5
$\begingroup$

Higher frequencies have a higher energy, simply stated by Planck's constant. When I am transmitting 100 watts on 500 kHz, and transmitting 100 watts on 5 GHz, is there an energy difference in the wave?

$\endgroup$
5
$\begingroup$

If the birthrate in Canada is 400 thousand babies per year, and the birthrate is Russia is 1680 thousand babies per year, is there a population difference in the nations?

Just as birthrate is a rate of stuff (births), power is a rate of energy. A watt is one joule per second, by definition. The frequency is irrelevant. To answer your question we need to know about time.

If you transmit 100 watts on 500 kHz for 1 second, and 100 watts on 5 GHz for 1 second, then each has transmitted the same energy:

$$ 100 \mathrm W \cdot 1 \mathrm s = 100 \mathrm J $$

It doesn't matter what the frequency is, or even that this is electromagnetic power. It could have been mechanical power. 100 watts of any kind for one second is 100 joules.

If by "the wave", you mean "per cycle", then our notion of time is something else. At 500 kHz ($500\cdot 10^3 \text{cycles}/\mathrm s$), each cycle is one 500,000th of a second, or 2 microseconds. Transmitting at 100W, each cycle then carries the energy:

$$ 100 \mathrm W \left ( \frac{1 \mathrm s}{500\cdot 10^3 \: \text {cycle}} \right ) = \frac{100 \mathrm W \cdot 2 \mu \mathrm s}{\text{cycle}} = \frac{200 \mu \mathrm J}{\text{cycle}} $$

At 5GHz ($5\cdot 10^9 \text{cycle}/\mathrm s$):

$$ 100 \mathrm W \left ( \frac{1 \mathrm s}{5\cdot 10^9 \: \text {cycle}} \right ) = \frac{100 \mathrm W \cdot 0.0002 \mu \mathrm s}{\text{cycle}} = \frac{0.02 \mu \mathrm J}{\text{cycle}} $$

Thus, the energy per cycle at higher frequencies is less, which makes sense because there are more of them per unit time.

As Kevin Reid explains, Planck's constant relates to the energy of individual photons, which is relevant to quantum physicists, but not to amateur radio operators.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Suppose you've got a Mack truck and a skinny guy on a moped both traveling at 25 MPH. Someone asks you, "what is the difference in their kinetic energies?" You reply: well if you divide their speeds by their lengths, the moped has a higher MPH per foot. That makes sense because its got more speed per unit length! The problem with that answer is that it leaves out the essential elements of KE: mass and velocity squared. Likewise your treatment of the question above contains none of the components of energy propagation in electromagnetic waves: the E-field and B-field. $\endgroup$ – user2338215 Jul 3 '14 at 21:11
  • $\begingroup$ @user2338215 Can you come up with some E-field and B-field that results in 100W but not 100J/s? If so, please let me know so I can rush to the patent office and cash in on my perpetual motion machine. $\endgroup$ – Phil Frost - W8II Jul 3 '14 at 21:30
  • $\begingroup$ It appears that you are saying that the definition for the Watt in terms of Joules provides great insight into the nature of energy transport in electromagnetic waves. Please explain. $\endgroup$ – user2338215 Jul 7 '14 at 13:33
  • $\begingroup$ @user2338215 the question was not, "please provide great insight into energy transport in electromagnetic waves." It was, "What is the energy difference between two equal-power sources?" which is unanswerable since energy is the time integral of power, and "time" was not part of the question. If you want to delve into subatomic physics you can, but I don't see much point in it when there are more fundamental issues with the question. $\endgroup$ – Phil Frost - W8II Jul 7 '14 at 14:04
  • $\begingroup$ The words used by the question asker were "is there an energy difference in the wave?" His words, not mine. Perhaps you have misread the question? Also, in no way am I suggesting that subatomic physics is required to address that question. I am just asking how the definition of the Watt, or the mathematical relationship between power and energy (your answer above), addresses the asker's question regarding electromagnetic waves? $\endgroup$ – user2338215 Jul 7 '14 at 14:12
4
$\begingroup$

Higher frequencies of electromagnetic radiation have higher energy per photon; “the wave” you transmit is made up of huge numbers of (coherent) photons — for example, at 5 GHz, 100 watts corresponds to $10^{25}$ photons per second. Therefore, the individual photons' energy is not all that relevant.

For the same power (energy per time), count and frequency are inversely proportional; at a higher frequency, there are fewer photons emitted, each carrying more energy. We can derive this from the Planck relation which contains Planck's constant ($ν$ = frequency):

$$E_{\text{photon}}=hν$$

This is a relation of individual photon energy $E_{\text{photon}}$ to frequency; to consider multiple photons we need to take further steps. First, let's relate a total energy by defining $n$ such that $nE_{\text{photon}}$ is the total energy of the transmission, and then using the Planck relation:

$$E_{\text{transmission}} = nE_{\text{photon}} = nhν$$

Second, the example you asked about is in power, not energy, so we need to introduce time; dividing both sides by a time interval $t$ gives us:

$$P_{\text{transmission}} = \frac{E_{\text{transmission}}}{t} = \frac{n}{t}hν$$

Note that $\frac{n}{t}$ is the rate of photons per time.

Then if we divide by $hν$ and rearrange we get:

$$\frac{P_{\text{transmission}}}{h\nu} = \frac{n}{t}$$

Plugging in your example values gives us values for $\frac{n}{t}$ of:

$$\frac{100\, \mathrm{watts}}{h \cdot 5\,\mathrm{GHz}} \approx 3\cdot 10^{25} \text{ photons per second}$$

$$\frac{100\, \mathrm{watts}}{h \cdot 500\,\mathrm{kHz}} \approx 3\cdot10^{29} \text{ photons per second}$$

On the other hand, if we want to find out the energy-per-photon of your transmission, the total power is irrelevant and we simply use the Planck relation by itself:

$$h \cdot 5 \,\mathrm{GHz} \approx 3.3×10^{-24} \,\mathrm{J}$$

$$h \cdot 500 \,\mathrm{kHz} \approx 3.3×10^{-28} \,\mathrm{J}$$

This is everything that Planck's constant tells us about the properties of your transmissions.


However, these figures are almost irrelevant to the practice of radio.

Often in discussions of the energies of photons the topic is photons of much higher individual energies, which become significant; for example, by being absorbed by individual atoms at specific energies.

The photons in the "radio" portion of the electromagnetic spectrum are comparatively so low-energy that these “quantum” effects are not usually of interest, as opposed to the net effect of many coherent photons making up a field which pushes electrons around in antennas.

The “energy in the wave” (provided we define such a quantity, either by choosing a duration in time or a volume of space) is entirely independent of the frequency of the wave.

| improve this answer | |
$\endgroup$
  • $\begingroup$ While this answer is correct in terms of particle energies and their numbers, it doesn't really address the question "is there an energy difference in the wave?". $\endgroup$ – user2338215 Jul 7 '14 at 16:48
  • $\begingroup$ @user2338215 I don't agree with your interpretation of the question, but I've expanded the answer to cover what I previously touched on in much more detail. $\endgroup$ – Kevin Reid AG6YO Jul 7 '14 at 18:02
0
$\begingroup$

To describe electromagnetic wave behavior it is necessary to use Maxwell's equations. Note that James Clerk Maxwell published these equations in 1865, long before any concept of quantum mechanics was put forth. His equations are part of classical electromagnetism, and do not involve photons or quantum theory, and don't require any knowledge of the particle nature of electromagnetism. But these equations are fundamental to describing the wave nature of radio waves. Note that working with them requires using partial differential equations.

\begin{eqnarray} \nabla \cdot \vec{E} &=& \frac{\rho}{\epsilon_0}\\ \nabla \cdot \vec{B} &=& 0 \nonumber \\ \nabla \times \vec{E} &=& - \frac{\partial B}{\partial t} \nonumber \\ \nabla \times \vec{B} &=& \mu_{0}\vec{J} + \mu_{0}\epsilon_{0}\frac{\partial E}{\partial t} \end{eqnarray}

Maxwell's equations describe mathematically how electromagnetic waves carry energy as they travel: it results from the energy density associated with both the electric field $\begin{equation} \vec{E} \end{equation}$ and the magnetic field $\begin{equation} \vec{B} \end{equation}$ components of electromagnetism. The rate of energy transport per unit area relates directly to your wave question, and can be derived with help from Maxwell's equations. The mathematical concept for energy flow in electromagnetic waves is known as the Poynting vector; which involves taking a cross product of the $\begin{equation} \vec{E} \end{equation}$ and $\begin{equation} \vec{B} \end{equation}$ components of the radio wave.

\begin{equation} \vec{S} = \frac{{\vec{E} \times \vec{B}}}{{\mu _0 }} \end{equation}

So the answer to your question comes down to whether the Poynting vector for otherwise-identical radio waves differs for a 500 kHz wave versus a 5 GHz wave.

A Google search will turn up a number of debates on the issue amongst physics aficionados, and I am relying on what appears to be the general consensus of that group. Based on their arguments, the answer appears to be YES, the Poynting vector contains a frequency-dependant term by virtue of the oscillating field components. So at various instants of time (shorter than the period of a cycle) the magnitude of the Poynting vectors might differ between waves of two different frequencies. However, taken in aggregate, averaged over multiple cycles, the net flow of energy averages out to be the same regardless a wave's frequency. Thus the short-term result does not violate the law of the conservation of energy in the long term.

Since most radio amateurs are not physicists, the practical answer is: "there is no difference". Our senses (particularly our ears) react only to actions lasting much much longer than the cycle period of a 5 GHz wave. Even our radio equipment generally extracts information at much lower bit rates than the carrier frequency being received. So from a practical standpoint radio operators observe and care only about the aggregate. But from a theoretical standpoint there is a subtle difference. Perhaps there is, or will be found, a practical or academic application for that subtle distinction. But regardless, I think yours was an excellent question!


POWER AND ENERGY

It is easy to confuse a definition with a mathematical model. To illustrate, suppose one asks about "power dissipation in a resistor at room temperature versus as the temperature approaches absolute zero." Without much thought one might quote:

$$ P = I^{2}R\, $$

Although we learned this relationship as Joule's Law, and it is very useful for estimating power dissipation in electronic circuits, when we apply it using the nominal room-temperature resistance value of a circuit component, it provides only an approximation of the power dissipated. Joule's Law as stated above is of little use in understanding how power is dissipated in materials as temperature varies - an obvious reality if you consider a resistor made of a superconductive material. That doesn't mean Joule's Law is wrong. It just means that it needs to be applied appropriately, to answer questions to which it applies. $P = I^{2}R$ simplifies the calculation of power dissipation by ignoring the physics describing how electrons actually travel and interact with the materials through which they move.

Similarly, the equation defining the joule:

$$ J = W\cdot\ s\, $$

While that definition works fine for converting between power and energy, it does not in any way explain how electromagnetic energy is carried, either in waves or in particles.

Waves are complicated things to describe mathematically. The math involved in describing electromagnetic waves goes well beyond algebra, or even basic calculus. Although radio amateurs may never need to work with Maxwell's Equations, valuable insight into wave behavior can be gained from examining the results of physicists' work. It is quite remarkable that the mathematical underpinnings for electromagnetism predate the invention of the first commercially practical incandescent light bulb.


OF PARTICLES AND WAVES

"An energy difference in the wave" pertains to the wave nature of electromagnetism, not the particle (photon) nature of that energy. But the question also mentions Planck's constant, which applies to the quantum nature of light. Therein lies a paradox. In physics the concepts of wave particle duality and the duality paradox have been the subject of much debate and study, but in the amateur radio community not so much.

The question acknowledges that radio waves are electromagnetic energy, and that electromagnetic energy can appear as particles (photons) or as waves. The wave particle duality paradox makes it necessary to address waves and particles separately because we never observe the wave and particle natures of energy at the same time. Sometimes radio energy looks like particles, and sometimes it looks like waves, but it never looks like particles and waves at the same time.

A previous answer addressed the energy associated with each particle of light. As a practical matter, you may never observe the effects of the different energies of electromagnetic particles of different frequencies. That difference in energy between particles of different frequency can be observed experimentally in the photoelectric effect - where in a carefully crafted experiment you might observe that some loosely-bound electrons get "knocked free" from the surface of a metal by 5 GHz photons, but not by 500 kHz photons even if you use a transmit power of 5000 GW! But your question addresses waves, not photons.

The approach taken in this answer (Maxwell's Equations) relies on classical electromagnetism, not quantum theory. But since quantum physics has a 100-year history of correctly predicting results in carefully conducted experiments designed to disprove it, we would be foolish to carelessly dismiss its results as not relevant. So some history might be in order to help illustrate why it is not appropriate to utilize quantum concepts to address the question that has been posed.

In the latter part of the 19th century classical physics had done such an excellent job of describing observed phenomena that it was widely believed that nothing new was left to be discovered. There were some pesky experimental results (the photoelectric effect, black body radiation curves to name but two) but they too would surely be explained in terms of classical physics in due time. Now, one-hundred-twenty-some-odd years later, we know that never did happen. Instead, quantum mechanics and relativity theory turned physics on its head. Adherents of classical theories worked feverishly to explain experimental results using the old ideas, but the new theories not only explained results that confounded classical theory, the new theories predicted never-observed phenomena (orbit of mercury, quantum entanglement, to name but two) that cemented the newer theories as accurate and reliable.

But what about classical electromagnetic theory? Are electromagnetic waves only to be understood as particles and quantized energy? The answer is no. Planck's relation does not describe waves, and is useless for providing insight into how waves operate. Maxwell's equations still do an excellent job of describing electromagnetic wave behavior, just as they did over 150 years ago.

If both classical and quantum equations work for their respective domains, then which equation should you then use? Here enters the largely misunderstood concept of wave-particle duality. Wave–particle duality is an ongoing conundrum in modern physics, and there are various explanations for it that involve particles and waves, only waves, neither-wave-nor-particle, and more. But for the question at hand, namely "...is there an energy difference in the wave?" we need only address the wave behavior. This is actually stated in the question. Thus it is just fine, actually mandatory, that we use the equations that describe the wave: Maxwell's famous equations. We can feel confident in this approach, in part, because based on the results of much experimentation, we do not observe electromagnetic particle behavior at the exact same instant at which we observe wave behavior - we need only deal with the equations that fit our task at hand. Quantum theory is relevant! But not in the context of the question asked.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Of course, if you are looking at timescales much less than an individual cycle, then you can't know the frequency (Gabor limit), so.... $\endgroup$ – Phil Frost - W8II Jun 28 '14 at 21:07
  • $\begingroup$ Now if the question were "can you determine the frequency of a signal given its properties at a single point in time" then the answer would be "no". But given the frequency (and other boundary conditions) of a signal, you can use Maxwell's equations to determine the wave's properties at any arbitrary point in time. In this question we were given frequencies a priori. $\endgroup$ – user2338215 Jun 28 '14 at 21:19
  • 2
    $\begingroup$ "Wave/particle duality" is not a "paradox". $\endgroup$ – Kevin Reid AG6YO Jun 30 '14 at 0:08
  • $\begingroup$ @AG6YO - Please take up that issue with Professor Stannard (open.edu/openlearn/science-maths-technology/science/…) and countless others. $\endgroup$ – user2338215 Jun 30 '14 at 0:27
  • $\begingroup$ It really isn't helpful or fair to down vote an answer when you haven't bothered to understand what it is saying. If you find an error then please do the whole community a favor by pointing it out. If you just don't like an answer, or it doesn't fit your preconceived notions, then you do everyone a disservice by attacking it without justification. $\endgroup$ – user2338215 Jul 6 '14 at 14:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.