# EM Radiation versus Transverse Waves

1. Aug 4, 2005

### DaTario

Electromagnetic wave are said to travel with velocity c, in vacuum and are also said to be a transverse wave. But consider this:

Let a positive charge Q be at point P of a three dimensional space.

Let this charge, at time t=0, begin an uniform circular motion with angular velocity OMEGA = 2 Pi rad/s and with radius R = 1 m. It moves just one turn to make things simpler.

Supose we (the observers) are very distant from P but in the plane of the particle's circular path. Let this distance be 10 light-seconds for instance.

At our position we will experience after t = 10 s, modifications on the electric field generated by the charged particle. It just so happens that in some part of the circular path the particle has come nearer to us, and in the other, the particle has gone farther, so that the field has presented not only transverse modifications but logitudinal modifications as well. Aren't these logitudinal modifications considered part of the radiation emitted by the particle ? Does these longitudinal dynamics of the field propagates with c also ?

Best Regards,

DaTario

2. Aug 4, 2005

### Antiphon

Yes.

But the logitudinal components (of the radiation) fall off with distance as $$1/r^3$$ so that
at any appreciable distance relative to the wavlength you only have the transverse
components which fall off as $$1/r^2$$.

3. Aug 4, 2005

### DaTario

Ok, but if it is just a matter of one component falling with distance more rapidly than the other, don't you think that the term "nearly transverse" should be used more often ? At least to shed some light on this aspect of the question ?

4. Aug 4, 2005

### Antiphon

Well, like so many other cases in Physics the terminology caters
to the well-versed not to the newcomers.

5. Aug 4, 2005

### DaTario

It is interesting to add here that, in the case of dipole emission, along the line conecting the two opposite charges, the field is considered zero. I understand it for there is a subtraction of field strenghs due to the positively and negatively charged particles' contributions to the field in this subspace, but it isn't precisely zero along this line.

Thank you,

Best Regards

DaTario

6. Aug 4, 2005

### Antiphon

It is not zero anywhere along the line.

The field of the positive charge points away from it.
The field of the negative charge points toward it.

When you add these, they are not zero anywhere in space.

7. Aug 4, 2005

### Hans de Vries

It's actually $$1/r$$ for the radiative (transversal) component and
$$1/r^2$$ for the generalized Coulomb (longitudinal) component.

The latter can have a small transversal component as well since it
points to the location were the emitting charge would have been if
it had continued to move in the same direction at the same speed.

(Liénard Wiechert potentials)

Regards, Hans

Last edited: Aug 4, 2005
8. Aug 4, 2005

### Antiphon

I stand corrected. I was thinking of radiated power instead of the fields when I wrote that.

9. Aug 4, 2005

### DaTario

I know it is not zero along the line, but as I remember, the polar diagram of this emission (supose the charges are located along a horizontal line) reminds the shape of the number eight. (8), so that, horizontally, this diagram suggests no field strengh.

Could you please confirm if you agree with this statement ?

10. Aug 4, 2005

### DaTario

Correct me if I am wrong in the following:

We can always apply Coulomb's Law to write the field at a given position.
If the charge is initially at the origin and we are look at the field at the end of the vector R. Then eu could write for static situation:
$$E = 9. 10^9 \frac{Q}{R^2} [\tex] in the direction of te vector R (as Q > 0). If the charge moves very very slowly, we may adopt the approximative stand point according to which the field has a static configuration at any given instant. (it seems important to this argument to work with an universe with finite volume V). But rigorously speaking if the charge was always in the position R0 and, at time t=0, it moves to the position R0 + dR, then we have to take into consideration the limitting velocity c, and the field's spatial configuration at a given instant can be roughly divided in three regions: (1) the already corrected field, the region where the field resembles that of an static charge at Ro + dR, (2) an old fashioned region, where the field resembles that on an static charge still located at R0, and (3) a "being corrected region, where we may face intermediary states between these two configurations ((1) and (2)). It is precisely the third region which represents the radiation. I am really not sure about the correctness of this reasoning, although it seems quite reasonable to me. Best regards, DaTario 11. Aug 5, 2005 ### Hans de Vries You're reasoning is right here. In the 1900 Emil Wiechert proves that the Lienard-Wiechert retarded potentials give the correct results in cases where the effects due to the finite speed of light can't be neglected. Calculations must be made by assuming that the potentials, both the electric and magnetic vector potential, propagate from the source with the speed of light.The fields E and H can be recovered by differentiating in the right way. The fields calculated for charges moving at speeds close to c give the same results as the Special Theory of relativity (note that this was five years before einsteins famous 1905 year ) Regards, Hans 12. Aug 5, 2005 ### DaTario Ok, but why is there so much emphasys in the acceleration of the charge as an important condition for it to emit radiation ? According to the arguments I gave before, I see no reason to treat differently one moving electron (with constant velocity) and another one with non zero acceleration. Does an electron moving with constant velocity have straight field lines ? I once have seen it but it seemed quite strange to me. 13. Aug 5, 2005 ### Antiphon You are thinking of the radiation pattern. There will indeed be no radiation component (of the transverse type) along the axis of the dipole. But in general, there will be a non-zero but tiny radial component even along this axis. For an osciallating dipole there will be brief moments where the two charges are located at the same place. In this moment there will be zero fields and this zero may propogate outwards. The acceleration topic is not trivial. But for the uniformly moving charge, the field lines are strait in the reference frame of the charge. In any other inertial frame they are bent and are accompanied by a magnetic field which is only not seen in the reference frame of the charge. Edit: Clarification: After Hans de Vries, yes the lines are strait but they are tilted with respect to one another. Last edited: Aug 5, 2005 14. Aug 5, 2005 ### Hans de Vries The radiative part (which decreases with 1/r) is proportional to the acceleration. No acceleration = no radiation. Below a copy from a document I wrote: ---------- The Lienard Wiechert potentials. Alfred Lienard (in 1898) and Emil Wiechert (in 1900) determined the potentials of an arbitrarily moving and accelerating point charge. It is proved that the potentials spread from the source with the speed of light. The equations for the electric potential V and the magnetic vector potential A are close to the classical ones: [tex]V(\vec{x},t) = \frac{q}{4 \pi \epsilon_0 r_{ret}} \left[ \frac{1}{(1-\beta \cos\phi)} \right] , \ \ \ \ \vec{A}(\vec{x},t) = \frac{q}{4 \pi \epsilon_0 r_{ret}} \left[ \frac{\vec{\beta}}{(1-\beta \cos\phi)} \right]$$

Where: $$\left\{ \begin{array}{lcl} \vec{\beta} & = & \mbox{speed vector of the charge: } \vec{v}/c \\ r_{ret} & = & \mbox{distance from the retarded charge.} \\ 1/(1-\beta \cos\phi) & = & \mbox{compression or 'shockwave' factor } \end{array} \right.$$

Note that we work with the retarded charge: The location where the
charge was at the moment when the potentials we measure left the
point charge. The only thing which needs some explanation is the
compression or 'shockwave' factor. It can become infinite in front
of the moving charge (psi = 0) when the speed goes to c. This effect
is equal to the shockwave building up in front of a plane which
approaches the speed of sound. We see that all the information that
is needed is: (1) the charge, (2) its location and (3) its speed.

The Electro Magnetic Fields

It is interesting that not the electric and magnetic fields E and B
are propagating away with the speed of light but rather the
potentials V and A. If we would assemble similar formula's for the
fields then we get incorrect results. We do get the EM fields back
however by carefully differentiating the potentials in space and
time. We get:

$$\vec{E}(\vec{x},t) \ \ = \ \ \frac{q}{4 \pi \epsilon_0 r^2_{ret}} \left[ \frac{(1-\beta^2) \ \vec{r}_{ph} }{(1-\beta \cos\phi)^3} \right] \ \ \ + \ \ \ \ \frac{q}{4 \pi \epsilon_0 r_{ret}} \left[ \frac{\hat{r}_{ret} \times (\vec{r}_{ph} \times \vec{a})} {c^2(1-\beta \cos\phi)^3} \right] ,$$

$$\vec{B}(\vec{x},t) \ \ = \ \ \hat{r}_{ret} \times \vec{E}$$

where: $$\left\{ \begin{array}{lcl} \vec{a} & = & \mbox{accelaration vector of the charge.} \\ \hat{r}_{ret} & = & \mbox{vector of length 1 from retarded charge towards } (\vec{x},t) \\ \vec{r}_{ph} & = & \mbox{the vector } (\hat{r}_{ret} -\vec{\beta}) \mbox{ from the phantom location to } (\vec{x},t) \end{array} \quad \right.$$

with $v \ll c$ this simplifies to:

$$\vec{E}(\vec{x},t) \ \ = \ \ \frac{q}{4 \pi \epsilon_0 r^2_{ret}} \ [ \hat{r}_{ret} ] \ \ \ - \ \ \ \ \frac{q}{4 \pi \epsilon_0 r_{ret}} \ [ \vec{a} / c^2 ]$$

We see that the first term is the standard Coulomb field. The
second term is the radiation term which is proportional to the
acceleration of the charge. It has a direction opposite to the field
which would cause the charge to accelerate. The charge opposes a
change in speed. It is this effect in general which causes
self-induction.

The 'Phantom' location of the charge.

Another interesting phenomena is the appearance of the 'phantom'
location in the general formula. This is the location were the
charge would be if it would have continued at the same speed after
the potentials left the charge at the retarded position. The force
from the generalized Coulomb field (the first term) points to the
phantom location. At the other hand, the radiation field has its E
and B components orthogonal to the vector coming from the retarded
position. We see this effect for instance if we look at the Sun. The
light comes from the retarded position while the gravitation seems
to come from the phantom position. Yes, in (Newtonian) gravitation
it is also the potential which spreads with the speed of light.

-----------

Regards, Hans

Last edited: Aug 5, 2005
15. Aug 5, 2005

### Hans de Vries

Yes, it has.

They are straight in any non-accelerating reference frame however not spherical
symmetric anymore but compressed with a factor $1/(1-v^2/c^2)$ in the direction
of the motion.

Regards, Hans

Last edited: Aug 5, 2005
16. Aug 5, 2005

### Antiphon

Hans, aren't these the classical potentials?

Could you elaborate on that? This sounds interesting.

17. Aug 5, 2005

### Hans de Vries

The terms between the square brackets are what makes them different from
the classical ones.

It's shown that only if you let V and A propagate away with c from the source
and then differentiate to obtain E and B you get the right results. This confirms
what we also see in QM: That the potentials and not the fields are the more
fundamental quantities.

Regards, Hans

18. Aug 5, 2005

### Antiphon

I'm comfortable with that. What I''m not comfortable with is that
you can't get a correct derivation resorting to the fields alone. This
seems like a formulational matter in setting up the proper equations.

That is, if I properly manipulate the field equations they should always
end up providing the same solutions as if I solve for potentials first
and differentiate.

19. Aug 5, 2005

### Hans de Vries

The differentiation is not the same in the two cases. A Δx connects potentials
which are emitted at different times. If you take E and B as propagating away
with c then you use quantities which are already pre-differentiated presuming a
source charge at rest. I do presume that, one can not recover the solutions
in the latter case, taken into account that the differentiation of E and B means
the loss of the "absolute values" of the potentials.

Regards, Hans