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Lorentz Factor

  1. Jul 13, 2012 #1
    Is it fair to call the lorentz factor the derivative of measured time relative to proper time? I've seen the lorentz factor equated twice now to [itex]\frac{dt}{d\tau}[/itex] and I wanted to know whether that was a legitimate way to look at it.

    Thank you
     
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  3. Jul 14, 2012 #2

    ghwellsjr

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    That's the way it is shown in the wikipedia article on Lorentz Factor, however, the article doesn't explain what t is. It is not a measured time but rather the arbitrarily defined coordinate time of a frame. As such, we usually use the reciprocal of gamma to show the time dilation of a clock moving at some speed v according to a particular previously selected frame. So it isn't generally the Proper Time on a moving clock (unless the clock is inertial) but rather the rate of ticking of the moving clock compared to the rate of ticking of the coordinate time of the frame.
     
  4. Jul 14, 2012 #3
    Sort of what I meant.

    For the latter part though, aren't the two statements equal? Assuming by 'compared to' you mean something equivalent to the rate of change between the two. If I measure object A as moving from point x to point x+dx in time t, and the object measures the time difference as [itex]\tau[/itex], the derivative dt/d[itex]\tau[/itex] would be equal to the lorentz factor performed on the object's instantaneous velocity during that period, no?
     
  5. Jul 14, 2012 #4

    ghwellsjr

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    When you're talking about the motion of the clock in a particular frame, don't you always refer to dx/dt as the speed? Would you ever try to figure out what dt/dx meant? In the same way, wouldn't it make more sense to talk about dτ/dt rather than dt/dτ?
     
  6. Jul 14, 2012 #5
    I don't think so, to the latter part, not the first part.

    [itex]\tau[/itex] is a constant from all points of reference, but t is frame-dependent. Wouldn't it make sense to be deriving with respect to the invariant?
     
  7. Jul 14, 2012 #6

    ghwellsjr

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    Wow, that's a novel way to mix up a definition. According to the wikipedia article on "invariant", it "is a property of a system which remains unchanged under some transformation". Since the tick rate of a clock is what changes when transformed to different frames, we cannot call it an invariant.
     
  8. Jul 14, 2012 #7
    But proper time is an invariant, thats why it's a useful tool. If an object follows a world-line that passes through two events, the time separation between two events that it measures is the proper time.

    It sort of makes sense to me why the derivative of measured time with respect to proper time is useful, it tells you (in a somewhat roundabout way) the instantaneous velocity of the object, but the roundaboutness might be useful in calculations.
     
  9. Jul 16, 2012 #8

    ghwellsjr

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    Yes, proper time is an invariant but I wasn't talking about proper time. I was talking about the tick rate of a clock moving in a frame compared to the tick rate of the coordinate time of the frame which is not an invariant. When you integrate the tick rate with respect to coordinate time over the coordinate time separation between the two events, then you get the invariant proper time. You cannot get the invariant proper time just from the coordinate time separation between the two events unless the clock is inertial but I assume it is not since in your next sentence, you mention instantaneous velocity, implying that it's speed is changing.
    It doesn't make any sense to me, can you explain exactly what you mean by this? You can start by explaining what you mean by "measured time" and how you measure it.
     
  10. Jul 16, 2012 #9
    That confusion was my mistake, I thought you were talking about something different.

    What I meant was that (and I still haven't taken the time to do any of this even semi-rigorously, so this is all just a mental image) the Lorentz factor is a function of velocity (I'm going to drop the relatives because I'll always be talking about an object and a single observer). Time dilation is obviously directly connected to the Lorentz Factor. It makes sense (non-mathematically, so far at least) to me that at any point you could measure the elapsed time and also measure the proper time (this is assuming you know the equations of motion) of the object. From that, as time elapsed becomes infinitesimal, it seems that you could define a ratio that is somewhat akin to saying 'how much faster is my time flowing than his'.

    In my purely mental image, it seems like this ratio would be the derivative [itex]\frac{dt}{d\tau}[/itex]

    Assuming everything I have said so far is correct, which I am prepared to be wrong about, it would then be a matter of reverse-working the dt/d[itex]\tau[/itex] in order to get the instantaneous velocity of the object at that point.

    I guess it's silly to be arguing for this method's usefulness because the confusion is whether this is a legitimate way to represent something that's known to be useful. My only real interest at the beginning was curiosity towards the origins of the lorentz factor in a non historical setting.
     
  11. Jul 16, 2012 #10

    ghwellsjr

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    I'm assuming your single observer is stationary in your selected frame.

    I'm assuming that the object is moving in some yet-to-be-determined way and that it is (or will be) far removed from the observer.

    I'm assuming that the object has an observable clock displaying proper time riding along with it.

    I'm assuming that the observer has an observable clock displaying coordinate time next to him.

    I'm assuming that the observer has to wait for the image of the object's clock to propagate to him at the speed of light.

    I'm assuming that the observer's calculation involves the delayed observation of the proper time on the object's clock and the non-delayed observation of the coordinate time on his own clock.

    If any of my assumptions are wrong, please advise.

    I'm not sure if you are envisioning the object moving around in 3D space or just 2D space or just 1D space--can you please clarify?

    Once we get the scenario ironed out, what calculation does the observer do to determine the instantaneous speed of the object?
     
  12. Jul 16, 2012 #11
    I was envisioning a textbook like scenario where you were given the equations of motion. Remember I was thinking this originally totally as an interesting educational point.

    I think we've gotten a little too literal. My original question was simply to verify the lorentz factor's legitimacy as a derivative, for my piece of mind as well as to further my intuitiveness with the equations of special relativity. I can't think of any reason why you would ever use it for calculations: as it requires derivatives of variables you can avoid entirely using the classical methods.

    I also just realized something else.

    If [itex]\gamma[/itex] is defined as [itex]\frac{dt}{d\tau}[/itex]

    Then the time dilation equation really just says [itex]t=\frac{dt}{d\tau}Δ\tau[/itex]

    Changing the time to infitesimal quantities and making it an integral of all the little changes, the time dilation equation just says [itex]t=\int\frac{dt}{d\tau}d\tau[/itex] from t1 to t2.
     
    Last edited: Jul 16, 2012
  13. Jul 16, 2012 #12

    ghwellsjr

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    This seems to be the exact opposite of what you are now saying.

    If you want to "verify the lorentz factor's legitimacy as a derivative" you should take it as dτ/dt=1/γ=√(1-β2). This means the instantaneous tick rate of a clock moving at the instantaneous speed β compared to the coordinate tick rate is equal to √(1-β2).

    So at β = 0.6, the moving clock is ticking at 80% of normal.

    You could also say that if a clock is ticking at 80% of normal, it has a speed of 0.6c.

    But it seems rather unusual, even weird, to say that if normal time is ticking at 1.25% of a moving clock, then that means the clock is moving at 0.6c.
     
  14. Jul 16, 2012 #13

    ghwellsjr

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    γ is not defined as dt/dτ. It's defined as 1/√(1-β2). See the wikipedia article on Special Relativity, under the section called "Metric and transformations of coordinates".

    The rest of your post doesn't make sense to me. Maybe you could show some examples of what you mean.
     
  15. Jul 16, 2012 #14
    I think we are arguing about different ways of saying the same thing. Directly, whether to say [itex]t'= \gamma t0 or \frac{t'}{\gamma} = t0[/itex]

    I hate LaTeX sometimes.
     
  16. Jul 16, 2012 #15

    ghwellsjr

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    I hate LaTex all the time. It doesn't work on my very old mobile device so I only use it when I quote someone else.

    I think you are saying that t'=γt0 is equivalent to t'/γ=t0 which is true since it is simple algebra (not that I know what it means, since you haven't defined your variables). But up until now you have been talking about a derivative and you can't do the same thing with derivatives. I tried to point this out in post #4.

    But since I can't tell what you are trying to say and you don't want to get precise, I have no way of knowing whether what you are saying is the same as the standard way of presenting and understanding time dilation and the Lorentz factor.

    But why are you pursuing this?
     
  17. Jul 16, 2012 #16
    I completely agree with you. I still have a math related question lingering but its small and rather tangential so I think I'll be able to answer it myself.

    You answered my original question but I got carried away in the details, so I'm sorry for that.

    Thank you for your help.
     
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