Calculating Frame Speed with Relativity and Rocket Events

[tony873004]

Relativity -- rocket problems

Homework Statement

Consider two rockets, each of rest length 100 meters. Rocket 1 is at rest in frame S and has its nose at x=0 and its tail at x=+100 meters. Rocket 2 is at rest in frame S’ and has its nose at x’=0 and its tail at x’=
-100 meters.

Now suppose that frame S’ moves with speed V in the x+ direction relative to frame S.

Event A: The nose of Rocket 2 passes the nose of Rocket 1 at time t_A = t'_A = 0

Event B: The tail of Rocket 2 passes the nose of Rocket 1 at time t_B = 2.5 microseconds in frame S.


Use the information about event B to calculate the speed V of S’ relative to S.

Homework Equations

v=d/t

$$L = \frac{{L_0 }}{\gamma } = L_0 \sqrt {1 - \frac{{v^2 }}{{c^2 }}}$$

$$T = T_0 \gamma = \frac{{T_0 }}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}$$

The Attempt at a Solution

v=100m / 2.5 microseconds = 40*10⁶ m/s

What doesn't make sense to me is that I used 100 m for d, yet there should be length contraction. So shouldn't I be using a different d? But how do I get that d if I don't know V?

 

[Doc Al]

What doesn't make sense to me is that I used 100 m for d, yet there should be length contraction.

Yes, there should be length contraction.

So shouldn't I be using a different d? But how do I get that d if I don't know V?

You won't be able to get the numerical value of d until you've solved for V, but you can certainly write the length contracted distance in terms of V and thus use it to solve for V.

 

[tony873004]

Thanks. I'm still confused though.

The length contracted distance in terms of V is

$$L = \frac{{L_0 }}{\gamma } = L_0 \sqrt {1 - \frac{{v^2 }}{{c^2 }}}$$

and v is
$$v = c\sqrt {\left( {1 - \left( {\frac{L}{{L_0 }}} \right)^2 } \right)}$$

So if I substitute the first formula into the 2nd, I get

$$\begin{array}{l}<br /> v = c\sqrt {1 - \,\left( {\frac{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}{{}}} \right)^2 } \,\,\,\, \Rightarrow \,\,\,\,\frac{v}{c} = \sqrt {1 - \,\left( {\sqrt {1 - \frac{{v^2 }}{{c^2 }}} } \right)^2 } \,\,\,\, \Rightarrow \,\,\,\, \\ <br /> \\ <br /> \,\frac{v}{c} = \sqrt {1 - \,\left( {1 - \frac{{v^2 }}{{c^2 }}} \right)} \,\,\, \Rightarrow \,\,\,\,\,\,\,\left( {\frac{v}{c}} \right)^2 - 1 = - \,\left( {1 - \frac{{v^2 }}{{c^2 }}} \right)\,\,\,\, \Rightarrow \,\,\,\,\,\, \\ <br /> \\ <br /> \,1 - \left( {\frac{v}{c}} \right)^2 = \,1 - \frac{{v^2 }}{{c^2 }}\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\frac{v}{c} = \frac{v}{c} \\ <br /> \end{array}$$

 

[gabbagabbahey]

Thanks. I'm still confused though.

The length contracted distance in terms of V is

$$L = \frac{{L_0 }}{\gamma } = L_0 \sqrt {1 - \frac{{v^2 }}{{c^2 }}}$$

and v is
$$v = c\sqrt {\left( {1 - \left( {\frac{L}{{L_0 }}} \right)^2 } \right)}$$

So if I substitute the first formula into the 2nd, I get

$$\begin{array}{l}<br /> v = c\sqrt {1 - \,\left( {\frac{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}{{}}} \right)^2 } \,\,\,\, \Rightarrow \,\,\,\,\frac{v}{c} = \sqrt {1 - \,\left( {\sqrt {1 - \frac{{v^2 }}{{c^2 }}} } \right)^2 } \,\,\,\, \Rightarrow \,\,\,\, \\ <br /> \\ <br /> \,\frac{v}{c} = \sqrt {1 - \,\left( {1 - \frac{{v^2 }}{{c^2 }}} \right)} \,\,\, \Rightarrow \,\,\,\,\,\,\,\left( {\frac{v}{c}} \right)^2 - 1 = - \,\left( {1 - \frac{{v^2 }}{{c^2 }}} \right)\,\,\,\, \Rightarrow \,\,\,\,\,\, \\ <br /> \\ <br /> \,1 - \left( {\frac{v}{c}} \right)^2 = \,1 - \frac{{v^2 }}{{c^2 }}\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\frac{v}{c} = \frac{v}{c} \\ <br /> \end{array}$$

Your first and second formulas represent the exact same relationship, so its no wonder you're not getting any new information about V. But you actually know two pieces of information here:

(1) The second rocket undergoes length contraction as seen in S:

$$L=L_0 \sqrt{1-\frac{V^2}{c^2}}$$

AND

(2) The rocket is moved a distance of $$L$$ in S, in a time $${\Delta}t=2.5 \mu s$$, while traveling at a speed $$V$$:

$$V=\frac{L}{{\Delta}t}$$

 

[tony873004]

Thanks. I think I got it. It's more algebra than I though it would be. Can someone double-check that I did this right?

 

[gabbagabbahey]

Thanks. I think I got it. It's more algebra than I though it would be. Can someone double-check that I did this right?

It's correct,although you seemed to be going in circle with your algebra for a little while there...it would have been easier for you to collect the terms that had V^2 on one side of the equation and the rest of it on the other right from the end of your second line of calcs...it probably would have saved you 3 lines or so of algebra.

 

[tony873004]

You're right. It saved me 3 lines. Thanks!... More relativity q's tommorow...

 

[tony873004]

New problem...
I get a different answer when I do the algebra the way you recommend.

I'm recomputing the answers using a more exact value for c:

first method:
sqr(299792458^2*100^2/((2.5e-6)^2*299792458^2+1)) = 39999964.3952457

second method:
sqr((100^2*(299792458)^2) / ((299792458)^2*(2.5e-6)^2+100^2) ) = 39648636.4200608

Assuming the first answer is correct, I can now compute the length of the rocket as viewed from frame S:

100 * sqr(1-39999964.3952457^2 / 299792458^2) = 99.1058843255855

Assuming the second answer is correct,
100 * sqr(1-39648636.4200608^2 / 299792458^2) = 99.1215910501519

In each case, the rocket contracts by about 0.9 meters, but the answers are different enough that it worries me.

And as a check, using these numbers to compute velocity, which was given in the problem as 2.5 microseconds:


First method:
99.1058843255855 / 39999964.3952457 = 2.47764931354201E-06

Second method:
99.1215910501519 / 39648636.4200608 = 0.0000025

I'm guessing that this implies that the 2nd method is correct, since I get back the exact value of the given time. But that would mean that I must have made a mistake in the algebra of the 1st method, although I can't find my error. Usually algebra errors give answers that are way off, not nearly-identical.

 

[gabbagabbahey]

First method:
99.1058843255855 / 39999964.3952457 = 2.47764931354201E-06

Second method:
99.1215910501519 / 39648636.4200608 = 0.0000025

I'm guessing that this implies that the 2nd method is correct, since I get back the exact value of the given time. But that would mean that I must have made a mistake in the algebra of the 1st method, although I can't find my error. Usually algebra errors give answers that are way off, not nearly-identical.

There is a slight error on your first calculation; when going from line 7 to line 8, you didn't multiply the 1 in the denominator by (100m)^2...It didn't affect your final answer much though because the first term in the denominator works out to be 1.875 X 10^7 m^2 which is much bigger than the 10^4 m^2 term. This is why it's usually best not to play around too much algebraically, there are more places to make an error.

 

[tony873004]

Thanks. No matter how many times I stared at that, I just couldn't see it. It's obvious now that you point it out.