Finding u and v for a Moving Rocket: Applying the Lorentz Transformation

[rashida564]

we are asked to find u and v. I thought by listing the known I can proceed to solve the question, but got no idea where I should start solving the question from.

[Moderator's note: Moved from a technical forum and thus no template.]

 

[Herman Trivilino]

Are you familiar with the Lorentz Transform?

 

[rashida564]

yeah, very familiar. But not sure if I can apply them here, and how can I apply them science there's an acceleration, (the rocket changed it's direction)

 

[bhobba]

Try looking at the word line of the rocket as in the twin paradox which it basically is (John Baez comes to the rescue):
http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_spacetime.html

Tons of other analysis about on the internet not necessarily using world lines, but its the way I tend to look at it.

Thanks
Bill

 

[PAllen]

yeah, very familiar. But not sure if I can apply them here, and how can I apply them science there's an acceleration, (the rocket changed it's direction)

The acceleration of the rocket is irrelevant, and taken to be instant, else the problem is intederminate. Further both S and S’ are inertial frames. No ‘frame’ that turns with the rocket is proposed. Note, I assume v is the relative speed of S and S’. Your excerpt only defined u not v.

Can you solve it now?

 

[rashida564]

I am thinking of calculating the invariant for space time both situations. By drawing two space-time diagram. one in S' and the other in S.
The one in S'. Δs'=2√x^2+c where x=ut', and for the other frame. ΔS=sqrt(1/25+t1'^2)+sqrt(17.64+t2'^2).
But got no idea what should I do next

 

[PAllen]

I am thinking of calculating the invariant for space time both situations. By drawing two space-time diagram. one in S' and the other in S.
The one in S'. Δs'=2√x^2+c where x=ut', and for the other frame. ΔS=sqrt(1/25+t1'^2)+sqrt(17.64+t2'^2).
But got no idea what should I do next

As @Mister T suggested in post #2, you should use the Lorentz transform to set up equations between event coordinates you know in S' and event coordinates you partially know in S. This will give you the ability to solve for u and v. Invariants, in this case, are not useful because they leave you with too many unknowns, and there aren't enough of them. You are given coordinate information, so that is what you must exploit.

 

[rashida564]

Can someone gives me hit to which events, I should apply this formula

 

[PAllen]

The rocket turnaround event, and the rocket arrival at S' origin after turnaround.

 

[rashida564]

I tried to solve to get the time. I now the coordinates of returning point in S'. Since it took one year to get there in S', then by symmetry it will take one more year to return. So I got coordinates of (0,2) in S' and S:(4,x). I solved using Lorentz transformation to get x=2 gamma. But got no idea what should I do next

 

[Herman Trivilino]

Since it took one year to get there in S', then by symmetry it will take one more year to return.

Assuming the return speed is the same!

Can you show us the work you did in applying the Lorentz transform equations?

 

[rashida564]

Assuming the return speed is the same!

Can you show us the work you did in applying the Lorentz transform equations?

won't the speed be the same in S'.
t=γ (t'+vx'/c^2)
So x' for the last point will be zero(it returned to the origin)
t' will be 2 by symmetry, since it will have the same speed in the frame
so t=γ t'=2γ

 

[PAllen]

won't the speed be the same in S'.
t=γ (t'+vx'/c^2)
So x' for the last point will be zero(it returned to the origin)
t' will be 2 by symmetry, since it will have the same speed in the frame
so t=γ t'=2γ

Yes, you have to assume symmetry in S’, else you don’t have enough info. But why not use the Lorentz transform for x instead of t, since you are given x values in S, not t values. Otherwise you are on right track. Can you give the S’ coordinates of the turnaround, in terms of u? This will be the other thing you need.

 

[PeroK]

won't the speed be the same in S'.
t=γ (t'+vx'/c^2)
So x' for the last point will be zero(it returned to the origin)
t' will be 2 by symmetry, since it will have the same speed in the frame
so t=γ t'=2γ

Although the question is not clear on this, I think you must assume that the rocket travels at the same speed, $u$, in frame $S'$ for both the outward and return journeys.

I suggest you need to define some notation for this. For eaxmple:

Let the turaround event be defined by $t'_1, x'_1$ in frame $S'$ and by $t_1, x_1$ in frame $S$.

Let the return event be defined by $t'_2, x'_2$ and $t_2, x_2$ respectively.

Now, you fill in what you know about these values and use the Lorentz Transfomation to find $u, v$.