How Does Time Dilation Affect Clocks at Different Speeds?

[MathJakob]

Using this equation

$$t\prime=t\sqrt{1-\frac{v^2}{c^2}}$$

If you were on a train circling the earth

How much would time slow down relative to the people on outside if you were going 50% the speed of light?

 

[mathman]

You can easily calculate it your self. (√3)/2

 

[ghwellsjr]

Using this equation

$$t\prime=t\sqrt{1-\frac{v^2}{c^2}}$$

If you were on a train circling the earth

How much would time slow down relative to the people on outside if you were going 50% the speed of light?

Not counting any effects from gravity, it would be √(1-0.5²) = √(1-0.25) = √(0.75) = 0.866

 

[MathJakob]

0.866

Yes this is what I thought but how is it read? 0.866% slower?

Because when if I use 90% of the speed of light wolfram says 0.435% so something isn't right.

 

[Bill_K]

Yes this is what I thought but how is it read? 0.866% slower?

Because when if I use 90% of the speed of light wolfram says 0.435% so something isn't right.

I miss the days when people could do simple arithmetic without the aid of Mathematica.

 

[DiracPool]

Not counting any effects from gravity, it would be √(1-0.5²) = √(1-0.25) = √(0.75) = 0.866

In addition to gravity, wouldn't there also be a factor of centripetal acceleration to be accounted for in moving around the earth?

 

[ghwellsjr]

Yes this is what I thought but how is it read? 0.866% slower?

Because when if I use 90% of the speed of light wolfram says 0.435% so something isn't right.

No, 0.866 is 86.6%. That means that a clock traveling at 0.5c (one-half the speed of light) according to an Inertial Reference Frame will tick more slowly than than the time of the IRF which is the same everywhere. So that means that if 1000 seconds of Coordinate Time transpires, then the clock will advance by 866 seconds. We are not comparing a moving clock to a stationary clock but rather to the time of the reference frame.

 

[ghwellsjr]

In addition to gravity, wouldn't there also be a factor of centripetal acceleration to be accounted for in moving around the earth?

Acceleration is not a factor in Special Relativity for determining Time Dilation, only the instantaneous speed. If the acceleration results only in a change in direction and not speed as is the case when going around the earth, the it has no effect on the Time Dilation.

 

[MathJakob]

No, 0.866 is 86.6%. That means that a clock traveling at 0.5c (one-half the speed of light) according to an Inertial Reference Frame will tick more slowly than than the time of the IRF which is the same everywhere. So that means that if 1000 seconds of Coordinate Time transpires, then the clock will advance by 866 seconds. We are not comparing a moving clock to a stationary clock but rather to the time of the reference frame.

But how? If 0.866 = 86.6% @ half the speed of light then how comes 90% the speed of light is 0.435 = 43.5% slower.

I thought the faster you went the slower time becomes relative to something else.

 

[ghwellsjr]

But how? If 0.866 = 86.6% @ half the speed of light then how comes 90% the speed of light is 0.435 = 43.5% slower.

I thought the faster you went the slower time becomes relative to something else.

If you had a clock that only advanced by 866 seconds in 1000 seconds, wouldn't you say it was slow?

 

[WannabeNewton]

In addition to gravity, wouldn't there also be a factor of centripetal acceleration to be accounted for in moving around the earth?

If we have a trajectory $x^{\mu}(\tau)$ in special relativity that describes a particle in an arbitrary accelerated motion and we are in a prescribed lab frame $O$ (in this case the Earth), then the notion of Lorentz boosting in this context means that at each instant $\tau$ we find an inertial frame $\bar{O}(\tau)$ that is instantaneously at rest with respect to the accelerating particle (i.e. an instantaneous rest frame, also called a momentarily comoving inertial frame) and Lorentz boost from $O$ to $\bar{O}(\tau)$ with velocity $\mathbf{v}(\tau)$ in the usual normal way.

As George mentioned, in this case we have a particle in uniform circular motion about the Earth (which is our lab frame) so the time dilation factor will remain constant throughout the process of applying a boost from the Earth (lab) frame to the instantaneous rest frame of the particle at each instant because the time dilation factor depends on the magnitude of the boost velocity which is constant for uniform circular motion.

 

[DiracPool]

Acceleration is not a factor in Special Relativity for determining Time Dilation, only the instantaneous speed. If the acceleration results only in a change in direction and not speed as is the case when going around the earth, the it has no effect on the Time Dilation.

Ok, forget centripetal acceleration, how about the centri-"fugal" acceleration/force on the train? This leads to a broader question. We send a ship to Mars with a rotating cabin in order to provide virtual gravity to the astronauts. Do these astronauts experience time dilation due to the virtual/artificial gravity as someone standing on the surface of the Earth would? Via the equivalence principle? Similarly, a train racing around the Earth at .5c...it would seem as if it would experience some significant centrifugal force. Am I off here? Of course, the gravity and rotation of the Earth would offset that some, but how much? Does anyone know how to set up an equation to determine that?

 

[Raze]

Yes this is what I thought but how is it read? 0.866% slower?

Because when if I use 90% of the speed of light wolfram says 0.435% so something isn't right.

Clocks "moving slower" means that the number of seconds the person at rest with respect to the clock reads will be less than the number of seconds the person who watches the clock move around the earth. The faster the person with the clock moves around the world, the LESS time that person will experience.

So, if the person on Earth experiences 10 seconds, then the person traveling at 0.5c will experience 10*0.866 seconds = 8.66 seconds. If the person is traveling at 0.9c, then while the person standing on Earth experiences 10 seconds, the person traveling will experience only 10*0.435 seconds = 4.35 seconds. Does that make any sense to you? The gamma factor is just the factor you use for transforming the time coordinate between the two coordinate systems.

If you want to know the percentage that a clock slows down, you have to define how you're measuring the speed of the clock. What does that even mean? Are these in units of /?Isn't that unit-less? So which one goes on top and which on bottom in that fraction?

Here is a better number to consider, I think:

(10 - .866)/10 = .9134 = 91.34% slower versus (10 - .435)/10 = .9565 = 95.65% slower. (whatever that means)So, (t - t')/t, rather than just t'/t.In the case of 0.5c, the clock on the spaceship travels 8.66 seconds for every 10 seconds the clock on Earth travels. For 0.9c, the clock on the spaceship travels 4.35 seconds for every 10 seconds the clock on Earth travels. Clearly the second one is the slower of the two moving clocks.

*I am not a physicist and am basing this response merely on what makes sense to me. Take it as you will.