Heliocentrism from a relativistic standpoint

[Austin0]

Britannica:


At the surface of the Earth closest to the Moon, the Moon's gravity is stronger than the centrifugal force. The ocean's waters, which are free to move in response to this unbalanced force, tend to build up a small bulge at that point. On the surface of the Earth exactly opposite the Moon, the centrifugal force is stronger than the Moon's gravity, and a small bulge of water tends to build up there as well.

http://scijinks.jpl.nasa.gov/weather/howwhy/tides/


The Moon's gravitational pull on the side of Earth nearest the Moon is strong enough to overcome the centrifugal force and pull the oceans toward the Moon. The Moon's gravity is tugging on the far side of Earth too, but because that side is farther away, the Moon's gravity is too weak to overcome the centrifugal force. Thus the oceans bulge on that side as well.

Aren’t these statements made as if the authors were speaking from the (accelerating and not part of a rotating) reference frame of the Earth?

Hi Saw

I have a question: Why would the gravitational pull on the moon side be acting in opposition to the centrifugal force?
Rather than acting in conjunction with the centrifugal force in opposition to the centripetal force of the Earth's gravity?
It does seem to be considered from a frame where the Earth is not rotating , otherwise the concept of centrifugal force would not make sense.

 

[D H]

The other alternative is what I was calling the Earth or the Sun frame. These two frames, as viewed from an inertial frame, are accelerating (just like the rotating frame) but (unlike the rotating frame) they do not have points at rest with each other ...

They're accelerating frames. Accelerating frames and rotating frames are different beasts. Did you read post #25? There is no centrifugal force in these frames because they are not rotating frames.

I say so because then, on that basis, we could still talk about gravity and centrifugal force as a way of describing tidal effects. We could thus explain, for example, the bulge of water on the surface of the Earth farther from the Sun as follows:

- From the reference frame of the Sun, as arising from the non-uniformity of the gravity field across distance: at that location the force is slightly weaker than the force at the center of the Earth (because of longer distance from the center of the Sun) and that is why the water is centripetally accelerated less.
- From the reference frame of the Earth, as arising from the fact that at that location centrifugal force overcomes gravity force.

The first explanation works even more clearly from an Earth centered frame. The latter explanation happens to work, but it is wrong. Suppose some alien spacecraft with the ability to violate Newton's first law gets stranded in the vicinity of the Earth. This spacecraft has the ability to steal linear momentum from a nearby object orbiting a star. The vehicle is stranded; what can it do? Simple: It can steal all of the Earth's linear momentum, making it plunge sunward. The Sun will still cause tides. In fact, the tides will get higher and higher as the Earth gets closer and closer to the Sun. The centrifugal force argument completely fails in this case. The tidal gravity explanation still works in this example.

 

[Saw]

I have a question: Why would the gravitational pull on the moon side be acting in opposition to the centrifugal force? Rather than acting in conjunction with the centrifugal force in opposition to the centripetal force of the Earth's gravity?

Hi Austin0, I don’t know how I have ended up here defending fictitious forces, when really I hate them. Everything is easier if you analyze things from an inertial frame or at least an almost inertial one for our purposes. Let me consider as such for this comment the Sun.

I visualize it as follows. The Earth has a tangential velocity in a straight line (inertial motion). If left alone, it would fly away. Now we introduce gravity. Think of one single pull from the Sun, as if that were possible. The composition of the vector tangential velocity and the vector centripetal velocity imparted by such pull would be a diagonal path. If the ocean water is closer to the Sun, then gravity force is stronger there and the second vector is of higher magnitude. This should result in a compounded vector with a steeper slope, closer to the center of the Sun. Now we allow for the fact that gravity is constantly there (BTW, with or without time gaps between its pulls?). The composition of all the little imaginary diagonal paths is the orbital motion, but in the case of the water closer to the Sun. But that’s the explanation I gave myself. Experts may correct if I went wrong.

Did you read post #25?

Yes, I read all your posts. They were very helpful, thank you.

Apparently, you are telling me that in, for example, the Earth frame I’m talking about, which

is neither an inertial frame nor a rotating frame. It is an accelerating frame with non-rotating axes.

,

fictitious forces arise due to the Earth's acceleration toward the Sun

It’s only that you do not reserve any role in that fictitious force for the concept of “centrifugal force” but note that

They are essentially tidal gravity forces.

Well, if you promise not to go into “Rant mode”, I’ll tell you frankly what your view is suggesting me.

I interpret that, contrary to the authors I have quoted, you are taking the easy-going way of explaining things in an accelerated frame.

For example, you are driving a truck with a tank half-full of water. You suddenly step on the brake and friction of the tyres with the ground makes the car decelerate. The water keeps going forward by inertia and spills over the cabin. However, if you still want to keep your accelerated truck as a reference, as if it were an inertial frame, even if it is not, in order to preserve the validity of Newton’s laws, you could say that a fictitious force equal to the mass of the water times its acceleration has pushed it forward. Here there’s no more complicated way of explaining things.

But what if, instead of stopping suddenly, you do it progressively, thus acquiring uniformly accelerated motion? The water, in my opinion, would form a constant bulge on the front part of the tank. Why? Because it would collide with the front wall of the tank and tend to recede, but then the truck’s speed would diminish more, thus reproducing the effect of collision with the front wall, and so on. The easy-going way of explaining this in the frame of the truck would be to say that there is a fictitious force making the water form precisely that bulge. Since here the attraction of the water towards the front wall of the tank is caused by the friction of the tyres with the ground, which has decelerated the tyres and communicated this new state of motion to the rest of the truck through its components and ultimately reached the front wall of the tank, we could say, mimicking your nomenclature, that in the truck frame the water has been subjected to a “contact tidal force”. And why does this force have this magnitude, which causes this effect and not another? Why does the water not spill over the cabin…? I suppose you know, I don’t, on the basis of that explanation. I would understand things better with the explanation of the quoted authors, who would take an apparently more complicated approach, but more explanatory in my view. They would say that in the truck frame the bulge of water is due to the net result of offsetting the contact decelerating force against a truck-fugal fictitious force.

I think this answers the example you have proposed:

The first explanation works even more clearly from an Earth centered frame. The latter explanation happens to work, but it is wrong. Suppose some alien spacecraft with the ability to violate Newton's first law gets stranded in the vicinity of the Earth. This spacecraft has the ability to steal linear momentum from a nearby object orbiting a star. The vehicle is stranded; what can it do? Simple: It can steal all of the Earth's linear momentum, making it plunge sunward. The Sun will still cause tides. In fact, the tides will get higher and higher as the Earth gets closer and closer to the Sun. The centrifugal force argument completely fails in this case. The tidal gravity explanation still works in this example.

Think of it in terms of my example: in yours gravity plays the part of the contact force in mine (contact with the ground and with the front wall of the tank) and a fictitious inertial force pulling away from the Sun plays the part of the truck-fugal force.

You say, “The Sun will still cause tides. In fact, the tides will get higher and higher as the Earth gets closer and closer to the Sun.” Of course, nobody discusses what happens, it must be the same for all frames.

“The centrifugal force argument completely fails in this case”. Just call it a Sun-fugal force and it should also work here.

 

[D H]

I visualize it as follows. The Earth has a tangential velocity in a straight line (inertial motion). If left alone, it would fly away. Now we introduce gravity. Think of one single pull from the Sun, as if that were possible. The composition of the vector tangential velocity and the vector centripetal velocity imparted by such pull would be a diagonal path. If the ocean water is closer to the Sun, then gravity force is stronger there and the second vector is of higher magnitude. This should result in a compounded vector with a steeper slope, closer to the center of the Sun. Now we allow for the fact that gravity is constantly there (BTW, with or without time gaps between its pulls?). The composition of all the little imaginary diagonal paths is the orbital motion, but in the case of the water closer to the Sun. But that’s the explanation I gave myself. Experts may correct if I went wrong.

That's just wrong. Note well: You did not invoke the centrifugal force in this explanation. More on centrifugal force below.

Here is the tidal gravity explanation. The gravitational acceleration toward the Sun experienced by some object located at a point \mathbf r' from the center of the Sun is, from the perspective of an inertial frame,

\mathbf a_{\text{inertial}} = - \,\frac{G M_{\text{sun}}}{r'^3}\mathbf r'

The Earth itself is accelerating toward the Sun. Because the oceans move more or less with the Earth, it is the relative acceleration with respect to the Earth that drives the tides. Denoting the position of the Earth with respect to the Sun as \mathbf R,

\mathbf a_{\text{relative}} =<br /> - GM_{\text{sun}}\left(\frac{\mathbf r}{r^3} - \frac{\mathbf R}{R^3}\right)

Finally, express the position of the object relative to the center of the Earth rather than the center of the Sun. Defining \mathbf r = \mathbf r&#039;-\mathbf R,

\mathbf a =<br /> -GM_{\text{sun}}\left(<br /> \frac{\mathbf R + \mathbf r}{||\mathbf R + \mathbf r||^3} -<br /> \frac{\mathbf R}{||\mathbf R||^3}<br /> \right)

Note that the position \mathbf r[/itex] and acceleration \mathbf a[/itex] are relative to the center of the Earth. The term GM_{\text{sun}}/R^3\,\mathbf R is a fictitious force.&lt;br /&gt; &lt;br /&gt; Because r&amp;lt;&amp;lt;R, to first order this becomes&lt;br /&gt; &lt;br /&gt; \mathbf a \approx -\frac {GM_{\text{sun}}}{R^3}(\hat r - 3 \cos\theta \hat R)r&lt;br /&gt; &lt;br /&gt; Here \hat r[/hat] and \hat R are unit vectors from the center of the Earth toward the object in question and toward the Sun. r and R are the magnitudes of the vectors \mathbf r and \mathbf R.&lt;br /&gt; &lt;br /&gt; &lt;blockquote data-attributes=&quot;&quot; data-quote=&quot;&quot; data-source=&quot;&quot; class=&quot;bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch&quot;&gt; &lt;div class=&quot;bbCodeBlock-content&quot;&gt; &lt;div class=&quot;bbCodeBlock-expandContent js-expandContent &quot;&gt; It’s only that you do not reserve any role in that fictitious force for the concept of “centrifugal force” &lt;/div&gt; &lt;/div&gt; &lt;/blockquote&gt;Physicists are very careful with nomenclature, Saw. Centrifugal force is a very specific thing: It is the fictitious force&lt;br /&gt; &lt;br /&gt; -m\boldsymbol \Omega \times \mathbf r&lt;br /&gt; &lt;br /&gt; where \boldsymbol \Omega is the rate at which the reference frame is rotating with respect to the remote stars. The centrifugal force is obviously not constant with respect to position. Assuming a constant rotation rate, the centrifugal force at a fixed point is constant over time. Compare this to the fictitious force due to the Earth as whole accelerating toward the Sun. This force is constant with respect to position but varies with respect to time. It doesn&amp;#039;t look anything like the centrifugal force.

 

[Saw]

Ok, it seems I've been misled by some texts. I'll study that. Regards.

 

[Saw]

Physicists are very careful with nomenclature, Saw. Centrifugal force is a very specific thing: It is the fictitious force

-m\boldsymbol \Omega \times \mathbf r

where \boldsymbol \Omega is the rate at which the reference frame is rotating with respect to the remote stars.

I'm studying your text. Shouldn't \boldsymbol \Omega be squared?

 

[D H]

Good catch. However, it is not omega squared. It is

-m\boldsymbol \Omega\times (\boldsymbol \Omega \times \mathbf r)

which reduces to

-m \Omega^2 r

in the special case of \mathbf r being normal to \boldsymbol \Omega.

 

[Saw]

it is not omega squared. It is

-m\boldsymbol \Omega\times (\boldsymbol \Omega \times \mathbf r)

which reduces to

-m \Omega^2 r

in the special case of \mathbf r being normal to \boldsymbol \Omega.

Ok, but according to the sources I am reading, if we make the analysis of the Earth-Sun system from a rotational frame (I talk now of acceleration instead of force):

- The acceleration due to the fictitious centrifugal force can be decomposed into two components: ω^2 * c (c being the distance between the center of mass of the Sun-Earth system and the Earth center) and ω^2 * r (r being the distance between the center of the Earth and the relevant point of the Earth where we're analysing the tidal effect).

- The component ω^2 * r cancels out somehow.

- So we are left exclusively with a "uniform" (the same size in all points of the Earth) centrifugal acceleration of intensity ω^2 * c.

Is that correct, in your opinion? Thanks.

 

[Saw]

Well, after deleting a last post with some blunder, I come back with a better understanding now.

Your analysis, DH, is clear:

Let us say M = mass of the Sun.
N and D are two points on the surface of the Earth separated from its center by the radius (r) of the Earth, N (from near) being the one facing the Sun and D (from distant) the one facing away from the Sun, at a given moment.
R = the distance from the center of the Sun to the center of the Earth.
CM = center of mass of the Sun-Earth system.
c = distance from the CM to the center of the Earth.

- In an inertial frame with origin at the CM (disregarding its acceleration due to other objects in the universe), objects at N and D would tend to be accelerated as follows:

* For N, a = - GM/(R^2 - r)
* For D, a = - GM/(R^2 + r)

Since the frame is inertial, these accelerations are due to "real" forces.

- In an accelerated frame with origin at the Earth center, we have to add to the real force a "fictitious" force pointing away from the Sun = + GM/R^2. Thus the accelerations would tend to be:

* For N, a' = - GM/(R^2 - r) + GM/R^2 = - net result, away from Earth center and toward the Sun
* For D, a' = - GM/(R^2 + r) + GM/R^2 = + net result, away from the Earth center and away from the Sun

However, all this is the analysis at a frozen moment, an ideal instant. It's not strange, hence, that the resulting net force is the same as if the Earth were falling in a straight line toward the Sun.

But here it is not that simple. To describe the kinematics from the simplest perspective, that of the inertial frame, the Earth is orbiting the CM, always with a fixed orientaton of its axes wrt the fixed stars: the center of the Earth is translating in a circle about the CM, whereas Points N and D are also translating in circles with the same radius (c), albeit different centers.

One solution could be what I tried to do in post #33. Why is that wrong? There I was taking the frame of the Sun, assuming that its motion around the CM is negligible and hence inexistent. But to be more precise, let's take the frame with origin at the CM (also disregarding its own acceleration due to gravitational interaction with the rest of the universe). Isn't the Earth's acceleration, wrt the CM and after a time lapse, the result of compounding the gravity force toward the Sun with the Earth's inertial tangential motion?

Another solution could be shifting to the perspective of a rotating frame and then transforming again back into the accelerated Earth frame or the inertial frame. I've got lost both in the rotating frame and in the way back. Does anyone know how to do it?

In any case, is the final result, wrt the Earth center, the following?

* For N, a' = - GM/(R^2 - r) + ω^2c
* For D, a' = - GM/(R^2 + r) + ω^2c