I Is time really moving backwards?

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The discussion centers around the nature of time and spacetime, questioning whether time can be perceived as moving backwards. It emphasizes that time is not a spatial dimension and cannot be compared to spatial movement, as the Earth is simply an object in space. The conversation also touches on the complexities of spacetime geometry and how mass influences its curvature, while clarifying that spacetime itself does not "move." Participants highlight the importance of understanding relativity through proper study rather than popularized interpretations, noting that spacetime is a useful concept for making predictions in physics. Ultimately, the dialogue reflects a desire for clearer explanations of these concepts in educational contexts.
  • #121
valenumr said:
But how does it make that transition without energy applied? A rocket doesn't lift ff magically.
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.

The rest of us are free to consider frames of reference where objects begin in states other than rest. This is exceptionally useful when considering collections of objects, not all of which begin at relative rest.
 
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  • #122
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.
I will disagree. It should be obvious that the rocket is undergoing an acceleration.
 
  • #123
valenumr said:
I will disagree. It should be obvious that the rocket is undergoing an acceleration.
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.

However, there are frames of reference where the result of the launch burn is to actually reduce the energy of the rocket.
 
  • #124
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.
I wasn't trying to cherry pick. In just think perhaps my use of the term energy turned this into a moot debate.
 
  • #125
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.
Possibly your understanding of the term "energy" needs some work. You do understand that it is frame-relative, right?
 
  • #126
jbriggs444 said:
Possibly your understanding of the term "energy" needs some work. You do understand that it is frame-relative, right?
Yes, absolutely.
 
  • #127
valenumr said:
Yes, absolutely.
So you agree that whether the rocket gains or loses kinetic energy at launch depends on the frame of reference that one adopts, right?
 
  • #128
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.
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?
 
  • #129
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?
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.

It takes a pretty big region to detect cosmic curvature.
 
  • #130
valenumr said:
If I understand your second point correctly, regarding inducing perpendicular spin to the velocity vector, it would induce a torque (say downward)?
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:
I suppose I could have said someone just spun a free floating pencil for the first case.
The pencil's axis still won't spin. Only points of the pencil that are off axis will.

valenumr said:
Observers would agree as to what actually had angular momentum, No?
By checking vorticity and seeing that it is nonzero, observers could agree that the pencil is spinning.
 
  • #131
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?
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.
 
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  • #132
valenumr said:
Is that last point definitive?
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.
 
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  • #133
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.
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.

Edit: Rockets do not push off from the Earth. They push off from their exhaust stream.
 
  • #134
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.
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.
 
  • #135
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?
valenumr said:
Not really.
Then you have a problem with classic Galilean Relativity. You should figure that out first, before trying to learn SR or GR.
 
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  • #136
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.
It's not that at all. I just think actual accelerations are theoretically identifiable amongst observers.
 
  • #137
valenumr said:
I just think actual accelerations are theoretically identifiable amongst observers.
Proper acceleration--what an accelerometer measures--is an invariant, yes.

But coordinate acceleration--the second derivative of position with respect to time--is not. That is because both "position" and "time" depend on your choice of coordinates.

Kinetic energy is also coordinate dependent, since it depends on coordinate velocity, the first derivative of position with respect to time, which is coordinate dependent for the same reason coordinate acceleration is.
 
  • #138
Note that the rocket on the pad before it fires its engines is already deviating from a geodesic path. Where is the energy transfer?
 
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  • #139
valenumr said:
It's not that at all. 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.

What is incorrect is that this is associated with an energy transfer. The force must be non-zero, but the power may be zero. Again, I recommend that you pause posting here and spend time to go through the math I posted earlier.
 
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  • #140
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.

You are confusing energy with force/acceleration and jumping back and forth between them.
 
  • #141
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.
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.
 
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  • #142
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.
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.
 
  • #143
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?
Einstein's happiest thought... Fair point.
 
  • #144
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.
Facepalm. 🤦‍♂️ I think the exhaust must end up with more mass than the fuel.
 
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  • #145
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.
 
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  • #146
valenumr said:
But where does the chemical energy released by burning rocket fuel go?
Into the exhaust.
 
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  • #147
Dale said:
Into the exhaust.
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.
It's more like I'm hashing up the work energy theorem.
 
  • #148
valenumr said:
It's more like I'm hashing up the work energy theorem.
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?
 
  • #149
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?
Absolutely. I appreciate everyone's discussion on the topic.
 
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  • #150
valenumr said:
Absolutely. I appreciate everyone's discussion on the topic.
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.

It's also a good exercise in classical, non-relativistic mechanics to analyse an elastic collision in one dimension and check that the change in total kinetic energy is the same in all inertial reference frames - even though each object may be seen as either accelerated or decelerated by the collision in different reference frames.
 
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