Why is the speed of light the same to any observer?

[Schrodinger's Dog]

The light is traveling in a perfectly straight line -- it only appears to be bending when we try to describe its path using coordinates that are not orthonormal.

You mean in reality/experimentation light is bent but in maths its traveling in a straight line. Seems like a contradiction to me. Can you explain gravitational lensing in terms that involve the light traveling in a straight line, and consistently model the experimental data?

The speed limit of the Universe as far as we know is c, in the same way that the fine-structure constant $\alpha = 7.297\,352\,5376 \times 10^{-3}$

It just is, I'm sure there are a million and one hypothesis as to why though.

 

[MeJennifer]

if you measure things in terms of Planck units, there remains no "why?". there is no property of free space, that is not a consequence of the units used to measure it, that determines the speed of propagation of EM (or whatever other fundamental interaction). still no question at hand.

In this respect how do you interpret the fine structure constant?

 

[rbj]

In this respect how do you interpret the fine structure constant?

i guess as the square of the elementary charge since Planck units say nothing about e and everything else in the expression for $\alpha$ is 1.

i know that it's true that historically Planck himself did not define a unit charge with the rest of the Planck units in 1899, but more recently a few authors (notably Michael Duff, but there are others) have essentially extended the idea to a unit charge (as such that when two charges, each of one unit, are spaced apart by one unit length, you get a unit of electrostatic force, just as it is done in cgs) and you get $q_P = \sqrt{4 \pi \epsilon_0 \hbar c}$. and i realize that such a definition is sort of arbitrary. i s'pose they could define the unit charge to be e, but then the Coulomb constant would be $\alpha$ (expressing simply why, given e=1, that the fine structure constant can be thought of as the coupling parameter for the strength of EM) and the shorthand expression for the fine structure constant that theoretical physicists prefer ($\alpha = e^2/(\hbar c)$) could no longer be used. (being an engineer, i like seeing the $4 \pi \epsilon_0$ in the expression because then i know it's correct in any system of units.)

so, for my pea-brained electrical engineering understanding of it, if $\alpha$ was measured to have changed (and, being a dimensionless parameter, there is an operational meaning to a changing $\alpha$), there are at least two ways to interpret the meaning of it. either all of the charged particles in the universe (if $\alpha$ is believed to be the same everywhere in the universe) have changed the amount of effective charge on them (if $4 \pi \epsilon_0$=1) or , if e=1, then the strength of the EM interaction, relative to the other fundamental interactions, has changed. at least that's my simple spin to it.

 

[Hurkyl]

Um, but Hurkyl. If we apply a locally orthonormal coordinate system, then remote velocities can exceed c, and appear to be traveling in arcs, can they not?

Yes, remote coordinate velocities can exceed c. That can happen in SR too (in a non-inertial chart)

They are still traveling in straight lines, though -- it's only when you replace the metric with the Euclidean metric on your coordinate chart that they are curved.

You mean in reality/experimentation light is bent but in maths its traveling in a straight line.

No, I mean that you're using the wrong notion of "straight line" -- you're using the one derived from the Euclidean metric, rather than the one derived from the actual metric on space-time.

 

[Schrodinger's Dog]

No, I mean that you're using an incorrect notion of "straight line" -- you're using the notion of "straight" given by the Euclidean metric, rather than the notion of "straight" given by the actual metric on space-time.

The shortest distance between two points you mean? What definition are you using may I ask?

 

[Hurkyl]

The shortest distance between two points you mean? What definition are you using may I ask?

Geometrically, its trajectory through space-time is a geodesic. Physically, it is traveling inertially.

 

[Schrodinger's Dog]

Geometrically, its trajectory through space-time is a geodesic. Physically, it is traveling inertially.

Well that's alright then. I can still say that light is bent by gravity though can't I?

 

[MeJennifer]

Well that's alright then. I can still say that light is bent by gravity though can't I?

Many do use such terminology, personally I stay away from it.

Something that follows a geodesic does not accelerate or bend in GR. Acceleration in GR is caused by EM or other forces.

Furthermore, one can argue if it is at all meaningful to speak of traveling light. Energy carriers like photons are always emitted and absorbed, we cannot possibly detect a free traveling photon.

 

[DaveC426913]

Furthermore, one can argue if it is at all meaningful to speak of traveling light. Energy carriers like photons are always emitted and absorbed, we cannot possibly detect a free traveling photon.

This jives with the concept that photons do not experience time. Their existence consists basically of a fixed, static line in space-time that ties their point of emission with their point of absorption.

 

[Antenna Guy]

This jives with the concept that photons do not experience time. Their existence consists basically of a fixed, static line in space-time that ties their point of emission with their point of absorption.

... and emission/absorbtion are instantaneous within a photon's frame - no matter how far it travels.

Regards,

Bill

 

[DaveC426913]

... and emission/absorbtion are instantaneous within a photon's frame - no matter how far it travels.

Regards,

Bill

Well, yeah, but I was hoping to explain it with a model that didn't make any reference to the passage of time (such as "instantaneous").

 

[Nickelodeon]

Furthermore, one can argue if it is at all meaningful to speak of traveling light. Energy carriers like photons are always emitted and absorbed, we cannot possibly detect a free traveling photon.

Presumably it still holds true that c = frequency x wavelength. Perhaps we should forget about c and concentrate on f and w. Hope that doesn't sound facitious.

 

[MeJennifer]

Presumably it still holds true that c = frequency x wavelength. Perhaps we should forget about c and concentrate on f and w. Hope that doesn't sound facitious.

Note that the measured frequency of an absorbed photon is not absolute but depends on the gravitational potential between the emitter and absorber.

 

[Schrodinger's Dog]

Many do use such terminology, personally I stay away from it.

Something that follows a geodesic does not accelerate or bend in GR. Acceleration in GR is caused by EM or other forces.

Furthermore, one can argue if it is at all meaningful to speak of traveling light. Energy carriers like photons are always emitted and absorbed, we cannot possibly detect a free traveling photon.

Yes I can see why, I tend to stay away from describing something that is happening visually in physics too.

 

[Phrak]

1) Light travels a path such that the interval between two events on the path is Maximal in the sense that the Lagrangian is maximal.

2) With light traveling in a direction x, under a change of coordinates u=x+ct and v=x-ct, the Lorentz transform obtains to a simple diagonal matrix ((d,0)(0,1/d)). The v coordinate of emmission of a photon and absorbtion are the same (in Minkowski space).

DaveC

This jives with the concept that photons do not experience time. Their existence consists basically of a fixed, static line in space-time that ties their point of emission with their point of absorption.

In the speed-of-light frame, along the trajectory of the photon, no time or distance intervens between emission and absorbtion. So the events of emission and absorbtion in this particular coordinate system are cojacent. One might be prompted to think that no photon is exchanged nor required at all, but that an exchange of energy is, instead, a direct interaction.

I was once told that Dirac did some investigation in this matter.

How all of this is consistent with point 1) is beyond me.

 

[neopolitan]

In case the OP is still interested:

My take on it is this. We could think of spacetime as free from units, if we liked, but there is a fundamental difference between timelike dimensions and spacelike dimensions, and to be able to discuss them in like terms we need a "currency conversion". The currency conversion is like the answer to the question "how many chunks of time are equivalent to one chunks of space?" The answer is pretty much 1. One chunk of time is equivalent to one chunk of space, which is why physicists like to use units like year and light-year, or Planck time and Planck length, where the exchange rate is 1:1.

If you chose other units, selected for their convenience in other realms (like working out how much time you have before the commercials are over and how far you need to run to put the kettle on), then you end up with a different exchange rate, such as 1:3x10^8

But really, the answer is still 1. One fundamental chunk of time equals one fundamental chunk of space.

--------------------------

Here is another way of looking at it. The pig exchange. Making the assumption that two pigs are identical, a pig from the US is worth one pig from Australia.

The exchange rate is therefore 1:1. But we find carrying around pigs difficult, and they are really difficult to push into one of those vending machine slots, so instead we have a pocket full of coins.

A pig's worth of coins in the US may amount to 0.0517 megadimes while a pig's worth of coins in Australia might amount to 15500000 millicents. The exchange rate then is 1:3E8

But really, the exchange rate is still 1 pig = 1 pig (relatively speaking).


cheers,

neopolitan