zztop

I offered a analysis in the locked post that addressed this issue. Introducing the third observer as a "third twin" is a bit misleading. As I pointed out in may last analysis, there is no way to arrange things with this set up where all three observers will agree that all three are the "same age" at the same time. The only way that three observers with different relative motions with respect to each other will agree on each other's respective clock readings is if they are co-located. In this scenario, two of the observers are co-located at the start and two at the end, but only one of them in the same observer in both cases. At no point are all three ever co-located.

 

The whole idea of "resolving" the twin paradox to to show how the two twins will agree at both the start and finish; that, while they may disagree as to what is happening during various points of the experiment, they both end up agreeing on the same end result.

 

In this respect (the only one that really matters), the so-called "triplet" paradox is even easier to explain. None of the observers change inertial frames, so all we have to do is examine it from the three different inertial frame.

 

The following diagrams present the three rest frames for the three observers Blue is our "stay at home" observer, Green starts co-located with him and then travels away, and Red starts some distance away, and passes Green on his way to joining the "stay at home" observer.

 

These diagram also show the light carrying visual information so we can note not only what each frame says the respective clock readings are, but what each observer would visually see as happening.

 

We start with Blue.

 

trip.gif

 

Green and Blue start with the same reading. As Green recedes, its clock ticks at a slower rate. And by the time it meets B Blue's Time reads 4 and Green's 3.46 (up to this point is doesn't matter what Red's time is, as it will be set to match Green's at this point.)

When Red meets Green, Blue visually sees Green's time as being somewhere between 2-2.5, and won't actually see Red and Green meet until his own clock reads ~6. Thus while Green's time goes from 0 to 3.46 while his clock goes from 0 to 4(time dilation) he sees Green's time tick from 0 to 3.46 while his clock goes from 0 to 6(Doppler shift).

 

From this point on We can ignore Green and concentrate on Red. Red goes from 3.46 to 6.92 while Blue goes from 4 to 8(time dilation), and Blue sees Red go from 3.46 to 6.92 while he goes from 6 to 8(Doppler shift).

No matter which way you examine it, Green leaves reading zero, transfers a clock reading of 3.46 to Red and then arrives at Blue reading 6.92.

 

Now we look at Red's rest frame.

trip2.gif

 

Both Blue and Green start a equal distance away and move towards him, with Green approaching faster, Both clocks tick slow, with Green's ticking the slowest.

He will visually see Green ticking the fastest, and only a short period of time passes between his seeing Green and Blue separate and his meeting up With Green. Green reads 3.46 at this moment and Red sets his clock accordingly.

From this moment on, he no longer cares what happens to Green.

At this moment Blue reads 5, but he visually sees a reading of ~2. and while Blue goes from 5 to 8 while he goes from 3.46 to 6.92(time dilation), in this same period, he sees blue tick from 2 to 8(Doppler shift). In either case, he meets up with Blue as his own clock reads 6.92 and Blue's reads 8

Once again, Blue and Green separate reading the same time, Red meets Green with Green transferring a clock reading of 3.46 to Red and Red and Blue meet up with blue reading 8 and Red 6.92

 

Lastly, Green's rest frame.

trip3.gif

 

Blue separates from Green with its Clock ticking slower. Red starts some distance from Green and approaches at a faster speed than Blue is receding.

When Red reaches Green, Greens clock reads 3.46, which is transferred to Red. Blue reads ~3 at this moment (time dilation), and Green sees Blue reading ~2.

Sometime after Greens clock reads 9, Red catches up to Blue. Red will have ticked from 3.46 to 6.92 while Blue goes from 3 to 8, both due to time dilation. Green does not See red catch Blue until his own clock reads ~14. Thus while his clock goes from 3.46 to 14, he sees Blue's clock go from 2 to 8 and Red's clock go from 3.46 to 6.92, both due to Doppler shift.

 

We start and end with the same results as in the last two frames.

 

All three observers agree upon the outcome and there is no "paradox".

I wish I could give you +10. The way it stands, I could only give you +1.

A few notes:

1. It is sufficient to do the analysis for half trip only. At the half trip, the travelling twin has clocked 3.46 and the stay at home twin has clocked 4. After that, symmetry takes over.

2. If one wants to take acceleration into account, for a more realistic analysis, I wrote these two sections for wiki long ago. The perspective of the travelling twin is most interesting.

Are these effects complimentary?

What do you mean "complementary"? The question doesn't make sense.

Are they similar in order of magnitude (I suppose that depends on the situation)?

GR defaults to Newtonian mechanics in the limit. This is well known

Does this temporal elongation cause internal forces, or does it merely appear elongated to an outside observer?

It causes internal strain. It is not an observer dependent artifact.

And what about the pancaking?

I already explained that this is a misleading, sloppy, unfortunate term used by Suskind.

I assume you refer to the pdf you linked a couple of posts ago?

I think you can do the math, so I'm not contesting any of it, but as far as I can tell:

- there is nothing about pancaking in there

- there is a gradient in the acceleration, so it says little about a situation with only acceleration (or at least negligible gradient).

 

I do have a question related to what is in there: how much does this result deviate from simply doing the calculation with Newtonian mechanics, which seems to do reasonably well for calculating the Schwarzschild radius? I'm not used to this Schwarzschild metric used.

It teaches you that the temporal distance between two endpoints of a rod increases as the radial coordinate decreases. This means that the rod gets elongated. As I already mentioned, one gets a similar (but not identical) result by using Newtonian mechanics.

Have you resolved this issue?

I still don't see how acceleration proper can make someone into a pancake without a spaceship that can turn you into a pancake regardless of whether it is near a black hole or not.

 

I admit I downvoted one of your posts, but only because you accused me of trolling, not for anything content related. I even commended you on you math.

Yes, I did. You need to be able to follow the proof.

 

 

Hmm. Is this true when I choose coordinates such that, say, the spatial coordinate [math]x^4 = ct[/math]? Not being a physicist I had always understood that a "point" in spacetime was an "event" as used in common parlance.

No, it is no longer true in 4-space but this is not what he was talking about. Please do not encourage the troll.

 

What makes you think I don't understand - I gave the figures for a SMBH and some rough ideas about other forces.

 

 

 

 

 

 

I am not talking about you.

I am pretty sure that the exact proof you posted relies on a second order series simplification - but frankly I cannot be bothered to go back and look; it should definitely come with a curly almost equals and a health warning.

No, my proof is exact, doesn't use any "simplification", there are no "curlies".

And if we are talking about neg-reps - I only give them out for rudeness and arrogance. I haven't bothered dishing one out for a while - but bombastically making sweeping claims and then backtracking by narrowing the application of those claims is always good to gather red marks from the groundlings

I wasn't talking about you.

Look, there isn't any dispute that the corners of the cube are events in spacetime while at the same time they are positions in our reference frame.

 

Events and positions are two different things. Get your facts straight.

The surface tides for a blackhole at the surface (EH) of a reasonable but undistinguished spiral galaxy (say 4million solar masses) would be a few tenths of a millimetre per second per metre *. So even two unconnected test masses would only move away from each very slowly and no spaghetification would occur outside the EH

 

* 6e-4 m/s^2 per metre

 

If I have my fermi estimations correct then at about 2e4 solar masses you would start to have problems (curl into the fetal position to avoid being stretched - although it might temporaraly fix a bad back) and at 10000 you would die. To be properly stretched at EH you would be down in the magnitude of 1000 solar masses and lower

Elongation exists for ANY BH (it exists for any radial fall towards a gravitating body). The effect can be shown not only with the GR formalism but also with the Newtonian one. The amount of elongation is related to the Schwarzschild radius (which is related to the gravitating body mass but [math]r_s[/math] is a more elegant way of calculating). I already posted the exact proof. Once again, I want to thank all the small - minded people that keep downvoting my posts. It gives a good tally of all the people that fail to understand the subject. If you do not understand , just ask, I am more than happy to explain.

 

 

There is a set of directions in the local coordinate system, just as there is a set of lengths. We seem to pretend to understand the length contraction part by ignoring any questions about the direction part of the system. The two must jibe with one another. Anything that affects length will also affect direction. So the question is "how do all of the directions in the local coordinated system remain coherent when we throw length contraction into the mix?"

 

This question can't really be answered by simply choosing between measuring or seeing. The effect of length contraction is a real effect caused by the requirement that spacetime has to be isotropic and time moves only in one direction.

You need o stop posting rubbish and trying to pass it as science. What I posted is textbook science.

An observer co-moving with the cube will not measure ANY length contraction, to this observer the cube will show no change.

A stationary observer will measure the cube (and its markings) contracted in the direction of motion.

The same stationary observer as above will "see" the cube rotated about an axis perpendicular on the direction of motion (see the Terrell-Penrose effect).

So, the issue is quite complex and depends on:

-the motion between cube and observer

-whether we are talking about "measuring" or "seeing"

Due to the Doppler effect, the light waves in the arms of MM interferometer, generated by the light source, are shortened in the direction of the source movement and elongated in the opposite direction.

These speed-dependent changes of the wavelengths in the arms of the interferometer, give in effect the null results of the MM experiment, and this is only when the light source and mirrors are not moving relative to each other.

This is in my opinion, clearly shown in the previously given link:

https://www.dropbox.com/s/9ljwu5wwsi0v9up/VerificationTheoryRelativity.xlsx?dl=0.

The claims that in the MM interferometer there is no the Doppler effect, testify only to the misunderstanding of this phenomenon.

repeating the same crank claims doesn't make them true

At the second attempt, it was. You needed a nudge towards using the gradient. A fact which, to me, makes your current stance all the more surprising.

 

Now I like to see your math about how a person in free fall, with his head and feat experiencing the same acceleration, will be pancaked.

I already posted it, u need to click on the link. "Pancaking" it is not illustrative of what happens (this is the term Suskind used) , elongation is much more appropriate.

about how a person in free fall, with his head and feat experiencing the same acceleration,

Based on the above, it is clear that you do not understand GR, the different parts of the body are experiencing different accelerations (both in terms of direction and in terms of value). Hence, the "spaghettification" and the "elongation" effects. If some of you spent as much time at dancing around the issue and posting (incorrect) prose as studying GR, you would have figured it out. None of you posted one equation. None.

Quite a few of us on this forum CAN follow the math, ZZ.

What's more,, we understand the math and don't try to 'hide' behind it.

First you claim there are no tidal effects within an event horizon ( again, its in black and white in posts #6 and #8 ), and then post a bunch of simplistic math to 'prove' that tidal effects are related to EH size. Which was our claim to begin with.

Yes, it is a simple proof that none of you was able to produce. It also happens to be correct

We don't claim that there is a preferred viewpoint, that only the far-away observer's frame is significant, and that Alice will never know what happens to Bob. Bob's frame of reference is equally valid, and he WILL know what is happening as he crosses the EH on his way to the only event in his future, the meeting with the ( possible ) singularity. The two views cannot be contradictory, and can be explained by considering the signal reaching Alice from the infalling Bob

Nothing to do with the subject being discussed, proving that you do not understand the subject, contrary to your claims.

Incidentally, you don't win arguments by down-voting your opponent.

That just makes you a sore loser.

Precisely what some of you did (in addition, you repeatedly resorted to ad-hominems). Thank you for making my point.

There is a compressive tidal force, or set of forces, on an object at right angles to the gravitational force (Newtonian), but it would assist with spaghettification, not pancaking.

Correct. This effect is not what my proof is about, My proof shows the elongation in the radial direction.

I want to thank all the that keep downvoting my posts, they demonstrate the mentality of this forum. Not worth wasting my time.

Right. There was no point in introducing it

 

 

The difference in the force on the two ends. I pointed that out all the way back in post #4, and others have pointed this out. Your wording implies that it would happen in a uniform field, and would break something apart outside the event horizon. Let's see the math, if that's your claim.

Has nothing to do with force, there is no such thing a gravitational gorce in GR. It is an intrinsic property of radial motion in a gravitational field .

Right, but there is a difference between "being on a space ship" and "being on a spaceship with a prop acceleration away from the black hole", the latter of which, as swansont pointed out, was not stipulated until your last post.

You both miss the point, the discussion is about the contribution of the BH gravitational field. Since you three share a misunderstanding, I suggested that I provide a rigorous proof as to how the gravitational acceleration (not the gradient) also deforms the objects. By itself.

200g is a condition you put in later, not part of the discuscussion in the post I responded to.

 

So absent this newly positioned goalpost, there is no effect from the acceleration.

The discussion is about the contribution of the BH gravitational field, not of the spaceship acceleration.

Since several of you seem not to understand the concept, I can give a rigorous proof that a rod falling radially into a BH gets stretched due to the gravitational field of the BH.

Only if the ship is accelerating away from the gravitating body with the equivalent of 200g.

That is precisely what I said.

And in an 800g gravitational field, that requires... a 1000g acceleration.

(Proper) acceleration is absolute, meaning that it isn't relative to the gravitating body.

Looks like (at least) three of you have some severe misconceptions about the fact that not only the acceleration gradient but the acceleration proper causes objects' deformation during radial freefall. A short mathematical explanation might help. Would you be interested in learning something new?

Firstly the electric field will always point towards the plate, that is, the direction of electrostatic force remains unchanged. The restoring force acts in opposite directions with respect to the displacement x. This means that if the electrostatic force attracts the ball due to opposite charges, a positive displacement towards the plate occurs. However a restoring force acts in the opposite direction. The displacement towards the plate occurs as long as F>kx. But at the extreme position, when kx> F, the ball comes back. This continues and causes a negative displacement.

This makes me think that the ball shows harmonic motion.

Yes I doubt whether the F is sinusoidal or not. But magnitude of F is dependent on x, and if we write it like this

F=kqQ/(l-x)^2, where l is distance of plate from mean position.

But if I put this value in the equation, it becomes so tedious. Therefore I assume that F=kqQ/d^2 sin wt

The correct equation is:

[math]m \frac{d^2x}{dt^2}+kx+\frac{a}{x^2}=0[/math]

The solution is:

this

The motion is not harmonic

Um, what? The ship will also be in freefall.

The ship as a 200g acceleration wrt to some frame of reference, in the absence of any gravitating body.

According to the Equivalence Principle, a body inside the spaceship is subjected to a hravitational acceleration of 200g.

You now add in a gravitating body that generates a gravitational field of 800g. The Equivalence Principle tells you that the effective acceleration exerted on the ship is now 1000g.

There is also no pancaking in the presence of the BH.

Incorrect, the EEP teaches you that one can get 1000g from the combined effects of a BH with 800g and a spaceship of 200g (opposite sense of vectors). In the absence of the BH the gravitational effect is only 200g. This thread is degenerating from science into trolling.

If you can't, there is no pancaking.

The point was generating the effect in the absence of the BH.

Yes, but would not the resistance depend on the thrust of the ship, in which case you'd get the same pancaking effect regardless of whether there is a black hole involved or not?

Sure, if you could find a spaceship capable of a 1000g+ thrust.

In case of the spaceship, the force compressing you depends entirely on the spaceship, and not on the bh. Any realistic spaceship would be freefalling with you, or perhaps resisting slightly.

.

The "resisting slightly" is what determines the amount of "pancaking".

More on topic: I think zztop was very clear about calculating for which size of bh the spaghettification happens inside or outside the eh

Yes, I did. This is why the language of physics is math.

Yes, I just re-read your post #6, and again in #8.

No reference to size of the EH at all.

Do you have selective memory ???

[math]r_s[/math] is the radius of the EH.

Sure, and your claim was that it occurs OUTSIDE the event horizon, while everyone else said it depends on the mass of the BH.

Or do I need to quote you ?

Can you follow the math that compares [math]r[/math] where the spaghettification occurs with the location of the EH? I think I had two very detailed posts on the subject. I also posted the wiki references that show the case where spaghettification occurs way outside the EH. I suggest that you re-read them.