Is there really no gravitational force at the center of a massive object?

[dst]

If you approached the OP from general relativity, you would find the hollow section to be a uniformly curved blob such that all the curvature added = 0 curvature and hence no "field"? How would time dilation be a factor here?

 

[Dale]

If you approached the OP from general relativity, you would find the hollow section to be a uniformly curved blob such that all the curvature added = 0 curvature and hence no "field"? How would time dilation be a factor here?

You would have no time dilation inside the shell, but you would have time dilation relative to points outside the shell. A photon emitted inside the shell would not be redshifted as it traversed the inside, but after it got outside the shell it would be redshifted as it went up to a distant observer. Sorry if I wasn't clear.

 

[DaveC426913]

You would have no time dilation inside the shell, but you would have time dilation relative to points outside the shell. A photon emitted inside the shell would not be redshifted as it traversed the inside, but after it got outside the shell it would be redshifted as it went up to a distant observer. Sorry if I wasn't clear.

I'm not sure that the notion of time dilation makes sense without comparing to another frame of reference.

I think I see what you mean though.

You would have no time dilation [measuring two points that are both] inside the shell, but you would have time dilation relative to points [inside the shell relative to points] outside the shell. A photon emitted inside the shell would not be redshifted as it traversed the inside, but after it got outside the shell it would be redshifted as it went up to a distant observer. Sorry if I wasn't clear.

 

[belliott4488]

I am wondering about something concerning gravitys effect on mass. Consider a body under a gravitational force. I have the impression that it "pulls" every atom with the same force (I am not taking change of gravitational force for the parts of the body further away from the gravitational source in consideration) in such a way that you won't feel anything (no compression or stretching). Is this correct?

This is a different question and really ought to be in its own thread.

The short answer, however, is that you are asking about what are called "tidal forces". They are generally negligible unless the physical extent of the object in question, i.e its size, is large with respect to the gradient of the field.

 

[belliott4488]

So, because
1] I like to beat things until they're not just dead, but parts start falling off and
2] I like to do graphics,

I've graphicized the original question.

In the graph, distance d is 100% an artifact of the illustrative representation, and has absolutely no real-world counterpart. There is no direct measurement I could do, even in principle (such as, say a gravitometer) that that would give me distance d. Correct?

Please explain. What is this graph supposed to represent? It doesn't look to me like the gravitational potential for either a solid sphere with a small hole drilled through it, or like a spherical shell, which are the two cases that we've discussed so far.

 

[DaveC426913]

Please explain. What is this graph supposed to represent? It doesn't look to me like the gravitational potential for either a solid sphere with a small hole drilled through it, or like a spherical shell, which are the two cases that we've discussed so far.

It is the potential for a sphere with a hollow centre where:
1] the diameter of the sphere is shown by the change in curvature from positive to negative. This can be seen to occur at the contour line second from the bottom.
2] the hollow centre is too small to be seen at this scale (the "me" dot eclipses it). At that tiny point, the curvature is flat (and parallel to the flat space outside the well).

 

[Loren Booda]

How do this thread's symmetry arguments for gravity compare to those for electromagnetism?

 

[DaveC426913]

Please explain. What is this graph supposed to represent?

I have updated the graphic for clarity. Happier?

 

[belliott4488]

I have updated the graphic for clarity. Happier?

Yes! Thanks.
Now that I get the figure, though, I'm still wondering what your question is. This shows the parabolic curve for the interior that I mentioned in an earlier post. It's flat at the center (where "me" is), so there's no gradient, i.e. no force. Is that a problem?

 

[belliott4488]

How do this thread's symmetry arguments for gravity compare to those for electromagnetism?

If you're referring to the arguments for the cancellation of the field, they're the same. For example, there is no E-M field inside a uniformly charged hollow sphere.

 

[capnahab]

No, because an atom is a fuction of the uncertainty principle. I was talking about being at the center of the singularity of a black hole. A very small space but I can visualize it. Let's get to the bottom of it. If you are at the center of the singlarity of a black hole, does it make any difference? Maybe I misunderstand the terminoligy. To the newbies, even the singlarity has over the diameter of the Plank length, no matter if the black hole has the weight of of three Suns or a three billion Suns.

I dare you to ask the astrophysictists the diameter of a 3 billion Sun black hole singularity. It is smaller than the diameter of the nucleas (can't spell it) of any atom.

 

[Andre]

At the centre of a very massive object, the gravitational force is zero. Is that net zero or is it gross zero?

What I mean is: If two people are pulling on my limbs with equal force in opposite directions, my net movement is zero, but I am under great stress. The net forces on me sum to zero, but the gross forces are that of two people pulling on me.

Is it the same with gravity? If I were at the center of the Earth, is there a test I could do to measure the absolute gravitational pull on me?

It seems counterintuitive, Dave? No force and hence no stress? But there is. Change that pulling in pushing. There is no net force but there is pressure. If you would be at the centre of the Earth there is no gravity indeed but you would still be under a tremendous amount of stress too, due to the sum of pressures of all overlying masses times their indivual gravity constant. That's why the 'iron' inner core is solid, despite the high temperatures, due to the shear pressue.

 

[Doc Al]

It seems counterintuitive, Dave? No force and hence no stress? But there is. Change that pulling in pushing. There is no net force but there is pressure. If you would be at the centre of the Earth there is no gravity indeed but you would still be under a tremendous amount of stress too, due to the sum of pressures of all overlying masses times their indivual gravity constant. That's why the 'iron' inner core is solid, despite the high temperatures, due to the shear pressue.

But that's not what's being discussed here. The scenario under discussion is: What's the gravitational field within a hollow cavity at the center of a spherically symmetric planet. There is none. (There's none even if you get rid of the cavity, of course--but then you have other effects: the weight of that mass pushing down on you.)