# Simultaneity of two light rays in different reference frames

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## Summary:

Do these two rays set off simultaneously if we move from train's frame to ground's frame?
I am studying the fact that two events that are simultaneous in a frame aren't (in general) simultaneous in another.

The lamp is equidistant from the two ends. When the light is switched on an observer on the train sees how both light rays hit the back and the front of the train simultaneously. This is not the case for an observer on the ground, who claims that the light ray going to the left hits the back before the light ray going to the right hits the front.

I am OK with this but I want now to focus on the simultaneity of the fire of both light rays. From the train's frame, both rays leave the lamp at the same time, one going to the right and the other to the left. My question now is:

Do they leave the bulb simultaneously from the ground's frame as they do wrt train's frame?

I'd say that the rays do leave at the same time as well, and this is my reasoning:

Let me be on the ground's frame and examine all the clocks in the train. I will realize that the clocks read different times depending upon their location because:

$$\bar t = -\gamma \frac{v}{c^2}x$$

Where ##\bar t## is the time elapsed in the train (according to the clocks in the train read by me, an observer located at the ground's frame). This is an illustration:

My point is that the master clock reads ##\bar t = 0##, so ##\bar t = 0 = t## which means that both light rays leave also at time zero (and thus simultaneously) from the ground's frame.

What do you think of my reasoning?

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Do they leave the bulb simultaneously from the ground's frame as they do wrt train's frame?
Yes. The two light rays being emitted from the bulb happens (roughly) at a single point in spacetime. There is never any disagreement about the number of events between frames.

The two light rays being emitted from the bulb happens (roughly) at a single point in spacetime.
Then the coordinates of the event are (0, 0, 0, 0) in both frames.

Orodruin
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The light pulses leaving the bulb are not different events, it is a single event. Every event is simultaneous with itself regardless of the frame.

vanhees71 and PeterDonis
The light pulses leaving the bulb are not different events
Why not?

Dale
Mentor
Why not?
Because they happen at the same place and the same time. That is the definition of a single event.

JD_PM
Orodruin
Staff Emeritus
Homework Helper
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Why not?
Do not be fooled by the nomenclature "event". In relativity it is a technical term referring to a given place at a given time. If the light signals are sent from the same place at the same time then they are by definition emitted from the same event.

Dale and JD_PM
Then the coordinates of the event are (0, 0, 0, 0) in both frames.
The choice of coordinates to label events is arbitrary. The main factor in choosing coordinates is convenience, and certainly choosing to label the event where the light is emitted from the bulb as ##(0,0,0,0)## is convenient. Just know that regardless of how either the train observer or the embankment observer chooses to label this event, they will always both agree that it was just one event.

JD_PM
pervect
Staff Emeritus
Suppose you have a red bulb, and a blue bulb. The blue bulb is moving at some velocity v relative to the red bulb. When the red bulb is at the same position in space as the blue bulb, they both emit flashes, the difference is that one bulb is moving relative to the other.

The speed of light does not depend on the state of motion of the source in special relativity, so the wavefront of the light flash from the red bulb is exactly the same wavefront as that emitted by the blue bulb.

vanhees71, PeroK and JD_PM
The speed of light does not depend on the state of motion of the source in special relativity
Yes.

At this point I am thinking of why we cannot accelerate light (let's say by gravity).

Actually we know that light has momentum ##p = \frac{E}{c}##. So we could naively think of Newtonian equation ##\frac{dp}{dt} = ma## and say: 'hey, we could consider an interval ##\Delta E## so that there's a change in energy which means there's a change in momentum which means there has to be a force accelerating light!' But it has been observed experimentally that relativistic momentum is a non-linear function of velocity (https://physics.stackexchange.com/questions/98750/why-doesnt-gravity-speed-up-light):

$$\vec p = \frac{m \vec v}{\sqrt{1 - \frac{v^2}{c^2}}}$$

So ##\frac{dp}{dt} = ma## doesn't hold I see...

But now I am interested in what GR can tell us on why light cannot be accelerated.

Dale
Mentor
I am interested in what GR can tell us on why light cannot be accelerated
Because nothing exerts a real force on light. Therefore it travels on a geodesic, specifically a null geodesic.

Because nothing exerts a real force on light.
What is a real force?

Would be enough to argue that as photons are massless, then ##\vec F = m\vec a= 0##?

I acknowledge I regard my own argument as not sufficiently convincing.

Dale
Mentor
What is a real force?
A real force is one that causes a deviation from a geodesic if it is not balanced by an opposing real force. This is the usual definition from Newton’s 2nd law, expressed appropriately for GR.

FactChecker
Gold Member
At this point I am thinking of why we cannot accelerate light (let's say by gravity).
Your example "let's say gravity" is a special case. Gravity changes the paths in spacetime which represent unaccelerated motion (geodesic). Since light follows the null geodesic paths, gravity does change the path of light, but it can not be called "acceleration". Light is still following the null geodesic unaccelerated paths in the gravity-modified spacetime. That is why there are effects such as gravitational lensing. If one ignores the effects of gravity, it may seem as though the light has been accelerated, but it has not. In the gravity-modified spacetime, the units of time and space have changed. In the special relativity geometry of spacetime, there is no such thing as a speed faster than c.

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pervect
Staff Emeritus
But now I am interested in what GR can tell us on why light cannot be accelerated.
Fundamentally, the reason light can't be accelerated is that in special relativity, it has a constant velocity, and GR is a generalization of SR (special relativity).

There's lots of little details, but that's the basics. The speed of light is only constant in a vacuum, so introducing media is one of those "little details". In a related note, one may need some discussion about how velocities are measured, the sort of velocity I am talking about is measured with local clocks and local rulers, and assumes that the speed of light is isotropic.

PeterDonis
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2019 Award
now I am interested in what GR can tell us on why light cannot be accelerated.
You have to be careful to distinguish two different questions here.

The first question relates to light, the actual stuff we observe in experiments. It is certainly possible to "accelerate" light in experiments, in the sense of making it propagate in a way that makes it move on timelike worldlines instead of null worldlines. For example, we can propagate it through a medium, or put it in a waveguide or a fiber optic cable.

However, note that I just said "makes it move on timelike worldlines instead of null worldlines". That brings us to the second question, which is: why can't we change which worldlines are null worldlines? Why does it have to be certain worldlines and no others that are null? And the answer to that is, because which worldlines are null worldlines (more generally, the light cone structure of spacetime) is inherent to the geometry of spacetime. And therefore, if we just let light travel freely in a vacuum, the worldlines on which it will move are determined by the geometry of spacetime.

JD_PM
vanhees71