Electric field of an accelerated charge

[ShayanJ]

I know that an accelerated charge produces a changing electric field and so propogates electromagnetic waves.I want to know the reason or the mathematical reasonings.
thanks

 

[James50]

Imagine a skipping rope with a person at either end. If you flick the rope while the other holds it still, a wave travels along it. This wave in the rope is analogous to electromagnetic radiation, and your hand is the accelerating charge. The greater the acceleration, the stronger the wave.

So if you have a charge, just sitting there, it will just have radial electric field lines which only represent a potential to do work. Now, accelerate that charge and a kink will develop in those lines. This kink, as it moves away from the charge (at the speed of light, no less!) is propagating energy. The kink will be largest in the direction perpindicular to the acceleration of the charge.

If you want to know the maths, check out Maxwell's equations and the Poynting vector on wikipedia. Be warned, you will need to know something about differential equations and operators!

 

[ShayanJ]

I meant how is it that an accelerating charge produces a changing electric field.
How that can be derived from maxwell equations?
thanks

 

[James50]

A charge is effectively defined as a source of an electric field. If that source accelerates, then the electric field must move with it [although the movement can only propagate at the speed of light]. The derivation involves manipulating Maxwell's equations using vector identities into the form of a wave equation. Check it out here http://en.wikipedia.org/wiki/Electromagnetic_radiation#Derivation

 

[santo35]

Imagine a skipping rope with a person at either end. If you flick the rope while the other holds it still, a wave travels along it. This wave in the rope is analogous to electromagnetic radiation, and your hand is the accelerating charge. The greater the acceleration, the stronger the wave.

So if you have a charge, just sitting there, it will just have radial electric field lines which only represent a potential to do work. Now, accelerate that charge and a kink will develop in those lines. This kink, as it moves away from the charge (at the speed of light, no less!) is propagating energy. The kink will be largest in the direction perpindicular to the acceleration of the charge.

If you want to know the maths, check out Maxwell's equations and the Poynting vector on wikipedia. Be warned, you will need to know something about differential equations and operators!

great...but i don't think all acclerating charges are capable of producing EM waves..coz if that was the case then all the electrons in plane current carrying wire shud produce EM waves...i think those charges which are ossilating in SHM can produce EM waves...which would thus cause harmonic variation in Electric and magnetic field ...
am i wrong ?

 

[Dale]

I meant how is it that an accelerating charge produces a changing electric field.
How that can be derived from maxwell equations?
thanks

Here you go

http://en.wikipedia.org/wiki/Liénard–Wiechert_potential
http://fermi.la.asu.edu/PHY531/larmor/index.html

 

[James50]

great...but i don't think all acclerating charges are capable of producing EM waves..coz if that was the case then all the electrons in plane current carrying wire shud produce EM waves...i think those charges which are ossilating in SHM can produce EM waves...which would thus cause harmonic variation in Electric and magnetic field ...
am i wrong ?

You are right and wrong - wrong to think that what you said is incorrect, but right to assume it would happen. Yes, all AC currents in normal wires produce EM radiation. However, at a frequency of 50Hz under a fairly low current, the energy propagated in the EM radiation isn't too great. How do you think, for example, radio and television transmissions are sent? Using an electronic oscillator - which is effectively just an AC current (at very high frequency) attached to a great big whopping transmitting aerial.

Remember - electromagnetic radiation (light, photons) is just a "harmonic variation in electric and magnetic field", as you put it. Light is just a an electric field (and magnetic, but it is considerably weaker) that points in one direction, then flips over and points in the other at a rate defined by its frequency. If you 'fired' light at something with charge, then it literally oscillates with the electric field of the light (assuming it's light enough to be moved quick enough!)

Of course, if you're talking about direct current, that's a little different because it is always going in one direction and is generally assumed to be at a constant current (depending on the layout of the circuit, and what you've got plugged in). Even in direct current you can get acceleration mind. But even assuming all direct current in a circuit is of constant current (direct just means unidirectional), you must still loop it back to its start point, it must still be accelerating accelerating around a centre point. Cyclotron radiation, for example, could be argued to be be DC or AC. It gives off an AC EM wave, but only because it travels DC-style around a centre.

 

[santo35]

then you mean to say that all acclerating charges are capable of proudcing EM waves? i mean like if i just apply a force to a positive charge and give you no information on what type of force or direction or magnitude i am giving to that charge ypu can still say that EM wave would be produced by that charge?

 

[James50]

They are not only 'capable' of producing EM waves, they always will produce EM waves when accelerated. As I've said, a charge is a source of an electric field. When you accelerate that charge, the electric field must move with it in a radial fashion (regardless of frame of reference), creating a 'kink' in the electric field lines. This kink is defined as EM radiation - it is a propagation of energy.
And to get any amiguity out of the way, in answer to your question:
Yes, always.

 

[RedX]

You are right and wrong - wrong to think that what you said is incorrect, but right to assume it would happen. Yes, all AC currents in normal wires produce EM radiation. However, at a frequency of 50Hz under a fairly low current, the energy propagated in the EM radiation isn't too great. How do you think, for example, radio and television transmissions are sent? Using an electronic oscillator - which is effectively just an AC current (at very high frequency) attached to a great big whopping transmitting aerial.

Would EM radiation of frequency 50 Hz get absorbed by everything? Is it true in general that the higher the frequency, the less absorption there is? I know that with 50 Hz there is less scattering (so that the sky is blue rather than red), but there should be more absorption.

Remember - electromagnetic radiation (light, photons) is just a "harmonic variation in electric and magnetic field", as you put it. Light is just a an electric field (and magnetic, but it is considerably weaker)

Aren't the magnetic and electric fields of light the same strength, in Gaussian units?

Of course, if you're talking about direct current, that's a little different because it is always going in one direction and is generally assumed to be at a constant current (depending on the layout of the circuit, and what you've got plugged in). Even in direct current you can get acceleration mind. But even assuming all direct current in a circuit is of constant current (direct just means unidirectional), you must still loop it back to its start point, it must still be accelerating accelerating around a centre point. Cyclotron radiation, for example, could be argued to be be DC or AC. It gives off an AC EM wave, but only because it travels DC-style around a centre.

But if you look at the electrons in a current individually, they are accelerating and crashing, but overall they have an average drift speed proportional to the current. When they accelerate and crash, doesn't this produce radiation?

 

[James50]

Would EM radiation of frequency 50 Hz get absorbed by everything? Is it true in general that the higher the frequency, the less absorption there is? I know that with 50 Hz there is less scattering (so that the sky is blue rather than red), but there should be more absorption.

I am not even sure what the question "Would EM radiation of frequency 50 Hz get absorbed by everything?" even means. There are many types of scattering, and there are many factors that determine how things scatter. You cannot generalise it by saying "higher frequencies scatter more" as that isn't necessarily true. And just because something may scatter more doesn't mean it has to be absorbed less (or vice versa!) The sky is blue due to Rayleigh scattering.

Aren't the magnetic and electric fields of light the same strength, in Gaussian units?

I think you should check out how Gaussian units actually work, and what their point is.

But if you look at the electrons in a current individually, they are accelerating and crashing, but overall they have an average drift speed proportional to the current. When they accelerate and crash, doesn't this produce radiation?

You are mixing up drift velocity in DC and a lack thereof in AC. It is an empirical law of the universe, accelerating charge produces EM radiation!

 

[RedX]

I think you should check out how Gaussian units actually work, and what their point is.

The relationship between the magnetic and electric fields is:

$$\vec{B}=\frac{1}{c}\vec{k} \times \vec{E}$$
where k is the direction of propagation.

But it doesn't make sense to me to compare fields of different units - that would be like comparing apples to oranges. So using a system of units where E and B have the same units, you'd get:
$$\vec{B}=\vec{k} \times \vec{E}$$
which says that the B-field and the E-field are of the same strength.

The B-field might not create a large force if the test charge on which it acts is slowly moving. But the fields themselves should be of the same strength.

 

[James50]

Again, you are misunderstanding the point of Gaussian units.

 

[RedX]

Again, you are misunderstanding the point of Gaussian units.

I'm not sure what you mean. Even in SI units, isn't the energy in the B-field the same as the energy in the E-field, i.e.:

$$\frac{1}{2} \epsilon_0 E^2=\frac{1}{2\mu_0}B^2$$