# Special relativity noob problem

1. Nov 16, 2008

### -Vitaly-

1. The problem statement, all variables and given/known data

Two observers A and B , in inertial frames, have identical clocks. B passes close to A
with a relative velocity 0.5c. t seconds after B has passed, A ﬂashes him a light signal.
2t seconds after receiving this signal (as measured on his own clock), B acknowledges by
a return light signal to A. When does A receive it?

2. Relevant equations
Lorentz transformations.

3. The attempt at a solution
I'm new to relativity, so this question may seem trivial to some of you. Nevertherless here is what I tried to do.

-Consider A to be stationary. With coordinate x=0. Then B is moving with V=0.5c relative to x. So B's coordinate wrt to A is x'=0.5cT/y (T time, of A reference frame). where y=(1-(0.5c)^2/c^2)^0.5=(sqrt3)/2

-At time (A reference frame) T=t, A flashes a light signal to B, which travels at c. So it's coordinate is c*(T-t), in A reference frame, because the light beam is kinda delayed by t (i.e. if you put T=t, the coordinate of the light beam is 0).

-Now, I need to calculate how long it takes in ref frame A for the light to get to B. So,
x'=c(T-t) => cT/sqrt(3)=c(T-t) => T=t*sqrt(3)/(sqrt(3)-1)

-So the light beam reached B, now B delays 2t time interval in B's frame (T'=2t). So I need to calculate how long it is in A frame.I start by writing down what T' is in T terms.
T'=(T-Vx/c^2)/(sqrt(3)/2)=(T-0.5x/c)*2/sqrt(3)=(2Tc-x)/(c*sqrt(3)), but x=0 (?not sure?,I'll try yes)
And this is equal to 2t. => 2t=2T/sqrt(3) => T=t*sqrt(3)

-So now I know that the total delay untill B flashes back from the start of the motion is (A reference frame) t+t*sqrt(3)/(sqrt(3)-1)+t*sqrt(3)=t(sqrt(3)-2)/(sqrt(3)-1)

-Now I need to find how long it takes for the light to get back to A from B.
This probably involves knowing the coordinate of B and the distance between the points. I don't know what to do next. And I have 16 of these questions to do till tomorrow (this is 3 and they get harder and harder ) Help please.
Thank you

2. Nov 16, 2008

### Staff: Mentor

No need to consider anything but A's viewpoint. Do it step by step.

According to A:
Where is B when A sends out his signal?
How long does the signal take to reach B?
Where is B when the signal reaches him?
How long does it take for B's signal to get to A?

3. Nov 16, 2008

### -Vitaly-

Sorry, I still didn't get the right answer. Here is what I tried to do,following your advice.

The answer is (3 + 2 √3)t

#### Attached Files:

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• ###### Homework.jpg
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Last edited: Nov 16, 2008
4. Nov 16, 2008

### Staff: Mentor

Note that I said that we can solve this by viewing things entirely from A's viewpoint. So there's (almost) no need for special relativity at all and certainly no need to use the Lorentz transformations.

For example: Where was B when A sent the signal? B was traveling at speed v = 0.5 c for t seconds, so his distance from A is D = vt = 0.5ct.

And so on.

Give it another shot.

5. Nov 16, 2008

### -Vitaly-

It's still not working

#### Attached Files:

• ###### Homework1.pdf
File size:
75 KB
Views:
67
6. Nov 16, 2008

### Staff: Mentor

7. Nov 16, 2008

### -Vitaly-

This doesn't include 1 part of the problem:
Where is B when A sends out his signal?
How long does the signal take to reach B?
Where is B when the signal reaches him?
How long does B delay before sending signal back in A's time frame?
How long does it take for B's signal to get to A?

This is exactly what I've done. If you have a look at the first diagram, then t1 corresponds to your 2nd line, t2 - 3rd, t3- 4th, t4- 5th

8. Nov 16, 2008

### Staff: Mentor

Ah, very good! I missed that. (D'oh! :uhh:) That's where some relativity comes in.

So I'd add another question to the pile, just after the one you added: Where is B when he sends his signal?

9. Nov 16, 2008

### Staff: Mentor

It looks like you made an error here:
X = 0.5c(t + t_2)
X = ct_2

So far, so good. But then somehow you get t_2 = 2t.

Last edited: Nov 16, 2008
10. Nov 16, 2008

### -Vitaly-

t2 is the time taken for the light beam from A to reach B.
So when they meet their coordinates are the same. So xb=xlight.
xb=vT (you told me not to use relativity), so xb=0.5c(t+t2), and xlight=ct2.
So t2=2t

Oops, I made an arithmetic mistake, 0.5t2=0.5t, lol
So t2=t

11. Nov 16, 2008

### -Vitaly-

I changed everything accordingly, but the answer is still wrong :(
See my attachment.

#### Attached Files:

• ###### Homework.pdf
File size:
72.2 KB
Views:
71
12. Nov 16, 2008

### -Vitaly-

So now I know t1=t, t2=t.

t3 is the delay of 2t in A reference frame.
I think of it as a time interval (some T2-T1), so

ΔT'=(t3-0)/sqrt(1-0.5^2) using Lorentz transformations and taking x=0
ΔT'=2t

Or do I need to use time dilation formula? What is it with my notation?

13. Nov 16, 2008

### Staff: Mentor

It looks like you misapplied the L.T. when you calculated the time delay according to A. Since 2t is the time elapsed on B's clock, A can apply time dilation. So: t_3 = (2t)γ.

14. Nov 16, 2008

### Staff: Mentor

This is the one you want:

$$\Delta t = \gamma(\Delta t' + v\Delta x'/c^2)$$

Δt' = 2t
Δx' = 0
Sure, you could just use time dilation, which is equivalent to the L.T. done right.

15. Nov 16, 2008

### -Vitaly-

Ok, using those gave me t3=4t/sqrt(3).

But the final answer is still wrong :(

16. Nov 16, 2008

### -Vitaly-

Ok, so what I do next is say that the coordinate of B is still xb=0.5c*(Overall time)
So xb=0.5c(t1+t2+t3)=
=0.5c(t+t+4t/sqrt(3))=c(t+2t/sqrt(3)).

So this distance the light has to travel back to A, so xlight=c*t4
c*t4=0.5c(t+t+4t/sqrt(3))=c(t+2t/sqrt(3)).

t4=t+2t/sqrt(3).

So the total time = t1+t2+t3+t4=
=t+t+4t/sqrt(3)+t+2t/sqrt(3)=3t+6t/sqrt(3)=t(sqrt(3)+2sqrt(3))

But the answer we were given is T=(3 + 2√3)t

Maybe it's a typo in their answer?

Last edited: Nov 16, 2008
17. Nov 16, 2008

### Staff: Mentor

Redo that very last step.

18. Nov 16, 2008

### -Vitaly-

Omg, I'm so stupid, I need to learn arithmetics again :D
Thank you very much for all your help.
I still have a few problems to solve, would it be ok if I post some other problems which I might need help with (maybe 1-2 hours later).
thanks again

19. Nov 16, 2008

### Staff: Mentor

Of course. Best to post in a new thread, to minimize confusion.

And you're welcome.