Energy of a particle in an orthoganally moving rest frame

[Pianodan]

Homework Statement

Two particles of equal mass M and equal net energy E are approaching each other at an angle of 90 degrees. Find the net energy of one particle in the rest frame of the other. Do not assume that the velocities are small compared to the speed of light.

Homework Equations

The Lorenz transformations and relativistic energy.

The Attempt at a Solution

I don't know. I've been bashing my head against this one for hours. It looks so simple, and yet I just can't make it work. I've been reading pages about invariant mass, relativistic addition formulas, 4-vectors, and so on, but I just can't seem to conceptualize it. This is a sample qualifying problem that I'm supposed to be able to solve in twenty minutes or less, and the exam is now two weeks away. Frustration is riding very high.

My guesses thus far start from the assumption that in the frame of the particle moving along the x axis, the components of the velocity of the particle moving on the y-axis is:

$\left\{-\gamma v, -v, 0\right\}$

and t' in the particle rest frame is $\gamma t$

Should I be trying to figure out the relative velocities of the two particles, and using that to compute a new gamma for E = gamma m c^2? If so, how? I've tried just taking the root of sums of the squares of the components, but it doesn't look right. If this isn't the right approach, what am I missing?

Also, please tell me I'm not crazy for trying to reenter this field after 13 years away.

 

[sgd37]

rotate by 90 degrees and boost to v

and you are crazy, and I'm not saying it to make fun of you

 

[mathfeel]

Write out the four-momentum of both particles. Then boost according to the rule of boosting a four-vectors.

 

[Dick]

1
My guesses thus far start from the assumption that in the frame of the particle moving along the x axis, the components of the velocity of the particle moving on the y-axis is:

$\left\{-\gamma v, -v, 0\right\}$

There is such a thing as a good kind of crazy. Now, why did you put a gamma on the x component of velocity?

 

[vela]

Lorentz transformations are your friends. They apply to not only (t, x, y, z) but to any four-vector, like, say, a particle's four-momentum.

 

[Pianodan]

@Dick - the gamma is there because I was assuming there was a relativistic contraction of the x component of the other particle's motion relative to the first one. I think now it should probably be 1/gamma, if that's even right.

@Everyone else - I appreciate the advice to write out the four vectors and transform, but I need more help - I just don't know how to do that.

I'll take a stab at the vectors. Call the particle moving on x A, and the one on y B, so we're looking for the energy of B in the A rest frame. (obviously, it'd be the equivalent problem, and the same final energy if you did it the other way around)

In the lab frame:

$v_{x} = \left\{1, v, 0, 0 \right\}$
$v_{y} = \left\{1, 0, -v, 0\right\}$

$p_{x} = \left\{E/c, p , 0, 0 \right\}$
$p_{y} = \left\{E/c, 0, p, 0\right\}$

where $p = \sqrt{E^{2} - m c^{2}}$

Since we are looking from the frame of a particle on the x axis, does that mean we can use the standard Lorentz transform shown http://en.wikipedia.org/wiki/Lorentz_transformation#Matrix_form" (my itex is improving, but I'm not going to type all THAT in!)

If I perform that multiplication, I get for the momentum of B in the rest frame of A:

$p' = \left\{\frac{\gamma E}{c}, - \beta \gamma, -p, 0\right\}$

However, I don't know how to extract a final energy scalar from this vector, or if this is even the right vector. Thanks for the encouragement so far. A little more help?

 

[vela]

The formulas weren't rendering because there were errors in the markup, like using \left with no matching \right.

I'll take a stab at the vectors. Call the particle moving on x A, and the one on y B, so we're looking for the energy of B in the A rest frame. (obviously, it'd be the equivalent problem, and the same final energy if you did it the other way around)

In the lab frame:

$v_{x} = \{1, v, 0, 0 \}$
$v_{y} = \{1, 0, -v, 0\}$

$p_{x} = \{E/c, p , 0, 0 \}$
$p_{y} = \{E/c, 0, p, 0\}$

where $p = \sqrt{E^{2} - m c^{2}}$

Good! You're just missing a negative sign for the y-component of p_y. Just a typo it seems.

Since we are looking from the frame of a particle on the x axis, does that mean we can use the standard Lorentz transform shown http://en.wikipedia.org/wiki/Lorentz_transformation#Matrix_form" (my itex is improving, but I'm not going to type all THAT in!)

If I perform that multiplication, I get for the momentum of B in the rest frame of A:

$p' = \{\frac{\gamma E}{c}, - \beta \gamma, -p, 0\}$

However, I don't know how to extract a final energy scalar from this vector, or if this is even the right vector. Thanks for the encouragement so far. A little more help?

Also, my formulas seem to have stopped working. What's up with that?

So far, so good. The x-component of your four-momentum is missing a factor of E/c. You still need to express $\gamma$ and $\beta$ in terms of E and M.

The energy is just the time component of the four-momentum.

 

[Pianodan]

The formulas weren't rendering because there were errors in the markup, like using \left with no matching \right.

Oops - that's what I get for trying to infer Latex markup instead of using the reference when I'm not quite ready.

Good! You're just missing a negative sign for the y-component of p_y. Just a typo it seems.

Yes indeed - I had it right on the paper.

So far, so good. The x-component of your four-momentum is missing a factor of E/c. You still need to express $\gamma$ and $\beta$ in terms of E and M.

The energy is just the time component of the four-momentum.

Oh, well, given that I don't need to bother with beta then, do I? So the answer for the final E in the rest frame of A is just:

$E' = \frac{\gamma E}{c}$

But since we know

$E = \gamma m c^{2}$

Then in regular units,

$E' = \gamma E = \frac{E^{2}}{m c^{2}}$

Is it really that simple?

 

[vela]

Yup, it's that easy if you stick to using energy and momentum. I find it's generally a good idea to avoid using velocities if you can.

Just thought I should mention the four-velocities are actually
\begin{align*}
v_A^\mu &= \gamma(c,v,0,0) \\
v_B^\mu &= \gamma(c,0,-v,0)
\end{align*}
If you Lorentz-transform those, you can see what the four-velocities are in the rest frame of A. You still have to do a little work after that to see what the three-velocity of B with respect to A is. You might want to try it just to see if you get the answer you think you should get.