Relativity - Oh dear, here we go again!
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Thanks, Mortfurd, but in that site I found this...
"From the pole point of view, the front gate closes just as the back of the pole enters. The surprising result is that the back gate is seen to close earlier from the pole framework, before the front of the pole reaches it. The gate closings are not simultaneous, and they permit the pole to pass through without hitting either gate."
This again appears to tell me that the pole is passing through the gate before it is open?The Bug-Rivet Paradox
And this was just plain dumb...{The Bug-Rivet Paradox}.
"From the pole point of view, the front gate closes just as the back of the pole enters. The surprising result is that the back gate is seen to close earlier from the pole framework, before the front of the pole reaches it. The gate closings are not simultaneous, and they permit the pole to pass through without hitting either gate."
This again appears to tell me that the pole is passing through the gate before it is open?The Bug-Rivet Paradox
And this was just plain dumb...{The Bug-Rivet Paradox}.
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Paulhoff
You can't expect perfection.
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You only prove to me why I am glad that humans are only on one planet in this universe.
But you still didn't get it right...
If Spaceman (1) travels for one second, he is going to be travelling a WHOLE lot further than 100 ft.
I think you mean "for one second in frame of reference of Spaceman (2)"
That is why I gave a distance, daaaaaaaaaaaaaaaa.
Also the questions were to be as simple as possible, so pull your head out, or do I have to tell you from where, and that where is yours, and how much force to pull with, and what angle to pull it out at, and it you should clean your head after pulling it out and am sure you can think of more BS.
Paul
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I knew you wouldn't let me down...
....then chomp your friggin little head right off!
You were having a little joke with your old friend weren't you Paulie?
Also the questions were to be as simple as possible, so pull your head out, or do I have to tell you from where [ass?], and that where is yours [around my head apparently], and how much force to pull with [it's going to require surgery - my head's too big and my arse too tight!], and what angle to pull it out at [180 degrees to the plane of the anus], and it you should clean your head after pulling it out [no, my ass is clean] and am sure you can think of more BS [how's this: try pushing yours in so far it emerges from between your teeth
....then chomp your friggin little head right off!
You were having a little joke with your old friend weren't you Paulie?
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That's interesting. But I'm not at all familiar with this concept (is it GR related?) You say "the same accelerating trajectory will have a different acceleration when measured from different reference frames". I admit that I have a little bit of a problem with this statement. Sure the acceleration will have different measurements if the other frames are also non-inertial. But the object that we measure will definately "feel" an acceleration that has one precise magnitude and one precise direction. And this is completely independant of the frames in which the measurement is made. If acceleration truly was relative, the reality of the inertial forces associated with it should also be relative.
A.P French, Special Relativity
I lost you again. a = dv/dt. The acceleration is the rate of change of velocity. For any frame that is inertial (no matter how rapidly these frames may change), if it is an inertial frame, acceleration is constant. I don't see the need to move from one inertial frame to the other to observe constant acceleration. One should be suffcient.Ziggurat said:
ETA: I may understand the need to change from one reference frame to the other in a cricular orbit. But remember, my question was within SR. I wanted a treatment of accelerating frames with SR. And I fail to see how the equations given apply to orbital trajectories. I actually fail to see how SR can solve such a problem.
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Ziggurat
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No, it's entirely special relativity (and only part of GR in the sense that SR is a subset of GR).
Sure, but at any given instance, its current velocity also defines a unique reference frame from which to judge the single acceleration that it feels. All other reference frames will observe different accelerations.
That's just it: it isn't independent. More below.
I'm not exactly sure what you mean by this, but momentum isn't linear with velocity in special relativity.
I lost you again. a = dv/dt. The acceleration is the rate of change of velocity. For any frame that is inertial (no matter how rapidly these frames may change), if it is an inertial frame, acceleration is constant. I don't see the need to move from one inertial frame to the other to observe constant acceleration. One should be suffcient.
It's not. Let's start with the fact that acceleration measured in different reference frames will be different. a = dv/dt. But dv depends on your reference frame. For example, let's say I'm standing still, Jeff is moving north from me at 0.6 c, and Suzy is moving south from me at 0.6 c. Now in Suzy's reference frame, she's standing still but I'm moving north at 0.6c. If we stop there we might naively conclude that dv should be reference-frame independent. But let's keep going: in her reference frame, how fast is Jeff moving? If dv is reference-frame independent (as it is classically), then Jeff should be moving north at 1.2c in Suzy's reference frame. But he isn't. He's moving at less than c. The difference in velocity between me and Jeff, which is 0.6 c in my reference frame, must be less than 0.4c in Suzy's reference frame. So dv is NOT reference-frame independent. Neither, as it turns out, is dt. So when you say a = dv/dt, there's an implicit assumption that you're measuring a from a particular reference frame, and if you change reference frames, the resulting quantity will not necessarily be the same.
When people talk about constant acceleration, they generally mean proper acceleration. That is, we measure instantaneous acceleration from the reference frame defined by the object's instantaneous velocity. This is the kind of trajectory in which the object will "feel" constant acceleration. But because the reference frame defined by the instantaneous velocity keeps changing, what appears as a constant acceleration to the object will NOT appear as a constant acceleration to a non-accelerating observer. Which makes sense: if you could maintain constant acceleration for long enough with respect to a fixed reference frame, you could go faster than light. But no matter for how long an object feels constant acceleration in its own non-inertial frame, it will never get faster than light.
I lost you again. a = dv/dt. The acceleration is the rate of change of velocity. For any frame that is inertial (no matter how rapidly these frames may change), if it is an inertial frame, acceleration is constant. I don't see the need to move from one inertial frame to the other to observe constant acceleration. One should be suffcient.
I guess I'll have to write the Lorentz transformations for acceleration. The sensible way to do it is to use four-vectors, but I'm going to stick with a simple boost. We have two frames, S and S', both inertial and we have some particle with accelerations a, a' in the respective systems. Let V be the relative velocity between both systems and v the velocity of the particle. Then
[latex]\footnotesize
\begin{align*}
a'_\parallel &= \frac{a_\parallel}{\Gamma^3 \left(1-\frac{\vec v\cdot \vec V}{c^2}\right)^3}, &
\vec a_\perp\,'&=\frac{\vec a_\perp}{\Gamma^2\left(1-\frac{\vec v \cdot \vec V}{c^2}\right)^2}+\frac{(\vec a\cdot\vec V)\vec v_\perp}{c^2\Gamma^2\left(1-\frac{\vec v\cdot \vec V}{c^2}\right)^3}
\end{align*}
[/latex]
These formulas should be in many SR books I have just derived them quickly, so they may be some mistakes. Parallel and perpendicular magnitudes are relative to V and Gamma = (1-V^2/c^2)-1/2. Everything comes from the realisation that
[latex]\footnotesize
\begin{align*}
\vec a &= \frac{\mathrm{d} \vec v}{\mathrm{d} t}, &
\vec a\,' &= \frac{\mathrm{d} \vec v\,'}{\mathrm{d} t'},
\end{align*}
[/latex]
You just have to write v' in terms of v and d/dt' in terms of d/dt.
Two different inertial frames measure different accelerations for the same particle. There is no GR here, really.
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Paulhoff
You can't expect perfection.
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Back from my travels. Very impressed with the quality and quantity of posts (thanks). Lots to catch up on. Also pleased that such a controversial subject hasn’t degenerated in to personal bickering, as it often does on this forum. I’m trying to take things slowly and trying to start with the basics. My previous post (#73) was to help clarify that time dilation and time delay are different and not related (sorry for my silly mistake - A ahead of clock B not behind). I’m still not fully confident that they’re not somehow being mixed up. I can’t see where Relativity accounts for the effects of time delay. Perhaps it doesn’t need to? In the twins scenario for instance, the time delay effects created by the twins separating would be totally reversed and negated when they get back together.
Talking about the infamous twins: As I understand it, Relativity says that it’s not just the acceleration from one frame to another that causes time dilation, but it’s also that one object is moving at a different speed relative to another. Given that there is no preferred frame, I don’t see how (after the period of acceleration) motion can be attributed to either frame. I would expect therefore that whatever time dilation applies to one frame would apply in equal measure to the other and would be the same overall. To say that twin A is moving away from twin B at a particular speed is no more valid than saying B is moving away from A at the same speed. I can understand how the different conditions of one twin accelerating, and the other not, could effect one twin differently than the other, but not when they are both in non-accelerated frames.
Talking about the infamous twins: As I understand it, Relativity says that it’s not just the acceleration from one frame to another that causes time dilation, but it’s also that one object is moving at a different speed relative to another. Given that there is no preferred frame, I don’t see how (after the period of acceleration) motion can be attributed to either frame. I would expect therefore that whatever time dilation applies to one frame would apply in equal measure to the other and would be the same overall. To say that twin A is moving away from twin B at a particular speed is no more valid than saying B is moving away from A at the same speed. I can understand how the different conditions of one twin accelerating, and the other not, could effect one twin differently than the other, but not when they are both in non-accelerated frames.
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Ziggurat
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People mix them up by mistake all the time. But they are indeed separate effects.
When performing calculations of what you observe, you do not need to account for it. When calculating what you see, or when determining what you observe from what you see, you do need to account for it. Since neither of those latter steps is necessary for introductory students, it's frequently ignored. But plenty of people have spent time thinking about it, and it produces all sorts of interesting optical illusions.
It can't. The mistake is thinking of the traveling twin as being in only one frame. But he isn't. He necessarily occupies at least two frames. Within each frame, the situation may look symmetric when considering only part of the journey, but the overall journeys of the two twins aren't symmetric from any frame.
Nope. Consider what happens if you adopt the reference frame of the outbound twin. The earthbound twin experiences time dilation, and the traveling twin doesn't for part of his journey. But if you STICK with this reference frame, what happens when the traveling twin heads back towards earth? The earthbound twin is still traveling away, and still time-dilated, in this outbound frame. But now the traveling twin is moving faster than the earthbound twin, and is now experiencing even greater time dilation than the earthbound twin. And if you do the calculations, you'll find that this increased time dilation for the return leg more than makes up for the lack of time dilation for the outbound leg, when viewed from the SINGLE reference frame of the outbound journey. So it's not that time dilation doesn't apply to both twins: it most certainly does. It's that you can't just change reference frames in the calculation without accounting for that change.
But let's step back for a moment and think about some geometry. What's the shortest distance between two points in space (we'll stick with Euclidean space for this)? It's a straight line. Any curved or bent line will be longer. Does the extra distance come from the bend itself? No, not really. Is distance calculated differently in different directions? No, it isn't. Now special relativity uses a different metric, and one of the properties of this metric is that the LONGEST time interval between two events is a straight line. Any curved line is actually shorter, not longer. This is a result of the exact same property of the metric that gives us time dilation. And it's true whether the two events we're talking about are in the same place or not (as long as they're time-like separated). Now we have two events: the twins leaving each other, and the twins coming together. If we pick the earth-bound twin's reference frame, he's stationary the whole time. If we pick the traveling twin's outbound frame, the earthbound twin is moving. But here's the thing: REGARDLESS of which inertial frame we pick, the earthbound twin's path from the first even to the second event is ALWAYS a straight line. And REGARDLESS of which inertial frame we pick, the traveling twin's path is NEVER a straight line. So in each and every reference frame, the traveling twin will always be observed to experience less time between events. If we pick an inertial reference frame in which part of the traveling twin's journey has less time dilation than the earthbound twin, we will find that this is more than made up for by the fact that the OTHER part of the traveling twin's journey (where he's moving in the other direction) has greater time dilation than the earthbound twin. Every time. But we don't need to do the calculation explicitly to know that, because all we need to know is that the earthbound twin has a straight worldline and the traveling twin doesn't.
Thanks, that will take some time (that I don’t have right now) to read and attempt to understand. In the meantime let’s make the scenario simpler by removing acceleration from the equation. Clocks A and B are moving relative to each other through space. At the point they pass each other they synchronise themselves to the same time. As they move apart, each becomes time dilated to the other in equal measure (?). I can’t see anything that makes the properties one any different from the other.
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Well, OK, but then you're basically just comparing the twins' whole trips, not each half separately while ignoring the brief period of heavy acceleration in the middle, which is what I thought you were trying to do.
The stationary twin (and likewise any other inertial observer) can ignore the period of acceleration, because during it, according to him, the traveling twin doesn't age much. But the traveling twin can't ignore the period of acceleration, because during it, according to him, the stationary twin ages a great deal.
Why do you want that?
I mean, in the absence of gravity, you can always just pick an arbitrary inertial reference frame and figure everything out relative to it. What's wrong with doing things that way? Does it merely feel like "cheating" to you, or is there some question you think really can't be answered by this method?
RussDill
Philosopher
In your example, they aren't different. The twin paradox is only interesting because the twins are brought back together again. If one of the clocks turns around and comes back to pass or meet the other clock, the time it displays will be behind the other clock.
The straight line vs curved line starting at the same point, and again meeting at the same point illustrates the twin paradox. Your above example looks at the paradox when the lines are shaped like a V, before one has turned around. Looking at one reference frame is just rotating the V so that one side of it points "up". You can rotate it so that either twin's path points "up". Its symmetrical. From either reference frame, time in the other is running slower. One or the other doesn't have to be right, it's just looking at the V (the diagram) from a different angle, aka, reference frame.
As soon the other twin turns around and returns, the diagram looks more like a D. The D isn't symmetrical. The two have taken decidedly different paths through spacetime.
RussDill
Philosopher
BTW, my favorite paradox for explaining all this stuff used to be the barn/pole paradox. But now, after it being mentioned in another thread, its definately the bug-rivet paradox.
I agree.
I would say “Yes” and I don’t understand your “No” (even followed by “not really“
. A straight line is a direct path between two points. A curved line is an indirect path between two points, and therefore a longer distance. The extra distance of a curve comes from the fact that it’s an indirect path.I agree.
So are you saying that a straight line is a shorter distance but a longer time interval, and a curved line is a longer distance but a shorter time interval? Perhaps it would help if you could clarify exactly what you mean by “time interval“.
I agree.
I agree. And the travelling twin is stationary except for the periods of acceleration.
Don’t really understand this. Is this because the travelling twin has undergone acceleration and the other hasn’t? If so, it seems that acceleration is the thing that causes time dilation.
So if there were three clocks A B and C involved: Clocks A and B Synchronise as they pass. A continues and passes clock C. C is moving in the same direction but faster than B. C Synchronises to A and continues passes B. All clocks always show the same time. Effectively there is no time dilation.
Let’s say that A, B and C are spaceships. B has two Synchronised clocks and gives one to A as he passes. This clock is rapidly accelerated to the frame of A. Then A passes and gives the clock to C. The clock is now accelerated to the frame of C. Due to the acceleration it has undergone the passed clock is time dilated compared to the clock that B kept.
I'm tempted to conclude that time dilation is caused by acceleration and not relative movements. I suspect however that the answer is that acceleration has somehow affected the movement so it also contributes to time dilation. This would seem to mean that movement somehow remembers the acceleration and this doesn't make sense to me.
Thank you Paul. I needed a good clean out.
I could do without the hole in the head though. Usually I have a pinhole there just in case some BS I hear turns out to have a kernal of truth. Now with that gaping hole, I might be susceptible to any short of crap that comes around.
