# Does light have “perpendicular inertia”?

I have a question after John Rennie's answer of this post Light & Observer moving perpendicular to each other.

Imagine a train station and a train passing very fast. There are two frames of reference: the train frame and the platform frame. We will look at the situation from above (from the sky view).

We turn on a laser in the platform (in the direction perpendicular to the direction of the train), creating a light ray. In the platform frame the light ray is vertical and in the train frame we can calculate the angle of the light ray to the vertical with Lorentz transformations.

In the same way, now we turn on the laser inside the train. In the train frame the light ray is vertical and in the platform frame we can calculate the angle of the light ray to the vertical with Lorentz transformations. So far so good.

Here come my questions. It seems that the direction of the light ray depends on the movement of the laser in the precise moment in which the light ray was created. But knowing that in both cases the light ray is created absolutely perpendicular, it seems like if the light ray had a kind of "inertia" in the sense of the Newton's first law, as throwing a ball through the window of the train. So, How is it possible? Why light "knows" how its source was moving? Why light does not move relative to some static fundamental medium (ignore this parenthesis: even though this medium could move relative to the space somehow)? How does it work? How this is explained? I think this leads to another question: What is the physical nature of light and through which physical tangible medium it is transmitted? (first it was thought that a real physical Ether existed, now is something like an unreal physical field modeled with mathematics). Thanks.

PS: This situation is used to informally obtain the Lorentz factor. It is assumed that the light ray (created inside the train) accompanies the train. I was just pondering this fact.

• I think you may find this article helpful: en.wikipedia.org/wiki/Aberration_(astronomy) – PM 2Ring May 1 '20 at 18:42
• Possible duplicate of Special Relativity and the Light Clock and which direction was the laser pointing? – John Rennie May 2 '20 at 6:23
• @JohnRennie glad you comment. Yes, it is about the same thing. There you say "When the light emerges from the light pen it carries on moving in the same diagonal direction.". So, you are saying that from the laser frame the beam was created in the vertical direction while from the other frame the beam was created in the diagonal direction and therefore has always been going in a straight line in the diagonal direction as opposed to have been created vertical and received a lateral "inertia"? From each frame the beam is created in a different direction?!? – Pepe Sospechas May 2 '20 at 18:03
• I don't know which is stranger: "perpendicular inertia" or "relative direction creation" xD Fuzzy relativity shit... – Pepe Sospechas May 2 '20 at 18:09

If you're talking about photons, then their momentum ($$p=\hbar k$$) or their wave vector:

$$k^{\mu} = (\omega/c, \vec k)$$

transforms like a 4-vector and that is it.

If your talking about a source emitting coherent light, then a laser (or a coherent antenna) has an plane whence the light is emitted. That plane has constant phase across it. If you boost into a frame moving parallel to that plane (which is perpendicular to the laser light, or radio signal), then the relativity of simultaneity induces a phase ramp across that plane that accounts for the beam direction.

Note that phase is manifestly a Lorentz scalar, everyone agrees on it:

$$\phi(\vec x, t) = \omega t - \vec k \cdot \vec x = (w/c,\vec k)(ct, \vec x) =k^{\mu}x_{\mu}= \phi(x_{\mu})$$

they just don't agree on $$t$$ and $$\vec x$$ (nor $$\omega$$ and $$\vec k$$).

What this answer means is that your premiss is wrong: if you solve the equation for a moving laser, you will find that light is not emitted perpendicularly to the opening. No one solves for a moving laser, though, it's not tractable. Hence, I mention the antenna. It is simpler to model, e.g. a dipole moving near $$c$$: the solution will have it a main lobe that is not perpendicular to the dipole element. Or, if you are familiar with a phase-array antenna, it is steered by putting a phrase ramp across the surface. The "tilt" in the time axis for different observers is exactly a phase ramp across the surface.

The phase is a Lorentz invariant: everyone agrees what it is. So in a frame in which the source is stationary, the phase is the same across the whole thing at any given time. For a boosted observer, their definition of "at any given time" is different for different positions along the surface, hence they see a phase ramp, hence the beam is steered off the normal.

There is no need to invoke "photon inertia" or ask how the light knows the source is moving: the source has to have physical extent to make a beam, and that means different frames have different definitions of "now" at different positions inside the source, and that accounts for the direction of emission.

• I have reformulated my question. Maybe I haven't explained myself properly. – Pepe Sospechas May 2 '20 at 0:47

I think the answer to your question can be found in the comments of this post: Photon: speed and mass.

To summarize, the answer is yes and no, mostly because the concept of inertia depends largely on how it's defined, which varies depending on the situation.

A photon on it's own has no inertia, because by most standard definitions, mass is a measure of inertia. Since a photon has no mass, by this definition it has no intertia. You can describe this in a more rigorous manner by saying that a photon is not an eigenstate of the mass operator, $$E^2 - |\vec{p}|^2$$, therefore it has no well-defined mass.

That being said, one can concoct situations in which we can define a 'mass' of a system containing a single photon, although what you are quantifying is the inertia of the system, not the inertia of the photon. Take for example the situation described in the comments from the link above:

Say you have a photon in a massless box. To accelerate that box to a speed $$v$$, one would need to apply an impulse equal to $$Ev/c^2$$, therefore the system has rest mass (and inertia) equal to $$E/c^2$$.

In terms of your confusion about the difference between the photon outside of the spaceship and the one inside, that is easier to understand by thinking about the photon outside of the spaceship (see http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/relmom.html). Basically photons have momentum $$|\vec{p}|=h\nu/c$$, so they can obey the Lorentz transformations accordingly. This is what leads to things like red-shift. The energy and momentum of the photon you measure depends on your reference frame, just as with all other things.

• I think you haven't read/understood my question – Pepe Sospechas May 1 '20 at 22:49

The problem is really similar as described in the mentioned post. When it is said there: light ray in vertical direction, some inertial frame is supposed for that. It could be a platform, or a train, or another ship. We then make the calculation for another inertial frame moving in the $$x$$ axis direction of the first frame.

It should be noted that the station platform belongs to the Earth's frame, that is moving at more than 100000 km/h to a solar based frame.

The very basis of special relativity is that not only the laws of mechanics but also of eletromagnetism are valid for all inertial frames. And that includes EM waves, that comes from Maxwell equations.

Imagine electric water kettle being turned on in a moving train. The water starts gaining mass. As it gains mass, it gains momentum. Imagine that the energy comes from a battery, which loses mass, so its momentum decreases. Total momentum of the whole train does not change, as there is no force pushing the train.

Now imagine a toy car in the place of the kettle, accelerating perpendicularly to the motion of the train. Total momentum of everything must not change, and battery still loses momentum. So toy car must gain longitudinal momentum as it accelerates perpendicularly.

How does the toy car know that it must gain longitudinal momentum? Well the energy that the car gains contains the longitudinal momentum that the car gains. Very simple.

What if the moving train takes energy from overhead electric wires, and stores the energy into the battery? Now the energy does not contain momentum, so the train does not gain momentum, although it gains mass. So it slows down.

Oh yes, the question involved light. Well, we could push a perpendicular light pulse from both sides, increasing its energy, and perpendicular momentum, and longitudinal momentum. I mean squeezing the light between mirrors.

Instead of overhead electric wires there could be a laser on the platform shooting perpendicularly to the tracks, and the train then collects that energy. Maybe that way it's easier to accept that the train slows down.