Time dilation (speed Vs. acceleration)

[Crowxe]

There's a common example to explain time dilation that states ...
"if you traveled at a speed close to the speed of light and came back in a round trip of 1 hour , you may find that on the departure frame of reference 2 hours had passed"

but that 2 hours trip can't be done without accelerating and decelerating, so if the whole trip was half an hour acceleration , half hour deceleration then the same in the opposite direction. will there still be time dilation ?

 

[vanhees71]

The time of a comoving clock is the proper time, i.e., for any world line
$$\tau=\int_{\lambda_1}^{\lambda_2} \mathrm{d} \lambda \sqrt{\eta_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}},$$
where $\lambda$ is an arbitrary world-line parameter, $\dot{x}^{\mu}=\mathrm{d} x^{\mu}/\mathrm{d} \lambda$, and the world line must be timelike, i.e., $\eta_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}$.

The definition, of course doesn't depend on the choice of $\lambda$. So particularly you can take $\lambda=t$, where $t$ is the coordinate time of an arbitrary inertial reference frame. Then the general formula boils down to
$$\tau=\int_{t_1}^{t_2} \mathrm{d} t \sqrt{1-\vec{v}^2}=\int_{t_1}^{t_2} \mathrm{d} t \frac{1}{\gamma}.$$
As you see, there's no difference in whether you are moving with constant velocity wrt. the reference frame or whether you are accelerated. The proper time $\tau$ is always given by this integral, and $\tau \leq (t_2-t_1)$ in any case, i.e., the moving observer always ages less compared to the situation where he stays at rest.

 

[Crowxe]

As you see, there's no difference in whether you are moving with constant velocity wrt. the reference frame or whether you are accelerated. The proper time $\tau$ is always given by this integral, and $\tau \leq (t_2-t_1)$ in any case, i.e., the moving observer always ages less compared to the situation where he stays at rest.[/QUOTE]

i presumed that acceleration may have the same effect of gravity (more gravity , time slows down) ! thank you for explaining and clearing that up

 

[Ibix]

so if the whole trip was half an hour acceleration , half hour deceleration then the same in the opposite direction. will there still be time dilation ?

Yes, although what you are describing is differential aging rather than time dilation. The effect will likely be weaker because you spent some of the time traveling slowly.

 

[Ibix]

i presumed that acceleration may have the same effect of gravity (more gravity , time slows down) ! thank you for explaining and clearing that up

It's only the velocity relative to the stay-at-home clock that matters. As you can see from vanhees71's post, acceleration does not appear in the maths.

 

[Crowxe]

Yes, although what you are describing is differential aging rather than time dilation. The effect will likely be weaker because you spent some of the time traveling slowly.

oops i

It's only the velocity relative to the stay-at-home clock that matters. As you can see from vanhees71's post, acceleration does not appear in the maths.

yeah , i had that misconception for long. even with using the average speed of the trip mentioned above.
the question popped in my head when i knew that GPS satellite are calibrated for less gravity due to the altitude although the experience zero gravity

 

[Ibix]

In a gravitational field what matters is the gravitational potential, not the acceleration. So distance from the centre of the Earth is the only thing that matters.

 

[Janus]

Just too clarify on what Ibex said. Assume you have a clock sitting at the surface of the Earth, and one 100 km above it, and stationary wit respect to the first clock. Neither clock travels with the rotating Earth (so we only have to concern ourselves with the gravitational time dilation).
Now consider a clock sitting at 2 Earth radii from a planet 4 times as dense as the Earth and a clock 100 km further away from the same planet.

The lower clock in both scenarios are experiencing 1g of acceleration the higher two clocks each experience a slightly lower acceleration. But the difference in acceleration between the first pair of clocks will be great than that for the second pair. However, the gravitational potential difference between the first pair is less than that between the second pair and so will the time dilation.
We can imagine a using a larger and larger mass for the planet while continuing to move the pair of clocks further away so that the lower clock always is at 1g. As you do so, the difference in acceleration felt by the two clocks will decrease and approach zero, but the potential difference and time dilation difference will increase.