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I am a 12th grade student from India. I was taught electrostatics before gravitation. I know that if in a uniformly charged sphere, a cavity is cut out unsymmetrically like towards an edge but still enclosed within the sphere, a non zero constant field would exist in the cavity. This doesn't violate Gauss's law as well. When a similar situation in gravitational, where a uniform mass density sphere has a cavity similar to the cavity above, a field does exist inside.

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However, I know that gravitational field has only sinks and no sources. Field lines appear from infinity and end on an object having mass. Inside the cavity, since its enclosed, infinity doesn't exist inside there. Then how does a field line originate inside there? This wasn't a problem in electrostatics as positive charges act as sources and negative charges act as sinks for the electric field. Also charges can induce, and form the necessary negative or positive charge on the surface for the field, so we can consider the field line inside it as arising from a positive charge and ending at an induced negative charge. But I'm pretty sure masses cannot induce, correct me if I am wrong.

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Please clarify my doubts. I am not very familiar with the General relativity and all that. My apologies if you face any difficulty in explaining to me.

Pritt Balagopal
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  • possible duplicate http://physics.stackexchange.com/questions/150238/gravitational-field-intensity-inside-a-hollow-sphere – Physicsapproval Mar 13 '17 at 06:39
  • +Physicsapproval No, this is not a duplicate of that question. My question deals with how a field line can arise there. – Pritt Balagopal Mar 13 '17 at 07:30
  • Since gravitation is a monopole system I don't think "field lines" is quite the right picture here. But does it resolve anything for you to treat it as if field lines originate/terminate on the test mass in the cavity? – Asher Mar 13 '17 at 16:04
  • +Asher In electrostatics, field lines start from positive charge to negative charge. But in gravitation, field lines end at mass, but where do they start? They can't start at mass right? They must start at "negative mass" which is impossible as far as I know. How do we account for this? – Pritt Balagopal Mar 14 '17 at 02:19

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Great question. Your charged sphere for the electrostatic analog does not have to be a conductor (no induced charge necessary). In other words, charge does not need to rearrange itself in order to satisfy Gauss's law (in this and all scenarios).

But this raises the interesting point, how can Gauss's law be satisfied if they are no apparent sinks? Think of two positive charges, if one the charges has a charge greater in magnitude, the E-field will naturally tend away from the charge that has a greater positive charge. However, like you aptly point out, the E-field will not have a sink near the surface since as one approaches the surface of the charge, (or surface of cavity) the E-field approaches infinity. Since Gauss's law cannot be violated, this lends itself to the idea of infinitesimals.

As you approach the surface of the cavity, the E-field will not diverge to infinity, but instead converge and approach a constant value. In some places near the surface, the E-field will diverge outwards, in other places the E-field will converge inwards. I know this is not the answer you wanted -- you wanted a more intuitive reason -- but Gauss's law cannot be violated, and if you were to do infinitesimal calculus for the cavity, (knowing its charge distribution) you would necessarily need to get a converging E-field, one that pointed inwards and outwards in some places. This is the same reason why the E-field near the surface of a spherically charged shell converges. (Note: I accidentally wrote this assuming you were asking about a positively charged non-conducting cavity, but the reasoning is the same for a cavity where gravitational fields are being examined).

H. Chandrasekaran
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When thinking about gravity in terms of field lines, you're right: all lines originate from infinity. However, they do not all terminate at the surface of your massive object. As you descend into your object, the gravitational field gets weaker: some field lines terminate, but others continue down in accordance with the amount of mass in each shell. The last field lines terminate in the center, resulting in zero field there.

In the cavity, some field lines have simply not terminated yet. They continue unhindered through the cavity and begin terminating when the reach the inner wall. There is no "induced mass" on the inner surface of the cavity.

I think some of your confusion came because you're comparing an electrical conductor to a gravitating mass. In conductors charges move freely, resulting in all charge accumulating on surfaces with a neutral interior. There is no such thing as a "gravitational conductor," gravitating mass is more like an insulator. A uniformly charged insulator would behave exactly as the gravitating mass you're considering.

Geoff Ryan
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  • My problem is not about where field lines end, its about where they START. They start from infinity, but infinite distance is not achievable inside the closed cavity. – Pritt Balagopal Mar 20 '17 at 11:20
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enter image description here

The astronaut in the cavity (she's wearing red) feels no net force from the spherically symmetric mass distribution around her, in green, which extends to the nearest surface of the blue sphere. However, she is attracted to the asymmetrical distribution of matter between the green & blue spheres. That is the gravitational field in the cavity.

rob
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  • Yes but can you give the intuition as to why the gravitational field needs to be uniform inside the cavity.... There is more mass on one side than other but the other side is nearer. Can a proof along the lines of the shell theorem be built here too. – Shashaank May 03 '20 at 07:16