How does gravitational field exert gravitational force on a body?

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SUMMARY

The discussion centers on the nature of gravitational fields and forces in Newtonian mechanics and general relativity. Participants clarify that in Newtonian mechanics, a gravitational field can be defined using the equation ##\mathbf F = m \mathbf g(r)##, where ##\mathbf g(\mathbf r)## represents the gravitational field. However, it is established that in general relativity, the concept of gravitational force is not applicable, as gravity is described through the curvature of spacetime rather than a force acting at a distance. The conversation highlights the complexities and misunderstandings surrounding these fundamental concepts in physics.

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  • Understanding of Newtonian mechanics and gravitational laws
  • Familiarity with general relativity and its principles
  • Knowledge of the mathematical representation of gravitational fields
  • Basic grasp of the stress-energy tensor in physics
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  • Study the mathematical formulation of Newton's law of gravitation
  • Explore the implications of the stress-energy tensor in general relativity
  • Learn about the differences between force fields and gravitational fields
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Students of physics, educators explaining gravitational concepts, and researchers interested in the foundational theories of gravity will benefit from this discussion.

adjurovich
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I am not sure if there is in fact a classical description of how it occurs, but I’d like to know if there isn’t too!
 
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In Newtonian mechanics, there is no "gravitational field". In general relativity, there is no "gravitational force".
 
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Hill said:
In Newtonian mechanics, there is no "gravitational field". In general relativity, there is no "gravitational force".
I am not sure about that, we can define the Newtonian gravitational field ##\mathbf F =m \mathbf g(r)##, where ##\mathbf F## is the force due to gravity on a mass ##m## and ##\mathbf g(\mathbf r)## the gravitational field.

Newton's law can even be made into a Gauss' law of gravity: ##\nabla\cdot \mathbf g=-4\pi G \rho(\mathbf r)## (together with ##\nabla \times \mathbf g = 0##).
 
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Hill said:
In Newtonian mechanics, there is no "gravitational field". In general relativity, there is no "gravitational force".
What do you use this equation for, then?

##g = \dfrac{F}{m}##

There is also gravitational field in classical mechanics, but Newton never truly understood gravity… but do we?
 
adjurovich said:
What do you use this equation for, then?

##g = \dfrac{F}{m}##

There is also gravitational field in classical mechanics, but Newton never truly understood gravity… but do we?
There is a gravitational field in Newtonian's mechanics. However this field is instantaneous and that already bothered Newton and many people after him.

There were mechanical explanations of gravity but none was very useful (aether theories were not any better either). The only one that became almost challenging was Le Sage's gravitational theory but even that violates thermodynamics. So I guess the best was just to keep Newton's law of gravitation as an axiom. There is a field and that fields pulls masses together. Period.
 
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Hill said:
In Newtonian mechanics, there is no "gravitational field". In general relativity, there is no "gravitational force".
I disagree. In Newtonian mechanics you certainly can define a gravitational field. It just doesn't have any interesting dynamics on its own.
 
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adjurovich said:
Newton never truly understood gravity… but do we?
Since in our best theory of gravity the gravitational field does not exert a gravitational force, what does the question in the title of the thread refer to?
 
I have asked about classical (more precisely Newtonian) description, not the relativistic one. You cannot expect from a high school student to know much about general theory of relativity?
 
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adjurovich said:
I have asked about classical (more precisely Newtonian) description, not the relativistic one. You cannot expect from a high school student to know much about general theory of relativity?
I see. Anyway, the modern answer to that question is, it does not. Here is "the answer" from Gravitation by Misner, Thorne, and Wheeler:
1713292509887.png
 
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This is essentially the same question as in your last thread, with "gravity" swapped for "electric".
 
  • #11
If you read both questions you can clearly see that they are referring to two totally different things. But since you seem to be very much interested into my posts, I won’t be explaining anything further. If you have nothing else to do than compare and analyze questions of a high school student, I highly suggest you to find a hobby. There is always an option to skip a question you don’t like? It’s better than spamming someone’s inbox and wasting your own precious time on writing useless replies
 
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adjurovich said:
It’s better than spamming someone’s inbox
He didn't spam your inbox. More likely you got a notification that he replied in your thread. Please be careful using the term "spamming" here. Thank you.
 
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adjurovich said:
I am not sure if there is in fact a classical description of how it occurs, but I’d like to know if there isn’t too!
That would be the second paragraph in post #5, to which you've yet to respond.
 
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  • #14
As others have mentioned, there are two major theories of gravity: Newtonian gravity and general relativity.

In Newtonian gravity, there is not a cause-and-effect relationship. Causes come before effects, and that doesn't happen in Newtonian gravity. A change in the sources occurs at the same time as a change in the field, not before. Furthermore, the field has no internal dynamics. So, while mass is the source of gravity, the relationship is not a causal one, which is weird.

In general relativity, there can be a cause-and-effect relationship in many cases. The field has interesting internal dynamics. The source of gravity is not just mass, but the full stress-energy tensor which includes mass, energy, momentum, pressure, and stress. The field is not a force field but represents the tidal effects of gravity. In the cases with cause-and-effect changes in the stress-energy tensor cause changes in the field, changes propagate at ##c##, and those cases include many realistic scenarios.
 
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