# Special Relativity Question

1. Oct 5, 2006

### Warr

The question is: Say a rod is travelling towards an observer O at a velocity v. The front of the rod is point A and the back is point B. First it wants me to calculate the apparent velocity of the rod to the observer.

So lets say at $$t_0=0$$, A is at O (ie x=0)

then when A is a distance d from O (x=-d), the time is $$t_2=\frac{-d}{v}$$

The time at which the observer sees A at a distance d is $$t_1=t_2+\frac{d}{c}=d\left(\frac{1}{c}-\frac{1}{v}\right)$$. (ie the time at which A is actually at d plus the time it takes the light to travel from x=-d to x=0)

therefore

When A gets to 0, the observer sees the object at the same time as it is actually there, $$t_0=0$$

therefore the difference in the time between when he sees it at x=d and x=0 is$${\Delta}t=t_0-t_1=0-d\left(\frac{1}{c}-\frac{1}{v}\right)=d\left(\frac{1}{v}-\frac{1}{c}\right)$$

so the apparent velocity $$v_{app}$$ is then

$$v_{app}=\frac{{\Delta}d}{{\Delta}t}=\frac{0-(-d)}{d\left(\frac{1}{v}-\frac{1}{c}\right)}=v\left(\frac{c}{c-v}\right)$$

What i'm confused about is whether I should this have a $$\gamma$$ term in it to account for special relativity?

The reason I think this is, is that if the rods system is O', then $$t_2$$ and consequenly $$t_1$$ should be instead $$t_2'$$ and $$t_1'$$ respectively, which would then make $${\Delta}t$$ instead be $${\Delta}t'$$

so that $${\Delta}t'=\frac{{\Delta}t}{\gamma}$$ which would then multiply my final solution by a factor of $$\gamma$$. (ie. $$v_{app}=v\gamma\left(\frac{c}{c-v}\right)$$)

which method is correct...if either

Last edited: Oct 5, 2006
2. Oct 5, 2006

### andrevdh

I am also confused. Special relativity is about the differences in observing the same event (a moving rod) in different IRFs that are experiencing relative motion. So the velocity of the rod, v, with respect to what reference frame is it??? How is this IRF moving with respect to the observer??? Only in this context do this question make sense to me.

Last edited: Oct 5, 2006
3. Oct 5, 2006

### HallsofIvy

Staff Emeritus
My question also. "a rod is travelling towards an observer O at a velocity v." Velocity v relative to whom? The only person mentioned is O and you are asked to find "the apparent velocity of the rod to the observer." The observer O? You were just told that it was v! If the velocity v is relative to something other than O, then we need to know the velocity of that frame of reference relative to O.

4. Oct 5, 2006

### Warr

I suppose v is with respect to the rod's frame, and the apparent velocity is the velocity that observer O measures. The exact wording of the question is:

Consider a stick with the two ends A and B and a proper length of L moving towards an observer O with velocity v. Choose the reference frame with the origin at the position of O and t=0 for the moment when the front of the stick (A) arrives at O. What is the apparent velocity $$v_{app}$$ of the stick?

Then there is just a picture that looks like this
v--->
B-----------------A ............. O

where (.....) is just empty space

Last edited: Oct 5, 2006
5. Oct 5, 2006

### P3X-018

That distance d ain't the same for for both the observers. If d is what O sees then the rod will see $d' = d/\gamma$. Isn't that what you are missing?

Last edited: Oct 5, 2006
6. Oct 5, 2006

### Warr

Well, I think i'm missing something. I don't think thats it, because A is on the rod, hence they are both in the same frame of reference.

7. Oct 5, 2006

### P3X-018

Sorry, I meant O, because I was refering to 2 different IRF's, one on the rod (where the rod is in rest) and one where the rod is moving.