# Muons experiment, strange logic

by Tantalos
Tags: experiment, logic, muons, strange
P: 46
 Quote by ghwellsjr Like I said, this is true if the twins always maintain a single velocity. But if you are going to use the "v" in the Lorentz transform to apply to the difference in speed between the twins, even if they are both "moving" (whatever that means), then don't you mean that in the two reference frames, one of them is stationary and the other is moving? Of course you don't have to do this but if you don't, you will have to be more precise in what you mean because there are an infinite number of ways to interpret what you are talking about.
I mean that if there are two reference frames A and B in relative movement to each other, then A is thinking that he is stationary and B is moving, but B is thinking the opposite, that he stationary and A is moving, since there is no preferred stationary reference frame.
So each one of them can make the age calculation of the other. A will come to the result that B will be younger than him, but B will come to a result that A should be younger than him. That's a contradiction.

 Quote by ghwellsjr Only if we don't put them in the same place. Just like we don't have to presume that an observer/object/clock is stationary in a Frame of Reference, we don't have to put any observer/object/clock at the origins and we certainly don't have to put both observers/objects/clocks at the same origin.
I agree, we don't need them to be in one place. The condition however is that the coordinate of their reference frames at time t = t' = 0 is x = x' = 0. The observers need not be at the origins and as long as v = const we can use the Lorentz transformation to calculate forward and back the space-time coordinates of some events at which they register the events. The events can happen anywhere at any time in any reference frame.
Knowing the space-time coordinates in one reference frame we can calculate the coordinates that registered an observer in another reference frame.

 Quote by ghwellsjr So if we make the origins of the two Frames of Reference be the turn-around event for the traveling twin, then we won't introduce any spurious offsets in the calculation of his aging during his trip.
Yes, but the travelling twin still needs to change his direction of movement in order to return to the other twin. Imagine what will see the stationary twin when he looks at clock of the travelling twin. When the twin moves with speed v then of course the stationary twin will see a lag between his clock and the moving clock. Now use the Lorentz formula for time coordinate to check what will happen when the travelling twin suddenly stops. The lag suddenly vanishes, the clocks will show the same time. It's because of violation of the v = const condition. Changing direction of movement is also a violation of this condition.

 Quote by ghwellsjr But we don't even have to do that, we can continue to use the same Frame of Reference in which the "traveling" twin was at rest during the outbound portion of his trip and during which the other twin was actually traveling away from him at some constant speed "v" throughout the entire scenario. Then at the turn-around event, the "traveling" twin has to go from rest to a speed greater than "-v" in order to catch up to the other twin. It doesn't matter which Frame of Reference we use to analyze the entire scenario of the Twin Paradox. The easiest one to use is the one they both start out at rest and in which only the traveling twin has a non-zero speed because the calculation is trivially simple but if you like to torture yourself, you can pick a different Frame of Reference and endure the more complicated calculations but you will end up with the same age difference in the two twins between the time they separate and the time they reunite.
This means that the travelling twin jumps from his reference frame to another that is moving in opposite direction. The catch is once again the condition t = t' = 0 at x = x' = 0 which is not met in this case.

 Quote by ghwellsjr It cannot according to Einstein's formulation of a Frame of Reference in his 1905 paper and he doesn't make that assumption. Where did you get that idea? In fact, if you look at the end of section 4 of his paper, you will see that he introduces the Twin Paradox and its solution to the world for the first time, although he doesn't call it that and he doesn't use twins, he uses a pair of clocks. And he assigns a single Frame of Reference to both clocks which he calls K, the stationary system.
This is mentioned in the last 4 lines at the end of page 4. Einstein here assumes a movement along a curved line, possibly a closed one, which of course require to change direction during that movement.
PF Gold
P: 4,784
 Quote by Tantalos I mean that if there are two reference frames A and B in relative movement to each other, then A is thinking that he is stationary and B is moving, but B is thinking the opposite, that he stationary and A is moving, since there is no preferred stationary reference frame. So each one of them can make the age calculation of the other. A will come to the result that B will be younger than him, but B will come to a result that A should be younger than him. That's a contradiction.
It would be a contradiction if you were claiming that two different Frames of Reference concluded that each twin was younger than the other at the end of the scenario, that is, after they reunited, but as long as they both remain at a constant speed with respect to each other, then it's no different than two different Frames of Reference concluding that one twin is at rest and only the other twin is moving. Do you see that as contradictory?
 Quote by Tantalos I agree, we don't need them to be in one place. The condition however is that the coordinate of their reference frames at time t = t' = 0 is x = x' = 0. The observers need not be at the origins and as long as v = const we can use the Lorentz transformation to calculate forward and back the space-time coordinates of some events at which they register the events. The events can happen anywhere at any time in any reference frame. Knowing the space-time coordinates in one reference frame we can calculate the coordinates that registered an observer in another reference frame.
Agreed, as long as by "observers" you mean the "twins".
 Quote by Tantalos Yes, but the travelling twin still needs to change his direction of movement in order to return to the other twin. Imagine what will see the stationary twin when he looks at clock of the travelling twin. When the twin moves with speed v then of course the stationary twin will see a lag between his clock and the moving clock. Now use the Lorentz formula for time coordinate to check what will happen when the travelling twin suddenly stops. The lag suddenly vanishes, the clocks will show the same time. It's because of violation of the v = const condition. Changing direction of movement is also a violation of this condition.
The lag that you are talking about does not disappear when the traveling twin turns around. The stationary twin will continue to see the traveling twin's clock running slow for most of the scenario until that lag gets used up and then he will see the clock running fast. However, the traveling twin sees this change in the stationary twin's clock at the moment of turn around and it is this difference that confirmes the same conclusion that an analysis based on a Frame of Reference comes to.
 Quote by Tantalos This means that the travelling twin jumps from his reference frame to another that is moving in opposite direction. The catch is once again the condition t = t' = 0 at x = x' = 0 which is not met in this case.
I agree you shouldn't do an analysis in which a twin jumps frames--so don't do that. Use just one frame and there will be no frame jumping, correct?
 Quote by Tantalos This is mentioned in the last 4 lines at the end of page 4. Einstein here assumes a movement along a curved line, possibly a closed one, which of course require to change direction during that movement.
Yes, of course. But you say this as if there is something wrong.
 P: 249 The egg came first if you buy into Darwinism, the step of evolution of an offspring would have come first, since after it hatched then you would say, "Now that is a chicken." Because, what laid the egg would not have been a chicken since it didn't evolve to a chicken yet.
PF Gold
P: 4,784
 Quote by John232 The egg came first if you buy into Darwinism, the step of evolution of an offspring would have come first, since after it hatched then you would say, "Now that is a chicken." Because, what laid the egg would not have been a chicken since it didn't evolve to a chicken yet.
P: 46
 It would be a contradiction if you were claiming that two different Frames of Reference concluded that each twin was younger than the other at the end of the scenario, that is, after they reunited, but as long as they both remain at a constant speed with respect to each other, then it's no different than two different Frames of Reference concluding that one twin is at rest and only the other twin is moving. Do you see that as contradictory?
No as long the conditions that the inventor of the Lorentz equations put on his equations are met. See also here http://www.physicsforums.com/showpos...3&postcount=18

 The lag that you are talking about does not disappear when the traveling twin turns around. The stationary twin will continue to see the traveling twin's clock running slow for most of the scenario until that lag gets used up and then he will see the clock running fast. However, the traveling twin sees this change in the stationary twin's clock at the moment of turn around and it is this difference that confirmes the same conclusion that an analysis based on a Frame of Reference comes to.
Of course the lag cannot suddenly disappear immediately when the travelling twin stops, but the Lorentz equation shows that it should disappear. Do you think Lorentz equations are wrong?

 I agree you shouldn't do an analysis in which a twin jumps frames--so don't do that. Use just one frame and there will be no frame jumping, correct?
But the jumping into a reference frame moving in opposite direction is the same scenario as turning around in one reference frame, thus they should give the same results.
Mentor
P: 17,530
 Quote by Tantalos But the jumping into a reference frame
I really dislike this terminology. It is sloppy language and almost always coincides with sloppy thinking and incorrect conclusions. I agree with ghwellsjr in his suggestion and recommendation. I also would point out that if you ever find yourself or someone else using the term "jumping into" or "jumping out of" a reference frame you can be very confident that a mistake has been made or is about to be made.

You cannot "jump into" a reference frame. An inertial reference frame has infinite spatial and temporal extent. You may be at rest in it or you may be moving in it. If you are non-inertial then over time whether or not you are at rest in a given inertial reference frame can change.

In one inertial frame the home twin is at rest, in this frame the travelling twin is never at rest. This frame concludes that the travelling twin is younger at the reunion.

In another inertial frame the travelling twin is initially at rest, in this frame the home twin is never at rest and the travelling twin is not at rest in the final leg. This frame concludes that the travelling twin is younger at the reunion.

In a third inertial frame the travelling twin is at rest on the final leg, in this frame the home twin is never at rest and the travelling twin is not initially at rest. This frame concludes that the travelling twin is younger at the reunion.
P: 46
 Quote by DaleSpam I really dislike this terminology. It is sloppy language and almost always coincides with sloppy thinking and incorrect conclusions. I agree with ghwellsjr in his suggestion and recommendation. I also would point out that if you ever find yourself or someone else using the term "jumping into" or "jumping out of" a reference frame you can be very confident that a mistake has been made or is about to be made.
Which conclusions are sloppy and incorrect (and why)?

 Quote by DaleSpam You cannot "jump into" a reference frame. An inertial reference frame has infinite spatial and temporal extent. You may be at rest in it or you may be moving in it. If you are non-inertial then over time whether or not you are at rest in a given inertial reference frame can change.
Ok, then don't jump, but you can change your speed and/or direction in your original reference frame to a speed that just happens to be the same as of an another third reference frame.

 Quote by DaleSpam In one inertial frame the home twin is at rest, in this frame the travelling twin is never at rest. This frame concludes that the travelling twin is younger at the reunion. In another inertial frame the travelling twin is initially at rest, in this frame the home twin is never at rest and the travelling twin is not at rest in the final leg. This frame concludes that the travelling twin is younger at the reunion. In a third inertial frame the travelling twin is at rest on the final leg, in this frame the home twin is never at rest and the travelling twin is not initially at rest. This frame concludes that the travelling twin is younger at the reunion.
When the travelling twin turns does he take the clock with him or he leaves it in his reference frame which is still moving forward?

 Related Discussions High Energy, Nuclear, Particle Physics 7 General Physics 3 General Physics 2 High Energy, Nuclear, Particle Physics 3 General Physics 6