# Relativity Questions and Frames (How fast...)

1. Sep 11, 2016

### RJLiberator

1. The problem statement, all variables and given/known data
The International Space Agency is designing a spaceship to reach the star Proxima Centauri, 4 cyrs (light years) away so that the on-board crew will age 4 years from departure to arrival. How fast must the ship travel?

2. Relevant equations

t(moving clock) = t(stationary clock)*sqrt(1-V^2/c^2)

t(between ship light clock ticks to us) = 2D/(sqrt(c^2-V^2)) = t(between ticks of our own light clock)/(sqrt(1-V^2/c^2))

3. The attempt at a solution

We must identify which frames we are in.

We are wondering how fast the ship must travel, V, to people on Earth.

The crew must age 4 years, so we take t' = 4 years, the time it takes for the crew to age (ship frame).

The ship must travel a distance of 4 c*yrs, which is in the Earth frame, so we say L = 4c*yrs.

This seems like insufficient information.

L = 4c*yrs
V = ?? unknown, need.
t = unknown
t' = 4 years
V' = ??
L' =???

I suppose, if I can say L' = 4 c*yrs, then I can say that the speed the ship is going at is c (speed of light), but then the equations start to break down so that seems like a dead end.

I know that the time of the clock on the ship must be a shorter time change than that of Earth frame.

Is there anything that I am stating as false?

2. Sep 11, 2016

### Andrew Mason

Use the Lorentz transformation:

$t' = \gamma(t - vx/c^2)$

Can you express t in terms of v and x? That would leave you with a quadratic equation to solve with only one unknown, v.

AM

3. Sep 11, 2016

### RJLiberator

$$t = t' \sqrt {1- \frac {v^2}{c^2}}+\frac {vx}{c^2}$$

where
$$\gamma = \frac{1}{ \sqrt {1- \frac {v^2}{c^2}}}$$

I'm not sure how there is only one unknown.

t' = 4 years.
v = unknown
x = 4 light years
but t = ???

4. Sep 11, 2016

### PeroK

Think about it this way:

If they set off at, say, v = 0.1c, then they will take 40 years in the Earth frame and, even with a bit of length contraction and time dilation, they will take longer than 4 years in the ship frame (not much less than 40 years in fact).

If they set of at v = 0.999c, then they will take just over 4 years in the Earth frame. Given significant length contraction and time dilation, they will take a lot less than this in the ship frame.

So, somewhere in between there must be a velocity, $v$, where the ship time will be exactly 4 years.

Everything depends on $v$ (including $t$ and $t'$), so there is only one independent variable.

5. Sep 11, 2016

### RJLiberator

Ah now with these two posts I am acquiring understanding!

So, as noted in my previous post:

$$t = t' \sqrt {1- \frac {v^2}{c^2}}+\frac {vx}{c^2}$$

Now, since t' = 4 years, the time the ship crew age.
and t is dependent on v via

$$t = 4yrs \sqrt{1-\frac{v^2} {c^2}}$$

we combine and see that

$$4yrs \sqrt{1-\frac{v^2} {c^2}} = 4yrs \sqrt {1- \frac {v^2}{c^2}}+\frac {v4lyrs}{c^2}$$

Now I try to solve this for v.

The problem is, by subtracting the LHS from both sides, I end up with v = 0.

Hm.

6. Sep 11, 2016

### PeroK

I've got no idea what that $x$ is doing there. Why not start with:

$t = D/v$

Where $D$ is the proper distance to the star? Just like you would in boring old classical mechanics!

7. Sep 11, 2016

### RJLiberator

The x came to be from the Lorentz Contraction transformation here.

I see that $$t = D/v$$ where D = 4 light years and v is what we want to find.

So,

t' = 4 years.
t = 4 light years/ v
v = what we want to find

If I use the time equation we get

$$t' = t \sqrt{1-\frac{v^2} {c^2}}$$

From here we plug in

$$4yrs = 4 lyrs / v (\sqrt{1 - \frac{v^2} {c^2}})$$

8. Sep 11, 2016

### RJLiberator

This leads to $$v = \sqrt{ \frac {c^2} {c^2+1}}$$

v = 0.9999 light years/yr ?

that doesn't seem to make sense to me.

9. Sep 11, 2016

### PeroK

I was going to say, in response to your previous post, that you may want to keep the equations algebraic for the time being and be careful about the units you use for this problem.

10. Sep 11, 2016

### RJLiberator

That's a good suggestion.

$$t' = t \sqrt{1-\frac{v^2} {c^2}}$$

Now divide both sides by t.

$$\frac{t'} {t} = \sqrt{1-\frac{v^2} {c^2}}$$

Square both sides

$$\frac{t'^2} {t^2} = 1-\frac{v^2} {c^2}$$

In putting t = D/v and solving for v we find

$$v^2 = \frac{1} {\frac{t'^2}{D^2}+{\frac{1} {c^2}}}$$

Square rooting both sides and letting D = 4 light years, t = 4 years we see an answer for v of

$$v = 0.9999 light years/year$$

hm

11. Sep 11, 2016

### PeroK

The best thing to do with that is to simplify it to:

$\frac{v}{c} = \frac{D}{\sqrt{c^2(t')^2 + D^2}}$

When you have a distance $D$ in $c.yrs$ it's good idea to aim for $D/c$ in your equations, which gives a unit of years. So:

$\frac{v}{c} = \frac{D/c}{\sqrt{(t')^2 + (D/c)^2}}$

Now you have dimensionless quantities on both sides of the equation. Try this and see what you get.

12. Sep 11, 2016

### RJLiberator

Aha.

I finally get it. (I think)

The units have confused me greatly. I need more experience with light years and the like. The key for me now was in understanding that a light year divided by c = year.

So, after messing with the algebraic part of the problem for a while, I see that

$$\frac{v} {c} = \frac{\frac{4lyrs} {c}} {\sqrt{16 yrs^2+\frac{16lyrs^2} {c^2}}}$$

This means
$$\frac {v} {c} = 0.707$$

Which means, the velocity of the ship is 0.707 c

13. Sep 11, 2016

### PeroK

I prefer $\frac{v}{c} = \frac{1}{\sqrt{2}}$

Then, you can precisely calculate $\gamma$ and double-check your answer by calculating $t'$ for this value of $v$.

Note that as you progress with SR, you'll find it more and more useful to express $v$ in this form.

14. Sep 11, 2016

### RJLiberator

I appreciate the help and the advice.