# If you were out in space, motionless

by taybot
Tags: motionless, space
 P: 26 If you were in a spaceship going the speed of light you wouldn't travel through time because you would be traveling through space instead. I got that. So if you were to sit out in space completely free from gravity, motionless, and you were wearing a watch, does that mean time would go faster? Since you aren't traveling through space at all, wouldn't that mean you are traveling through time at light speed and wouldn't that mean your watch would go faster?
Mentor
P: 17,202
 Quote by taybot If you were in a spaceship going the speed of light you wouldn't travel through time because you would be traveling through space instead.
No, you wouldn't be going the speed of light because you have mass.

 Quote by taybot So if you were to sit out in space completely free from gravity, motionless, and you were wearing a watch, does that mean time would go faster? Since you aren't traveling through space at all, wouldn't that mean you are traveling through time at light speed and wouldn't that mean your watch would go faster?
In in any inertial frame the rate of any watch's proper time wrt coordinate time would be given by:
$$\frac{d\tau}{dt}=\frac{1}{\gamma}=\sqrt{1-\frac{v^2}{c^2}}$$

This expression indeed achieves a maximum of 1 for v=0.
 P: 26 not sure what that formula means but since you wrote indeed, I take it that the watch will tick faster if you are out in space, motionless... So how much faster? I figure it wouldn't be that much faster than normal time that passes for us, right? Because already if you are sitting somewhere, you're accelerating through space, just not accelerating that much I figure, so probably your watch wouldn't tick that much faster if you were out in space, motionless. Is that right?
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P: 17,202
If you were out in space, motionless

 Quote by taybot not sure what that formula means ... So how much faster?
Sorry, I don't know I can answer you in a satisfactory manner then. The answer to "how much faster" is given by the formula.
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 not sure what that formula means but since you wrote indeed, I take it that the watch will tick faster if you are out in space, motionless...
What do you mean by motionless ? I think you are failing to take into account that phenomena like clocks slowing refer to observations from different frames of reference. You would not notice anything as long as your motion is uniform ( unaccelerated).

But someone travelling at uniform velocity relative to you would notice your clock apparently slowing.
 Sci Advisor P: 2,795 You can never be motionless with respect to every reference frame. In some reference frame you are motionless, and in others you are moving. I don't think your question can obtain a satisfactory answer because one of your assumptions is wrong.
 P: 26 I read this book, The Fabric of the Cosmos and he says you can be motionless if you jump into a shaft that has air removed from it. Even if you can't be motionless, you would be falling, pretty much motionless. Closer to motionless than when you are sitting in front of your computer, because if you're doing that, you're accelerating. Anyways, let's say you jump into that airless shaft and you are falling, motionless, or almost motionless. And if you're wearing a watch, would it tick faster, or wouldn't it? Edit: I looked at Dale's answer again and he says: "The answer to "how much faster" is given by the formula." So at least I know your watch would indeed tick faster. I was just wondering if it would tick twice as fast, or less. I don't know how to do that equation. Thanks!
P: 3,967
Quote by DaleSpam
In in any inertial frame the rate of any watch's proper time wrt coordinate time would be given by:
$$\frac{d\tau}{dt}=\frac{1}{\gamma}=\sqrt{1-\frac{v^2}{c^2}}$$

This expression indeed achieves a maximum of 1 for v=0.

 Quote by taybot ... I looked at Dale's answer again and he says: "The answer to "how much faster" is given by the formula." So at least I know your watch would indeed tick faster. I was just wondering if it would tick twice as fast, or less. I don't know how to do that equation. Thanks!
The watch on your wrist always appears to tick at the same rate whether you are motionless far out in space or on the surface of a neutron star or in a rocket travelling at 0.999c relative to the Earth. It is the rate of other clocks moving relative to you or located at different heights in a gravitational field that appear to be running faster or slower than your own watch.

Using the equation given by Dale, the rate of a clock moving at at 0.866c relative to you is:

$$\frac{d\tau}{dt}=\frac{1}{\gamma}=\sqrt{1-\frac{v^2}{c^2}} = \sqrt{1-0.866^2} \approx 1/2$$

In other words the clock moving at 0.866c relative to you appears to be ticking twice as slow as your watch, but this factor of 2 only happens at this precise velocity relative to you. If the other clock was moving at 0.99498c relative to you, then the other clock would appear to be running 10 times slower than your own watch because:

$$\frac{d\tau}{dt}=\frac{1}{\gamma}=\sqrt{1-\frac{v^2}{c^2}} = \sqrt{1-0.99498^2} \approx 1/10$$

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