1
$\begingroup$

I'm getting into some confusion with relativity.

Consider a rocket and the Earth as two observers, and then the rocket travels to a distant star. Is the proper time taken measured in the frame of the rocket or the Earth?

As far as I know the rocket is at rest relative to the event, and the Earth is moving at a velocity $v$ relative to the event, so maybe I'm just getting confused about the definitions of proper time and dilated time.

If the Earth measures the dilated time then is the time it measures given by $t=t_0\gamma$, with $t_0$ being the proper time? And the proper time would be the time as measured by the rocket, so simply $t_0=d/v$?

$\endgroup$
  • $\begingroup$ I think you're misunderstanding, because that would contradict everything I've been taught in the last few weeks. $\endgroup$ – ODP Dec 8 '14 at 18:30
  • $\begingroup$ Are you asking about the twin paradox? $\endgroup$ – RedGrittyBrick Dec 8 '14 at 18:32
  • $\begingroup$ No, I have a homework question giving the speed a rocket travels and and the distance, and the question is asking what time the Earth measures it to take. So my question is whether the Earth measures t=d/v or a dilated time. $\endgroup$ – ODP Dec 8 '14 at 18:41
  • $\begingroup$ @RedGrittyBrick - proper time is a standard technical term in relativity, it's always defined with respect to events on a particular worldline, and it just means the time that would be measured between the events by an ideal clock which has the same worldline. There is also a notion of "proper distance" along spacelike worldlines, although no physical object could have a spacelike worldine unless there were particles that moved faster than light ('tachyons'). $\endgroup$ – Hypnosifl Dec 8 '14 at 21:21
  • $\begingroup$ Is the proper time taken measured in the frame of the rocket or the Earth? -- The answer is "yes." This question is confused in thinking that there is one special measurement that qualifies as "proper time". $\endgroup$ – David Hammen Dec 8 '14 at 23:07
4
$\begingroup$

I think you're kind of misunderstanding the notion of what proper time and dilated time is.

Special relativity, at base, is a theory of four dimensional spacetime. It is always clearest when you think of things in terms of different points in spacetime, and paths between those points.

So, if you have your spaceship travelling from earth to mars, you can think of three spacetime events -- the rocket leaving earth (point O), the rocket arriving at mars (point M), and the time when the people on Earth observe the rocket as arriving at mars (let them subtract out the light travel time) (point E).

There are two paths through spacetime here, the path taken by the rocket (OM), and the path taken by the people on Earth (OE). Each of these paths have a proper time associated with them. They are both "proper" times. The point, though, is that they won't be the SAME time interval -- they will be dilated RELATIVE to each other.

$\endgroup$
  • $\begingroup$ I agree that proper time is always defined with respect to worldlines and there can be different proper times between the same pair of events along different worldlines. But in this case if the two events are "rocket leaves Earth" and "rocket arrives at distant star", only one object in the problem (the rocket) has both events on its worldline--you can't really talk about the proper time on Earth between these events, only the coordinate time (or equivalently the proper time between the event of the rocket departing and an event on Earth that's simultaneous in Earth's frame with the arrival). $\endgroup$ – Hypnosifl Dec 8 '14 at 21:26
  • $\begingroup$ @Hypnosifl: sure, which is why I have to posit the third event of "arrival of rocket is observed on Earth". There are two paths with the same starting point but different endpoints, and these two paths have different proper times. $\endgroup$ – Jerry Schirmer Dec 8 '14 at 21:28
  • $\begingroup$ By "observed" do you mean when light from the event arrives on Earth? If so, then to avoid confusion for Olly Price I want to clarify that this is different from the event E on Earth that is simultaneous in the Earth's frame with the rocket arriving. Also, Olly should understand that the dilated coordinate time in the Earth's frame between the rocket departing and arriving is equal to the proper time on Earth's worldine between the departure and E, not the proper time on Earth's worldine between the departure and people on Earth seeing light from the rocket's arrival. $\endgroup$ – Hypnosifl Dec 8 '14 at 22:04
  • $\begingroup$ @Hypnosifl -- I specify that they should subtract the light travel time. $\endgroup$ – Jerry Schirmer Dec 9 '14 at 1:26
  • $\begingroup$ Sorry, I responded to your comment without realizing that "I have to posit the third event..." was referring to part of your original answer. $\endgroup$ – Hypnosifl Dec 9 '14 at 2:01
1
$\begingroup$

"If the Earth measures the dilated time then is the time it measures given by $t=t_0 \gamma$, with $t_0$ being the proper time? And the proper time would be the time as measured by the rocket, so simply $t_0 = d/v$?"

If $d$ is meant to be the distance from Earth to the star in their own mutual rest frame, then $d/v$ would be the coordinate time in that same frame for a rocket moving at $v$ to move from Earth to the star, i.e. $t=d/v$ where $t$ is the time in the Earth frame. If $t_0$ is the proper time along the rocket's worldine from the event of leaving Earth to arriving at the star, then the correct distance/velocity relation would be $t_0 = d' /v$, where $d'$ is the distance between the Earth and the Star in the rocket's rest frame, which is shorter than the distance between them in the Earth/Star rest frame, due to length contraction. In general, whenever you want to use an expression of the form time = distance/velocity, you have to make sure that all three quantities are defined in the same frame (and keep in mind that proper time between events on an inertial observer's worldline is identical to the coordinate time between those events in the observer's rest frame, so $t_0$ is the coordinate time in the rocket's rest frame). Since you said this is homework, I'll let you play around with the equations to see how it can be true that $t=d/v$, $t_0 = d'/v$, and that $t=t_0 \gamma$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.