[Bored chemist]

Microwave can produce observable electric spark on sharp metal objects.

You can also do that with high voltage DC.
So what?
It's not got a lot to do with EM radiation, or with photons.

[Bored chemist]

Do you realise that 1 photon per second is not a limit because you can have one photon per week?

How long does a photon with 1 THz frequency last, from beginning to its end?

Forever.

What if the transmitter is only powered up for 1 second?

Then you can not tell if the  frequency was 999,999,999,999.5 Hz or 1,000,000,000,000.5Hz

I'm not sure how adding that amount of uncertainty can compensate for the reduction of photon duration from forever to 1 second. Is this really the currently accepted theory?

You can think of it in terms of Fourier analysis, or you can just consider the "counting error".
If my clock ticks exactly once a  second and I count the ticks in one  minute, I will not always get 60 ticks.
Sometimes, if I start about half a second after the minute,  I will get just 59.

If the "tick" has a finite duration, I can get 61 ticks in a minute if I catch the tail endo of one and the start of another

This isn't a matter of "currently accepted theory"; it's just common sense.

So, if the plate is kept at 0K, is the photoelectric effect by uv light still observed?

You can not get something to 0K.

Do you understand that the photoelectric effect is not the same as the thermoelectric effect?

[hamdani yusuf (OP)]

I'm not sure how adding that amount of uncertainty can compensate for the reduction of photon duration from forever to 1 second. Is this really the currently accepted theory?

It's the consequence of wave mathematics and fourier analysis. By switching off after 1 second you have terminated the low frequency spectrum at 1 Hz.

We were talking about 1 THz transmitter rated at 1 Watt, attenuated using cascaded filters to produce a single photon. There should be around 1 trillion cycles of electromagnetic waves.
What do you mean about termination of low frequency spectrum?

[hamdani yusuf (OP)]

Here is an example of a Gaussian pulse and its Fourier transform.

https://www.gaussianwaves.com/2014/07/generating-basic-signals-gaussian-pulse-and-power-spectral-density-using-fft/
This kind of pulse can be generated using electronic circuits. The value of σ can be adjusted. The vertical axis of frequency domain is not probability. It's magnitude instead.




https://slidetodoc.com/fourier-series-the-fourier-transform-what-is-the/

Shorter light pulses have broader spectra F(w) f(t) t w t w Duration of a light pulse: Width of the spectrum: So: The shorter the pulse, the broader the spectrum! Or inversely, to make a shorter light pulse you need to use light waves with a broader range of frequencies. 29

Making the pulse shorter doesn't make the frequency becomes more uncertain. It makes the frequency has a broader range.

[hamdani yusuf (OP)]

Here are some more examples. Perhaps we can see some patterns.

[alancalverd]

Making the pulse shorter doesn't make the frequency becomes more uncertain. It makes the frequency has a broader range.

Therefore your attempt to determine the "frequency" of the pulse is subject to greater uncertainty.
"Frequency" only has meaning for a truly repetitive phenomenon. Don't confuse it with the temporal half-width of a single pulse. See, for example, the frequency domain analysis of a delta function, that you have just shown.

[hamdani yusuf (OP)]

Making the pulse shorter doesn't make the frequency becomes more uncertain. It makes the frequency has a broader range.

Therefore your attempt to determine the "frequency" of the pulse is subject to greater uncertainty.
"Frequency" only has meaning for a truly repetitive phenomenon. Don't confuse it with the temporal half-width of a single pulse. See, for example, the frequency domain analysis of a delta function, that you have just shown.

If we are forced to represent the curve in frequency domain into a single real number value, then yes, there will be some uncertainty need to be considered. It can be said to have a mean frequency with some standard deviation to represent the uncertainty, hence we need two numeric values. But it loses the information about the shape of the curve.

[alancalverd]

The frequency analysis of a delta function has no mean, unless you think that ∞/2 is a number!

Photon interactions are as close as you can get to delta functions.

Once again: we need two distinct mathematical models to describe the behavior of electromagnetic radiation. There is no reason why they should be interchangeable. Conceptually, particle interactions with a single energy,  spatially distributed as described by wave propagation, seem to fit the bill.

[hamdani yusuf (OP)]

The frequency analysis of a delta function has no mean, unless you think that ∞/2 is a number!

The curve goes in both positive and negative direction, hence the mean is 0.

[alancalverd]

But E = hf and E > 0 for a photon!

More to the point, there is no such thing as a negative frequency. Fact is that it doesn't matter where you put the "center" frequency, the spectrum of a delta function is infinite, but the energy of a photon is absolutely defined.

[hamdani yusuf (OP)]

But E = hf and E > 0 for a photon!

More to the point, there is no such thing as a negative frequency. Fact is that it doesn't matter where you put the "center" frequency, the spectrum of a delta function is infinite, but the energy of a photon is absolutely defined.

You can use online calculators such as wolfram alpha to show that the width of the curve in time domain is proportional to the height of the corresponding curve in frequency domain. So, if the time domain signal is infinitesimally thin, then the frequency domain curve is infinitesimally low. I'd like to elaborate further, but I'm afraid that I'll have to do it in new theory section.

[alancalverd]

Beware of dividing 0 by 0 and coming up with whatever number you like! Remember Wolfram is at best a discrete  approximation to a continuous function and

0/x = 0 ∀ x,  so Lim_x→0 (0/x) = 0

x/x = 1 ∀ x,  so Lim_x→0 (x/x) = 1

x/0 = ∞ ∀ x,  so Lim_x→0 (x/0) = ∞

[Bored chemist]

Perhaps you can see my point about

You can think of it in terms of Fourier analysis, or you can just consider the "counting error".

which is enough to show that a finite transmission has an imprecisely defined frequency.

[hamdani yusuf (OP)]

Beware of dividing 0 by 0 and coming up with whatever number you like!

That's precisely why Alex Kusenko rejected analysis of the Blackbird, which made him agree to take the wager, and eventually cost him 10k. 

Remember Wolfram is at best a discrete  approximation to a continuous function

For simple enough equations, WolframAlpha can solve them analytically, which means that it leaves no uncertainty. One side of the equation is just a symbolic representation of the other side.

Algebraic equations consist of two mathematical quantities, such as polynomials, being equated to each other. Solving equations yields a solution for the independent variables, either symbolic or numeric. In addition to finding solutions to equations, Wolfram|Alpha also plots the equations and their solutions.

https://www.wolframalpha.com/examples/mathematics/algebra/equation-solving/

[hamdani yusuf (OP)]

“The greatest enemy of knowledge is not ignorance, it is the illusion of knowledge.”
-Stephen Hawking.

We can't be sure if a model is accurate just because it works. Humans have been using fire for millenias before they know what it really is. They have been using arrows before they know about aerodynamics and simpler mechanics. They have been using boats long before they know about theory of buoyancy.

[Bored chemist]

We can't be sure if a model is accurate just because it works.

No, but we can be sure a model is wrong if it doesn't work..

So we can rule out lots of possible "new  models".

[alancalverd]

If a model works, it is by definition accurate to the extent that it predicts what actually happens.

The problem with 0/0 is that the answer depends on the initial condition of an implicit function. You can't guess the initial condition from the endpoint.

You don't need a model to use a product. We have all sorts of useful devices whose physics is not fully understood - cross-country skis are a particularly good example of a clever piece of engineering that is between 2000 and 10000  years old but still not well modelled!

I don't think we have any illusions about the actual behavior of electromagnetic radiation, and most of us are humble enough to recognise that there isn't a useful single model that predicts everything. But we do have two models that work extremely well.

Returning to 0/0, you can see the problem with the downwind Blackbird car. The motive power comes from the difference between wind speed and ground speed, so once the car is moving, you should be able to reduce the wind speed to zero and the car will keep going.....a perpetual motion machine.....Wrong initial condition, wrong model! 
   

[hamdani yusuf (OP)]

Returning to 0/0, you can see the problem with the downwind Blackbird car. The motive power comes from the difference between wind speed and ground speed, so once the car is moving, you should be able to reduce the wind speed to zero and the car will keep going.....a perpetual motion machine.....Wrong initial condition, wrong model!

Not really. The force due to speed difference must overcome friction first before it can produce acceleration.
If the speed difference is exactly 0, what's the direction?

[alancalverd]

Nothing to do with initial acceleration. I have assumed that the vehicle is moving at ground speed x, and the wind then drops from y to zero. Does the car continue to extract energy from the difference between wind speed (y = 0) and ground speed (x)?

[hamdani yusuf (OP)]

More to the point, there is no such thing as a negative frequency.

https://www.linkedin.com/pulse/significance-negative-frequency-fourier-transform-samaksh-kr-sharma

Negative frequency doesn't make much sense for sinusoids, but the Fourier transform doesn't break up a signal into sinusoids, it breaks it up into complex exponentials.

These are actually spirals, spinning around in the complex plane:


Spirals can be either left-handed or right-handed (rotating clockwise or counterclockwise), which is where the concept of negative frequency comes from. You can also think of it as the phase angle going forward or backward in time.

In the case of real signals, there are always two equal complex exponentials rotating in opposite directions, so their real parts combine and imaginary parts cancel out, giving you a real sinusoid as the result. Depending on the phase of the two spirals, they could cancel out leaving a purely real sine wave or a real cosine wave or a purely imaginary sine wave, etc.

The negative and positive frequency components are both necessary to produce the real signal, but if you already know that it's a real signal, the other side of the spectrum doesn't provide any extra information, so it's often hand waved and ignored. For the general case of complex signals, you need to know both sides of the frequency spectrum.

The meaning of negative frequencies is just mathematical(not physical) similarly to the imaginary part of a complex signal. In real world, the negative frequency does not exists and the spectral content on negative frequencies must be added to the spectral content at the positive frequencies, to save energy. The concept of negative frequencies is used broad wide namely in this sense, in the connection with complex numbers.