Conservation of relavistic energy

[Samuelriesterer]

A Google search for "Lorentz transformation for momentum and energy" turns up e.g.

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/vec4.html

Yes I saw this page but I don't understand it much. I've scoured the web and haven't found anything that explains it well. How would I go about solving for the velocities. I would assume:

For the rest frame:

$\frac{v1}{\sqrt{1 - v1^2}} = \frac{p1}{m1}$
<same for v2>

For the detector frame, use the relavistic velocity equation:

$v1' = \frac{.8 + v1}{1+.8v1}$
<same for v2'>

 

[jtbell]

That approach should work, although as I said it uses more algebra and arithmetic. Be careful about + and - signs on the velocities. Also, this method is sensitive to roundoff errors. Don't round off any numbers until you get to the very end. Ideally, keep intermediate results in your calculator as you go along.

 

[Samuelriesterer]

That approach should work, although as I said it uses more algebra and arithmetic. Be careful about + and - signs on the velocities. Also, this method is sensitive to roundoff errors. Don't round off any numbers until you get to the very end. Ideally, keep intermediate results in your calculator as you go along.

Boy you were right! I worked out all of this yesterday and was way off with the conservation, I thought I was doing something wrong. But I did what you said and left the figures in my calculator and it all worked out! I think I learn more from this forum then in my class :)

Now for the last bit, the pions move off in a direction perpendicular to the kaon. I would assume that the momentum stays the same in the rest frame but the velocities in the detector frame would have to be converted to the y axis:

$u1'_y = \frac{u_y}{\gamma (1- \frac{u_x v}{c^2})}$
<same for u2'_y>

I am having a hard time though identifying the variables:

I have v_kaon = 0, v1, and v2 in the rest frame (which I assume remain the same in this question)
and in the detector frame: v_kaon' = 0.8c, v1', and v2'

 

[jtbell]

Now for the last bit, the pions move off in a direction perpendicular to the kaon. I would assume that the momentum stays the same in the rest frame but the velocities in the detector frame would have to be converted to the y axis:

In the kaon frame, the magnitudes of the pion momenta are the same as before, but their directions are now along the +y and -y directions.

In the detector frame, the pions each come out at some angle between the x and y directions. In order to "boost" a velocity that is "originally" along the y-direction, using a relative velocity along the x-direction, you need to use a "vectorized" version of the velocity-addition formula. I've seen such a thing, but don't remember where offhand. Try Googling for something like "relativistic velocity vector addition."

This is where the Lorentz-transformation method becomes really nice. On the page I linked before, it was written in four-vector-and-matrix format. The boost is still along the x-direction, so the Lorentz transformation matrix is the same here. You just apply it to a different energy-momentum four-vector.

(disclaimer: I haven't actually worked out this part yet...)

 

[Samuelriesterer]

I'll do a little digging and see what I can find. Thanks