Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Special relativity states:

  • The speed of light in a vacuum is always $c$, regardless of the velocity of the observer.
  • The laws of physics are the same for all observers in uniform motion.

These two statements that we know as "special relativity" indicate that $c$ (speed of light) does not follow the same laws of physics as the observer.

  • The laws of physics are the same for all observers in uniform motion.

  • The observer is (anything that never travels at the velocity $c$).

  • The laws of physics are the same for (anything that never travels at the velocity $c$).

  • The laws of physics are not the same for anything that does travel at the velocity $c$.

This suggests that when we use our physics to measure and predict light, what we are predicting and measuring is only the disturbance in our physical laws, caused by the passing of something that does not exist within our physical laws. As an analogy, the electromagnetic wave is to the photon as thunder is to lightning. Our physical laws enable us to measure and predict the disturbance in our reality that we know as electromagnetic wave light, but the cause of this electromagnetic wave exists under entirely different laws of physics and therefore a different reality.

share|improve this question

closed as off-topic by ACuriousMind, Colin McFaul, Jerry Schirmer, Kyle Kanos, Ilmari Karonen Jul 22 at 15:50

  • This question does not appear to be about physics within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

3  
What's wrong with just saying that a ray of light is not a valid observer? –  Jerry Schirmer Jul 21 at 4:57
5  
It doesn't rule out anything about the physics of light. It just tells us that we cant' look at physics <b>from the light's perspecive</b>. It is still 100% consistent to talk about the influences of light rays, and the interaction of the electromagnetic field with matter. There is just no observer that will be able to look at an electromagnetic wave and say that it is static. This makes sense, because it would otherwise break Maxwell's equations. –  Jerry Schirmer Jul 21 at 5:22
2  
Comment to the latest edit (in particular on its form): Mathematics is a constructive language that allows to write (true) statements (within a given theory). Rhetoric is a form of language, and you write statements, but has not much else in common with mathematics/physics...rhetorical questions does not seem the best way to understand mathematical/physical concepts. As John Rennie, I will not write anymore because it seems that there is no constructive improvement. Also, to speak the truth, this whole argument is not so interesting (at least for me). –  yuggib Jul 21 at 13:53
4  
I'm voting to close since this question appears to be off-topic because it is about nothing well-defined. You keep throwing words around like "the laws of physics", "existence" and "the photon's universe" without ever providing a definition of them, and since your understanding of these words seem to differ significantly from that of every physicist, the question is unanswerable. To me, it makes as much sense as asking "Why does blue taste of unicorns?". –  ACuriousMind Jul 21 at 23:56
2  
Given that this question has drastically changed from its initial incarnation, the answers to this question appear to be no longer relevant and this question should be closed. If you have new questions, based on the original question, then ask a new one but include a link to this one. –  Kyle Kanos Jul 22 at 14:51

5 Answers 5

To make progress we need to be clear what we mean by the laws of physics and observer.

A law of physics is just some set of equations that we use to predict what happens. So if for example we're trying to describe how charges interact with light our set of equations, i.e. our law of physics, would be Maxwell's equations.

But to write down Maxwell's equations and solve them we need to choose a set of coordinates. We need an origin and time and distance vectors. Choosing these chooses a frame of reference, and this then allows us to assign $(t,x,y,z)$ positions to things that happen. The act of observation is just the recording of these positions.

So far so good. Now suppose that you and I are making measurements of some system, e.g. an electron, and we're moving relative to each other at some high velocity near $c$. The laws of physics, Maxwell's equations in this case, are the same for both of us but that does not mean we observe the same thing. If the electron is stationary in my frame Maxwell's equations tell me it will have an electrostatic field and that's what I will measure. In your frame the electron will be moving and Maxwell's equations tell you it has both electric and magnetic fields and that's what you will measure. So we observe different things even though we're using the same laws of physics.

What special relativity tells us is that my observations and your observations are linked by a Lorentz transformation. If I take my observations and Lorentz transform them into your frame they will be the same as what you observe. This is always the case for all physicists moving at any speed relative to us of less than $c$. This is what we mean when we say the laws of physics are the same for all observers. It does not mean they all observe the same thing, only that their observations are all equivalent in that they are linked by a Lorentz transformation.

Now on to the key part of your question. What happens in a frame of reference moving at a speed of $c$ relative to us?

Well, there's a get out clause. If you take my example above as a reference we can never observe an electron moving at the speed of light relative to us, because accelerating an electron to $c$ would require infinite energy. So while we can ask what measurements are made in a reference frame traveling at 0.9$c$ or 0.999999999$c$, the question what measurements are made in a frame travelling at $c$ is meaningless - no such measurements can be made. Just as well really, because the Lorentz transformations become singular at a relative velocity of $c$ so we couldn't do the calculation anyway.

But of course light travels at $c$. However this doesn't mean the laws of physics are different for light, it just means we can't write down Maxwell's equations in a reference frame travelling at the speed of light relative to us. But that's OK because light can't make measurements so there's nothing to use Maxwell's equations to predict. Only we massive things can make measurements, and we can't attain relative speeds equal to $c$.

Going back to your question, your last point 4 states:

The laws of physics are not the same for anything that does travel at the velocity $c$.

The trouble with this is that the obvious interpretation is:

The laws of physics do not apply to anything that does travel at the velocity $c$.

which is obviously wrong because the behaviour of light is well described by Maxwell's equations, and as we've seen Maxwell's equations are the same in all (sub-luminal) reference frames. On the other hand if you wrote instead:

Maxwell's equations cannot be written down in a reference frame travelling at a relative velocity of $c$.

then this I would have to agree with. But while this may seem startling to non-physicists, for us physicists well that's kind of obvious and it's just how the universe is.

share|improve this answer

I will expand my comment above into an answer, but I will not comment further on it to avoid the usual very long discussions of your posts.

In my opinion, you are trying to argue on a logical level, but it is not clear if you have enough knowledge of logical theories to do so on a mathematical/physical level.

Without entering too much into details, a mathematical theory consists of a collection of objects and symbols that define the formal language, a collection of sentences in the formal language (the axioms) and a collection of inference rules that tells how formal sentences can be combined to obtain other sentences. This rigorous approach is often relaxed, and more discursive (informal) proofs and statements are allowed (however formalizable, i.e. reducible to formal statements and proofs).

The connection with your question is the following: special relativity is not a complete logic theory by itself, it is only a set of axioms (call this set $SR$), independent of the others, within some more complete logical theory. They are independent in the sense that there is no other axiom $P\notin SR$ such that $P\implies Q$, $Q\in SR$. These axioms are more precisely schemes, in the sense that they are rules that apply to the objects of the logical theory, but do not define any ad hoc new object.

In particular "light", "photon", "laws of physics" etc. are not defined by special relativity. Their existence is not related to special relativity, the latter just gives some rules that these objects, provided they are correctly formalized and belong to the theory, must satisfy.

You say that your opinions on the object you call "light" and its "laws of physics" are suggested by special relativity, but as you put them they seem to me completely unrelated to it. Even if you were able to make your ideas into statements that are formally acceptable in some logical theory, you have also to guarantee that at least part of your objects satisfy the axioms and schemes of $SR$, if you would like to have a theory that contains special relativity, and that is not unrelated to it. If you are able to do so, and obtain by means of your theory (at least a part of) the predictions that the usually accepted physical theories are able to predict, and in addition something different, that is verifiable experimentally, then you would have every physicist's attention. If else you are only doing philosophical speculation, metaphysics or call it whatever you like, but not mathematics/physics.

In conclusion: Since you are looking for something that is not defined in the usually accepted physical theories, your question is mathematically meaningless as it stands. This is unless you define precisely what the "laws of physics", "light" and "observables" are in your theory, which are the objects and axioms, and thus their predictive power. You should then be able to investigate why, within your theory, "light and observables have different laws of physics", because the latter would be then a meaningful sentence (and hopefully true, or at least not false).

share|improve this answer
    
In theory, you wouldn't be able to "predict" something that happened in a universe without time. –  Derek Roberts Jul 21 at 23:30

OK, just one more try to end this stupid question.

There IS a way to formulate physics using light rays as your basis: a double null coordinate system${}^{1}$. If you have a ray moving in the $+x$ direction, define the two coordinates

$$2\xi = t + x\;\;\quad\quad\quad2\eta = t - x$$

Then, the metric becomes

$$ds^{2} = -4d\xi \,d\eta + dy^{2} + dz^{2}$$

All measurements can be expressed in this coordinate system. Individual light rays moving only in the $x$ direction are identified by their constant value of $\xi$ or $\eta$, and rather than measuring the evolution of "time", they see spatial variation through identification of surfaces of varying $\eta$ intersecting with them.

Somehow, I think you're going to be completely unsatisfied with this answer, because you're out to yell about how physics is wrong and physicists are arrogant idolaters with false preconceptions about the universe, and it would be obvious if people just THOUGHT about it, man.

${}^{1}$ In string theory, they call these coordinates light-cone coordinates. I like this terminology better, as it's clearer and more elucidating, at least to me.

share|improve this answer
    
+1 short and concise. –  Phonon Aug 9 at 19:50

Special relativity states:

... I'll select and discuss the given statements in some particular order (which may be called "in order of simplicity of discussion") ...

[...] The observer is (anything [...])

Right. Synonymous to "observer" or "anything", in the context of the theory of relativity, there are also the descriptions "material point" or "principal identifiable point" or "participant".

Of particular relevance is Einstein's foundational guideline that:

All our well-substantiated space-time propositions amount to the determination of space-time coincidences [such as] encounters between two or more recognizable material points.

Thereby

  • the identity of distinct recognizable observers in coincidences, and
  • the distinctiveness of coincidence events due to different subsets of identifiable observers

are presumed self-evident. Arguably then any such coincidence event is in turn considered observable, and any of its observations (by any particular identified observer) is in turn considered a coincidence (of that observer with the corresponding signal front).

The speed of light [...]

It is to be noted that

  1. In this context of (geometry and kinematics of) the theory of relativity, "light" means any signal front of signals exchanged between observers; and

  2. The statement under consideration is meaningful (reproducible) only as far as the notion of (how to measure) "speed" has been defined (operationally) in the first place.

Einstein expressed this requirement (for the particular notion of "simultaneity") quite memorablý:

We thus require a definition of simultaneity such that this definition supplies us with the method by means of which, in the present case, he can decide by experiment whether or not both the lightning strokes occurred simultaneously.

As long as this requirement is not satisfied, I allow myself to be deceived as a physicist (and of course the same applies if I am not a physicist), when I imagine that I am able to attach a meaning to the statement of simultaneity.

(I would ask the reader not to proceed farther until he is fully convinced on this point.)

[...] establish an inertial frame

That's concerning the primary (and especially difficult) method of measurement to be defined (within the General Theory of Relativity):

How to determine whether any two given participants (who were never coincident with each other, but "separated") were at rest to each other?

In other words:
How to establish (joint membership of separated observers together in) one "inertial frame"; if that's possible at all.

Only once this definition has been selected, and the determination of an "inertial frame" did actually succeed, it is possible to speak of (i.e. define and determine) "distance" between participants (or at least: "distance ratios"), "simultaneity" of indications of separated participants, and finally "speed".

The speed of light in a vacuum is always $c$

This statement follows as a theorem, given the usual, suitable definition of "distance between participants $A$ and $B$" as "$c/2$ ping duration", provided $A$ and $B$" were at rest to each other throughout.
(Concretely: the symbolic, literal, non-zero coefficient "$c$" which is part of the distance definition is identified unambiguously as signal front speed; a.k.a. "speed of light in a vacuum".)

The speed of light in a vacuum is always $c$ regardless of the velocity of the observer.

No: this applies only to observers who are involved in the determination of a "speed" value (let's say as "starting gate $A$", and as "finish line $B$") and who were therefore at rest to each other (a.k.a. jointly members of the same "inertial frame", a.k.a. in equal "uniform motion" rather than being individually "accelerated" with respect to some/any "inertial frame").

The laws of physics are the same for all observers in uniform motion.

Certainly the method how to determine whether any given observer had been "in uniform motion", or not, is the same/reproducible (by anyone who likewise understands/uses the basic notions of "identifiable observers" and their "observable coincidences").

The observer is (anything that never travels at the velocity $c$).

At least:

  • any identifiable observer (who could "carry a signal") cannot have been measured travelling faster than the signal front speed $c$, and

  • members of any two "inertial frames" cannot have measured their mutual speed (of their mutual "uniform motion") as the signal front speed $c$ (otherwise the prerequisite determination of "mutual rest" of observers/participants "within each inertial frames" would have failed in at least case to begin with).

The laws of physics are not the same for anything that does travel at the velocity $c$.

This formulation is confusing anything (observer, participant) which/who is considered observable and identifiable with any "signal itself" (which may be called "nothing but the signal having been exchanged between observers").
But yes, certainly the methods of measurement used by observers, based on their ability of mutually observing and recognized each other's signals, can not meaningfully be "used by signals themselves".

Therefore, conversely, it is in this sense meaningful

that Light and Observers have different laws of physics

However, that's perfectly consistent with RT, derived from "blatantly obvious" notions as sketched above.

share|improve this answer

Your notion seems to be based on the thinking that light is a bunch of photons, and a photon is some kind of weird particle that travels at the speed of light, like some tiny spaceship. Then you ask, how can this tiny spaceship violate physical laws? What makes it so special?

But a photon isn't a particle in any classical sense. It's not like a tiny spaceship, or a "particle" of dust, or anything else. In fact, virtually anything that you might associate with "particle" from classical experience is probably wrong. For example, as demonstrated by the double-slit experiment, a photon does not travel from point A to point B like a particle.

A photon is not a particle: it's a quantum. That is, it's a parcel of energy that can't be divided.

A photon, not being a classical particle, can't observe anything. Asking what a photon observes makes about as much sense as asking what your voice observes as it travels from your mouth to a listener's ear. We can talk about your voice's effects on the air, and the listener's ear, and what these objects might observe: but your voice, the wave traveling in the air, it isn't an object. It's just a system of behavior that describes the effects we will observe on objects.

share|improve this answer
    
Excitations of quantum fields, like photons (and indeed every other fundamental particle), are the things we call particles. These things are what particle physics studies. Photons are particles. And photons travel at the speed of light. And this is OK and not at all a problem. Their path between two non-identical places takes zero proper time and finite time as measured by all observers and this is OK and not at all a problem. –  dmckee Jul 22 at 1:44
    
@dmckee yes, you call them particles, but no one else does. The word "particle" in the mind of anyone who is not a physicist invokes a very different kind of thing. Then, these people get all confused when a physicist's "particle" isn't anything like the "particle" they know. You can't do a double-slit experiment with dust particles. They aren't waves in any sense. A dust particle can be divided in half with a knife. The conceptual overlap between your "particle" and a classical "particle" is very, very small. –  Phil Frost Jul 22 at 11:12

Not the answer you're looking for? Browse other questions tagged or ask your own question.