What is the speed of a photon traveling along the sine function?

[somega]

On the image you can see a photon starting at point A at t=0.

The photons travels along the sine function and arrives point C.

I knot that this takes T=λ/c.

But this is the time for a object traveling directly from the origin to point C and not along the sine wave!

If the photon travels along the sine wave with speed c it will take longer than T for the photon to arrive at point C.

Where is my mistake?

 

[PeroK]

View attachment 259582

On the image you can see a photon starting at point A at t=0.

The photons travels along the sine function and arrives point C.

I knot that this takes T=λ/c.

But this is the time for a object traveling directly from the origin to point C and not along the sine wave!

If the photon travels along the sine wave with speed c it will take longer than T for the photon to arrive at point C.

Where is my mistake?

A photon, as a relativistic massless particle, in simple terms travels in a straight line. It doesn't travel along a sine-shaped path!

Electromagnetic radiation is the propagation of sinusoidal electro-magnetic fields. That is a different model of light altogether.

 

[Nugatory]

Although we often say "photon" in the relativity forum because it's easier than saying "flash of light" or "light signal", that's incorrect - photons don't move or travel or follow paths or anything that you're thinking they do. If you reword your question to say "flash of light" it will be clear where your mistake is - a flash of light always moves in a straight line unless you bounce it around with a mirror (or bend its path with a lens or prism, but then it's not moving in a vacuum).

 

[Dale]

Where is my mistake?

Here:

The photons travels along the sine function

It does not travel along the sine function (unless maybe you have laid out some fiber optic cable in the shape of a sine function)

 

[Janus]

Rather than thinking in terms of the photon traveling along the sine wave, you need to visualize the sine wave as traveling at c, like this:

This particular image is 1 wavelength long. The wave travels at c to the right.
The time it takes for a wave to traverse the width of the image, or 1 wavelength is $T = \frac{\lambda}{c}$
Thus the frequency is $f = \frac{c}{\lambda}$

 

[vanhees71]

A photon, as a relativistic massless particle, in simple terms travels in a straight line. It doesn't travel along a sine-shaped path!

Electromagnetic radiation is the propagation of sinusoidal electro-magnetic fields. That is a different model of light altogether.

A photon doesn't travel along any line, let alone a straight line. It's a Fock state of a massless spin-1 quantum field, describing the electromagnetic field within QED. It is a completely wrong picture to envisage this as a localized particle. A photon does not even have a position observable in the usual sense. It simply cannot be localized.

I have also no clue, why a photon should travel along a sinusoidal line to begin with.

 

[robphy]

In the OP’s posted graph of a classical electromagnetic wave, the “y”-axis should be the “y-component of the electric field”. As noted by others, the wave disturbance travels along the x-axis at speed c (with no displacement components along the y or z directions).

The correct graph is neither a spatial diagram nor a Spacetime diagram.

 

[vanhees71]

Then the correct labeling would be $x$ for the abscissa and $E_y$ for the ordinate of the coordinate system.

 

[PeroK]

A photon doesn't travel along any line, let alone a straight line. It's a Fock state of a massless spin-1 quantum field, describing the electromagnetic field within QED.

Yes, but what is it within the context of a B-level thread or even I-level undergraduate physics? Until one knows what a Fock state is, your explanation doesn't help.

 

[Orodruin]

Yes, but what is it within the context of a B-level thread or even I-level undergraduate physics? Until one knows what a Fock state is, your explanation doesn't help.

Until one knows what a Fock state is it is better not to talk about photons at all as it will invariably lead to misconceptions. At least they should not be talked about as ”little balls of light”.

 

[vanhees71]

Yes, but what is it within the context of a B-level thread or even I-level undergraduate physics? Until one knows what a Fock state is, your explanation doesn't help.

May be, but then one shouldn't give socalled "explanations" which are utterly wrong either!

 

[somega]

Does a photon have a frequency at all? Or is it the electromagnetic signal which has a frequency? (And the electromagnetic signal is made of many many photons.)

 

[PeroK]

Does a photon have a frequency at all? Or is it the electromagnetic signal which has a frequency? (And the electromagnetic signal is made of many many photons.)

If you search online for "photon frequency" what do you find?

 

[PeterDonis]

a photon

This thread is in the relativity forum, not the quantum physics forum, so the concept of "photon" has no meaning here; the relevant theory of electrodynamics here is classical electrodynamics, not quantum electrodynamics.

Thread closed.