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Proper time

  1. Feb 24, 2010 #1
    A very basic question, perhaps, but I am starting from basics and checking all my understanding.

    In Relativity is τ (tau), the proper time experienced by an observer adjacent to a clock in an inertial frame of reference, an invariant quantity?
    And if not, in what way can it vary?
     
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  3. Feb 24, 2010 #2

    JesseM

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    Re: τ

    Yes, the proper time between any two events on an observer's (or clock's) worldline is frame-invariant. It does not even need to be an inertial observer, proper time along the worldline of a non-inertial observer is invariant too--after all, there must be a definite truth about how much he has aged between any pair of events on his worldline (say, the event of his leaving Earth and the event of him returning), you can't have one frame predicting that at the second event he'll be young and another frame predicting he'll be an old man, since all frames must agree in their predictions about localized events which occur at a single point in spacetime.
     
  4. Feb 24, 2010 #3

    Meir Achuz

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    Re: τ

    In any frame, [tex]\tau^2=t^2-x^2[/tex], and is invariant Even if x and t change, tau doesn't. If you are adjacent to the clock so x=0, then tau=t.
    x, t, and tau usually stand for time differences here.
     
  5. Feb 24, 2010 #4

    Fredrik

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    Re: τ

    I realize that this is more than the OP is asking for, but I hope someone finds it useful.

    I prefer to define proper time as a property of a smooth timelike curve. One of the axioms of both SR and GR is that a clock measures the proper time of the curve in spacetime that represents its motion. If we define proper time as what a clock measures, then there's no elegant way to state that axiom.

    Let's say that the curve is [itex]C:[a,b]\rightarrow M[/itex], where M is the spacetime manifold. Then the proper time of that curve is defined as

    [tex]\tau(C)=\int_a^b\sqrt{-g_{C(t)}(\dot C(t),\dot C(t))}dt[/tex]

    That C is timelike just means that the thing under the square root is positive for all t. The definition doesn't use any coordinate systems, and is also independent of parameterization. What this means is that if [itex]\phi:[a,b]\rightarrow [c,d][/itex] is a smooth strictly increasing function and [itex]C=B\circ\phi[/itex], then [itex]\tau(B)=\tau(C)[/itex].

    The definition of proper time can of course also be expressed using a coordinate system [itex]x:U\rightarrow\mathbb R^4[/itex], where U is an open subset of M. The components of the velocity vectors in x are

    [tex]\dot C(t)^\mu=\dot C(t)x^\mu=(x^\mu\circ C)'(t)[/tex]

    so if we define [itex]C^\mu=x^\mu\circ C[/itex], then

    [tex]\dot C(t)^\mu=C^\mu'(t)=\dot C^\mu(t)[/tex]

    The last step simply defines what we mean by [tex]\dot C^\mu(t)[/tex]. We have

    [tex]g_{C(t)}(\dot C(t),\dot C(t))=g_{C(t)}(\partial_\mu,\partial_\nu)\dot C^\mu(t)\dot C^\nu(t)=g_{\mu\nu}(C(t))\dot C^\mu(t)\dot C^\nu(t)[/tex]

    and therefore

    [tex]\tau(C)=\int_a^b\sqrt{-g_{\mu\nu}(C(t))\dot C^\mu(t)\dot C^\nu(t)}dt[/tex]

    If we choose (M,g) to be Minkowski spacetime, we can choose the coordinate system to be a global inertial frame. Minkowski spacetime is the set [itex]\mathbb R^4[/itex] equipped with a manifold structure and a metric. We can identify the global inertial frames with functions of the form

    [tex]\mathbb R^4\ni x\mapsto\Lambda x+a\in\mathbb R^4[/itex]

    where [itex]\Lambda[/itex] is a linear operator that satisfies [itex]\Lambda^T\eta\Lambda=\eta[/itex], where

    [tex]\eta=\begin{pmatrix}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}[/tex]

    (I'm identifying linear operators the corresponding matrices in the standard basis. See this post if you need to understand that better). One of those functions is the identity map I, defined by I(x)=x for all x in [itex]\mathbb R ^4[/itex]. So we can take the coordinate system x mentioned earlier to be the identity map. I hope I'm not causing too much confusion by renaming the coordinate system to I, and the curve to x

    Note that the definition [itex]C^\mu=x^\mu\circ C[/itex] introduced earlier now turns into [itex]x^\mu=I^\mu\circ x[/itex], and that the function on the right is just what we would have called [itex]x^\mu[/itex] anyway (even without the earlier definition of [itex]C^\mu[/itex]).

    The components of the metric in this coordinate system are [itex]g_{\mu\nu}(x(t))=\eta_{\mu\nu}[/itex] (for all t), so

    [tex]\tau(x)=\int_a^b\sqrt{-\eta_{\mu\nu}\dot x^\mu(t)\dot x^\nu(t)}dt=\int_a^b\sqrt{-\eta_{\mu\nu}\frac{dx^\mu(t)}{dt}\frac{dx^\nu(t)}{dt}}dt[/tex]

    This is often expressed as

    [tex]\tau=\int_a^b\sqrt{-\eta_{\mu\nu}dx^\mu dx^\nu}=\int_a^b\sqrt{dt^2-dx^2-dy^2-dz^2}[/tex]
     
    Last edited: Feb 24, 2010
  6. Feb 25, 2010 #5
    Re: τ

    Fredrik, thank you and Wow!
    I think I followed that as far as paragraph two but as you so rightly surmise the rest is beyond me. :rolleyes:

    And thank you Meir and Jesse.

    Now if I am understanding this, would it be correct to say:

    τ is the timelike space-time interval for a particular clock, not the frame, because the frame only provides the coordinate system, while the clock is present at an event that has coordinates that can be specified within that frame?

    So how does Proper time relate to τ?

    Is proper time the rate(?) at which time passes within a frame as measured from within that frame?
    Or is proper time the measure of time passing within a frame, measured from within that frame?

    And is coordinate time that which is measured from another frame, from which, using LT, the proper time for the subject frame can be calculated?
     
  7. Feb 25, 2010 #6
    Re: τ

    I don't quite understand it,but don't you think you neglect a 'c' before dt?
    [tex]\[
    \sqrt {\left( {dt} \right)^2 - \left( {dx} \right)^2 - \left( {dy} \right)^2 - \left( {dz} \right)^2 }
    \]
    [/tex]
    is definitely a wrong expression since dt and dx have different dimensions,the former [T],the latter [L]

    And it seems that the formula you guys give is the interval in Minkoswi space-time, not the proper time
     
  8. Feb 25, 2010 #7

    JesseM

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    Re: τ

    In the usual notation tau is nothing more than the proper time parameter rather than a spacetime interval, but there may be cases where it's used differently--are you looking at a particular book or website? Anyway, if a clock is moving inertially, then the timelike spacetime interval between two points on its worldline will be equal to the proper time the clock experiences between those points, so it's really only when dealing with clocks moving non-inertially that you have to worry about the difference between the two.
    Proper time is just the amount of time elapsed on a physical clock. If a clock is moving inertially, then the proper time between two events on its worldline is the same as the coordinate time between those events in the clock's own rest frame (remember, coordinate time in an inertial frame is defined in terms of the readings on a set of clocks at rest in that frame, so if the clock whose proper time you're interested is also at rest in some frame then it'll be at rest right next to one of these coordinate clocks, so naturally both keep time with one another). Likewise, if a clock is moving inertially, then in a frame where the clock is moving at velocity v, if the coordinate time between two events on its worldline is t then the proper time the clock experiences between those events is t*squareroot(1 - v2/c2), that's the physical meaning of the time dilation equation. But again, proper time is more general than either of these descriptions since you can talk about proper time for a non-inertial clock too.

    In the case of a non-inertial clock, if you know its velocity as a function of time v(t) in an inertial frame, and you know the times t0 and t1 of the two events on its worldline, then what you basically do is break up its non-inertial worldline into a bunch of segments lasting an infinitesimal time dt in that frame and treat the clock as if it was moving inertially during each segment, so the proper time it accumulates on each segment is dt*squareroot(1 - v(t)2/c2)...then you can integrate over all the segments to get the total proper time [tex]\int_{t_0}^{t_1} \sqrt{1 - v(t)^2/c^2} \, dt[/tex]
     
    Last edited: Feb 25, 2010
  9. Feb 25, 2010 #8

    Fredrik

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    Re: τ

    I realize this overlaps with Jesse's answer, but I was already almost finished when I saw his post.

    I'm afraid the most general definition can only be stated in the language of differential geometry, but if we restrict ourselves to Minkowski spacetime, things get easier. This is a definition that you will understand: We define Minkowski spacetime as the set [tex]M=\mathbb R^4[/tex] equipped with the usual vector space structure and a bilinear form [itex]g:M\times M\rightarrow\mathbb R[/itex], defined by

    [tex]g(x,y)=x^T\eta y[/tex]

    The x and the y on the left are vectors in M, and the x and the y on the right are the 4×1 matrices of components of those vectors in the standard basis. This definition isn't manifestly coordinate independent, because we're referring to a basis, and a basis defines a coordinate system. But what's important here is that the quantity on the left is coordinate independent. (We could have have used a different basis on the right to define the same thing).

    If we use the convention to write the row and column indices of [itex]\eta[/itex] as subscripts, and the row indices of x and y as superscripts, then the definition of matrix multiplication tells us that the right-hand side can be expressed as

    [tex]\eta_{\mu\nu}x^\mu y^\nu[/tex]

    Now consider a curve

    [tex]x:[a,b]\rightarrow M[/tex]

    The proper time of that curve is defined as

    [tex]\tau(x)=\int_a^b\sqrt{-g(x'(t),x'(t))}dt[/tex]

    where x' is the derivative of x. And if we again use the same notation for a member of M and its matrix of components in the standard basis, we have

    [tex]\tau(x)=\int_a^b\sqrt{-x'(t)^T\eta x'(t)}dt=\int_a^b\sqrt{-\eta_{\mu\nu}x^\mu'(t)x^\nu'(t)}dt=\int_a^b\sqrt{dt^2-dx^2-dy^2-dz^2}[/tex]

    I should probably mention that even though proper time is a property of a curve, it's quite uncommon to use a notation like [itex]\tau(x)[/itex] that shows that explicitly. The name of the curve is usually suppressed. You will usually see something like this instead:

    [tex]\tau=\int_{t_0}^t\sqrt{dt^2-dx^2}=\int_{t_0}^t\sqrt{1-\left(\frac{dx}{dt}\right)^2}dt=\int_{t_0}^t\sqrt{1-v^2}dt=\int_{t_0}^t\frac{1}{\gamma}dt[/tex]

    [tex]\frac{d\tau}{dt}=\frac{1}{\gamma}[/tex]

    Proper time is a property of a curve, defined by that integral. The symbol used for proper time is [itex]\tau[/itex]. A clock measures the proper time of the curve in spacetime that represents its motion. If you prefer to call the property of the curve "[itex]\tau[/itex]" and define proper time be "what clocks measure", then the axiom turns into "proper time=[itex]\tau[/itex]", but now it's camouflaged to look like another definition instead of like an axiom.

    I'm not sure what either of those statements mean, or if there's a difference between them, but a clock that's stationary in an inertial frame will display the coordinate time of every event on its world line (plus some constant, if it wasn't set to agree with coordinate time at one of those events).
     
    Last edited: Feb 25, 2010
  10. Feb 25, 2010 #9

    Fredrik

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    Re: τ

    I'm using units such that c=1. I highly recommend it. It's really annoying to have to keep track of the c's, and as you noticed yourself, it's quite easy to restore them in the final result if you need them there.
     
  11. Feb 26, 2010 #10
    Re: τ

    The spacetime interval (dS) in terms of space can be written as:

    [tex]dS = c\ d\tau = \[
    \sqrt {\left( {c\ dt} \right)^2 - \left( {dx} \right)^2 - \left( {dy} \right)^2 - \left( {dz} \right)^2 }
    \]
    [/tex]

    Dividing both sides by c give the invariant interval in terms of proper time:

    [tex] d\tau = \[
    \sqrt {\left( {dt} \right)^2 - \left( {dx/c} \right)^2 - \left( {dy/c} \right)^2 - \left( {dz/c} \right)^2 }
    \]
    [/tex]

    and using c=1 gives the shorthand version of proper time given by Fredrik.
     
  12. Feb 26, 2010 #11
    Re: τ

    OK,I understand
     
  13. Mar 2, 2010 #12
    Thank you Gentlemen, Thank you.

    And I am sorry Frederik but I am not a mathematician and you lost me once again in your first statement!

    But you have answered my, rather badly expressed, question; and I now see how these different terms are related.
     
  14. Mar 19, 2010 #13
    Having taken my time and pondered what you have said, I believe I would be right to say that If one says that there is a stationary observer, with a standard clock at the origin of the coordinates of an Inertial Frame of Reference, then the time scale would be proper time.
    And that any distances measured between stationary clocks in that frame would be measured as proper distances.
     
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