# SR problem

1. Mar 21, 2005

### Pengwuino

Ok i think the book is trying to noobify me by tricking me.... ok heres the problem.

Ok so Earth is the origin

Object A moves in the positive X direction at 0.900c measured from the earth
Object B moves in the positive X direction at 0.800c measured from the earth

Object B is further along on the X axis as object A so B will soon overtake object B.

Observers on the Earth witness A overtake B at a relativistic speed of 0.100c

With what speed is A overtaking B if observed by A?

Shouldnt it be the same, 0.100c?

2. Mar 21, 2005

### polyb

NO! You are not doing a Gallilean transform of coordinates. This is a Lorentzian transform. The earth frame is the rest frame. You need to think of what velocity observer A is seeing B going at as well as what observer B is seeng A go at. Just try to keep track of your reference frames so you dont get lost.

3. Mar 21, 2005

### Pengwuino

So is it .36c?

I used u'x = (ux - v)/(1-(ux * v)/c^2)

Ok thats not right either... how in the world :D

Last edited: Mar 21, 2005
4. Mar 21, 2005

### HallsofIvy

Staff Emeritus
Earth sees A moving away at 0.8c so A sees Earth moving away at 0.8c. If B is movi g away from Earth at speed v (relative to A) then its speed, relative to earth is
(0.8c+ v)/(1+ 0.8v/c2)= 0.9c. Solve for v.

5. Mar 21, 2005

### polyb

How this: Let us assume that all three observers will start measuring their times and distances simultaniously starting at an abritrary time and place. So, from the earth observers pointof view we have:

at $$T_{0}=0$$, just to make the math easier, spaceship A is at point $$X_{A0}$$ and spaceship B is at $$X_{B0}$$, and when B passes A we'll have $$T_{f}$$ where A is at $$X_{Af}$$ and B is at $$X_{Bf}$$. Let us go ahead and say that $$X_{Af}=X_{Bf}$$ to make things simpler.

Now from the earth observers point of view we already know that $$\frac {X_{Bf}-X_{B0}}{T_{f}}- \frac {X_{Af}-X_{A0}}{T_{f}} = 0.1c$$

The question you need to ask yourself is what from the point of view of ship A what $$X_{Af}-X_{A0}$$ will be measured and what $$T_{f}$$ will be measuered as well as what $$X_{Bf}-X_{B0}$$. The Lorentz factor for ship A will yeild the correct result. Now the same goes for observer B as well except the Lorentz factor will be different. This may seem like a long way of doing this but it is imperitive that you properly peg down what is being measured in what frame as well as the basis for what is being measured. Once you see who is measuring what you'll be able to see the easier way of doing so I hope this helps a little and at best you'll have to do a little algebraic manipulation. I can't overemphasize the idea keeping track of who is measuring what relative to the rest frame, once you get this down you'll find it easier when you start looking into GR. Good Luck and I hope this helps a little!
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HallsofIvy,
Are you saying that Euclid was a voyer with your signature?:rofl:

6. Mar 21, 2005

### HallsofIvy

Staff Emeritus
"HallsofIvy,
Are you saying that Euclid was a voyer with your signature?"

No, that's the first line of a well known poem by Edna St. Vincent Millay:

Euclid alone has looked on Beauty bare.
Let all who prate of Beauty hold their peace,
And lay them prone upon the earth and cease
To ponder on themselves, the while they stare
At nothing, intricately drawn nowhere
In shapes of shifting lineage; let geese
Gabble and hiss, but heroes seek release
From dusty bondage into luminous air.
O blinding hour, O holy, terrible day,
When first the shaft into his vision shone
Of light anatomized! Euclid alone
Has looked on Beauty bare. Fortunate they
Who, though once only and then but far away,
Have heard her massive sandal set on stone.

7. Mar 21, 2005

### polyb

I know, I was just having a little fun! Thanks for posting it!

Though I have to admit that the line alone can give one a nefarious impression of Euclid! :rofl: