Special Relativity Andromeda galaxy Question

[IntellectIsStrength]

The Andromeda glaxaxy is two million light-years from earth, measured in the common rest frame of the Earth and Andromeda. Suppose you took a fast spaceship to Andromeda, so it got you there in 50 years measured on the ship. If you sent a radio message home as soon as you reached Andromeda, how long after you left Earth would it arrive, according to timekeepers on earth? Note that radio waves travel at the speed of light.

_____

How do I go about this question?
With my own calculations I got a negative imaginary number. What I did was I found out the speed of the spaceship, and I got it to be 40000c. I'm not sure if this is right. Any help would be greatly appreciated.

 

[dicerandom]

The spaceship cannot possibly travel faster than light, that's one of the consequences of Special Relativity's postulate that all inertial observers see the speed of light as constant, therefore its velocity relative to the Earth-Andromeda system must be some fraction of c. You need to use the proper Lorentzian time and distance formulas.

 

[emptymaximum]

with a speed of 40000c you've either found a spaceship with warp drive, or you're wrong.
my assumption is the latter.

how did you find this speed?

 

[IntellectIsStrength]

Thanks for the replies.
For the speed I used a light-years (ly) equation.
ly = cy (y is years)
c= ly / y
c= 2 000 000 / 50 years
c= 40 000
My hunch is also that this is wrong but I don't know what else to do.
dicerandom, do you mean Ls = gamma Lm? How do I use that in this situation? I need the speed for the gamma; do I not?

 

[emptymaximum]

no. c is a constant. it is ≈ 300 000m/s (in vaccuum).
also, you used thetime as measured ON THE SHIP. the ships clock runs slow from the PoV (Point of View) of the earth. how much slower?
that is up to you to find out.
note that the distance to Andromeda is also from the PoV of the earth.

 

[IntellectIsStrength]

the ships clock runs slow from the PoV (Point of View) of the earth. how much slower?

It is slower by a factor of gamma, right? The equation is t = gamma t' (t' being the 50 years). But I need a speed to solve for the gamma. How can I find out the speed of the spaceship?

 

[emptymaximum]

use the lorentz transformation for velocity.

 

[IntellectIsStrength]

Is there any other way of doing it? My class hasn't learned the Lorentz Transformation.

 

[dicerandom]

Your factor should be $$1/\gamma$$, not $$\gamma$$. This is assuming that your book is using the standard definition of $$\gamma$$:

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

You then have the following transformation equations:

$$\Delta t^\prime = \frac{\Delta t}{\gamma} = \sqrt{1-v^2/c^2} \Delta t$$
$$\Delta x^\prime = \gamma \Delta x = \frac{\Delta x}{\sqrt{1-v^2/c^2}}$$

Think of the problem this way. You have two events, one of which is that the spaceship departed Earth and the second is that the spaceship arrived at the andromeda galaxy. Think about where and when both of these events occurred in both refrence frames and I think you'll find that you can use the equations above to reduce the problem to a system of two equations with two unknowns.

 

[IntellectIsStrength]

Thanks dicerandom but I have never come across the transformation equations that you mentioned. So shouldn't here be an alternative method of doing this problem?

 

[emptymaximum]

special relativity problem and you haven't come across these equations?
are you sure?

 

[dicerandom]

There are many ways to solve it, I chose the method by which I first learned the theory since I thought perhaps your book followed a simmilar approach. Your book should have some simmilar equations to transform intervals and events between different refrence frames, read through the relevant section and use the equations that they provide.

What book are you using?

 

[IntellectIsStrength]

Can I use the Galilean transformations for this?

 

[IntellectIsStrength]

Oh nevermind dicerandom you changed your equations... I understand the ones you put up now... Thanks.. So let me give it a try.

 

[dicerandom]

Oh nevermind dicerandom you changed your equations... I understand the ones you put up now... Thanks.. So let me give it a try.

Sorry, I screwed up the TeX the first time around :)

 

[dicerandom]

I just tried working the problem out and it occurred to me that you need the full event transformations, not just the interval transformations. That was my mistake, sorry, I didn't pay enough attention when thinking it through the first time.

Here are the equations you'll need:

$$t^\prime=\gamma\left(t-\frac{v}{c^2}x\right)$$
$$x^\prime=\gamma(x-vt)$$

$$t=\gamma\left(t^\prime+\frac{v}{c^2}x^\prime\right)$$
$$x=\gamma(x^\prime+vt^\prime)$$

 

[IntellectIsStrength]

dicerandom I'm not understanding this equation:
$$t^\prime=\gamma\left(t-\frac{v}{c^2}x\right)$$

Isn't the time equation just t' = t / gamma ?

 

[dicerandom]

When you're transforming elapsed intervals of time between refrence frames, yes, that's correct. That equation is for transforming events between refrence frames, it's more general as it allows for both temporal and spatial seperation. You can derive the first equation from that one.

 

[Tide]

Obviously, the spaceship would have to travel very close to the speed of light so 4 MILLION years would be a very good approximation to the elapsed time for an earthbound observer.

 

[emptymaximum]

and in the 2 million years on the way there, there's a good bet they'll invent warp drive and beat you there anyways :)

 

[IntellectIsStrength]

lol... okay
I apologize dicerandom it's late at night here and my brain is just not functioning properly I guess. And I simply just don't understand the question.
What are my givens?
x = 2 000 000 ly
t' = 50 years
c = 3 x 10^8 m/s

correct?
So what is the first step from here on?

 

[dicerandom]

That's right :)

You also know that x' is 0 (when the spaceship gets to andromeda, there's no separation between them). Using all that you can solve the last equation I gave for v, which you can then use to solve for the elapsed time on Earth. Be warned that v is extremely close to the speed of light, I had to take it out to a good number of decimal places before I saw that it was indeed less than c.

 

[IntellectIsStrength]

Alright, so my givens are:
x = 2 000 000 ly
x' = 0
t' = 50 years
c = 3 x 10^8 m/s
But that last equation you gave, $$x=\gamma(x^\prime+vt^\prime)$$
What's that gamma doing there? I only know the equation x = x' + vt
By the way, anytime you want to give up on helping me just go ahead and do it... I understand

 

[dicerandom]

It's no problem, I struggled with this stuff for months the first time I learned it :)

x = x' + vt is the normal Galallean transformation, the version with the gamma in it is the corrected version for Special Relativity.

 

[IntellectIsStrength]

But we never learned any other version so how can I be obligated to use them? Is there any way to solve it using simple Galilean equations or so?

 

[dicerandom]

No, the Galilean transformations do not account for the fact that observers who are moving relative to one another experience time differently. As Tide mentioned you can get the correct answer to a very good degree of accuracy by just saying that the ship's velocity is basically c, but I don't think that's the point of the problem. You'd said earlier that the interval equations (the first set I posted) looked familliar to you, perhaps there's a way to do this problem using those equations that I missed.

Like I said earlier though, these two sets of equations are very closely linked. If you haven't seen the ones I've posted yet you will very soon.

 

[jtbell]

You're using distance in light-years (ly) and time in years (yr) so the natural unit of velocity is ly/yr. Note that in these units the speed of light is c = 1 ly/yr. It simplifies the arithmetic a bit.

 

[robphy]

Here's one way to do it WITHOUT explict use of the Lorentz Transformations.
Draw a http://www.google.com/search?q=spacetime+diagram":

• parallel lines E and A, representing the Earth and andromeda worldlines... vertical lines in the Earth frame.
• a timelike segment from event O (origin) on line-E to event P (planet) on line-A.
• a lightlike segment from event P to event R (reception) back on line-E

You want the timelike length of OR.
Locate the event on line-E (call it Q) that the Earth says is simultaneous with P.
Observe that OR=OQ+QR.
I'll leave it to you to determine OQ and QR.

 

[dicerandom]

Here's one way to do it WITHOUT explict use of the Lorentz Transformations.
Draw a http://www.google.com/search?q=spacetime+diagram":

• parallel lines E and A, representing the Earth and andromeda worldlines... vertical lines in the Earth frame.
• a timelike segment from event O (origin) on line-E to event P (planet) on line-A.
• a lightlike segment from event P to event R (reception) back on line-E

You want the timelike length of OR.
Locate the event on line-E (call it Q) that the Earth says is simultaneous with P.
Observe that OR=OQ+QR.
I'll leave it to you to determine OQ and QR.

I like the idea of using a geometrical approach, but I must confess that I'm having a hard time seeing how this will work without knowing the relative speed of the two frames and therefore what slope the O-P line should be (other than Tide's observation that it must be very close to c->45 degrees). The only thing I can think of is drawing the top hyperbola which is the locus of points that have a spacetime interval of 50 from the origin, then use the intersection of that hyperbola with the A line as your point P, but I don't think that's what you had in mind.

Am I missing something here?

Edit: On second thought, you'll never get the kind of accuracy from a piece of paper and a ruler to distinguish the true world line from the lightlike line, so I suppose it doesn't matter?

 

[robphy]

Think "triangles"... legs as spatial and temporal components, hypotenuses as the invariant spacetime-intervals. You can write down an explicit algebraic answer in terms of your given quantities... without first explicitly calculating the outgoing velocity of the traveller.

 

[dicerandom]

Think "triangles"... legs as spatial and temporal components, hypotenuses as the invariant spacetime-intervals. You can write down an explicit algebraic answer in terms of your given quantities... without first explicitly calculating the outgoing velocity of the traveller.

Oh, of course, I see it now. I could've used that trick on the GRE ;)