# Special Relativity: Trip to Andromeda by rocket

1. Dec 20, 2014

### QuantumCurt

1. The problem statement, all variables and given/known data

This is from chapter 1 (2nd edition) of Taylor and Wheeler's 'Spacetime Physics.'

It's a four part problem, and I didn't have any troubles with parts a, b, or c. I'm stuck on part d. I'm supposed to determine if we could make it to Andromeda within a lifetime.

Assume that the Andromeda galaxy is 2 million light years distant from Earth. Treat both Earth and Andromeda as points, and neglect any relative motion between them.

d. Trip 3: Now set the rocket time for the one-way trip to 20 years, which is all the time you want to spend getting to Andromeda. In this case, what is your speed as a decimal fraction of the speed of light?
Discussion: Solutions to many exercises in this text are simplified by using the following approximation, which is the first two terms in the binomial expansion
$$(1+z)^n~\approx~1+nz$$
if
$$z\ll1$$

Here n can be positive or negative, a fraction or an integer; z can be positive or negative, as long as its magnitude is very much smaller than unity. This approximation can be used twice in the solution to part d.

2. Relevant equations

$$(spacetime~interval)^2=(difference~in~time~readings)^2-(difference~in~space~readings)^2$$
$$I^2=(\Delta t)^2-(\Delta x)^2$$

3. My attempt at a solution

Following the above equation, I assumed that $\Delta t=20~yrs$ and that $\Delta x=2.00\times10^6~yrs$

This gives me

$$I^2=(20~yrs)^2-(2.00\times10^6~yrs)^2$$

Which led me to

$$I^2=-4\times 10^{12}~ yrs$$

Which is a negative number. This solution didn't make sense to me.

Then I tried calculating it directly by converting the distance of $2.00\times 10^6~yrs$ into meters, and 20 years into seconds. This gave me

$$2.00\times 10^6~yrs ~ \times ~ \frac{31,536,000~s}{1~yr}~\times~3.00\times 10^8~\frac{m}{s}=1.89216\times10^{22}~m$$

And

$$20~yrs~\times~\frac{31,536,000~s}{1~yr}=630,720,000~s$$

Then dividing distance by time I get $3.00\times10^{13}~ m/s$

Which is faster than c, and clearly wrong.

Since they mention the binomial expansion, I'm assuming that I have to incorporate it in this problem somehow. I'm not seeing how to go about doing that. Do I need to somehow incorporate the difference between the Earth frame and the rocket frame in calculating my time interval? Is the space interval going to be different than the 2 million years? Any help would be much appreciated. :)

2. Dec 20, 2014

### TSny

In which frame of reference is the time interval 20 yrs? In which frame of reference is the distance 2 million light years?

When calculating the spacetime interval between two event using $I^2=(\Delta t)^2-(\Delta x)^2$, do $\Delta t$ and $\Delta x$ need to be measured in the same inertial frame?

Sample problem 1-2 of chapter 1 is good preparation for your problem.

3. Dec 20, 2014

### QuantumCurt

The time interval 20 yrs is in the frame of reference of the rocket. Since event 1 (leaving Earth) and event 2 (arriving at Andromeda) both occur at the same point (at the rocket) in the frame of reference of the rocket, could I then say that the space interval is zero?

This would mean that $I=20~yrs$ in the frame of the rocket, correct?

4. Dec 20, 2014

### TSny

Yes, that's right.

5. Dec 20, 2014

### QuantumCurt

Now, is it correct that the spacetime interval for both the Earth frame and the rocket frame are the same?

The space interval from the Earth frame is 2 million years. Do I need to use this and the spacetime interval to find the time interval from the Earth frame?

6. Dec 20, 2014

### TSny

Yes!! :)

Yes, but I would recommend solving the resulting equation for $\Delta t$ in terms of the symbols $\Delta x$ and $I$ without plugging in any numbers yet.

Or, you could first express $\Delta t$ in terms of the rocket speed $v$ and $\Delta x$ and then use this in the expression for the spacetime interval to get an equation for $v$.

7. Dec 20, 2014

### QuantumCurt

So could I write this expression for $\Delta t$ like this?

$$I^2=(\Delta t)^2-(\Delta x)^2$$
$$\Delta t^2=I^2+\Delta x^2$$
$$\Delta t=[I^2+\Delta x^2]^{1/2}$$

Then use the approximation for the binomial expansion to get

$$\Delta t=I^2+\frac{1}{2}\Delta x^2$$

Then I'd use this along with the space interval from the Earth frame to find the velocity?

Last edited: Dec 20, 2014
8. Dec 20, 2014

### QuantumCurt

Or would I want to reverse $I^2$ and $\Delta x^2$ to get $$\Delta t=\Delta x^2+\frac{1}{2}I^2$$

Since I is much smaller than x?

9. Dec 20, 2014

### TSny

OK, up to here.

This is not a correct application of the approximation (1+z)n ≈ 1+nz. Note the 1 in this expression.
You can see that the left hand side of your equation is in years while the right had side is in (years)2.

I suggest that you stick with $\Delta t=[I^2+\Delta x^2]^{1/2}$ and use this to find an expression for $v$ in terms of $I$ and $\Delta x$ before using any approximations.

10. Dec 20, 2014

### QuantumCurt

Since $v=\frac{\Delta x}{\Delta t}$, I can just write this expression as $$v=\frac{\Delta x}{[I^2+\Delta x^2]^{1/2}}$$

Then I can do some factoring and get

$$v=\frac{\Delta x}{[I^2(1+\frac{\Delta x^2}{I^2})]^{1/2}}$$

Then

$$v=\frac{\Delta x}{I(1+\frac{\Delta x^2}{I^2})^{1/2}}$$

Then use an approximation to obtain

$$v=\frac{\Delta x}{I(1+\frac{1}{2}\frac{\Delta x^2}{I^2})}$$

Am I on the right track here? Would the $\Delta x^2$ and the $I^2$ become $\Delta x$ and $I$ since I'm taking a square root?

11. Dec 20, 2014

### TSny

Which would be better to factor out of the denominator: $\small I$ or $\small \Delta x$? Which is a small quantity: $\frac{\Delta x}{I}$ or $\frac{I}{\Delta x}$?

12. Dec 20, 2014

### QuantumCurt

Ahh...that makes sense. So I'm going to get this for my expression

$$v=\frac{\Delta x}{[\Delta x^2(1+\frac{I^2}{\Delta x^2})]^{1/2}}$$
$$v=\frac{\Delta x}{\Delta x(1+\frac{I^2}{\Delta x^2})^{1/2}}$$
$$v=\frac{1}{(1+\frac{I^2}{\Delta x^2})^{1/2}}$$
$$v=\frac{1}{1+\frac{1}{2}\frac{I^2}{\Delta x^2}}$$

Does $\frac{I^2}{\Delta x^2}$ become $\frac{I}{\Delta x}$ since I'm taking a square root?

If I use $\frac{I^2}{\Delta x^2}$ I get a numerical answer of .99990001. If I use $\frac{I}{\Delta x}$ I get .999995. The answer given is $1-5\times 10^{-11}=.99999999995$

I'm guessing there's still another step here. What am I missing or doing wrong? Thanks for the help. :)

13. Dec 20, 2014

### TSny

Note that you can write this as $$v= \left ( 1+\frac{I^2}{\Delta x^2} \right )^{-1/2}$$ Now try the approximation.

Last edited: Dec 20, 2014
14. Dec 20, 2014

### QuantumCurt

Ah, there it is.

$$v= \left ( 1+\frac{I^2}{\Delta x^2} \right )^{-1/2}$$

$$v=1-\frac{1}{2}\frac{I^2}{\Delta x^2}$$

$$v=1-\frac{1}{2}\frac{(20~yrs)^2}{(2.00\times10^6~yrs)^2}$$
$$v=1-\frac{1}{2}\frac{400}{4.00\times10^{12}}$$
$$v=1-5\times10^{-11}=.99999999995$$

It's interesting that this comes out directly as a percentage of c. With the other parts in this specific problem, I had to find the velocity in terms of m/s, and then take it as a ratio with c to find the final answer.

Thank you so much for the help. It was very much appreciated. :)

15. Dec 20, 2014

### TSny

OK. Good work!

16. Dec 20, 2014

### Staff: Mentor

The easiest way to do this problem is to make use of the Lorentz Transformation:
$$\Delta x = \gamma(\Delta x'+v\Delta t')$$
where the primes refer to the rocket frame.

Chet

17. Dec 20, 2014

### QuantumCurt

I figured that would be the case, but I actually haven't gotten to that in this book yet. I've used the Lorentz Transformation in the honors project I did a couple semesters ago, but it won't be introduced in this book for a couple more semesters yet.