Observer on Earth watching a spaceship traveling at c/2

[dainceptionman_02]

i have just started on relativity, so be simple. you are going to travel 1 lightyear at half the speed of light. therefore, it would take 2 years to travel 1 lightyear. on Earth, you would see me going half the speed of light, so it should take two years as well?
i calculated gamma as being 0.866, so the time measured on Earth would be 1.15 times the time as measured on the ship. so you would measure 2.3 years on Earth versus 2 years on the ship. but this question is so simple, because you are seeing 0.5c and i am seeing 0.5c, and both calculate the time takes to go to one lightyear as 2 years...

 

[dainceptionman_02]

so on Earth, the viewers would have to set a wait time at 2.3 years for this to finish?

 

[FactChecker]

i have just started on relativity, so be simple. you are going to travel 1 lightyear at half the speed of light.

In relativity, you must always be clear about who is observing and measuring every time and distance. Any two observers moving relative to each other will not agree on those measurements.

UPDATE: Add to that "simultaneous". Two observers moving relative to each other will not agree on what two events, spatially separated in the direction of motion, are "simultaneous".

 

[dainceptionman_02]

forgive me, i don't have an instructor to beat this answer into me

 

[jbriggs444]

i have just started on relativity, so be simple. you are going to travel 1 lightyear at half the speed of light. therefore, it would take 2 years to travel 1 lightyear.

If we are using the Earth frame of reference to state these figures then indeed, it takes 1 year to cover the one light year at half the speed of light.

If we are using the space ship frame then... the statement makes no sense. The space craft is motionless in its own frame. It is not travelling at half the speed of light. It would never "travel" any distance.

But wait. In the space ship frame, its target is approaching at half the speed of light from an initial distance 0.866 light years away. It will take 1.73 years of space ship time to wait for the arrival.

 

[vanhees71]

i have just started on relativity, so be simple. you are going to travel 1 lightyear at half the speed of light. therefore, it would take 2 years to travel 1 lightyear. on Earth, you would see me going half the speed of light, so it should take two years as well?
i calculated gamma as being 0.866, so the time measured on Earth would be 1.15 times the time as measured on the ship. so you would measure 2.3 years on Earth versus 2 years on the ship. but this question is so simple, because you are seeing 0.5c and i am seeing 0.5c, and both calculate the time takes to go to one lightyear as 2 years...

I am not sure, whether I understand your question right. I guess you have the situation that you travel from Earth to a point that is at rest wrt. the Earth all the time and being at a distance of 1 Ly. For an observer, using his proper time, you need of course 2y, because you travel with 0.5c.

From your point of view you stay at rest and the target moves with 0.5c toward you. At the initial time, the distance for you is length-contracted, i.e., $L'=L/\gamma$ thus the time you need as measured from your point of view $t'=L'/v=t/\gamma$.

Also the observer on Earth can calculate, how much time you'll measure since he knows that for you his time is dilated by a factor $\gamma$, i.e., he considers $t=\gamma t'$ or $t'=t/\gamma$.

So you come to the same conclusion using different kinematical effects from the point of view of different observers.

 

[Mister T]

you are going to travel 1 lightyear at half the speed of light. therefore, it would take 2 years to travel 1 lightyear. on Earth, you would see me going half the speed of light, so it should take two years as well?

The distance of one lightyear is frame dependent. If it's one light year in Earth's rest frame then it's less than one light year by a factor of 1.15 in the traveler's rest frame.

 

[PeroK]

forgive me, i don't have an instructor to beat this answer into me

From where are you learning SR?

 

[phinds]

so the time measured on Earth would be 1.15 times the time as measured on the ship. so you would measure 2.3 years on Earth versus 2 years on the ship.

As has already been pointed out, this is incorrect. You do NOT measure 2 years on the ship. Length contraction means you on the ship see the distance as being 2LY/1.15 and the time of travel is similarly contracted.

 

[PeterDonis]

i have just started on relativity, so be simple.

The actual simplest way to work problems is with math, not words. Let's see if that helps here.

In the frame in which the Earth, the starting point, and the destination point are both at rest, the ship leaves Earth at time $t = 0$, and Earth has spatial coordinate $x = 0$, so the starting event (i.e., the point in spacetime at which the journey starts) has coordinates $(x, t) = (0, 0)$. In this frame, the destination has spatial coordinate $x = 1$ (we're using units in which the speed of light is $1$, so distances are in light years and times are in years), and the ship arrives there in 2 years, so the ending event (i.e., the point in spacetime at which the journey ends) has coordinates $(x, t) = (1, 2)$.

To obtain the coordinates of these events in the ship's frame, we do a Lorentz transformation with $v = 0.5$. The Lorentz transformation equations are:

$$
x' = \gamma \left( x - v t \right)
$$

$$
t' = \gamma \left( t - v x \right)
$$

where $\gamma = 1 / \sqrt{1 - v^2} = 1.15$. (Note that this is not the value of gamma you said you got in your OP; your OP value of gamma is the reciprocal of this. This seems to have confused you--see further note below.)

We can see immediately that, since both events have $x = v t$ (they must because they are on the ship's worldline), we have $x' = 0$ for both events. The first event also has $t' = 0$ (because we chose the spacetime origin of both frames to be that event, which makes things simple), and the second event has $t' = 1.732$.

What does $t'$ mean, physically? That should be evident if you consider what coordinates in the ship's frame mean: they mean times according to the ship's clock. So $t' = 1.732$ is the elapsed time on the ship's clock during the journey. Note that $\gamma$ is then the ratio of the Earth frame elapsed time to the ship frame elapsed time; in your OP you appear to be thinking of it the other way around.

 

[PeterDonis]

you are going to travel 1 lightyear at half the speed of light

Note that in post #10 just now, I used the 1 light year distance in the Earth frame, and therefore used the 2 year elapsed time as the elapsed time in the Earth frame. However, you appear to be thinking of it as a distance in the ship frame. But in the ship frame, you don't "travel" at all. The Earth and the destination move; you don't.

If, however, you really did intend to specify the scenario as the trip taking 2 years according to the ship's clock, then your statement in the OP that it would take 2.3 years in the Earth frame is indeed correct. However, the distance traveled would be longer in the Earth frame. To see how that works, let's do the math with this new set of specifications.

So now we have the following: in the ship frame, the journey starts at coordinates $(x', t') = (0, 0)$ and ends at coordinates $(x', t') = (0, 2)$.

Now we transform back to the Earth frame, which requires the inverse Lorentz transform to the one we did in post #10:

$$
x = \gamma \left( x' + v t' \right)
$$

$$
t = \gamma \left( t' + v x' \right)
$$

The starting event is still at the spacetime origin, $(x, t) = (0, 0)$. The "journey's end" event is at $(x, t) = (1.15, 2.3)$. So in this frame, a distance of 1.15 light years was covered, in 2.3 years (same speed of 0.5). We still have that $\gamma$ is the ratio of the Earth frame time to the ship frame time; but now we see that it is also the ratio of the Earth frame distance to the ship frame distance. In other words, in the ship frame, the "trip distance" is length contracted, just as the "trip time" is time dilated.

 

[Vanadium 50]

This seems to be needlessly complicated. I think the question you have is not the question you aksed. Going by the title as the question, it is entirely a question of light travel time: no time dilation at all. The rocket is moving at c/2 in the Eartyh's frame:

Case 1: moving toward the earth. At t=0, say it is 2 ly away. We see it at t=2. At t=2, it is 1 ly away, and we see it at t=3. So we see it moving at c. We do not observe it moving at c - "observe" ,,eams to correct for light travel time.

Case 2: moving away from the earth. At t=0, say it is 1 ly away. We see it at t=1. At t=2, it is 2 ly away, and we see it at t=4. So we see it moving at c/3. The same distinction of see vs. observe as above applies.

More complicated in-between scenarios have answers that are more complicated and in-between.

Note that if these are moving clocks. this analysis will not tell you what the clocks read at any place or time.