valenumr
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I'm still working through it.Dale said:If a force is applied perpendicular to velocity then there is a deviation from a geodesic without energy transfer. As I showed above
I'm still working through it.Dale said:If a force is applied perpendicular to velocity then there is a deviation from a geodesic without energy transfer. As I showed above
I am still trying to reconcile how anything (macroscopic) can alter course without an energy input.Dale said:If a force is applied perpendicular to velocity then there is a deviation from a geodesic without energy transfer. As I showed above
I am guessing here that taking one's foot off the accelerator, putting the vehicle in neutral and turning the steering wheel counts as an "energy input" in your book?valenumr said:I am still trying to reconcile how anything (macroscopic) can alter course without an energy input.
Did you not follow the explanation above? If so, perhaps start with pointing out the place where you first get confused. We can go from there.valenumr said:I am still trying to reconcile how anything (macroscopic) can alter course without an energy input.
I would recommend pausing until you can finish working through it.valenumr said:I'm still working through it.
Maybe you are confusing energy with force?valenumr said:I am still trying to reconcile how anything (macroscopic) can alter course without an energy input.
That's not a good analogy. Friction dissipates the vehicles energy.jbriggs444 said:I am guessing here that taking one's foot off the accelerator, putting the vehicle in neutral and turning the steering wheel counts as an "energy input" in your book?
We've given you good analogies. But you've ignored them all.valenumr said:That's not a good analogy. Friction dissipates the vehicles energy.
I missed this earlier. No I do not claim that. The external force is indeed required.valenumr said:So you both claim that a fundamental principal of GR, an "observer" in "free fall" along a geodesic can deviate from that path without any external force applied? Ok.
Perhaps I am using the wrong term from the outset, but I also don't see how energy, momentum, and force can be extricated. I'm not saying that energy is used up, just transferred.A.T. said:Maybe you are confusing energy with force?
Going around in circles. I repeat: from what to what? From one billiard ball through a thread to another billiard ball?valenumr said:Perhaps I am using the wrong term from the outset, but I also don't see how energy, momentum, and force can be extricated. I'm not saying that energy is used up, just transferred.
valenumr said:Perhaps I am using the wrong term from the outset, but I also don't see how energy, momentum, and force can be extricated. I'm not saying that energy is used up, just transferred.
From one object to another.jbriggs444 said:Going around in circles. I repeat: from what to what?
So we have these two billiard balls with a thread between them. Please identify the energy transfer.valenumr said:From one object to another.
Look, if you want to idealize a frictionless surface and perfect string, you are modeling a system where nothing loses energy. But if one ball gains energy, the other must lose some.jbriggs444 said:So we have these two billiard balls with a thread between them. Please identify the energy transfer.
Neither ball gains energy. Neither ball loses energy. Both orbit at constant speed. Neither follows a geodesic. Your claim is that because a geodesic is not followed, energy must be transferred. But when challenged on the point, you have no answer.valenumr said:Look, if you want to idealize a frictionless surface and perfect string, you are modeling a system where nothing loses energy. But if one ball gains energy, the other must lose some.
Gravity is indeed a fictitious force in GR, the same way that centrifugal and coriolis are fictitious forces in classical mechanics (and GR). A fictitious force is one that causes proper acceleration, as opposed to coordinate acceleration; if you're not already clear about this essentiall distinction you'l want to look for some of our many threads about proper acceleration.valenumr said:But is not gravitational acceleration in GR a fictitious force? An object traveling along a geodesic is a rest frame? Or is that totally a misunderstanding on my part?
Well, this is the crux.. I'm saying an object deviating from it's geodesic must necessarily change energy. That is all.jbriggs444 said:You keep dodging your own scenario. You have a moving object subject to an external force. What makes you think it must gain energy?
And that is wrong. Again.valenumr said:Well, this is the crux.. I'm saying an object deviating from it's geodesic must necessarily change energy. That is all.
Well, I see your point. It doesn't change energy, but perhaps increases in kinetic energy and loses potential energy or vice versa.jbriggs444 said:And that is wrong. Again.
See subsequent edit in the post you quote. A viable claim would be:valenumr said:Well, I see your point. It doesn't change energy, but perhaps increases in kinetic energy and loses potential energy or vice versa.
But how does it make that transition without energy applied? A rocket doesn't lift ff magically.valenumr said:Well, I see your point. It doesn't change energy, but perhaps increases in kinetic energy and loses potential energy or vice versa.
You are assuming a prior state of rest in your chosen frame of reference. That assumption is not dictated by anything physical. It is pure personal prejudice.valenumr said:But how does it make that transition without energy applied? A rocket doesn't lift ff magically.
I will disagree. It should be obvious that the rocket is undergoing an acceleration.jbriggs444 said:You are assuming a prior state of rest in your chosen frame of reference. That assumption is not dictated by anything physical. It is pure personal prejudice.
Sure. And this is an example of an object where its deviation from a geodesic is pretty manifestly associated with an expenditure of chemical energy into kinetic. Cherry-picked examples do not help prove your general assertion.valenumr said:I will disagree. It should be obvious that the rocket is undergoing an acceleration.
I wasn't trying to cherry pick. In just think perhaps my use of the term energy turned this into a moot debate.jbriggs444 said:Sure. And this is an example of an object where its deviation from a geodesic is pretty manifestly associated with an expenditure of chemical energy into kinetic. Cherry-picked examples do not help prove your general assertion.
Possibly your understanding of the term "energy" needs some work. You do understand that it is frame-relative, right?valenumr said:I wasn't trying to cherry pick. In just think perhaps my use of the term energy turned this into a moot debate.
Yes, absolutely.jbriggs444 said:Possibly your understanding of the term "energy" needs some work. You do understand that it is frame-relative, right?
So you agree that whether the rocket gains or loses kinetic energy at launch depends on the frame of reference that one adopts, right?valenumr said:Yes, absolutely.
Just out of curiosity... Is that last point definitive? I know we can see creature of space on large scales, but would thinks look curved or flat locally even on soghettifying scales?Nugatory said:Gravity is indeed a fictitious force in GR, the same way that centrifugal and coriolis are fictitious forces in classical mechanics (and GR). A fictitious force is one that causes proper acceleration, as opposed to coordinate acceleration; if you're not already clear about this essentiall distinction you'l want to look for some of our many threads about proper acceleration.
Thinking in terms of frames can be very confusing until you're clear on what a frame is and which quantities are frame-dependent and which are not. Energy, velocity, coordinate acceleration, momentum are all frame-dependent.
Every object is in a rest frame (because all objects are always in all frames). That frame in which an object is at rest may or may not be inertial. An inertial frame is only inertial within a region of spacetime small enough that gravitationa tidal effects cannot be detected; the global inertial frame of special relativity assume the complete absence of any gravitational effects.
For any fixed curvature, and for any particular experimental sensitivity, there is a local space-time region that is small enough so that the the effects of that fixed curvature are no longer detectable at that experimental sensitivity.valenumr said:Just out of curiosity... Is that last point definitive? I know we can see [curvature?] of space on large scales, but would thinks look curved or flat locally even on soghettifying scales?
No. What I am saying is more drastic: that what you are intuitively thinking of when you say "inducing perpendicular spin" is not possible. The notion of "perpendicular" that you are intuitively using does not work for a body with nonzero vorticity.valenumr said:If I understand your second point correctly, regarding inducing perpendicular spin to the velocity vector, it would induce a torque (say downward)?
The pencil's axis still won't spin. Only points of the pencil that are off axis will.valenumr said:I suppose I could have said someone just spun a free floating pencil for the first case.
By checking vorticity and seeing that it is nonzero, observers could agree that the pencil is spinning.valenumr said:Observers would agree as to what actually had angular momentum, No?
Not really. I think it would still be observable that the rocket is gaining kinetic energy. I mean, ultimately it is trading off propellent, but the acceleration effects are observable. It's hard to put into words, but even comparing multiple inertial frames, the rocket is gaining kinetic energy, not the Earth it is pushing off from.jbriggs444 said:So you agree that whether the rocket gains or loses kinetic energy at launch depends on the frame of reference that one adopts, right?
Yes. Locality is defined by the scale small enough that tidal effects are not detectable. No matter how strong the tidal effects are, we can make them arbitrarily small by limiting our consideration to a sufficiently small region.valenumr said:Is that last point definitive?
Great! We finally have a disagreement that we can sink our respective teeth into.valenumr said:Not really. I think it would still be observable that the rocket is gaining kinetic energy. I mean, ultimately it is trading off propellent, but the acceleration effects are observable. It's hard to put into words, but even comparing multiple inertial frames, the rocket is gaining kinetic energy, not the Earth it is pushing off from.
But the fact that the astronauts on the rocket experience the acceleration? And especially when we get into free space. I suppose we can argue the propellent feels an acceleration too though.jbriggs444 said:Great! We finally have a disagreement that we can sink our respective teeth into.
The rocket is NOT unambiguously gaining energy. There is nothing about the size of the Earth that matters. It is all about frames of reference written with pencil on paper.
One useful exercise in classical physics involves a cart, a spring and a ball bearing that it shoots rearward. The experiment can be analyzed in any inertial frame you choose. Whether the cart gains energy, loses energy or remains at the same energy changes with the choice of reference frame. But energy and momentum are always conserved.
You can choose the reference frame after you run the experiment. So it obviously cannot influence the experimental results.
jbriggs444 said:So you agree that whether the rocket gains or loses kinetic energy at launch depends on the frame of reference that one adopts, right?
Then you have a problem with classic Galilean Relativity. You should figure that out first, before trying to learn SR or GR.valenumr said:Not really.
It's not that at all. I just think actual accelerations are theoretically identifiable amongst observers.A.T. said:Then you have a problem with classic Galilean Relativity. You should figure that out first, before trying to learn SR or GR.
Proper acceleration--what an accelerometer measures--is an invariant, yes.valenumr said:I just think actual accelerations are theoretically identifiable amongst observers.
That is definitely true. You can simply attach an accelerometer and thereby identify the actual (proper) acceleration.valenumr said:It's not that at all. I just think actual accelerations are theoretically identifiable amongst observers.
That wasn't the question. It was about the frame dependence of KE changes. Even two inertial frames will see different KE changes for the same object, which has a frame invariant proper acceleration.valenumr said:It's not that at all. I just think actual accelerations are theoretically identifiable amongst observers.
So an accelerometer at the center of the Earth would read 0 and an accelerometer on the rocket would read non-zero, clearly identifying that the Earth continues on its geodesic and the rocket is the thing that is unambiguously departing from its geodesic.valenumr said:Not really. I think it would still be observable that the rocket is gaining kinetic energy. I mean, ultimately it is trading off propellent, but the acceleration effects are observable. It's hard to put into words, but even comparing multiple inertial frames, the rocket is gaining kinetic energy, not the Earth it is pushing off from.
Even in free space, one could argue the frame of the rocket versus propellent I suppose. But it is still the rocket that is defying it's normal geodesic.
Right, I fully understand that part. But where does the chemical energy released by burning rocket fuel go? I think this is my point of confusion.Dale said:So an accelerometer at the center of the Earth would read 0 and an accelerometer on the rocket would read non-zero, clearly identifying that the Earth continues on its geodesic and the rocket is the thing that is unambiguously departing from its geodesic.
However, whether it is gaining or losing KE depends on the reference frame. In a reference frame where ##\vec v## is in the opposite direction as the nose is pointing then ##P=\vec F \cdot \vec v < 0## so the rocket is losing KE, not gaining it. At the same time, in a frame where ##\vec v## is in the same direction as the nose is pointing then ##P=\vec F \cdot \vec v>0## and the rocked is gaining KE. And in a frame where the rocket is momentarily at rest then ##P= \vec F \cdot 0= 0## and for that moment the rocket is neither gaining nor losing energy in that frame.
Einstein's happiest thought... Fair point.jbriggs444 said:Note that the rocket on the pad before it fires its engines is already deviating from a geodesic path. Where is the energy transfer?
Facepalm.valenumr said:Right, I fully understand that part. But where does the chemical energy released by burning rocket fuel go? I think this is my point of confusion.
Into the exhaust plume. Its velocity (and therefore energy) will be different in the different frames, and the change in energy of the rocket plus the change in energy of the plume will equal (give or take losses to vibration and noise etc) the energy released by the fuel.valenumr said:Right, I fully understand that part. But where does the chemical energy released by burning rocket fuel go? I think this is my point of confusion.
Into the exhaust.valenumr said:But where does the chemical energy released by burning rocket fuel go?
Dale said:Into the exhaust.
It's more like I'm hashing up the work energy theorem.A.T. said:That wasn't the question. It was about the frame dependence of KE changes. Even two inertial frames will see different KE changes for the same object, which has a frame invariant proper acceleration.
You are confusing energy with force/acceleration and jumping back and forth between them.
I agree. I don’t think the issue you are running into here has anything to do with relativity per se. However, it seems like you may have gotten it now. Are you now comfortable with the objections raised?valenumr said:It's more like I'm hashing up the work energy theorem.
Absolutely. I appreciate everyone's discussion on the topic.Dale said:I agree. I don’t think the issue you are running into here has anything to do with relativity per se. However, it seems like you may have gotten it now. Are you now comfortable with the objections raised?
Suppose two airplanes take off from the same airport, one flying East and the other West. In an inertial reference frame where the Earth is rotating, then the one flying East speeds up at take-off and the one flying West slows down at take-off.valenumr said:Absolutely. I appreciate everyone's discussion on the topic.