Roger's Equations

If energy is conserved, friction is ignored, and the antarctic is 3000 m higher than the north pole, wouldn't the train come popping out of the ocean on the north pole side or need braking? i.e. higher potential energy at the south than north pole.

You Wrote: "If energy is conserved, friction is ignored, and the antarctic is 3000 m higher than the north pole, wouldn't the train come popping out of the ocean on the north pole side or need braking? i.e. higher potential energy at the south than north pole."

Yes, which is why I wrote "We'll also pretend the water from the North Pole wont pour into our tunnel and we'll try to forget that the South Pole is at an elevation of over 3000 meters (thanks to the ice) and the North Pole is at sea level." Seriously, of all the assumptions I made, that was the one that bothered you?

The radius of the Earth is 6,380 km so 3 km is roughly 0.47 % of that. So the gravitational force felt at the South Pole would be 1.0047 times larger than at the North Pole.

Sorry--tried to forget but just couldn't!

I was asking for a lot of assumptions. No air, no water, no magma, no iron core, no friction, no deviation from spherical, etc. Inevitably the common sense side of your brain will step in and say "hang on a sec".

Still, the result is interesting. I always hated the general solutions of differential equations, but once you get the hang of boundary conditions, they are really flexible. We can assume an initial velocity, no initial velocity, starting from R, starting from somewhere else, etc.

I never cease to be amazed how some folks read a problem and base their answers to include factors excluded in the OP!

There has been a loooong discussion on a well known BBS in the windy city over a question involving a somewhat similar.

In a similar idealized hole through the earth what would be the air pressure at the center? None of the replies considered the effect of gravity if any.

Pardon me, but if the 'train` is in free fall, whence cometh the

"accelleration felt by the passengers" ????

Its true that the first half of the trip from the South Pole to the North Pole (or vice-versa) is in free fall. However, the second you pass the center of the Earth, you are no longer in free fall, the train begins decelerating until it finally stops at the North Pole (or South Pole). This is where the passengers "feel" the acceleration, or to be more colloquial, the deceleration.

Hi Roger, nice puzzle!

I'm afraid I don't agree with your statement "However, the second you pass the center of the Earth, you are no longer in free fall, the train begins decelerating until it finally stops at the North Pole (or South Pole)."

The only time that the train is not in free fall is when it is "caught" at the South- or North-Pole. In gravitation theory a free fall is where no other force than gravity influences an object's dynamics and it does not matter whether the object is "falling" upwards or downward.

Another way to put it, the passengers, in the idealized situation that you sketched, will be weightless all the way until they are "caught" at the station for embarking/disembarking.

Regards, Jorrie

Jorrie,

This is a tough one, I guess I have three thoughts on this:

1. The original post indicated that the passengers were in freefall and therefore felt no acceleration. Perhaps I could have better responded by pointing out that an object in free fall does experiences 1g of acceleration, so the conclusion wasn't valid regardless as to whether the premise was correct.

2. Since the train is falling for only half the trip and is rising once it passes the center of the Earth, I don't think its accurate to say the train (and passengers) are in freefall the entire time.

3. Freefall in the general sense implies a constant acceleration of g, although this is not the scientific definition. Its worth pointing out that the train does not feel a constant acceleration but rather an acceleration that varies from 1g to -1g sinusoidally over the course of the trip.

I feel that the only time the passengers would not be subject to any G forces would be when their velocity is not changing i.e when they are passing through the centre of the earth, I think that as they are increasing speed on the downward part of the trip and losing it on the upward part the g forces on them would vary in a sinusoidal manner.

of course the forces on the train carriage would vary in the same manner so they would need seat belts to avoid floating about.

.

While hunting for information on this subject I came upon an interesting point, the gravity train does not have to travel through the centre of the Earth, if a straight tunnel was cut between London and Glasgow and equipped with frictionless rails the train could still make the trip in 42 minutes propelled only by gravity

Couldn't London to Glasgow be speeded up by making the tunnel a catenary?

No the only way it could be speeded up is by re-making the Earth with more dense material

second thoughts, all these calculations have been based on a uniform density sphere but of course the density of the Earth increases as you go down so there might be some advantage in taking a different route but the mathematics of it are beyond me.

Hi Syphrum

Perhaps I should have said that it definitely could be speeded up by taking a different route, although the optimum would not be a catenary (because of the variable density o fthe Earth).

Take an extreme case - a level distance of 10-metres. If the path is straight, the time will be 81-minutes (just like between any other two places). If we allow the path to go down five metres, then bend across for the 10-metres, and then come up, the time will be just over three seconds. Of course, on this small scale gravity is almost invariant, so the optimum will actually be very close to a catenary...

Regards

Fyz

syhprum, you wrote: "I feel that the only time the passengers would not be subject to any G forces would be when their velocity is not changing i.e when they are passing through the centre of the earth, ..."

No, passengers in an idealized, zero friction, free falling 'train' are not subject to any g-forces in the usual sense of the word - they experience weightlessness. The fact that they are accelerating/decelerating towards/away from the center of Earth does not make it any different from an astronaut drifting around in free space in a weightless condition.

The reason for this is that their train couches are accelerating at precisely the same rate as their bodies, so they feel a complete lack of force. Interestingly, the train itself, because of its (presumably) considerable length, will experience tidal gravity effects. The effects will put some 'stretching' strain on the train, except when it passes the centre, when the net strain will be compressive!

Jorrie

Jorrie,

There is a difference between the passengers on the train and a person in orbit. The person in orbit feels a relatively constant acceleration. They are constantly falling towards the Earth at 1g or close to it. The people on the train experience a change in acceleration from 1g to -1g. Still, your point probably still stands.

I think we all agree the people and the train are accelerated from 1g to -1g, that is certain. The question being discussed is "what would the passengers "feel"".

Forgetting our free-fall discussion, which is just a vocabulary thing anyway, from a relativity standpoint, does the fact that the acceleration is changing, at a varying rate, make any difference? I don't know the answer to this but I'm pretty sure we are in relativity territory and you would know.

Syphrum said "subject to". He also recognised that they would feel weightless throughout. I think he used the words correctly.

Hi Roger, your "2. Since the train is falling for only half the trip and is rising once it passes the center of the Earth, I don't think its accurate to say the train (and passengers) are in free fall the entire time." is a bit off standard definition.

Wikipedia defines "free-fall" as follows: "Free fall in its strictest sense is the condition of acceleration which is due only to gravity. In other words, the objects undergoing free fall experience only one force: their own weight."

The Moon is in a constant condition of free fall around Earth and the Earth/Moon around the Sun. The fact that they are in elliptical orbits means that at times they are moving farther from the source of gravity, just like you 'train' moves away from Earth's center for the 2nd half of the trip.

Free fall does not imply a constant acceleration. It happens whenever there is only gravity working on a body, however weak it is!

Jorrie,

I would think by now you would have known the first source I checked after your comment was Wikipedia. I love that site. However, this is one time where I have to disagree with it, as much as it hurts me to do so (forgive me Wiki). Wiki got the definition verbatim from Websters and I think Websters has it wrong.

Falling is falling. I'm just not convinced that shooting a cannon straight up and saying the cannonball is in "freefall" while its rising is correct (even though the only force acting on it is gravity).

Here is a link that seems to support wiki's definition, until the second bulleted point contradicts it:

http://www.glenbrook.k12.il.us/gbssci/Phys/Class/1DKin/U1L5a.html

An alternate definition that can be found is

"The ideal falling motion of a body that is subject only to the earth's gravitational field."

Note the "falling" part of the definition, which I believe Webster "implied".

I think (hope) the confusion is purely semantic. Whatever it's origin, there should be no question about the following*: the passengers will feel weightless (just like being in a satellite or in a freely-falling object, however widely or narrowly defined).

P.S. Just like the feeling, if you resolve the acceleration of an orbiting satellite, you will see that the time must be the same as you'd take when orbiting the earth at sea level. The input energy requirement, of course, would be quite different...

*as implied by Jorrie

Fyz

That's quite a definitive statement! Again I point out that a satellite experiences a constant acceleration (around 1g) whereas the acceleration of the train varies, so its not legit to simply say whats true for one is true for the other. Clearly they are not the same situation.

Now it may very well be the case that in both instances there is weightlessness, but one scenario does not prove the other, since they are not equivalent. That's why I reject the term "freefall", as it is being used to describe two different situations and makes them seem to be the same.

I don't mind you telling us so positively that the passengers will feel weightlessness, I suspect you're correct, but please offer a better explanation as to why.

That was not an explanation, just a likeness. The explanation is the same for all cases, however: in none of the cases is there any force opposing gravity - the objects simply accelerate as gravity dictates. The effect of "weight" is provided by forces between parts of the object (compressive or tensile) and some external constraint; as no such forces are present, no weight is experienced.

I'm convinced that they are weightless during the trip. I agree that the seat never applies a force to the body since they are both in freefall. I just would like to know if you feel anything when accelerated. I suspect the answer in no, but I'd like an explanation.

Hi Roger, I think Physicist has covered it pretty well and besides, in my nick of the woods it's almost bedtime, so I will stay out of serious technicalities...

In Newton's gravitation, the inertial mass and the gravitational mass of objects are identical, hence all objects fall with the same acceleration in the same gravitational field. This explains why one does not feel forces while free falling in vacuum (except the inevitable tidal forces...)

One of your other queries was (paraphrased): "Falling through Earth would cause variable acceleration while orbits are at constant acceleration." Not quite: elliptically orbiting objects experience variable acceleration all the way and also weightlessness all the way! It is essentially the same thing as falling through Earth, or shall we make it "through the asteroid"?

Finally, the semantical difficulty with "free fall" one can ignore, because it makes no difference to any predictions. The calculations turn up the same answers - if done correctly!

Jorrie

You Wrote: One of your other queries was (paraphrased): "Falling through Earth would cause variable acceleration while orbits are at constant acceleration." Not quite: elliptically orbiting objects experience variable acceleration all the way and also weightlessness all the way! It is essentially the same thing as falling through Earth, or shall we make it "through the asteroid"?

The variation in acceleration is much much larger than experienced by the train as compared to a spacecraft in an elliptical orbit.

That said, I agree that there would be weightlessness for the people on the train during the trip.

I did say they would need seat belts to avoid floating about!

Sorry syphrum, I skim-read and missed that. The difficulty remains to convey this truth to Roger P.

Regards

Fyz

Most of the details are worked out, we are in a very exact discussion right now because I was probably incorrect in my statement that the passengers would "feel" 1g. Of course, if I'm to accept being wrong, I want at least a correct answer with a quantitative explanation, which is where the conversation is right now. My objections are:

• Saying falling through the Earth is the same as being in orbit is not correct. One has a varying acceleration, the other a constant one.
• Something that is rising is not in freefall. I don't care what websters ill conceived definition is.
• Something in freefall certainly experiences a force, otherwise it wouldn't accelerate. So the passengers experience force, but do they "feel" it.
• Weightlessness means that if you stood on a scale, you wouldn't weigh anything. I'm not convinced that weighlessness implies we wouldn't feel anything.

An interesting discussion, I hope we get to the bottom of it.

You certainly would feel something, at least initially - pretty sick. But you wouldn't "feel" the acceleration or weight except by observing your velocity against other objects. If the train had no windows, and there was no air resistance, there would be nothing for you to see or hear.

This weightlessness is but one of the precepts of general relativity - but actually one that was pretty-much accepted under Newtonian mechanics. Remember, that the feeling of having weight or acceleration is a result of the forces between your body and the surrounding world and forces between different parts of your body. There is no known mechanism (other than observation of external objects) that could distinguish this situation from that of being in a satellite.

"You Wrote:"There is no known mechanism (other than observation of external objects) that could distinguish this situation from that of being in a satellite."

Is that correct? This does seem to be the root of the discussion. There is a difference between da/dt=f(t) and da/dt=0. Is there nothing in the natural world that would notice the difference? The situations may be instantaneously equivalent, but otherwise are not the same. Is this never noticed by nature ever? (This is definitely relativity territory, I wish Jorrie was around)

I agree though that the objects would be weightless. I'm pretty close to convinced we wouldn't feel anything either (I agree with the no windows explanation you offered). I would really like an equation or something that showed the above statement to be true though.

Hi Roger, this thread has a momentum that takes it places faster than what one can read and/or type!

You asked: "... There is a difference between da/dt=f(t) and da/dt=0. Is there nothing in the natural world that would notice the difference? The situations may be instantaneously equivalent, but otherwise are not the same. Is this never noticed by nature ever? ..."

There are many situations in gravitation that are only momentarily (or in technical terms, rather "locally Lorentz") equivalent. One example is the equivalence between uniform acceleration and gravity - it is only true for 'infinitesimal time and space intervals'. Make the time and space intervals larger and one can detect tidal gravity in a gravitational field and not in uniform acceleration.

Falling 'through' a massive body has completely different temporal characteristics than falling towards such a body. However, in a very small lab (relative to the size of the body) and in a 'short time', you will not be able to distinguish between the two! Add to this that what you feel on the "seat of your pants" is pretty instantaneous and you have little chance!

Darn! I think this came out badly and I must go to bed... Sorry...

Jorrie

You wrote "Darn! I think this came out badly and I must go to bed... Sorry..."

I'm not sure what you mean, your explanation seems to make sense, but if you see an error in your explanation, let me know.

So it seems that what is unique about this trip for the passengers is that every molecule in their bodies is accelerated at the same rate meaning their are no normal forces at work that the passengers would "feel". Thus from the moment the motion starts to the moment the motion stops, the passengers feel nothing.

This is significant if we put it this way:

Imagine if the train had no windows. People get strapped into their seats and the lights go out. Strapped in, the passengers can feel they're hanging by the straps, since the blood rushes to their front (the seats and the train face the hole). Pitch black and suddenly the blood pressure goes away, the passengers feel nothing......for 42 minutes, then suddenly they feel as though they are sitting with their back to the ground, like on a rocket before take off, the lights come on, they leave the train and they are on the other side of the world.

Ok, lets build one.

Hi Jorrie

Guilty as charged for neglecting various effects, including the forces between the parts of the body due to the slope in the gravitational field. In this respect, I plead that I was comparing the situation with an earth-facing satellite, where the 'vertical' slope of the inertial acceleration is of similar magnitude to the slope of the field (and, needless to say, the effects are additive).

I also neglected local gravitational effects between the occupants and consequent deformation in the sides of the tunnel (would you call these tidal??). Then there are gravitational waves (someone may care to calculate how many atto-meters of height might this lose in a typical transit), eddy-current induced drag on the residual direction-finding magnet in our sculls...
Maybe we'd become sensitised to some of this if we evolved in micro-gravity for a few aeons?

Jorrie was responding to my question regarding nature and certainly wasn't suggesting the passengers would feel anything. I agree with both of you now, the passengers wouldn't feel anything, unless the train was very very long.

Think of it this way, when the front of the train reaches the center of the Earth, it experiences no acceleration due to gravity, while the back of the train experiences some acceleration due to gravity, meaning the train will act like its being pushed from behind. For someone sitting on the train, that will create a normal force between their seat and themselves, which would grow stronger the closer to the front of the train they sat. If you made the train long enough, it would be a noticeable effect. We could probably figure out how long a train it would take to produce a noticeable effect. No doubt much much too long to even be remotely realistic.

Hi Roger, your "... when the front of the train reaches the center of the Earth, it experiences no acceleration due to gravity, while the back of the train experiences some acceleration due to gravity, meaning the train will act like its being pushed from behind. ..." is not quite what would happen, me thinks.

A rigid train will accelerate at the rate of its center of mass. Since the acceleration linearly increases with distance from the center, passengers at the rear end will fall slightly too fast for the train and will experience something like when the train is braking.

Passengers near the front will fall slower than the centre of mass and will experience something like acceleration of the train. Passengers sitting on the center of mass will be truly weightless. This happens during the whole trip and not just near the center of Earth.

The effect is extremely small, in the order of 0.16 μg per meter distance from the centre of mass. Real micro-gravity! In essence, it is the opposite of normal tidal gravity outside of a massive body, which tends to stretch everything in the radial direction. Here the train will tend to be compressed radially.

Jorrie

Jorrie,

I agree about the center of mass and that would occur the whole trip. How did you get 0.16 μg per meter? I expect that value to vary during the trip, but I could be wrong. I'm realling enjoying this discussion, its not often I get to think about da/dt.

Your equation shows linear variation of total gravitation forces with radial position. Polar radius is about 6356km => about 0.16 ug/m

Fyz

You Wrote "Your equation shows linear variation of total gravitation forces with radial position. Polar radius is about 6356km => about 0.16 ug/m"

Yes thats true, the gravitational force does decrease linearly with respect to distance according to the equation:

F=(⁴/₃)πρGmr

so

F/r=(⁴/₃)πρGm=.16 μg/m

Ok, I'm caught up.

So a 150 Meter train would have a maximum of 12 μg in normal force for passengers in the very front (speeding up) and the very rear (braking), and for everyone inbetween less and at the dead center of the train, 0g. Thats interesting because it means that particles from the center of the train would be pulled towards the front and back (depending on what side of the center you start from) on the way down and pulled towards the center of the train on the way up (past the center).

The train that is supposed to be the longest in the world is 7300 meters long. Half would be 3850 meters leading to a maximum normal force experienced at the extremes of the train of 616 μg or .006 g, so even if they used the longest train in the world, the maximum force felt by any of the passengers during the trip do to tidal effects would be .006 g.

Roger, you wrote in reply to Physicist: "Thats interesting because it means that particles from the center of the train would be pulled towards the front and back (depending on what side of the center you start from) on the way down and pulled towards the center of the train on the way up (past the center)."

Not quite. Radial tidal gravity inside a solid massive body always works towards the center, because a body in free fall is squeezed, not stretched! This happens for downward and upwards free fall. Outside of a massive body radial tidals work the other way round (stretching), also for both up and down movement. In the Space Station the radial tidals amount to about 22 μg/m, away from the center of mass.

Think about it like this: the side of the train nearest to Earth's center will experience less gravitation and the side farthest more gravitation, no matter which way they move. Looking at it in other ways tends to bend the brain a bit!

Jorrie

PS: There is also a tiny tidal gravity squeeze from the sides in-wards, so particles originally distributed evenly in a container free falling down a frictionless mine shaft will tend to migrate towards the centre of mass of the container.

I wrote "Not quite. Radial tidal gravity inside a solid massive body always works towards the center, because a body in free fall is squeezed, not stretched!".

In order to avoid confusion, "towards the center" refers to the falling body's, not Earth's.

Tidal gravity is confusing enough; it does not need any help from sloppy writing!

Jorrie

Jorrie,

You Wrote: "Think about it like this: the side of the train nearest to Earth's center will experience less gravitation and the side farthest more gravitation, no matter which way they move. Looking at it in other ways tends to bend the brain a bit!"

I think the confusion is coming from my original phrase "first half of the trip" what I saying was the "first half of the one way trip" not the "first half of the round trip".

Yes, this is precisely my point and the danger of treating this as a traditional freefall question. The side of the train closest to the Earth's center for the first half of the trip to the North Pole (21 minutes) becomes the side of the train farthest from the center for the second half (21 minutes). Then on the return trip it switches again, but not because of the reverse motion, but because on the way back to Antarctica the train passes the center again, so what was closest to the center on the train becomes farthest from the center on the train and vice-versa.

Jorrie,

I definitely agree now that the particles on the train would travel towards the center of the train regardless of which side of the center of the Earth the train was on. I need an equation to understand why.

You Wrote: "Radial tidal gravity inside a solid massive body always works towards the center, because a body in free fall is squeezed, not stretched!"

Could you provide a link to your website or paste the equation so I can understand this more quantitatively. (My copy of Relativity for Engineers is currently at the office and I am not for a couple of days)

Thanks,

Roger

Roger, you asked: "Could you provide a link to your website or paste the equation so I can understand this more quantitatively".

My website does not deal with gravitation inside solid bodies, but you should have a quick look at the download from my web page on tidal gravity. Combined with your own Blog's gravitational equations, you should be able to construct one.

In my GMT+2 timezone it is a bit late on a New Year's Eve to produce an equation, but I will look at in the New Year...

Happy New Year when it comes in your zone!

Jorrie

Hi Roger, you asked: "... or paste the equation so I can understand this more quantitatively."

You are better with the fundamental equations than me, but here is my simplistic 'engineering view'.

Inside a homogeneous, spherical Earth, the gravitational acceleration at distance r from the center is, replacing 4πρ/3 with M/R³, given by: a = -GMr/R³, where M is the mass of Earth, R Earth's radius and G the usual gravitational constant.

This can also be written as: a = -GM/R² x r/R = r/R g, where g is the usual -9.8 m/s². Since this is linear in r, it is obvious that for a change Δr in r, the change in acceleration Δa is given by: Δa = Δr / R g.

So, if one considers the train's COM at radial distance r from Earth's center, and the half-length of the train as scalar L, then Δr will be L for the higher end and -L for the lower end. It is easy to see that there will be a 'g-force' working towards the center of the train. The train's rigidity prevents compression, but particles will migrate to the train's center, unless constrained.

In the relativistic view, the spacetime geodesic of the lower end of the train has less spacetime curvature than the geodesic of the train's center and the higher end has more spacetime curvature. Hence the tendency for free falling objects to move towards the train's center of mass.

Jorrie

Hi Roger

Looking at things a slightly different way from Jorrie (but with much the same results), the sign of the gradient does not change at the centre of the Earth, what changes is the sign of the average. So the force is always towards the centre of the train. An easy way to look at this is to imagine the train stationary at the centre of the Earth. It is obvious then that all forces must be towards the centre of the train.
BTW, the hole (tunnel) through the Earth means that the Earth does not exert any inward squeeze relative to the tube. There will be a rather smaller lateral component due to the mass of the train; however, the occupants will not feel this, only their own similarly small inward gravitational compression.

One more note - the equations we have used all assume a uniform Earth. Actually, the density of the core is considerably greater than the average, so the compression will be lower than stated near the surface and higher towards the centre. This does not, of course, affect the acceleration near the surface (as this is already based on an average), but it means the reduction in gravitation will be less marked near the surface - so the overall time would be shorter than calculated.

Regards

Fyz

Correction:

I wrote "the hole (tunnel) through the Earth means that the Earth does not exert any inward squeeze relative to the tube's axis". Although this is correct in a sense, it is misleading. In fact, the removal of the tube means that there will be a small net outward radial pull that grows with distance from the centre (within the tube).

[I suspect that all g-force variation within the hole would be zero if the hole was spherical]

Happy 2007 everyone.

Fyz

You Wrote: "the sign of the gradient does not change at the centre of the Earth"

I'm assuming when we are talking about the gradient, we are talking about the gradient of the potential, which is the force. That being the case, you're correct that sign doesn't change when the train passes the center, however direction of the force does.

Still, I agree that the pull on particles in the train would be towards the center for the entire trip, I can see that now (the example with the train centered at the center of the Earth made that clear).

So if you are talking about another gradient, can you provide an equation to help me understand?

Roger

I'm talking about the gradient of the gravitational force. If we call the distance north of the centre of the earth "x", then, within the (assumed uniform) earth, the potential changes as x^2. The force changes as x, and the gradient of the force (dF/dx) is constant. Because the train moves solely under the action of gravity, the net apparent force on the whole train is zero. All that we can see is the gradient of the force - which is developed as a result of the 0.16-ug/m gradient in the gravitational field. Hope this is now clearer

Regards, and a happy new year

Fyz

You Wrote: "and the gradient of the force (dF/dx) is constant"

But the Force is a vector. How can we be taking the gradient of a vector field?

I have to go, but hopefully we can pick this up tomorrow. I'm just looking for a clear mathematical explanation and I'm not seeing it (not that I doubt it is true).

Happy New Year!

The gradient of a vector field is in general a tensor2. However, in this particular case (and if we confine our attention to the North-South line passing through the centre of the Earth), the rotational symmetry means that only the single component of the tensor along the polar axis relevant. (Because the vector is always along the axis, both it and its gradient can just as easily be treated as signed scalars).

If you Google for vector derivative, you will find a number of links. Many of them present the tensor in terms of independent named coordinates (e.g. grad, div, curl etc.), which I personally find tiresome - but it is possible you might find this presentation more familiar.

BTW, was it clear that Jorrie (and I following him) were referring to the gradient of the gravitational field, and that ug stood for micro-EarthGravities?

Best regards

Fyz

Little did I realise when I jokingly suggested taking this rail trip to confirm I would really be upside down when I reached Australia that I would trigger this esoteric this discussion of gravity.

Best of new years wishes to you all

Peter

Don't mistake the vagueness for an esoteric conversation. If any of us had a clear answer in all of this, it wouldn't be going on 50 posts. I think I'm going to have a follow up blog entry on some the relativistic things we were discussing towards the end. Hopefully with a little research I can get a better understanding of this then I have, not the motion of the train, that we've had nailed down, but rather what happens to a nonpointlike object in freefall.

I agree this is not esoteric - we are considering the train and its occupants strictly from the viewpoint of Newtonian mechanics (even though some of the language we use may be modern - after all, Newtonian mechanics is fully relativistic in mechanical terms - it just doesn't satisfy the "special" constraint about the speed of light).

My perception is that your difficulty is mainly because you are aware of the implications of Vector gradients, and so have not been confident in the treatment of the (single-dimensional) movement along the Earth's axis as a (signed) scalar.

Please let me know if/when this is inaccurate; or which of the tensor formulations you are comfortable with, in which case I can generate a full expression (though you should not be surprised to find that only one of the tensor2 terms is relevant for movement along the axis (the relevant non-diagonal terms are zero)

Fyz

I'm uncomfortable because we are using words instead of equations to describe the situation within the train.

Jorrie made the statement "Radial tidal gravity inside a solid massive body always works towards the center, because a body in free fall is squeezed, not stretched!"

Now that's a pretty general statement which must have some equation associated with it. It's not that I doubt Jorrie, in fact its the opposite, I believe him completely, but I can't learn it myself if I can't see the math.

As for gradients of vectors, yes that is making me uncomfortable as well. An equation would clear much of this up. Specifically an equation describing the tidal forces within an free falling object that is not pointlike.

OK, I'll integrate Jorrie's simply stated derivative, so that you can see everything in vector format. But I won't do it here, because the available space is getting so narrow you may not be able to read it. Instead, it will go to the current end of the thread.

Fyz

Sounds good, thanks.

Hi Roger, you're quote from my post "Radial tidal gravity inside a solid massive body always works towards the center, because a body in free fall is squeezed, not stretched!" must obviously be taken in context - it is only valid inside a massive body, not outside it, where it is the opposite in the radial direction.

Look again at my simple, yet correct equation posted in #57 above, for the acceleration gradient inside a homogeneous, spherical body: Δa = Δr / R.

I'm too lazy to write the full differential equation and will gladly leave it to Fyz.

Jorrie

Jorrie,

You Wrote "Δa = Δr / R. "

I'm having trouble with your equation above because one side has units of m/s² and the other is unitless.

Roger, you wrote: "I'm having trouble with your equation above because one side has units of m/s² and the other is unitless."

Right!. When I copied and pasted from my post #57, I missed the "g".

It should be Δa = Δr / R g, solving the units issue (g is the constant -9.8 m/s²)

Jorrie

You Wrote: "It should be Δa = (Δr / R) g

Ok, so if I read this equation, it tells me that the person in the second seat is accelerated more than the person in the first seat, which to me would say that the motion isn't toward the middle of the train but toward the front of the train for the person in the second seat. But you said earlier that the acceleration is always towards the center of mass of the train, which I'm not seeing. (I believe you, I just don't see it here, which means I'm missing something or this isn't the equation I need.

Hi Roger, you wrote, "Ok, so if I read this equation, it tells me that the person in the second seat is accelerated more than the person in the first seat, which to me would say that the motion isn't toward the middle of the train but toward the front of the train for the person in the second seat."

Ok, let's first try ambiguous wording again and then put it in a simple equation (Physicist has done it in a formal way). You have to consider things relative to the trains center of gravity (c.o.g.), and not relative to the c.o.m., as physicist has pointed out and which I used wrongly before!

That's effectively what I said in post #57. Let the trains c.o.g. be at r from Earth's center. The rigid train will accelerate more than the passenger in the front of the train, with seat 1 pushing more than seat 2 and so on... If any passenger releases a free particle, it will migrate to the c.o.g. of the train and oscillate around that point.

Let us analyze this in a one-dimensional coordinate system with the trains c.o.g. permanently at rest at the origin and the x-axis aligned with the center of the tunnel. The rigid train and its seats do not accelerate in this frame. Earth's centre accelerates in oscillating fashion in the range -R ≤ x ≤ R with acceleration:

a_e = -9.8 x_e / R m/s², as shown in post #57. Here x_e is the trains instantaneous coordinate and R Earth's radius. A particle's free fall acceleration will be:

a_p = -9.8 (x_p- x_e) / R + a_e = -9.8 x_p / R m/s², which is always towards the origin, i.e., negative for positive x and visa versa.

Note: x_p- x_e is the signed distance of the particle relative to the center of Earth, used to get the particle's acceleration relative to Earth. Add Earth's own coordinate acceleration and we have the particle's coordinate acceleration.

I have not had chance to study your post #80 or Physicist's reply (I started on this about 10 hrs ago but got distracted by work), but I'm sure that those, together with this one (unless I goofed), should put this issue to rest.

Jorrie

Reply to #56 - don't know whether anything I can add will help, but there's no reason we can't differentiate force (one component, as pointed out by physicist) wrt distance. Then as force = k*x, d/dx(force) = k, i.e. rate of change of gravitational field with distance is constant thru the tunnel.

Can compare this with the tidal effect outside a body of mass M at distance r. Then gravitational field varies as r^-2 and roc of field, and tidal effect, varies as r^-3. The power of 3 explains why the influence of the Moon on tides on Earth is greater than the Sun's, despite Moon's much smaller mass.

We can definitely take the derivative of the Force with respect to distance, but I don't think it answers our question.

The derivative of the force with respect to r:

F=-(⁴/₃)πρGmr

^dF/_dr=^d/_dr[-(⁴/₃)πρGmr]

^dF/_dr=-(⁴/₃)πρGm

which gives us units of kg/s²

I'm thinking that since tidal effects are related to distance, that there should be distance in the units of the answer, which in the above there is not. This may be flawed thinking though.

Reply to Roger, #73 - it's OK, dF/dr is newton/m. Newton = kg*m*s^-2, so newton/m

and kg*s^-2 have same dimensions. Reason distance isn't there is because within the tunnel gradient of force (wrt distance), hence tidal effect, is constant. Different situation outside, as my earlier posting.

You are absolutely correct, my bad.

So dF/dr gives units of F/m.

Although it unlikely that a usable one will ever be built I can think of two devices that have the basic elements, ie a long straight evacuated tube that does not follow the curvature of the Earth.

One is the LIGO gravity wave detector installation and the other the Stanford electron beam linear accelerator.

I wonder if the designers of these devices have considered gravity train effects

LIGO is another thread. Does anyone seriously believe it can detect near-planar gravitational waves?

Guest wrote: "LIGO is another thread. Does anyone seriously believe it can detect near-planar gravitational waves?"

The answer is a positive yes, but LIGO needs a reasonably strong gravitational wave to detect it, something that may not happen in a hurry. In my mind, the idea of LIGO is more to test the technology than observation, but, should a suitably strong grav-wave happen to pass by, it will be nice.

The planned space-based LISA system should be able to routinely detect gravitational waves.

Jorrie

Jorrie

I tried some quick sums, but they got too messy without approximations. When I made different approximations (to allow a solution), the net observable effects following passage of a plane wave appeared to change direction for the different approximations. This suggests either that the effects are very small indeed, or that I am looking in the wrong place. Can you give me an accessible reference for a proper solution? Thanks in advance if you can.

Fyz

Hi Fyz, you asked: "Can you give me an accessible reference for a proper solution? Thanks in advance if you can."

Apart form the Wiki-articles on gravitational waves (gw's) (which you surely have read), the only other semi-accessible paper that I could find (and follow somewhat) is: Gravitational Waves: An Introduction, from arxiv.org, 1999.

Hope it helps - it's an intimidating subject; even Einstein struggled with gw's for a long time, first announcing them, much later denouncing them and eventually accepting them again!

Jorrie

Jorrie

Thanks

For once I had neglected to revisit Wikipedia - silly of me (the current article wasn't there when I first looked). It confirmed that the potential generators that I was considering (which were the rotationally and cylindrically symmetric configurations appropriate for a gravitational collapse) gave null outputs

Looking further, I found that the expected frequency of detectable events (massive collisions within measurable range) was one-per-sixty years - which is more than I'd thought, but not that likely to occur in my lifetime. However, I remain to be convinced that the radiative output from such events will not confuse the situation - either because there is a short-wave radiation burst that disturbs the mirrors, or because the total mass of radiation passing the Earth at the same time is more than sufficient to account for the observed effects.

Unfortunately, I fear that Pulsars are not close enough to provide a detectable signal - even taking into account the possibility of coherent detection*. (If they were, adequate data is probably already available)

*Several of these signals have correlated stabilities that are sub 1E-15Hz, so detection bandwidth would be limited only by the measurement time.

Fyz

I too believe we are straying into the territory of general relativity although the instrumentation to measure any effects would be challenging due to the low velocity relative to c.

As the velocity is the same as orbiting satellites it may be measurable as I believe these effects have to be taken into account with the GPS system

Roger,

Jorrie is right: just add some direction to the acceleration and gravitational vectors and you will fine out that passengers are in perfect weightless condition.

This is used in old mine shafts to have a short time of weightless conditions for experiments, for a low price.

It is also done in the parabolic flight: fall straight to the earth and start pulling up in time, else you have a problem.

I have another problem to solve: the air-water bottle rocket: how high will it go depending on the amount of water and the air pressure. You can leave out the friction.

Thanks Gwen, I agree that the passengers are weightless during the trip. The Pressure problem sounds interesting.

I wrote: "However, the second you pass the center of the Earth, you are no longer in free fall, the train begins decelerating until it finally stops at the North Pole (or South Pole). This is where the passengers "feel" the acceleration, or to be more colloquial, the deceleration."

Although I believe my freefall point to be correct, the conclusion I reach is definitely wrong. Passengers will either feel the acceleration for the entire trip or feel no acceleration for the entire trip. The passengers are weightless the entire trip, though they are accelerated, from 1g to -1g during the trip.

So to the original statement I responded to which said "Pardon me, but if the 'train` is in free fall, whence cometh the "accelleration" felt by the passengers" ????

I should have responded that a person in freefall experiences acceleration due to gravity, though it is worth pointing out that once the train crosses the center of the Earth, it is no longer falling.

Regarding the deceleration on the exit side of the tunnel (after the train passes the center of the earth), will the momentum of the train be adequate to carry it all the way to the north pole exit. It would seem that gravity would be pulling it back to the center. Not to mention that the train must travel an additional 3000 m at the north.

Maybe I'm missing something mentioned in a previous post if so please direct me to it.

Thank You

I doubt that 3-km extra height at the South pole is significant compared with the other things we know we are neglecting, but we would of course either need to provide a means for lifting the train at the South pole or have to get out af the train at about sea level and climb.

Neglecting other losses, the momentum generated in falling would of course be sufficient to bring the train back to the original height (on whichever side).

Fyz

Roger,

Thanks for sharing this entertaining and educational problem. I have pondered this very scenario since childhood without answer or explanation until now.

Here are equations for a gravity-train inside a uniform spherical Earth, ignoring the effect of the hole that forms the tunnel. They are just formalisations of the derivative originally given by Jorrie.

Gravitation vector = -1.5E-6*[x,y,z] (m/s/s)
where [x,y,z] is the position relative to the centre of the Earth (in metres)

The hole train is allowed to accelerate together under this field, so the acceleration of every part of the train may be written:
Acceleration = -1.5E-6*[x0,y0,z0] (m/s/s)
where [x0,y0,z0] (m) is the position of the centre of gravitation of the train
note that the centre of gravitation here is not here the same as the centre of mass, because the gravitational field is non-uniform.

Therefore, the perceived gravity (the difference between the gravitational force and the actual acceleration) will be:

Apparent gravitation = -1.5E-6*[(x-x0),(y-y0),(z-z0)] (m/s/s)

Note that this is a force towards the centre of gravitation of the train, and is indistinguishable from the force on a train that is sitting stationary and unconstrained at the centre of the earth.

The forces would be modified because in "theoretical practice" the train is in a cylindrical tunnel. This will remove the effective inward gravitation (due to the Earth) near the centre of the train, and marginally reduce the inward gravitation everywhere along the axis of the tunnel. The radial forces on the train will be outwards. I don't propose to show the maths here, because the integrations inside a cylinder get a bit messy (and I can't be bothered to complete them, because they add nothing to the generality).

In case of interest, the formal representation of Jorrie's gravitational gradient (in tensorial array format) would be
-1.5E-6, 0.0000 , 0.0000
0.0000 , -1.5E-6, 0.0000
0.0000 , 0.0000 , -1.5E-6
{The spurious precision and commas are to allow the columns to align.}
(units are m/s/s/m, or 1/s/s)

N.B. that the zero off-diagonal terms allow you to treat the whole thing as a scalar if you are looking at motion along any single direction.

Fyz

Hi Fyz, great theoretical treatment of the problem, thanks!

The only thing I am unclear about is your paragraph "The forces would be modified because in "theoretical practice" the train is in a cylindrical tunnel. This will remove the effective inward gravitation (due to the Earth) near the centre of the train, and marginally reduce the inward gravitation everywhere along the axis of the tunnel. The radial forces on the train will be outwards."

I know it is the part that you said adds nothing to the generality, but a few more word on it would be appreciated. I can understand that the linear tunnel through Earth disturbs the homogeneity and could reduce the radial gravitational forces along the axis of the tunnel. But "The radial forces on the train will be outwards"???

Jorrie

Hi Jorrie

I don't really want to present the integrals in a formal form (it's too much like hard work, and I've more than enough of that running my development teams). But let's see if the following sketch (that does not need any extra algebra, but does require some 2.5-D visualisation) will do:

First, consider a sphere with a spherical hole inside it. If we calculate the gravitation that would be due the complete sphere (no hole) and the gravitation that would be due to the "missing" sphere that forms the hole, we can easily see that the gravitation inside the spherical hole is uniform and parallel to the axis that joins the centres of the holes.

Now remove the remainder of the cylinder with the above-defined axis and that only just contains the missing sphere. Consider the gravitation at any point where the missing cylinder and the missing sphere would have touched. In the radial direction, the newly removed material would all have created a force that attracted particles at the selected point towards the axis. As there was previously no net gravitational attraction towards the axis, removing this material must result in effective gravitational repulsion from the axis.

Of course, this only shows that material at the edge of the tunnel will necessarily be repelled from the central axis. But it is possible to construct a number of thought experiments that give similar results for other locations.

Now, if you can be bothered, it is not all that difficult to do the integration for an off-axis point inside an infinite cylindrical skin - you can start by integrating the straight lines to give an inverse linear law (my untidy version went with the recycling the other day).
Again, this is not a complete proof of the present case on its own, but examining the form of the law for rings at different heights will easily show that finite cylinders provide more outward forces than infinite ones. [Or you could be really masochistic and do the complete finite integrations - but I can't see the point myself].

Hope that works for you

Fyz

Hi Fyz, thanks I understand your reasoning and agree. What was still confusing me, although it might just be the vagueness of language, was your statement in a previous post: "The radial forces on the train will be outwards"

I have read "radial" as along the train/tunnel axis, which did not make sense. I now realize that you meant radial from the tunnel centerline outwards.

Jorrie

Sorry - it was ambiguous - I was referring to the tunnel, which has an axis and radii. Once I'd mentioned axis, I forgot to define what I meant by radius. Mea culpa.

Fyz

An addendum: the above argument (argument1) demonstrated that the radial component of gravitation is outwards at the edge of the tunnel. The following argument is not intended to be conclusive, but indicates why I consider it likely that the radial force is outwards throughout the interior of the tunnel.

An object on (or outside) the surface of an open cylinder would clearly be attracted towards the cylinder's axis. Considering the sphere as an assemblage of cylinders, it is obvious from argument1 that at least some of the cylinders are exerting gravitation with an outward radial component. So, as we approach an open cylinder, there will be gravitational attraction towards its axis. As we pass through the surface, the attraction remains towards the axis. However, at some point it reverses. A second reversal is plausible, but seems unlikely. In the absence of a subsequent reversal, the radial force near the axis of the tunnel must be repulsive. This would support my intuition (or pig-headedness) that the radial gravitation is outwards throughout the tunnel - but it is wordy and is not a proof.

N.B. The situation in the plane of a ring is much simpler, as the reversal occurs as you pass through the ring (the integration being much more straightforward at that point)

Shamefacedly

Fyz

I have to confess that I've been concerned about the comments in this post. Mostly because I don't understand many of them. I've decided to write down what I think I know about this problem. Please let me know if I've made any mistakes.

The force the train experiences as it travels through the Earth is:

F =-(⁴/₃)πρGmr

The position of the train as a function of time can be expressed as:

r(t) = RCos((4πρG/3)^1/2 t)

thus,

t(r)= [Cos^-1(r/R)]/(4πρG/3)^1/2

The velocity of the train as a function of time can be expressed as:

v(t)= -R(4πρG/3)^1/2Sin((4πρG/3)^1/2t)

The acceleration of the train as a function of time can be expressed as:

a(t)= -R(4πρG/3)Cos((4πρG/3)^1/2t)

thus,

a(r)= -(4πρG/3)r

For a train length L, the tidal acceleration for a point on the train can be expressed as:

a_T(x)= a(r)-a(r+x)

where x is the distance from the center of the train (-L/2<x<L/2)

a_T(x)= a(r)-a(r)-a(x)

a_T(x)= (4πρG/3)x

thus,

a_Tmax= ±(4πρG/3)L/2 = ±[L x (7.7 x 10^-7 s^-2)]

Like I've said, please let me know if I've made any errors in the calculations above.

Thanks,

Roger

Hi Roger

So long as we don't disturb the train from the axis, this method will work perfectly*.

Other than going around the houses to get from the first line
F= -4/3.pi.roh.G.m.r
to
a(r)= -4/3.pi.roh.G.r
it looks fine up to that point to me.

However, either your definition of tidal force is different from mine or you've swapped the terms.
I would have ^*

aT(x) = a(r+x) - a(r) = a(x) = -4/3.pi.roh.G.m.x

This expression is identical to the attraction of the train as a whole towards the centre of the Earth - indicating the attraction towards the centre of the falling object. Another way of viewing this is that, if we dropped two objects starting from different points, they would 'oscillate' around each other with the same period as the whole - scarcely surprising if we consider the top and bottom points of their individual motions (perhaps oscillation based on these end points is the explanation we should have used initially??)

I'm not clear what 'aTmax' is intended to convey, so can't comment here.

*To make the linear relation clear, I would normally choose a variable different from r, but it doesn't matter here.

^*It's also been expressed in terms of the Earth's gravity and radius, equivalent to:
aT(x) = -gn/R.x
I wrote it as a vector in MKS units - but it's essentially the same expression if we confine the movement to the axis.

Regards

Fyz

Roger - reply to #80, all looks OK to me. Only comment is on basic equation

F =-(⁴/₃)πρGmr. As discussion is limited to tunnel thru Earth, I find it simpler to use F =-mg/R*r as it's a bit shorter and the constants are more intuitive (to me anyway). Different of course if we're considering any spherical body.

Also has anybody made any progress on question raised by Syhprum in #12 - what's the path giving minimum time of travel between any 2 points on surface? Physicist mentioned a catenary but I don't whether that was a guess or calculated. Looks like a problem of calculus of variations. I might have a go at it sometime but I'm not too sure of success! Clearly the path must lie in a plane thru the 2 points and the centre of the earth, but that's not taking it very far.

Hi Codemaster,

The catenary was a starting point based on long-term memory, as it is the quickest path in a uniform gravitational field.

Regarding progress: - inside a uniform sphere, the quickest path is a hypercycloid - the curve generated by a point on the circumference of a wheel rolling inside the sphere's circumference (and subject to the planar constraint you described above).

Fyz

When the train is running in the hypercycloid tunnel is it in 'freefall' or does it have to be restrained by its rails ?

Clearly it needs to be restrained perpendicular to its motion (rails or whatever) - as does the train running along a linear track between London and Glasgow.

Fyz

You Wrote "F =-(⁴/₃)πρGmr. As discussion is limited to tunnel thru Earth, I find it simpler to use F =-mg/R*r as it's a bit shorter and the constants are more intuitive (to me anyway). "

I think you meant to say F=-mgr/R

Reply to Roger #88 - it's the same thing, or was meant to be. R is on the bottom, r on top. I kept the r at the end of the expression as it's the variable, and to keep it similar to your original. Perhaps I should have put F =-(mg/R)*r to make it clearer.

Cheers

Ok, I see it. So I agree that F=-(mg/R)r

and with Jorrie's post a while back that said

a(r)=-g(r/R)

thus

a_T(x) = a(r+x) - a(r)

which Fyz correctly pointed out in an earlier post I had gotten reversed, gives

a_T(x) = a(x) = -g(x/R) for (-L/2 ≤ x ≤ L/2)

where x is the distance from the center of the train

Hi Roger, you wrote: "... gives a_T(x) = a(x) = -g(x/R) for (-L/2 ≤ x ≤ L/2) where x is the distance from the center of the train".

Bingo!

It still pays to be patient, whether you're teaching or learning!

Wrote: "It still pays to be patient, whether you're teaching or learning!"

Yes, it took me a while but I finally got it. I always feel most comfortable when I can look at the equations and see where they are coming from. The great thing is I have a better understanding of how the moon (and sun) effects the tides, how gravity tidal forces can rip apart a meteor, etc.