Relativity discussion from “What is Time?” thread

[HydrogenBond]

This discussion brings us to the question, what is actually dilating during time dilation? It appears everything in the reference feels it; not just clocks.

 

Here is a thought experiment. We have a ship moving near C, that is showing time dilation. We have the technology, to quantum tunnel from stationary reference, and skip the energy barrier, that was needed for the ship to achieve the final velocity.

 

We materialize inside the ship, directly from stationary reference. What will happen to our matter and all its force fields? Does the ship slow with an equal and opposite reaction to us equilibrating? Is there a virtual flash from the ship that converges onto us due to transfer of relativistic mass, since we start with none, and the ship has plenty. Does this alter the time dilation of the ship because of energy transfer?

[PerfectLiquid]

Alternately, the thought experiment could involve the “home” observer (it’s no longer appropriate in this version to call him “Earthbound”, so let’s just call him “Homer”) or the “traveler” (let’s call him “Rover”) experiencing no outside forces. Imaging Homer is in a motorless ship in deep space, experiencing practically no gravitational force. Rover is in an identical ship positioned 200 km behind an Earth-radius and mass spheroid being towed by a ship accelerating at about 9.2 m/s², an arrangement that results in Rovers ship freefalling toward the sphere with at precisely the same acceleration as it and the ship.

 

Both Homer and Rover experience a period of weightlessness. After a 2 year out-and-back trip, Homer and Rover compare clocks, and discover that Homer’s clock reads about 65.62 days later than Rover’s.

 

Notice I’ve used the phrase “nearly identical” a lot in these examples. Though not important to the outcome of the experiments or the theory of relativity, bonus points to whoever explains what’s not quite identical about the forces experienced by Homer and Rover.

 

Hi CraigD,

 

Homer should be experiencing no force, since he is neither accelerating nor in a gravitational field. Rover on the other hand is in a gravitational field and is accelerating linearly due to the towing arrangement. The force that Rover experiences should be 9.2 times his mass. The time dilation that he experiences would be a combination of that produced by the gravitational field he is in PLUS that due to his linear speed (that is constantly increasing). Doesn't the difference attributable to his speed point towards a preferred frame of reference? Otherwise, there would only be an apparent difference (relativistic symmetry) versus a real difference (asymmetry)? This view will obviously not be universally shared. ;)

 

Regards,

Al

[modest]

Imaging Homer is in a motorless ship in deep space, experiencing practically no gravitational force.

 

Interestingly, If you start Homer and Rover collocated and hold Homer stationary against an artificial and uniform gravitational field and end the thought experiment with Homer and Rover again collocated and stationary then Homer ends up the younger twin (as the Principle of Extremal Aging would demand). You are probably aware, but this is Einstein's original solution to the twin paradox entirely from the accelerating twin's perspective (giving the 'traveling' twin a single non-inertial frame of reference) with GR.

 

A translation of a paper he wrote in 1918 is strikingly similar to your thought experiment:

 

Dialog about Objections against the Theory of Relativity - Wikisource

 

I'd very strongly recommend it to anyone who has pondered the asymmetry of the two twins and the symmetry of acceleration and being stationary in a gravitational field.

 

~modest

[Michael Mooney]

Interesting thought experiment there, Craig, but I'm still on about the ontology of time, so I'll post a very simple thought experiment on that (a repeat, actually) in the other... "what is it?" thread.

 

But I'll go for the bonus points you offered here, as per:

Notice I’ve used the phrase “nearly identical” a lot in these examples. Though not important to the outcome of the experiments or the theory of relativity, bonus points to whoever explains what’s not quite identical about the forces experienced by Homer and Rover.

 

I touched on it, I think, above with:

...I do understand the concept of momentary co-moving inertial frames (as to how a lot of quick "snapshots" factor out acceleration artificially.)

In my reference above, I meant "momentary co-moving reference frames." My study of same revealed the calculus like nature of analyzing a string of theoretical "instants" which factor out acceleration but this math device does not claim that there is *no difference in accelleration* between the "co-moving" bodies moving at "the same velocity" yet having different clock readings. There are always sleight differences in acceleration between bodies even at *approximately* the same velocity.

Is this what you were getting at?

Michael

[CraigD]

Homer should be experiencing no force, since he is neither accelerating nor in a gravitational field. Rover on the other hand is in a gravitational field and is accelerating linearly due to the towing arrangement. The force that Rover experiences should be 9.2 times his mass.

We should, here, describe in a bit more detail what we mean by “experiencing a force”.

 

What we mean is that some part of us, or some external measuring device we can interact with in our windowless room, experiences a different force than some other part, and either accelerates relative to it, or experience an opposite force so that it doesn’t accelerate. Stepping (or strapping himself into) a bathroom scale attached to the “floor” or any wall of his room tells Homer he’s experience zero force. Rover doing the same tells him nearly the same thing (or within the precision limits of a bathroom scale, precisely the same thing). Without some information from outside of their rooms neither Homer nor Rover can tell – nearly – if he is the one being towed on a long space trip, or the one being left unaccelerated in space.

Notice I’ve used the phrase “nearly identical” a lot in these examples. Though not important to the outcome of the experiments or the theory of relativity, bonus points to whoever explains what’s not quite identical about the forces experienced by Homer and Rover.

I’ll claim the bonus points for my own question (a bit of a cheat that ;)):

 

The way Rover can tell he’s being accelerated by gravity at 9.2 m/s², and Homer can tell he’s not, is to precisely measure his weight standing on each floor/wall of his room. If he’s being accelerated by the tug ship described in post #17, his weight on the forward and rearward facing surfaces won’t be quite zero. Lets say the rooms are cubes 10 m on each edge, walls of uniform thickness and density, Rover’s oriented as described above. A 100 kg Rover (let’s assume his center of mass is 1 m minus the thickness of the scale, putting him it 4 m from the room’s center of mass) on his forward or rear wall will weigh

[math]100 \cdot u \left(\frac1{6570000^2} - \frac1{(6570000+4)^2} \right) \dot= \, 0.0011 \,\mbox{N}[/math]

, where [math]u[/math] is the Earth-mass sphere’s standard gravitational parameter in m³/s²

, 6570000 is the sphere’s Earth-ish radius in meters plus 200000 m distance between it and Rover's ship

- an undetectably small weight for a bathroom scale, but possibly not beyond the precision of a very sensitive spring-type scale (usually more precise balance scales won’t work for this measurement), or some other, more cunning experiment. Thinking-outside-the-box (no pun intended :)) bonus points for any one who can think up a cunning experiment that Rover could do using ordinary, stuck-in-a-spaceship-for-two-years materials on hand to determine if he’s freefalling in a gravitational field.

 

Homer, or course, weights the same nearly zero wherever he weighs himself.

 

Notice that if we revert to the first version of the thought experiment, the situation is reversed – Rover can detect he’s on a uniformly accelerating spaceship because he weight the same thing on the floor as he does standing on tall platform next to the ceiling, while Homer can detect that he’s on Earth because he weights slightly less next to the ceiling than on the floor.

The time dilation that he [Rover] experiences would be a combination of that produced by the gravitational field he is in PLUS that due to his linear speed (that is constantly increasing).

True. I’ve been ignoring the gravitational time dilation in these thought experiments because

1. it complicates things
   
2. it’s small - For the 2 year (Rover time) trips we’ve been discussing, about 0.05 seconds =
   [math]\left( 1 -\sqrt{1-\frac{2u}{6370000 \cdot c^2}} \right) \cdot 74790000 [/math]
   , where [math] 6370000 [/math] is the Earths radius in meters
   , [math]c[/math] the speed of light in m/s, exactly 299792458
   which hardly compares to the couple of months difference due to velocity time dilation.
   

Doesn't the difference attributable to his speed point towards a preferred frame of reference?

No. This is the real essence of relativity – the part of it that’s really relative. This post, however, is a bit overlong, as is my post-writing timeout, so I’ll save another batch of thought experiments addressing it for a later post.

[watcher]

No. This is the real essence of relativity – the part of it that’s really relative. This post, however, is a bit overlong, as is my post-writing timeout, so I’ll save another batch of thought experiments addressing it for a later post

 

you don't have to if you can just clarify your intentions. it would be much appreciated, since your thought experiments was in effect trying to simulate two things subjected to same exact conditions except that they are in a different location.

 

are you telling us that there are other factors aside from motion that causes time dilation? and if i am right in my understanding, are saying that time dilation happens by virtue of two objects simply located on two different point in space?

[freeztar]

Thinking-outside-the-box (no pun intended ;)) bonus points for any one who can think up a cunning experiment that Rover could do using ordinary, stuck-in-a-spaceship-for-two-years materials on hand to determine if he’s freefalling in a gravitational field.

 

Wouldn't he have crashed into the "Earth mass spheroid" in 2 years time with a negative acceleration of 0.6m/s^2? I'm not sure how to calculate that...

 

As far as a cunning experiment, I'll give it a try. Perhaps he could position lasers and matching receptors on opposing ends of the spacecraft. Depending on the precision of the receptors, there should be a slight difference between the two when the beams are synchronously pulsed. The beams should bend slightly towards the direction of the gravity well, per GR.

 

I'm not sure if that qualifies as "stuck-in-a-spaceship-for-two-years material", but the science is sound. :eek_big:

[CraigD]

Wouldn't he have crashed into the "Earth mass spheroid" in 2 years time with a negative acceleration of 0.6m/s^2? I'm not sure how to calculate that...

The acceleration of gravity of an 1 Earth size and mass spheroid 200 km above its surface, which is about 6370 km from its center, is [math]\frac{6370^2}{(6370+200)^2} \dot= \, 0.94 \,\mbox{g} \dot= \, 9.2 \,\mbox{m/s}^2[/math], so no, the distance between the ship and the tug sphere accelerating at about 9.2 m/s² should remain constant. We could alter the sphere slightly, increasing its mass or decreasing its diameter, to make the tug/tugged system’s acceleration exactly 1 g, without having a significant impact on the thought experiment.

 

In practice, I’m certain such a system would have to have some sort of active control system adjusting the tug ship’s acceleration to keep the tugged ship at a constant distance – a minor challenge, I’d think, for folk who can drag Earth masses around at such accelerations.

 

Gravity tugs similar to the one described, but much, much smaller and slower, may be realized in the near future, because they address the problem of how to maneuver “flying gravel pile” and other hard-to-get-a-grip-on SSSBs, either to prevent catastrophic Earth or other collisions, or to exploit them for their resources.

As far as a cunning experiment, I'll give it a try. Perhaps he could position lasers and matching receptors on opposing ends of the spacecraft. Depending on the precision of the receptors, there should be a slight difference between the two when the beams are synchronously pulsed. The beams should bend slightly towards the direction of the gravity well, per GR.

I don’t think this would work, because, like the whole ship, the laser senders and receivers are in freefall toward the tug mass – though I might be missing something subtle.

 

What I had in mind is much simpler, requiring no more than a couple of machine nuts or similar masses and some thread or dental floss. Released very carefully in the center of the ship, assuming no excessive air motion, it should align itself with the direction of acceleration via Gravity-gradient stabilization. The same dynamic principle accounts for the orientation of the fictional “integral trees” in Niven’s science fiction novel of the same name, one of my favorites. :thumbs_up

 

These thought exercises are way off the topic of relativity, but so much fun I couldn’t resist. :embarass:

[PerfectLiquid]

[quote name='CraigD;278462

...

True. I’ve been ignoring the gravitational time dilation in these thought experiments because

1. it complicates things
   
2. it’s small - For the 2 year (Rover time) trips we’ve been discussing' date=' about 0.05 seconds =
   [math']\left( 1 -\sqrt{1-\frac{2u}{6370000 \cdot c^2}} \right) \cdot 74790000 [/math]
   , where [math] 6370000 [/math] is the Earths radius in meters
   , [math]c[/math] the speed of light in m/s, exactly 299792458
   which hardly compares to the couple of months difference due to velocity time dilation.
   

---Quote (Originally by PerfectLiquid)---

Doesn't the difference attributable to his speed point towards a preferred frame of reference?

---End Quote--

No. This is the real essence of relativity – the part of it that’s really relative. This post, however, is a bit overlong, as is my post-writing timeout, so I’ll save another batch of thought experiments addressing it for a later post.

 

Hi CraigD,

 

Masterful treatment of the problem. However, WRT the preferred frame of reference question, since we see time dilation occurring in the case of differences in speed (as in your example), this dilation is not symmetrical. In a truly relativistic sense, Homer appears to be moving away from Rover at the same speed as the reverse (i.e., Rover relative to Homer). But only Rover's clock reflects a slower rate of ticking. If it was truly relative, both parties should perceive the slowing down effect, but only one does. This suggests to me that the speed effect is relative to a background value (ideally stationary space-time to make the calculations simpler). This would be consistent with the asymmetrical effect on time dilation of a gravitational field (relative to a background of gravity-free space).

 

Regards,

Al

[modest]

Hi CraigD,

 

Masterful treatment of the problem. However, WRT the preferred frame of reference question, since we see time dilation occurring in the case of differences in speed (as in your example), this dilation is not symmetrical. In a truly relativistic sense, Homer appears to be moving away from Rover at the same speed as the reverse (i.e., Rover relative to Homer). But only Rover's clock reflects a slower rate of ticking. If it was truly relative, both parties should perceive the slowing down effect, but only one does. This suggests to me that the speed effect is relative to a background value (ideally stationary space-time to make the calculations simpler). This would be consistent with the asymmetrical effect on time dilation of a gravitational field (relative to a background of gravity-free space).

 

Regards,

Al

 

In the most relativist, Machian sense, Rover can consider himself at rest while Homer moves back and forth slowing Homer's clock. The equivalence principle establishes that an accelerating frame of reference is the same as a gravitational field.

 

Rover can consider himself to be in an accelerating frame of reference (moving back and forth relative to other things), or he can think of himself as at rest in a gravitational field. These are, by the equivalence principle, equivalent. Trying to make the twin paradox as symmetric and Machian as possible, we can choose the latter coordinate choice and say that Rover, the planet-sized mass that is near him, and the tow-ship pulling it, are all being held in place by the work of the tow-ship's engine against a uniform gravitational field causing the rest of the universe (including Homer) to free fall with the field at 9.2 m/s² (the strength of the field everywhere).

 

If this doesn't makes sense then check out: The Twin Paradox: The Equivalence Principle Analysis (as I am usually quite awful at describing things)

 

Of course, when the tow-ship turns the mass around and starts pulling Rover back toward Homer then Rover will need to switch the direction of the field meaning from Rover's perspective Homer will start free-falling back toward him.

 

When considered in this way, Rover would calculate Homer's clock to be slowed because of relative velocity exactly as Homer did. Rover has been, after all, motionless while Homer had relative velocity—first accelerating away from him then toward. It would seem then that velocity time dilation is still symmetric and the paradox remains unresolved. But, using this coordinate choice Rover must now take into consideration the time dilation caused by the gravitational field.

 

In a gravitational field a clock which is higher in the field and has greater gravitational potential will tick faster. The greater the distance between the clocks, the greater the difference in their rate. Rover must now calculate Homer's time dilation due to velocity (the net effect of which is to slow Homer's clock relative to his own) and the changing clock rates from the difference in gravitational potential (the net effect of which is to increase the rate of homer's clock relative to his own.

 

It turns out when calculated that the total time dilation from the field speeds up Homer's clock by twice the factor that velocity slows it. Rover will then agree that he will be the younger twin. For a different explanation about half way down the page:

Dialog about Objections against the Theory of Relativity - Wikisource

And a proof of the math:

Advanced university physics - Google Books

 

 

 

 

 

Both Homer and Rover experience a period of weightlessness. After a 2 year out-and-back trip, Homer and Rover compare clocks, and discover that Homer’s clock reads about 65.62 days later than Rover’s.

 

I get 28.89 days.

 

Using the relativistic rocket equation from:

The Relativistic Rocket

Using:

[math]t = \left( \frac{c}{a} \right) \times \frac{ e^{(a \tau /c)} - e^{-(a \tau /c)}}{2} [/math]

If I follow your thought experiment correctly you have a = 9.2, proper time, T, is 2 years giving the trip 4 equal segments of T=0.5. Rover starts at rest relative to Homer and ends at rest giving (in meters and seconds accelerating for 1/2 a year at 9.2):

[math]t = \left( \frac{(299792458)}{(9.2)} \right) \times \frac{ e^{((9.2)(15778463)/(299792458))} - e^{-((9.2)(15778463)/(299792458))}}{2} [/math]

[math]t = 16402294.83 \ seconds[/math]

Times 4 to complete the trip,

[math]t_{F} = 16402293.83 \times 4 = 65609175.31 \ sec = 759.37 \ days[/math]

Minus 2 years of proper time:

[math]t_{F} - \tau = 28.89 \ days[/math]

 

If Rover has constant 9.2 m/s² acceleration for 2 years of Rover's proper time and ends up at rest where he started (at rest) then I'm pretty sure I get Homer's clock ahead by ~29 days.

 

~modest

[UncleAl]

The momentum four-vector is conserved. The clock that travels through the most space accumulates the least time. Acceleration is irrelevant and relative velocity can be arbitrarily small.

 

The Twin Paradox

unaccelerated twin paradox in SR

The Light Cone: The Twin Non-Paradox

Twin Paradox

 

Twin Paradox: One twin travels relativistically, one twin stays home. They reunite. The traveling twin aged much less. The twin who travels through more space accumulates less time; also true for an orbit. Interval sqrt(t^2 - x^2 - y^2 - z^2) between the two events, expressed in inertial coordinate system (t,x,y,z), is conserved. Given the invariant interval, the larger sqrt(x^2 + y^2 + z^2)is the smaller sqrt(t^2) must be.

 

The ratio by which the two aged when they are again local is identical in all reference frames: ratio = sqrt(t^2 - x^2 - y^2 - z^2)/t (units of c=1).

 

Three identical clocks as kits and not constructed until the experiment is running. Each clock has a short toggle switch. Individual spaceships carry a kit each. Set up the experiment.

 

CLOCK 1: Our clock sits stationary in our inertial reference frame with its toggle sticking out. Touch the toggle and "off" state goes "on" or "on" state goes "off." Build it from parts just before needed, in the "off" state, zeroed.

 

CLOCK 2: In a spaceship traveling at 0.999c relative to our inertial frame and positioned far to our left. Clock 2 was built after all acceleration ceased during setup, set to zero, "off" state. It skims past Clock 1 (our clock) in vacuum free fall, toggles touch, both Clocks 1 and 2 are "on" and locally synchronized by touching. Elapsed time accumulates in each clock.

 

CLOCK 3: In a spaceship traveling at 0.999c relative to our inertial frame of reference, but 180 degrees counter in direction to Clock 2, far far to our right. It was built after all acceleration ceased during setup, set to zero, "off" state.

 

An arbitrary time after Clocks 1 and 2 synchronize and turn "on" by touching, Clocks 2 and 3 brush past each other, both in vacuum free fall, touching toggles. Clock 2 is now "off," Clock 3 is now "on." Write down the elapsed time in "off" Clock 2. The spaceship with Clock 3 returns over the path taken by the spaceship with Clock 2.

 

CLOCK 1: Our clock. It sits stationary in our inertial reference frame with a little toggle sticking out. Clock 3 vacuum free falls past, toggles touch. Clocks 3 and 1 are off. Write down elapsed times. No clock accelerated while "on" or while existing.

 

BOTTOM LINE: Send results by radio. Numbers on paper don't change. Throughout the entire run three clocks were passive observers in vacuum free fall with zero acceleration.

 

Compare elapsed times. Elapsed times #2+#3 does not equal #1, the local stationary reference frame summation. The sum of #2+#3 elapsed time is about 4.5% that than of #1's accumulated elapsed time. The Twin (Triplets) Paradox obtains without any clock having been accelerated.

 

--

Uncle Al

UNDER SATAN'S LEFT FOOT

Vote a 10 for doing the experiments!

[Little Bang]

I have no way to perform this experiment but I'm sure there are some who can if I can make this attachment thing work.

[Little Bang]

Sorry about that attachment. It took me awhile to get it big enough to read.

[freeztar]

If I understand your question correctly, there should be no change in distance between A and B, so there would not be a shift either way, like the Earth and Moon. Or maybe I misunderstood the question?

[modest]

 

The first example says "mirror A is stationary with mirror B" and "the distance between them is increasing". I'm having trouble making sense of that.

 

In the second example, I don't think there would be any doppler shift. The only frequency change should be due to time dilation. B observes a higher frequency than A emits and A observes a slower frequency than B emits. The two shifts would cancel for either individual observer such that A would observe the returning beam to have the same frequency as the beam he emits.

 

You can think of it in terms of a gravitational field. Think of B being lower in a gravitational well while A is higher. B's clocks will run slow by [math]\sqrt{1-r^2 \omega^2/c^2}[/math].

 

~modest

 

 

 

 

EDIT:

 

If I understand your question correctly, there should be no change in distance between A and B, so there would not be a shift either way, like the Earth and Moon. Or maybe I misunderstood the question?

 

I think, as far as doppler shift goes, that is correct.

 

But, also, the moon is always having a different (greater) velocity than the earth. From our perspective here on earth the clocks there should run a tad slow from time dilation. The frequency of a laser is essentially an accurate clock so that someone on the moon emitting a laser would have it arrive here with slightly less frequency than when it left.

 

~modest

[Little Bang]

Ok, the first example. A and B each have a gyroscope and circuitry to keep them parallel even though B continues to move in a straight line away in the direction of the arrow thus the distance between them increases.

 

Modest, in my opinion I think you are absolutely correct . The doppler shift experienced by B is due to it's changing frame of reference ( Delta t ) and links doppler shift to my hypothesized earlier posts suggesting that gravity and Inertia are also due to changing frames of reference.

[Michael Mooney]

Micheal, you ask valid ontological questions, none of which can be fully explained. Let me ask you a question. Go to here, http://hypography.com/forums/physics-and-mathematics/20793-relativity-discussion-from-what-time-thread.html#post279488, Post number 32, example two and explain with your ontological view why B sees the light it receives as blue shifted.

 

I checked it out and found it way too specific a question for the present discussion.

(Examining the DNA of a specific tree, so to speak, while the overview of the forest is my present interest here.)

 

I am not an expert on red/blue shift anyway, tho I do know that the whole topic is presently being challenged by certain "Whackos" like Alton Arp. (So name-called by our dear Modest, tho Dr. Arp's credentials are quite impressive.)

See Halton Arp's discoveries about redshift

 

Gotta go.

Michael