Recent content by Paulo Figueiredo

You are right. Now I understand my error. Thank you very much.

Let's see: O and A are initially in the same place (t=t'=x=x'=0). Correct? Then A moves with a velocity c/3. It was as if O moves farway from A at that velocity. The event 1 is the event concerning at t1=2 sec. Then x1 (distance between O and A, measured by O) is 200,000 Km. I don´think that...

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) ?

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

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, 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...

Thank you very much. Tomorrow I will think better.

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 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...

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...

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

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...

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.

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).

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...