How can EM waves maintain their energy over infinite distance?

[k9b4]

EM field strength dies quickly with distance, what's so special about going up and down that allows EM waves to maintain their energy over infinite distance?

 

[Simon Bridge]

EM waves retain their total energy in accordance with the law of conservation of energy.

 

[k9b4]

EM waves retain their total energy in accordance with the law of conservation of energy.

What? Strength of electric field dies out quickly with distance. Why does the strength of an electric field going up and down not die out quickly with distance?

 

[Simon Bridge]

The electric field in the EM waves propagates because it is being generated by a changing magnetic field. In the static case, there is no changing magnetic field to generate more electric field. But it is difficult to see what you are asking - in general, the amplitude of an EM wave will decrease with distance from the source.

 

[k9b4]

The electric field in the EM waves propagates because it is being generated by a changing magnetic field. In the static case, there is no changing magnetic field to generate more electric field. But it is difficult to see what you are asking - in general, the amplitude of an EM wave will decrease with distance from the source.

How does a changing magnetic field generate a changing electric field and vice versa?

 

[Nugatory]

How does a changing magnetic field generate a changing electric field and vice versa?

The relationship between the electrical and magnetic fields at any given point in space is given by Maxwell's equations. Google for "Maxwell equations", but be prepared to deal with a fair amount of vector calculus.

 

[k9b4]

The relationship between the electrical and magnetic fields at any given point in space is given by Maxwell's equations. Google for "Maxwell equations", but be prepared to deal with a fair amount of vector calculus.

But I am looking for a physical explanation, not a mathematical one.

 

[e.bar.goum]

But I am looking for a physical explanation, not a mathematical one.

The physical explanation is just going to be Maxwell's equations written out as a sentence. The physical explanation you want is encoded in the mathematics.

 

[Drakkith]

The EM wave spreads out as it travels, which reduces both the field strength and the energy of any section of the EM wave. Total energy of the wave remains the same, however.

 

[Orodruin]

The physics explanation is the mathematical one, simply because physics is based on quantitative predictions using mathematical models. In some cases it might be possible to describe the mathematical framework in conceptual terms, but it is always going to be a likeness, not a precise statement.

 

[Simon Bridge]

I was building Drakkith's answer, but the question is more about propagation so anticipating...

But I am looking for a physical explanation, not a mathematical one.

... mathematics is the language of physics - mathematical explanations are the physical ones. All the others are artistic answers.

Imagine we can suddenly switch on a point charge. (Worry about how we may do that later.)
We do this at t=0, relativity means that it takes some time for the electric field to reach farther out. It takes 1s to reach 1 light-second away and so on (electric fields travel at the speed of light.)
This is a time-varying electric field, and the change in the field is what travels out.

Any changing electric field is accompanied by a changing magnetic field and vice versa.
(That's actually two of Maxwell's equations.)

All this is pretty inadequate but it may help you get a glimmering.

 

[k9b4]

The physical explanation is just going to be Maxwell's equations written out as a sentence. The physical explanation you want is encoded in the mathematics.

... Which I cannot interpret because I lack knowledge of mathematics
... Which is why I come to a physics forum for help

 

[NTW]

I propose a 'mental experiment', a sort of 'literary answer'.

Let's imagine a point source in space, emitting light for 1/300000 s. The light will propagate as a spherical shell 1 km thick, and the energy contained in that shell will be the same, one second or one hour after the flash... The difference will be the energy density, but not the absolute value of that energy, that will remain constant.

 

[Simon Bridge]

Which I cannot interpret because I lack knowledge of mathematics
... Which is why I come to a physics forum for help

... and we cannot help you unless you are prepared to acquire the knowledge ;)
You'll just have to accept that EM waves travel through space is a fact of Nature which has been empirically confirmed.

What do you need to know for?

 

[k9b4]

The light will propagate as a spherical shell

It wouldn't be perfectly spherical though would it? Because the point charge in oscillating in one plane?

You'll just have to accept that EM waves travel through space is a fact of Nature which has been empirically confirmed.

I do accept that this happens. I am asking why it happens.

What do you need to know for?

For my own understanding

 

[Orodruin]

... Which I cannot interpret because I lack knowledge of mathematics
... Which is why I come to a physics forum for help

Without knowledge of mathematics, you will have to accept that certain things can only be described with analogues which are more or less flawed in nature. Some are going to be good enough to get an intuition about the particular phenomenon, but others will not. If you really want to understand on a more fundamental level, it is necessary to learn mathematics.

Compare this to reading a poem by some great artist. If you do not speak the language, you will need a translation and more often than not translated poems simply are not the same as the original and several subtleties will be lost in translation.

 

[NTW]

It wouldn't be perfectly spherical though would it? Because the point charge in oscillating in one plane?

(...)

It doesn't matter, perfectly spherical or not, for the purpose of the mental experiment.

 

[k9b4]

I was building Drakkith's answer, but the question is more about propagation so anticipating...
... mathematics is the language of physics - mathematical explanations are the physical ones. All the others are artistic answers.

Imagine we can suddenly switch on a point charge. (Worry about how we may do that later.)
We do this at t=0, relativity means that it takes some time for the electric field to reach farther out. It takes 1s to reach 1 light-second away and so on (electric fields travel at the speed of light.)
This is a time-varying electric field, and the change in the field is what travels out.

Any changing electric field is accompanied by a changing magnetic field and vice versa.
(That's actually two of Maxwell's equations.)

All this is pretty inadequate but it may help you get a glimmering.

Thanks, this is more what I was looking for. I know that the change in the field is what is traveling.

But why is it that an unchanging electric field does not affect things an infinite distance away, but a changing electric field does?

It doesn't matter, perfectly spherical or not, for the purpose of the mental experiment.

But it would not be perfectly spherical right? Because if the point charge is oscillating vertically, then directly above the point charge there would be no change in electric field?

EDIT: There would be change in electric field, but the change would be oscillating in the direction of travel, like a sound wave right?

 

[Simon Bridge]

...an unchanging electric field does not affect things an infinite distance away, ...

... this is false.
Where are you getting these ideas from?

if the point charge is oscillating vertically, then directly above the point charge there would be no change in electric field?

Note: if you had an oscilating electric charge, and it was oscillating vertically, then that implies there is gravity. You don't want to include gravity, that's an un-needed complicaton. But we do need a label for the oscillation axis and "oscillation axis" is a bit much to write. Let's say the charge bobs up and down on the z-axis. We can pick the +z direction as the direction of the first displacement.

Some distance along the +z axis, would be "directly above" in your description.
An observer there would have an electric charge getting closer and then further away.
The electric field strength varies with distance.
This means the electric field strength is getting bigger and smaller with time.

 

[Nugatory]

If you want a perfectly spherical wavefront, symmetrical in all directions, you'll have to generate it in a different wa... But this is a digression from the original question about how electromagnetic radiation propagates so should be discussed in a different thread.

 

[k9b4]

... this is false.

So do the not-oscillating electrons on Earth affect the electrons on the moon?

My hand is not repelled from my desk until I get very close to it.

 

[Orodruin]

So do the not-oscillating electrons on Earth affect the electrons on the moon?

My hand is not repelled from my desk until I get very close to it.

This is because your hand is overall neutral and forces on positive and negative forces cancel. If you somehow removed all the electrons from the solar system, the electrical forces between the remaining nuclei woul be orders of magnitude stronger than the gravitational forces.

 

[k9b4]

This is because your hand is overall neutral and forces on positive and negative forces cancel. If you somehow removed all the electrons from the solar system, the electrical forces between the remaining nuclei woul be orders of magnitude stronger than the gravitational forces.

Oh yeah, I didn't think of that.

So then, could you please explain, why, if we have an electron surrounded by atoms, does it not affect far away things. But as soon as we start oscillating that same electron, far away things are now affected?

 

[Drakkith]

Oh yeah, I didn't think of that.

So then, could you please explain, why, if we have an electron surrounded by atoms, does it not affect far away things. But as soon as we start oscillating that same electron, far away things are now affected?

It does affect distant objects. But electrical charges rarely have the chance to separate to the extent that you would notice their effect at great distances since strong electric fields cause ionization and results in the neutralization of the charges.

 

[Simon Bridge]

So do the not-oscillating electrons on Earth affect the electrons on the moon?

In principle - in principle, this is the case yes.
By definition an electric field is defined by it's effects on charges. If it has no effect then the field is zero.
You are observing that moving the electrons about vigorously has a more noticeable effect than more static charges - for instance, we can use the effect to send radio signals to the moon.

If you just push uniformly on one end of a long metal bar, there is no movement at the other end unless the force is strong enough to push the whole thing... but if you tap lightly on the end of the bar, the the other end moves slightly. The slight movement is detectable as sound. Sometimes a small action can have a bigger effect than a large one.

What the radio receiver detects is the change in the position of the charges in the circuit which is responds to as a changing electric current. How is a static field supposed to make those changes - the electrons in the receiver are pulled one way by the field, and another way by their atoms, resulting in a different equilibrium than if the field were not there.

I have already told you that artistic descriptions are inadequate to answer your questions - all you are noticing is that the answers are inadequate.
Well... you were warned.

It would help us make better answers if you told us what the information was for.

 

[k9b4]

It would help us make better answers if you told us what the information was for.

What do you mean for?

 

[Dale]

why, if we have an electron surrounded by atoms, does it not affect far away things. But as soon as we start oscillating that same electron, far away things are now affected?

There is nothing special about an electron. The important thing is charge density and current density. If you have a net charge then you will get a field from static charges. If you have a net current then you will get a field from the current.

The reason "why" for your question above is that in the first case you have no charge or current density, but in the second case you have a current density.

 

[russ_watters]

Why not just consider the photon to be a particle and say that it must travel forever or until it hits something in accordance with Newton's first law? Just like any other object in motion.