Why is the speed of light a constant?

[greg_rack]

Approaching for the first time to these "higher level" topics is mind-blowing, and indeed I cannot understand why is the speed of light a constant... why doesn't it vary relatively to the emitter state of motion? And isn't it affected by gravity(I know it is affected in the sense of a spacetime curvature due to gravity)?

 

[fresh_42]

Approaching for the first time to these "higher level" topics is mind-blowing, and indeed I cannot understand why is the speed of light a constant...

It isn't, it depends on the medium. What you probably refer to is the speed of light in vacuum.

... why doesn't it vary relatively to the emitter state of motion?

This is exactly the subject of special relativity and distinguishes classical and relativistic physics, Newton versus Einstein. Why? This cannot be answered. It just is.

And isn't it affected by gravity(I know it is affected in the sense of a spacetime curvature due to gravity)?

Yes.

 

[PeterDonis]

And isn't it affected by gravity

The coordinate speed of light can be affected by gravity. But it can also be affected by choosing non-inertial coordinates in flat spacetime.

The measured speed of light in any local experiment will always be $c$, even in the presence of gravity.

 

[PeterDonis]

Yes.

I don't think it's that simple. See my response to the OP just now.

 

[phyzguy]

Several of your questions boil down to, "Why is the universe the way it is?" No one can answer that question. We all got here the same way you did. Wait until you start studying quantum mechanics!

 

[strangerep]

Approaching for the first time to these "higher level" topics is mind-blowing, and indeed I cannot understand why is the speed of light a constant... why doesn't it vary relatively to the emitter state of motion?

Although we don't have a totally satisfying "why" answer, we do know a few things about how this notion of a constant limiting relative speed is a necessary consequence of more fundamental assumptions about non-accelerated (inertial) observers.

I tried to outline some of this in an earlier thread, starting at post #8 therein. You might wish to read that to avoid rehashing too much old ground.

 

[anuttarasammyak]

why is the speed of light a constant... why doesn't it vary relatively to the emitter state of motion?

The speed of light in vacuum is a familiar observation of the upper limit for the speed at which conventional matter and information can travel according to SR. If it could change with the emitter state of motion, it should contradict with the statement that it is an example of the upper limit of the speed.

Why there exist upper limit in the speed ? Because of principle A which I do not know even whether it exists.
Why principle A ? Because of principle B.
Why principle B ? Because of principle C.
Why ...
Such a almost infinite repeats of why-because might go on in future, might not.

 

[Helena Wells]

It isn't, it depends on the medium. What you probably refer to is the speed of light in vacuum.

This is exactly the subject of special relativity and distinguishes classical and relativistic physics, Newton versus Einstein. Why? This cannot be answered. It just is.

Yes.

The speed of light is not affected by gravity. Just the path of light gets bigger and it takes light more time to cross it.

 

[PeterDonis]

The speed of light is not affected by gravity.

It's a bit more complicated than that. See post #3.

 

[Helena Wells]

It's a bit more complicated than that. See post #3.

#3 doesn't make any sense.

 

[Ibix]

#3 doesn't make any sense.

Yes it does. Do you understand the difference between coordinate speed and the locally measured speed?

 

[Helena Wells]

Yes it does. Do you understand the difference between coordinate speed and the locally measured speed?

There isn't coordinate speed, there is only local speed.

 

[PeterDonis]

There isn't coordinate speed

If you mean that coordinate speed is not something that is directly measured, but only calculated, yes, that is true.

If you mean that there is no such quantity period, that is false.

 

[Ibix]

There isn't coordinate speed, there is only local speed.

Really? What do you call $dx^i/dt$ then?

 

[kent davidge]

What do you call $dx^i/dt$ then?

Derivative of $x^i$ with respect to $t$

 

[Ibix]

Derivative of $x^i$ with respect to $t$

Ask a stupid question, get a stupid answer, as my old dad used to say...

 

[etotheipi]

the locally measured speed?

I fear I may be asking a similarly stupid question... but are 'locally measured' velocity components another way of saying the (spatial) components of the four velocity $\frac{dx^i}{d\tau}$? So the 'local' bit just implies time derivatives taken w.r.t. a clock moving with the thing whose velocity you are measuring?

 

[PeterDonis]

the 'local' bit just implies time derivatives taken w.r.t. a clock moving with the thing whose velocity you are measuring?

No. The thing whose velocity we are measuring is light. You can't have a clock moving along with light.

Locally measured speed means the speed of the thing being measured, as measured by clocks and rulers that are carried along with the observer who is doing the measuring, at the event where the observer's worldline and the worldline of the thing being measured coincide.

 

[etotheipi]

Locally measured speed means the speed of the thing being measured, as measured by clocks and rulers that are carried along with the observer who is doing the measuring, at the event where the observer's worldline and the worldline of the thing being measured coincide.

Ah OK, so 'locally measured' could refer to an observer in any acceptable state of motion (not just one moving with the thing being measured... which would of course not make sense anyway if that thing is light!), so long as the event where the measurement is taken is at the intersection of the worldlines of the thing being measured and the chosen observer.

 

[Ibix]

When you measure speed locally you typically adopt a local orthogonal frame, using Cartesian coordinates on a small "near enough" flat bit of spacetime. Then $dx^i/dt$ is the velocity.

But if you measure speed over long distances there isn't typically a Cartesian coordinate system available, and there's a degree of flexibility over how you define time and space anyway, so there isn't a "the" speed, there's just a rate of change of your chosen spatial coordinates with respect to time, a.k.a. the coordinate velocity.

The canonical example is the surface of the Earth. You can measure your speed on a running track with a ruler and a clock. But your speed over longer distances is (in general) difficult to specify in a single unique way. If the coordinates we are using are latitude and longitude, for example, then your velocity definitely isn't just the rate of change of your coordinates.

 

[fresh_42]

Locally measured speed means the speed of the thing being measured, as measured by clocks and rulers that are carried along with the observer who is doing the measuring, at the event where the observer's worldline and the worldline of the thing being measured coincide.

So this is the speed they measure in experiments where they e.g. slow down light to an almost complete halt?

 

[Ibix]

So this is the speed they measure in experiments where they e.g. slow down light to an almost complete halt?

No - that's basically a high refractive index (edit: or a trick where you measure average speed of a diverging or dispersive wave packet). It's just that, on scales where spacetime curvature is apparent, "distance" and "time" aren't uniquely definable so "speed" isn't uniquely defined either.

 

[PeterDonis]

this is the speed they measure in experiments where they e.g. slow down light to an almost complete halt?

That speed is not the speed of light in vacuum. Only the speed of light in vacuum is invariant.

 

[Nugatory]

Would it be wrong to say that locally measured speeds are the coordinate speeds measured by a nicely behaved e.g. Cartesian frame?

It may be better to phrase it the other way: If the coordinate speed is $c$ (with appropriate homogeneity and isotropy) then you are using a nicely behaved coordinate system.

 

[anuttarasammyak]

There isn't coordinate speed, there is only local speed.

Here I observe local speed c. There he observes local speed c. I do not admit his observation value c as right one because his observing condition suffers from gravity in his place, e.g. time dilation and length contraction. He does not admit my observation value c as right one because my observing condition suffers from gravity in my place, e.g. time contraction and length dilation. Both of them construct their coordinate speed from reinterpretation of local observations and influence of gravity in DISTANT places.

 

[BiGyElLoWhAt]

Several of your questions boil down to, "Why is the universe the way it is?" No one can answer that question. We all got here the same way you did. Wait until you start studying quantum mechanics!

^this
Maybe it's just where I am (as I've been on a history kick lately), but I feel like the OP would benefit from looking at Reyleigh-Jean law and the ultraviolet castrophe and into the Michelson-Morsely experiment, to see HOW we discovered the constancy of the speed of light, and that it just kind of seems to be that way, therefore GR. =D

 

[etotheipi]

I wanted to check that I understand this correctly...

So an inertial observer will always measure a coordinate speed of $c$, and an accelerating observer who uses a (infinite) series of instantaneous inertial frames will then also always measure $c$. But if an accelerating observer uses accelerating coordinates, then they won't necessarily measure $c$...

...I "found" an equation for one dimensional motion (in Rindler coordinates, which I don't claim to really understand either ) which says$$v_{\text{light}} = c \left(1+ \frac{ax}{c^2} \right)$$if $a$ is the frame acceleration and $x$ is the position of the light. So in the case that we make a 'local' measurement (described in #18) where the constraint is that the event of the measurement is where the worldlines coincide (i.e. $x = 0$), we do again obtain $v_{\text{light}} = c$. That seems to be consistent with post #3. Is it vaguely correct? Also, if that equation is right, I wondered if someone could point me to a derivation or something? Thanks

 

[jbriggs444]

I wanted to check that I understand this correctly...

So an inertial observer will always measure a coordinate speed of $c$

I suspect that you have learned enough to have figured this all out yourself. But if not, here is my take.

If you are measuring a coordinate speed you do no need a separate notion of an "observer". You just read the speed off from the coordinates. Speed = change in position coordinate divided by change in time coordinate. Whether you get a result of c or not depends entirely on the coordinates.

But perhaps you wish for the inertial observer to be working in flat space-time and using the [unique up to rotation] set of non-rotating coordinates in which he is at rest at the origin. Then yes, these are inertial coordinates and the coordinate speed of light will be c.

and an accelerating observer who uses a (infinite) series of instantaneous inertial frames will then also always measure $c$.

If you graft a series of instantaneous tangent coordinate systems together then you will get a non-inertial coordinate system (and one that may not cover all of space and time reversibly). If you measure the speed of light against this non-inertial coordinate system the procedure is, in painful detail:

1. Light pulse is emitted somewhere. Find a clock reading on the observer's wristwatch so that the emission event is simultaneous in the observer's then-tangent inertial frame with that wristwatch tick event. [Hope that this clock reading is unique]

2. Find the position coordinate of the light emission event in that tangent inertial frame.

3. Let a short time elapse on the observer's wristwatch.

4. Find the position coordinate of the event on the world line of the light pulse which is simultaneous in the observer's new tangent inertial frame with the observers new wristwatch tick.

5. Compute the difference in the position coordinates (taken from two different tangent coordinate systems).

6. Divide by the elapsed proper time on the observer's wristwatch.

The result is not guaranteed to be c. The change in inertial coordinate system between the start event and the end event can jigger the results.

However, if the light speed is being measured close to the observer, the effects of the coordinate system change due to the observer's proper acceleration get smaller. It's like a coordinate system rotation -- the closer you are to the origin, the smaller the effects of rotation become. In the limit, you get perfectly accurate agreement with c.

 

[etotheipi]

Thanks, that's really helpful. Is there then no practical difference between a series of instantaneous inertial coordinate systems and an accelerated coordinate system?

(I found some discussion here about uniformly accelerated frames and as far as I can tell (I think I am slightly out of my depth ) they say the metric will have a top left component $g_{00} = (1+ Ax)^2$ which means then that the differential "local time" goes like $d\tilde{t}= (1+ Ax) dt$ and the locally measured speed is $\tilde{v} = \frac{dx}{d\tilde{t}} = v/(1+ Ax)$ if $v$ is the coordinate speed. That appears to give the same equation as above, and I think it corresponds to the measurement process you outlined?)

 

[PeterDonis]

If you graft a series of instantaneous tangent coordinate systems together then you will get a non-inertial coordinate system

One particular kind of non-inertial coordinate system, yes. But not the only possible kind.

Is there then no practical difference between a series of instantaneous inertial coordinate systems and an accelerated coordinate system?

The particular kind of non-inertial coordinate system that @jbriggs444 described is basically intended to accomplish this as much as possible. But, as noted above, this is not the only possible kind of non-inertial coordinate system.

I found some discussion here about uniformly accelerated frames

The uniformly accelerated coordinates described there are Rindler coordinates with the spatial origin $x = 0$ at the worldline of the observer with proper acceleration $A$. The paper has some useful discussion, but it is rather odd that it never even mentions Rindler coordinates. (It's also rather odd that the paper never mentions that the scenario in section VII is the Bell spaceship paradox--reference [7] is Bell's paper on that scenario.)