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Suppose there is a laser beam powerful enough to burn through iron aimed at a piece of iron. You observe this event while you are in the same frame as the piece of iron and the laser-beam generator. In this frame, there is a certain part of space that you know that the light is traveling through.

Now say that you get in a rocket that travels a few meters away from the laser beam in a direction perpendicular to eventually at a constant speed arbitrarily close to the speed of light. As you do so, the area that you knew the light was traveling through contracts in the direction that you are traveling in.

Say that the amplitude that you think that the light has decreases to the point where the light would no longer be carrying enough energy to burn through the piece of iron. If you see the iron stop being burned by the laser, our universe is seriously weird. I don't think this would happen. Would you see the laser continue to burn through the iron even though it does not seem to you to have the energy necessary to do so? Would this mean that, instead of having the length contract, that there is such a thing as absolute distance?

(If we suppose that the amplitude you perceive the laser to have remains constant regardless of what inertial frame you are in, then the laser would appear to have a constant amplitude even as objects around it continued to contract in the direction you are traveling through. This would mean that the laser would have to seem to affect more and more of space as you traveled faster and faster, so it seems it would have to seem to burn through more and more objects as you went faster, which doesn't seem right to me.)

Does this contradict the theory of relativity? Is there an error in here somewhere?


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Are you confusing amplitude and wavelength/frequency? – bdesham Oct 25 '11 at 17:10
Sounds like bdesham is right. The amplitude of an EM field is not a length, it's a field strength. My guess without having done any calculations: The EM field might get weaker, but since the area contracts as well, the beam is focused better so power might stay the same? – Lagerbaer Oct 25 '11 at 17:18
EM field amplitude transforms as a rank 2 tensor, so it's definitely not invariant. I think the power transmitted is invariant, though ($P = E/t$, and both $E$ and $t$ are timelike components of four-vectors). I'll come back to this and write up a proper answer later, when I have time, if nobody else has gotten to it first. – David Z Oct 25 '11 at 19:33
I think he is thinking of light as a literal wave, the way people draw it on paper, so that length contraction perpendicular to the beam would reduce the height of the peaks. – Colin K Oct 26 '11 at 6:32
@ColinK: I don't think so--- this would be absolutely silly--- I am almost certain that he is thinking of the fact that the beam is deflected sideways when you move perpendicularly, so it isn't shining head-on, but at an angle. When you shine a flashlight (or laser) on a wall at an angle, as opposed to head on, it gets dimmer. This is a geometric effect of the beam spreading out, it is not as obvious that this spreading out doesn't apply in this case. I answered this version of the question. – Ron Maimon Oct 26 '11 at 7:02
up vote 3 down vote accepted

You mean frequency, not amplitude--- you mean chasing the light until it is too weak to burn through the iron. But then the iron is rushing toward the light, and the relative motion of the iron and the light is what determines the impact energy, and this is unchanged in any frame.

EDIT: Perpendicular motion

Now that you said what you meant--- you meant perpendicular motion. Then the beam is slanting down, and like a flashlight shining at an angle, you assume that it covers more area and is reduced in intensity. This is just not true. The reason a flashlight gets dimmer at an angle is that the same number of photons are hitting more area, because a given angle-spread at the emitter gets turned into more area at a further distance.

The laserbeam is just tilted by your motion, not spread in angle, so it hits the exact same area when you are moving, in fact, a smaller area because of the Lorentz contraction. The intensity goes up not down, but the atoms Lorentz contract just the same, so that the number of photon collisions per atom stays the same. Each collision is physical, so there is no mystery why it should be invariant.

The frequency of the light is also increaseded by your motion. But the relative motion of the light and the atom is unchanged, as I said above.

EDIT: Amplitude decreasing

The amplitude of light is not a length, and does not extend in physical space. It's an internal thing. A high amplitude wave can extend over a big, or small, area, which is independent of the amplitude. There is no relation between space and "amplitude space".

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see my comment on the question – Colin K Oct 26 '11 at 6:33

There are several wrong things with your relativity. I will refer to the figure below and consider two reference frames: the lab frame where the radiation field has components $\bf{E}$ and $\bf{B}$, and the rocket frame, where, as we will see, the radiation field has components $\bf{E}'$ and $\bf{B}'$ (notice that the magnetic field is drawn along the $y$ axis even if not indicated ). The rocket is moving with velocity $\bf{v}$ with respect to the lab frame.

enter image description here

Now, a couple of facts. First: the electric and magnetic fields for radiation are perpendicular to each other and to the direction of movement. They are also equal so one can write \begin{equation} \bf{E} = \bf{B}\times \hat{\bf{n}} \end{equation} where $\hat{\bf{n}}$ is a unitary vector along the direction of movement. Two: the Lorentz transformation for the electric and magnetic fields are not the ones implied by your guess "amplitude that you think that the light has decreases" neither those stated in "If we suppose that the amplitude you perceive the laser to have remains constant". Because the electric and magnetic fields are components of the electromagnetic field strength tensor, their transformations are a bit more interesting.

Actually, the fields in the direction of motion $\bf{v}$ don't change \begin{equation} E'_x = E_x , \quad B'_x=B_x = 0 \end{equation} while the fields perpendicular change as \begin{equation} E'_z = v\gamma B_y,\quad B'_y=\gamma H_y . \end{equation} The remaining components are unchanged. That this is correct you can check by looking at any book on electrodynamics. Moreover you can check that the new electric field again has the same magnitude than the new magnetic field $\gamma B$.

So, first conclusion: since $\gamma>1$ the amplitude of the field seen from the rocket is larger and not smaller (or even equal) than the amplitude of the field seen from the lab frame.

The thing is that the new wave is travelling at an angle $\alpha$. The angle is given by the ratio \begin{equation} \tan \alpha = \frac{E'_z}{E_x} = v\gamma \end{equation} Notice that, as it should be, if you are moving with a speed very close to $c$, gamma is huge and the angle is almost 90 degrees.

And here comes the important point, the total flux of radiation through some area, if the radiation incides with an angle $\alpha$ is \begin{equation} F = I \sin^2\alpha \end{equation} where $I$ is the intensity of the laser. In the rocket frame \begin{equation} I \propto B'^2 = B^2\gamma^2 \end{equation} and \begin{equation} \sin^2\alpha = \frac{1}{1+v^2\gamma^2} = \frac{1}{\gamma^2} \end{equation} So the total flux of energy through the screen is invariant. This makes sense. If it wasn't so you could design an experiment in which you have a machine collecting energy and after some quantity has been recolected, it kills a cat and turns itself off. If the flux over the whole surface wasn't invariant you would have, after some time, a frame with a dead cat and a frame with a cat alive. That can't be.

The flux per unit area increases though. This is of course consistent with the fact that the screen length in the $x$ direction decreases and also, obviously, the width of the laser beam. But notice that the width of the beam has nothing to do with the amplitude of the electric field.

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It is not good to use E and B fields transformation laws for this sort of thing, because the energy momentum photon laws are simpler. – Ron Maimon Oct 26 '11 at 14:37
My answer is as "good" as it can be using a particular set of tools. Different proofs of the same problem will highlight different aspects and in any case, that is not even the point. Another thing that's simpler is p-forms and yet most people still study the classical Maxwell equations. In the OP's message there is no mention of photons and plenty of mention to the amplitude of a laser. The question seemed to me to correspond to a typical course on classical electrodynamics and special relativity. Thus, I provided the answer in that spirit. – whistles Oct 26 '11 at 14:59
I wrote that before I understood the OP's confusion. You are right--- this is the best way. +1. – Ron Maimon Oct 26 '11 at 21:02

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