Transforming Relativistic Velocities

  • Thread starter Thread starter psholtz
  • Start date Start date
  • Tags Tags
    Relativistic
psholtz
Messages
133
Reaction score
0

Homework Statement


My question concerns what happens when one is transforming relativistic velocities in more than one dimension.

The Lorentz velocity addition formula is well known. Suppose there is an inertial reference frame F' moving at speed v in the +x direction with respect to a "stationary" lab frame F, and suppose that in F' the speed of a projectile, in the x-direction, is given by u_x'. Then the speed of this projectile, as measured in F, will simply be:

u_x = \frac{u_x' + v}{1 + \frac{u_x' v}{c^2}}

That's all well and good, but suppose that we don't launch the projectile in F' precisely parallel to the x-axis. Suppose rather, that we launch it at some angle to the x'-axis, say \theta'. The question is, what velocity is measured now in the frame F?

Homework Equations



u_y' = u' \sin \theta '

u_x' = u' \cos \theta '

The Attempt at a Solution


My "naive" attempt at a solution would be to take the "vertical" component, on the y-axis, and transform it "as is", and take the x-component and subject that to the Lorentz velocity addition law. In other words:

u_y = u_y' = u' \sin \theta '

u_x = \frac{u' \cos \theta' + v}{1 + \frac{v u' \cos \theta' }{c^2}}

The problem w/ this approach is in the limit as either u' or v approaches c.

Suppose, for the sake of illustration, that v approaches c. Then by the above formula, we would have:

u_x = c

as expected. But the problem is that there is still this u_y component which is finite and non-zero, and when we add that in, by Pythagoras, we get an answer for the velocity in F which is in excess of c.

Apparently, I would think that as the u_x component increases, then the u_y component must somehow decrease... but I'm wondering how to specify this quantitatively..??
 
Physics news on Phys.org
psholtz said:

My "naive" attempt at a solution would be to take the "vertical" component, on the y-axis, and transform it "as is"


Even the "vertical" components need to be transformed. They will transform in a way somewhat similar to the x-component of velocity, but they will have an additional lorentz-factor tacked on. I'd recommend you google the words: special relativity velocity transformation.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top