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Old Jun15-09, 07:00 PM                  #1
Somerschool

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Falling down an infinitely deep well

Back in 1921, Theodor Kaluza published his idea of a very tiny curled up fifth dimension, which I will refer to as the "w" dimension. In a space consisting solely of the x and w dimensions, space is cylindrical, and a ray traveling at the speed of light along x looks like a light wave, while the same ray traveling in the w direction looks like a particle. Changing the direction the ray travels produces all the distortions of length and time we recognize as the Lorentz transformations. Einstein was intrigued by the possibilities of such a curled-up dimension, but the idea never really caught on in his lifetime.

I'm trying to think through the mathematics of Kaluza-space, and need some mathematical help. I've worked out the formula for a ray traveling at the speed of light through an xw space where the "w" dimension gets larger as one travels along the x axis. (In such a case, xw forms a cone rather than a cylinder.) If the cone is long and skinny enough (i.e., about a light year from the vertex to a diameter much smaller than an electron) the formula for the observable motion of a ray is indistinguishable from d=1/2 a*t^2 (for small elapsed time "t") and from v=c (for very large t).

I'd like to compare my formula for displacement as a function of time with what the basic theory of relativity would predict, but it turns out to be hard to find the formula for the motion of a particle subjected to a constant force like that of gravity. I am NOT looking for the formula for a rocket that "accelerates" at a constant rate (within its own frame of reference), but for the path of a rock thrown down an "infinitely deep well" which just happens to have a physically impossible constant gravitational pull all the way down.

Does anybody happen to know this formula off hand? If not, can someone derive it for me?
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Old Jun17-09, 08:08 PM                  #2
Somerschool

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Re: Falling down an infinitely deep well

I posted the same question to "Ask the Physicist." Here's his answer:

"This problem is not as simple as it seems. I have consulted some friends who are good with relativity and may have an answer for you later."

Since there aren't that many infinitely deep wells around, let me spell out an alternate scenario that produces the same formula:

"Suppose we could build a gravity generator that lets us create a powerful gravitational field without using mass. Now build a spaceship and mount the gravity generator on a long pole out in front of it. Turn the gravity generator 'on' and the spaceship 'falls' towards the generator... but the pole pushes the generator along as it goes. The ship travels as if it were falling down an infinitely deep hole where the gravity never changes. What formula describes the ship's motion?"
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Old Jun17-09, 08:58 PM                  #3
DrGreg

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Re: Falling down an infinitely deep well

Originally Posted by Somerschool View Post
...which just happens to have a physically impossible constant gravitational pull all the way down.
Originally Posted by Somerschool View Post
Suppose we could build a gravity generator that lets us create a powerful gravitational field without using mass.
If you ask about things that are impossible, no-one can answer.
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Old Jun18-09, 09:22 AM                  #4
Somerschool

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Re: Falling down an infinitely deep well

Perhaps I should have gone to a math forum. They deal with impossible things for a living.

But, seriously, folks--I'm trying to use a word picture to describe a straightforward physics question. The fact that we can't (yet) build a gravity generator doesn't change the physics or the math.

What formula describes displacement (d) as a function of time (t) for a body subjected to an unvarying force?

If this was on your graduate school physics final you'd at least TRY to answer it! If it can't be answered, you'd explain why. I can't believe a one-sentence question can stump this forum.
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Old Jun18-09, 09:27 AM                  #5
Somerschool

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Re: Falling down an infinitely deep well

I don't need to tell the relativity experts that an "unvarying force" will produce varying acceleration as mass increases near the speed of light, but folks who don't do Lorentz transformation on a regular basis might need that clue.
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Old Jun18-09, 09:45 AM                  #6
Cyosis

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Re: Falling down an infinitely deep well

You want to find the relativistic kinematic equations, in particular displacement, for an object in a constant gravitational field? Just solve F=dp/dt, with F a constant then solve for v and integrate once more.
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Old Jun18-09, 11:15 AM                  #7
Al68

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Re: Falling down an infinitely deep well

Originally Posted by Somerschool View Post
Perhaps I should have gone to a math forum. They deal with impossible things for a living.

But, seriously, folks--I'm trying to use a word picture to describe a straightforward physics question. The fact that we can't (yet) build a gravity generator doesn't change the physics or the math.

What formula describes displacement (d) as a function of time (t) for a body subjected to an unvarying force?

If this was on your graduate school physics final you'd at least TRY to answer it! If it can't be answered, you'd explain why. I can't believe a one-sentence question can stump this forum.
Displacement relative to what? Displacement is the distance between two objects. Identifying the reference object or reference frame will make the answer easy.

The simple answer to your question is d = (c2/a) (sqrt[1 + (at/c)2] - 1), where a is proper acceleration, F/m.

Or d = .5at^2, where a is coordinate acceleration, or for velocities small relative to c.

Either way, d is the distance between two objects, there is no such thing as displacement for a single object without any reference to another.
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Old Jun18-09, 11:31 AM                  #8
Somerschool

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Re: Falling down an infinitely deep well

Al, thank you! I THOUGHT it ought to be a simple question.

Here's a word problem that clarifies the frame of reference:

Identical twins Bill and Bob shake hands at their secret lab, then Bob gets into his UltraPod (TM) and zooms off into space. The UltraPod is subject to a constant force that always accelerates it in the same direction. What is the formula for the distance of the UltraPod from Bill (d) as a function of time (t) from Bill's frame of reference?
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Old Jun18-09, 12:07 PM                  #9
DaleSpam

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Re: Falling down an infinitely deep well

Hi Somerschool,

I agree with Al68's answer, but you should realize that it is simply the relativistic rocket equation, which you rejected as an answer a priori. If you really see some difference in what you are asking and a relativistic rocket then you will have to be more specific. Perhaps it would help if you specified the metric.
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Old Jun18-09, 12:17 PM                  #10
Somerschool

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Re: Falling down an infinitely deep well

I'm asking questions here because my math is too rusty to answer them on my own. I need all the help I can get.

It seems to me like the relativistic rocket formula should be different from the "falling down an infinitely deep hole" formula. The force applied to the relativistic rocket changes (from the perspective of the observer at rest) over time, because time dilates as velocity increases. I'm looking for the formula for motion where the force remains constant, from the perspective of the observer at rest.

Which formula did you just give me, Al68?

If it makes things easier (which it shouldn't), the formula I'm seeking should also describe the path of a photon inside the event horizon of a black hole. Like that helps.
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Old Jun18-09, 12:30 PM                  #11
Cyosis

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Re: Falling down an infinitely deep well

The differential equation to solve was listed in post #6, with p being relativistic momentum. Secondly for a relativistic rocket F is simply equal to ma, which is constant as well. You've to remember that for occupants in the rocket there is no such thing as increasing mass when speeding up. As for the black hole question, I don't think think gravity is constant within the event horizon.
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Old Jun18-09, 12:40 PM                  #12
DaleSpam

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Re: Falling down an infinitely deep well

I believe that you want the motion of a particle under the influence of a constant coordinate force rather than a constant proper force, and it sounds like you actually want it in flat spacetime.

Re the black hole and photon. You are talking about the Schwarzschild metric, which has no "infinitely deep" solution. All geodesics that cross the event horizon hit the singularity in a finite amount of proper time. That said, here is a nice paper on geodesics in the Schwarzschild metric. The equations are not at all the same.
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Old Jun18-09, 01:07 PM                  #13
Somerschool

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Re: Falling down an infinitely deep well

I only doggie paddle!

It's not so much that I WANT the answer in "flat spacetime." What I really want is to compare the "infinitely deep well" function the formula you get when you treat gravity as a tightly coiled up Kaluza dimension, which is:

d^2 + 2*d*R -c^2*t^2 = 0

Where R is a REALLY big distance constant (on the order of 1 light year).
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Old Jun18-09, 01:37 PM                  #14
DaleSpam

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Re: Falling down an infinitely deep well

Originally Posted by Somerschool View Post
IWhat I really want is to compare the "infinitely deep well" function the formula you get when you treat gravity as a tightly coiled up Kaluza dimension
The problem is that AFAIK there is no "infinitely deep well" metric, so you should probably take Cyosis' approach in flat spacetime.
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Old Jun18-09, 02:11 PM                  #15
Al68

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Re: Falling down an infinitely deep well

Originally Posted by Somerschool View Post
I'm asking questions here because my math is too rusty to answer them on my own. I need all the help I can get.

It seems to me like the relativistic rocket formula should be different from the "falling down an infinitely deep hole" formula. The force applied to the relativistic rocket changes (from the perspective of the observer at rest) over time, because time dilates as velocity increases. I'm looking for the formula for motion where the force remains constant, from the perspective of the observer at rest.

Which formula did you just give me, Al68?
I gave you the formula for a constant force applied. An object in freefall has zero applied force and zero proper acceleration. Its distance relative to a local inertial reference will remain constant, neglecting gravitational non-uniformity.

The coordinate acceleration of an object in freefall depends on the reference frame, but is normally relative to a gravitational source, and so cannot be constant over an indefinite time period.

In your example of a ship with a gravity generator on a pole the coordinate acceleration of the ship relative to the gravity source would be zero, and the displacement of the ship would remain constant relative to the gravity source. That's the extent of the answer unless there are other reference objects defined.
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Old Jun18-09, 02:18 PM                  #16
DrGreg

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Re: Falling down an infinitely deep well

Somerschool

I'm sorry my previous answer (#3) was a little flippant. I suppose my point was that you had already conceded that both your scenarios were impossible, and the short answer is that there are no equations to describe impossible situations.

The answer to the question you asked in post #8 is a lot easier. It's exactly as Al68's first answer:
LaTeX Code: \\left( d - \\frac{c^2}{a} \\right)^2 - c^2t^2 = \\left(\\frac{c^2}{a} \\right)^2
(dc2/a)2c2t2 = (c2/a)2 for the benefit of IE6 users who can't read the above equation
What confused me was that in your original post #1, you said "I am NOT looking for the formula for a rocket that "accelerates" at a constant rate (within its own frame of reference),..,". But that is the answer to your restated question. I suppose the missing piece of the jigsaw is that a constant force applied to Bob's constant mass moving in a straight line does produce constant proper acceleration (i.e. as measured by Bob), but not constant coordinate acceleration (as measured by Bill). Under these circumstances, Bob and Bill both measure the same force, which is a bit surprising given that Bob and Bill disagree over almost every other measurement. There's no obvious explanation for this (that I know of), it's just something that drops out of the mathematics.

Note that the equation you gave in post #13 can be rewritten as
LaTeX Code: (d + R)^2 - c^2t^2 = R^2
which is the same as mine/Al68's with R = −c2/a. (About one light year if −a is the "acceleration due to gravity" on the earth's surface.)

As others have said, an "infinitely deep well" doesn't seem to make much sense in general relativity (GR). You also should realise that something falling under gravity isn't being accelerated by a force, according to GR. It's not accelerating at all relative to itself. But something that is hovering at a constant altitude is accelerating upwards, relative to itself, and experiencing a constant upward force.
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