Fe gravity as it relates to the speed of light

By mixing them, all I meant is we're starting to introudce, at your impetus, GR into an issue that can be resolved by SR alone. I think trying to consider accelerated reference frames is muddying the issue.

You had argued that you know you aren't standing still if you felt acceleration. Well you "feel" acceleration all the time just by living on the surface of the earth. It's no different than if the floor were racing up at you at 9.8 m/s^2 as the UA hypothesis contends. However, you are not accelerating along a radial direction of the earth. So just because you feel acceleration, that doesn't mean you aren't standing still.

We're starting to mix GR and SR when it really isn't needed. But if you think that is proof of motion then you just proved the UA hypothesis correct because I feel the ground pushing up against my body.

Special Relativity cannot handle an accelerated frame of reference. When the flight based twin turns around, his line of simultaneity with the earth based twin jumps. This jump could be thousands of years. If he calculates he earth based twin's rate of ageing at any point during the trip (either direction), it is always slower than his own. The whole thing you read in books about how the flight based twin accelerates and that breaks the symmetry is just hand waving. The best way to look at it is that the flight based twin jumps from one intertial reference frame to another.

Quote from: emailking on September 24, 2007, 06:33:11 AM

Quote from: Brennan on September 24, 2007, 03:59:30 AM

When you reach a high velocity you undergo time dilation due to this you are experiencing less time than a stationary observer so you measure your acceleration wrongly (your length perception will also be out).

There is nothing "wrong" about the acceleration or length measurements from the high velocity observer. These measurements are relative. According to the observer, he is standing still and it is you that are approching the speed of light.

Ah, one of those posts that appears to argue, when it actually agrees.

I'm not using GR's set of accelerated iFoRs. Again, acceleration is not relative.

The FE-based observer knows of his or her motion relative to the "left-behind" observer as the UA has accelerated the FE-based observer--and the FE-based observer knows it.

Quote from: Gulliver on September 24, 2007, 11:23:49 AM

I really don't see a difference. The traveling twin can detect either acceleration or the jump in iFoRs. Either breaks the symmetry.

I agree the symmtry breaks. But it doesn't break because of the acceleration. It breaks because the reference frame jumps. i.e. you turn around.

Quote

I'm not using GR's set of accelerated iFoRs. Again, acceleration is not relative.

It most certainly *is* relative.

Let's say someone is moving 99% the speed of light relative to me and accelerating. He exerpeinces 1g of effective gravity in his own reference frame. Someone (an intertial observer) moving the same speed as him at that instant measures his acceleration to be 9.8 m/s^2. Yet I measure his acceleration to be much less than that.

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The FE-based observer knows of his or her motion relative to the "left-behind" observer as the UA has accelerated the FE-based observer--and the FE-based observer knows it.

Are you talking at the start of the trip? because that acceleration is irrelevant. It's the turning around part that matters. It doesn't matter what you do or do not "know" either. I like the example of two streams going back and forth and observers synchonizing their clocks and such. No acceleration of any kind occurs in this example, yet you still have the twin apradox effect.

Quote from: Gulliver

Tell me the reason you contend that acceleration doesn't break the symmetry. We know that acceleration affects time dilation. Some two observers can tell by their clocks that symmetry is broken with nothing more than acceleration. Your frame jumps are off topic. We're dealing with constant acceleration.

Ok there is a broken symmetry between the two when one accelerates, true. But it has nothing to do with the explanation for why the flight bound twin ends up younger. Even with constant acceleration the entire time, this is still true. The flight bound twin sees the earth bound twin age more slowly than him for the first 1/4 and last 1.4 of the trip. The middle half of the journey is just the turnaround stretched out.

Quote from: Gulliver

You misunderstand what we mean by relativity. The observers all agree and come up with the same values for all observers. You know he's been under constant acceleration. You all know the effect on the FE-bound observer's time dilation. You calculate the same values accordingly. The knowledge exists to resolve who is moving and who is accelerating and who feels what acceleration. That means acceleration is not relative.

That someone is accelerating is not relative in SR.

The value of the acceleration IS relative.

So yes, acceleration is relative.

Quote from: Gulliver

The observers all agree and come up with the same values for all observers.

But they don't. They may agree to do this, but it isn't possible. Yes with knowledge of SR you can work out what you think someone else is observing, but you observe what you observe

Quote from: Gulliver

I'm talking about the moment after the "left-behind" observer is left behind. No, it's not just the turning around part that matters. It most certainly does matter what you know. That's the basis of even the twin paradox. We know that they started at the same chronological age, for example.

I'm sorry but you're wrong on all counts there. The acceleration around the earth isn't relevant. Only the turnaround (or the acceleration involved with it if you strethc it out) is relevant here. The symmetry still exists for the first 1/4 of the trip for example, even though one twin has been accelerating the whole time. For that entire part of the trip, he sees the earth bound twin ageing more slowly than he ages.

"The issue in the general relativity solution is how the traveling twin perceives the situation during the acceleration for the turn-around. This issue is well described in Einstein's twin paradox solution of 1918[1]. In this solution it was noted that from the viewpoint of the traveller, the calculation for each separate leg equals that of special relativity, in which the Earth clocks age less than the traveller. For example, if the Earth clocks age 1 day less on each leg, the amount that the Earth clocks will lag behind due to speed alone amounts to 2 days. Now the accelerated frame is regarded as truly stationary, and the physical description of what happens at turn-around has to produce a contrary effect of double that amount: 4 days' advancing of the Earth clocks. Then the traveler's clock will end up with a 2-day delay on the Earth clocks, just as special relativity stipulates."

Isn't that relative?

Just because you can calculate for something doesn't mean it's magically not relative to individuals. Just because I can understand someone's motive for killing doesn't make the value the same for someone else trying to kill.

You're both right, except Gulliver.

Quote from: Gulliver

No. The values are the same. Each knows of the FE-based observer's acceleration.

Yes, they both know of it. But they measure different values. Look, you can't accelerate at 9.8 m/s^2 relative to an inertial observer forever. If you could, you'd get higher than the speed of light. Acceleration doesn't have some special definition in relativity. It's still the rate of change of velocity. The transformation becomes more complicated, but the values ar enot the same as would be the case in Newtonian mechanics.

Let me be clear about this. If you and I are inertial observers in 2 different inertial frames of reference we will measure DIFFERENT accelerations for some accelerating body.

Huh? We just agreed that the acceleration breaks the symmetry. You also failed to address how I'm wrong regarding using what we know in the Twin Paradox. You argued that it's not what we know, but what we observe.

BTW, your intelligent responses are very intriguing.

"Tell me the reason you contend that acceleration doesn't break the symmetry. We know that acceleration affects time dilation. Some two observers can tell by their clocks that symmetry is broken with nothing more than acceleration. Your frame jumps are off topic. We're dealing with constant acceleration."

Ok there is a broken symmetry between the two when one accelerates, true. ...

Quote from: Gulliver

Why would you claim that the inertial observer would measure the acceleration greater than 9.8 ms-2? Both observers see and understand the "felt" acceleration on the FE remains at 1g. Neither believes that the FE is traveling faster than light. Neither believes that the FE accelerates at a constant 1g in any iFor. That's a sophomoric error about acceleration. Acceleration must certainly does have a special definition in relativity. Just consider that velocities don't add linearly and you'll see your error.

I never said it would be more than 9.8 m/s^2. In fact it's less. What aren't you getting about this? Nothing Do you agree a comoving observer with the accelerated frame always measures 9.8 m/s^2? Yes Do you also agree if a non-accelerated observer always measured an acceleration of 9.8 m/s^2 that the object would eventually be traveling faster than light?? Can't happen

As an object gains speed (relative to say, me) it's mass increases. I can keep applying the same force, but the acceleration *I* observe will decrease. The acceleration observed by the object is always the same.

I understand how to add velocities. That's why it works this way, and not your way.

Quote from: Gulliver

Let's be sure I understand you. Do you recant one of the above?

No I don't. They're referring to two different things. You're forcing me to admit there's a break in the symmetry the moment one accelerates and yes of course that's true. This is really more by the definition of symmetry than any useful physical insight. That "symmetry break", however, has nothing to do with the twin paradox effect or its resolution/explanation. The key point of symmetry breaking there is the turnaround.

Quote from: Gulliver

Oh, and before you reply please recall the Twin Paradox under the circle path at constant speed of .9c throughout. Even though they pass each other at .9c they see that their ages now differ, and differ yet again on each circuit.

It's the same problem formulated differently. The turnaround still constitutes the middle half of trip. Don't be confused by the constant speed overall. There is acceleration and deceleration along each direction.

"My point is that the Twin Paradox survives your inherent requirement of deceleration back to zero velocity relative to the non-traveling twin."

No it doesn't. The speed along the diameter of the circle goes to zero halfway through the trip.

"You're right. My mistake. I don't know how I misread your statement. Sorry. I would like to know what you think my way of adding velocities is."

Well it may be correct if you just misread what I said. I had no idea how you were doing but I knew it had to be wrong because I was right.

"It's hard to keep track of what you're saying when I have to guess under what conditions your statements apply. I maintain that acceleration breaks the symmetry and allows both observers to know the proper acceleration. If there is a proper acceleration, then acceleration is not relative."

We're going to be talking past each other forever on this.

All I'm saying is that the initial acceleration and final deceleration are unimportant...they don't even have to occur. It's the turnaround that matters.

"Actually it's at the .25 and .75 of a circuit. My point is that the Twin Paradox survives your inherent requirement of deceleration back to zero velocity relative to the non-traveling twin."

I don't think you agree with me. (And I am right.) The part of the trip that matters is [0.25-0.75.]. During the first 1.4 and last 1/4 of the journey, the circling observer sees time dilation (not speedup) for the stationary observer. Just think of a straight and back trip. All you're doing is stretching it out along the perpendicular so that the overall speed can stay the same.

Quote from: emailking on September 24, 2007, 05:30:20 PM

"My point is that the Twin Paradox survives your inherent requirement of deceleration back to zero velocity relative to the non-traveling twin."

No it doesn't. The speed along the diameter of the circle goes to zero halfway through the trip.

There is no "The" diameter of a circle, so you're not making sense.

Your point was that it was the turnaround that made the paradox work. I gave you an example where there is no point of turnaround. You haven't answered the challenge.

Quote from: emailking on September 24, 2007, 05:27:01 PM

"You're right. My mistake. I don't know how I misread your statement. Sorry. I would like to know what you think my way of adding velocities is."

Well it may be correct if you just misread what I said. I had no idea how you were doing but I knew it had to be wrong because I was right.

"It's hard to keep track of what you're saying when I have to guess under what conditions your statements apply. I maintain that acceleration breaks the symmetry and allows both observers to know the proper acceleration. If there is a proper acceleration, then acceleration is not relative."

We're going to be talking past each other forever on this.

All I'm saying is that the initial acceleration and final deceleration are unimportant...they don't even have to occur. It's the turnaround that matters.

"Actually it's at the .25 and .75 of a circuit. My point is that the Twin Paradox survives your inherent requirement of deceleration back to zero velocity relative to the non-traveling twin."

I don't think you agree with me. (And I am right.) The part of the trip that matters is [0.25-0.75.]. During the first 1.4 and last 1/4 of the journey, the circling observer sees time dilation (not speedup) for the stationary observer. Just think of a straight and back trip. All you're doing is stretching it out along the perpendicular so that the overall speed can stay the same.

This point remains: The observers know which of them is accelerated. Acceleration is not relative. They can determine who has been accelerated, then they calculate the same values. Why would they do any differently? Surely, you're not suggesting that they lie to themselves, are you?

Quote from: Gulliver on September 24, 2007, 06:38:18 PM

Quote from: emailking on September 24, 2007, 05:27:01 PM

"You're right. My mistake. I don't know how I misread your statement. Sorry. I would like to know what you think my way of adding velocities is."

Well it may be correct if you just misread what I said. I had no idea how you were doing but I knew it had to be wrong because I was right.

"It's hard to keep track of what you're saying when I have to guess under what conditions your statements apply. I maintain that acceleration breaks the symmetry and allows both observers to know the proper acceleration. If there is a proper acceleration, then acceleration is not relative."

We're going to be talking past each other forever on this.

All I'm saying is that the initial acceleration and final deceleration are unimportant...they don't even have to occur. It's the turnaround that matters.

"Actually it's at the .25 and .75 of a circuit. My point is that the Twin Paradox survives your inherent requirement of deceleration back to zero velocity relative to the non-traveling twin."

I don't think you agree with me. (And I am right.) The part of the trip that matters is [0.25-0.75.]. During the first 1.4 and last 1/4 of the journey, the circling observer sees time dilation (not speedup) for the stationary observer. Just think of a straight and back trip. All you're doing is stretching it out along the perpendicular so that the overall speed can stay the same.

This point remains: The observers know which of them is accelerated. Acceleration is not relative. They can determine who has been accelerated, then they calculate the same values. Why would they do any differently? Surely, you're not suggesting that they lie to themselves, are you?

*Because* it has no relevance until the situation changes. Theoretically, the fliught bound observer could accelerate off to infinity forever and he always always always sees a time-dilated earth bound twin. The earth observer sees the exact same thing. This symmetry is *never* broken.

If you want me to admit that Bob concludes he's in an accelerated frame of reference and Alice claims she's in an inertial frame then fine. That's true. I said that before and don't ever remember saying otherwise.

Absolutely not. We already went over this, and you said I was right.

Quote from: emailking on September 24, 2007, 04:27:12 PM

I never said it would be more than 9.8 m/s^2. In fact it's less. What aren't you getting about this? Nothing Do you agree a comoving observer with the accelerated frame always measures 9.8 m/s^2? Yes Do you also agree if a non-accelerated observer always measured an acceleration of 9.8 m/s^2 that the object would eventually be traveling faster than light?? Can't happen

As an object gains speed (relative to say, me) it's mass increases. I can keep applying the same force, but the acceleration *I* observe will decrease. The acceleration observed by the object is always the same.

I understand how to add velocities. That's why it works this way, and not your way.

You're right. My mistake. I don't know how I misread your statement. Sorry.

Quote from: Gulliver on September 24, 2007, 05:19:29 PM

Quote from: emailking on September 24, 2007, 04:27:12 PM

I never said it would be more than 9.8 m/s^2. In fact it's less. What aren't you getting about this? Nothing Do you agree a comoving observer with the accelerated frame always measures 9.8 m/s^2? Yes Do you also agree if a non-accelerated observer always measured an acceleration of 9.8 m/s^2 that the object would eventually be traveling faster than light?? Can't happen

As an object gains speed (relative to say, me) it's mass increases. I can keep applying the same force, but the acceleration *I* observe will decrease. The acceleration observed by the object is always the same.

I understand how to add velocities. That's why it works this way, and not your way.

You're right. My mistake. I don't know how I misread your statement. Sorry.

There. At least I thought you were conceding the point. But read it again. A co-moving inertial observer will always measure an acceleration of 9.8 m/s^2 for Bob. (Bob himself of course measures no acceleration for himself as he does not move in his own reference frame.) Alice measures a constantly decreasing acceleration for Bob. If she didn't, his speed would eventually exceed c, which you agreed is impossible.