Special Relativity: Velocity Addition

[Dgray101]

Homework Statement

Two objects fly toward you, one from the east with speed u, and the other from the west with speed v. Is it correct that their relative speed, as measured by you, is v+u? Or should you use the velocity addition formula, V=(u+v)/(1+uv/c2)? Is it possible for their relative speed, as measured by you, to exceed c?

Homework Equations

It is more conceptual then anything, at least that's supposed to be it's purpose.

The Attempt at a Solution

The wording of the problem is very confusing to me... but here is an explanation.

If I consider the object from the other frame to be a moving frame S' then the velocity formula needs to be used, especially when their speeds are approaching C. Secondly it is possible for their relative speed, as measured by me, to be greater then C. This is not to mean that they themselves are exceeding the speed of light, but the speed of one particle RELATIVE to the other particle ACCORDING to me is greater then c. To measure an object according to myself and relative to myself, with a speed greater than c is not permissible due to Einsteins second postulate.

 

[Nugso]

Well, the second formula, the one with the Lorentz transformation that is, is always correct. [tex]V+U[/tex] can be considered correct only when [tex]v,u<<c[/tex]And no, I don't think it is possible for their relative speed to exceed c. Assume that both u and v are moving at the speed of light. Even in that case the relative speed would be c.

 

[vela]

Homework Statement

Two objects fly toward you, one from the east with speed u, and the other from the west with speed v. Is it correct that their relative speed, as measured by you, is v+u? Or should you use the velocity addition formula, V=(u+v)/(1+uv/c2)? Is it possible for their relative speed, as measured by you, to exceed c?

Homework Equations

It is more conceptual then anything, at least that's supposed to be it's purpose.

The Attempt at a Solution

The wording of the problem is very confusing to me... but here is an explanation.

If I consider the object from the other frame to be a moving frame S' then the velocity formula needs to be used, especially when their speeds are approaching C.

It's not clear exactly what you're talking about here.

Secondly it is possible for their relative speed, as measured by me, to be greater then C. This is not to mean that they themselves are exceeding the speed of light, but the speed of one particle RELATIVE to the other particle ACCORDING to me is greater then c. To measure an object according to myself and relative to myself, with a speed greater than c is not permissible due to Einsteins second postulate.

This is the correct answer to the question.

 

[Dgray101]

Vela if you were to answer the question how would you have said it? I just want a different view on it. Our professor is HORRRIBLE at teaching. He is a relativist but he is god damn horrible at teaching in lecture, everything I have learned has been self taught either from the book or the internet. If you can elaborate or clean up my response that would be great. P.S This isn't for marks, this is just assigned problems he suggested for the upcoming midterm.

 

[Chestermiller]

You doped it out perfectly. You measured one object's speed relative to your frame of reference, and got u. You measured the other object's speed relative to your frame of reference, and got v in the opposite direction. So, as reckoned by you, their relative speed is u+v.

Chet

 

[vela]

In the first sentence you wrote, what is "the other frame"? There are three inertial frames of reference in this scenario. I can guess at what you mean, but I think what you wrote assumes that we know the notation or conventions your class is using. The second part, I think, is fine. If you want to elaborate a bit more, you could explain that the distance between the two objects as measured by you decreases at a rate given by u+v.

 

[Dgray101]

Sorry Vela I did use the text notation. It becomes a little problematic as our professor teaches out of one textbook but he assigned us a different textbook for the class, so I find myself in autopilot now.(makes no sense at all...). Anyways... if there are more comments feel free to shed some light on me.

 

[voko]

I do not disagree with Chestermiller and vela, but I would like to add my two cents. The whole idea of "their relative speed as measured by you" is weird. "Their relative speed" is well defined and is given by the relativistic formula. But when you measure "their relative speed", you do not measure it directly. You measure their speeds (velocities) relative to yourself, then you infer what their relative speed is. And that becomes a matter of convention. You could say, well, I know they are moving fast, so I should use the relativistic formula to find their relative speed. That is fine and the would agree with their own measurements. But if you use "their" relative speed to predict when and where they will meet, your prediction will be wrong. For a correct prediction, you need to use u + v as explained by vela.

 

[vanhees71]

I must disagree with the answer in the original posting somewhat. It is, of course, right that when you measure the velocity of one particle relative to you (assuming you are in an inertial frame for simplicity) to be [itex]\vec{v}_1[/itex] and another's particle velocity relative to you to be [itex]\vec{v}_2[/itex], you can calculate the vector [itex]\vec{v}_1-\vec{v}_2[/itex], and the modulus of this vector can be up to [itex]2c[/itex], but that doesn't have any physical meaning.

The relative velocity between two particles is defined as the one particle's velocity in the rest frame of the other particle (assuming that at least one of the particles has finite mass and can be at rest; if both particles are massless, special care is necessary to define the relative velocity), and this proper definition of "relative velocity", which is used in the definition of the cross section in scattering processes for instance, can never lead to relative velocities faster than the speed of light in vacuo.

 

[voko]

you can calculate the vector [itex]\vec{v}_1-\vec{v}_2[/itex], and the modulus of this vector can be up to [itex]2c[/itex], but that doesn't have any physical meaning.

As explained by vela, the modulus of the vector thus found has to do with the rate of change of the distance between two objects in the observer's frame. As I explained earlier, this vector has direct physical significance for the observer, allowing the observer to predict whether the objects will meet and, if so, where and when, in the observer's frame.

The fact that this vector does not represent the velocity of a physical object does not per se mean that it has no physical meaning.

 

[Dgray101]

The wording of the problem is weird I totally agree. My professor picks out problems that confuse the entirety of the classroom. So it would be improper then to use the relativistic formula over just simply U+V?

 

[Dgray101]

Sorry guys I am still a little unclear as to specifically why it is wrong to use the relativistic velocities compared to the U+V formula :S

 

[Dgray101]

I mean I sort of get why the U+V makes sense, but what I don't quite understand is why it is not permissible for this problem to specifically use the relativistic transform.

 

[Dgray101]

Let me pose my question this way in particular " You know the answer is U+V, explain why you would not want to use the relativistic formula for velocity addition."

 

[Chestermiller]

I totally agree with Voko's assessment in #8. If you are going to correctly predict when and where they meet, you better be using u + v. In this problem everything is being reckoned from, and referenced to, your frame of reference.

Chet

 

[vela]

The wording of the problem is weird I totally agree. My professor picks out problems that confuse the entirety of the classroom. So it would be improper then to use the relativistic formula over just simply U+V?

I thought the wording of the problem was fine.

Sorry guys I am still a little unclear as to specifically why it is wrong to use the relativistic velocities compared to the U+V formula :S

I mean I sort of get why the U+V makes sense, but what I don't quite understand is why it is not permissible for this problem to specifically use the relativistic transform.

Let me pose my question this way in particular " You know the answer is U+V, explain why you would not want to use the relativistic formula for velocity addition."

That's what you're supposed to figure out. Suppose u/c=0.5 and v/c=0.5. What does u+v=c mean? What does the quantity ##\frac{u+v}{1+uv/c^2} = 0.8 c## represent physically?

 

[Dgray101]

Alright, Refined Answer:

The quantity U+V is what would be used in this case. Since we are concerned with their relative speeds as measured by me, then it is true that the rate of change in their distance is V+U since both velocities are measured according to me. If we were to use the relative velocity formula we are calculating the relative velocity of an object according to myself, and relative to myself which in this case it is not possible for the speeds to be greater than c because according to Einsteins second postulate nobody can observe an object moving relative to their frame with a speed greater then c.