# Time measured by a car-Special relativity

1. Mar 8, 2017

### Paulo Figueiredo

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

I wish a help for the following problem:
Consider the situation: a car A is moving away from a stationary observer O with velocity c/3. When the point A is at a distance 200.000 Km of the O a light signal is sent from the observer O to the car A. The question is: how long will take the signal to reach the car A (in the time measured by A) ?
(note: assume that c=300.000 Km/s)

2. Relevant equations
Lorentz 's Formulas

3. The attempt at a solution
My solution: t'=(1-c/3*100.000/c^2)/(sqrt(1-1/9)=0,9428 seconds

2. Mar 8, 2017

### TSny

Your solution does not look correct. Can you explain why you set up your equation as you did? Also, be careful with the units. I recommend including the units in your calculation.

3. Mar 8, 2017

### Staff: Mentor

As reckoned from O's frame of reference, at what distance and time does the signal arrive at car A?

4. Mar 11, 2017

### Paulo Figueiredo

I had used the following formula: t=1/sqrt(1-v^2/c^2), with v=100000Km/s and c=300000Km/s.
Then, t=1/sqrt(8/9) =0,943 seconds.

5. Mar 11, 2017

### PeroK

So, the distance of $200,000km$ is irrelevant?

6. Mar 11, 2017

### TSny

Paulo,

In your post #1 you have

t'=(1-c/3*100.000/c^2)/(sqrt(1-1/9)=0,9428 seconds

Are you sure you want to use 100.000 km here? See PeroK's remark.

7. Mar 11, 2017

### Staff: Mentor

In the stationary observer's frame, $ct=200 +\frac{c}{3}t$
So, t=300/c and x = 300

8. Mar 11, 2017

### TSny

Chet, Paulo's numerical notation is to use a period for a comma and a comma for a decimal point. I was initially confused by this.

9. Mar 11, 2017

### Staff: Mentor

Yes. I was confused by this too. In any event, his approach to solving the problem was incorrect. Everything can be determined the in the stationary observer's frame, and then transformed to the moving observer's frame.

10. Mar 11, 2017

### TSny

Yes. It appears that Paulo is trying to do this, but he has a mistake in the distance the light travels in the stationary frame. Unfortunately, he has not explained in words why he set up his equation as he did.

11. Mar 11, 2017

### Paulo Figueiredo

Sorry for the numerical notation: in Portugal we use often the "." with the meaning of the ",".
I was trying to use the following Lorentz formula: t' = (t-vx/c^2)/(sqrt(1-v^2/c^2)), where (x,t) are the coordenates of the event , measured by O and (x',t') are the coordenates of that event measured by A.
So, in the stationary frame (measured by O), I think that t=1 and x=(300,000-200,000)=100,000 . I make this because , in the aplication of that formula it was necessary that for t=0, xO=xA...
I have seen another relation: Δt'=Δt*sqrt(1-v^2/c^2), where Δt is, in this case, 1 (=300,000/c) and v/c=100,000/c=1/3). Then , the time measured by A is 1*sqrt(8/9)=0.943seconds

Last edited: Mar 11, 2017
12. Mar 11, 2017

### TSny

I need to go slowly. Is the question asking for the time, t', of a single event, or is it asking for the time interval, Δt', between two specific events?

13. Mar 11, 2017

### Paulo Figueiredo

I think that we can interpret Δt' as the interval between the moment when the ray of light is sent (Event 1) and the moment when the ray of light reach the car A (Event 2).

14. Mar 11, 2017

### PeroK

Can you calculate the time and position of these two events in O's reference frame?

15. Mar 11, 2017

### Paulo Figueiredo

In O's reference frame, i think that in Event 1: t=0 sec; x=0 Km, and in the event 2: t=1sec and x=300,000 Km.

16. Mar 11, 2017

### PeroK

In your original post you have that $x = 200,000 Km$.

More importantly, can you see why you cannot have event 1 at $t=0, x=0$?

17. Mar 11, 2017

### Paulo Figueiredo

I think i'm confused. There are two things diferentes. I denote by x the distance between the ray of light and the observer O. So, at the instante t=0, x=0.
In my original post, 200,000 Kms is the distance between O and A in the instante t=0...

18. Mar 11, 2017

### PeroK

Okay, yes, $x= 300,000Km$ when the light reaches A.

How are you going to calculate the events 1 and 2 in A's frame?

19. Mar 11, 2017

### Paulo Figueiredo

In the event 1: x=0; t=0
In the event 2: t=x/c=(x-200,000)/(c/3)⇒xc/3=(x-200,000)c⇔x/3=x-200,000⇔x=300,000 (Kms). Then t=300,000/300,000=1 sec

20. Mar 11, 2017

### PeroK

That looks like the calculation in O's frame!

In A's frame event 1 must be at some point $x' < 0$ and some unknown time $t'$.

Note that in order to use the Lorentz Transformation, the origins of the two frames must coincide. That means that A and O must be in the same place when you set their clocks to $t=0$ at $x=0$ and $t' = 0$ at $x' = 0$ respectively.