The physical interpretation of the Minkowski spacetime diagram

Discussion in 'Physics & Math' started by geordief, Dec 21, 2017.

  1. geordief Registered Senior Member

    Messages:
    625
    Along the North South axes we measure off on the graph paper equal distances for equal physical quantities of [ct] and along the East West it is the same for lengths in space in the one (x) direction.

    So what do these measurements of [ct] actually mean physically?

    If we were conducting a scientific experiment ,and the experimenter is situated at the origin how does he or she make these two measurements so as to uniquely quantify a point on the graph?

    Is "t" counted by ,theoretically any regular cyclic event such as someone very regularly tapping their finger or a tap dripping very regularly (or a good clock ,obviously)?

    Then after a number of cycles of this "time keeper" how is [ct] measured? Do you use a mirror and divide by two after you have measured the distance to it in exactly the same way as you have measured distances along the x-axis?

    If I have correctly described that physical process can I next ask about how the spacetime interval itself would be physically measured in a way corresponding to a [ct,x] point on the graph?
     
  2. Google AdSense Guest Advertisement



    to hide all adverts.
  3. Mike_Fontenot Registered Member

    Messages:
    23
    One thing that might help is to choose units of time and distance so that the speed of light "c" is numerically equal to one. For example, let the length unit be "lightyears", and the time unit "years". Then the two axes can be interpreted as ordinary distance and time. That's what I do in my use of Minkowsky diagrams in some Twin "paradox" scenarios on my webpage here:

    https://sites.google.com/site/cadoequation/cado-reference-frame
     
  4. Google AdSense Guest Advertisement



    to hide all adverts.
  5. arfa brane call me arf Valued Senior Member

    Messages:
    5,417
    So, conceptually you're multiplying the speed of light by a year and calling it a lightyear but then you're only saying light travels one lightyear in one year. You set c to 1 this way but you don't forget it's a velocity.

    Linguistically, you should really have a "lightspeedyear" and then contract it to "lightyear". Wonder if you could use indices? (just kidding).
     
  6. Google AdSense Guest Advertisement



    to hide all adverts.
  7. geordief Registered Senior Member

    Messages:
    625
    What I am asking is:For any two points on a Minkowski spacetime diagram (let's just choose[ct,x]=[0,0] and[c,0]** )is there a physical experiment corresponding to those two events that would show that the diagram was correct(that it was a faithful map of what had happened;in this case someone or something not moving for one tick of a clock)?

    I accept the validity of the spacetime diagram but,like a doubting Thomas am looking for the physical evidence to tie it down to actual events and to be able to say that I have measured along both axes and have verified that the said event did indeed occur at that point in spacetime.

    How would I set up my experimental apparatus for the simple example I have given (or any other)?

    **ie t=1 and so ct=c
     
  8. arfa brane call me arf Valued Senior Member

    Messages:
    5,417
    Unfortunately there is only one part of a Minkowski diagram with physical significance, it's the only part of the whole thing where you can measure or detect anything.
    Can you guess where it is?

    Which is not to say Minkowski space isn't mathematically significant, it predicts things too, at least for objects with constant relative velocity, an uncommon coincidence (objects accelerate generally, in our universe).
     
  9. geordief Registered Senior Member

    Messages:
    625
    Yes,I can ;-)

    Thanks for the edit.I was about to ask for a clarification.
     
  10. Q-reeus Valued Senior Member

    Messages:
    2,655
    http://math.ucr.edu/home/baez/physics/Relativity/SR/acceleration.html

    That link, among many similar that could be furnished, is NOT an invitation to start arguing on fine points.

    Please Register or Log in to view the hidden image!

     
  11. arfa brane call me arf Valued Senior Member

    Messages:
    5,417
    Minkowski space (and the diagrams; Penrose diagrams are a more general representation of "future" and "past" events), can be considered a composition of quotient spaces. If you quotient out a concept of "measurement in finite time intervals" you get "everything that I can label as an event in my simultaneous space", a co-concept. You can quotient out that space too, leaving just you at a moment of time in a "pointlike" part of space, i.e. you are the only event.
     
  12. Confused2 Registered Senior Member

    Messages:
    475
    The muon experiment confirms Einstein's prediction about spacetime.
    http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/muon.html

    For what follows it is essential to understand that the muon pilot sees the counting stations fly past his/her cockpit window, as far as he is concerned the counts occur in the same place (though not at the same time). In the Earth frame the counting stations are 10 kilometers apart. Again. In the Earth frame the counting stations are 10 kilometers apart.

    For some (unknown) reason the Hyperphysics link just plucks time dilation from nowhere.
    Actually predicting time dilation from the invariant spacetime interval we get...

    Where s is the invariant spacetime interval
    s²=-x²-y²-z²+c²t²
    Using lower case for the Earth frame and upper case for the muon frame...
    where v is the muon velocity in the Earth frame we have x=vt,y=0,z=0 so
    s²=-v²t²+c²t²

    In the muon frame we have
    S²=-X²-Y²-Z²+c²T² where X=0,Y=0,Z=0
    so
    S²=c²T²
    since S²=s²
    c²T²=-v²t²+c²t²
    Or T=t√(1-v²/c²)
    The time dilation is a consequence of the invariant spacetime interval - as predicted.
     
  13. arfa brane call me arf Valued Senior Member

    Messages:
    5,417
    From Baez et al.
     
  14. Confused2 Registered Senior Member

    Messages:
    475
    Whereas I give the invariant spacetime interval as s²=-x²-y²-z²+c²t² I think Minkowski changes sign and substitutes ic (where i is the square root of -1) for the 'c' constant I use.
    Hence he quotes something like
    s²=x²+y²+z²+(ic)²t²
    which may be not be tremendously helpful unless you are fluent in 'i' ... the square root of -1.
     
  15. Confused2 Registered Senior Member

    Messages:
    475
    If you imagine someone going East is on a Camel and a person going North is on a horse...
    The person going East travels 300 camel footsteps and the person going North travels 400 horse footsteps - how far apart are they? We'd have to establish a shared unit of length to find that out. Without knowing the length of a horse footstep and the length of a camel footstep in our chosen set of units we can't know how far apart they are.
    If the final distance is r (in metres) and a horse footstep is p metres and a camel footstep is q metres we'd get the final distance r (in metres) is
    r²=(400p)²+(300q)²
    Ideally we'd measure the distance North and East in some common unit (probably metres) and kind'a hide the p and q conversion factors.
    In the 'simple' invariant spacetime interval no attempt is made to hide the conversion factor between distance and time, we have x,y,z in metres and c in metres/second giving (say) s²=x²+y²+z²-c²t²
    For convenience you can agree a set of unit where c=1 giving (say)
    s²=x²+y²+z²-t²
    Or you can agree a set of unit where c=i giving (say)
    s²=x²+y²+z²+t²
     
  16. Mike_Fontenot Registered Member

    Messages:
    23
    When the units are chosen so that c = 1, the presence of c in equations is there only for dimensional correctness. So you can leave out the factors of c in any of your equations, which makes them simpler, and they will still be numerically correct. That's what I do in most of the equations that I discuss and use in my webpage (whose link I gave in my previous post).

    When I plot Minkowsky diagrams, the axes I use are distance X and time T ... I don't use cT, since c = 1, and there is no reason to include it. I plot the X axis vertically, and the T axis horizontally, which is the opposite of the normal convention. There are two reasons I reverse the usual convention: first, in most twin "paradox" type scenarios, the world line of the traveler goes out and back at least once, and so the X variable first increases, and then decreases, at least once, and in some scenarios, multiple times. But the T variable just keeps increasing, sometimes for quite a while, and it is more convenient and familiar to have that longer axis be the horizontal axis, not the vertical axis. The second reason is that, in most of the scenarios, the variable T takes on more of the role of independent variable, and X has more the role of dependent variable, and we are all more accustomed to plotting the independent variable horizontally, not vertically.

    In the simple case of an instantaneous turnaround, the world line of the traveler in the Minkowski diagram will initially be a straight line sloping upward with a slope equal to the relative speed v (which is always less than 1 in units where c = 1). Along this line, the traveler's age "t" can be marked off as a parameter. Then, at the turnaround point (where X has reached its maximum value for the trip), the world line changes to a straight line sloping downward, with slope equal to -v. And on this Minkowski diagram, the lines of simultaneity for the traveler, before and after the turnaround, can be plotted as straight lines of slope 1/v and -1/v.

    The case of finite accelerations (specifically, 1g accelerations) can be plotted similarly, but with a curving world line during the segments of acceleration. This is all discussed in detail in the webpage.
     
  17. arfa brane call me arf Valued Senior Member

    Messages:
    5,417
    Well I'm having a thing with numbers at the moment. There is a difference between "setting" c to 1, and calling it the number 1. But yes then that means they disappear from both sides of the equations.

    Ok so you still draw a Minkowski diagram with ct meaning lightyears or lightseconds, or what have you. That is to say, you choose (an interval of time) for the basis. An interval of time isn't a number, it maps to a number. There's only one way to map the number to the interval if you mean the time domain.
     
    Last edited: Dec 24, 2017
  18. Mike_Fontenot Registered Member

    Messages:
    23
    No, the c factors aren't just canceled out, because different terms on the right-hand-side have different numbers of c factors in them. So after all the c's are erased (not cancelled) from the equation, different terms on the right-hand-side of the equation will have different dimensions (i.e., some of the terms will have dimensions of time, and others will have dimensions of distance). So the equation is no longer dimensionally correct. But it IS numerically correct, so the equation gives us everything we need.

    No. I don't use c in the Minkowski diagram at all. The horizontal axis is the home twin's age T, and the vertical axis is the location of the traveling twin in the home twin's spatial coordinate X.
     
  19. Mike_Fontenot Registered Member

    Messages:
    23
    But in this case, that difference is EXACTLY what I'm choosing to ignore ... I'm treating c as if it IS just a constant number, and it is therefore a factor that can be ignored (and "erased") in all terms of an equation. Likewise, I treat the velocity v as if it is just a dimensionless number, with a value less than 1 and greater than -1. (More precisely, you can write "v = beta c", and beta actually IS a dimensionless number. But by ignoring "c" everywhere it appears in an equation, "v" and "beta" then get treated as though they are synomynous, because they ARE numerically equal.) Since v is a variable number and not a constant number, it can't be ignored in the equation like c can, but it can be considered to be just a dimensionless number. The result is that the equations become much simpler, and the analysis of twin "paradox" type scenarios becomes much easier and quicker.
     
    Last edited: Dec 24, 2017
  20. Neddy Bate Valued Senior Member

    Messages:
    1,482
    Interesting approach on that webpage. I'm not sure I follow everything, because I am used to the convention where the variables are x y z t and x' y' z' t' and you have changed things quite a bit. In the case of "Idealized Instantaneous Velocity Changes," would there be some way to reformulate the "turn around" so that instead of t=-20 changing to t=+20 it could be v changing to -v, or something else like x changing to -x? The turnaround represents a change in coordinate system, so something like that would make more sense to me. If I see t change like that, it makes me think that the time displayed on the t clock has changed.
     
  21. Neddy Bate Valued Senior Member

    Messages:
    1,482
    A Minkowski diagram is just a graphical representation of the Lorentz transformation equations. So, any physical experiment that is in agreement with the Lorentz transformations, should also be in agreement with a Minkowski diagram of those same transformation equations. I don't think there is really a ct dimension, for example. I think time is just a t dimension, and the ct axis is chosen to make the graphic work. I do think there are some folks who take the Minkowski diagram more literally than I do, though.
     
  22. Confused2 Registered Senior Member

    Messages:
    475
    It may be lever but is it atually learer?
     
  23. geordief Registered Senior Member

    Messages:
    625
    Surely it is the diagram is just the open book on which you show the possible relationships and the Lorentz transformation is just the one we are especially interested in.

    If ct is a length and x is a length ,then it is fair game to chart one against the other

    Taking spacial measurements from the origin are we not comparing the distance traveled by an object with the distance traveled by a beam of light which left simultaneously?

    That is the physical experiment I asking about. Is is feasible to show this experimentally?

    I can see that the Lorentz transformations are far more interesting but I am just trying to get the fundamental basics and geometry of the graphing method clear in my own head.
     

Share This Page