# Special Relativity: Trip to Andromeda by rocket

## 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. :)

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

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?

TSny
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Yes, that's right.

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?

TSny
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Now, is it correct that the spacetime interval for both the Earth frame and the rocket frame are the same?
Yes!! :)

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?
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 $v$.

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?

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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?

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

$$I^2=(\Delta t)^2-(\Delta x)^2$$
$$\Delta t^2=I^2+\Delta x^2$$
$$\Delta t=[I^2+\Delta x^2]^{1/2}$$
OK, up to here.

Then use the approximation for the binomial expansion to get
$$\Delta t=I^2+\frac{1}{2}\Delta x^2$$
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.

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?

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

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. :)

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

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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. :)

TSny
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OK. Good work!

Chestermiller
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