Would there be centrifugal force if I were alone in the universe? [duplicate]
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This question already has an answer here:
Is rotational motion relative to space?
7 answers
When I'm rotating, I feel centrifugal force.
But if I were the only one in the universe and rotating, wouldn't I just kinda be still (since I'm not rotating with respect to anything) or would there just be a force pulling me out?
What if there were another person a billion miles away, this should not change anything, right?
reference-frames centrifugal-force machs-principle
edited Aug 27 '18 at 10:07
Qmechanic♦
105k121881202
asked Aug 26 '18 at 23:34
Jake B.Jake B.
22933
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marked as duplicate by sammy gerbil, John Rennie, Kyle Kanos, stafusa, Qmechanic♦ Aug 28 '18 at 14:49
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
|
show 3 more comments
$begingroup$
This question already has an answer here:
Is rotational motion relative to space?
7 answers
When I'm rotating, I feel centrifugal force.
But if I were the only one in the universe and rotating, wouldn't I just kinda be still (since I'm not rotating with respect to anything) or would there just be a force pulling me out?
What if there were another person a billion miles away, this should not change anything, right?
reference-frames centrifugal-force machs-principle
edited Aug 27 '18 at 10:07
Qmechanic♦
105k121881202
asked Aug 26 '18 at 23:34
Jake B.Jake B.
22933
$endgroup$
marked as duplicate by sammy gerbil, John Rennie, Kyle Kanos, stafusa, Qmechanic♦ Aug 28 '18 at 14:49
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
5
$begingroup$
Before considering writing an answer, I first want to ask you if you're aware of the Kerr metric, the solution for a spacetime with just a rotating, eternal black hole.
$endgroup$
– Alfred Centauri
Aug 27 '18 at 0:14
1
$begingroup$
@AlfredCentauri That's an excellent point.
$endgroup$
– Aiman Al-Eryani
Aug 27 '18 at 9:56
2
$begingroup$
More on Mach's principle.
$endgroup$
– Qmechanic♦
Aug 27 '18 at 10:10
3
$begingroup$
You are a composite system from many elements, they can rotate to eachother.
$endgroup$
– peterh
Aug 27 '18 at 13:42
5
$begingroup$
Possible duplicate of Is rotational motion relative to space?
$endgroup$
– sammy gerbil
Aug 27 '18 at 23:05
|
show 3 more comments
$begingroup$
This question already has an answer here:
Is rotational motion relative to space?
7 answers
When I'm rotating, I feel centrifugal force.
But if I were the only one in the universe and rotating, wouldn't I just kinda be still (since I'm not rotating with respect to anything) or would there just be a force pulling me out?
What if there were another person a billion miles away, this should not change anything, right?
reference-frames centrifugal-force machs-principle
edited Aug 27 '18 at 10:07
Qmechanic♦
105k121881202
asked Aug 26 '18 at 23:34
Jake B.Jake B.
22933
$endgroup$
This question already has an answer here:
Is rotational motion relative to space?
7 answers
When I'm rotating, I feel centrifugal force.
But if I were the only one in the universe and rotating, wouldn't I just kinda be still (since I'm not rotating with respect to anything) or would there just be a force pulling me out?
What if there were another person a billion miles away, this should not change anything, right?
This question already has an answer here:
Is rotational motion relative to space?
7 answers
reference-frames centrifugal-force machs-principle
reference-frames centrifugal-force machs-principle
edited Aug 27 '18 at 10:07
Qmechanic♦
105k121881202
asked Aug 26 '18 at 23:34
Jake B.Jake B.
22933
edited Aug 27 '18 at 10:07
Qmechanic♦
105k121881202
asked Aug 26 '18 at 23:34
Jake B.Jake B.
22933
edited Aug 27 '18 at 10:07
Qmechanic♦
105k121881202
edited Aug 27 '18 at 10:07
Qmechanic♦
105k121881202
edited Aug 27 '18 at 10:07
Qmechanic♦
105k121881202
105k121881202
asked Aug 26 '18 at 23:34
Jake B.Jake B.
22933
asked Aug 26 '18 at 23:34
Jake B.Jake B.
22933
asked Aug 26 '18 at 23:34
Jake B.Jake B.
22933
22933
marked as duplicate by sammy gerbil, John Rennie, Kyle Kanos, stafusa, Qmechanic♦ Aug 28 '18 at 14:49
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by sammy gerbil, John Rennie, Kyle Kanos, stafusa, Qmechanic♦ Aug 28 '18 at 14:49
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
5
$begingroup$
Before considering writing an answer, I first want to ask you if you're aware of the Kerr metric, the solution for a spacetime with just a rotating, eternal black hole.
$endgroup$
– Alfred Centauri
Aug 27 '18 at 0:14
1
$begingroup$
@AlfredCentauri That's an excellent point.
$endgroup$
– Aiman Al-Eryani
Aug 27 '18 at 9:56
2
$begingroup$
More on Mach's principle.
$endgroup$
– Qmechanic♦
Aug 27 '18 at 10:10
3
$begingroup$
You are a composite system from many elements, they can rotate to eachother.
$endgroup$
– peterh
Aug 27 '18 at 13:42
5
$begingroup$
Possible duplicate of Is rotational motion relative to space?
$endgroup$
– sammy gerbil
Aug 27 '18 at 23:05
|
show 3 more comments
5
$begingroup$
Before considering writing an answer, I first want to ask you if you're aware of the Kerr metric, the solution for a spacetime with just a rotating, eternal black hole.
$endgroup$
– Alfred Centauri
Aug 27 '18 at 0:14
1
$begingroup$
@AlfredCentauri That's an excellent point.
$endgroup$
– Aiman Al-Eryani
Aug 27 '18 at 9:56
2
$begingroup$
More on Mach's principle.
$endgroup$
– Qmechanic♦
Aug 27 '18 at 10:10
3
$begingroup$
You are a composite system from many elements, they can rotate to eachother.
$endgroup$
– peterh
Aug 27 '18 at 13:42
5
$begingroup$
Possible duplicate of Is rotational motion relative to space?
$endgroup$
– sammy gerbil
Aug 27 '18 at 23:05
5
5
$begingroup$
Before considering writing an answer, I first want to ask you if you're aware of the Kerr metric, the solution for a spacetime with just a rotating, eternal black hole.
$endgroup$
– Alfred Centauri
Aug 27 '18 at 0:14
$begingroup$
Before considering writing an answer, I first want to ask you if you're aware of the Kerr metric, the solution for a spacetime with just a rotating, eternal black hole.
$endgroup$
– Alfred Centauri
Aug 27 '18 at 0:14
1
1
$begingroup$
@AlfredCentauri That's an excellent point.
$endgroup$
– Aiman Al-Eryani
Aug 27 '18 at 9:56
$begingroup$
@AlfredCentauri That's an excellent point.
$endgroup$
– Aiman Al-Eryani
Aug 27 '18 at 9:56
2
2
$begingroup$
More on Mach's principle.
$endgroup$
– Qmechanic♦
Aug 27 '18 at 10:10
$begingroup$
More on Mach's principle.
$endgroup$
– Qmechanic♦
Aug 27 '18 at 10:10
3
3
$begingroup$
You are a composite system from many elements, they can rotate to eachother.
$endgroup$
– peterh
Aug 27 '18 at 13:42
$begingroup$
You are a composite system from many elements, they can rotate to eachother.
$endgroup$
– peterh
Aug 27 '18 at 13:42
5
5
$begingroup$
Possible duplicate of Is rotational motion relative to space?
$endgroup$
– sammy gerbil
Aug 27 '18 at 23:05
$begingroup$
Possible duplicate of Is rotational motion relative to space?
$endgroup$
– sammy gerbil
Aug 27 '18 at 23:05
|
show 3 more comments
4 Answers
4
active
oldest
votes
$begingroup$
What you're basically asking about is Mach's principle. You might want to read the presentation of an example like this in Einstein's paper "The foundation of the general theory of relativity," which can be found in translation online.
According to general relativity, the answer to your question is yes, there would be a centrifugal force. However, there are alternative theories of gravity that are more Machian. One is called Brans-Dicke gravity. The original paper on this theory, C. Brans and R. H. Dicke, "Mach’s Principle and a Relativistic Theory of Gravitation," Physical Review 124 (1961) 925, starts off with a simple and readable physical example in the same style as the ones we've been discussing.
Brans-Dicke gravity was very successful in setting off a lot of interesting tests of general relativity back in the 1970's. The tests were all consistent with general relativity. Brans-Dicke gravity involves a parameter that makes it go over to general relativity in a certain limit. The value of that parameter has been constrained by solar system tests to have a value so close to the GR value that BD gravity is no longer considered to be much of a contender.
So the answer to your question is yes, there would be a centrifugal force. Rotational motion is not relative.
As a side note, I think a lot of people are turned off by discussions of Mach's principle, because they're often introduced in terms of "suppose the universe were almost empty," which is counterfactual. That makes it sound like useless philosophizing. But BD gravity is a real theory. The stuff about an empty universe is just motivation.
Alfred Centauri pointed out in a comment the existence of:
the Kerr metric, the solution for a spacetime with just a rotating, eternal black hole.
If Einstein had known about this solution circa 1920, he would have been disturbed and upset, because his ambition was for GR to turn out more Machian than it really was. In fact, he was upset by the Schwarzschild solution for similar reasons.
answered Aug 27 '18 at 1:03
Ben CrowellBen Crowell
51.2k6156302
$endgroup$
1
$begingroup$
Does Mach's principle require the existence of "the distant stars"? This question is probably be meaningless, but: does your answer still apply if there is truly only one object in the universe? I ask because my intermediate mechanics instructor (Dan Kleppner) asked this very question at the end of a lecture. He said "Some would say there is rotation relative to the distant stars." He shrugged, and left the room.
$endgroup$
– garyp
Aug 27 '18 at 3:26
1
$begingroup$
Didnt Thirring show that if other masses.would rotate your rotation.would only be relatively to them? All in standard GR.
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– lalala
Aug 27 '18 at 5:44
$begingroup$
@garyp: You might want to read the Brans-Dicke paper. It's very readable. Mach's principle is vague, but Brans-Dicke gravity is a theory that makes predictions.
$endgroup$
– Ben Crowell
Aug 27 '18 at 13:07
2
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It’s very confusing that the question asks about centrifugal forces, and your answer says “yes, there would be a centripetal force”—which is it? It seems that it must be either yes, there is a centrifugal force as suggested, or no, the force would be centripetal, instead. And if it is somehow neither or both of those things, I think the answer should explain that.
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– KRyan
Aug 27 '18 at 17:06
1
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@KRyan: The OP's original question said "centripetal," and Qmechanic's edit changed it to "centrifugal." I think Qmechanic's edit actually improves the question, because it's more logical to talk about the centrifugal force, which in Newtonian mechanics is a fictitious force. (In GR, we don't really have gravitational or inertial "forces" at all.) I repeated "centripetal" in my answer without thinking carefully about it. I'll change it to "centrifugal." For completeness, there is also a Coriolis force.
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– Ben Crowell
Aug 27 '18 at 21:38
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show 1 more comment
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How did you start rotating and how do you know in the first place that you are? :)
I'd image you start alone, but to rotate you have to accelarate for which you have you create thrust that will emit particles in the other direction what in turn means you are no longer alone...
answered Aug 27 '18 at 9:52
Martin RauscherMartin Rauscher
1791
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3
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I'm guessing the idea is you're rotating if you feel centrifugal force.
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– Aiman Al-Eryani
Aug 27 '18 at 12:09
2
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You may be rotating your tail, as cats do (or other limb :) )
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– Ruslan
Aug 27 '18 at 17:01
1
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Why do you assume OP would start with no rotating?
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– JiK
Aug 27 '18 at 20:55
add a comment |
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As always, the real answer is that we don't know. We have no way of conducting the experiment. There are arguments both ways. So, we don't know.
The argument in favour of centripetal force is that things moving around in circles in not a natural movement and you need to use a force to make it happen.
The argument against is that it seems reasonable to assume that the universe as a whole cannot have an angular momentum (momentum of rotation). If you are the only object in the universe, then you cannot have angular momentum, cannot rotate. So, no force.
Here, the question has chanced a bit and become "If you are the only being in the universe, is it possible for you to rotate?"
This might seem like a different question, but they are connected by the definition of "rotate".
Ponder this: If you were the only object in the universe, how would you know if you were rotating?
And the answer is: "If you need a force to keep your arms and legs in place." With no remote "fixed" stars, this is the only way know if you are rotating.
This gives us two alternatives:
- There is no force. Which means there is no rotation. OR,
- There is a force. Which means that rotation is possible.
Since we can't do experiments, we can't tell which of these would be the case.
Adding another person to the universe, regardless how far away, changes everything. Now the two of you can have opposite angular momenta so that the total momentum of the universe is zero.
And you can measure you momentum by looking at the other person.
answered Aug 27 '18 at 9:15
Stig HemmerStig Hemmer
21724
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3
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"it seems reasonable to assume that the universe as a whole cannot have an angular momentum"... Does it? Because if the universe currently has zero angular momentum and I'm not at the dead center of it, I can easily give it angular momentum by linearly accelerating.
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– Sneftel
Aug 27 '18 at 14:55
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@Sneftel To accelerate, you have to give something else momentum in the opposite direction.
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– Acccumulation
Aug 27 '18 at 18:09
1
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@Acccumulation He's making a point about reference frames. Angular momentum is not reference-frame independent. If you think a far-away object is still and not rotating, and then you accelerate relative to it, you'll think it has linear (and thus angular!) momentum in the opposite direction.
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– Jahan Claes
Aug 27 '18 at 20:23
add a comment |
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The pseudo-forces such as centripetal acceleration are the result of using a rotating frame of reference. They apply in all cases, even if there are no other objects. Note that they are pseudo-forces. They're not actually forces, but rather mathematical results of using rotating frames which happen to appear to act as forces.
So the real question would be what equations of motion make sense in your universe. Do they line up with the equations of motion of an inertial frame (aka Netwonian frame), or do they line up with the equations of motion in a rotating frame.
Because you start from the declaration that you are alone in the universe and rotating, it is reasonable to assume that the equations of motion you use to predict the path of objects will be those of a rotating frame. That's simply because you used a word which has that meaning.
Now if you were the only object in the universe, and were roughly a point-mass, it would be enormously difficult to measure your rotation, but that's another question entirely.
answered Aug 27 '18 at 5:12
Cort AmmonCort Ammon
23.2k34776
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4
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centripetal forces are not usually a pseudo-force. For example if your rotating frame follows two weights at each end of a string and those are rotating around their barycenter the force that opposes the centrifugal force would be tension forces of the string.
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– Taemyr
Aug 27 '18 at 8:35
1
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Your second sentence is pretty misleading. Remember that in classical mechanics we assume the metric (flat Euclidean space). When we then do calculation in non-inertial frames, we see the pseudo-forces arrive. The whole point of GR is that the metric itself is a result of the distribution of matter, so shouldn’t assume the metric to answer OP’s question. Sure in GR we have Minkowski space, which is as Euclidean as it gets. Minkowski space is a great model for the space between stars, but we don’t know what the universe would look like were there no matter.
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– Andrea
Aug 27 '18 at 13:27
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Well, a point mass can't rotate by definition. The things like quantum angular momentum that we ascribe to putatively point objects (nuclear particles) is an observed value, not necessarily an actual rotation.
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– Carl Witthoft
Aug 27 '18 at 18:32
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
What you're basically asking about is Mach's principle. You might want to read the presentation of an example like this in Einstein's paper "The foundation of the general theory of relativity," which can be found in translation online.
According to general relativity, the answer to your question is yes, there would be a centrifugal force. However, there are alternative theories of gravity that are more Machian. One is called Brans-Dicke gravity. The original paper on this theory, C. Brans and R. H. Dicke, "Mach’s Principle and a Relativistic Theory of Gravitation," Physical Review 124 (1961) 925, starts off with a simple and readable physical example in the same style as the ones we've been discussing.
Brans-Dicke gravity was very successful in setting off a lot of interesting tests of general relativity back in the 1970's. The tests were all consistent with general relativity. Brans-Dicke gravity involves a parameter that makes it go over to general relativity in a certain limit. The value of that parameter has been constrained by solar system tests to have a value so close to the GR value that BD gravity is no longer considered to be much of a contender.
So the answer to your question is yes, there would be a centrifugal force. Rotational motion is not relative.
As a side note, I think a lot of people are turned off by discussions of Mach's principle, because they're often introduced in terms of "suppose the universe were almost empty," which is counterfactual. That makes it sound like useless philosophizing. But BD gravity is a real theory. The stuff about an empty universe is just motivation.
Alfred Centauri pointed out in a comment the existence of:
the Kerr metric, the solution for a spacetime with just a rotating, eternal black hole.
If Einstein had known about this solution circa 1920, he would have been disturbed and upset, because his ambition was for GR to turn out more Machian than it really was. In fact, he was upset by the Schwarzschild solution for similar reasons.
answered Aug 27 '18 at 1:03
Ben CrowellBen Crowell
51.2k6156302
$endgroup$
1
$begingroup$
Does Mach's principle require the existence of "the distant stars"? This question is probably be meaningless, but: does your answer still apply if there is truly only one object in the universe? I ask because my intermediate mechanics instructor (Dan Kleppner) asked this very question at the end of a lecture. He said "Some would say there is rotation relative to the distant stars." He shrugged, and left the room.
$endgroup$
– garyp
Aug 27 '18 at 3:26
1
$begingroup$
Didnt Thirring show that if other masses.would rotate your rotation.would only be relatively to them? All in standard GR.
$endgroup$
– lalala
Aug 27 '18 at 5:44
$begingroup$
@garyp: You might want to read the Brans-Dicke paper. It's very readable. Mach's principle is vague, but Brans-Dicke gravity is a theory that makes predictions.
$endgroup$
– Ben Crowell
Aug 27 '18 at 13:07
2
$begingroup$
It’s very confusing that the question asks about centrifugal forces, and your answer says “yes, there would be a centripetal force”—which is it? It seems that it must be either yes, there is a centrifugal force as suggested, or no, the force would be centripetal, instead. And if it is somehow neither or both of those things, I think the answer should explain that.
$endgroup$
– KRyan
Aug 27 '18 at 17:06
1
$begingroup$
@KRyan: The OP's original question said "centripetal," and Qmechanic's edit changed it to "centrifugal." I think Qmechanic's edit actually improves the question, because it's more logical to talk about the centrifugal force, which in Newtonian mechanics is a fictitious force. (In GR, we don't really have gravitational or inertial "forces" at all.) I repeated "centripetal" in my answer without thinking carefully about it. I'll change it to "centrifugal." For completeness, there is also a Coriolis force.
$endgroup$
– Ben Crowell
Aug 27 '18 at 21:38
|
show 1 more comment
$begingroup$
What you're basically asking about is Mach's principle. You might want to read the presentation of an example like this in Einstein's paper "The foundation of the general theory of relativity," which can be found in translation online.
According to general relativity, the answer to your question is yes, there would be a centrifugal force. However, there are alternative theories of gravity that are more Machian. One is called Brans-Dicke gravity. The original paper on this theory, C. Brans and R. H. Dicke, "Mach’s Principle and a Relativistic Theory of Gravitation," Physical Review 124 (1961) 925, starts off with a simple and readable physical example in the same style as the ones we've been discussing.
Brans-Dicke gravity was very successful in setting off a lot of interesting tests of general relativity back in the 1970's. The tests were all consistent with general relativity. Brans-Dicke gravity involves a parameter that makes it go over to general relativity in a certain limit. The value of that parameter has been constrained by solar system tests to have a value so close to the GR value that BD gravity is no longer considered to be much of a contender.
So the answer to your question is yes, there would be a centrifugal force. Rotational motion is not relative.
As a side note, I think a lot of people are turned off by discussions of Mach's principle, because they're often introduced in terms of "suppose the universe were almost empty," which is counterfactual. That makes it sound like useless philosophizing. But BD gravity is a real theory. The stuff about an empty universe is just motivation.
Alfred Centauri pointed out in a comment the existence of:
the Kerr metric, the solution for a spacetime with just a rotating, eternal black hole.
If Einstein had known about this solution circa 1920, he would have been disturbed and upset, because his ambition was for GR to turn out more Machian than it really was. In fact, he was upset by the Schwarzschild solution for similar reasons.
answered Aug 27 '18 at 1:03
Ben CrowellBen Crowell
51.2k6156302
$endgroup$
1
$begingroup$
Does Mach's principle require the existence of "the distant stars"? This question is probably be meaningless, but: does your answer still apply if there is truly only one object in the universe? I ask because my intermediate mechanics instructor (Dan Kleppner) asked this very question at the end of a lecture. He said "Some would say there is rotation relative to the distant stars." He shrugged, and left the room.
$endgroup$
– garyp
Aug 27 '18 at 3:26
1
$begingroup$
Didnt Thirring show that if other masses.would rotate your rotation.would only be relatively to them? All in standard GR.
$endgroup$
– lalala
Aug 27 '18 at 5:44
$begingroup$
@garyp: You might want to read the Brans-Dicke paper. It's very readable. Mach's principle is vague, but Brans-Dicke gravity is a theory that makes predictions.
$endgroup$
– Ben Crowell
Aug 27 '18 at 13:07
2
$begingroup$
It’s very confusing that the question asks about centrifugal forces, and your answer says “yes, there would be a centripetal force”—which is it? It seems that it must be either yes, there is a centrifugal force as suggested, or no, the force would be centripetal, instead. And if it is somehow neither or both of those things, I think the answer should explain that.
$endgroup$
– KRyan
Aug 27 '18 at 17:06
1
$begingroup$
@KRyan: The OP's original question said "centripetal," and Qmechanic's edit changed it to "centrifugal." I think Qmechanic's edit actually improves the question, because it's more logical to talk about the centrifugal force, which in Newtonian mechanics is a fictitious force. (In GR, we don't really have gravitational or inertial "forces" at all.) I repeated "centripetal" in my answer without thinking carefully about it. I'll change it to "centrifugal." For completeness, there is also a Coriolis force.
$endgroup$
– Ben Crowell
Aug 27 '18 at 21:38
|
show 1 more comment
$begingroup$
What you're basically asking about is Mach's principle. You might want to read the presentation of an example like this in Einstein's paper "The foundation of the general theory of relativity," which can be found in translation online.
According to general relativity, the answer to your question is yes, there would be a centrifugal force. However, there are alternative theories of gravity that are more Machian. One is called Brans-Dicke gravity. The original paper on this theory, C. Brans and R. H. Dicke, "Mach’s Principle and a Relativistic Theory of Gravitation," Physical Review 124 (1961) 925, starts off with a simple and readable physical example in the same style as the ones we've been discussing.
Brans-Dicke gravity was very successful in setting off a lot of interesting tests of general relativity back in the 1970's. The tests were all consistent with general relativity. Brans-Dicke gravity involves a parameter that makes it go over to general relativity in a certain limit. The value of that parameter has been constrained by solar system tests to have a value so close to the GR value that BD gravity is no longer considered to be much of a contender.
So the answer to your question is yes, there would be a centrifugal force. Rotational motion is not relative.
As a side note, I think a lot of people are turned off by discussions of Mach's principle, because they're often introduced in terms of "suppose the universe were almost empty," which is counterfactual. That makes it sound like useless philosophizing. But BD gravity is a real theory. The stuff about an empty universe is just motivation.
Alfred Centauri pointed out in a comment the existence of:
the Kerr metric, the solution for a spacetime with just a rotating, eternal black hole.
If Einstein had known about this solution circa 1920, he would have been disturbed and upset, because his ambition was for GR to turn out more Machian than it really was. In fact, he was upset by the Schwarzschild solution for similar reasons.
answered Aug 27 '18 at 1:03
Ben CrowellBen Crowell
51.2k6156302
$endgroup$
What you're basically asking about is Mach's principle. You might want to read the presentation of an example like this in Einstein's paper "The foundation of the general theory of relativity," which can be found in translation online.
According to general relativity, the answer to your question is yes, there would be a centrifugal force. However, there are alternative theories of gravity that are more Machian. One is called Brans-Dicke gravity. The original paper on this theory, C. Brans and R. H. Dicke, "Mach’s Principle and a Relativistic Theory of Gravitation," Physical Review 124 (1961) 925, starts off with a simple and readable physical example in the same style as the ones we've been discussing.
Brans-Dicke gravity was very successful in setting off a lot of interesting tests of general relativity back in the 1970's. The tests were all consistent with general relativity. Brans-Dicke gravity involves a parameter that makes it go over to general relativity in a certain limit. The value of that parameter has been constrained by solar system tests to have a value so close to the GR value that BD gravity is no longer considered to be much of a contender.
So the answer to your question is yes, there would be a centrifugal force. Rotational motion is not relative.
As a side note, I think a lot of people are turned off by discussions of Mach's principle, because they're often introduced in terms of "suppose the universe were almost empty," which is counterfactual. That makes it sound like useless philosophizing. But BD gravity is a real theory. The stuff about an empty universe is just motivation.
Alfred Centauri pointed out in a comment the existence of:
the Kerr metric, the solution for a spacetime with just a rotating, eternal black hole.
If Einstein had known about this solution circa 1920, he would have been disturbed and upset, because his ambition was for GR to turn out more Machian than it really was. In fact, he was upset by the Schwarzschild solution for similar reasons.
answered Aug 27 '18 at 1:03
Ben CrowellBen Crowell
51.2k6156302
edited Aug 27 '18 at 21:39
answered Aug 27 '18 at 1:03
Ben CrowellBen Crowell
51.2k6156302
answered Aug 27 '18 at 1:03
Ben CrowellBen Crowell
51.2k6156302
answered Aug 27 '18 at 1:03
Ben CrowellBen Crowell
51.2k6156302
51.2k6156302
1
$begingroup$
Does Mach's principle require the existence of "the distant stars"? This question is probably be meaningless, but: does your answer still apply if there is truly only one object in the universe? I ask because my intermediate mechanics instructor (Dan Kleppner) asked this very question at the end of a lecture. He said "Some would say there is rotation relative to the distant stars." He shrugged, and left the room.
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– garyp
Aug 27 '18 at 3:26
1
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Didnt Thirring show that if other masses.would rotate your rotation.would only be relatively to them? All in standard GR.
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– lalala
Aug 27 '18 at 5:44
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@garyp: You might want to read the Brans-Dicke paper. It's very readable. Mach's principle is vague, but Brans-Dicke gravity is a theory that makes predictions.
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– Ben Crowell
Aug 27 '18 at 13:07
2
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It’s very confusing that the question asks about centrifugal forces, and your answer says “yes, there would be a centripetal force”—which is it? It seems that it must be either yes, there is a centrifugal force as suggested, or no, the force would be centripetal, instead. And if it is somehow neither or both of those things, I think the answer should explain that.
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– KRyan
Aug 27 '18 at 17:06
1
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@KRyan: The OP's original question said "centripetal," and Qmechanic's edit changed it to "centrifugal." I think Qmechanic's edit actually improves the question, because it's more logical to talk about the centrifugal force, which in Newtonian mechanics is a fictitious force. (In GR, we don't really have gravitational or inertial "forces" at all.) I repeated "centripetal" in my answer without thinking carefully about it. I'll change it to "centrifugal." For completeness, there is also a Coriolis force.
$endgroup$
– Ben Crowell
Aug 27 '18 at 21:38
|
show 1 more comment
1
$begingroup$
Does Mach's principle require the existence of "the distant stars"? This question is probably be meaningless, but: does your answer still apply if there is truly only one object in the universe? I ask because my intermediate mechanics instructor (Dan Kleppner) asked this very question at the end of a lecture. He said "Some would say there is rotation relative to the distant stars." He shrugged, and left the room.
$endgroup$
– garyp
Aug 27 '18 at 3:26
1
$begingroup$
Didnt Thirring show that if other masses.would rotate your rotation.would only be relatively to them? All in standard GR.
$endgroup$
– lalala
Aug 27 '18 at 5:44
$begingroup$
@garyp: You might want to read the Brans-Dicke paper. It's very readable. Mach's principle is vague, but Brans-Dicke gravity is a theory that makes predictions.
$endgroup$
– Ben Crowell
Aug 27 '18 at 13:07
2
$begingroup$
It’s very confusing that the question asks about centrifugal forces, and your answer says “yes, there would be a centripetal force”—which is it? It seems that it must be either yes, there is a centrifugal force as suggested, or no, the force would be centripetal, instead. And if it is somehow neither or both of those things, I think the answer should explain that.
$endgroup$
– KRyan
Aug 27 '18 at 17:06
1
$begingroup$
@KRyan: The OP's original question said "centripetal," and Qmechanic's edit changed it to "centrifugal." I think Qmechanic's edit actually improves the question, because it's more logical to talk about the centrifugal force, which in Newtonian mechanics is a fictitious force. (In GR, we don't really have gravitational or inertial "forces" at all.) I repeated "centripetal" in my answer without thinking carefully about it. I'll change it to "centrifugal." For completeness, there is also a Coriolis force.
$endgroup$
– Ben Crowell
Aug 27 '18 at 21:38
1
1
$begingroup$
Does Mach's principle require the existence of "the distant stars"? This question is probably be meaningless, but: does your answer still apply if there is truly only one object in the universe? I ask because my intermediate mechanics instructor (Dan Kleppner) asked this very question at the end of a lecture. He said "Some would say there is rotation relative to the distant stars." He shrugged, and left the room.
$endgroup$
– garyp
Aug 27 '18 at 3:26
$begingroup$
Does Mach's principle require the existence of "the distant stars"? This question is probably be meaningless, but: does your answer still apply if there is truly only one object in the universe? I ask because my intermediate mechanics instructor (Dan Kleppner) asked this very question at the end of a lecture. He said "Some would say there is rotation relative to the distant stars." He shrugged, and left the room.
$endgroup$
– garyp
Aug 27 '18 at 3:26
1
1
$begingroup$
Didnt Thirring show that if other masses.would rotate your rotation.would only be relatively to them? All in standard GR.
$endgroup$
– lalala
Aug 27 '18 at 5:44
$begingroup$
Didnt Thirring show that if other masses.would rotate your rotation.would only be relatively to them? All in standard GR.
$endgroup$
– lalala
Aug 27 '18 at 5:44
$begingroup$
@garyp: You might want to read the Brans-Dicke paper. It's very readable. Mach's principle is vague, but Brans-Dicke gravity is a theory that makes predictions.
$endgroup$
– Ben Crowell
Aug 27 '18 at 13:07
$begingroup$
@garyp: You might want to read the Brans-Dicke paper. It's very readable. Mach's principle is vague, but Brans-Dicke gravity is a theory that makes predictions.
$endgroup$
– Ben Crowell
Aug 27 '18 at 13:07
2
2
$begingroup$
It’s very confusing that the question asks about centrifugal forces, and your answer says “yes, there would be a centripetal force”—which is it? It seems that it must be either yes, there is a centrifugal force as suggested, or no, the force would be centripetal, instead. And if it is somehow neither or both of those things, I think the answer should explain that.
$endgroup$
– KRyan
Aug 27 '18 at 17:06
$begingroup$
It’s very confusing that the question asks about centrifugal forces, and your answer says “yes, there would be a centripetal force”—which is it? It seems that it must be either yes, there is a centrifugal force as suggested, or no, the force would be centripetal, instead. And if it is somehow neither or both of those things, I think the answer should explain that.
$endgroup$
– KRyan
Aug 27 '18 at 17:06
1
1
$begingroup$
@KRyan: The OP's original question said "centripetal," and Qmechanic's edit changed it to "centrifugal." I think Qmechanic's edit actually improves the question, because it's more logical to talk about the centrifugal force, which in Newtonian mechanics is a fictitious force. (In GR, we don't really have gravitational or inertial "forces" at all.) I repeated "centripetal" in my answer without thinking carefully about it. I'll change it to "centrifugal." For completeness, there is also a Coriolis force.
$endgroup$
– Ben Crowell
Aug 27 '18 at 21:38
$begingroup$
@KRyan: The OP's original question said "centripetal," and Qmechanic's edit changed it to "centrifugal." I think Qmechanic's edit actually improves the question, because it's more logical to talk about the centrifugal force, which in Newtonian mechanics is a fictitious force. (In GR, we don't really have gravitational or inertial "forces" at all.) I repeated "centripetal" in my answer without thinking carefully about it. I'll change it to "centrifugal." For completeness, there is also a Coriolis force.
$endgroup$
– Ben Crowell
Aug 27 '18 at 21:38
|
show 1 more comment
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How did you start rotating and how do you know in the first place that you are? :)
I'd image you start alone, but to rotate you have to accelarate for which you have you create thrust that will emit particles in the other direction what in turn means you are no longer alone...
answered Aug 27 '18 at 9:52
Martin RauscherMartin Rauscher
1791
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3
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I'm guessing the idea is you're rotating if you feel centrifugal force.
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– Aiman Al-Eryani
Aug 27 '18 at 12:09
2
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You may be rotating your tail, as cats do (or other limb :) )
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– Ruslan
Aug 27 '18 at 17:01
1
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Why do you assume OP would start with no rotating?
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– JiK
Aug 27 '18 at 20:55
add a comment |
$begingroup$
How did you start rotating and how do you know in the first place that you are? :)
I'd image you start alone, but to rotate you have to accelarate for which you have you create thrust that will emit particles in the other direction what in turn means you are no longer alone...
answered Aug 27 '18 at 9:52
Martin RauscherMartin Rauscher
1791
$endgroup$
3
$begingroup$
I'm guessing the idea is you're rotating if you feel centrifugal force.
$endgroup$
– Aiman Al-Eryani
Aug 27 '18 at 12:09
2
$begingroup$
You may be rotating your tail, as cats do (or other limb :) )
$endgroup$
– Ruslan
Aug 27 '18 at 17:01
1
$begingroup$
Why do you assume OP would start with no rotating?
$endgroup$
– JiK
Aug 27 '18 at 20:55
add a comment |
$begingroup$
How did you start rotating and how do you know in the first place that you are? :)
I'd image you start alone, but to rotate you have to accelarate for which you have you create thrust that will emit particles in the other direction what in turn means you are no longer alone...
answered Aug 27 '18 at 9:52
Martin RauscherMartin Rauscher
1791
$endgroup$
How did you start rotating and how do you know in the first place that you are? :)
I'd image you start alone, but to rotate you have to accelarate for which you have you create thrust that will emit particles in the other direction what in turn means you are no longer alone...
answered Aug 27 '18 at 9:52
Martin RauscherMartin Rauscher
1791
answered Aug 27 '18 at 9:52
Martin RauscherMartin Rauscher
1791
answered Aug 27 '18 at 9:52
Martin RauscherMartin Rauscher
1791
answered Aug 27 '18 at 9:52
Martin RauscherMartin Rauscher
1791
1791
3
$begingroup$
I'm guessing the idea is you're rotating if you feel centrifugal force.
$endgroup$
– Aiman Al-Eryani
Aug 27 '18 at 12:09
2
$begingroup$
You may be rotating your tail, as cats do (or other limb :) )
$endgroup$
– Ruslan
Aug 27 '18 at 17:01
1
$begingroup$
Why do you assume OP would start with no rotating?
$endgroup$
– JiK
Aug 27 '18 at 20:55
add a comment |
3
$begingroup$
I'm guessing the idea is you're rotating if you feel centrifugal force.
$endgroup$
– Aiman Al-Eryani
Aug 27 '18 at 12:09
2
$begingroup$
You may be rotating your tail, as cats do (or other limb :) )
$endgroup$
– Ruslan
Aug 27 '18 at 17:01
1
$begingroup$
Why do you assume OP would start with no rotating?
$endgroup$
– JiK
Aug 27 '18 at 20:55
3
3
$begingroup$
I'm guessing the idea is you're rotating if you feel centrifugal force.
$endgroup$
– Aiman Al-Eryani
Aug 27 '18 at 12:09
$begingroup$
I'm guessing the idea is you're rotating if you feel centrifugal force.
$endgroup$
– Aiman Al-Eryani
Aug 27 '18 at 12:09
2
2
$begingroup$
You may be rotating your tail, as cats do (or other limb :) )
$endgroup$
– Ruslan
Aug 27 '18 at 17:01
$begingroup$
You may be rotating your tail, as cats do (or other limb :) )
$endgroup$
– Ruslan
Aug 27 '18 at 17:01
1
1
$begingroup$
Why do you assume OP would start with no rotating?
$endgroup$
– JiK
Aug 27 '18 at 20:55
$begingroup$
Why do you assume OP would start with no rotating?
$endgroup$
– JiK
Aug 27 '18 at 20:55
add a comment |
$begingroup$
As always, the real answer is that we don't know. We have no way of conducting the experiment. There are arguments both ways. So, we don't know.
The argument in favour of centripetal force is that things moving around in circles in not a natural movement and you need to use a force to make it happen.
The argument against is that it seems reasonable to assume that the universe as a whole cannot have an angular momentum (momentum of rotation). If you are the only object in the universe, then you cannot have angular momentum, cannot rotate. So, no force.
Here, the question has chanced a bit and become "If you are the only being in the universe, is it possible for you to rotate?"
This might seem like a different question, but they are connected by the definition of "rotate".
Ponder this: If you were the only object in the universe, how would you know if you were rotating?
And the answer is: "If you need a force to keep your arms and legs in place." With no remote "fixed" stars, this is the only way know if you are rotating.
This gives us two alternatives:
- There is no force. Which means there is no rotation. OR,
- There is a force. Which means that rotation is possible.
Since we can't do experiments, we can't tell which of these would be the case.
Adding another person to the universe, regardless how far away, changes everything. Now the two of you can have opposite angular momenta so that the total momentum of the universe is zero.
And you can measure you momentum by looking at the other person.
answered Aug 27 '18 at 9:15
Stig HemmerStig Hemmer
21724
$endgroup$
3
$begingroup$
"it seems reasonable to assume that the universe as a whole cannot have an angular momentum"... Does it? Because if the universe currently has zero angular momentum and I'm not at the dead center of it, I can easily give it angular momentum by linearly accelerating.
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– Sneftel
Aug 27 '18 at 14:55
$begingroup$
@Sneftel To accelerate, you have to give something else momentum in the opposite direction.
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– Acccumulation
Aug 27 '18 at 18:09
1
$begingroup$
@Acccumulation He's making a point about reference frames. Angular momentum is not reference-frame independent. If you think a far-away object is still and not rotating, and then you accelerate relative to it, you'll think it has linear (and thus angular!) momentum in the opposite direction.
$endgroup$
– Jahan Claes
Aug 27 '18 at 20:23
add a comment |
$begingroup$
As always, the real answer is that we don't know. We have no way of conducting the experiment. There are arguments both ways. So, we don't know.
The argument in favour of centripetal force is that things moving around in circles in not a natural movement and you need to use a force to make it happen.
The argument against is that it seems reasonable to assume that the universe as a whole cannot have an angular momentum (momentum of rotation). If you are the only object in the universe, then you cannot have angular momentum, cannot rotate. So, no force.
Here, the question has chanced a bit and become "If you are the only being in the universe, is it possible for you to rotate?"
This might seem like a different question, but they are connected by the definition of "rotate".
Ponder this: If you were the only object in the universe, how would you know if you were rotating?
And the answer is: "If you need a force to keep your arms and legs in place." With no remote "fixed" stars, this is the only way know if you are rotating.
This gives us two alternatives:
- There is no force. Which means there is no rotation. OR,
- There is a force. Which means that rotation is possible.
Since we can't do experiments, we can't tell which of these would be the case.
Adding another person to the universe, regardless how far away, changes everything. Now the two of you can have opposite angular momenta so that the total momentum of the universe is zero.
And you can measure you momentum by looking at the other person.
answered Aug 27 '18 at 9:15
Stig HemmerStig Hemmer
21724
$endgroup$
3
$begingroup$
"it seems reasonable to assume that the universe as a whole cannot have an angular momentum"... Does it? Because if the universe currently has zero angular momentum and I'm not at the dead center of it, I can easily give it angular momentum by linearly accelerating.
$endgroup$
– Sneftel
Aug 27 '18 at 14:55
$begingroup$
@Sneftel To accelerate, you have to give something else momentum in the opposite direction.
$endgroup$
– Acccumulation
Aug 27 '18 at 18:09
1
$begingroup$
@Acccumulation He's making a point about reference frames. Angular momentum is not reference-frame independent. If you think a far-away object is still and not rotating, and then you accelerate relative to it, you'll think it has linear (and thus angular!) momentum in the opposite direction.
$endgroup$
– Jahan Claes
Aug 27 '18 at 20:23
add a comment |
$begingroup$
As always, the real answer is that we don't know. We have no way of conducting the experiment. There are arguments both ways. So, we don't know.
The argument in favour of centripetal force is that things moving around in circles in not a natural movement and you need to use a force to make it happen.
The argument against is that it seems reasonable to assume that the universe as a whole cannot have an angular momentum (momentum of rotation). If you are the only object in the universe, then you cannot have angular momentum, cannot rotate. So, no force.
Here, the question has chanced a bit and become "If you are the only being in the universe, is it possible for you to rotate?"
This might seem like a different question, but they are connected by the definition of "rotate".
Ponder this: If you were the only object in the universe, how would you know if you were rotating?
And the answer is: "If you need a force to keep your arms and legs in place." With no remote "fixed" stars, this is the only way know if you are rotating.
This gives us two alternatives:
- There is no force. Which means there is no rotation. OR,
- There is a force. Which means that rotation is possible.
Since we can't do experiments, we can't tell which of these would be the case.
Adding another person to the universe, regardless how far away, changes everything. Now the two of you can have opposite angular momenta so that the total momentum of the universe is zero.
And you can measure you momentum by looking at the other person.
answered Aug 27 '18 at 9:15
Stig HemmerStig Hemmer
21724
$endgroup$
As always, the real answer is that we don't know. We have no way of conducting the experiment. There are arguments both ways. So, we don't know.
The argument in favour of centripetal force is that things moving around in circles in not a natural movement and you need to use a force to make it happen.
The argument against is that it seems reasonable to assume that the universe as a whole cannot have an angular momentum (momentum of rotation). If you are the only object in the universe, then you cannot have angular momentum, cannot rotate. So, no force.
Here, the question has chanced a bit and become "If you are the only being in the universe, is it possible for you to rotate?"
This might seem like a different question, but they are connected by the definition of "rotate".
Ponder this: If you were the only object in the universe, how would you know if you were rotating?
And the answer is: "If you need a force to keep your arms and legs in place." With no remote "fixed" stars, this is the only way know if you are rotating.
This gives us two alternatives:
- There is no force. Which means there is no rotation. OR,
- There is a force. Which means that rotation is possible.
Since we can't do experiments, we can't tell which of these would be the case.
Adding another person to the universe, regardless how far away, changes everything. Now the two of you can have opposite angular momenta so that the total momentum of the universe is zero.
And you can measure you momentum by looking at the other person.
answered Aug 27 '18 at 9:15
Stig HemmerStig Hemmer
21724
answered Aug 27 '18 at 9:15
Stig HemmerStig Hemmer
21724
answered Aug 27 '18 at 9:15
Stig HemmerStig Hemmer
21724
answered Aug 27 '18 at 9:15
Stig HemmerStig Hemmer
21724
21724
3
$begingroup$
"it seems reasonable to assume that the universe as a whole cannot have an angular momentum"... Does it? Because if the universe currently has zero angular momentum and I'm not at the dead center of it, I can easily give it angular momentum by linearly accelerating.
$endgroup$
– Sneftel
Aug 27 '18 at 14:55
$begingroup$
@Sneftel To accelerate, you have to give something else momentum in the opposite direction.
$endgroup$
– Acccumulation
Aug 27 '18 at 18:09
1
$begingroup$
@Acccumulation He's making a point about reference frames. Angular momentum is not reference-frame independent. If you think a far-away object is still and not rotating, and then you accelerate relative to it, you'll think it has linear (and thus angular!) momentum in the opposite direction.
$endgroup$
– Jahan Claes
Aug 27 '18 at 20:23
add a comment |
3
$begingroup$
"it seems reasonable to assume that the universe as a whole cannot have an angular momentum"... Does it? Because if the universe currently has zero angular momentum and I'm not at the dead center of it, I can easily give it angular momentum by linearly accelerating.
$endgroup$
– Sneftel
Aug 27 '18 at 14:55
$begingroup$
@Sneftel To accelerate, you have to give something else momentum in the opposite direction.
$endgroup$
– Acccumulation
Aug 27 '18 at 18:09
1
$begingroup$
@Acccumulation He's making a point about reference frames. Angular momentum is not reference-frame independent. If you think a far-away object is still and not rotating, and then you accelerate relative to it, you'll think it has linear (and thus angular!) momentum in the opposite direction.
$endgroup$
– Jahan Claes
Aug 27 '18 at 20:23
3
3
$begingroup$
"it seems reasonable to assume that the universe as a whole cannot have an angular momentum"... Does it? Because if the universe currently has zero angular momentum and I'm not at the dead center of it, I can easily give it angular momentum by linearly accelerating.
$endgroup$
– Sneftel
Aug 27 '18 at 14:55
$begingroup$
"it seems reasonable to assume that the universe as a whole cannot have an angular momentum"... Does it? Because if the universe currently has zero angular momentum and I'm not at the dead center of it, I can easily give it angular momentum by linearly accelerating.
$endgroup$
– Sneftel
Aug 27 '18 at 14:55
$begingroup$
@Sneftel To accelerate, you have to give something else momentum in the opposite direction.
$endgroup$
– Acccumulation
Aug 27 '18 at 18:09
$begingroup$
@Sneftel To accelerate, you have to give something else momentum in the opposite direction.
$endgroup$
– Acccumulation
Aug 27 '18 at 18:09
1
1
$begingroup$
@Acccumulation He's making a point about reference frames. Angular momentum is not reference-frame independent. If you think a far-away object is still and not rotating, and then you accelerate relative to it, you'll think it has linear (and thus angular!) momentum in the opposite direction.
$endgroup$
– Jahan Claes
Aug 27 '18 at 20:23
$begingroup$
@Acccumulation He's making a point about reference frames. Angular momentum is not reference-frame independent. If you think a far-away object is still and not rotating, and then you accelerate relative to it, you'll think it has linear (and thus angular!) momentum in the opposite direction.
$endgroup$
– Jahan Claes
Aug 27 '18 at 20:23
add a comment |
$begingroup$
The pseudo-forces such as centripetal acceleration are the result of using a rotating frame of reference. They apply in all cases, even if there are no other objects. Note that they are pseudo-forces. They're not actually forces, but rather mathematical results of using rotating frames which happen to appear to act as forces.
So the real question would be what equations of motion make sense in your universe. Do they line up with the equations of motion of an inertial frame (aka Netwonian frame), or do they line up with the equations of motion in a rotating frame.
Because you start from the declaration that you are alone in the universe and rotating, it is reasonable to assume that the equations of motion you use to predict the path of objects will be those of a rotating frame. That's simply because you used a word which has that meaning.
Now if you were the only object in the universe, and were roughly a point-mass, it would be enormously difficult to measure your rotation, but that's another question entirely.
answered Aug 27 '18 at 5:12
Cort AmmonCort Ammon
23.2k34776
$endgroup$
4
$begingroup$
centripetal forces are not usually a pseudo-force. For example if your rotating frame follows two weights at each end of a string and those are rotating around their barycenter the force that opposes the centrifugal force would be tension forces of the string.
$endgroup$
– Taemyr
Aug 27 '18 at 8:35
1
$begingroup$
Your second sentence is pretty misleading. Remember that in classical mechanics we assume the metric (flat Euclidean space). When we then do calculation in non-inertial frames, we see the pseudo-forces arrive. The whole point of GR is that the metric itself is a result of the distribution of matter, so shouldn’t assume the metric to answer OP’s question. Sure in GR we have Minkowski space, which is as Euclidean as it gets. Minkowski space is a great model for the space between stars, but we don’t know what the universe would look like were there no matter.
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– Andrea
Aug 27 '18 at 13:27
$begingroup$
Well, a point mass can't rotate by definition. The things like quantum angular momentum that we ascribe to putatively point objects (nuclear particles) is an observed value, not necessarily an actual rotation.
$endgroup$
– Carl Witthoft
Aug 27 '18 at 18:32
add a comment |
$begingroup$
The pseudo-forces such as centripetal acceleration are the result of using a rotating frame of reference. They apply in all cases, even if there are no other objects. Note that they are pseudo-forces. They're not actually forces, but rather mathematical results of using rotating frames which happen to appear to act as forces.
So the real question would be what equations of motion make sense in your universe. Do they line up with the equations of motion of an inertial frame (aka Netwonian frame), or do they line up with the equations of motion in a rotating frame.
Because you start from the declaration that you are alone in the universe and rotating, it is reasonable to assume that the equations of motion you use to predict the path of objects will be those of a rotating frame. That's simply because you used a word which has that meaning.
Now if you were the only object in the universe, and were roughly a point-mass, it would be enormously difficult to measure your rotation, but that's another question entirely.
answered Aug 27 '18 at 5:12
Cort AmmonCort Ammon
23.2k34776
$endgroup$
4
$begingroup$
centripetal forces are not usually a pseudo-force. For example if your rotating frame follows two weights at each end of a string and those are rotating around their barycenter the force that opposes the centrifugal force would be tension forces of the string.
$endgroup$
– Taemyr
Aug 27 '18 at 8:35
1
$begingroup$
Your second sentence is pretty misleading. Remember that in classical mechanics we assume the metric (flat Euclidean space). When we then do calculation in non-inertial frames, we see the pseudo-forces arrive. The whole point of GR is that the metric itself is a result of the distribution of matter, so shouldn’t assume the metric to answer OP’s question. Sure in GR we have Minkowski space, which is as Euclidean as it gets. Minkowski space is a great model for the space between stars, but we don’t know what the universe would look like were there no matter.
$endgroup$
– Andrea
Aug 27 '18 at 13:27
$begingroup$
Well, a point mass can't rotate by definition. The things like quantum angular momentum that we ascribe to putatively point objects (nuclear particles) is an observed value, not necessarily an actual rotation.
$endgroup$
– Carl Witthoft
Aug 27 '18 at 18:32
add a comment |
$begingroup$
The pseudo-forces such as centripetal acceleration are the result of using a rotating frame of reference. They apply in all cases, even if there are no other objects. Note that they are pseudo-forces. They're not actually forces, but rather mathematical results of using rotating frames which happen to appear to act as forces.
So the real question would be what equations of motion make sense in your universe. Do they line up with the equations of motion of an inertial frame (aka Netwonian frame), or do they line up with the equations of motion in a rotating frame.
Because you start from the declaration that you are alone in the universe and rotating, it is reasonable to assume that the equations of motion you use to predict the path of objects will be those of a rotating frame. That's simply because you used a word which has that meaning.
Now if you were the only object in the universe, and were roughly a point-mass, it would be enormously difficult to measure your rotation, but that's another question entirely.
answered Aug 27 '18 at 5:12
Cort AmmonCort Ammon
23.2k34776
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The pseudo-forces such as centripetal acceleration are the result of using a rotating frame of reference. They apply in all cases, even if there are no other objects. Note that they are pseudo-forces. They're not actually forces, but rather mathematical results of using rotating frames which happen to appear to act as forces.
So the real question would be what equations of motion make sense in your universe. Do they line up with the equations of motion of an inertial frame (aka Netwonian frame), or do they line up with the equations of motion in a rotating frame.
Because you start from the declaration that you are alone in the universe and rotating, it is reasonable to assume that the equations of motion you use to predict the path of objects will be those of a rotating frame. That's simply because you used a word which has that meaning.
Now if you were the only object in the universe, and were roughly a point-mass, it would be enormously difficult to measure your rotation, but that's another question entirely.
answered Aug 27 '18 at 5:12
Cort AmmonCort Ammon
23.2k34776
answered Aug 27 '18 at 5:12
Cort AmmonCort Ammon
23.2k34776
answered Aug 27 '18 at 5:12
Cort AmmonCort Ammon
23.2k34776
answered Aug 27 '18 at 5:12
Cort AmmonCort Ammon
23.2k34776
23.2k34776
4
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centripetal forces are not usually a pseudo-force. For example if your rotating frame follows two weights at each end of a string and those are rotating around their barycenter the force that opposes the centrifugal force would be tension forces of the string.
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– Taemyr
Aug 27 '18 at 8:35
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Your second sentence is pretty misleading. Remember that in classical mechanics we assume the metric (flat Euclidean space). When we then do calculation in non-inertial frames, we see the pseudo-forces arrive. The whole point of GR is that the metric itself is a result of the distribution of matter, so shouldn’t assume the metric to answer OP’s question. Sure in GR we have Minkowski space, which is as Euclidean as it gets. Minkowski space is a great model for the space between stars, but we don’t know what the universe would look like were there no matter.
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– Andrea
Aug 27 '18 at 13:27
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Well, a point mass can't rotate by definition. The things like quantum angular momentum that we ascribe to putatively point objects (nuclear particles) is an observed value, not necessarily an actual rotation.
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– Carl Witthoft
Aug 27 '18 at 18:32
add a comment |
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centripetal forces are not usually a pseudo-force. For example if your rotating frame follows two weights at each end of a string and those are rotating around their barycenter the force that opposes the centrifugal force would be tension forces of the string.
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– Taemyr
Aug 27 '18 at 8:35
1
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Your second sentence is pretty misleading. Remember that in classical mechanics we assume the metric (flat Euclidean space). When we then do calculation in non-inertial frames, we see the pseudo-forces arrive. The whole point of GR is that the metric itself is a result of the distribution of matter, so shouldn’t assume the metric to answer OP’s question. Sure in GR we have Minkowski space, which is as Euclidean as it gets. Minkowski space is a great model for the space between stars, but we don’t know what the universe would look like were there no matter.
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– Andrea
Aug 27 '18 at 13:27
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Well, a point mass can't rotate by definition. The things like quantum angular momentum that we ascribe to putatively point objects (nuclear particles) is an observed value, not necessarily an actual rotation.
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– Carl Witthoft
Aug 27 '18 at 18:32
4
4
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centripetal forces are not usually a pseudo-force. For example if your rotating frame follows two weights at each end of a string and those are rotating around their barycenter the force that opposes the centrifugal force would be tension forces of the string.
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– Taemyr
Aug 27 '18 at 8:35
$begingroup$
centripetal forces are not usually a pseudo-force. For example if your rotating frame follows two weights at each end of a string and those are rotating around their barycenter the force that opposes the centrifugal force would be tension forces of the string.
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– Taemyr
Aug 27 '18 at 8:35
1
1
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Your second sentence is pretty misleading. Remember that in classical mechanics we assume the metric (flat Euclidean space). When we then do calculation in non-inertial frames, we see the pseudo-forces arrive. The whole point of GR is that the metric itself is a result of the distribution of matter, so shouldn’t assume the metric to answer OP’s question. Sure in GR we have Minkowski space, which is as Euclidean as it gets. Minkowski space is a great model for the space between stars, but we don’t know what the universe would look like were there no matter.
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– Andrea
Aug 27 '18 at 13:27
$begingroup$
Your second sentence is pretty misleading. Remember that in classical mechanics we assume the metric (flat Euclidean space). When we then do calculation in non-inertial frames, we see the pseudo-forces arrive. The whole point of GR is that the metric itself is a result of the distribution of matter, so shouldn’t assume the metric to answer OP’s question. Sure in GR we have Minkowski space, which is as Euclidean as it gets. Minkowski space is a great model for the space between stars, but we don’t know what the universe would look like were there no matter.
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– Andrea
Aug 27 '18 at 13:27
$begingroup$
Well, a point mass can't rotate by definition. The things like quantum angular momentum that we ascribe to putatively point objects (nuclear particles) is an observed value, not necessarily an actual rotation.
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– Carl Witthoft
Aug 27 '18 at 18:32
$begingroup$
Well, a point mass can't rotate by definition. The things like quantum angular momentum that we ascribe to putatively point objects (nuclear particles) is an observed value, not necessarily an actual rotation.
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– Carl Witthoft
Aug 27 '18 at 18:32
add a comment |
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Before considering writing an answer, I first want to ask you if you're aware of the Kerr metric, the solution for a spacetime with just a rotating, eternal black hole.
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– Alfred Centauri
Aug 27 '18 at 0:14
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@AlfredCentauri That's an excellent point.
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– Aiman Al-Eryani
Aug 27 '18 at 9:56
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More on Mach's principle.
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– Qmechanic♦
Aug 27 '18 at 10:10
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You are a composite system from many elements, they can rotate to eachother.
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– peterh
Aug 27 '18 at 13:42
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Possible duplicate of Is rotational motion relative to space?
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– sammy gerbil
Aug 27 '18 at 23:05