A simple observation regarding the equivalence of acceleration and gravity

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  • #1
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This is not an attempt at refuting the validity of relativity, i adore Einstein like most girls love their fave rock star.

To me, there is a glaring discrepancy between Einstein and Newton and it behooves me, out of shear respect, to make this right in my mind.

Take the very simple explanation that Einstein gave, regarding the equivalence of gravity and acceleration, remember, the one with the little man in the room. Okay, i know there have been problems with precisley proving his rightness, but if you've ever driven in a vehicle, you've experienced the effect he is talking about, so there must be more then a little accuracy to his observation. Also, Newton, in his genius ascribed gravity to all masses. So, if the acceleration of the room accounts for the rooms gravitational pull, then musn't we also acclerate the little man in the opposite direction to account for his own gravitational pull? What happens then?

okay, i'll tell ya. As the man acclerates, his resistance to being accelerated in the opposite direction by the room increases, as does the kinetic energy or force experienced by both, and yes, this happens even if the man is accelerating very little compared to the room. Our experience of gravity is not so. Otherwise, by sometime next year i might weigh 2000 lbs or more, without actually having grown an inch.

Also, interestingly enough, if both the room and the object start off as the same mass and accelerate at exactly the same rate, neither would move, yet both would experience all the effects of acceleration.

I know it sounds elementary, but i am finding this very problematic in my reasoning. Any ideas from the rest of you bored geniuses?
 

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  • #2
DaveC426913
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Can you please elaborate on this:

"As the man acclerates, his resistance to being accelerated in the opposite direction by the room increases..."

I don't think you're understanding what's happening; I think we should break it down a bit.


Under a constant acceleration, you will feel a constant force. Drive in a car, starting from zero, keeping the acceleration constant. You will remain pushed back in your seat by the same amount for the duration. It's not like by the time you reach 60mph, you are being flattened...
 
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  • #3
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If i push something along in an attempt to accelerate it, it's resistance to being accelerated is it's mass, but if it is being accelerated the opposite direction as i'm pushing, would i not have to push harder then?
 
  • #4
JesseM
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So, if the acceleration of the room accounts for the rooms gravitational pull, then musn't we also acclerate the little man in the opposite direction to account for his own gravitational pull?
What do you mean when you say "the acceleration of the room accounts for the rooms gravitational pull"? Einstein says there is an equivalence between what is felt in a room at rest in a gravitational field, like one sitting on Earth (where the gravity felt inside the room is due to the Earth's gravitational pull, not the room's), and what is felt in a room that is being accelerated at a constant rate in deep space with no significant sources of gravity. If there's a man in the accelerating room in deep space, then as long as his feet are on the floor, he's being accelerated at the same rate and in the same direction as the room, not the opposite direction.
 
  • #5
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okay bear with me. the room's acceleration towards the man accounts for the rooms gravitaional pull on the man.

the man must have his own gravitational attraction towards the room, non?
that is why we would then have to accelerate him towards the roomas well. otherwise the attractive force can only be ascribed to the room , and not to the man inside of it.

why it must be in the opposite direction should be clear.
 
  • #6
JesseM
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okay bear with me. the room's acceleration towards the man accounts for the rooms gravitaional pull on the man.
No, it accounts for the feeling that there is an external gravitational field pulling on both the room and the man, as there would be in the case where the room + man were at rest on Earth (in which case the external gravitational field would be the Earth's).
 
  • #7
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i really need help with this.
 
  • #8
JesseM
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i really need help with this.
Can you say what it is about my explanation that you don't understand/disagree with? When I feel gravity standing in a room on Earth, this is because of the Earth's gravitational pull, not the room's, isn't it?
 
  • #9
JesseM
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all of these responses are haughty and dismissive and none of them helpful at all.
I'm really not trying to be haughty, I'm just explaining why I don't think your argument makes sense. I can't help further unless you clarify why you think my response is wrong or why you think I'm missing the point you're trying to make. I hope you would not reject any response that dismisses your argument (in a reasoned way, not an unthinking 'dismissive' way)--surely the fact that all these physicists haven't seen a problem with the equivalence principle for almost a century should lead you to take seriously the idea that there might be some fundamental error in your reasoning?
 
  • #10
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Seems like a blind application of ideas heard in SR (and which only apply there for special reasons) to a more general situation (curved spacetime).

Girl, should an accelerated object gain kinetic energy (or speed, or inertial mass) if measured only by an observer who also accelerates (as to remain equidistant from the object)?
 
  • #11
JesseM
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Seems like a blind application of ideas heard in SR (and which only apply there for special reasons) to a more general situation (curved spacetime).
Seems like a misunderstanding of the equivalence principle from GR, not ideas from SR...what ideas are you thinking of?
 
  • #12
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clearly, the model was included in GR, not SR, but i think i get what he is saying and the reason the inertial mass increases is because the man is contained by the room, he is not merely observing it.

So we all know the story , i won't rehash it.

The problem is that in Einsteins model the man is completely passive. The force he experiences is due entirely to the room's acceleration. According to Newton, the man must not be passive. He must have his own force, which is why he must have his own rocket pack to accelerate him the direction of the floor of the room. This is the opposite direction the room is accelerating in.

If i am pushing on something in an attempt to accelerate it, and it starts accelerating back at me, the force required to move it increases. Also the resistance i would experience from said object would increase.
 
  • #13
JesseM
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The problem is that in Einsteins model the man is completely passive. The force he experiences is due entirely to the room's acceleration. According to Newton, the man must not be passive. He must have his own force, which is why he must have his own rocket pack to accelerate him the direction of the floor of the room. This is the opposite direction the room is accelerating in.
When you say "according to Newton", can you be specific about which Newtonian law you're thinking of? In Newtonian physics, the only force on the man would be the force of the floor pushing his feet in the direction the room was accelerating, he wouldn't need a separate force from a rocket pack or something. Imagine instead of a man, you have a spring, and when no forces are acting on it the spring has a natural relaxed length L. If one end of the spring is attached to a surface, like the floor of the room, which is accelerating in the direction of the spring's unattached end, and this is the only force acting on the spring, then the spring will become compressed to a length shorter than the relaxed length L, according to Newtonian physics. Do you disagree? If not, then something analogous is happening with the man, he feels heavier than in zero G because his body is being slightly compressed by the acceleration in a similar way, he has to maintain tenseness in his muscles or the compression would cause his joints to bend and make him fall over.
 
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  • #14
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i know that it seems ridiculous that no one else noticed it before me. In fact for a while i figured that someone else must have pointed this out before. That doesn't automatically mean i'm wrong.

i don't mean to get so irritated, i just am frustrated. and i am very bad at expressing myself constructively, i'm sorry for that. i'm just totally hung up on this stupid thing and it seriously keeps me up at nights, no lie. i really love physics that much.
 
  • #15
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The squishing of the spring bit. That would be proportional to the rate of acceleration, OR how fast it is nearing C. What's the point? Other than to get me off track and make me look a fool.

Does the squishiness of the spring have anything to do with what i'm saying?
 
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  • #16
JesseM
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Yes, JesseM, i disagree. The only way the spring would compress is if you accelerated the room nearing the speed of light.
I wasn't talking about Lorentz contraction in relativity, a frame-dependent phenomenon where the length of an object moving at close to light speed relative to me will be measured to shorten in my rest frame (even though its length will remain unchanged in the object's own rest frame), but rather about real physical compression of the spring as measured in its own rest frame due to being accelerated by the floor, a phenomenon that can be explained in purely Newtonian terms with no reference to relativity.
girlwonder said:
You are still missing my point.

Newton figured that you have to ascribe gravity to ALL masses. In Einstein's
model the man is passive, and the force he feels can only be attributed to the room. What force can we then attribute to him? NADA.

If we want to attribute a force to him, then we must accelerate him towrds the room.
I think the problem here is a misunderstanding of Newtonian physics, rather than of relativity. It's true that all masses gravitate, but the gravitational force is not the only one at work in Newtonian physics, and for small objects like springs and rooms the gravitational force can be treated as negligible. For example, if the room is being accelerated by some external force like a rocket, and the floor of the room is pushing on the bottom of the spring, this has nothing to do with the gravitational force of either the room or the spring, it's due to the electromagnetic force between the atoms of the floor and the atoms of the part of the spring in contact with the floor. This force, pushing the spring away from the floor if they are mashed together (either because the spring is pushed downwards or because the floor is pushed upwards), is far more powerful than the gravitational force between the spring and the floor pulling them towards each other. And there are similar forces between atoms within the spring throughout its entire length, which is why the spring pushes outward when you compress it to shorter than its relaxed length.

Have you studied the mathematical version of Newtonian physics, where the motion of an object can be calculated by summing the various force vectors acting on it, or is your understanding more conceptual? If you have an understanding of the use of force vectors I can show you a simple example to explain why, in Newtonian physics, a spring attached to an accelerating platform will indeed become compressed. If your understanding is just conceptual, I definitely encourage you to take a course in basic Newtonian mechanics or pick up a textbook to learn the basics yourself, trying to reason about physics in purely verbal terms can often lead to confusion.
 
  • #17
JesseM
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The squishing of the spring bit. That would be proportional to the rate of acceleration, or how fast it is nearing C.
See above--I'm not actually talking about anything relating to relativity, just ordinary Newtonian squishing.
girlwonder said:
Does the squishiness of the spring have anything to do with what i'm saying?
The point I'm making is that I think the reason we "feel" our weight in a gravitational field is somewhat similar to why the string becomes compressed, our body is being somewhat squashed together and so we have to maintain tension in our muscles to avoid having our joints buckle. In fact, almost all the ways in which an accelerating observer would notice obvious "gravity-like" phenomena, including feeling heavy and weighing something on a scale and noticing that dropped objects seem to accelerate towards the ground, can be explained in purely Newtonian terms; relativistic effects, like the fact that clocks closer to the floor tick slightly slower than clocks higher above it (which mirrors gravitational time dilation in a gravitational field), would be too small to be noticeable except with very specialized equipment. And the Newtonian explanation could neglect gravitational forces entirely, since we're talking about a room being accelerated in deep space away from any gravity wells, and gravitational interactions between the room and objects within it would be negligible compared to electromagnetic interactions, since the mass of the room and the stuff inside it is tiny compared to the mass of the planet-sized objects with noticeable gravitational fields.
 
  • #18
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Indeed i am studying the entire volume of the Principia.

And i'm not saying that the acceleration of the room IS gravity. Niether was Einstein.

Still the man can not be passive. If he has attributable force he must be moving towards the floor, just as the floor is moving towards him.
If he moves towards the floor, even at a very small rate of acceleration, at the same time the floor moves towards him, the force will increase over time, as both the room and the man accelerate against each other.

The only way to make the force constant is to make the man passive, with no motion or force attributable to him.

Although i am more of an idea girl, i have a respectable understanding of vectors, so shoot away.
 
  • #19
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girlwonder, I think I understand your argument. According to Newton's theory of gravitation, there is a force exerted on the man BY the earth AND a force exerted on the earth BY the man.

It seems that you are wondering why Einstein's thought experiment included the fact that there is a gravitational force acting on the man but there is no gravitational force acting on the earth by the man.

I haven't studied relativity formally (I will this semester) but I do know a bit about the general principles and the thought experiment you are mentioning (man in the chest). I think the answer may lie in the fact that the man and the room being accelerated upward SIMULATES one external gravitational field--there is no actual massive body which is exerting a gravitational force on the man (I think you know this too).

A Newtonian model would say that the force experienced by the man on Earth is GMm/r^2 downward, and the same force would be experienced by the earth from the man (upward). The thought experiment only observes the forces acting on the man from his reference frame. If the man drops a ball, it will appear to him to fall as if there is a gravitational field.

So, if we think about the Earth in a room, and we accelerate the Earth so that its acceleration is the same as that provided by the man's gravitational force, we are now simulating the attractive force on the Earth by the man. It seems to me that with Einstein's thought experiment, you would have to consider each body (Earth or man) separately like this if you wanted to imagine how the Earth is affected by the man's gravitational force, and you can't consider both of them in the same situation. Essentially, the man in the room being accelerated is simulating ONE external gravitational field (of the Earth), so you can't try and use the same situation to simulate two simultaneous gravitational fields. Instead, you then think about the entire Earth accelerating as a result of the gravitational force by the man on the Earth.

Maybe I got it, maybe not?
 
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  • #20
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JesseM, I don't know how to show diagrams or math on this cotton picking thing. You are talking about that little diagram that shows how two vectors in different directions combine in that nifty diamond shape with a line through it right?
 
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  • #21
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gabee, there is definitley two gravitational forces to be considered here. And there is no reason not to consider them together in the same thought experiment. There is a man and a room, related to each other by the containment of the man inside of the room. There is the gravity of the man and the gravity of the room, and the man would experience a combination of both.

The man can not be passive, unless you are willing to forfeit the idea that he has his own attractive force.

No, it does not add up.

Einstein said that further exploration of this equivalence of gravity and acceleration would lead to completed theory of gravitation. One he never completed.

It should be made to add up.
 
  • #22
JesseM
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Indeed i am studying the entire volume of the Principia.
Well, as I said in a PM, the treatment in the Principia is a lot more complicated than what you'd find in modern textbooks, so I'd definitely recommend supplementing your reading with a modern textbook on classical physics.
girlwonder said:
Still the man can not be passive. If he has attributable force he must be moving towards the floor, just as the floor is moving towards him.
If the room is accelerating upwards (due to being pushed from below by a rocket, say), and the man is simply floating above the floor with no forces at all acting on him, then the floor will accelerate towards him--to an observer standing on the floor, who treats the floor as if it were at rest, it'll look just as if the man is falling towards the floor at an increasing speed, exactly the same as if the floor were at rest in a gravitational field and the man were falling towards it. Once the man actually hits the floor and gets stuck on it, the reason he feels his weight has to do with the compression of his body which I talked about before...but see the spring example below for more on that.
girlwonder said:
If he moves towards the floor, even at a very small rate of acceleration, at the same time the floor moves towards him, the force will increase over time, as both the room and the man accelerate against each other.
When you say "the force will increase over time", what specific force are you talking about? The gravitational force between the man and the floor? Again, if the floor is being accelerated upwards by a non-gravitational force such as a rocket attached to the bottom, the effects of the gravitational force from the man acting on the floor will be so miniscule in comparison that you can ignore them. Likewise, once the main is actually touching the floor, the electromagnetic force from the floor pushing upwards on the soles of his feet will be vastly, vastly stronger than the gravitational force pulling the man towards the floor.
girlwonder said:
Although i am more of an idea girl, i have a respectable understanding of vectors, so shoot away.
OK, it's usually assumed in Newtonian physics that the spring force obeys [URL [Broken] law[/url], or F = kx, where x is the distance the spring has been squashed or stretched away from it relaxed rest length in the absence of external forces, and k is a constant known as the spring constant. The vector points outward when the spring is compressed, inward when it's stretched. To avoid having to calculate the dynamics of each point along the spring as one end is accelerated, let's just imagine we have a single 6 kg ball which starts out at rest 20 meters above the room's floor before it begins to accelerate, and that there is a force between the ball and the floor which behaves just like that of an idealized spring whose rest length is 20 meters, and with a spring constant k of 4 Newtons/meter = 4 kg/s^2 (the way this idealized force works is actually not so different from the way the electromagnetic force between atoms in a solid works--see the 'worked example' on this page, for example). So, for example, if the ball is pushed down until it's 17 meters above the floor, an 3-meter displacement, then it experiences a force upwards of (3 m)*(4 kg/s^2) = 12 kg*m/s^2. Since its own mass is 6 kg, and a = F/m, this means that when let go it will begin to accelerate upwards at a rate of 12/6 = 2 m/s^2 (of course the rate of acceleration will decrease as its distance from the floor gets closer to 20 m).

Now, if you start out with both the floor and the ball at rest, and then begin to accelerate the floor upwards at a rate of 10 m/s^2 towards the ball, decreasing their distance, the force repelling the ball will begin increasing and start to push the ball away from the floor, and initially you'll probably get some oscillation as the distance between the ball and the floor keeps varying due to the force between them changing and the floor constantly accelerating in the direction of the ball. But if you have some friction in the air in the room, eventually you'll reach an equilibrium where the "spring force" between the floor and the ball is just right to keep the ball accelerating in the direction of the ceiling at 10 m/s^2 too, so that it remains at a stable height above the accelerating floor. Using F = ma, we can see that for a 6 kg ball to be accelerating at 10 m/s^2, the force must be 60 kg*m/s^2. Setting this equal to kx, with k = 4 kg/s^2, we get x = 15 meters. So, when the floor begins accelerating upwards and the ball stops oscillating and reaches an equilibrium it will have moved from its original resting height at 20 meters above the floor to a new height of 20 - 15 = 5 meters above the floor.

Calculating the behavior of a more realistic spring, whose mass is distributed along its entire length instead of concentrated at one end, and which may experience uneven compression and thus unequal force of tension along its length, would be a lot more complicated, but the end result would be the same--if the spring is attached to the floor, and you accelerate the floor at a constant rate, then the spring will eventually reach an equilibrium where it's not oscillating, and where its length is compressed from what it was when the floor was not accelerating and the spring was relaxed.
 
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  • #23
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okay this is futile.

i think you secretly understand what i'm saying you are just bored.
i don't even get what the hell a bunch of springs has to do with it.

Einstein was right in his model and i think he also would appreciate my input in this case. However he is dead and i am not, so you're stuck with just me. Burn me all you want. He wasn't afraid of modifying Newton and he would do it now from the grave if he could.
 
  • #24
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Seems like a misunderstanding of the equivalence principle from GR, not ideas from SR...what ideas are you thinking of?
Kinetic energy depends on relative velocity. Hence, in SR, it may usually be said that "energy increases as speed increases", or even "mass dilates with acceleration" (as things speed up, it gets more difficult to force them faster still). If one were to naively combine this with the equivalence of gravity and acceleration, you'd get "anything resisting gravity (like a person standing on a planet) must be getting heavier each second (requiring increasing force to support them)", which isn't correct since, despite that the person is accelerating, the person is not increasing velocity relative to the ground (and so their weight upon the ground remains constant).


The man would be completely passive, with no gravitational force attributable to him. This is against Newton however you slice and dice it. If Newton was right the man must have his own gravitational force.

That is why he has to have his own little rocket pack to accelerate him in the direction of the room.

Why is this so hard to explain, or are you just being difficult?

- I think people here are genuinely trying to understand your question so as to help you.

- He shouldn't need a rocket if the room (or elevator) has a floor (since it will exert an upward force onto him).

- The usual introductory examples for GR would assume that the second body (the man in your example) has neglegible mass, since doing otherwise (ie. considering the mutual interaction), correctly according to GR, requires solving difficult problems that make no sense without high level mathematics.
 
  • #25
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gabee, there is definitley two gravitational forces to be considered here. And there is no reason not to consider them together in the same thought experiment. There is a man and a room, related to each other by the containment of the man inside of the room. There is the gravity of the man and the gravity of the room, and the man would experience a combination of both.

The man can not be passive, unless you are willing to forfeit the idea that he has his own attractive force.

No, it does not add up.

Einstein said that further exploration of this equivalence of gravity and acceleration would lead to completed theory of gravitation. One he never completed.

It should be made to add up.

I don't think you can consider both forces--this seems to me to be a common error a lot of people make when dealing with Newton's second and third laws. When you are considering an object falling to earth, what forces are acting upon it? Only gravitational force. Similarly, there is only one force acting on the earth at the same time--the gravitational force from the apple. However, you cannot add these two forces up or subtract them, because they are acting on different bodies!

Also, the room has no gravity--the room is just a container of the man to give him his own reference frame. (I assume you probably mean the gravity of the earth.) The simulated "gravity" that the man in the room feels is apparent because the room is being accelerated.

It looks as though the thought experiment is only designed to simulate one gravitational field at a time.
 
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  • #26
JesseM
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okay this is futile.

i think you secretly understand what i'm saying you are just bored.
I really don't. Like I said, all the main phenomena observed in a room with gravity would be duplicated in an room accelerating in deep space, and you can explain these phenomena in Newtonian terms. Contrary to what you seem to be saying, in this Newtonian analysis you don't need to apply a separate force downward on the objects in the room in order to replicate these pseudo-gravitational phenomena.

Could it be that when you talk about a downward force, you're just talking about the Newtonian law that "every action has an equal and opposite reaction", so that if the room pushes upward on the bottom of my feet with some force F, my feet must be pushing back downward on the room with an equal and opposite force -F? If that's what you're talking about, then this is true, but if we assume the room is much more massive than the man, then according to F = ma the man's acceleration will be affected much more by the force F applied on him than the room's acceleration will be affected by the force -F applied on it. For the purposes of the thought-experiment, we can consider the room's acceleration to be totally constant, either because its mass is much larger than that of anything inside it, or because the thing accelerating it (a rocket, say) is adjusting its force to compensate for the internal forces from objects in the room, in order to keep the room's acceleration perfectly even.

But if this is not what you're talking about when you refer to an opposite force, please try to explain further instead of giving up--it might help if you give some simple numerical example like mine with the spring, so I could see more clearly why you think there needs to be an opposite force on the man to get a certain pseudo-gravitational effect.
girlwonder said:
i don't even get what the hell a bunch of springs has to do with it.
Like I said, I think the reason you "feel" heavy in a gravitational field has to do with your body being compressed slightly, like the spring. But for another point of relevance, think of standing on a spring scale. How does the scale measure your weight? By measuring how much the springs in the scale compress when you stand on the scale, compared with when there is nothing on the scale. And if you were on a platform in space accelerating at 1G, standing on a scale would give the same reading as it does no Earth--this situation would be a lot like my example, where you have the bottom of the spring on the ground and a large mass placed at the top, so my analysis helps show why the spring compresses on an accelerating platform just like it does in a gravitational field (if you like, I could repeat my analysis of the same situation with the ball held up by the springlike force, except that instead of being above a platform accelerating at 10 m/s^2 it could instead be held above the ground of a planet whose gravitational acceleration was 10 m/s^2 -- in this case the ball and springlike force would behave just the same, with the ball dropping from its relaxed 0-G height of 20 m to a compressed height of 5 m above the ground).
girlwonder said:
Einstein was right in his model and i think he also would appreciate my input in this case. However he is dead and i am not, so you're stuck with just me. Burn me all you want. He wasn't afraid of modifying Newton and he would do it now from the grave if he could.
I'm really not trying to "burn" you, just to explain why your arguments don't make sense to me. Again, it would definitely help if you gave more specifics, like an example of a specific phenomena observed in a gravitational field (like my example of standing on a spring scale and getting a certain reading) that you think would not be replicated on an accelerating platform unless you applied an opposite force to the man. If a Newtonian analysis shows that every specific phenomenon seen in a gravitational field is replicated on an accelerating platform, without the need for any additional outside forces on the objects besides the upward force from the platform, wouldn't this show there's a conceptual flaw in your argument? Since your argument is just a verbal one, and yet it's something that no physicist has brought up for almost a century, please be open to the possibility that you're just making a conceptual mistake that would disappear when you do a mathematical analysis.
 
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  • #27
pervect
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okay this is futile.

i think you secretly understand what i'm saying you are just bored.
i don't even get what the hell a bunch of springs has to do with it.

Einstein was right in his model and i think he also would appreciate my input in this case. However he is dead and i am not, so you're stuck with just me. Burn me all you want. He wasn't afraid of modifying Newton and he would do it now from the grave if he could.

It's rather early in the morning, and I've skimmed over your most recent responses, but I don't understand what you're trying to say either.

If we consider the man to be suspended from the ceiling of the elevator by a rope, there is only one force to consider - the tension in the rope holding him up.

If instead of being suspended from a rope, we have the man stand on a pillar on the floor, there is still only one force to consider - in this case it is a compressive force in the pillar (in the rope it was a tension force).

If you think there is more than one force, please try to explain how you would measure it with some force-measuring instrument.

It seems to me pretty clear that there is only one force involved here. There are two different viewpoints - an inertial viewpoint, and a non-inertial viewpoint, which are basically different coordinate systems.
 
  • #28
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girlwonder, i can't really understand what the point your trying to make is. Can you please explain it better?
 
  • #29
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Okay, i've been warned about my direspectful remarks. Sorry JesseM, really i like you and appreciate what you have to say,
the problem is definitley with my delivery.

here goes a simpler, better explanation.

I'm tired of the man in the room thing. Let's call the room a solid hollow object A, and call the man solid object B. To simplify things i will specify that i am not refering to the ACTUAL gravity of A or B, rather, i will refer to the SIMULATED gravitational attraction of either by their respective acceleration.

So to begin with B is at rest inside of A.

If i wanted to simulate the gravitational attraction of A towards B, i may accelerate A towards B. At the point where A touches B the kinetic energy of A would be transfered to B through the process described by JesseM. As long as the rate of acceleration was constant, so would the SIMULATED gravitational force be constant.

Conversely, if i wanted to simulate the gravitational attraction of B towards A, i may accelerate B towards A. At the point where they touch the kinetic energy of B would be transfered to A, through said process, (thanx JesseM) and as long as the rate of acceleration was constant, so would the SIMULATED gravitational force be costant.

However, we know that the resultant gravitational attraction between two bodies is a combination of the gravitational forces of both bodies.

So if object B is contained by object A and we accelerate them towards each other at respective constant rates, the kinetic energy at the point where they touch would indeed increase over time.

Better?
 
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  • #30
DaveC426913
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...To simplify things i will specify that i am not refering to the ACTUAL gravity of A or B, rather, i will refer to the SIMULATED gravitational attraction of either by their respective acceleration.


If i wanted to simulate the gravitational attraction of A towards B
: light blub goes on in head : Waitaminnit! The Principle of equivalence has nothing to do with gravitational attraction between the two objects!

The Principle of Equivalence assumes the two objects have inconsequuential mass, and thus inconsequential gravtitational attraction.

The gravitational force experienced by object B inside object A, is from a source completely external to both.

Is it possible that this is the source of your confusion?
 
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  • #31
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Oh my goodness, DaveC426913 (can i just call you Dave?), i think we are on the same page!

However, object A and B are in very deep space and there is no external source of gravity, i should have stated that.

Also, what is accelerating each object is also unspecified.

In order to simplify things i did specify that the ACTUAL gravity of A and B
was not being refered to, but rather the SIMULATED effect of gravity due to their respective acceleration.

i would avoid assuming the inconsequential nature of each objects mass and gravity, i merely avoided refering to them in order to draw attention to the resultant simulation.

Anyway, if B is contained by A and both are accelerating towards each other at a constant rate, wouldn't you agree that at the point where they touch the kinetic energy would increase over a period of time?
 
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  • #32
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object A and B are in very deep space and there is no external source of gravity [...] i would avoid assuming the inconsequential nature of each objects mass and gravity
Let's skip this A inside B thing (I do now suspect you misunderstood the equivalence principle.. plus, if you were inside a hollow mass, then the gravitational forces would tend to cancel out); are you happy with the scenario where (in deep space away from external gravitational sources) some two objects A & B (each having mass) happen to (initially) be stationary a finite distance apart from each other?

In the Newtonian picture, the two objects will just attract (accumulating kinetic energy) and collide. According to GR there is no attractive force whatsoever.. but space-time is twisted so strangely that the only way that the objects could have seemed initially stationary (to us distant observers) is if they had really always been moving straight towards each other on a collision course. (Note, the kinetic energies here don't really increase, but they are very ambiguous for any observer to measure.)

Now, say we put a stout rod between the objects, so that they keep a small distance apart. According to Newton the rod is pushing each object with a force to exactly cancel the gravitational pull on that object (resulting in no motion or acceleration).

In the GR picture there is no gravitational pull, but the rod is still pushing on each object with the same force as before. This causes each object to accelerate outwards. However, space-time (i.e., in this case, the definition of "outwards") is curved around in such a way that, despite their individual accelerations, the objects don't get any further apart.

The objects of course can't tell the two pictures apart, since being accelerated (as in an elevator) feels just the same as standing stationary against a massive planet's gravity. (Although they might detect the space-time curvature, by some clever scheme to send probing signals around the region.)
Anyway, if B is contained by A and both are accelerating towards each other at a constant rate, wouldn't you agree that at the point where they touch the kinetic energy would increase over a period of time?
As for kinetic energy in my last two pictures, the two objects each remained a constant distance from other, so each should concede that the other has zero relative velocity (and therefore, continually, no kinetic energy in the other's frame of reference).
 
  • #33
DaveC426913
Gold Member
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Anyway, if B is contained by A and both are accelerating towards each other at a constant rate, wouldn't you agree that at the point where they touch the kinetic energy would increase over a period of time?
I don't know what this has to do with the EP.

I need to ask for clarification (even if it means we repeat a lot of what we've covered): Are you looking for understanding of the main principle of EP, or are you looking at add some additional elements to examine the implications?

Are you satisfied with EP in and of itself?
 
  • #34
JesseM
Science Advisor
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here goes a simpler, better explanation.

I'm tired of the man in the room thing. Let's call the room a solid hollow object A, and call the man solid object B. To simplify things i will specify that i am not refering to the ACTUAL gravity of A or B, rather, i will refer to the SIMULATED gravitational attraction of either by their respective acceleration.
The problem is that the equivalence principle is not about simulating the attraction between the room and the person, it is specifically assumed that the mass of the room and the person are both totally negligible, and thus that their own gravity is negligible as well. (and even if it weren't, why would you need to 'simulate' their gravity, when their own gravity will be exactly the same regardless of whether they're in deep space or on Earth?) The equivalence principle is about simulating the effects of a gravitational field caused by some gigantic external mass, like a planet. If you accelerate the room at 1G in deep space, then the effects experienced inside will be virtually identical to those experienced if the room is sitting at rest on the surface of the Earth. The reason the planet has to be gigantic in comparison to the room is that the equivalence principle specifically assumes you are only making measurements in a small enough region of spacetime that tidal forces can be treated as negligible--tidal forces have to do with the fact that the gravitational field actually varies over space and time, like the fact that the Earth's gravitational pull is noticeably stronger on the surface than it is millions of miles away, or that in a sufficiently large room you might notice that the path of dropped objects is not perfectly parallel since they're both moving on converging straight lines towards the center of the Earth. For the equivalence principle to work, you have to assume that the region of spacetime you're doing your experiments in is very small compared with the distances and times in which these kinds of effects would be noticeable--the equivalence principle is a limit principle in the calculus sense, you must either take the limit as the volume of the room (and the time-interval in which the measurements are made) goes to zero, or else maybe you could take the limit as the size of the source of the gravitational field goes to infinity. Either way, if either the room or the person inside had enough mass so that their own gravitational effects were non-negligible, this would obviously imply measurable tidal forces (since different points inside the room would experience different gravity due to different distances from the person/sides of the room).

If what you're basically concerned about is the fact that gravitational attraction should be reciprocal, then maybe we could think of it this way. Suppose we have a room on Earth in which we're doing experiments, and then I want to know if I would get precisely the same results if the room were instead in deep space being accelerated at 1G by the rocket. Because gravity is reciprocal, when I move around in the room on Earth, this actually has some slight gravitational effects on the Earth itself--if I climb a ladder to the top of the room my body's pull on the Earth is weaker, while if I stand near the floor my body's pull on the Earth is stronger. Of course because the Earth's mass is so huge, my movements will only affect it to a tiny degree, but these small movements of the Earth would nevertheless have a slight effect on the gravity experienced within the room, which would in principle be measurable (assuming for the sake of the argument that there were no other objects on Earth moving around and affecting its movements too). So if we want to have a really accurate simulation of conditions on Earth by accelerating the room at 1G, it seems we can't just accelerate the room at precisely 1G for all time--we'd have to very slightly alter the rate of acceleration as the person inside moved around to simulate these sorts of tiny changes in the Earth's gravitational pull due to the Earth moving around in response to the person's movements.

Would this be a reasonable example of what you're trying to argue, or am I still missing the point? If it is an OK example, then my answer would be that, again, the equivalence principle is a limit principle, the effects of the person's movement on the entire Earth (and thus on the gravitational field of the Earth as measured inside the room) will go to zero in the limit as the room and everything inside it become smaller and smaller.
 
  • #35
JesseM
Science Advisor
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Anyway, if B is contained by A and both are accelerating towards each other at a constant rate, wouldn't you agree that at the point where they touch the kinetic energy would increase over a period of time?
If the main thing you're concerned about is two objects pushing against each other, here's another example. Suppose we have two giant cubes--planet-sized, or at least asteroid-sized--mutually attracted by gravity, with a surface of each cube pressed against the other's surface. If something is trapped in between the two surfaces, it will experience a tremendous force from each side, squashing it flat as a pancake.

Now, as I said before, the equivalence principle is understood only to work as you zoom in on a small region of spacetime, so that tidal forces become negligible--the jargon for this would be something like "general relativity reduces to special relativity locally". So obviously when thinking about the equivalence principle, we can't consider a region large enough to encompass the full mass of each cube--the gravitational force would vary considerably in force and direction along a line from the center of one cube to the center of the other, for example. But you might think that even if you zoomed in on a small region where the surfaces of each cube were pressed against each other, there would be a problem with the equivalence principle because the force still wouldn't be uniform--an object squashed in the middle is experiencing a strong force in one direction from cube A, and a strong force in the opposite direction from cube B. The important thing to realize is that these equal and opposite forces in your small region are not gravitational. In your small region, the gravitational force should be pretty much uniform--in fact, if the cubes are equal in size and mass, there will be no significant gravitational force in that region. Instead, the force pressing the squashed object together is electromagnetic, it's the force that keeps the material that the cube is made out of basically rigid, so that if you push strongly on one side, the other side must move along with it. The average gravitational force throughout each cube is pulling each one strongly towards the other one, and individual regions which themselves are not experiencing much gravity are pulled along with the rest due to these electromagnetic bonds between atoms.

So, if you looked at the small chunks of each cube contained within your small region, and then you transported these two chunks into an identically-shaped region of deep space with no matter outside the region to create gravity, and you accelerated the region in such a way as to duplicate whatever gravitational force was experienced in the original region (again, if the cubes were identical in size and mass the gravity in the original region would be negligibly different from zero, so you wouldn't have to accelerate the second region in deep space at all), and you duplicated the electromagnetic forces on the boundary of the original region which were causing the two chunks of cube within that region to be pressed against each other so forcefully, then in this case there would be no significant difference between what would be experienced within this second region in deep space and the original region which was surrounded by those giant cubes. So there is no problem with the equivalence principle here--as long as you pick a small region within a gravitational field, then if you look at all the objects and non-gravitational forces both within and on the boundary of that region, then if you duplicate the same objects and non-gravitational forces in a similar region in deep space, and accelerate it to match the gravitational forces felt in the original region within the gravitational field, then measurements made within either region will give all the same results. And remember, because your first region within the gravitational field must always be small enough so that the gravity within the region is essentially uniform (no tidal forces), this means the second region in deep space just needs a single uniform acceleration, you never need to accelerate different parts of it in different ways to duplicate gravitational forces in different directions.
 

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