# Problem on special relativity

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1. May 7, 2015

### Abrahamlure

Hi! First of all, sorry for my english, I'll try to do my best translating the exercise. I've been working on it this evening but I'm having some troubles:

Problem:
On the Star Wars film, there's a starship traveling at a velocity of 5c. This is nonsense, but we can assume this if we suppose that people on Star Wars had no idea of what they were saying and compare distances and time on different frames without any cares.

a) Suppose that a starship travels from Earth to a close star, which is 1000 light years away. If the pilot, who doesn't know about relativity, uses his own time and this distance, from the earth and without moving, for calculate his velocity, at which velocity will he say he travelled?

b) What will be his velocity seen from the earth?

My solution:

a) I found this section pretty easy, which makes me think that there's a mistake at some point. What I did is to find the distance on meters, by using that the star was 1000 light years away. Then, we knew the time he has been traveling, so we do v=distance/time to find the velocity we were asked to find.

b) I don't know how to start here. The only thing we've done so far on relativity is the Lorentz transformation. Could someone enlighten me with some hint or a proposal of resolution?

Thanks a lot for your time.

2. May 7, 2015

### phinds

When you take nonsense as your starting point, what difference does it make what the "answer" is? No, you cannot "assume this" since it is nonsense and if it is the basis of your question, you are asking "if the laws of physics don't apply, what do the laws of physics say about <any nonsense you care to state>?"

3. May 7, 2015

### Abrahamlure

Hi phinds. I think that the paragraph of Star Wars on the exercise has nothing to do there. It's only saying that on the exercise, if you don't know what relativity is, it could happen to obtain velocities >c. In fact, on the a) section we obtain a v>c.

Last edited: May 7, 2015
4. May 7, 2015

### phinds

This is nonsense. The universe does not care what you know or I know or some Star Trek writer knows. The universe does what it does and it does not allow objects with mass to travel at c, much less faster than c.

5. May 8, 2015

### wabbit

The formulation of the problem is confusing.

I think you are saying, let $d$ be the distance travelled by the ship as measured by earth, let $t'$ be the duration of that travel in the ship's proper time, what is the value of the ratio $v^{pilot}=d/t'$ which the pilot calls his velocity?

I get $v^{pilot}=d/t\cdot t/t'=v\cdot t/t'$, where $v$ is the more usual velocity of the ship relative to earth and $t$ is the travel duration as measured by earth. The ratio $t/t'$ is obtained with the usual relativistic formula.

Also, I doubt the pilot would have got his license if he flunked SR, but that "pilot velocity" sounds like a useful metric in his occupation - he is using the space coordinates on his standard issue chart of the galaxy, and his proper time, which seems a reasonable way to track his progress along his route.

Last edited: May 8, 2015
6. May 9, 2015

### Abrahamlure

Hi, thanks for your answer. Then are you saying that the value $v^{pilot}=d/t'$ can be used to find $v^{earth}$ using that $v^{pilot}=d/t\cdot t/t'=v\cdot t/t'$? Thank you.

7. May 9, 2015

### wabbit

Right - t and t' being time intervals during which the velocity is approximately constant.

Note that $v^{pilot}$ is not a velocity in the usual sense since it mixes space and time coordinates belonging to different charts, while $v$, which you also call $v^{earth}$ is a plain vanilla relative velocity.

Last edited: May 9, 2015
8. May 10, 2015

### Abrahamlure

Yes, that's what I've noticed too. If we do $v^{pilot}=1000 light years/2 weeks$, we can't use the obtained value to find $v^{earth}$, because $v^{pilot}$ is not a well-calculated velocity. It doesn't matter?

9. May 10, 2015

### wabbit

The formula with t/t' would not apply to a "normal" velocity, it is just a direct consequence of the definition of that $v^{pilot}$ and isn't dependent on whether you call it a velocity or a warpicity or a turbodrivimetry:)

So what result did you find?

10. May 10, 2015

### Abrahamlure

Oh! okay, I understand it now. I obtained
Oh, okay, I understand it. On the section a) I obtained $v^{pilot}=7.81\cdot 10^{12}$ m/s. From here, to solve b), I was thinking on use $\displaystyle v^{earth}=\frac{v^{pilot}\cdot t'}{t}=\frac{7.81\cdot 10^{12}\cdot t'}{t}$, with $t'=1209600$s (two weeks). Thank you for all your help, I'm trying to understand how time and space works when objects move closer to a light speed velocity. Now I have to find what $t$ is, right?

How did you find the formula $v^{pilot}=d/t\cdot t/t'=v\cdot t/t'$?

11. May 10, 2015

### wabbit

No, you have to find t/t'. These are both time interval measurements, between two given events on the spaceship worldline, but measured in two different frames. Think Lorentz transform.
I didn't find the formula, I assumed $v^{pilot}=d/t'$ is what you meant by that (fairly unclear) sentence about what the pilot was calling his velocity.

12. May 10, 2015

### Abrahamlure

Okay, then I think that $\displaystyle\frac{t}{t'}=\frac{1}{\gamma}={\sqrt{1-{(v^{pilot}/c)}^{2}}}$. Am I right? Thank you again, wabbit. I'm learning more from this exercise than from my professor.

13. May 10, 2015

### wabbit

Let's pretend you never wrote that disgraceful last formula (which incidentally gives an imaginary number as a result) and try again. What velocity goes into gamma ? Didn't you just say $v^{pilot}$ is not a true velocity ?

14. May 10, 2015

### Abrahamlure

Hahahaha, I'm sorry. Right, $v^{pilot}>c$ so the number would be imaginary. The velocity that goes into gamma, at least on my notes, is the velocity seen from the earth.

15. May 10, 2015

### wabbit

Right. So you get an equation relating $v, \sqrt{1-v^2/c^2}$ , and $v^{pilot}$. Solve this equation for v and you get your result.

16. May 10, 2015

### Abrahamlure

Wabbit, thank you very much. You helped me a lot.

17. May 10, 2015

### wabbit

No problem, glad I could help.