# Reconciliation of Maxwell's Equations with Relativity

1. May 16, 2008

### Usaf Moji

Let's say you're in a very fast car that can accelerate from zero to, eventually, 0.7c. I understand that as the car moves faster and faster, the driver will observe the light that hits him in the face to be of higher and higher frequency. This seems consistent with things becoming narrower in the direction of travel. But it seems inconsistent with things appearing to happen more slowly in time.
According to Maxwell, the rate of change of the E field with respect to space must be equal and opposite to the rate of change of the B field with respect to time. So, the driver sees light of higher frequency as he goes faster, and thus greater rate of change of E with respect to space; therefore, I would expect him to also see a greater rate of change of B with respect to time.

But if B is changing faster with respect to time, doesn't this contradict the idea that the observer sees things around him happening slower in time?

What am I missing here?

2. May 16, 2008

### Mentz114

Your expectation is unfounded. The equation of a plane EM wave travelling in the z direction can be written

$$E_x = E_{0x}cos( \omega t + kz)$$ for the electric part

$$B_y = B_{0y}cos( \omega t + kz)$$ for the magnetic part.

If you receive at a boosted source, the frequency changes for both identically.

Last edited: May 16, 2008
3. May 16, 2008

### Staff: Mentor

Not sure why you're picking on B. The same considerations would apply to E and it didn't seem to bother you that the rate of change of that field appeared faster.

In any case, the relativistic Doppler effect has two elements: the standard increase in frequency due to the frequency source approaching the observer (as in sound) and a decrease in frequency due to time dilation. The first effect is always greater than the second, at least for the longitudinal Doppler effect. See: Relativistic Doppler Shift

[Mentz114 beat me to it! ]

4. May 16, 2008

### Usaf Moji

Ok, but what I meant was, that for a Maxwell EM wave, the following conditions are true:
$$\frac{\partial{E}}{\partial{z}}$$ = - $$\frac{\partial{B}}{\partial{t}}$$

and

$$\frac{\partial{B}}{\partial{z}}$$ = - $$\epsilon_{o}\mu_{o}\frac{\partial{E}}{\partial{t}}$$

where z is the spatial direction of propagation, and t is time.

So, my point was, where the frequency of light increases, the rate of change of both the E and B fields with respect to space should increase, as expected.

But the (magnitude of the) rate of change of both the E and B fields with respect to time should also increase to satisfy Maxwell - but if this happens, things would seem to be happening faster in time (time contraction) rather than time dilation.

So does this mean that the effects of space contraction would cancel the effects of time dilation - i.e. that if left to relativity alone, without resort to doppler, the wavelength of the light wouldn't seem to change at all? Ugg, so confused!

Last edited: May 16, 2008
5. May 16, 2008

### Mentz114

Usaf,

I can't understand your problem, sorry.

I think if this equation is boosted ( actually the source and/or receiver is boosted ) in the z direction)

$$E_x = E_{0x}cos( \omega t + kz)$$

it will go to

$$E_x = E_{0x}cos( \gamma \omega t + \gamma kz)$$

so the velocity

$$\frac{\omega}{k} \rightarrow \frac{\gamma \omega}{\gamma k}$$

remains constant ( k has dimension reciprocal length ).

This doesn't look right, maybe an expert can help.

M

6. May 16, 2008

### lbrits

You are treating $$E$$ and $$B$$ as scalar quantities. Perhaps you should treat them as components of a tensor, keeping in mind that they themselves transform into eachother under Lorentz transformations.

7. May 16, 2008

### Mentz114

My post above is naive.

To boost a light source we have to write the plane wave as a wave vector and apply a Lorentz boost. I did the calculation, and then found it is nicely laid out here -

http://en.wikipedia.org/wiki/Wave_vector

The wave vector for a light plane wave moving in the x-direction is ( all vectors transposed to save space)

$$\left( \frac{\omega}{c}, k_x, 0, 0 \right)$$

Boosted in the x direction this gives -

$$\left( \gamma \left( \frac{\omega}{c}-\beta k_x\right), \gamma\left( k_x-\beta\frac{\omega}{c}\right), 0, 0 \right)$$

and if we let

$$\frac{\omega'}{c} \equiv \gamma \left( \frac{\omega}{c}-\beta k_x\right)$$

then with $$c=\frac{\omega}{k}$$

$$\omega' = \gamma\left(\omega - v_xk\right) = \omega\gamma\left( 1 - \beta)$$

as expected.

Last edited: May 16, 2008