Was Einstein lucky when not considering twin paradox as paradox?

  • #51
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It's possible that the confusion in that paper is an accurate reflection of the confusion of physicists (including Einstein himself) in the early days of relativity. But just because people were confused about it in the past doesn't mean that we need to confuse ourselves in the same way.

The OP was asking about the perception of the twin paradox in the early days (starting with Einstein's paper). If discussing this issue still causes confusion today, it shows that probably not everything is as clear-cut here as it is sometimes portrayed. Only a continued discussion of confusing issues can lastly lead to full clarification, not their suppression.
 
  • #52
stevendaryl
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The OP was asking about the perception of the twin paradox in the early days (starting with Einstein's paper). If discussing this issue still causes confusion today, it shows that probably not everything is as clear-cut here as it is sometimes portrayed. Only a continued discussion of confusing issues can lastly lead to full clarification, not their suppression.

I'm not talking about suppression. I'm talking about intentionally introducing misconceptions, and then trying to clear them up. I don't see that that's helpful.
 
  • #53
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In the English translation On the Electrodynamics of Moving Bodies at the top of page 11, the final paragraph of §4.

"Thence we conclude that a balance-clock at the equator must go more slowly, by a very
small amount, than a precisely similar clock situated at one of the poles under otherwise identical conditions."

(I assume you realise that circular motion, around the equator in this case, is non-inertial motion.)

You possibly misunderstood this paragraph. Einstein's argument is as follows here:

1) a circle can be approximated by a polygon (by making the sides infinitesimally small)
2) a polygon is a piece-wise linear curve, so we have a piece-wise constant velocity vector that only changes in direction (and thus leaves the time dilation factor constant)
3) so for a circle of circumference C, the time dilation is the same as for a straight line of length C (assuming the speed v is the same).

So rather on the contrary, for a circular orbit there is a time dilation as it can be approximated by a (piece-wise) inertial motion (for which Einstein knows how to derive the time dilation).
 
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  • #54
stevendaryl
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if the reference frame is linearly accelerating then one wouldn't need to do anything with newton's laws and they would take their simplest form anyway. objects that move under the influence of the same force as the Reference frame would appear to stand still or move uniformly and objects that would appear to accelerate would be the once that move with different accelerations than the reference frame. anyways, we would have just a shift in perception of what's accelerating and what's not and not a change in laws.

I don't think that's correct. If you do Newtonian physics using an accelerated coordinate system, then Newton's laws don't hold.

In an inertial coordinate system, you have:

[itex]m \frac{d^2 x^j}{dt^2} = F^j[/itex] (F = ma)

Now, switch to a new coordinate system [itex]x'^j = x^j + A^j t^2[/itex]

In this new coordinate system, you have:

[itex]m \frac{d^2 x'^j}{dt^2} - m A^j = F^j[/itex]

This does not have the same form. You could try to restore it to the same form by moving the constant acceleration [itex]m A^j[/itex] to the other side:

[itex]m \frac{d^2 x'^j}{dt^2} = F'^j = m A^j + F^j[/itex]

so you have a new "ficititious force" [itex]m A^j[/itex]. But this new force DOESN'T obey Newton's laws. In particular, it doesn't obey the third law, "For every action, there is an equal and opposite reaction". If that's interpreted to mean that whenever there is a force on one object, that object exerts an equal and opposite force, then that's false for fictitious forces. The mass [itex]m[/itex] has a force [itex]mA^j[/itex] exerted on it, but it doesn't exert an equal and opposite force on anything. Momentum is not conserved in this new coordinate system.
 
  • #55
stevendaryl
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You possibly misunderstood this paragraph. Einstein's argument is as follows here:

1) a circle can be approximated by a polygon (by making the sides infinitesimally small)
2) a polygon is a piece-wise linear curve, so we have a piece-wise constant velocity vector that only changes in direction (and thus leaves the time dilation factor constant)
3) so for a circle of circumference C, the time dilation is the same as for a straight line of length C.

So rather on the contrary, for a circular orbit there is a time dilation as it can be approximated by a (piece-wise) inertial motion (for which Einstein knows how to derive the time dilation).

What you're saying is always true. You can always approximate the time dilation for any noninertial motion by breaking it up into small segments, and approximate those segments by constant-velocity segments.

That prescription gives rise to the following formula for computing proper time for noninertial motion:

The proper time for taking a path [itex]x(t)[/itex] from time [itex]t=t_1[/itex] to [itex]t=t_2[/itex] is

[itex]\int_{t_1}^{t_2} \sqrt{1-\frac{(\frac{dx}{dt})^2}{c^2}} dt[/itex]
 
  • #56
Dale
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Inertial motion is what Special Relativity is based on
This is not quite correct. Special relativity is based on inertial frames, not inertial motion. There is an important difference between the two. Even in the first paper, Einstein's "on the electrodynamics of moving bodies", it was clear how to correctly treat non inertial motion (see section 4). To the OP's question it wasn't luck and it was not unaddressed by Einstein.

EDIT: I see stevendaryl made the same point first!
 
  • #57
Dale
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I already added this in my last example as you can see, but this doesn't help.
I didn't see the "last example" you are referring to, but it does completely resolve the issue. The accelerating twin is the one which interacts with the external field or the other particle.
 
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  • #58
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What you're saying is always true. You can always approximate the time dilation for any noninertial motion by breaking it up into small segments, and approximate those segments by constant-velocity segments.

Still, it is only a geometrical argument. Each of the twins could describe the motion of the other this way. So how do we single out one of the twins on this basis?
 
  • #59
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I don't think that's correct. If you do Newtonian physics using an accelerated coordinate system, then Newton's laws don't hold.

In an inertial coordinate system, you have:

[itex]m \frac{d^2 x^j}{dt^2} = F^j[/itex] (F = ma)

Now, switch to a new coordinate system [itex]x'^j = x^j + A^j t^2[/itex]

In this new coordinate system, you have:

[itex]m \frac{d^2 x'^j}{dt^2} - m A^j = F^j[/itex]

This does not have the same form. You could try to restore it to the same form by moving the constant acceleration [itex]m A^j[/itex] to the other side:

[itex]m \frac{d^2 x'^j}{dt^2} = F'^j = m A^j + F^j[/itex]

so you have a new "ficititious force" [itex]m A^j[/itex]. But this new force DOESN'T obey Newton's laws. In particular, it doesn't obey the third law, "For every action, there is an equal and opposite reaction". If that's interpreted to mean that whenever there is a force on one object, that object exerts an equal and opposite force, then that's false for fictitious forces. The mass [itex]m[/itex] has a force [itex]mA^j[/itex] exerted on it, but it doesn't exert an equal and opposite force on anything. Momentum is not conserved in this new coordinate system.
as I mentioned the new coordinate system would be just shifted compared to the old one, your fictious force ##mA^j## would only be present from the old coordinate systems point of view.
but if you fix your new reference frame is accelerating compared to the old one then from the new RF point of view (witch you suppose to be stationary) the old RF is accelerating and you would have fictious forces if you look at the old one from the new RF.
But the laws you can state the same as in the old one.

while when the RF is rotating you would clearly have laws involving rotational forces right away, and if you step out of it you would have newton's ordinary laws. you would not be able to compensate and have only linear acceleration in the rotating RF-laws.

so clearly there is a sense of linear symmetry in our universe but not symmetry between rotational and linear reference frames, this is a consequence of the way the universe is setup and how things are moving.
 
  • #60
Dale
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I'm saying that the case of a noninertial clock is a deduction from Einstein's paper.
Furthermore, it is a deduction which he explicitly makes.
 
  • #61
Dale
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but how does one make sure that a force is or is not fictional?
One uses an accelerometer.

if the reference frame is linearly accelerating then one wouldn't need to do anything with newton's laws and they would take their simplest form
This is not correct. In a linearly accelerating reference frame there is a fictitious force. Accelerometers at rest in linearly accelerating reference frames register the acceleration.

both of the twins must make their measurements from their point of view and since in both cases the laws of physics take their simplest form, both of them are correct in assuming that their reference frame is inertial and that the other one is accelerating.
This is simply false. An accelerometer aboard the traveling twin's ship registers a sharp spike halfway through the journey. He knows that his rest frame is not inertial and there are no "laws of physics take their simplest form" which would explain that accelerometer reading.
 
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  • #62
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But how does one make sure that a force is or is not fictional? one has to find the source of the force or one has to find other reference frames in which those forces disappear

Yes, that is pretty much it. Of course finding such a reference frame is a purely mathematical exercise - I just need to find a coordinate transformation that makes the coordinate accelerations go away.

If we observe a force that accelerates all objects equally regardless of their mass (as does centrifugal and coriolis force) that's a fairly solid hint that we're dealing with a fictional force and that such a coordinate transformation exists. Indeed, the only non-fictional force with that property is Newtonian gravity - and GR eliminates that special case by providing a mathematical framework in which it is also a fictional force.
 
  • #63
stevendaryl
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as I mentioned the new coordinate system would be just shifted compared to the old one, your fictious force ##mA^j## would only be present from the old coordinate systems point of view.

I know that's what you said, but it's not true. That term is in the equations, either as a "force" term (on the F side of F=ma) or as an acceleration term (on the ma side of F=ma). Neither choice leaves Newton's laws unchanged. Either you have to modify the notion of "acceleration" to include terms due to noninertial, non-Cartesian coordinates, or you modify the third law, and allow for forces without a "reaction" counterpart.

but if you fix your new reference frame is accelerating compared to the old one then from the new RF point of view (witch you suppose to be stationary) the old RF is accelerating and you would have fictious forces if you look at the old one from the new RF.

I'm sorry, but that's just not true. In terms of [itex]x^j[/itex], you have:

[itex]m \frac{d^2 x^j}{dt^2} = F^j[/itex] (no fictitious forces)

In terms of [itex]x'^j[/itex], you have:
[itex]m \frac{d^2 x^j}{dt^2} = F^j + m A^j[/itex] (fictitious forces are present)

Some frames have fictitious forces, and some frames do not. The ones that do not are the inertial frames.
 
  • #64
Dale
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as I mentioned the new coordinate system would be just shifted compared to the old one, your fictious force ##mA^j## would only be present from the old coordinate systems point of view.
No, this is not true. The fictitious force is intrinsic to the coordinate system itself and not required to be in reference to any other coordinate system's point of view. That fictitious force must be included to match the observations of motion within that frame alone.

but if you fix your new reference frame is accelerating compared to the old one then from the new RF point of view (witch you suppose to be stationary) the old RF is accelerating and you would have fictious forces if you look at the old one from the new RF.
This is also not true. The lack of fictitious force in the inertial reference frame is inherent to the frame itself and does not require any reference to any other frame. The fictitious force must be absent to match the observations of motion within that frame alone. If you include it you would get the wrong motions.

EDIT: or in other words: "What stevendaryl said".
 
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  • #65
stevendaryl
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Still, it is only a geometrical argument. Each of the twins could describe the motion of the other this way. So how do we single out one of the twins on this basis?

You don't. You approximate each twin's path by a bunch of little constant-velocity paths, and you compute the proper time for each path using the formula: [itex]\delta \tau = \sqrt{\delta t^2 - \frac{1}{c^2} \delta x^2}[/itex] You add up [itex]\delta \tau[/itex] for each segment, and that gives you how much each twin ages along his path. There is no singling out of one twin over the other.

There is a choice that must be made, which is to pick an inertial coordinate system for measuring [itex]\delta x[/itex] and [itex]\delta t[/itex]. But every inertial coordinate system will give the same value for [itex]\delta \tau[/itex]. It's exactly like computing the length of a line segment in Euclidean geometry. The length is given by: [itex]\delta L= \sqrt{\delta x^2 + \delta y^2}[/itex]. You can choose any Cartesian coordinate system you like to measure [itex]\delta x[/itex] and [itex]\delta y[/itex], and you will get the same answer for [itex]\delta L[/itex].
 
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  • #66
PeterDonis
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if you have only 2 point particles an nothing else, how do you determent witch one is accelerating?

Easy: attach an accelerometer to each particle. The one whose accelerometer reads nonzero for some portion of the trip is the one who accelerated.

DaleSpam and stevendaryl have correctly pointed out that, strictly speaking, for one of the particles to accelerate in the above sense (i.e., for its accelerometer to read nonzero at some point), there must be other "stuff" present in the scenario. The "point particle" whose accelerometer reads nonzero has to exchange momentum with something (for example, a rocket exhaust). But you don't have to know any of the details of how that happens to know which particle accelerated from the accelerometer readings.
 
  • #67
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It may be merely a student exercise today, but exactly what we are discussing here was a serious issue for Einstein already before he published his GR, which was discussed by leading scientists at the time. I quote from the Wikipedia article
Starting with Paul Langevin in 1911, there have been various explanations of this paradox. [..]
I gave you the link to his paper; you can check for yourself that, contrary to Wikipedia's suggestion*, it was not considered to be an existing paradox. Instead it was an original, non-paradoxical example of predictions based on what Einstein later named "special relativity".

*About Wikipedia, I'm not sure that the person who wrote that intended to make your claim, it may be just poor phrasing. And you can use the back-in-time feature of Wikipedia to find different opinions. ;)
 
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  • #68
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[..] In retrospect, General Relativity was not needed to describe things from the point of view of an accelerated coordinate system. That description is derivable from SR alone. And that description has terms that are "gravity-like", but all within SR. GR is only needed if you want to describe real gravity, due to the presence of massive objects.
"In retrospect"? Why do you think that this was not understood at that time? It had been straightforward in classical mechanics to describe things from the point of view of an accelerated coordinate system, so it seems unlikely that this wasn't understood from the start in SR. (sorry for going slightly off topic, but it fits rather well in the discussion here).
 
  • #69
stevendaryl
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"In retrospect"? Why do you think that this was not understood at that time? It had been straightforward in classical mechanics to describe things from the point of view of an accelerated coordinate system, so it seems unlikely that this wasn't understood from the start in SR. (sorry for going slightly off topic, but it fits rather well in the discussion here).

You're certainly right, that noninertial frames came up in Newtonian mechanics. So where did the idea that GR was necessary to handle an accelerated reference frame come from?

I think that part of it is the insistence on relativity. Although Newtonian mechanics also satisfied a principle of (Galilean) relativity, I don't think that it played that much role in the teaching and application of the subject. Nobody bothered (as far as I know) to try to write Newtonian mechanics in a way that treated all coordinate systems equally. I don't think that the latter was developed until after GR (the Newton-Cartan formulation of Newtonian physics).
 
  • #70
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Easy: attach an accelerometer to each particle. The one whose accelerometer reads nonzero for some portion of the trip is the one who accelerated.

DaleSpam and stevendaryl have correctly pointed out that, strictly speaking, for one of the particles to accelerate in the above sense (i.e., for its accelerometer to read nonzero at some point), there must be other "stuff" present in the scenario. The "point particle" whose accelerometer reads nonzero has to exchange momentum with something (for example, a rocket exhaust). But you don't have to know any of the details of how that happens to know which particle accelerated from the accelerometer readings.
you would have to calibrate the accelerometers differently in order for both of them to be zero if they accelerate differently, because if you assume that your accelerating RF is actually stationary then you calibrate your accelerometer to read zero in your reference frame, don't you?

and there is another thing, if you change acceleration in a linear fashion then your laws are not the simplest anymore and maybe its the way to think about the twin paradox.
 
  • #71
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also an accelerometer doesn't measure acceleration in free fall only when you stand on the earths suface does it do so, you can see that from the link to Wikipedia http://en.wikipedia.org/wiki/Accelerometer
 
  • #72
Dale
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you would have to calibrate the accelerometers differently in order for both of them to be zero if they accelerate differently, because if you assume that your accelerating RF is actually stationary then you calibrate your accelerometer to read zero in your reference frame, don't you?
Although calibration is necessary for real devices, for the purpose of thought experiments it is generally a detail which is glossed over. We simply assume ideal measuring devices such as rods, clocks, and accelerometers.

In principle, it is not a bad assumption. If you miscalibrate the accelerometer then you will detect violations of the conservation of momentum which are not accounted for. You will not be able to get experiments to match the known laws of physics. So a miscalibrated accelerometer will be something which is experimentally detectable in the end.

also an accelerometer doesn't measure acceleration in free fall only when you stand on the earths suface does it do so, you can see that from the link to Wikipedia http://en.wikipedia.org/wiki/Accelerometer
Yes. This is why in relativity free-fall frames are inertial and frame attached to the earth's surface is non-inertial.
 
  • #73
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Although calibration is necessary for real devices, for the purpose of thought experiments it is generally a detail which is glossed over. We simply assume ideal measuring devices. In principle, it is not a bad assumption. If you miscalibrate the accelerometer then you will detect violations of the conservation of momentum which are not accounted for. You will not be able to get experiments to match the known laws of physics. So a miscalibrated accelerometer will be something which is experimentally detectable in the end.
what I mean is that if you for example take the equivalence principle, take an elevator witch you can think of as the rocket of the traveling twin.

if the elevator is accelerating with uniform acceleration it acts just like if it were in free fall.
so even in an accelerated RF (elevator) one would not measure acceleration even with a "accelerometer", the thing is that the rocket of the traveling twin travels with different accelerations not only with different speeds, so that that might break the symmetry somehow.
im not yet sure how but I think its the right way to go.
 
  • #74
stevendaryl
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what I mean is that if you for example take the equivalence principle, take an elevator witch you can think of as the rocket of the traveling twin.

if the elevator is accelerating with uniform acceleration it acts just like if it were in free fall.

That's not true. If you drop a ball inside an elevator that is accelerating upward with uniform acceleration, then the ball will drop to the floor. If the elevator is accelerating downward, then the ball will rise to the ceiling. But if the elevator is in freefall, then the ball will just drift where you dropped it.

An elevator in freefall is completely different from an elevator with uniform acceleration.

The equivalences are: An elevator in freefall near a planet is equivalent to an inertial elevator in empty space. An elevator at rest on a planet is equivalent to an elevator accelerating upward in empty space.

An accelerometer cannot distinguish between freefall and inertial motion. So, from the point of view of General Relativity, they are both inertial. An accelerometer cannot distinguish between an accelerating elevator and an elevator at rest in a gravitational field, so from the point of view of GR, they are both noninertial.

[QUOTEso even in an accelerated RF (elevator) one would not measure acceleration even with a "accelerometer"[/QUOTE]

That's not true.

As DaleSpam said, freefall is considered inertial motion from the point of view of General Relativity, precisely because an accelerometer would show no acceleration.
 
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  • #75
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As DaleSpam said, freefall is considered inertial motion from the point of view of General Relativity, precisely because an accelerometer would show no acceleration.
I forgot to mention:
if the elevator (rocket) accelerates under the influence of an gravitational field (at the turnaround) then what would we have ????

an inertial frame that accelerates ??

even from the point of view of special relativity, this has to be a IRF since the laws of stay unchanged in their simplest form.
 

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