Why is the speed of light...

[Samm]

What is Cherenkov radiation?

 

I think it's where the electrons or other charged particles travel faster through a medium than the speed of light in that medium. And then this produces radiation through some mechanism that I don't understand.

 

Edit: On further research it has to be a dielectric medium. But I don't really know what that means either.

Edited January 24, 2012 by Samm

[DrRocket]

The question you really want to ask is why is the speed of light the same in any inertial frame of reference? That is independent of the motion of the source.

 

Because Maxwell said so.

[Tres Juicy]

Because Maxwell said so.

 

 

If Maxwell said to jump under a bus....

[ajb]

If Maxwell said to jump under a bus....

 

DrRocket is quite correctly referring to Maxwell because of his equations of electromagnetism. In fact, this has already been discussed.

[Tres Juicy]

DrRocket is quite correctly referring to Maxwell because of his equations of electromagnetism. In fact, this has already been discussed.

 

 

I know, I just found it funny

[DrRocket]

If Maxwell said to jump under a bus....

 

I would find one of the many deserving souls around here and toss him under the bus.

[michel123456]

I would find one of the many deserving souls around here and toss him under the bus.

A "no motion" bus I hope.

[DrRocket]

A "no motion" bus I hope.

 

yeah -- in its rest frame

Edited January 29, 2012 by DrRocket

[1=1]

In my opinion, eventually we will have to get to the basic rules of the universe, such as 1=1, which is more logical, or something along the lines of a atom being 100 000 fm. There will be something that just is, there will be no reason it is as it is, it just is.

[StringJunky]

We don't have a model for a photon in a rest frame, and no real way to test one if we did. So it's a big problem.

 

Up to now, I've understood it that there is no time in the frame of the photon and therefore arrives at it's destination instantly, so, to ask: "What do things look like travelling at that velocity?" is moot....it's all arrivals without the journeys in between in that frame. Is this incorrect?

Edited January 29, 2012 by StringJunky

[D H]

Up to now, I've understood it that there is no time in the frame of the photon and therefore arrives at it's destination instantly.,

You'll see that written in many places, but it is essentially nonsense. There is no such thing as a rest frame of a photon. The idea doesn't make sense.

 

At age 16 Einstein started thinking about what it would be like to be riding along with a beam of light. He realized back then that things got a bit paradoxical, maybe even nonsensical. Realizing that this concept was nonsense was what finally let Einstein see relativity ten years later.

[StringJunky]

You'll see that written in many places, but it is essentially nonsense. There is no such thing as a rest frame of a photon. The idea doesn't make sense.

 

At age 16 Einstein started thinking about what it would be like to be riding along with a beam of light. He realized back then that things got a bit paradoxical, maybe even nonsensical. Realizing that this concept was nonsense was what finally let Einstein see relativity ten years later.

 

Thanks, so, how do you deal with it...just ignore it because it's not a sensible question if one actually understood the math?

[ajb]

Thanks, so, how do you deal with it...just ignore it because it's not a sensible question if one actually understood the math?

 

You don't exactly ignore it, it is a fundamental result of special relativity. More mathematically this is equivalent to the the Lorentz group being non-compact. That is the parameter in the Lorentz transformations, v is strictly less than c. Thus you cannot by a Lorentz transformation start from some inertial frame to find another inertial frame of reference in which the photon is at rest.

 

One can though consider more general transformations and end up with frame dependent effects. For example, one could consider light-cone coordinates. Here we have effects due to the coordinate system we pick. Importantly we do not have that the speed of light is equal to c. This is a quite generic effect of not picking inertial frames, which we are of course free to pick.

 

The key thing in special relativity is that we have a set of (globally) preferred frames, the inertial ones. Most of the statements made in special relativity explicitly refer to things in inertial frames and things that are invariant between such frames.

[PeterJ]

In my opinion, eventually we will have to get to the basic rules of the universe, such as 1=1, which is more logical, or something along the lines of a atom being 100 000 fm. There will be something that just is, there will be no reason it is as it is, it just is.

We are already there in metaphysics, where just the first four integers and one imaginary number are required for the universe.

[yaik7a]

I think this depends on your view on faster then light speeds and if they're attainable too. If you think you can't go faster then light then any large number above 300'000 is fine but if you think you can go faster then light then it has to remain 300'000.

[PeterJ]

You don't exactly ignore it, it is a fundamental result of special relativity.

Erm. Sorry, but what is a fundamental result? The fact that we cannot say anything about time for photons?

[ajb]

Erm. Sorry, but what is a fundamental result? The fact that we cannot say anything about time for photons?

 

It is a result, or maybe rather a postulate depending on how you approach the subject, that there are no inertial frames of reference in which the speed of light is not measured to be c. This means we cannot talk about "time as measured by a photon" or similar statements.

[PeterJ]

I'm afraid I can't see why we have to decide on this issue on the basis of a theory that has nothing to say about it.

 

If photons are out of time then is this not interesting enough to talk about?

 

 

 

 

 

 

 

 

 

 

[swansont]

I'm afraid I can't see why we have to decide on this issue on the basis of a theory that has nothing to say about it.

 

If photons are out of time then is this not interesting enough to talk about?

 

To talk about it, we need a model. We don't have one. We can only discuss how photons behave in the frames of reference we can be in, because that's the model we have.

[PeterJ]

Okay. I see what you're saying. Physics will never have anything to say about this. It's another of those interesting philosophical problem beyond the reach of science, to be discussed on a different forum.

 

But surely if this is the case then it means that I was right to start with, that from the photons perspective it is timeless or outside time, and this would be the reason why we cannot talk about it in terms of our Minkowskian model. No?

[ajb]

But surely if this is the case then it means that I was right to start with, that from the photons perspective it is timeless or outside time, and this would be the reason why we cannot talk about it in terms of our Minkowskian model. No?

 

Define what you mean by "timeless".

 

If it is a statement like: the affine parameter describing a null geodesic cannot by an affine reparametrisation be understood as the time measured by a standard clock moving along the null geodesic, then yes, photons are timeless.

 

The problem is that the space-time interval for null geodesics, that is the paths of light is always zero. This messes things up.

 

For a time-like geodesics, that is the paths of massive particles moving in just a gravitational field (or no field) one can reparametrise things so that the parameter describing the geodesic is the local time as measure by some clock following that geodesic. One can talk about time as "felt" by any massive test particle.

[PeterJ]

Right. Got that. I think we are just differing on the words.

 

Is the space-time interval zero? I thought that as one parameter went to zero the other went to infinity.

[ajb]

[math]ds^{2} = dx^{2}- c^{2}dt^{2}[/math]

 

along some path.

 

For null geodesics this is always zero. So you cannot use s (or - s) to parametrise a null geodesic.

[DrRocket]

For a time-like geodesics, that is the paths of massive particles moving in just a gravitational field (or no field) one can reparametrise things so that the parameter describing the geodesic is the local time as measure by some clock following that geodesic. One can talk about time as "felt" by any massive test particle.

 

In fact the parameter required in this instance is just arc length, with respect to the natural metric of spacetime. So what is involved is just the usual trick of parameterizing a curve by arc length. So, one can say that for a timelike curve the length of the curve is the propertime associated with that world line.

 

This in fact applies to any time-like curve in spacetime, whether or not it is a geodesic. If one also notes that with a metric of signature +,-,-,- that geodesics actually maximize arc length (quite different from the Riemannian case) one has a quick resolution of the "twin paradox" (the "non-traveling twin" has a world line that is a geodesic while the traveling twin has a non-geodesic world line, so the non-traveling twin's world line has the greater proper time).

 

I guess you might stretch this to say that the proper time associated with any segment of a null geodesic is zero, but I don't know that you get anything useful out of that stretch.

Edited January 30, 2012 by DrRocket

[ajb]

This in fact applies to any time-like curve in spacetime, whether or not it is a geodesic.

 

Yes, of course.

 

 

I guess you might stretch this to say that the proper time associated with any segment of a null geodesic is zero, but I don't know that you get anything useful out of that stretch.

 

I doubt this is a useful concept, but is as close to "photons being timeless" as I think we can get.