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where the proper time is invariant why change (differential) in proper time $d\tau$ is not zero? $\Delta \tau=\tau_f-\tau_i$ as i know. $d(invariant)=0$ note to comment: action $S=-m_oc^2\int_C d\tau$

Invariance always applies between some set of conditions. Things can be invariant in time, invariant over changes in position or over changes of coordinate systems and so on. In relativity things described as "invariant" without a descriptive clause are general things that all observers can agree upon. The claim that "proper time is invariant", means that the proper time between two events (four points in space time) is something that all observers can agree upon, not that the measured proper time is zero.

something that all observers can agree upon like speed of light but this does not means that the measured in speed of light is not zero $dc\not= 0$,

ok guys i got the point: invariance is not a synonym for being a constant. thanks

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Hello antoni, I think you should formulate your question better. The differential in $d\tau$ probabily doesn't mean variation along the transformation under which you declare invariance. –  Bzazz Jan 21 '13 at 22:44
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Invariance always applies between some set of conditions. Things can be invariant in time, invariant over changes in position or over changes of coordinate systems and so on.

In relativity things described as "invariant" without a descriptive clause are general things that all observers can agree upon. The claim that "proper time is invariant", means that the proper time along a particular path between two events (four points in space time) is something that all observers can agree upon, not that the measured proper time is zero.

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Not quite: the proper time along a particular space-time path is invariant. The proper time is path dependent - this is the resolution of the twin paradox! –  Michael Brown Jan 22 '13 at 0:38
    
@MichaelBrown Uhm...yes. That's roughly what I was visualizing, but I have clearly writen in incorrectly. You words are exactly what was required to repait it. Thanks. –  dmckee Jan 22 '13 at 0:47
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Invariance is not a synonym for being constant. The differential of a constant is zero, but a number is an invariant if it is constant with respect to change of reference frame. So, if I have a particle attached to a harmonic oscillator, and I undergo a galilean transformation, I'll see that the particle's acceleration remains unchanged. Therefore, the particle's acceleration is an invariant with respect to Galilean transformations. However, over time, the acceleration changes, so even though acceleration is an invariant in this case it changes over time.

The proper time $\tau$ is an invariant with respect to Lorentz transformations in special relativity, because the proper time separating two spacetime events (also given a path between them) is the same regardless of whatever Lorentz transformation of coordinates you do. However, in some other reference frame with time coordinate $t$, $\frac{d\tau}{dt}$ may be nonzero. The usual case you can look at would be when a particle with proper time $\tau$ and velocity $v$ (with respect to your reference frame with time coordinate $t$ and velocity $0$), and you can figure out that $\frac{d\tau}{dt}=\frac{1}{\sqrt{1-v^2/c^2}}=\gamma$, giving you the formula for time dilation.

See also: http://en.wikipedia.org/wiki/Proper_time#Derivation

(Disclaimer: Not an expert on this stuff! And, I'm only referring to special relativity, because that's all I've studied.)

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In special relativity, the proper time between events $(t_1, \mathbf{x}_1)$ and $(t_2, \mathbf x_2)$ is given by

$\Delta\tau_{12}^2 = -(t_2-t_1)^2 + (\mathbf x_2-\mathbf x_1)^2$.

This quantity is has the same value as measured by any two inertial observers. To prove this, we note that the coordinates of these events as measured by a different inertial observer differ by a Poincare transformation, and such a transformation leaves the proper time difference $\Delta\tau_{12}$ the same (I'll leave the proof of this fact to you). This is true whether or not $\Delta\tau_{12}=0$.

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