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I Velocity through spacetime

  1. Oct 6, 2018 #1
    I have read that the velocity of a body through spacetime is equal to 'c'.
    Can I therefore draw a spacetime velocity diagram (triangle) with sides: "v" velocity through space.
    "c". ------------------ time
    "c" ------------------ spacetime?

    runningc
     
  2. jcsd
  3. Oct 6, 2018 #2

    PeterDonis

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    Have you read it in an actual textbook or peer-reviewed paper?

    No. The "velocity through spacetime" ##c## is the length of the object's 4-velocity vector. You can draw a triangle in velocity space with this as one side, and the second side being the spatial velocity ##v##; but the third side will not be ##c##.
     
  4. Oct 6, 2018 #3

    robphy

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    Here's a spacetime diagram
    upload_2018-10-7_7-24-50.png
    In trigonometric form,
    [itex]v/c\quad\ =\tanh\theta=\frac{opp}{adj}[/itex]
    [itex]\gamma\qquad \ =\cosh\theta =\frac{adj}{hyp} [/itex]
    [itex]\gamma(v/c)=\sinh\theta =\frac{opp}{hyp} [/itex]

    (I updated the diagram by indicating the rapidity [itex]\theta[/itex] (the Minkowski-analogue of the angle).
    By the way, the rapidity of the Minkowski-right-angle is infinity.
    I also added the interpretation of "tangent" as "opposite over adjacent", etc...

    Also note: the tangent-line to the Minkowski-circle is Minkowski-perpendicular to the radius.)
     
    Last edited: Oct 7, 2018
  5. Oct 7, 2018 #4
    Thank you; I was afraid it might not be that simple. I read it in "Why does E= mc^2" by Cox and Forshaw.
    RC
     
  6. Oct 7, 2018 #5
    Thanks very much, although I not quite sure how much it will actually help me in what I am trying to do as the diagram has 'c' and 'v' at right angles. I am working on an idea which seems to suggest that, despite Minkowsky, the Angle is variable and cannot exceed 90 deg.
    RC
     
  7. Oct 7, 2018 #6

    Ibix

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    Do be aware of PF rules on personal theories. In summary, take it to a journal - we do not discuss them here.

    The diagram @robphy posted shows a four velocity (diagonal arrow, length ##c##) with its time component (vertical line, length ##\gamma c##) and its spatial component (horizontal line, length ##\gamma v##). And ##c^2=(\gamma c)^2-(\gamma v)^2##. There are a number of other things shown on there relating to other frames, but those are the relevant bits.

    The angle robphy mentions is also called the rapidity, and is a way of representing velocities and gamma factors. It does, of course, vary with velocity.
     
  8. Oct 7, 2018 #7

    robphy

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    I forgot to mention rapidity and draw it in the diagram. It's updated now. Thanks.
     
  9. Oct 7, 2018 #8
    This sounds Brian Greenish.

    Cheers
     
  10. Oct 7, 2018 #9

    PeterDonis

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    Indeed, Brian Greene has used this in at least one of his pop science books. A key red flag is that he has not used it in any actual textbook or peer-reviewed paper.
     
  11. Oct 8, 2018 #10
    I've often seen this pop-sci factoid go something like this:

    "Your speed through spacetime is always ##c##. When you're at rest, your movement is entirely through time. As your speed through space increases, your speed through time decreases so that your speed through spacetime remains ##c##."

    First of all, "speed through time" is word salad, isn't it? But if we have to label something "speed through time," we should consider this equation for the squared norm of the four-velocity:

    ##c^2 = (\gamma c)^2 - (\gamma v)^2##

    Of course, ##v## is speed ("through space"), so maybe ##c## should be "speed through time"? But if so, it doesn't decrease as ##v## increases; it's a constant. Or maybe if we want to call ##\gamma v## "speed through space" (celerity magnitude), then we can call ##\gamma c## "speed through time," but now "speed through time" actually increases as "speed through space" increases, and it is in fact always the greater of the two (because ##c > v##).

    Am I missing something?

    Seems like the point is to convey the idea of "time components," which is fine. But can't this be done without implying that the geometry of spacetime is Euclidean? One can at least emphasize that it's only an analogy, that it isn't quite true.
     
  12. Oct 8, 2018 #11

    Ibix

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    I have to say I've started to think that "four-velocity" is a bit of a misnomer. Velocity in the intuitive sense is coordinate 3-velocity, and the 4-velocity is more like the 4-d extension of a unit direction vector than it is like an extension of 3-velocity.
     
  13. Oct 8, 2018 #12
    I'd point out, though, that while the 4-velocity (in natural units) is indeed always a unit vector, it's also necessarily timelike and future-pointing, whereas unit 4-vectors may be either timelike (future- or past-pointing) or spacelike.
     
  14. Oct 8, 2018 #13

    PeterDonis

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    I would agree with this. Unfortunately misnomers are common in physics. "Relativity" itself is arguably a misnomer, as Einstein often said.
     
  15. Oct 8, 2018 #14

    Matterwave

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    I've always hated this description of things...by this logic is the "speed" of a photon through space time 0? Does my "speed through space time" change if I choose to parameterize time-like curves using something other than the proper time and stop normalizing my 4-velocities to c? What in the heck is speed through time???
     
  16. Oct 8, 2018 #15

    kith

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    We need to divide by ##\gamma^2## in order to have ##v^2## isolated: [tex]\frac{c^2}{\gamma^2} = c^2 - v^2.[/tex] If we rearrange this, we get the equation [tex]c^2 = \frac{c^2}{\gamma^2} + v^2[/tex] which could be written as [tex]c^2 = v_{\text{time}}^2 + v_{\text{space}}^2[/tex]
    where ##0 \leq v_{\text{time}} \leq c##. So there's indeed an equation which "shows" the alleged trade-off.

    Note that the equation can be derived by dividing the equation for the timelike spacetime interval ##(\Delta s)^2 = (c \Delta \tau)^2= (c\Delta t)^2 - (\Delta x)^2## by ##(\Delta t)^2##. From this, we see that [tex]v_{\text{time}} = \frac{c}{\gamma} = \frac{c \Delta \tau}{\Delta t}[/tex] which I think is what's meant by the informal notion of "the speed through time".

    I also haven't seen this explicitly spelled out anywhere. The concept of the "fixed speed through spacetime" originates from Epstein's book "Relativity Visualized" which doesn't contain the formulas. It does contain some numerical examples and special spacetime diagrams which involve proper time instead of coordinate time. The book is carefully argued and he makes the status of the "fixed speed through spacetime" concept clear by calling the chapter where he introduces it "The Myth". The level of the book is elementary; I recommend it to @runningc.
     
    Last edited: Oct 8, 2018
  17. Oct 8, 2018 #16

    I see!

    Thanks for showing that. I think it's a bizarre and unhelpful way to conceptualize the relationship between ##\gamma## and ##\beta## (it treats the frame-dependent ##\Delta x## and the Lorentz-invariant ##c \Delta \tau## as a "pair"), but now at least I understand what the heck "speed through time" is supposed to mean.
     
  18. Oct 9, 2018 #17

    Ibix

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    Yes, although this is partly convention. There's nothing stopping you defining ##\tau'=-\tau## and hence defining a past-directed four-velocity. And there's an analogous concept for spacelike worldlines using proper distance instead of proper time - it's just less physically interesting because nothing travels on them and we wouldn't call it a velocity.
    There's an xkcd cartoon about a guy who invents a time machine and uses it to tell Franklin to label his electrical charges the other way around. I think we need a list...
     
  19. Oct 9, 2018 #18

    martinbn

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    I don't know if this is one reason, but in geometry the tangent vector to a curve is often called the velocity of the curve and its length the speed. Now, worldlines are curves and a convenient parametrization for timelike worldlines is one with tangent vector having length ##c##. So, it makes sense to say that every observer has speed (in the above sense) ##c## in space-time. This will only bring confusion in popular texts, where the typical reader wouldn't have any geometry background.
     
  20. Oct 9, 2018 #19
    Sorry, but I can't see what significance the equation has:
    (a). It simplifies to: c^2= c^2 (b). It only has one variable. i.e. 'v'.
     
  21. Oct 9, 2018 #20

    Ibix

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    The right hand side is an explicit statement of the modulus-squared of the four-velocity. This is indeed ##c^2## whatever the value of ##v##.
     
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