Special Relativity: Finding Particle Speed

[Geoff]

Hello.
I am doing a special relativity question.
I got a particle traveling at some speed v. It travels 20000 light years from reference of galaxy in 30 years from the reference of the particle.
I need to find the speed of the particle.
Well, i tried to use time dilation and length contraction formulas but i end up with 3 unknown and 2 equations.
So after thinking a little more i made up another equation:
30 years=Dialated length/velocity.
Of course the units were correct years into seconds and so on.
Then i bassicle get 2 v's, one inside the gamme and the other one on the bottom of the fraction, but the answer i get is C. So i made a mistake somewhere...any ideas?
Thx

 

[robphy]

Can you show some details of your work?
If you can think trigonometrically (using rapidities), the answer and its method of solution should become obvious.

 

[Doc Al]

So after thinking a little more i made up another equation:
30 years=Dialated length/velocity.

Nothing wrong with this approach. Can't tell where your mistake was unless you show more details. (I assume by "dilated length" you actually mean contracted length: the galactic distance traveled as measured in the particle frame.)

 

[Geoff]

well here is what i got:
Let say that time taken to go the distance in the frame of reference of a particle is T. and it is in seconds.
The contracted distance is not known to us but we do know that proper distance is D in meters.
Then we have
T=(D/sqrt(1-v^2/c^2))/v
then:
vT=D/sqrt(1-v^2/c^2)
so we square it all
we got
v^2*T^2=d^2 - d^2*c^2/v^2
now...
we need to solve for v, but to do that i will need to use power of 4...and i think that that is a good indicator that i am wrong...if anyone seeas a mistake in above logic please tell me other i will solve it out...but that seems about right that i do have a mistake in there.

 

[Geoff]

scratch my last response i made a mistake in there, but i am stil getting c...
here is how
assume t being time, and d being proper distance.
t=d*gamma/v
t^2v^2=d^2*(1-v^2/c^2)
t^2v^2+d^2v^2/c^2=d^2
v^2(t^2+d^2/c^2)=d^2
now since t^2 is so much smaller then d^2/c^2 it is insignificant
then when u solve u get c for the speed.
I think i might have messed something up
note that t is the time from the perpective of the partice, d is the distance from the inertial reference,and i need to know velocity of the particle.
Any ideas where i made a mistake?
note that proper distance is 23 000 light years and relative time is 30 years, that is why it is insignificant and does nothing

 

[lightgrav]

well, Tv = d/gamma would lead to T^2 v^2 = d^2 ( 1 - v^2 / c^2) ...

you're deleting the only epsilon that distinguishes its speed from c.
(yea, it is really fast, apparently, but not quite that fast!)
just keep a few more digits.

 

[Geoff]

well sig figs will not affect it at all, i mean the diffrence between the 2 values is something like 10 ^10...there is no way to keep sig figs there...u sure that that is correct?
Basicly here is the initial problem, maybe i interpreted it incorrecly?
a particle travels 23 000 light years proper distance in 30 years from the particles frame of reference.
i need to find it;s speed, when i did as i said above it gives me practically almost c, without any use for sig figs since they are at a 10 ^10 off the mark...

 

[lightgrav]

closer to 10^6 . write v/c = d(1-e)

 

[Geoff]

how did u get that v/c equation? and what is e?
sry for my stupid question:(

 

[lightgrav]

that was a suggestion for you to try ... e , epsilon? is 0 << e << 1 ,
so you can expand sqrt(1-e) = 1 - e/2 , that sort of thing . yours is tc/d .

 

[Geoff]

how did u get there exectly?, because i don;t see how u can isolate v/c by itself...
i mean the sqrt(1-e) = 1 - e/2 that is an estimate right?
and why do u say e=tc/d should it not be v^2/c^2?

 

[Doc Al]

Here's how I'd do it:
$$L_0 / \gamma = v t$$

Rewrite in terms of $\beta = v/c$:
$$(L_0/c) \sqrt{(1 - \beta^2)} = \beta t$$
or:
$$(L_0/(ct)) \sqrt{(1 - \beta^2)} = (\beta)$$

Square:
$$(L_0/(ct))^2 (1 - \beta^2) = \beta^2$$

Now just solve for $\beta$, saving your approximations for the last step.

 

[robphy]

Here's my trig/rapidity solution:
cT=c(proper time)="hypotenuse of a [Minkowski]-right triangle"
L=(distance between earth-at-rest and galaxy-at-rest in Earth frame)="opposite leg"=cT*sinh(rapidity)=$cT*\sinh(\theta)$
$\beta=\tanh(\theta)$
So,
$$\beta=\tanh\left(\sinh^{-1}\left(\displaystyle\frac{L}{cT}\right)\right)$$
[add]
Spacetime Trigonometry:
$\beta=\tanh(\theta)$
$\gamma=\cosh(\theta)$
$\beta\gamma=\sinh(\theta)$
Quiz: what is the interpretation of $\exp(\theta)$?
[/add]