2
$\begingroup$

Electromagnetic waves have electric (and magnetic) fields that oscillate spatially and with time. But light, moving at the universal speed limit, is a "space-like" object according to relativity since it does not experience time.

My understanding is light's time dilation is the limit when time would stop for it's frame.

But if light doesn't move through time how does it oscillate in time?

I suspect I am looking at this incorrectly. Perhaps the oscillation in our time frame is easily justified when approached properly?

$\endgroup$
  • $\begingroup$ This is far from a complete answer: the oscillation of light arises as a result of how it propagates, not the other way around. Changes in the electric field induce a magnetic field, and changes in the magnetic field induce an electric field - if you imagine both of these fields as oscillating, and treat the fluctuations as the cause of one another, it should be apparent that it is this interplay between fields that 'moves' energy through the fields as fast as the fields can react through cause and effect - the speed of light itself. $\endgroup$ – Xeren Narcy Mar 3 '15 at 4:46
2
$\begingroup$

For a closer understanding, you must be more precise and avoid any misconception, and in your question there are 2 of them:

  • light's movement is not spacelike but lightlike.

  • for time dilation, observed time and proper time may not be confused: if you say that light does not experience time, you are talking about the (hypothetical) clock of light, that means its proper time. But time dilation is the dilation of the proper time of an object for an observer. If you want, you may say that the hypothetical proper time for a photon moving from Sun to Earth is zero, and for us its hypothetical proper time appears dilated to a time period of 8 minutes.

Based on these explanation, the answer to your question is simple - it is relativity:

  • From the hypothetical point of view of the photon which is outside of spacetime, no time is passing on its path between Sun and Earth, Sun is situated immediately adjacent to Earth. The spacetime interval is zero.

  • For observers in spacetime, this zero time lapse is "stretched" into the form of a wave (8 minutes).

  • Try intuitively to understand oscillations and electromagnetism as marks of vacuum permeability and permittivity of spacetime rather than as marks of light (Both are true, but often the first aspect is neglected).

$\endgroup$
1
$\begingroup$

It is a very common mistake to think that going faster and faster gets you closer and closer to traveling at the speed of light. "I know I can never get there, but surely I am closer to it than I was."

Frames of reference that are faster and faster than yours see the universe as more and more foreshortened and time dilated. Surely light's frame of reference sees distance and time shrunk to 0.

You have hit upon one problem with this reasoning - It leads to a wrong answer. Light oscillates as it travels. In your frame of reference, there is a distance and a time from crest to crest.

Another problem with this line of reasoning is that going faster and faster does not get you closer to catching light. No matter how fast you go, light passes you at the same speed.

The correct answer is that no frame of reference describes the universe from the point of view of light. It is very natural to ask "What does the universe look like if I go the speed of light?" It isn't at all satisfying that the answer is "You can't go the speed of light." And this is why the mistake is so common.

$\endgroup$
  • $\begingroup$ This explains well why we will always see light oscillate in time, and space. I suppose a follow up I have is how time and space being shrunk to zero for light gives rise to these oscillations for us non-zero mass things? $\endgroup$ – user129818 Mar 3 '15 at 15:50
0
$\begingroup$

One cannot speak of the reference frame of light because that would mean it's speed would be 0, but light moves at $v=c$ in all frames by the postulates of special relativity.

$\endgroup$
0
$\begingroup$

(Adding to the existing answers)

Light must move at $c$ (locally) in every inertial reference frame. In a reference frame belonging to light, this rule would break, so such a reference frame does not exist / is not valid for consideration.

Consider that to transmit light from point A to point B, it has to cross some physical distance. This 'crossing' takes time from the point of view of any inertial frame of reference (or someone who is standing still).

However to light, this 'crossing' is instantaneous. It is not immediately clear why this should result in any oscillations... To assist with this, an image from this article:

Electromagnetic Wave

Where $\vec B$ is the magnetic field, $\vec E$ is the electric field and $\vec v$ is the direction of "travel".

Without going neck deep in math, the shapes of the red and blue curves are sinusoidal. Notice how they are exactly aligned. Other 'alignments' (polarizations) are possible but this is the simplest to picture.

What you need to understand is the $x$ axis here is actually starting and ending in two different times - $x$ is both a distance and a duration (when using suitable units in terms of $c$). What this diagram shows is how the electric and magnetic fields change over time and space during the propagation of light.

The fields increase / decrease in strength in a perpendicular direction of travel, due to the other field changing. This self-influence is what travels at light speed and makes up light (rather the energy carried by this self-influence). But in order for this self-influence to be possible, the fields need to be changing in strength in a particular way through space and time.

This gives rise to oscillations for observers that can't move with light / at light speed.

$\endgroup$
0
$\begingroup$

As others have pointed out, you can't travel at c, so the question in its present form is moot. But you can look at what happens at the limit, as your speed approaches c, and at that limit the frequency is zero. Its a bit like - no exactly like - saying that x/x = 1. This is not true if x=0 as 0/0 is undefined, but we can still talk about the limiting case.

This result - the frequency drops to zero - is not as counter intuitive as it first appears. As you get closer and closer to c, the measured frequency of the light decreases, giving the limiting case zero frequency. But the energy of the photon decreases proportionally to its frequency (Planck's law) so as its frequency decreases so does its energy. At the limiting case (traveling at c) the energy of the photon has decreased to zero so there's no E or M fields to measure.

$\endgroup$

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.