# Does a photon need a reference?

• B
Gold Member
Suppose we have two spaceships a lightminute apart from each other. Ship A sends a photon to ship B. To my knowledge the photon will arrive at ship B a minute after is was departed from ship A.

The photon will travel away from ship A with speed c, and will travel towards ship B with speed c.

Suppose that ship A will start moving away from ship B a relativistic speed ten seconds after it sent the photon. What reference does the photon have to determine its speed? It should keep moving toward ship B with speed c, and it should keep moving away from ship A with speed c, but the photon wouldn't keep track of where the spaceships are, right?

Related Special and General Relativity News on Phys.org
Ibix
Nothing physical needs a reference. It just does what it does.

You need reference points if you are going to make measurements, like measuring the speed of light. But why should a light pulse care about that?

anuttarasammyak
Gold Member
In IFRs of ship A, ship B, ship B after thrust and in any other IFRs, the photon has speed c. A simple world.

Last edited:
Gold Member
So ship A undergoes time dilation and space contraction I think? Since the spaceships are equivalent, why shouldn't ship B undergo those?

Ibix
So ship A undergoes time dilation and space contraction I think? Since the spaceships are equivalent, why shouldn't ship B undergo those?
There isn't an absolute answer to this. If you use a frame where A is stationary and B is moving, B is length contracted and time dilated. If you use a frame where B is stationary and A is moving, A is length contracted and time dilated.

vanhees71 and entropy1
anuttarasammyak
Gold Member
The photon has speed c in all IFRs. But its momentum and energy vary in every IFR, e.g. micro wave, infrared red, blue, ultraviolet, X-ray and gamma-ray.

Gold Member
So, to be precise, the photon has speed c, with respect to any frame of reference?

anuttarasammyak
Gold Member
Yes, that's what Einstein stated for IFR in SR at the beginning of the last century. Even in GR local speed of light or photon is constant c in any FR.

Gold Member
So, to be precise, the photon has speed c, with respect to any frame of reference?
Yes, that's what Einstein stated for IFR in SR at the beginning of the last century. Even in GR local speed of light or photon is constant c in any FR.
Doesn't that suggest that light and spacetime are very related?

anuttarasammyak
Gold Member
Light or photon happens to have zero mass so SR says their speed coincides with the maximum speed of propagation, constant c. This maximum speed was named as speed of light in history, but later other zero mass particles with speed c were found. Welcome to the world of relativity !

Gold Member
To return to my question, if FR A and FR B have relative speed 0, or FR A and FR B fly apart at some relativistic speed, either way the photon moves between ship A and ship B at speed c. If we only have FR A and FR B as reference, the photon corresponds to both of them. So to account for a change in relative speed of ship A and ship B, for the photon to maintain its speed c, it needs that time dilation and length contraction occurs. However, these are relative too, right?

jbriggs444
Homework Helper
2019 Award
To return to my question, if FR A and FR B have relative speed 0, or FR A and FR B fly apart at some relativistic speed, either way the photon moves between ship A and ship B at speed c. If we only have FR A and FR B as reference, the photon corresponds to both of them. So to account for a change in relative speed of ship A and ship B, for the photon to maintain its speed c, it needs that time dilation and length contraction occurs.
You need one more thing: relativity of simultaneity.

The light will be measured to move at c in any inertial frame. To make this work, the time of emission, the time of reception and the distance of travel may all be different than those reckoned in a different inertial frame.

hutchphd and vanhees71
Ibix
So, to be precise, the photon has speed c, with respect to any frame of reference?
No.

With respect to any inertial frame, yes. One of your ships is accelerating, so care is needed when you consider its measurements, to make sure you are using only one inertial frame for your analysis (or using the more complex maths for handling non-inertial frames).

etotheipi, vanhees71, entropy1 and 1 other person
Gold Member
With respect to any inertial frame, yes. One of your ships is accelerating, so care is needed when you consider its measurements, to make sure you are using only one inertial frame for your analysis (or using the more complex maths for handling non-inertial frames).
In #11 I changed this to IFR's A and B. Sorry for the misunderstanding. (And thanks for the clarification )

Gold Member
If TD and LC are "in the eye of the beholder" (are relative), how can they actually happen? And in this case, how can a photon choose between moving at speed c either away from A or towards B, if TD and LC are not actually happening? And if TD and LC are actually happening, relative to what?

PeterDonis
Mentor
2019 Award
If TD and LC are "in the eye of the beholder" (are relative), how can they actually happen?
If you define "time dilation" and "length contraction" to be the frame-dependent things that those terms have been used to mean in this thread, then they don't actually happen. They are just frame-dependent things.

how can a photon choose between moving at speed c either away from A or towards B, if TD and LC are not actually happening?
The photon moves on a particular null worldline through spacetime. That's all it has to choose. That null worldline, plus the timelike worldlines of A and B, are sufficient to determine any actual observable.

if TD and LC are actually happening
If you define "time dilation" and "length contraction" to be things that actually happen--i.e., things that are actual observables, and hence must be frame invariant--then you need to tell us what definition you are using and what invariants these terms refer to.

Gold Member
If you define "time dilation" and "length contraction" to be the frame-dependent things that those terms have been used to mean in this thread, then they don't actually happen. They are just frame-dependent things.

The photon moves on a particular null worldline through spacetime. That's all it has to choose. That null worldline, plus the timelike worldlines of A and B, are sufficient to determine any actual observable.

If you define "time dilation" and "length contraction" to be things that actually happen--i.e., things that are actual observables, and hence must be frame invariant--then you need to tell us what definition you are using and what invariants these terms refer to.
If the photon's speed is expressed through TD and LC, and these effects are relative to the observer, then speed c of this photon is relative to the observer, right? I mean, if the speed of light relative to any observer is constant, you get TD and LC.

However, if we experience for instance TD in satellites, this is actually happening (following standard relativity).

So if the speed difference between ship A and ship B is symmetrical, if A stays younger than B, B should stay younger than A also, from their pov, right? This seems to me a contradiction (of which the solution has to lie in acceleration I think).

PeterDonis
Mentor
2019 Award
If the photon's speed is expressed through TD and LC
It isn't.

You have things backwards. You cannot reason correctly about relativity if you try to use TD and LC as your starting point. TD and LC are frame-dependent quantities that, if you really care about them (and you shouldn't), you can derive once you have correctly analyzed a scenario. But you cannot start with them if you want to correctly analyze a scenario.

if we experience for instance TD in satellites, this is actually happening
It is unfortunate that the term "time dilation" is used to refer to two different things: a frame-dependent quantity, and an invariant that should more properly be called "differential aging". The latter is what you are referring to with respect to satellites. (Note that satellites orbiting the Earth cannot be analyzed using SR to begin with, since spacetime curvature cannot be ignored, and there is no such thing as a global inertial frame in which you can analyze their motion, so the SR concept of TD doesn't even apply.)

vanhees71
PeterDonis
Mentor
2019 Award
if the speed difference between ship A and ship B is symmetrical, if A stays younger than B, B should stay younger than A also, from their pov, right?
Again, you cannot correctly analyze a relativity scenario if you start from TD and LC alone. But that is what you are trying to do.

It is true that if, in addition to TD and LC, you include relativity of simultaneity in your analysis, you can correctly analyze relativity scenarios. You are failing to include relativity of simultaneity in the above quote; putting it in will fix the particular problem you think you see in that particular example.

In general, though, I think it is better to just give up the idea of using TD and LC as starting points, period. The best starting point for any analysis of a relativity scenario is spacetime geometry and invariants.

Nugatory
Mentor
If the photon's speed is expressed through TD and LC....
It isn't . Time dilation and length contraction are irrelevant to the movement of the flash of light, we only need them when we are comparing calculations made using different inertial frames.

An analogy: I draw a line on sheet of paper. That's a real physical thing: some of the paper fibers are soaked in ink and some aren't, and we all agree about which these are.

Someone says "I'm not looking at the sheet of paper right now, can you tell me where the line is?"
I reply "Sure. Let's put the origin at the bottom left hand corner of the sheet of paper, run the y axis up the left edge and the x axis along the bottom edge. The line is described by ##y=2x## - it starts at the bottom left-hand corner and climbs two units vertically for every unit horizontally"
And then someone else says "Well, I'm going to put the origin at the same place, but will use the x'-axis and y'-axis which are parallel to the edges of the table. And I say the line is described by ##y'=x'## - it climbs only one unit vertically for every unit horizontally". (Of course what's really going on is that the sheet of paper is not square to the tabletop and it's the exact same line either way - nothing has changed about which fibers are inked).
We continue this discussion, and we realize that for any point anywhere on the sheet of paper, when I say that it's at ##(x,y)## they say that it's at ##(x',y')## and always ##x'>x## and ##y'<y##.... and.... wait for it.... We have just discovered "x contraction" and "y dilation"!

etotheipi, Ibix and vanhees71
Mister T
Gold Member
What reference does the photon have to determine its speed?
The photon does not measure its own speed. Nothing can measure its own speed. The speed of the photon is something that's measured by each of the space ships.

When you drive your car at a speed of, say. 90 km/h, what you are actually measuring is the speed of the roadway relative to your car. And of course, that's the same as the speed of the car relative to the roadway. But without the roadway the measurement would be impossible.

vanhees71, Nugatory and Ibix
Ibix