Order of Operations, Lorentz Transformations & Superposition

1. Jul 15, 2010

univox360

I am wondering about the order of operations concerning the Lorentz transformation of fields and the superposition of fields.

I was given a problem:

Two positively charged electrons start at the origin and then travel along the x axis at a constant speed v in opposite directions. Calculate the electric field in the (far) radiation zone.

My approach was to first transform the electric potential field from each electron's rest frame to the lab frame. Then, in the lab frame I added the two fields together via superposition. Afterward, I calculated the electric field making approximations for the (far) radiation zone.

Is there something fundamentally wrong with this approach? Why?

2. Jul 15, 2010

starthaus

$$E= \frac {Q}{4 \pi \epsilon_0 r^2}* \frac {1- \beta^2}{(1- \beta^2 sin^2 \theta) ^{3/2}}$$

where
$$\beta = v/c}$$ and $$\theta$$ is the angle formed between the observer-charge axis and the direction of charge motion (the x-axis in your case).

So:

$$E_1= \frac {Q}{4 \pi \epsilon_0 r_1^2}* \frac {1- \beta_1^2}{(1- \beta_1^2 sin^2 \theta_1) ^{3/2}}$$

$$E_2= \frac {Q}{4 \pi \epsilon_0 r_2^2}* \frac {1- \beta_2^2}{(1- \beta_2^2 sin^2 \theta_2) ^{3/2}}$$

Now:

$$\beta_1=v/c=-\beta_2$$

So:

$$E_1= \frac {Q}{4 \pi \epsilon_0 r_1^2}* \frac {1- \beta^2}{(1- \beta^2 sin^2 \theta_1) ^{3/2}}$$

$$E_2= \frac {Q}{4 \pi \epsilon_0 r_2^2}* \frac {1- \beta^2}{(1- \beta^2 sin^2 \theta_2) ^{3/2}}$$

Add the two together and you get your result. Is this what you got?

Last edited: Jul 15, 2010
3. Jul 15, 2010

univox360

I transformed the 4-vector potential due to both charges, not the electric field directly.

4. Jul 15, 2010

starthaus

You should be getting the same answer.

5. Jul 15, 2010

univox360

Well, yes they must eventually come to the same answer. But I am more concerned with whether or not the order of operations is important.

6. Jul 15, 2010

starthaus

Can't tell without seeing your calcs. Post them and we'll talk.