General Relativity, Acceleration, Gravity...

[boooeee]

Actually acceleration was not the answer to that paradox. You could get around the acceleration by using three ships. The REAL problem people had was switching inertial frames (away-from-earth to towards-earth). (switching frames is like acceleration, but not the same)

I don't want to turn this thread into another discussion of the twin paradox, but I was wondering how this worked. It seems that somebody would have to change velocities at some point in a twin paradox scenario. How does having three ships allow you to avoid acceleration?

 

[uruk]

Time dialation has also been observed in linear accelerators and cyclotrons and particle smashers. (As well as spatial dialation.)

 

[Dr. Fendetestas]

I don't want to turn this thread into another discussion of the twin paradox, but I was wondering how this worked. It seems that somebody would have to change velocities at some point in a twin paradox scenario. How does having three ships allow you to avoid acceleration?

The idea is that the travelling twin can jump from a ship getting away from his brother to a ship coming back home. It's just a way of changing frames instantaneously, without acceleration.

In other words, even in the idealised situation of no acceleration and instantaneous turning back, the situation is not symmetrical, because the travelling twin needs two different reference frames (one is the ship getting away and the other the ship coming back). Acceleration is not important to the paradox.

 

[Ziggurat]

I don't want to turn this thread into another discussion of the twin paradox, but I was wondering how this worked. It seems that somebody would have to change velocities at some point in a twin paradox scenario. How does having three ships allow you to avoid acceleration?

The idea there is that you have one clock stationary on earth, one clock moving away from earth at a rapid speed, and one clock coming towards earth at a rapid speed. Sync the earth-bound clock and the outbound clock when the outbound clock is on earth (or passes earth, if you had it moving already), then sync the inbound clock to the outbound clock as they pass. Then you've got a record of the time taken for that outbound-inbound combo, but nobody ever actually accelerated. The point being, acceleration doesn't matter.

I'll go back to what I said earlier, which is that the math is unambiguous. Here's the metric:

s^2 = x^2 + y^2 + z^2 - (ct)^2

For a curved path, then, we integrate ds along the path. And we get that with our differential metric:

ds^2 = dx^2 + dy^2 + dz^2 - c^2*dt^2

Now the three components of velocity v can be written as

vx = dx/dt, dy = dy/dt, vz = dy/dt

Or alternatively,

dx = vx*dt, dy = vy*dt, dz = vz*dt

Substitute this into our differential metric, and we get:

ds^2 = (vx^2 + vy^2 + vz^2 - c^2)*dt^2 = (|v|^2-c^2)*dt^2

So velocity is used to calculate the proper time experienced along a time-like trajectory in space-time, but acceleration gets ignored. It never enters directly, it only affects what we use for velocity. In the case of a clock in a centrifuge, we don't even need to know that, since |v| doesn't change, only the direction of v.

What the metric is really telling is is that the LONGEST path between two points in space-time is a straight line. Curved lines are shorter, and to get a curved line, yes, you need acceleration, but it's not the acceleration per se that shortens the time - as the example of two centrifuges with the same linear speed but different radii and therefore different acceleration shows.

 

[Hellbound]

The idea is that the travelling twin can jump from a ship getting away from his brother to a ship coming back home. It's just a way of changing frames instantaneously, without acceleration.

In other words, even in the idealised situation of no acceleration and instantaneous turning back, the situation is not symmetrical, because the travelling twin needs two different reference frames (one is the ship getting away and the other the ship coming back). Acceleration is not important to the paradox.

I'd argue that this is a non-argument. You can't "avoid acceleration" because the moving twin has to change his velocity (in direction at least) which is, by definition, acceleration. Without acceleration you don't end up with two twins, you end up with one twin and one gaseous twin/ship molecule nebula. If you ignore physical constraints and allow things such as a sudden velocity change without acceleration, then of course all bets are off. But you're redefining other physical laws to fit into a constrained (and physically impossible) scenario.

Even in the jump from one ship to another, the change cannot be instaneous. the moving twin keeps his initial velocity until he contacts the second ship, at which point that ship accelerates him. With a speed difference of near c, that will result in a highly efficient matter to energy conversion. Assuming that you measure the age of the particulate matter that results from said conversion as it travels back to earth, you still can't avoid acceleration (as this particulate matter was accelerated as well).

Quite simply, if I define the impossible to be possible, then yeah, you don't have to deal with acceleration. But this becomes similar to applying quantum mechanics to macro-scale objects...trying to remove acceleration yet still allow a velocity change is trying to stretch SR into areas it was never meant to cover. SR does not apply to accelerating frames of reference, as accelerating frames of reference are by definition distinguishable from non-accelerating frames. You can label it as "changing frames", which is true. When the ship accelerates it is involved in constantly changing it's frame of reference, and thus outside the realm of SR.

Even assuming the ship-to-ship jumo, you STILL have the initial acceleration (when the ship left Earth) and the decceleration (when the ship returns), thus breaking the equivalence of the two frames.

 

[garys_2k]

As with all of relativity, it's easy to confuse oneself and make it more complex. However, it's all very simple, if you can get yourself into the right Zen-like state. Einstein is rightly hailed as great because, while many, many people worked on relativity, he was especially good at being a kind of boddhisatva for the rest of us.

I think you're doing this kind of self-confusion. There's no dishonor in that; we all go through that when understanding relativity.

Guilty as charged! "Self-confusion" is an excellent term for what I'm feeling, thank you. I just didn't realize what a leap it was (for me, at least) to grok GR. Special Relativity seems easy compared to this, but I do think some daylight is starting to penetrate!

But it's really easy. You are correct that all observers have to agree on the laws of physics. Consider Fred and George, each in spaceships that are going neck-and-neck. Each is sitting at the stern, and each has clocks at the front that are emitting light pulses. When a light pulse is emitted, George starts to accelerate. So, Fred will measure the light pulse as hitting George earlier than it hits Fred because, by accelerating, George will have shortened the distance it had to go. George also has to see it earlier, because he will also measure himself as having reduced the distance, and the speed of light is constant for all observers.

I have all of this, yes. Both observers would agree that the light hit George first. But I need some clarification on this next part, please:

Since the clock is held at a constant distance while George is accelerating, every light pulse will have to come earlier, and it has to happen so long as George is accelerating. The only way this can happen is for George to see the clock as running faster.

So, is that like seeing the light from the clock blueshifted? I would assume that George would see it blueshifted (he's plowing headlong into the photons), correct?

Now, how does that light-hitting-George-first/clock-running-faster/MAYBE-blueshift thing make it seem that (in the case of one ship, light clock at the front and observed from the back) make it look like the ship is getting elongated? Or is it squashed shorter?

 

[teck49]

the LONGEST path between two points in space-time is a straight line. Curved lines are shorter

Sounds good, but if that is so,

then wouldn't a cicumference of a circle be shorter than its diameter?

 

[Dr. Fendetestas]

I'd argue that this is a non-argument. You can't "avoid acceleration" because the moving twin has to change his velocity (in direction at least) which is, by definition, acceleration.

[...]

You are right, of course. I don't like the 'ship jumping' story myself. Some people use it to illustrate an instantaneous velocity change, but I don't see the point. I just mentioned it because someone explicitly asked for it. The point is that you don't need a finite acceleration time to break the simmetry, the fact that one of the twins needs two reference frames is the really important bit.

 

[boooeee]

Re: Re: Re: General Relativity, Acceleration, Gravity...

Edit to add: I just found the equation for gravity-induced time change:

t' = t/sqrt(1-2GM/rc^2)

where t is time measured very far from any massive bodies.

It looks like we have that "r" term for distance, and there is no equivalent acceleration term that could be substituted for it. We could substitute g = 9.8 m/s^2 for G, as long as we also use an r = 6.38 x 10^6m, but you have to have both terms.

So, is there no "right" equivalent absolute time dilation on an accelerating ship?

Actually, you can approximate it with a pretty simple formula.

Define g as the acceleration of the ship, h as the height distance between the two clocks, and c as the speed of light.

A clock that is h meters above the floor will run at a rate that is (1 + gh/c^2) faster than a clock on the floor. Plugging in real numbers, a clock on the floor will record one hour, while a clock on top of a three meter bookshelf will record an extra ( 9.8 * 3 / 300000000^2 ) = 3.3x10^-16 hours. Or about an extra second every 350 billion years.

There is a great explanation of all this in the last chapter of Volume 2 of the Feynman Lectures on Physics. It is the most clear explanation of general relativity that I have ever come across.

 

[Ziggurat]

Sounds good, but if that is so,

then wouldn't a cicumference of a circle be shorter than its diameter?

Let me clarify slightly, since my original statement is a bit too general:

The longest time-like path between two points in space-time is a straight line.

(by time-like I mean that ds^2 is negative at all points along the path)

There is some applicability to space-like curves as well, but it's more complicated. As you note, a half-circumference of a circle (I assume that's what you meant) is longer than the diameter. However, if you take points with the same time coordinate but different space coordinates and draw a straight line between them, that length WILL be longer than if you draw the same line but now lift the midpoint out of the equal-time plane.

For example, put a point at t=0,x=0 and one at t=0,x=1 light second. The straight-line distance between them is 1 light second. But if you draw the line between them as two segments, from t=0,x=0 to t=0.4sec,x=0.5lightsec to t=0,x=1lightsec, then the distance there is 0.6 light seconds - the curved path is now shorter.

So yes, there's some dependence on HOW you curve the line. For space-like separations, you have to curve it in the time-like direction to shorten it, while curving in the space-like direction lengthens it. But for time-like paths, ALL curved time-like paths between two points are shorter than the straight time-like line between them.

 

[boooeee]

The idea is that the travelling twin can jump from a ship getting away from his brother to a ship coming back home. It's just a way of changing frames instantaneously, without acceleration.

In other words, even in the idealised situation of no acceleration and instantaneous turning back, the situation is not symmetrical, because the travelling twin needs two different reference frames (one is the ship getting away and the other the ship coming back). Acceleration is not important to the paradox.

Thanks. I guess I would still call that acceleration since the twin's velocity is still changing.

Although non-instantaneous acceleration may not be needed for the paradox, I think it definitely helps resolve the paradoxical nature of the problem.

For me, the confusion was always when you considered the travelling twin's point of view. When he is travelling away from his twin, he observes his twin to age slower. When he is travelling toward his twin, he still observes his twin to age slower. So, how is it when he arrives back home that he observes his twin to have aged much faster? The answer lies at the halfway point when he turned his ship around. While he is firing his thrusters in the opposite direction, that is when he observes his twin aging much more rapidly than he is. The reason is that from the travelling twin's point of view, his twin is very high up in a gravitational field, where clocks run faster.

By introducing instantaneous acceleration, this implies that from the travelling twin's point of view, his other twin's age jumps instantaneously at that point.

 

[Ziggurat]

By introducing instantaneous acceleration, this implies that from the travelling twin's point of view, his other twin's age jumps instantaneously at that point.

Sure, but that's not really a problem, not if you draw it all out in space-time diagrams. In that case, all you're doing is tilting the traveling twin's space axis (the line which defines all space at the same time for him), and that line then intersects the earth-bound twin at a much later point along the earth-bound twin's trajectory (instant aging). Part of the difficulty for some people here is that what the twin SEES and what he OBSERVES are different. There's a major discontinuity in what he observes, but that's actually fine, there's no problem with that that needs any resolving. The only thing that we can't permit a discontinuity in is what he sees, but there's never a discontinuity in what he sees, even with instantaneous velocity change (infinite acceleration).

 

[garys_2k]

For me, the confusion was always when you considered the travelling twin's point of view. When he is travelling away from his twin, he observes his twin to age slower.

Correct. He is looking back at photons that left at time intervals shorter than they look on his clock. So, yes, the earth-bound twin's clocks all seem to be runing slow.

Originally posted by boooeee
When he is travelling toward his twin, he still observes his twin to age slower.

Here I disagree. On the trip back he's plowing into photons that had left his twin brother's clock at time intervals that appear to be MUCH shorter than his own. Therefore, the traveller would say that the earth-bound twin's clock is running fast.

Originally posted by boooeee
So, how is it when he arrives back home that he observes his twin to have aged much faster? The answer lies at the halfway point when he turned his ship around. While he is firing his thrusters in the opposite direction, that is when he observes his twin aging much more rapidly than he is. The reason is that from the travelling twin's point of view, his twin is very high up in a gravitational field, where clocks run faster.

No. Nothing of note happens during the time his thrusters take him from relativistic speeds to zero (relative to earth), except that he sees the clock on earth speeding up. If he stops his rockets when he's stationary with respect to earth he'll note that both clocks are running at the same speed (but a residual error in absolute time exists, because there are still a lot of photons on the way).

 

[Ziggurat]

Here I disagree. On the trip back he's plowing into photons that had left his twin brother's clock at time intervals that appear to be MUCH shorter than his own. Therefore, the traveller would say that the earth-bound twin's clock is running fast.

No, this is incorrect. You are confusing what the twin observes with what the twin sees. They are NOT the same thing. The traveling twin SEES the earthbound twin's clock running fast on the return journey due to doppler shift, but he corrects for that when figuring out what he OBSERVES, and what he observes is the earthbound twin's clock is running slow on the inbound journey. Relativistic time dilation is something you observe, not see, and it is always independent of the direction of relative motion. What you see has doppler effects piled on top, and that IS dependent on the direction of relative motion.

It's easy to get tripped up on this distinction, but it's absolutely critical.

 

[boooeee]

garys_2k - What Ziggurat said.

My mind always locks up when I try to consider both relativistic effects and doppler style effects at the same time. I'm still working my mind around Ziggurat's earlier post on the discontinuity only occurring in what is observed, not seen.

 

[Dr. Fendetestas]

garys_2k - What Ziggurat said.

My mind always locks up when I try to consider both relativistic effects and doppler style effects at the same time. I'm still working my mind around Ziggurat's earlier post on the discontinuity only occurring in what is observed, not seen.

It's quite simple. At the 'turn' event, the line of simultaneity of the travelling twin changes, so he measures a sudden age jump in the stay at home twin. However, if he were seeing him on an on board TV, he wouldn't see his brother suddenly turning into an old man. Think about this. If you still have a problem, we will explain it more thoroughly.

 

[garys_2k]

OK, for the twin thing, here's how I've looked at it ever since I made up the spacetime diagram. But this is the first time I've heard of a distinction between "observed" and "seen" in this context.

Setup:

Two brothers, Able and Bob. Able's an astronaut and plans on going to a rock one light year from earth. Bob stays home. Able will go to the rock at almost the speed of light, will stop there for one minute to have a look around and return at the same speed.

From Bob's point of view:

Bob watches Able blast off and get almost to light speed. He sees the outbound trip take just over a year, 366 days to be exact. When he sees Able get to the rock he turns on a big illuminated "Congratulations!" sign. After a minute goes by he sees Able restart his rocket and return. This trip also takes 366 days.

At the end of the trip Bob has aged two years, two days and one minute. Able has aged two days and one minute.

During the trip Bob can watch Able's dashboard clock. It moved slowly on both the outbound and return trips. In fact, when he sees Able reach the rock only a single day had elapsed on the ship's clock, whereas 366 days had elapsed on earth. The same thing happened on the way back -- another day elapsed on the ship while 366 days elapsed on earth. Able's dashboard clock only ran at the correct (to Bob, on earth) speed for the minute that Able was waiting at the rock.

>>>>> Question: Is the above sceario possible?

From Able's point of view:

Able watches Bob receed from view out his back window. He notices that Bob's clock run's slowly, too, and that only one day goes by on Bob's clock during the journey. What's equally strange, though, is that the distance he's travelling has been compressed. What had looked like a light-year's distance has been dilated down to a single light-day. The trip takes him what feels like (and is, according to his dashboard clock) only one day to make.

>>>>> Question: Is that possible? If so, his clock and earth's appear to Able to be synchronized.

OK, assuming it could read one day, the story continues:

At the rock, while stopped, Able now notes that both Bob's earth clock and his (still?) keep the same time. He sees light from one day after he left and the earth clocks bear that out. He waits a minute and sets off on his return journey.

Now he starts to run into light that had left earth in the intervening 365 days. Time seems to rush by on earth as he watches, and about a third of the way back he finally gets to see Bob's "Congratulations!" greeting.

He continues on, all the while taking what seems to him to be a single day (distance dilation has made this a one light-day trip, instead of a one light-year trip) but he can see what looks like a sped-up movie going on back on earth. When he finally arrives he's seen 365 days (most of his outbound trip) + 366 days (his return trip) worth of time go by on earth. The minute he waited at the rock took place at an equal pace for both and was part of the duration of his journey from both points of view.

Able finds his brother to have aged, most of it taking place on the way back. They both agree that Able has aged only two days and one minute, and Bob has an extra two years tacked onto his age.

So, where did I miss the boat?

 

[epepke]

So, is that like seeing the light from the clock blueshifted? I would assume that George would see it blueshifted (he's plowing headlong into the photons), correct?

A blueshift does happen. It's exactly like looking at pulses of light, only at a much finer level.

Now, how does that light-hitting-George-first/clock-running-faster/MAYBE-blueshift thing make it seem that (in the case of one ship, light clock at the front and observed from the back) make it look like the ship is getting elongated? Or is it squashed shorter?

I think I see what you're saying, but I'm not quite sure, so please correct me if I'm wrong. To make it simple, let's say that George's ship is under constant acceleration. Ignore Fred. George tries to measure his ship by sending out a pulse of light from the stern of the ship to a mirror at the prow and times how long it takes to go forward and return, then calculates the length of his ship by that time multiplied by c. He knows the time t that it will take, by his clock, when he is not accelerating. The time he measures it as taking will be slightly longer than t, so he could conclude, in a sense, that his ship is slightly longer (at least the part that's to the front of him).

Similarly, if he were at the prow of the ship and sent a light pulse to a mirror at the stern, he would conclude that, in a sense, the sternward part would look shorter.

If, however, he measured it with a stick, he would of course get the same result as if he had not been accelerating.

Of course, under constant acceleration, this apparent lengthening and contraction would be constant, which is how I justified saying that the distance would be constant and result in the faster clock at the prow.

It gets worse. Let's say that he tried to measure the width of the spaceship by the same means. Because he's accelerating, the path of the light to him would be curved, so he'd have to point the light slightly forward of the surface normal.

Also because of the curvature, he will actually see it as longer if he looks forward, if he compares it with a standard perspective transformation. I've attached a crappy little diagram to show this.

By now it should be ovbious why, while SR is nice and simple and linear, GR is nonlinear and complex.

 

[69dodge]

Originally posted by garys_2k
OK, for the twin thing, here's how I've looked at it ever since I made up the spacetime diagram. But this is the first time I've heard of a distinction between "observed" and "seen" in this context.

"See" means literally see: you "see" something when light from it enters your eye. "Observe" means that you correct for the light propagation time: if you "see" something happening now, but you know that it's a light year away, then you "observe" that it happened a year ago.

Two brothers, Able and Bob. Able's an astronaut and plans on going to a rock one light year from earth. Bob stays home. Able will go to the rock at almost the speed of light, will stop there for one minute to have a look around and return at the same speed.

From Bob's point of view:

Bob watches Able blast off and get almost to light speed. He sees the outbound trip take just over a year, 366 days to be exact. When he sees Able get to the rock he turns on a big illuminated "Congratulations!" sign. After a minute goes by he sees Able restart his rocket and return. This trip also takes 366 days.

Does B turn on the sign 366 days after A left, which is when he knows A is landing on the rock, or does he wait until he actually sees A land on the rock, which takes an extra year because the rock is a light-year away?

and similar sorts of questions about some of the rest of the scenario...

 

[garys_2k]

re the twin paradox : Egads! I do need to rethink this. Thanks for the fresh eyes look, people, it is appreciated.

Yes, of COURSE slap_forehad Bob wouldn't be able to see Able reach the rock when I thought he would. Hmm, I'll have to pull out the spacetime diagram again, I'd thought I'd plotted light cones from both sides but obviously I either blew it or forgot how it looked.

I figured I'd messed up something when I realized that Able would see Bob's clock agreeing with his on the way out. No can do, Bob's clock should look like it's running much slower.

A background question: At very high speeds the traveler should see his clock appear to read time normally, but he should perceive his motion to be fantastic, right? I mean, from his point of view he should be able to cover what we see as a light year's distance in a day, correct?

re the GR acceleration/gravity time dilation: Yes, it is much more complex than the SR speed dialation problem. As is clear, I >thought< I had a reasonable handle on the SR side, so this is a tougher nut to crack.

I again want to thank everyone that's helping with this!