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Time with no gravitational force

  1. Sep 24, 2011 #1
    Hello everyone,

    I was doing my Physics homework and a question arose from it: if a mass affects time there is a difference in the time measured near different masses. However if there was no mass at all?
    Let's pretend this is possible just for theoretical purposes, a perfectly flat space-time. In this situation time would not pass at all.
    So how can light travel through such regions? (I know there is not such a thing in the Universe but intergalactical space may have very little space-time distortion.)

    Just to confirm if my understanding is correct: time would pass slower near the Sun than near the Earth? (imagine they are both static, they do not have either translation or rotation)

    Thank you very much!
     
  2. jcsd
  3. Sep 24, 2011 #2

    WannabeNewton

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    Why do you say time will not pass at all? [itex]ds^{2} = -dt^{2} + dx^{2} + dy^{2} + dz^{2} [/itex] and [itex]\tau = \int_{\gamma }(-g_{\mu \nu }dx^{\mu }dx^{\nu })^{1/2}[/itex] so in the rest frame of a test particle [itex]\Delta \tau = \int_{t_{i}}^{t_{f}}dt = \Delta t[/itex] so there is a passage of proper time.
     
  4. Sep 24, 2011 #3
    1. This should go in the Homework section.
    2. Yes, time does pass slower near the Sun than near the Earth(however, both amounts of time dilation are so small they are irrelevant, we aren't near neutron stars or black holes here)

    I really do not know how time would pass in such a region. That is really beyond me. There are plenty of people here who know more about this stuff than I do.
     
  5. Sep 24, 2011 #4
    Hi WannabeNewton,

    I couldn't understand perfectly the equation you posted but I got the idea. So actually my reasoning was misleading, time would be pass extremely fast in a space-time without gravity.
    Is this difference in time taken into account when calculating distances in light-years?

    xeryx35, this question was not from my homework, actually the way I wrote made it confusing.

    Thanks!
     
  6. Sep 24, 2011 #5
    When you say extremely fast, it's somewhat meaningless. Anyone can travel through time, its given by your temporal velocity which comes from the velocity 4-vector, [itex]u_{\alpha}=\langle u_0, \vec{u} \rangle [/itex]. Here we can look at [itex]u_0[/itex], which equals [tex]u_0=\gamma c=\frac{c}{\sqrt{1-\frac{v^2}{c^2}}}[/tex] where [itex]v^2[/itex] is the magnitude of the 3-velocity. This can also be written
    [tex]
    \gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{dt}{d\tau}
    [/tex]
    where [itex]\tau[/itex] is the proper time. This says you can move through time at a maximum speed of 1 sec/sec, which is essentially what you are experiencing now relative to the rest of us and everyday experiences. But as you seemed to have picked up, this depends on the gravitational field one is immersed in, so in flat space, you are moving through time at essentially the rate you are moving through it right now since the curvature is practically negligible.
     
  7. Sep 24, 2011 #6
    Flat spacetime is a gravitational field. its commonly expressed as [itex]\eta_{\alpha\beta}[/itex]. In Einstein's relativity, there are sill solutions to the field equations that determine the dynamics of gravity even in the absence of sources of gravitation. This is similar to Maxwell's equations, for instance [itex]\vec{\nabla}\cdot\vec{E}=\rho /\epsilon_0[/itex] still has solutions even when [itex]\rho=0[/itex], that is, there are "free-field solutions". In fact there are many solutions that satisfy the vacuum equations, for instance, the solution describing the spacetime around our star, Sun. I hope this puts some things in order for you.
     
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