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QuantumCurt

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## Homework Statement

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.

## Homework 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. :)