Gravity: Force or Distortion of Spacetime?

[Ibix]

If you drop an object inside the black hole it will feel the huge tidal forces from the black hole.

Note, though, that tidal effects are a manifestation of spacetime curvature (at least, as they are modeled in general relativity). So I'd be careful calling them forces if you're not modelling gravity as a force.

 

[Lunct]

Exactly. We assume that gravity is a force for systems with low masses (like our Earth and moon) . But for heavy objects (like stars etc.) the gravity is so strong that it curvatures the space time. So you can assume that the gravity is coincided with the space-time geometry. However, other forces may be occur in such events. If you drop an object inside the black hole it will feel the huge tidal forces from the black hole.

Is that like gravitational waves?

 

[PeterDonis]

We assume that gravity is a force for systems with low masses (like our Earth and moon) . But for heavy objects (like stars etc.) the gravity is so strong that it curvatures the space time.

No, that's not the point Orodruin was making, and it's not correct. In GR, gravity is not a force, period. It's spacetime curvature. But other interactions--electromagnetism, weak, and strong--are still forces in GR.

 

[maria_phys]

Note, though, that tidal effects are a manifestation of spacetime curvature (at least, as they are modeled in general relativity). So I'd be careful calling them forces if you're not modelling gravity as a force.

Hmm, ok I agree with you in general. But assuming that I am an astronaut and I'm traveling directly to the black hole. The effect of the tidal forces wouldn't be the same like the system Earth moon. I will stressed out or shrink in one direction. A force has a direction.

 

[maria_phys]

No, that's not the point Orodruin was making, and it's not correct. In GR, gravity is not a force, period. It's spacetime curvature. But other interactions--electromagnetism, weak, and strong--are still forces in GR.

Newtonian dynamics works really good for objects with low masses. Take for example the gravitational force between Earth and Moon. The space-time curvature is so negligible that has no meaning. In general I agree with you. Yes, in GR is not force. I said that before. But I've tried to clarify when GR has a meaning.

 

[maria_phys]

Is that like gravitational waves?

No gravitational waves are something else.

 

[PeterDonis]

other forces may be occur in such events. If you drop an object inside the black hole it will feel the huge tidal forces from the black hole

Tidal gravity is spacetime curvature in GR. It is not a force. An object that is affected by tidal gravity might have internal forces/stresses inside it, yes, but those are not due to tidal gravity; they are due to the non-gravitational forces between the object's parts (electromagnetic forces between the atoms) that are resisting the effects of tidal gravity.

The effect of the tidal forces wouldn't be the same like the system Earth moon. I will stressed out or shrink in one direction.

You will be stretched in one direction, and squeezed in the other two orthogonal directions.

A force has a direction.

In GR, a force is something that causes proper acceleration (weight). Tidal gravity does not do that. So tidal gravity in GR is not a force. It is, as I said above, spacetime curvature.

Newtonian dynamics works really good for objects with low masses.

The difference between Newtonian gravity and GR is not in which domain each one "works really good". GR also "works really good" for objects with low masses, in fact it works even better than Newtonian gravity does--it has to, because Newtonian gravity is just an approximation to GR, and leaves out effects that GR includes. Those effects are very, very small for objects with low masses, but they are not zero.

Take for example the gravitational force between Earth and Moon. The space-time curvature is so negligible that has no meaning.

You are incorrect. Tidal gravity is spacetime curvature, and the tidal gravity of the Earth and Moon on each other is easily observable.

 

[maria_phys]

Thank you for the clarification. So, the only forces in GR are those between atoms (electromagnetism, weak and strong force)?

they are due to the non-gravitational forces between the object's parts (electromagnetic forces between the atoms) that are resisting the effects of tidal gravity.

Can the strong tidal forces from a black hole change the structure of an atom?

 

[Herman Trivilino]

Newtonian dynamics works really good for objects with low masses. Take for example the gravitational force between Earth and Moon. The space-time curvature is so negligible that has no meaning.

The tides are explained by spacetime curvature and it's not negligible. Just ask anyone who lives near the sea. They are also explained by Newton's model. What's negligible is the quantitative differences between their predictions. But qualitatively the two models are very different.

When tidal gravity is large the two models make very different quantitative predictions and its the spacetime curvature model that makes the more accurate predictions.

My point is that the Newtonian model doesn't have its own separate realm of validity. Einstein's model is perfectly valid everywhere that Newton's model is valid.

 

[PeterDonis]

the only forces in GR are those between atoms (electromagnetism, weak and strong force)?

Those forces aren't just between atoms. But yes, anything that produces proper acceleration, and therefore counts as a "force" in GR, will be due to one of those three interactions.

Can the strong tidal forces from a black hole change the structure of an atom?

Tidal gravity is not a force in GR, as I said before. Strong enough tidal gravity could in principle break atoms apart, yes--because the force between the electrons and the nucleus is not strong enough to hold the atom together against the effects of sufficiently strong tidal gravity (meaning, the effects of sufficiently strong spacetime curvature).

 

[Orodruin]

Exactly. We assume that gravity is a force for systems with low masses (like our Earth and moon) .

No, not exactly. I did not say this. In Newtonian gravity, gravity is a force. In GR it is an effect of curved spacetime. Whether the masses are small or not does not matter.

The space-time curvature is so negligible that has no meaning.

This is not true, it keeps the Moon orbiting the Earth.

But I've tried to clarify when GR has a meaning.

GR is applicable in all situations where Newtonian gravitation is applicable and then some. It works perfectly well for weak fields.

 

[maria_phys]

Tidal gravity is not a force in GR, as I said before. Strong enough tidal gravity could in principle break atoms apart, yes--because the force between the electrons and the nucleus is not strong enough to hold the atom together against the effects of sufficiently strong tidal gravity (meaning, the effects of sufficiently strong spacetime curvature).

Thank you very much all for your replies. According to your post can we have quarks moving freely, Inside the black hole ? Could the strong tidal gravity overcome the strong force between the quarks?

 

[PeterDonis]

Could the strong tidal gravity overcome the strong force between the quarks?

When the tidal gravity is strong enough, yes.

 

[maria_phys]

When the tidal gravity is strong enough, yes.

Interesting. So, assuming that quarks are moving freely in an environment with very strong tidal forces and gravity, can the shape (and properties) of them be changed?

 

[PeterDonis]

assuming that quarks are moving freely in an environment with very strong tidal forces and gravity, can the shape (and properties) of them be changed?

I'm not sure what you mean by the "shape" of a quark, or "properties". Tidal gravity can't change things like the quark's mass or color charge.

 

[maria_phys]

I'm not sure what you mean by the "shape" of a quark, or "properties". Tidal gravity can't change things like the quark's mass or color charge.

That exactly what I meant. Thank you.

(By the word "shape") A particle could be transform into a string, inside a back hole because of the huge tidal forces?

 

[PeterDonis]

A particle could be transform into a string, inside a back hole because of the huge tidal forces?

What do you mean by a "string"?

 

[maria_phys]

What do you mean by a "string"?

I mean, if the tidal force can stressed out the quark, transform it into a string..

 

[PeterDonis]

if the tidal force can stressed out the quark, transform it into a string..

Again, what do you mean by a "string"? A quark doesn't have a shape to begin with; it's not a little sphere.

 

[maria_phys]

Yes, I thought about that after I posted my question.

Forget about the string. Having a quark moving freely inside the black hole, could tidal forces stretch it out?

 

[PeterDonis]

Having a quark moving freely inside the black hole, could tidal forces stretch it out?

Stretch what? Once again, a quark is not a little sphere. It doesn't have a shape. I have said that several times now. Please read my posts again.

 

[maria_phys]

The "shape" has no meaning assuming that quarks has no inner structure. I'm talking about the case of quarks consists of strings.

 

[PeterDonis]

I'm talking about the case of quarks consists of strings.

Do you mean strings as in string theory?

 

[maria_phys]

Do you mean strings as in string theory?

Exactly.

 

[PeterDonis]

Exactly.

Ok, then what does "stretching" mean in string theory terms? (Hint: this is not a "B" level question and the obvious "B" level answer is not correct.)

The short answer is that the simple conceptual model you are using is not valid. A longer answer would be an "A" level discussion, for which it does not appear that you have the background. But briefly, in string theory, spacetime itself is an emergent phenomenon, so tidal gravity is also an emergent phenomenon. In other words, spacetime, and consequently tidal gravity, are "made of" strings in string theory; they're not things that can do things to strings. So your question is not even well-defined in string theory terms.

 

[Buzz Bloom]

I experience gravitational "force" with an additional different meaning than what has been discussed here so far. I directly feel an upward force on the bottom of my feet when I stand on a floor.

 

[pervect]

I experience gravitational "force" with an additional different meaning than what has been discussed here so far. I directly feel an upward force on the bottom of my feet when I stand on a floor.

You can feel that same "force" on your feet in an accelerating elevator (usually called Einstein's elevator) as well as by gravity. Sometimes the analogy of the "force" you feel when you're in a car and the car accelerates is also used to illustrate this point. The physical experience (the pressure you feel on your feet) is the same when you're standing on a floor in a gravitational field, or when you're standing on the floor in Einstein's elevator. In Newtonian physics at an introductory level, what you feel in Einstein's accelerating elevator is typically regarded as an pseudo force. People are generally cautioned about the difference about pseudo forces and real forces, but rather than repeat or attempt to explain these cautions I'm going to take a different route.

Rather than repeating cautions, or going into the underlying mathematics (which is great, if one has the necessary background) I'll focus on some simpler physical experiments that one can do on Einstein's elevator that illustrate why the model of the "gravity" in the elevator being "a force" is at best incomplete. Suppose there are two clocks on such an elevator, one on the floor, the other on the ceiling. The two clocks can exchange light signals, and if the elevator accelerates at a uniform rate, the propagation delay of the exchange of these light signals is constant. This constant propagation delay allows the rates of the clocks on the floor and ceiling to be compared.

Famously, when one compares the rates of the two clocks on Einstein's elevator in this manner, the rates of the two clocks are found to be different. This was noticed by Einstein (as far as I know he was the first to notice this), and rightly seen as an obstacle to describing a model of gravity that was compatible with Special relativity. The "force" model of gravity doesn't lead us to the expectation that the clocks will run at different rates due to their elevations - but that's what the analysis shows.

Going into the details of these calculations is beyond the scope of what I want to write for this short post, but I hope it helps illuminate some of the concepts involved. I'll also add that the effect, which has been termed "gravitational time dilation" has been confirmed for clocks on the Earth's surface in an actual gravitational field, for instance the famous Harvard tower experiment. As far as I know a comparison of clocks on an accelerating elevator is just a "thought experiment" rather than something that's been directly tested. While the direct test of this nature hasn't been done, many other tests of the principles involved have been carried out. See the usual FAQ references on experimental tests of the principle of equivalence for more.

 

[Buzz Bloom]

You can feel that same "force" on your feet in an accelerating elevator (usually called Einstein's elevator) as well as by gravity.

Hi pervect:

The point I was hinting at was that there is a perception of a gravitational force on your feet which is directly experiential in the real world in which we live. It does not depend on which model we use to understand the properties of this "force". In that sense, there is an undeniable gravitational "force" in the real world.

By the way, in the thread

https://www.physicsforums.com/searc...e+beam+interruption.&o=date&c[user][0]=547865​

there is a discussion of how a person in a ten meter elevator can distinguish whether the elevator is on the Earth's surface, or it is in space accelerating at 1 g. The method involves measuring tidal force with an apparatus that could be constructed with current technology, but might take a very long time to accumulate sufficient statistics.

Regards,
Buzz

 

[Herman Trivilino]

I experience gravitational "force" with an additional different meaning than what has been discussed here so far. I directly feel an upward force on the bottom of my feet when I stand on a floor.

Even in the Newtonian approximation that's not the force of gravity that you feel. Or measure. It's a contact force exerted on you by the floor. And you exert an equal but opposite force on the floor ala Newton's Third Law.

If the floor (and you) are in free fall the gravitational interaction between you and planet Earth is still present, but you can't feel or measure it locally, by for example, standing on a bathroom scale.

 

[Buzz Bloom]

If the floor (and you) are in free fall the gravitational interaction between you and planet Earth is still present, but you can't feel or measure it locally, by for example, standing on a bathroom scale.

Hi Mister T:

I agree. When I am in free fall, I do not directly experience gravitational force, nor do I experience any acceleration from the gravitation force. I experience the force when I am stationary with respect to the gravitating source due to something preventing me from accelerating, like the floor.

Regards,
Buzz