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Do you think Relativity theory can explain all the paradoxes?

  1. Nov 25, 2013 #1
    Most of the paradoxes are solved by either "Relativity of simultaneity" or "Born rigidity"
    And there is a 3rd tool that i know is "Minkowsky spacetime diagram"
    But i do not understand how clear/imaginable it will be when you represent a paradoxical event on Mink Sp Diagram (MSD)
    When i asked a question which can neither be solved with "Relativity of simultaneity" or "Born rigidity" some people represented it on a Minkowsky spacetime diagram and claimed "paradox solved!"
    Sometimes they try to question my experiment like asking "how can the time be synchronised/ how can you separate a distance like that/ it is only a thought experiment so there won't be a clear answer etc" I think they were trying to skip the question.

    I'll give you that paradox

    "Imagine you are standing on earth and there is another planet (Say Planet X) 4 light years away. Both planets are relatively at rest to eachother (What i mean here is there is no relative motion between these planets, here both planets are floating in space, not revolving around the sun)

    Both the planets are synchronised in time by sending light pulses so that they can keep 2013 on both planets.

    Suppose there has been an agreement for conducting an experiment between these planets. Scientists on Planet X accelerate a proton in a 'particle accelerator experiment' until the proton reaches a velocity of 99.99999999% c. When it is 2014 on planet x (on earth too) they eject the proton towards the earth.
    Just imagine that this proton has an 'inbuilt clock' in it (just imagine).
    As soon as the proton leaves the accelerator, a stopwatch which which is attached to it activates.
    Scientists on earth expect the proton reaching the earth on 2018 + some months.
    Unfortunately a scientist (Robert) on earth dies in 2017 and proton reaches the earth on 2018.
    The proton will have only aged an year or so.

    (This is an event that has been happened and there is no way it can be undone)

    Now imagined everything from the "protons rest frame"

    What proton see is that as soon as it leaves the accelerator, planet x moves away from it with a velocity of 99.99% c and the earth moves towards it with the same velocity
    Its clock will be running normally and it sees earth to be time dilated. Earth will reach at the proton after 4 years.
    When earth reaches the proton, the earth will have aged only about 1 year (2015) and 4 years pass for proton.
    Lucky from protons view the scientist Robert is alive and healthy

    This is what we gets from a protons reference frame, both predictions are according to relativity and both are real. But it's a paradox.
    In the first event from earths rest frame the scientist Robert dies before seeing the proton and in the protons rest frame of reference The scientist is alive!
    So how do relativity solve this paradox?
    How can MS Diagram can solve this paradox?
    I agree they can represent it on a diagram but it doesn't solve the paradox.
    If it solves how can i visualize it, i mean i need a picture with definite result, a result that is agreeable for all observers
  2. jcsd
  3. Nov 25, 2013 #2


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    If the two planet clocks are synchronized with one another, then the traveling clock is synchronized with neither - it's moving relative to both.

    You have also confused yourself somewhat by talking about the "proton's rest frame"; you'll hear people use that phrase all the time, but it is more accurate to speak of "a frame in which the proton is at rest". The difference becomes important when you're dealing with accelerating objects such as your proton. Go back and solve this problem from the the point of view of an observer who was moving at 99.999999% of the speed of light relative to the two planets (so that after it's accelerated, the proton is at rest relative to that observer) and you'll see what I mean.
  4. Nov 25, 2013 #3
    I knew someone would come up with things like (acceleration or rest frame inorder to skip the topic.)
    I have already told you everything. The proton is accelerating only when it is in the accelerator, time during that part is not considered here. Also i meant what you said, i can say either in protons rest frame or, the frame in which the proton is at rest
  5. Nov 25, 2013 #4


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    The protons clock will not have the same elapsed time as the clocks on the planets.

    Take the advice you've been given so many times and draw a Minkowski diagram. There is no paradox.
  6. Nov 25, 2013 #5


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    So, like I said, try solving the problem in a frame in which the post-acceleration proton is at rest.

    When the clock on Earth and on planet X reads 2014, our scientist Robert has three years to live, and will die when those clocks read 2017; the proton will reach earth when those clocks read 2018.

    Let's arrange things so that a clock carried by an observer who is at rest relative to the proton (after it's been accelerated) also reads 2014 at the moment that that observer passes planet X and the proton leaves planet X. According to that observer, what do the synchronized earth and planet X clocks read? What will they read and what will the moving observer's clock read when Robert dies? When the proton reaches Earth?
  7. Nov 25, 2013 #6

    Good "paradox". Congratulations :smile: (The quote marks are there for not to make people angry. )

    Let's correct a small error: At velocity 0.9999 c the factor of change is 70.7. (I use the smaller one of the two velocities given.)

    Travel time according to earth clock: 1.001 times four years=4.004 years (1.001 times the travel time of light)
    4.004 years divided by 70.1 = 20.8 days

    About 21 days is the time measured by the proton's clock.

    So you might say someting like: For 21 days time the proton sees a time dilated earth that is aging at rate 1/70.1 of the protons aging rate.
    Last edited: Nov 25, 2013
  8. Nov 25, 2013 #7
    You have assumed a time dilation factor F of about 4 to 1 when the actual time dilation factor is about 70,711 to 1. That is a huge difference. For the time dilation factor to be 4:1 the speed of the proton would have to be about 0.866c which is not so convenient if we want to assume v ≈ c. It is unfortunate that you did not check this, because it means we cannot directly use your original figures.

    I will go with a gamma factor of 4 and a velocity of 0.866c.

    Once the proton is moving towards Earth, the simultaneity will be different in its new rest frame and and it will calculate the date is 2014 on planet X and 2017 (+ some months) on Earth. This is calculated using the simultaneity equation, which is L*v/c^2 where L is the distance between planet X and Earth as measured in their mutual rest frame. This works out as 4*0.866 = 3.46 approximately which means the time on Earth is roughly 3.5 years ahead of the time planet x. Now it takes L/v*1/F = 4/0.866*1/4 = 1.15 years to travel to Earth as measured by the proton. At the start of its journey the time on Earth was 2017 + 0.46 years so proton arrives at Earth at 2018 + 0.6 years approximately.

    According to observers on Earth, the proton left planet X at time 2014 and takes L/v = 4/0.866 = 4.6 years approximately to travel, so it arrives and time 2018 + 0.6 years approximately.

    The proton observer and the Earth observers therefore agree on the arrival time of the proton at Earth and everyone agrees Robert was dead before the proton arrived. No paradox.


    Now lets do it again, very roughly, using your original velocity very close to the the speed of light and an enormous time dilation factor. At the start of the proton's journey, the time on planet X is 2014 and the time on Earth is 2018 using the simultaneity equation L/v ≈ 4. Due to the extreme time dilation it takes very little time to travel to Earth as measured by the proton, so by its reckoning, it arrives at time 2018 (+ a little bit).

    According to the Earth observers the proton leaves planet X at time 2014 and at close to the speed of light, it takes just over 4 years to travel so it arrives at Earth at time 2018 (+ a little bit).

    No paradox.
  9. Nov 25, 2013 #8
    I find there no use of adding an observe with the proton, anyway i assume so. From the observer traveling with the protons rest frame/the frame in which they are at rest watches the time on both planets dilated in the same rate. When earth hits the observer/proton, the earth time will be only about an year or even days (i didn't calculate how much it is dilated)

    Instead of saying moving observe, i would like to say the frame in which the observes is at rest, so that i can remove length contraction.
    In my observers/protons rest frame, when earth hits the observer, Robert will be alive and there is no use in saying what does the observers clock read when Robert dies because the experiment stops as soon as earth and observer meets.
    When the proton reaches earth? I've told you everything already
  10. Nov 25, 2013 #9
    this scenario mixes up different measures of time/length. At rest the distance between planets is 4ly. in motion with the proton its nearly the inverse of the proper time it measured (99.99%c).

    Anything that happens local to and at rest with this clock that is magically synched to planet x's can be thought of as invariant, that is the simultaneity of events is invariant i.e. robert dies & clock displays 2017.

    You shouldn't be surprised that the proton magically thinks to itself "I see that Robert died when his clock read 2017, though I measured it as happening much much sooner, and remarkably much much closer to me. When Earth gets here, maybe Roberts successor could help me with the calculations for those length & time measurements I've made , opps i mean once I arrive there. I always forget about my accelerometer measurements. Hmmm... most notable, I arrived after Robert dies, just my coordinate mapping has these simultaneous happenings, like roberts death and the clock reading 2017 so close together, it looks like I arrived just a short time but traversed a great distance after robert died. And during that interval of the accelerometer measurements my coordinate map gets all swirly where I assume continuity of proper time/length throughout that duration. Perhaps its all a matter of spacetime position(interval), as opposed to the presumption the measure of time and length are truly independent. I though I already learned geometry in grade three! "
  11. Nov 25, 2013 #10


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    No. Robert dies before the proton reaches his location and this will always be so. No transformation can alter that fact.

    I just saw your last post. It is scientifically completely wrong and very unpleasant also.
    Last edited: Nov 25, 2013
  12. Nov 25, 2013 #11
    In whose frame?
    If you are considering this in a frame in which the proton is at rest, then Robert won't die when earth hits proton. That is because earth has undergone time dilation because of its velocity.
    I dont want to deviate this thread, so I'm not ready for an arguement with you
  13. Nov 25, 2013 #12


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    Putting aside my distaste - if the earth clock shows 2017 when Robert dies then all observers agree on that fact. This is because the time elapsed on a clock is the proper length of its spacetime path, which is a geometric invariant and independent of the frame of the observer.
  14. Nov 25, 2013 #13
    What I'm talking is not about whether Robert dies or not, I'm talking about the order of the events that happen.
    I'm forced to repeat again.
    Let say again, proton is in rest and earth is moving towards it from a distance of 4 light years. Can you agree on this car and tell me yourself what will proton observe when earth reaches it????
    I have already given you the experimental setup and if you give me an answer, then we can solve this paradox step by step

    From protons rest frame, earth is 4light years away
    As soon as the proton leave the accelerator, time on earth and Planet x was 2014.
    When earth reach proton after 4 years, what will be the time on earth?????????
  15. Nov 25, 2013 #14


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    The earth will not be 4 light years away in the proton's frame. It will be much less. This is basically the same well know phenomenon as muons being created in the upper atmosphere by cosmic rays. Their half life should allow only 600 meters of air to pass them before half of them decay. Instead, essentially all reach the ground. The reason is that in the muon frame, the atmosphere is only a few meters thick.

    In this case, earth will reach the proton in much less than 4 years (per the proton). Earth clocks will be running slow (per the proton), but the earth clocks will be out of synch with other planet (per the proton). Syllogism: if two clocks are in synch per earth/planet, they are out of synch for the proton - period. So the proton expects that earth clocks will read 2018 on arrival despite running slow, and earth only having a short distance to travel because its clocks are set 'way earlier' than planet x clocks (by the very process that earth and planet x use to synchronize them per their frame).
  16. Nov 25, 2013 #15

    Jonathan Scott

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    Whenever something changes its velocity (speed or direction), the relationship between apparent times of distant events and local clocks instantly changes as well, following the rules of the Lorentz transformation. This is similar to the way in which changing the direction of travel in a 2-D plane instantly changes the definition of whether some other travelling object is "beside", "ahead" or "behind" the observer in the direction of travel.

    This means for example that a fast-moving object (such as your example proton) does NOT see the time at a distant location as being the same as that seen by someone in the rest frame of the emitting apparatus.

    The Lorentz transformation is mathematically a sort of "complex rotation" of the coordinate system, and transforms one point of view into another. It is mathematically provable that this cannot lead to any paradox, regardless of the complexity of the example.
  17. Nov 25, 2013 #16


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    Perhaps another way to understand this false paradox is to focus on what would be visually 'seen' by the proton, then how it would reasonably interpret this.

    As of when it leaves the accelerator, it sees earth approaching from (e.g. 1 ly) away, at a large fraction of c. The earth clock is visually seen to be blueshifted and running fast. However, accounting for light delay, this rapid running image must be considered to have originated before the proton left the accelerator. By simple logic and calculation, the proton sees that only the last few months on the earth clock were emitted after the proton left the accelerator, at which earth has nearly arrived. By incorporating light delay, and compensating for Doppler, the proton concludes that the earth clock has been running slow (since it left the accelerator; it would be analyzed as running fast while the proton was in the accelerator - but that can be ignored because we can make the acceleration phase arbitrarily short). However, visually, the earth clock is seen to elapse 4 years while the proton is in transit - almost all being Doppler blue shifted history from before the proton left. The visual speed of the earth clock is uniform and fast the whole journey.
  18. Nov 25, 2013 #17
    explain how this massive proton slows down, define arrives and consider the issue with calling something simultaneous while it is accelerating, since "moments" are frame dependent. Oh and where the proton hits Earth, is it same place where Robert dies / earth clock is.
  19. Nov 25, 2013 #18
    I'll give you another thought experiment, Instead of proton, use Jack and instead of earth use Jill, both are 20 years old and are 4LY away. If Jack moves towards Jill (Please remove Acceleration), how old will be Jack when he reaches Jill?
    Okay, just answer what will be their age when they both meet together?
  20. Nov 25, 2013 #19


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    Your velocity is not consistent with the rest of your analysis so I'm going to change that velocity to 97% c. This will result in the proton taking one year of its Proper Time to traverse four light-years in the Earth-Planet rest frame.

    Although the earth time is dilated in the proton's rest frame during its trip, the proton cannot see the Time Dilation of the earth. Instead, you have to use the Doppler analysis to see what the proton actually sees and this is the key to your misunderstanding. Instead of seeing time on the earth going four times slower, the proton sees time on the earth going 8.1 times faster. This means that every 45 days, the proton will see another year go by on the earth. The proton starts by seeing the earth time 4 years earlier than its own and after 360 days of travel, it sees the earth progress through 8 years and then 5 days later it arrives at the earth.

    Not in the frame in which the proton is at rest during the trip. Earth will start out 0.97 light-years away from the proton and it approaches the proton at 0.97c and arrives at the proton one year later.

    Not even close. The earth ages 3 months, a month and a half before 2018 and a month and a half after. Remember, the earth's time is dilated by a factor of 4. Only 1 year passes for the proton.

    Wrong again. In the frame in which the proton is at rest during the trip, Robert died over three and a half years before the proton left the planet. When the proton reaches 315 days into the trip it sees the earth at year 2017.

    Here is a spacetime diagram depicting the scenario in the earth/planet rest frame. The earth is shown as the thick blue lines with dots marking its years, the planet is shown in red and the proton in black:


    The thin blue lines show the image of the earth time as it propagates to the planet and the proton. At the beginning of the trip in 2014, the planet and the proton see the earth as it was in 2010. During the trip, the proton sees the earth advancing in time at a rate of 1 year for each 45 days of the proton's time. Fifty days before the trip is over, the proton sees the year 2017 on the earth.

    Here is a closeup of the important events tranformed into the frame in which the proton is at rest during the trip:


    As you can see, nothing changes for any observer, either for what they see or for anything that happens to them.

    Does this make it all clear for you?

    Attached Files:

    Last edited: Nov 25, 2013
  21. Nov 25, 2013 #20


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    Not in the proton's rest frame. In that frame Earth will reach the proton in a much shorter time.

    No, a very short time will have passed for the proton, and a very very very much shorter time still will have passed for the Earth.

    By relativity of simultaneity: in the proton's rest frame while it is traveling, the event of it leaving Planet X is simultaneous with an event a very very very short time before 2018, *not* with an event in 2014. (The event of the proton leaving Planet X is simultaneous with 2014 to an observer on Earth and an observer on Planet X, but they are moving relative to the proton so their simultaneity is different.) So even though the Earth only ages a very very very small amount during the experiment, the proton still arrives on Earth in 2018, Earth time, and all observers agree on this.

    It does if you represent it correctly. If there are multiple ways of representing a scenario in relativity, they must all be consistent. I suspect you have made an error in drawing the diagram, but I can't tell for sure unless you post the diagram you think is correct.

    Edit: I see ghwellsjr posted a correct diagram.
  22. Nov 25, 2013 #21


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    You cannot ignore the acceleration. All you are doing here is creating a paradox by making impossible assumptions, proving only that garbage in = garbage out.
  23. Nov 25, 2013 #22


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    Oh come on. You refuse to learn the basic elements of what a 100 years of experiments have shown. "Both 20 years old" means nothing without a qualifier. If we remove acceleration, then each has always seen the other moving fast.

    - If they are both 20 years old per Jill, then when Jack reaches Jill, Jack will be younger.
    - If they are both 20 years old per Jack, then when Jill reaches Jack, Jill will be younger.

    There is no content to a statement such both 20 years old now without additional information.
  24. Nov 25, 2013 #23


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    Try making that calculation before you assume that it will be of no use. It is essential to resolving the apparent paradox.
  25. Nov 25, 2013 #24


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    This one is no different. It is solved by relativity of simultaneity.

    OK, all of this is fine. Due to the inconsistency I used the .9999 c value below rather than the higher number above (just too many digits to keep track of). Let the unprimed frame be the inertial frame where the planets are at rest and let the primed frame be the inertial frame where the proton is at rest during its trip.

    No, with γ=70.71 the proton will have aged 0.05657 years.

    All of this is fine.

    No, earth reaches the proton in 0.05657 years in the proton's frame.

    In this amount of time the earth will have aged only 0.0008 years.

    Nope. You forgot to account for the relativity of simultaneity. In the primed frame, due to the relativity of simultaneity, planet earth's clocks are ahead by 3.9996 years at t'=0. So in the primed frame scientist Robert is already dead before the proton even leaves. In fact, Robert died 70.684 years before the proton's launch in the primed frame.

    No, you did not apply relativity correctly. The second prediction neglected the relativity of simultaneity. There is no paradox if you do the math correctly and include relativity of simultaneity.
    Last edited: Nov 25, 2013
  26. Nov 25, 2013 #25


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    The way I interpret the above is that you don't understand space-time diagrams, and thus don't find them convincing.

    With a large (and unfortunate) side order of "if I don't understand it, it must not be true". But I'll ignore this for now, there's not much that can be done about it, except to try to point out that if you don't understand something, it doesn't mean it's false. It doesn't mean its true, either. It means you don't understand it, and you need to study it more to get anywhere.

    Space-time diagrams are not really terribly complex. They are essentially no more complex than a graph of position vs time. This should already be familiar to most physics students.

    It would be possible to go into more effort here. For instance, to point out how the notion of time can be represented by a timeline, and the notion of a single dimension of space could also be represented on a corresponding "space-line".

    But I'm not sure if it helps any - the question in my mind is "what don't you understand about space-time diagrams". It seems like a waste of effort to go into more about the above if you already understand the notion that you can make a graph of position vs. time.

    Another important feature about space-time diagrams I can think of is the notion of one-one correspondence. This is also a basic notion in math. Applied to space time diagrams, the notion of one-one correspondence says this: for every event in space-time, there is one, and only one, point on the space-time diagram. For every point on the space-time diagram there is one, and only one, event.

    This is an important property of space-time diagrams - it's really their key feature. A related idea is that two space-time diagrams are equivalent if they represent the same space-time. I suppose some careful language usage is necessary here. The space-time diagram is a sort of a "map", it represents the space-time but is not the space-time. Two space-time diagrams are equivalent if they represent the same space-time. They may look different visually, but if there is a 1:1 representation of the space time on each of the diagrams, there is also a 1:1 representation between the different looking diagrams, such that every point on one diagram has a unique point on another diagram.

    Is this part clear as well? Any issues here?

    Maybe the third part is where the confusion arises - I don't know. This part is where we get into the issues of assigning specific coordinates. The third part says that if you have a space-time diagram, you can attach labels to it for your own convenience. These labels are coordinates.

    To every point on a 2-dimensional space-time diagram, we need to attach two labels - the "time of the point as used by some specific observer" and "the position of the point as used by some specific observer".

    Every observer will have a different labelling system, so we need to keep track of which observer we are using when we use coordinates if we have more than one obsever.

    Is this part clear? The two-dimensional part of it might actually require a whole lot of math to justify in detail - is that the issue? I.e. we need only assign one coordinate to represent points on a line, two coordinates to represent points on a plane, three coordinates to represent points on a cube, and 4 coordinates to represent a point in a 4 dimensional space-time. It seems natural to me, though if you go into the detailed math, the "why" of this can get complex.

    Most of the time though, for simplicity, we deal with 2 dimensional space-times (one space + 1 time dimension), and all we need are 2 dimensional diagrams, which we can draw on a piece of paer.

    Because there are 1:1 correspondences between the points on two equivalent space-time diagrams, there are also 1:1 correspondences between the coordinates associated with the points on two equivalent space-time diagrams.

    This one-one correspondence is given explicitly by the Lorentz transform.

    The feature that I expect may be confusing you can now be explained.

    We assume we have two obsevers, O and O'. Observer O, assigns coordinates (x,t) to events, Observer O', assigns coordinates (x', t') to events.

    On any space-time diagram the set of points that are simultaneous to obsever O must have the feature that they share the same t coordinate. This will be a line on the space-time diagram.

    We can also draw, on a space-time diagram, the set of points that are simultaneous to obsever O', the set of points which have the same t' coordinate. The key feature of relatiivty is that they are two different lines, they are not the same line. We can choose any equivalent representation of the space-time diagram we like. In all representation, the set of points that are simultaneous for observer O will be a different set of points than the set of points that are simultaneous for observer O.
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