Thornton and Marion, Chapter 2, Prob. 22

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Homework Statement



This is a four part problem -- the only issue I have is on part (c) so I'll condense the question:

If the equation of motion of a particle of charge q in an electric field and magnetic field is

qv_y B \hat{\mathbf{i}} + \left( qE_y - qv_xB \right) \hat{\mathbf{j}} + qE_z \hat{\mathbf{k}} = m \frac{d}{dt} \left( v_x \hat{\mathbf{i}} +v_y \hat{\mathbf{j}} + v_z \hat{\mathbf{k}}\right),

obtain expressions for v_x \left(t\right) and v_y \left(t\right). Show that the time averages of these velocity components are

\left\langle v_x\left(t\right)\right\rangle = \frac{E_y}{B}

and

\left\langle v_y\left(t\right)\right\rangle = 0

(Show that the motion is periodic and then average over one complete period.)

Homework Equations





The Attempt at a Solution



Solutions to the DE are

v_x \left(t\right) = \frac{E_y}{B} + C_1 \cos \left(\omega t\right) + C_2 \sin \left(\omega t\right)

and

v_y \left(t\right) = -C_1 \sin \left(\omega t\right) + C_2 \cos \left(\omega t\right),

where \omega is the cyclotron frequency. Now, clearly the velocity functions are periodic with period 2\pi/\omega. What do they mean "time averages," and what does the \left\langle \right\rangle notation mean?

Perhaps,

\left\langle v_x\left(t\right)\right\rangle = \frac{ v_x\left(0\right) + v_x \left(2\pi/\omega\right)}{2},

like an arithmetic mean, but I'm just guessing based on the hint (and that doesn't yield the correct answer). Do they mean average velocity?
 
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Hi union68! :smile:
union68 said:
Solutions to the DE are

v_x \left(t\right) = \frac{E_y}{B} + C_1 \cos \left(\omega t\right) + C_2 \sin \left(\omega t\right)

and

v_y \left(t\right) = -C_1 \sin \left(\omega t\right) + C_2 \cos \left(\omega t\right),

where \omega is the cyclotron frequency. Now, clearly the velocity functions are periodic with period 2\pi/\omega. What do they mean "time averages," and what does the \left\langle \right\rangle notation mean?

"Time average" means the average value over time (and the <> notation just means average, or expectation value) …

in this case, theoretically you'd integrate over a whole period, and then divide by the period.

However, in this case it's a bit obvious that the average of a cos or sin over the period is 0, so that just leaves you with … ? :wink:
 
Ah, yes. Integrating over the period and dividing by the period gives you the correct answer. The cosines and sines vanish.

However, what exactly is this procedure though? I guess I've never seen integration related to an average.
 
union68 said:
However, what exactly is this procedure though? I guess I've never seen integration related to an average.

uhh? :redface: … how else would you define an average of something that varies continuously? :smile:

(and how do you think root-mean-square voltage etc is defined? :wink:)
 
After a quick Google search I found some explanation. I have never seen continuous averaging before.

I have no idea what RMS voltage is either -- I'm a math major dabbling in physics.

Thank you for the help.
 
To solve this, I first used the units to work out that a= m* a/m, i.e. t=z/λ. This would allow you to determine the time duration within an interval section by section and then add this to the previous ones to obtain the age of the respective layer. However, this would require a constant thickness per year for each interval. However, since this is most likely not the case, my next consideration was that the age must be the integral of a 1/λ(z) function, which I cannot model.
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