Abstract Special Relativity, particle at rest and one moving

[Floris Meersschaert]

Homework Statement

In frame S particle 1 is at rest and particle 2 is moving to the right with velocity u. Now consider a frame S 0 which, relative to S, is moving to the right with velocity v. Determine the value of v such that the two particles appear in S' to be approaching each other with equal and opposite velocities.

Homework Equations

u'=(u-v)/(1-uv/c^2)
-u1'=u2'

The Attempt at a Solution

Particle 1, as it doesn't have any initial velocity, will be moving at the velocity of frame S' but in opposite direction (-v). Particle 2 however, does have initial velocity so the total equation would be looking like this:
v=(u-v)/(1-uv/c^2)
Solving for v by multiplying the right denominator by v would get:
v-(uv^2)/c^2=u-v
we will get:
2v-(uv^2)/c^2-u
Entering this into the quadratic equation and using taylor expansions would leave us with
v=1/2u
However, this looks like Newtonian physics and if we replace u with any number, we would not get the particles to have equal speeds but in opposite direction.

 

[kuruman]

Hi Floris Meersschaert and welcome to PF.

v=(u-v)/(1-uv/c^2)

Is this the velocity of particle 2 in the primed frame? I don't think so. The symbol v is the unknown velocity.

 

[Floris Meersschaert]

Hi Floris Meersschaert and welcome to PF.Is this the velocity of particle 2 in the primed frame? I don't think so. The symbol v is the unknown velocity.

Thank you for your response.
"v" indeed is not the velocity of particle 2 in the primed frame. (u-v)/(1-(uv)/c^2) however is. The velocity of the particle in the primed frame should be equal but opposite to the velocity of particle 1 in the primed frame. The velocity of particle 1 in the primed frame = -v
However, in the notes, I mention that the velocity of particle 1 (u1') is equal but opposite to the velocity of particle 2 (u2'). Hence, I get the equation that v=(u-v)/(1-(uv)/c^). However, through solving for v, I get a supposedly wrong answer.

 

[kuruman]

However, through solving for v, I get a supposedly wrong answer.

What is your supposedly wrong answer in the form $v = ...~$? I don't see that in your post.

 

[Floris Meersschaert]

Yes, it is in the very last line. v = 1/2 U. Plugging this back into the formula's with u approaching c would give me the velocity of particle 2 to be c and the velocity of particle 1 still to be -1/2 c, therefore no equal magnitude.
Once again, thank you for your swift response

 

[kuruman]

I thought you had to do the calculation relativistically. If the particles do not travel at relativistic speeds, then the answer is trivially obvious. If one particle is at rest relative to the lab and the other is moving with velocity u, an observer moving at (½)u will see the first particle moving away from him at speed (½)u and the other towards him with the same speed. Are you sure your calculation is supposed to be non-relativistic?

 

[Floris Meersschaert]

I thought you had to do the calculation relativistically. If the particles do not travel at relativistic speeds, then the answer is trivially obvious. If one particle is at rest relative to the lab and the other is moving with velocity u, an observer moving at (½)u will see the first particle moving away from him at speed (½)u and the other towards him with the same speed. Are you sure your calculation is supposed to be non-relativistic?

That is my whole concern. The answer I get is non-relativistic but as we're doing special relativity right now, I suppose the calculation has to be relativistic as shown by my calculations. Therefore, I hope you can help me with trying to figure out where I went wrong.

 

[kuruman]

Therefore, I hope you can help me with trying to figure out where I went wrong.

You went wrong when, instead of solving the quadratic equation, you expanded the series. I can help. As a starting point, please post two equations showing the velocity of particles 1 and 2 in the primed frame.

At this point I suggest that you invest some time to learn how to use LaTeX if you are not familiar with it. It's a great tool for writing equations and a good skill to have. To access the user's guide click the "LaTeX" link next to the question mark on the left at the bottom of the message area.

 

[Floris Meersschaert]

Thanks a lot!
The velocity of particle 1 in the primed frame will be $u'_1 = \frac {u_1 - v} {1 - \frac {u_1*v}{c^2}}$
$u_1 = 0$ as it is at rest in the regular frame. This way, the formula will be: $$u'_1 = \frac {0 - v} {1 - \frac {0*v} {c^2}}$$
This will end up in $u'_1 = \frac {-v}{1} = -v$
The velocity of particle 2 in the primed frame will be $u'_2 = \frac {u_2 - v} {1 - \frac {u_2*v}{c^2}}$
$u_1 = u_1$ as the velocity of particle 2 is not defined as a quantity or another letter. The formula will therefore stay the same so: $$u'_2 = \frac {u_2 - v} {1 - \frac {u_2*v} {c^2}}$$
As $-u'_1=u'_2$ we can say that: $$v = \frac {u_2 - v} {1 - \frac {u_2*v} {c^2}}$$
With some algebra, I found out that: $$ \frac {u_2*v^2}{c^2} - 2v + u_2 = 0$$
Using the quadratic formula: $x = \frac {-b \pm \sqrt{b^2 -4ac}} {2a}$ we get: $$v = \frac {2 \pm \sqrt{4 - 4*\frac{u_2}{c^2}*u_2}}{\frac{2u_2}{c^2}}$$
I hope this is clear enough and otherwise I'd be more than glad to hear when clarification is needed.

 

[kuruman]

Except for an additional cosmetic cancellation of factors of 2 in the numerator and denominator, that's what I got. This is impressive not only because you got it right, but also because it seems you are well on your way to mastering LaTeX.

Which of the two solutions do you think is the answer and why?

 

[Floris Meersschaert]

Thank you! I would say the solution with the minus would be correct as otherwise we would get a value over the speed of light which is not possible. Is this correct?

 

[kuruman]

I would say the solution with the minus would be correct as otherwise we would get a value over the speed of light which is not possible. Is this correct?

Absolutely correct. Keep up the good work.

 

[Floris Meersschaert]

Awesome. So I'll write out my final calculations. To correct for the cosmetic cancellation of 2, as you said, I would end up with: $$v = \frac {1 - \sqrt {1- \frac {u_2^2}{c^2}}}{ \frac {u_2}{c^2}}$$
Then, we get:
$$v = \frac {c^2 - \sqrt {c^4 - {c^2}{u_2^2}}}{u_2}$$
However, I would not know how to proceed from this point on without using taylor expansions...

 

[kuruman]

Proceed to where? I thought the problem asked you to find $v$ and you did. Is there another part to this problem?

 

[Floris Meersschaert]

Alright! I supposed there would be a simplified solution but if that's in then I thank you for your time!

 

[kuruman]

You could make it look prettier by rewriting it as $$v/c=\frac{1-\sqrt{1-\beta^2}}{\beta}$$where $\beta = u/c$. In this form it is immediately obvious why $u/c$ is greater than 1 with a plus sign in front of the radical.

 

[Floris Meersschaert]

Thank you, I appreciate all your help and your time!