Is time really moving backwards?

[Dale]

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.

 

[valenumr]

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.

 

[valenumr]

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.

 

[valenumr]

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.

 

[Ibix]

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.

 

[Dale]

But where does the chemical energy released by burning rocket fuel go?

Into the exhaust.

 

[valenumr]

Into the exhaust.

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.

 

[Dale]

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?

 

[valenumr]

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.

 

[PeroK]

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.

 

[darth boozer]

"The closest I can make is to observe that Monday morning comes at different times for frames in motion with respect to the Earth."

Monday morning comes between 11 and 13 hours (depending on time of year) earlier in New Zealand than in UK, both of which are not in motion with respect to the Earth or each other (ignoring plate techtonics).

 

[Ibix]

Monday morning comes between 11 and 13 hours (depending on time of year) earlier in New Zealand than in UK, both of which are not in motion with respect to the Earth or each other (ignoring plate techtonics).

A more precise statement of the point I was making is that frames in relative motion have different notions of simultaneity. That means that what one frame calls "now" another frame will call partly in the past, partly in the future, and partly now. This is not the same concept as different locations that share a notion of simultaneity having different names for the same time, which is what your example is.

 

[jbriggs444]

A more precise statement of the point I was making is that frames in relative motion have different notions of simultaneity. That means that what one frame calls "now" another frame will call partly in the past, partly in the future, and partly now. This is not the same concept as different locations that share a notion of simultaneity having different names for the same time, which is what your example is.

To make this a bit more concrete, suppose that I have a Zoom conference at 8:00 am EST and my colleague in London joins the meeting at 1:00 pm GMT.

Per my time coordinate system (EST) we both join the conference simultaneously at 8:00 am EST.
Per my colleague's time coordinate system (GMT) we both join the conference simultaneously at 1:00 pm GMT.

From the perspective of time zones, we both agree that we both joined simultaneously [to within 20 milliseconds or so anyway]. Relativity of simultaneity is not a problem.

 

[etotheipi]

From the perspective of time zones, we both agree that we both joined simultaneously [to within 20 milliseconds or so anyway].

I'm not sure I understand this remark as written; certainly the spatially separated clocks will appear desynchronised to an observer with relative speed $v$ to the Earth by $\delta t = Lv/c^2$, and if $v \sim c$ then this is $\delta t \sim L/c \sim 10^{-2} \, \mathrm{s}$ between the UK-clock and the USA-clock. Is that what you mean by 'from the perspective of time zones'?

 

[jbriggs444]

I'm not sure I understand this remark as written; certainly the clocks will appear desynchronised to an observer with relative speed $v$ to the Earth by $\delta t = Lv/c^2$, and if $v \sim c$ then this is $\delta t \sim L/c \sim 10^{-2} \, \mathrm{s}$ between the UK-clock and the USA-clock. Is that what you mean by 'from the perspective of time zones'?

The two joins to the Zoom conference are off by only 20 milliseconds at most, no matter what frame one chooses. Not the 5 hours that the time zone offset calls for.

At a relative velocity of only 1000 miles per hour or so and nearly at right angles to the displacement, the actual discrepancy between a Washington, DC frame and a London England frame's assessment of the delta T between the joins would be far less than even the 20 milliseconds.

[Just checked. I'm seeing 79.8 ms RTT to London right now from Nashua, New Hampshire. Not bad. It's within 50% of light speed]

 

[PeterDonis]

A more precise statement of the point I was making is that frames in relative motion have different notions of simultaneity.

Even more precisely, inertial frames in relative motion have different notions of Einstein clock synchronized simultaneity.

The qualifiers are important, as will be apparent in a moment.

This is not the same concept as different locations that share a notion of simultaneity

But the notion of simultaneity that is shared by people at different locations on the Earth, that let's them all call into a Zoom call at the same time even though they are in widely different time zones, is not the same notion of simultaneity that inertial frames have. It is not based on Einstein clock synchronization. It can't be, because it is impossible to use Einstein clock synchronization to establish a global notion of simultaneity for a rotating family of observers. (In more technical language, it is impossible to use Einstein clock synchronization to establish a global notion of simultaneity for a family of observers whose worldlines form a congruence with nonzero vorticity. Observers at rest on the rotating Earth are such a family of observers.)

certainly the spatially separated clocks will appear desynchronised to an observer who is using the Einstein clock synchronization notion of simultaneity for the inertial frame in which he is at rest

See the qualifier in the quote above. Clocks at rest on the rotating Earth and using the common time standards that we all use for our clocks will appear desynchronized in any inertial frame that is moving relative to the center of the Earth, if we use that frame's notion of simultaneity. The clocks we use on Earth are simply synchronized using a different notion of simultaneity (roughly, the notion of simultaneity of the inertial frame in which the center of the Earth is at rest, with the clock rate adjusted to the rate on the geoid).