# I Journey time relative to an observer on Earth and on a spaceship

#### Arman777

Gold Member
Summary
time dilation
Let us suppose a spaceship moving from earth to another star that is 10ly away with a speed of 0.99c.

Calculate the of years spaceship takes to reach its destination (a) in the rest frame of an observer on Earth and (b) as perceived by a passenger on board the ship

For (a) I find that $t_0 = x/v$ which is 10.1 years. However for part (b) I am finding t as 71.6 years. However it seems wrong to me. Is it true or not ?

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#### Ibix

(a) is correct. (b) is wrong. How did you go about it?

#### Arman777

Gold Member
I used the classical equation $$t = \frac{t_0} {(1 - (v/c)^2)^{1/2}}$$. However I guess I put the time values in different places. But putting t = 10.1 year does not make sense somehow. Whats the reason behind that ?

#### Nugatory

Mentor
I used the classical equation...
That’s not the right formula - you left out a reciprocal somewhere, effectively multiplying by $\gamma$ when you should be dividing.

However, the best thing to do is to forget about the time dilation formula for now and instead use the Lorentz transformations:

The two relevant events are “ship leaves earth” and “ship arrives at destination”.

When we use the frame in which the earth is at rest to assign coordinates to these events we get $(t=0,x=0)$ for the “leaves” event and $(t=10.1,x=10)$ for the “arrives” event. The difference between the times is $10.1-0=10.1$ so that’s how many years the trip takes using that frame in which the earth is at rest.

Now use the Lorentz transformations to find the coordinates assigned to these two events using the frame in which the ship is at rest (if you do it right, the x coordinate will be zero at both events). The difference between the t coordinates will be the elapsed time using that frame in which the ship is at rest.

And then look at what you’ve just done - you’ve derived the time dilation formula from first principles! (At least if you don’t plug in the actual numerical values until you’re done with the algebra).

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#### Dale

Mentor
And then look at what you’ve just done - you’ve derived the time dilation formula from first principles!
I also favor this approach for this very reason. The time dilation formula automatically appears whenever it is valid, and you avoid errors due to incorrectly using it when it doesn’t apply.

#### Arman777

Gold Member
Using the lorentz transformation is actually good idea. I understand it I guess. Thanks

#### George Jones

Staff Emeritus
Gold Member
Let us suppose a spaceship moving from earth to another star that is 10ly away with a speed of 0.99c.

Like @Nugatory and @Dale, I prefer Lorentz transformations over a rote application of the time dilation formula, which I find too easy to screw up. I also like a method that uses invariance of the spacetime interval, which I think is the key concept in relativity.

In these methods, the first thing to do is to identify the key events. In this case, key event 1 is the coincidence of the spaceship and the Earth, and key event 2 is the coincidence of the spaceship and the star.

I think that a lot of students' problem are caused by failure to identify clearly spacetime key spacetime events for a particular problem.

There are two frames of reference - the frame of reference in which the Earth is at rest, and the frame of reference of the spaceship. Arbitrarily label one of the frames as unprimed and one as primed - it doesn't matter which is which. In what follows, I label the Earth's frame as the unprimed frame, and the spaceship's frame as the primed frame.

In the Earth's frame, the spatial distance between the two key events is $\Delta x =10$, and the elapsed time in the Earth's between the events is $\Delta t = \Delta x/v$, as in a).

In the spaceship's's frame, the spatial distance between the two key events is $\Delta x' = 0$, since the spaceship is coincident with both events, and the spaceship doesn't move in its own frame. The elapsed time In the spaceship's's frame between the events is $\Delta t'$.

Invariance of the spacetime interval, a fundamental property of spacetime, gives
$$\left( \Delta x \right)^2 - \left( \Delta t \right)^2 = \left( \Delta x' \right)^2 - \left( \Delta t' \right)^2 .$$

From the above, the elapsed time in the spaceship's frame is
\begin{align} \left( \Delta t' \right)^2 &= \left( \Delta x' \right)^2 - \left( \Delta x \right)^2 + \left( \Delta t \right)^2 \nonumber \\ &= 0^2 - \left( \Delta x \right)^2 + \left( \Delta x/v \right)^2 \nonumber \\ &=\left( \Delta x \right)^2 \left( \frac{1}{v^2} - 1 \right) \nonumber . \end{align}

#### Ibix

Using the lorentz transformation is actually good idea. I understand it I guess. Thanks
I see you've got to the answer, and I too recommend using the Lorentz transforms until you know when you can get away with using special cases like time dilation.

Nevertheless, you could have used it here. You just had the $t$ and $t_0$ the wrong way round (either in the formula or in your interpretation of which was associated with which clock).

"Journey time relative to an observer on Earth and on a spaceship"

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