Problem Of Length Contraction At Cern Due To Spe.relativity

Discussions on classical and modern physics, quantum mechanics, particle physics, thermodynamics, general and special relativity, etc.

Problem Of Length Contraction At Cern Due To Spe.relativity

Postby mahesh on January 30th, 2018, 6:21 am 

1) When I was in school & teacher was teaching Spe relativity to us. I had asked one question to the teacher. If I fixed four pulleys A, B, C & D in such a way that they would create rectangle & one thread is mounted on that pulleys & that thread is revolve on the pulleys with some velocity then what will happen. Length of thread would decrease due to length contraction as inter molecular distance get contracted (as per SR) but centers’ of four pulleys would remain at same point. How would this possible? My teacher said that velocity of thread is not much. So, this problem would not be created in actual life.
Now, same problem can be asked for CERN.

2) In CERN, 2808 bunches of protons are rotated in circular path, initial velocity of rotation is V1 = 0.9993 C at 25 GeV & final velocity of rotation V2 = 0.99999999 C at 6.5 TeV..
So, value of Y changes from Y1={(1-V1^2/C^2)}^-0.5 = 26.7308 to Y2 = 7071.068. This is very big change in Y. So, distance between any two bunches of protons will get decrease by 264.5288 times when velocity of protons increases in accelerator.
This will decrease the total circular path of proton from 27000 m (27 km) to just 102.0683 m (0.102 km).
& radius from 4299.363m to just 16.253 m.
How is this possible as accelerating magnets are at same position for rest observer?
mahesh
Member
 
Posts: 61
Joined: 21 Mar 2016


Re: Problem Of Length Contraction At Cern Due To Spe.relativ

Postby JMP1958 on January 30th, 2018, 1:15 pm 

The bunches in CERN are not connected to each other. They are being accelerated by the magnets which are in the lab frame and thus the distance between the bunches is maintained as being an equal distance in the lab frame. What this mean that if you were to view things from the frame of the bunches, they would see the distance between themselves and adjacent bunches as increasing.
But there is a additional factor that comes into play when it comes to the frame of the bunches, it is a rotating frame. So even when they are no longer being accelerated by the magnets, and thus maintaining a constant speed relative to the lab frame, they are still in an accelerated frame. Dealing with measurements in an accelerating frame is quite a bit more complex than dealing with them from an inertial frame. For example, as viewed from the accelerated frame, two clocks, separated along the line of acceleration will run at different rates even if there they are both at rest with respect to each other.

As a result, How each bunch views is neighbors depends upon their relative positions in the accelerator. But if you were to go to all the trouble to work it out, you would find no conflict with the Lab frame.

In a way, your question is similar to Bell's spaceship paradox. In it you have two spaceships under equal acceleration, one behind the other and attached by a string. Since they are both accelerating the same, the distance between them stays constant, but since they are gaining velocity the string should undergo length contraction and break.
The solution revolves around the term "under equal acceleration". According to which frame? The inertial frame the ships are accelerating with respect to or the ships' frame itself.
Which of these frame use apply the "equal acceleration" under determines the outcome.
If it is applied under the inertial frame, the ships maintain a constant distance in this frame and the string contracts and breaks. As seen in the ships frame, the acceleration of the two ships is not equal, the distance between them increases, and the string breaks.
If it is applied to the ship frame, then in the ship frame, the distance between the ships remains constant and the string doesn't break, while view from the inertial frame, the acceleration of the ships is not equal, both the distance between them and the string length contract, and the string doesn't break.

In either case both frame agree as to whether the string breaks or not.
User avatar
JMP1958
Member
 
Posts: 58
Joined: 02 Jul 2016
BurtJordaanBraininvat liked this post


Re: Problem Of Length Contraction At Cern Due To Spe.relativ

Postby mahesh on January 31st, 2018, 4:00 am 

Means you are accepting that due to increase in velocity distance between bundle of protons decreases from 27000/2808 =9.615 m to 9.615/264.528 =0.036m for scientist of CERN but number of bundles in circular path is fixed i.e. 2808.
Means, circumference path for 2808 bundle to rotate in CERN will decreases to 0.036 x 2808 =101.088 m (from 27000m)
Now, circular path will decrease but magnet of CERN in circular path remain at same position.
How is this possible?
Scientist in CERN will see the circular path of proton bundles decreases (because distance between two bundles decreases) but path creating magnets remain at same position.
mahesh
Member
 
Posts: 61
Joined: 21 Mar 2016


Re: Problem Of Length Contraction At Cern Due To Spe.relativ

Postby JMP1958 on January 31st, 2018, 2:06 pm 

mahesh » January 31st, 2018, 1:00 am wrote:Means you are accepting that due to increase in velocity distance between bundle of protons decreases from 27000/2808 =9.615 m to 9.615/264.528 =0.036m for scientist of CERN but number of bundles in circular path is fixed i.e. 2808.
Means, circumference path for 2808 bundle to rotate in CERN will decreases to 0.036 x 2808 =101.088 m (from 27000m)
Now, circular path will decrease but magnet of CERN in circular path remain at same position.
How is this possible?
Scientist in CERN will see the circular path of proton bundles decreases (because distance between two bundles decreases) but path creating magnets remain at same position.


No. The distance between the photon bunches does not decrease in the Lab frame, because they are being accelerate by the magnets in the Lab frame, which also maintains their separation as measured in the Lab frame. This is the equivalent of the Bell ship scenario where the Ships maintain a constant acceleration relative to the inertial frame and the string breaks.
However with CERN there is nothing equivalent to the string between the bunches trying to hold them equally spaced in their frame. So all that happens is that in the Lab frame, the distance between bunches remains constant, but in the frame of the bunches themselves, it does not (keeping in mind the additional complications caused by the bunches being in a rotating frame).
User avatar
JMP1958
Member
 
Posts: 58
Joined: 02 Jul 2016


Re: Problem Of Length Contraction At Cern Due To Spe.relativ

Postby mahesh on February 1st, 2018, 3:29 am 

I just consider frame with two moving adjacent protons
Means, you say that distance between two bundles of protons remain 9.615m at velocity V1=0.9993 C & remain same 9.615m at final velocity of V2 = 0.99999999 C even when Y changes from Y1={(1-V1^2/C^2)}^-0.5 = 26.7308 to Y2 = 7071.068.
Where complete moving co-ordinate system get contracted by 264.5288 times for co-ordination system at rest in velocity direction.
Means, co-ordinate points get contracted but particles bundles will remain at same distance. How is it possible?
mahesh
Member
 
Posts: 61
Joined: 21 Mar 2016


Re: Problem Of Length Contraction At Cern Due To Spe.relativ

Postby BurtJordaan on February 1st, 2018, 9:21 am 

mahesh » 01 Feb 2018, 09:29 wrote:Means, co-ordinate points get contracted but particles bundles will remain at same distance. How is it possible?

There are 3 different inertial frames in your simplified problem: the inertial frame of the two magnets, and each proton in its own "instantaneous" inertial frame. Why? Because you are talking about an accelerator. If the centrifugal forces are ignored, as JMP said, you are looking at a Rindler accelerated coordinate system, where the definitions of simultaneity are different for each of them. The simple Lorentz transformation between inertial frames that you are using then don't work. See e.g. https://en.wikipedia.org/wiki/Rindler_coordinates.

It boils down to the distance between the protons remains constant in the magnet frame, but increases in the frames of each of your two protons. That's why JMP reminded you about the 'breaking string' between two rockets that suffer the same acceleration in the lab frame.

Bring in the centrifugal forces and the the problem gets much, much more complex. See e.g. http://math.ucr.edu/home/baez/physics/Relativity/SR/rotatingCoordinates.html
User avatar
BurtJordaan
Forum Moderator
 
Posts: 2595
Joined: 17 Oct 2009
Location: South Africa
Blog: View Blog (9)


Re: Straight & Narrow Path

Postby Faradave on February 1st, 2018, 12:39 pm 

There are of course, linear accelerators, which may limit the acceleration zone to a small portion of the path. For the rest of the trip, until hitting the wall at the far end, the particles see a contracted path, while the observers at rest with the lab see it longer. Such is Special Relativity. Spatial separation is a matter of opinion, the interval traversed is not.
User avatar
Faradave
Active Member
 
Posts: 1667
Joined: 10 Oct 2012
Location: Times Square (T2)


Re: Straight & Narrow Path

Postby BurtJordaan on February 1st, 2018, 3:07 pm 

Faradave » 01 Feb 2018, 18:39 wrote:There are of course, linear accelerators, which may limit the acceleration zone to a small portion of the path.


The only linear accelerators of note (that I know of) accelerate particles for the whole length of the linear tunnel and then collide them with something or other. I suppose this is the whole point of an accelerator - getting the kinetic energy as high as possible at the other end. Sometimes they are then fed into a storage ring where there is only centrifugal acceleration, or they are fed into a circular accelerator, like CERN.
User avatar
BurtJordaan
Forum Moderator
 
Posts: 2595
Joined: 17 Oct 2009
Location: South Africa
Blog: View Blog (9)


Re: High on Energy

Postby Faradave on February 1st, 2018, 3:32 pm 

Agreed. For high energy physics, the goal is usually to accelerate as much as possible prior to collision. For therapeutic applications, sometimes particle beams of particular (limited) energies are required.

"The two major types of radiation beams are photons and electrons. Both are produced by the linear accelerator. In general, photons (X-rays) travel completely through a body or tissue. Electrons penetrate to only a defined depth."

"Proton-beam therapy is a form of cancer radiotherapy that uses positively charged particles to kill tumor cells. Unlike traditional radiotherapy using X-rays, protons can be programmed to deposit most of their energy in the tumor, minimizing damage to surrounding, healthy tissue."

In any case, there is nothing to stop them from producing an unaccelerating beam at relativistic speed for the sake of simplifying the problem.

Spatial separation is a matter of opinion (observer perspective). Interval separation is not.
User avatar
Faradave
Active Member
 
Posts: 1667
Joined: 10 Oct 2012
Location: Times Square (T2)


Re: Problem Of Length Contraction At Cern Due To Spe.relativ

Postby BurtJordaan on February 2nd, 2018, 12:18 am 

OK, but the OP's confusion seems to have stemmed from his misuse of the standard Lorentz transformations for multiple particles being accelerated linearly or circularly over some distance. In this case you have to be careful using the spacetime interval, because it may be different for different observers - it depends on the path that a particle takes between two events.
User avatar
BurtJordaan
Forum Moderator
 
Posts: 2595
Joined: 17 Oct 2009
Location: South Africa
Blog: View Blog (9)


Re: Problem Of Length Contraction At Cern Due To Spe.relativ

Postby mahesh on February 2nd, 2018, 4:39 am 

1) You are true that the path of two protons bundle is circular but that circle is huge circle of length 27 km & 9m or 0.036 m distances are very less. So locally we can consider this acceleration as linear acceleration. (Nothing is completely linear in the world)
2)You say that
"the distance between the protons remains constant in the magnet frame, but increases in the frames of each of your two protons."
How does length in prime frame increases ? I think, this is not possible.
3)For multiple frame, let one observer is platform & observing multiple moving train on different track then also he will see that all every train get contracted due to its different velocity. So, multiple frame is not the problem.
mahesh
Member
 
Posts: 61
Joined: 21 Mar 2016


Re: Problem Of Length Contraction At Cern Due To Spe.relativ

Postby BurtJordaan on February 2nd, 2018, 7:46 am 

mahesh » 02 Feb 2018, 10:39 wrote:2) How does length in prime frame increases ? I think, this is not possible.

In linear acceleration, where all protons along the tunnel suffer equal proper acceleration, each proton 'observer' sees the others as having increasing proper distances from itself. The accelerator frame observes the protons as being at constant distance from each other.
Look at https://en.wikipedia.org/wiki/Bell%27s_spaceship_paradox, where one can clearly see these effects.

3)For multiple frame, let one observer is platform & observing multiple moving train on different track then also he will see that all every train get contracted due to its different velocity. So, multiple frame is not the problem.

This is not the same as the accelerator scenario that you asked about, so comparing them will surely lead you to false conclusions. The answer is completely given in the link above.
User avatar
BurtJordaan
Forum Moderator
 
Posts: 2595
Joined: 17 Oct 2009
Location: South Africa
Blog: View Blog (9)
mahesh liked this post


Re: Problem Of Length Contraction At Cern Due To Spe.relativ

Postby mahesh on February 5th, 2018, 3:53 am 

This is the only possible solution but this has some problem. Value of Y is so much that distance between protons will be much more than track or tunnel length (for non-contracted also) in proton frame. For example:-
Let, consider that King has palace & his office is at a distance 27000m.
His convoy contains 2808 vehicles & when they move from palace to office, they maintain equivalent distance 9.615 m.
So, when first vehicle reach the office, last vehicle start from palace.
Just consider that velocity of vehicles is V2 = 0.99999999 C .
then if king see our vehicles then nearer vehicle will be at 9.619x7071.068 = 68016.603 km.
& 27000 km track will shrink to 3.818 m.
How is this possible? This track is so small that even two vehicles will not stand on it.
1)There are other kinematics problem also arises.
mahesh
Member
 
Posts: 61
Joined: 21 Mar 2016


Re: Problem Of Length Contraction At Cern Due To Spe.relativ

Postby BurtJordaan on February 5th, 2018, 5:34 am 

mahesh » 05 Feb 2018, 09:53 wrote:This is the only possible solution but this has some problem. Value of Y is so much that distance between protons will be much more than track or tunnel length (for non-contracted also) in proton frame.

Even if they fill the whole length of the tunnel with bunches (per tunnel frame), which they actually don't, you agreed that the proton bunches stay at constant length and hence will always physically fit into the tunnel. However from any one of the many bunch frames, they won't. This is due to the simultaneity differences between tunnel and bunch frames. For a linear case, the resolution is exactly as per the old 'ladder paradox'.
User avatar
BurtJordaan
Forum Moderator
 
Posts: 2595
Joined: 17 Oct 2009
Location: South Africa
Blog: View Blog (9)
Faradave liked this post


Re: Problem Of Length Contraction At Cern Due To Spe.relativ

Postby mahesh on February 6th, 2018, 3:59 am 

Ok, means King (particle) will not see any other vehicle on track, not even nearest vehicle (particle).
mahesh
Member
 
Posts: 61
Joined: 21 Mar 2016


Re: Problem Of Length Contraction At Cern Due To Spe.relativ

Postby mahesh on February 7th, 2018, 4:42 am 

1) Means, in proton bundle frame, front proton bundle will accelerate more & back proton bundle will de-accelerate to increase the distance in prime frame & to keep same distance in non-prime frame. This only possible when forces acting on each proton will be different in the prime frame.
(also remember that in a ring acceleration front accelerating & back de-accelerating is not possible because front & back is relative. front can become back & back can become front)
2)Means, on prime frame there is only one vehicle on track & on non-prime frame 2808 on track. King will see only his vehicle & his office and palace in prime frame.
mahesh
Member
 
Posts: 61
Joined: 21 Mar 2016


Re: Problem Of Length Contraction At Cern Due To Spe.relativ

Postby BurtJordaan on February 7th, 2018, 6:28 am 

There are not one, but multiple simultaneous inertial frames, one for each bundle, in which each bundle is momentarily at rest. And yes, although the bundles suffer the same proper acceleration, each of them will observe the ones ahead as suffering more (coordinate) acceleration and ones behind less acceleration than itself.

BTW, these are tough concepts to grasp, especially for someone seemingly still struggling to get a grip on elementary SR.
User avatar
BurtJordaan
Forum Moderator
 
Posts: 2595
Joined: 17 Oct 2009
Location: South Africa
Blog: View Blog (9)
Faradave liked this post


Re: Problem Of Length Contraction At Cern Due To Spe.relativ

Postby JMP1958 on February 7th, 2018, 2:41 pm 

mahesh » February 7th, 2018, 1:42 am wrote:1) Means, in proton bundle frame, front proton bundle will accelerate more & back proton bundle will de-accelerate to increase the distance in prime frame & to keep same distance in non-prime frame. This only possible when forces acting on each proton will be different in the prime frame.
(also remember that in a ring acceleration front accelerating & back de-accelerating is not possible because front & back is relative. front can become back & back can become front)
2)Means, on prime frame there is only one vehicle on track & on non-prime frame 2808 on track. King will see only his vehicle & his office and palace in prime frame.


There actually is a relative "front" and "back" for the bunches in the ring accelerator. This is because they are accelerating. The rule for dealing with accelerated frame are not as simple as the ones that can be used with inertial ones.

Below we have a ring accelerator shown as the circle.
ringAcc.png

Shown are three of the bunches (red dots). If we consider the middle bunch, it is under acceleration along two vectors, Tangentially along the circle (grey arrow) and centripetally toward the center of the ring( blue arrow), the resultant acceleration experienced will in a direction somewhere in between (green). When you are in an accelerating frame, clocks behave differently than they do in an inertial frame. Clocks in the direction of the acceleration run fast (compared to your own clock) and the further they are from you in that direction, the faster they run. Conversely, clocks in the opposite direction run slow, and again, the further away, the slower they run. This applies toall clocks regardless of whether they share the accelerated frame with you or not. Each bunch will have its own separate dividing line.

In the image, this is shown by the dashed purple line which divides the two regions. Note that fewer of the other bunches would be in the running slow camp vs. the running fast camp. (clocks in the lab frame would have the additional time dilation due to their relative velocity in addition to this.) The point being that sine each of the other bunches has a different position relative to the the dividing line on any bunch, that bunch will see different behaviors for the other bunches and a clear distinction between bunches "in front" and bunches "behind".
As already mentioned, dealing with acceleration in SR is a much more complex issue than when working in inertial frames.

As far as the King's convoy goes, it seems that you are using the argument of incredulity. Just because you feel that it is incredulous that a large number of cars can fit between the start and ends points of the trip in one frame, but not even one car in another, does not make it untrue. As pointed out, this is a simultaneity issue. Imagine that we where to put a number of clocks strung along the road and synchronized to each other in the road frame as the front and rear of each car passes a clock, the time on it is noted.

If we switch to the frame of the cars, these clocks are no longer in sync with each other. This and time dilation combine and insure that each of these clocks still read the exact same time on them as when the front and rear of a particular car passes it as it did according to the road frame.

In essence, the rules of Relativity prevent physical contradictions between frames.
User avatar
JMP1958
Member
 
Posts: 58
Joined: 02 Jul 2016
maheshralfcis liked this post


Re: Problem Of Length Contraction At Cern Due To Spe.relativ

Postby mahesh on February 14th, 2018, 6:29 am 

Sorry, due to office work I can not give reply in time.
Ok, now, we do your mathematics for king paradox & consider that king vehicle is at the front of the convoy & even it crosses front point on the track, his vehicle will move in straight line.
1)Let, on track of 27000 m, at front point B & starting points A, timer clock is fitted with initial reading 0 second & consider that king vehicle is just on the front of the track at point B'.
Now, let, origin of moving frame for king is at B (end of track) then
king will see co-ordinate of A & B as B (0,0,0,0) & A{-27000/y ,0,0, -y(vx)/c2} at the end of track.
This is because t’=y{t-(vx)/c2}
Now, for king , time at A will be zero when t-(vx)/c2=0 in moving frame.
Means, t=(vx)/c2 = -(v.27000) /c2
Means, after time dt=0- (-v.27000) /c2 = (v.27000) /c2 in moving frame is the time co-ordinate of point A will become zero for king.
Equivalent time in vehicle frame is dt’= ydt=y(v.27000) /c2
Means, Co-ordinate of A after time dt’ for king will be A(-27000/y-vdt’,0,0,0)
---------------------------------------------------
Means , For King , time co-ordinate of A & B are zero when X=-27000/y-vdt’ for A & X=0 for B.
Now, paradox is solve if
Convoy length = 27000 =(0)-( -27000/y-vdt’)=extended length of track in king frame to co-relate time t=0 at point A & B
i.e. 27000 = 27000/y+v. y(v.27000) /c2
i.e. 1=1/y +y.(v2/c2)=Y{(1/y2)+(v2/c2)}
i.e. 1=Y
So, this will not solve our paradox of king
---------------------------------------------------
Please, tell me, where is my calculation wrong?
mahesh
Member
 
Posts: 61
Joined: 21 Mar 2016


Re: Problem Of Length Contraction At Cern Due To Spe.relativ

Postby mahesh on February 27th, 2018, 6:15 am 

Means, if time is synchronize in non-prime frame i.e. t'1=t'2
then observer in prime frame see that length get contracted
i.e. Length L = L'/y & time is not same at all points for example t1=0 at B then t2= -y(v.27000)/c2 at point A
------------------------------------------------------------ --------------
But if I try to synchronize time in prime frame for points A & B i.e. t1 =t2 for observer
then Length L=L'/y +y(v2.L')/c2
i.e. L/y= L'
means, frame where we synchronize time, length in that frame is reduces for equating to length in other frame.
i.e. L= y. L'
--------------------------------------------------------------------------
Means, Length in prime frame & non-prime frame may be equal
but for observer in prime frame can not find both as equal even time in non-prime frame is synchronize
or time in prime frame is synchronize.
i.e. t1=t2
or t'1=t'2
then also L = L' is not possible for observer to observe.
------------------------------------------------------------------------
This will create complicated paradox. I like this.
mahesh
Member
 
Posts: 61
Joined: 21 Mar 2016


Re: Problem Of Length Contraction At Cern Due To Spe.relativ

Postby mahesh on March 5th, 2018, 3:50 am 

Now, consider Paradox similar to Bell spaceship paradox :-
Consider two similar metal blocks A & B are elevated on magnetic field track & pushed in x-direction with same magnetic force F.
String of length L is tide between these two blocks.
As same force F is acting on A & B, both will accelerate with same acceleration 'a" & distance between both remain same.
Now, as given in old space ship paradox,
Strain is produce in the string & it break.
This strain will create stress or force f which breaks the string.
Now, before string break:-
Force acting on block A will be (F-f) & on block B will be (F+f) in X-direction where f is stress in string.
Means, acting forces on blocks will not remain same in x-direction then acceleration of both block cannot remain same.
Means, to keep same acceleration of blocks A & B, we have to apply different forces on each block A & B
Even there is no string tied in between blocks A & B then also to keep distance between two block same & same constant acceleration. It is necessary to apply different forces on A & B.
For example, one cabin is falling in earth gravity then person on earth will see that distance of everything in the cabin in direction of gravity contracted due to accelerated velocity. To keep distance same in cabin, we have to apply force in opposite direction & create opposite acceleration in cabin.
From above it is very clear that we have to apply different forces on each proton bundle to keep distance between them same in CERN frame.
My question is :- 1)How can it possible to apply different forces on each proton bundle to keep distance between them constant because number of bundles are 2808 & have a huge velocity?
2)If A is 1st protons bundle & B is last proton bundle then we have to apply more force on A & less force on B. Now, in circular path A will come closer to B in opposite direction. So, we have to apply more force on B & less force on A to keep distance between each protons bundle same.
& FA more than FB and FB more than FA cannot be possible by any magnet.
3) Why can we require different forces for same acceleration?
mahesh
Member
 
Posts: 61
Joined: 21 Mar 2016



Return to Physics

Who is online

Users browsing this forum: No registered users and 9 guests