Does something weigh less 1km under the ground?

[K^2]

I can tell you with certainty that it's not a function of gravitational strength. In a constant gravitational field, for example, time dilation will vary as you move along the field, despite field itself staying constant.

But I'm not sure it can be written as a function of potential either. Potential enters into rest mass, which is given by the norm of the 4-momentum. That norm will have contributions from both the time dilation/contraction and space dilation/contraction effects.

I'm going to take a look at some references, see if I can work out the equation at least for the perfectly uniform perfectly spherical body with no rotation and no pressure terms.

 

[Dickfore]

If you are in the center of the Earth, you would feel no gravitational force. However, as you move towards the Earth's surface, you would feel a certain gravitational force pulling you towards the center. Therefore, you need to do work against the gravitational forces to get to the Earth's surface. This means that the Earth's surface is surely at higher gravitational potential \varphi than the center of the Earth.

Finally, the potential at infinity is chosen to be zero. This means that the gravitational potential is negative everywhere. So, because the potential is smaller in the center than at the surface, it means it is actually greater by absolute value, but more negative.

According to the formula:

<br /> g_{00} = 1 + \frac{2 \, \varphi}{c^{2}}, \; |\varphi| \ll \frac{c^{2}}{2}<br />

and the relation between proper time and the zeroth component of space-time displacement:

<br /> d\tau = \frac{\sqrt{g_{00}}}{c^{2}} \, dx^{0}<br />

it means that the more negative \varphi is, the slower the proper time passes (smaller g_{00}. So, time passes slower in the center of the Earth.

 

[AJ Bentley]

I feel an uncontrollable urge to say 'When you find you're in a hole, stop digging'



(Sorry 'bout that)

 

[K^2]

According to the formula:

<br /> g_{00} = 1 + \frac{2 \, \varphi}{c^{2}}, \; |\varphi| \ll \frac{c^{2}}{2}<br />

That's the bit that's new to me. Do you have a link to any place that derives it?

 

[Dickfore]

Yes. L.D. Landau, E. M. Lifgarbagez, The Classical Theory of Fields (Course of Theoretical Physics vol.2), Fourth Revised English Edition, Eqn.(87.12)

 

[K^2]

So it is an approximation, rather than general result. I did not catch that from previous post. That makes more sense.

Certainly good enough for this problem, though. And thank you for the ref. to derivation. Good reading.

 

[Dickfore]

Of course it's an approximation, there is no meaning to define a gravitational potential in General Relativity.

 

[K^2]

Good point.

 

[DrGreg]

Of course it's an approximation, there is no meaning to define a gravitational potential in General Relativity.

Skimming through my copy of Rindler seems to suggest that, for some GR coordinates (static or stationary, I think), then there is such a thing as relativistic potential given not approximately but exactly by

g_{00} = e^{2 \varphi/c^2}​

I haven't read this in detail so I might be misunderstanding something.

 

[Dickfore]

Might be true, because, when solving the Einestein's equations for the Schwartzshild metric one makes a similar exponential approximation.

However, the case of a static field, strictly speaking can be achieved only in the presence of a single body, since two-body problems are time-dependent. Stationary fields have components g_{0 \alpha} which are of the same order of magnitude as g_{0 0} and, since they get multiplied by v_{\alpha}/c. contributed to the next order in the Lagrangian of a particle. Therefore, this gravitational potential does not really represent potential energy of anything.

 

[JDługosz]

Maybe this will help: Imagine only a dense iron core, with the mantle and crust missing. You are on a tower 4000 miles from the center. As you descend, you get heavier because you get closer to the dense core.

Add a fluffy crust a few feet thick. When you move from the "outside" to the "inside" of this outer sphere, you decrease your weight by the gravity contributed by this light fluffy shell. Are there conditions such that the latter effect is smaller than the former? sure; if the crust is light enough and/or the core dense enough.

Continental rocks literally float on the mantle. So I would in fact expect that effect to occur, but can't tell without further information which change dominates.

 

[diligence]

Imagine falling into the earth. As you fall there is a semiphere pulling you up.

Incorrect

 

[diligence]

With all due respect, that simply is not true at all.

With all due respect, you don't know what you're talking about.

 

[diligence]

Look up the Gauss' Theorem. You seriously aren't helping yourself by arguing something from pure ignorance and refusing to look up something that actually explains the physics behind it.

Ding Ding Ding

 

[Dickfore]

Gauss's Law is only true for forces that diminish with the square of the distance. So, the other explanations were true as well.

 

[D H]

Since gravitation is a 1/r² force, Gauss' law most certainly does apply. What it tells you is something else. In particular, it tells you that for a spherical mass in which density is a function of radial distance only, the gravitational force at some point inside the mass increases with depth if the local density is less than 2/3 of the mean density of all of the material beneath the point in question.

Since the density of the Earth's crust is less than half of the mean density of the Earth (2.7 g/cc versus 5.52 g/cc), gravitational force increases with depth in the Earth's crust.

 

[yuiop]

There is a metric that describes gravity inside a spherical body. Using that metric, you'd be able to derive time dilation. I think it should be maximum at the center, but I'm not sure.

Hi K^2, in this thread https://www.physicsforums.com/showthread.php?t=458218 in the relativity forum I give the metrics for outside a shell, inside the material of a shell and inside a vacuum cavity enclosed by the shell. Basically it agrees with your conclusion here.

 

[yuiop]

Maybe this will help: Imagine only a dense iron core, with the mantle and crust missing. You are on a tower 4000 miles from the center. As you descend, you get heavier because you get closer to the dense core.

Add a fluffy crust a few feet thick. When you move from the "outside" to the "inside" of this outer sphere, you decrease your weight by the gravity contributed by this light fluffy shell. Are there conditions such that the latter effect is smaller than the former? sure; if the crust is light enough and/or the core dense enough.

Continental rocks literally float on the mantle. So I would in fact expect that effect to occur, but can't tell without further information which change dominates.

This seems reasonable. If we consider the Earth plus its atmosphere, then the gravitational force at the top of the atmosphere is less than the force at the bottom of the atmosphere because the density of the atmosphere is significantly less than the density of the Earth and the inverse square law dominates. Consider a sphere of radius r1 with constant density p1 enclosed by a shell extending from r1 to r2 with constant density p2. If have done the calculations correctly, then the force on a particle at radius x (where r1<=x<=r2) is equal to the force on a particle at r1 when p2/p1 = r1/x. If p2 < p1*r1/x then the force at x inside the shell is greater than at the surface (r2) of the shell.

 

[Redbelly98]

Just a note, this thread is mostly 6 months old. Not that it isn't worth continuing the discussion.

A derivation of the "2/3 density" result D H mentioned is in Post #6 of this other recent thread:
https://www.physicsforums.com/showthread.php?t=463706

Also, a plot of g vs. depth (and altitude)

And the accompanying image of density as you go down to the center:

Note: horizontal scales are different for the two plots!

Full-sized image of g-vs-depth plot: http://upload.wikimedia.org/wikipedia/commons/8/86/EarthGravityPREM.jpg

 

[Rap]

I think what palladin was saying was that for a person below ground, there is some mass above them which pulls them upward, less mass below them to pull them downward, therefore they will weigh less, and that is a true and valid argument, which does not deny Gauss' theorem. Palladin is right to say that each element of the shell exerts gravitational attraction on any mass inside the shell, but wrong to say that the gravitational field is non zero inside the shell, since the field is the sum of the attractions due to all elements of the shell, which sum to zero at every point inside the shell.

 

[D H]

I think what palladin was saying was that for a person below ground, there is some mass above them which pulls them upward, less mass below them to pull them downward, therefore they will weigh less, and that is a true and valid argument, which does not deny Gauss' theorem.

That apparently is what palladin was saying, and it is a false argument.

A correct statement is that at some depth below the surface of a spherical mass distribution, the spherical shell of mass above that depth contributes absolutely nothing to the gravitational force. The mass overhead does not pull them upward. All that matters is the sphere of below than the depth in question. In other words, with r=R-d, where R is the radius of the Earth and d is the depth below the surface,

g(r) = \frac{GM(r)}{r^2}With this, and with a model of density inside the Earth such as the Preliminary Reference Earth Model (http://geophysics.ou.edu/solid_earth/prem.html ), one can investigate the behavior of g inside the Earth. Note that for a spherical mass distribution the derivative of mass wrt radial distance is related to the local density via

\frac{dM(r)}{dr} = 4\pi r^2 \rho(r)

I'll define the mean density at some radial distance r from the center as

\bar{\rho}(r) \equiv \frac{M(r)}{V(r)} = \frac{M(r)}{4/3\pi r^3}

With this, differentiating gravitational force respect to radial distance r yields

\aligned<br /> \frac{dg(r)}{dr} &amp;=<br /> \frac{G dM/dr}{r^2} - 2\frac{GM(r)}{r^3}\\[6pt]<br /> &amp;= \frac{G 4\pi r^2 \rho(r)}{r^2} - 2\frac 4 3 \pi G \bar{\rho}(r)\\[6pt]<br /> &amp;= 4\pi G\left(\rho(r) - \frac 2 3 \bar{\rho}(r)\right)<br /> \endaligned

In words, gravitational force increases with depth if 2/3 of the mean density of the stuff at greater depths exceeds the local density at the depth in question, decreases otherwise. There are two points inside the Earth where marked changes in density makes the difference 2/3 \bar{\rho}-\rho change from positive to negative with increasing depth. The core-mantle boundary (the D" zone) marks the transition from the rocky material of the mantle to the molten iron of the outer core. In the transition zone at the top of the inner mantle, the mantle rock changes crystalline structure and hence changes density. (The point in the lower mantle where this density difference changes from negative to positive is of no special interest.)

 

[Rap]

That apparently is what palladin was saying, and it is a false argument.

A correct statement is that at some depth below the surface of a spherical mass distribution, the spherical shell of mass above that depth contributes absolutely nothing to the gravitational force. The mass overhead does not pull them upward.

What I am saying is that both arguments are correct: Let's assume that the Earth is sphere with density that is a function of radius only. If you go underground, and then take a plane that passes through you, perpendicular to a line connecting you with the center of the earth, then that plane will divide the Earth into two parts - above you and below you. The net force that you experience will be the sum of the upward force due to the mass above you and downward force due to the mass below you, which will be less than the force you would experience if you were at the surface.

You could also take a sphere centered at the center of the earth, passing through you when you are underground. This will again divide the Earth into two pieces, the inner ball and the outer shell. The outer shell will have a thickness equal to the distance you are underground. The net force you experience will be the sum of the downward force from the mass of the inner ball, and the force from the mass of the outer shell which will be zero. This will be less than the force you experience at the surface, and equal to the force that you calculate by the first method.

 

[yuiop]

What I am saying is that both arguments are correct: Let's assume that the Earth is sphere with density that is a function of radius only. If you go underground, and then take a plane that passes through you, perpendicular to a line connecting you with the center of the earth, then that plane will divide the Earth into two parts - above you and below you. The net force that you experience will be the sum of the upward force due to the mass above you and downward force due to the mass below you, which will be less than the force you would experience if you were at the surface.

Both arguments are not correct because the conclusion of your partition plane argument contradicts the conclusion obtained by many here, that the force of gravity at the surface can be less than the force of gravity lower down if the density increases towards the centre. The partition plane argument fails to take into account that the inverse square (Gauss) law for the measurement at the surface.

 

[Subductionzon]

Rap, if you were familiar with Newton's shell method for calculating gravitational force you would realize it does not matter what is above you. This model assumes that the density is the same at anyone radius around the globe. If you do the calculations you will find that the mass above you cancels itself out. If you were within a hollow shell the force of gravity would be zero everywhere within that sphere.

 

[D H]

What I am saying is that both arguments are correct: Let's assume that the Earth is sphere with density that is a function of radius only. If you go underground, and then take a plane that passes through you, perpendicular to a line connecting you with the center of the earth, then that plane will divide the Earth into two parts - above you and below you.

You could do that, but who would want to do that? This involves solving for the gravitational acceleration due to two spherical caps with nonuniform density. Doable, but yech.

The net force that you experience will be the sum of the upward force due to the mass above you and downward force due to the mass below you, which will be less than the force you would experience if you were at the surface.

That is not necessarily true, and it is definitely not true for the Earth.

Take a look at the plot in post #69. As one proceeds from the surface toward the center of the Earth, gravitational acceleration initially increases. The acceleration reaches a local max of just over 10 m/s² at the bottom of the transition zone (the 670 km discontinuity). At this point gravitational acceleration starts a slight decline, reaching a local minimum about 800 km into the lower mantle. At this point it starts increasing again, reaching a global max of about 10.7 m/s² at the core-mantle boundary. Only then does gravitational acceleration start a monotonic decrease to 0 at the center of the Earth.

Also note that the gravitational acceleration inside the Earth remains greater than the surface value until one has gone more than half the distance to the center.

 

[Rap]

Both arguments are not correct because the conclusion of your partition plane argument contradicts the conclusion obtained by many here, that the force of gravity at the gravity at the surface can be less than the force of gravity lower down if the density increases towards the centre. The partition plane argument fails to take into account that the inverse square (Gauss) law for the measurement at the surface.

That is not necessarily true, and it is definitely not true for the Earth.

I stand corrected. My argument did not take into account the inverse square law for the different distances. I will amend it to say simply that if you are underground and take a plane passing through you, perpendicular to the radius, you will have divided the Earth into two parts, what is above you and what is below you. The net force you experience will be the sum due to the mass below you (downward) and the mass above you (upward). If you have a sphere centered at the center of the earth, passing through you, the net force you experience will be equal to the force of the inner ball only, since the force from the outer shell will be zero. This force will be equal to that calculated by the first method. I don't mean to say that the first method is mathematically convenient, only that it is true.