any linear function satisfying

$\frac{d^2 t}{ds^2} = 0$

Next, we make the equations 4-D instead of 3-D by making $t$ into a coordinate:

$x^0 = t, x^1 = x, x^2 = y, x^3 = z$
$F^0 = 0, F^1 = F_x, F^2 = F_y, F^3 = F_z$

Now define the "velocity" with respect to the parameter $s$ as follows:

$U^\mu = \frac{d x^\mu}{ds}$

In terms of $U^\mu$, we have:

$m \frac{d}{ds} U^\mu = F^\mu$

Then, to make it covariant, we just have to replace $\frac{d}{ds}$ by the covariant notion of the directional derivative with respect to the parameterized path $x^\mu(s)$:

$m D_s U^\mu = F^\mu$

or in terms of connection coefficients:

$m ( \frac{d}{ds} U^\mu + \sum_{\nu \lambda} \Gamma^\mu_{\nu \lambda} U^\nu U^\lambda) = F^\mu$

This equation has the same form in any coordinate system (although the values of the coefficients $\Gamma^\mu_{\nu \lambda}$ are different in different coordinate systems.

 

[CKH]

There is a covariant way of expressing $\vec{F} = \frac{d\vec{p}}{dt}$. First, introduce a scalar parameter, $s$, and a trivial equation of motion for $t$ itself:

This equation has the same form in any coordinate system (although the values of the coefficients $\Gamma^\mu_{\nu \lambda}$ are different in different coordinate systems.

Thanks. I wish I understood your equations better (the notations are quite unfamiliar), but perhaps I get the point. The equation while not written for a particular coordinate system has embedded in it a parameterized coordinate system. In order to use the equation for a coordinate system you plug in some coefficients? It does not seem that any finite set of scalar "coefficients" would be enough. So I have some more questions:

Don't you need a complete mapping of spacetime to some other spacetime?

How would that mapping be described? In other words, given an arbitrary coordinate system, how do you find these "coefficients"?

What are these coefficients in a inertial frame?

Is the force in that last equation (they right side) a real force or does it include fictitious force?

 

[CKH]

That's a vague philosophical question of little interest to me.

You are mistaken that it is a philosophical question. It is a physical question. Fictitious force fields are called "fictitious" for a real physical reason. They exist only in the frame in which they are measured. They are entirely accounted for by the frame's proper acceleration and have no independent cause nor independent existence. That is, fictitious forces have no frame-independent reality, any more than your shadow has a separate reality from yourself . Your shadow is entirely caused by your own position and motion relative to the sun, it has no independent existence. That fact (for both your shadow and the fictitious forces that accompany your non-inertial motion) is easily verified by experiment. We are easily disabused of the notion that our shadow exists without us, we should be as easily disabused of the notion that fictitious forces exist around us in our environment.

 

[stevendaryl]

Thanks. I wish I understood your equations better (the notations are quite unfamiliar), but perhaps I get the point. The equation while not written for a particular coordinate system has embedded in it a parameterized coordinate system.

I don't think that's quite the way I would put it. The point is that the equations only vary from one coordinate system to another through the "connection coefficients" $\Gamma^\mu_{\nu \lambda}$. Those coefficients have different values in different coordinate systems. Essentially, these coefficients record the "fictitious forces" associated with the coordinate system.

In order to use the equation for a coordinate system you plug in some coefficients? It does not seem that any finite set of scalar "coefficients" would be enough.

For an specific coordinate system, there are only finitely many connection coefficients, of the form $\Gamma^\mu_{\nu \lambda}$. There are 64 coefficients, in general, although in many cases, you find many of those are zeros and others are repeats. Except that the coefficients can vary from location to location, so in that sense, there are infinitely many parameters.

So I have some more questions:

Don't you need a complete mapping of spacetime to some other spacetime?

No, you don't need anything more than the connection coefficients associated with a specific coordinate system. This is equivalent to knowing how vectors transport from place to place (that is, knowing which vectors over there are parallel to which vectors over here.)

How would that mapping be described? In other words, given an arbitrary coordinate system, how do you find these "coefficients"?

The connection coefficients can be found through parallel transport, which can be determined empirically (at least, if you make a few assumptions about what is "force-free" motion).

What are these coefficients in a inertial frame?

The special fact about Cartesian coordinates in an inertial frame is that the connection coefficients are all zero. It's not enough that the frame be inertial--spherical coordinates have nonzero connection coefficients even in an inertial frame.

Is the force in that last equation (they right side) a real force or does it include fictitious force?

No, the force on the right is "real" forces, and all the "fictitious" forces are captured by the term

$m \Gamma^\mu_{\nu \lambda} U^\nu U^\lambda$

Fictitious forces are all velocity-dependent, even though that isn't always apparent.