Time measured by a car-Special relativity

[Paulo Figueiredo]

Homework Statement

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)

Homework Equations

Lorentz 's Formulas

The Attempt at a Solution

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

 

[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.

 

[Chestermiller]

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

 

[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.

 

[PeroK]

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.

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

 

[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.

 

[Chestermiller]

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

 

[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.

 

[Chestermiller]

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.

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.

 

[TSny]

Everything can be determined the in the stationary observer's frame, and then transformed to the moving observer's frame.

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.

 

[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

 

[TSny]

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.

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?

 

[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).

 

[PeroK]

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).

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

 

[Paulo Figueiredo]

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

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.

 

[PeroK]

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.

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

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

 

[Paulo Figueiredo]

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

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

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...

 

[PeroK]

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...

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?

 

[Paulo Figueiredo]

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?..

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

 

[PeroK]

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

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.

 

[Paulo Figueiredo]

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.

Okay. I think I understand that. But, can´t we make a "variable change", making x'=x-200,000, where x represents the distance between A and O ? I think this, because in the instante t=t'=0, the distance between A and O is 200,000 Kms. The meaning of x' would be the distance between the ray of the light and O (measured by O).
Then, using the LorentzTransformation, in the event 2, for the observer A, we will have x'=300,000-200,000=100,000 and so t'=(1-c/3*100,000/c^2)/(sqrt(1-100,000^2/c^2))=0.9428.

 

[PeroK]

Okay. I think I understand that. But, can´t we make a "variable change", making x'=x-200,000, where x represents the distance between A and O ? I think this, because in the instante t=t'=0, the distance between A and O is 200,000 Kms. The meaning of x' would be the distance between the ray of the light and O (measured by O).
Then, using the LorentzTransformation, in the event 2, for the observer A, we will have x'=300,000-200,000=100,000 and so t'=(1-c/3*100,000/c^2)/(sqrt(1-100,000^2/c^2))=0.9428.

No, that won't work. In general, you need to be more careful about times and measurements when you switch from one frame to another. Let me tell you the two ways to do the problem.

1) Use the Lorentz Transformation. For this you need a common origin, so you must have $t=0, x=0; t'=0, x'=0$ at the event where A is at O. You can't choose $t=0$ later than this, like you can in classical physics.

2) Analyse the problem from A's frame, with O moving away from A at $c/3$. This means that you need to calculate the distance that O is from A when the signal is emitted (and it won't be 200,000km).

My advice is to do it both ways and check you get the same answer!

 

[Paulo Figueiredo]

No, that won't work. In general, you need to be more careful about times and measurements when you switch from one frame to another. Let me tell you the two ways to do the problem.

1) Use the Lorentz Transformation. For this you need a common origin, so you must have $t=0, x=0; t'=0, x'=0$ at the event where A is at O. You can't choose $t=0$ later than this, like you can in classical physics.

2) Analyse the problem from A's frame, with O moving away from A at $c/3$. This means that you need to calculate the distance that O is from A when the signal is emitted (and it won't be 200,000km).

My advice is to do it both ways and check you get the same answer!

I made this: 1)calculate the (x;t) for the observer A when the signal is emitted: (x,t)=(-200,000; 2/3). For the observer O, I have found t'=0.9428 sec. It is right?
2) I have calculate the (x,t) for A when the light reach A: (x,t)=(-300,000;1). For O, I've found t'=1,414. Then , for O the time between the emission of the signal and the moment when it reaches A is 0,472...
I think that is wrong!
Can you tell me the resolution of the problem?

 

[PeroK]

I made this: 1)calculate the (x;t) for the observer A when the signal is emitted: (x,t)=(-200,000; 2/3). For the observer O, I have found t'=0.9428 sec. It is right?

I'm confused by this. By convention you would use $(t, x)$ for coordinates in O's frame and $(t', x')$ for coordinates in A's frame.

You should have the co-ordinates of both events in O's frame. You can simply calculate those from the information given:

$t_1 = 2s, x_1 = 0, t_2 = 3s, x_2 = 300,000km$

You then use the Lorentz Transformation to get the coordinates in A's frame.

2) I have calculate the (x,t) for A when the light reach A: (x,t)=(-300,000;1). For O, I've found t'=1,414. Then , for O the time between the emission of the signal and the moment when it reaches A is 0,472...
I think that is wrong!

Perhaps try using the Lorentz Transformation. This is not right at all!

 

[Paulo Figueiredo]

I'm confused by this. By convention you would use $(t, x)$ for coordinates in O's frame and $(t', x')$ for coordinates in A's frame.

You should have the co-ordinates of both events in O's frame. You can simply calculate those from the information given:

$t_1 = 2s, x_1 = 0, t_2 = 3s, x_2 = 300,000km$

You then use the Lorentz Transformation to get the coordinates in A's frame.
Perhaps try using the Lorentz Transformation. This is not right at all!

Sorry, but I don' understand. Why t1=2s ? At the moment of the emission of the ligth, t1 is not equal a 2/3 sec (because the distance between O e A is 200,000 Km)? And why t2=3s?

 

[PeroK]

Sorry, but I don' understand. Why t1=2s ? At the moment of the emission of the ligth, t1 is not equal a 2/3 sec (because the distance between O e A is 200,000 Km)? And why t2=3s?

I'm going offline now, so my last post for today:

As I said before, you have to start at $t=0$ when A leaves O. In O's frame, A is moving at $c/3$ and it takes $2s$ to get to $200,000km$ etc.

 

[Paulo Figueiredo]

I'm going offline now, so my last post for today:

As I said before, you have to start at $t=0$ when A leaves O. In O's frame, A is moving at $c/3$ and it takes $2s$ to get to $200,000km$ etc.

Thank you very much. Tomorrow I will think better.

 

[Paulo Figueiredo]

Thank you very much. Tomorrow I will think better.

Perok,
Thank you.
I've made the following calculus:
t1=2; x1=200,000. Then, using Lorentz Transformation, t'1=(2-100,000*200,000/c^2)/sqrt(1-(100/300)^2)=2*sqrt(8)/3;
t2=3;x2=300,000. Then, t'2=sqrt(8).
So, Δt=t'2-t'1=sqrt(8)/3=sqrt(8/9)=0.9428 sec.
That is the result of my original post, but it's a coincidence.
In fact, if the distance initial between O and A were a, with the same ratiocination, I deduce that t'=((ct-a)/v-v(x-a)/c^2)/γ, where γ=sqrt(1-v^2/c^2).
It was coincidence, because c=300,000; t=1,a=200,000, v=100,000 and so, (ct-a)/v=1.
I think that it is correct.

 

[PeroK]

Perok,
Thank you.
I've made the following calculus:
t1=2; x1=200,000. Then, using Lorentz Transformation, t'1=(2-100,000*200,000/c^2)/sqrt(1-(100/300)^2)=2*sqrt(8)/3;
t2=3;x2=300,000. Then, t'2=sqrt(8).
So, Δt=t'2-t'1=sqrt(8)/3=sqrt(8/9)=0.9428 sec.
That is the result of my original post, but it's a coincidence.
In fact, if the distance initial between O and A were a, with the same ratiocination, I deduce that t'=((ct-a)/v-v(x-a)/c^2)/γ, where γ=sqrt(1-v^2/c^2).
It was coincidence, because c=300,000; t=1,a=200,000, v=100,000 and so, (ct-a)/v=1.
I think that it is correct.

That's not what I get. It looks like you forgot the gamma factor in the Lorentz transformation.

In fact, I didn't bother to calculate $t'_1, t'_2$. Instead, I did:

$t'_1 = \gamma(t_1 - \frac{vx_1}{c^2}), t'_2 = \gamma(t_2 - \frac{vx_2}{c^2})$

Hence:

$\Delta t' = t'_2 - t'_1 = \gamma((t_2 - t_1) - \frac{v(x_2-x_1)}{c^2})$

 

[Paulo Figueiredo]

That's not what I get. It looks like you forgot the gamma factor in the Lorentz transformation.

In fact, I didn't bother to calculate $t'_1, t'_2$. Instead, I did:

$t'_1 = \gamma(t_1 - \frac{vx_1}{c^2}), t'_2 = \gamma(t_2 - \frac{vx_2}{c^2})$

Hence:

$\Delta t' = t'_2 - t'_1 = \gamma((t_2 - t_1) - \frac{v(x_2-x_1)}{c^2})$

I think that is a equivalent expression. In the situation in study, i think that t2-t1=x/v-a/v=(x-a)/v, and x2-x1=x-a.
Then, Δt'=γ((x-a)/v-v(x-a)/c^2)=γ((ct-a)/v-v(x-a)/c^2), where γ=1/sqrt(1-v^2/c^2).

 

[PeroK]

I think that is a equivalent expression. In the situation in study, i think that t2-t1=x/v-a/v=(x-a)/v, and x2-x1=x-a.
Then, Δt'=γ((x-a)/v-v(x-a)/c^2)=γ((ct-a)/v-v(x-a)/c^2), where γ=1/sqrt(1-v^2/c^2).

Nevertheless, 0.9428s is not the correct answer.

 

[Paulo Figueiredo]

Nevertheless, 0.9428s is not the correct answer.

Maybe not. If you get another answer I will apreciate to know.
Thank you very much

 

[PeroK]

Maybe not. If you get another answer I will apreciate to know.
Thank you very much

Why don't you try finishing it off? I get:

$\Delta t' = t'_2 - t'_1 = \gamma((t_2 - t_1) - \frac{v(x_2-x_1)}{c^2}) = \gamma(1s - \frac{x_2}{3c})$

Note that I haven't used any numbers such 200,000 at all. Also, note that $x_2 = 1$ light-second $= 1c s$

 

[Paulo Figueiredo]

Why don't you try finishing it off? I get:

$\Delta t' = t'_2 - t'_1 = \gamma((t_2 - t_1) - \frac{v(x_2-x_1)}{c^2}) = \gamma(1s - \frac{x_2}{3c})$

Note that I haven't used any numbers such 200,000 at all. Also, note that $x_2 = 1$ light-second $= 1c s$

I think that the problem is the value of x1. x2=1c s -Okay-(it is approximately 300,000 Km). But x1=200,000 Km (2c/3 c s) ?

 

[PeroK]

I think that the problem is the value of x1. x2=1c s -Okay-(it is approximately 300,000 Km). But x1=200,000 Km (2c/3 c s) ?

Ah, okay. But $x_1 = 0$.