# How does Maxwell equation suggest that the speed of light is the same

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1. Nov 29, 2013

### kakolukia786

Hi guys,

I am reading griffiths electrodynamics. I have a question. How does Maxwell equation suggest that the speed of light is same for all observer? I know Maxwell's equation gives a value of C. How can we infer that it is same for all observer. Special relativity just assumes that the speed of light is same but how do we prove this theoretically?

Thanks

2. Nov 29, 2013

### Staff: Mentor

We don't. You cannot prove your postulates theoretically. All you can do is to perform experiments which will either validate or falsify the postulates.

3. Nov 29, 2013

### dvf

(Quibble) ... which will or will not falsify the postulates. I think that "validate" is too strong, as it might be interpreted as a synonym of "prove".

4. Nov 29, 2013

### kakolukia786

5. Nov 29, 2013

### Staff: Mentor

We can't prove it, but there are only two ways to get an observer-dependent speed of light out of Maxwell's equations. One would be to introduce the observer's velocity into the equations somehow; and the other (adopted between 1861 and 1905 by most physicists including Maxwell himself) is to make an additional assumption about the existence of a "luminiferous aether". Either approach leads to various unpleasant asymmetries (Einstein mentions these in the introduction to his 1905 paper, but he was hardly alone among his contemporaries in his awareness of and distaste for them).

Thus, we get an invariant speed of light unless we make additional and otherwise unnecessary assumptions. That's not a proof that the speed of light is independent of the speed of the observer, but it's a fairly strong hint that we should consider that possibility - and then experimental observations will settle the question.

6. Nov 30, 2013

### Staff: Mentor

An experimental confirmation of a prediction of a theory does more than just "not falsify" a theory, it requires that all future theories must agree with the validated theory in the limit that applies to that particular experiment.

For example, when calculating orbital mechanics for rocket missions etc. people use Newtonian gravity rather than General Relativity. This is because Newtonian gravity is a valid theory for the rocket mission, i.e. it is experimentally validated in this limit and the more complete theory of GR reduces to Newtonian gravity for this purpose.

Classical mechanics remains valid in the classical limit, despite the development of more general theories that work outside the domain of validity of classical mechanics.

Last edited: Nov 30, 2013
7. Nov 30, 2013

### Staff: Mentor

The fact that a theory is valid in a certain domain of applicability does not mean it is The Way The World Really Is, which I doubt we can ever Really Know.

8. Dec 2, 2013

### PhilDSP

Greater clarity on what you first asked can be gained by looking at the deeper history of the subject. What we call the Maxwell Equations were intended by Maxwell to apply only in the case where the charge was not moving with respect to the laboratory or measurement device. He had worked a bit at developing equations that could be used for moving charges but found that the situation was too complicated to quickly arrive at a clear solution for that case.

Hendrik Lorentz found the working solution for moving charges from which the Lorentz Transformation arose. In essence the LT is the mathematical apparatus that would have otherwise, in a different form, been inserted into the Maxwell Equations to give solutions for moving charges. It's the fact that the light propagation factor c in the Maxwell Equations is unchanged by the LT that leads one to believe that light propagates at the same speed in any kinematic situation.

Much more detail on this is available in the book "Electrodynamics from Ampere to Einstein" by Olivier Darrigol.

Last edited: Dec 2, 2013
9. Dec 6, 2013

### Meir Achuz

c appears as a constant in Maxwell's equations, related to the ratio of electric charge as it appear in magnetism, and as it appears in static electricity. If Galileo's principle of relativity is extended to electromagnetism (of which he knew little), then that constant must be the same in all 'inertial systems'.
When EM wave propagation is derived from Maxwell's equations, the wave speed turns out to be that constant.
So it is the extension of relativity to include EM that requires an invariant speed of light.

10. Dec 8, 2013

### Rena Cray

The speed of light is not c. This is simplistic. We can pick out a particular direction where the phase velocity is c, minimally,in vacuum. In all other directions the phase velocity exceeds c.

You must understand that when the typical poster says the velocity of light is c, they are speaking of some idealized experimental arrangement.

The best way to see this is in the propagation of electromagnetic radiation down a rectangular wave guide. The velocity is greater than c down a wave guide. However, the primary waves have velocity c. This means our focus is predisposed is focused on the direction down the axis of the wave guide that tells use that the velocity is less than c.

It is a good idea not to convolute the speed of light with the constant c.

Light has no particular velocity. But c seems immutable.

Last edited: Dec 8, 2013
11. Jan 2, 2015

### leroile

Maxwell's equations indicate that the speed of light is $$v=\frac{1}{\sqrt{\epsilon_{0}\mu_{0}}}$$ Here is the full proof of this formula.

12. Jan 2, 2015

### alva

A very silly question.

How do you measure the speed of light ?
You measure distance using rulers.
You measure time using clocks.

What devices would you use for measuring the speed of ligth? At rest or for a moving observer.

No "doopler relativistic formulae", please. A device that could really measure, not infer, the speed of light.

I apologize if my question does not seem to fit into this thread.

13. Jan 2, 2015

### ghwellsjr

You start with your clock and your light source together and you place a mirror some measured distance away so that when you turn on the light it will go the distance and reflect back to you. You note the time on the clock when you turn the light on and you note the time when you see the reflection get back to you. Then you take double the distance and divide that by the time the light traveled from when you turned on the light until you see the reflection. It won't matter if you are at rest or moving as long as you are not accelerating.

14. Jan 2, 2015

### phinds

And to add to ghwellsjr's description, this has been done MANY times over many years by many people.

"c" is now a DEFINED value, no longer subject to further refinement by more and better measurements.

15. Jan 2, 2015

### Staff: Mentor

I like Nugatory's explanation, but to expand/compare:

The speed of light and speed of sound are similar in that the most basic equations that utilize them don't discuss the issue of frame dependent speed -- they're just constant. The difference is that sound has a known medium upon which it travels at that constant speed, but light doesn't. This leads one along a line of logic that says that if there is no medium for it, then there is no way to describe a scenario where Maxwell's equations produce difference speeds for different observers -- so all observers must see the same speed for it. That's the critical insight that Einstein had -- recognizing that oddity was a reality.

16. Jan 3, 2015

### alva

Im not talking about "my clock" and "when I turn the light on". The source of light might be moving respect to my laboratory.
Could you use two photodiodes located 1m apart connected to a scope, each one to one channel of the scope and measure the time between the signals received ? (Yes/No, please).

17. Jan 3, 2015

### ghwellsjr

No. That method would add other factors, like the propagation times of the signals in the cables, but what you could do is have two photodiodes aimed in opposite directions back to back and then let the light from the source start your clock when it hits one photodiode and then travel to a mirror one meter away and stop the clock when the reflection hits the other photodiode.

You can only measure the round trip speed of light. Trying to measure the one way speed of light requires you to have synchronized clocks. The two way measurement requires just one clock,

18. Jan 3, 2015

### Ookke

How about this idea? Light (as EM wave) doesn't need medium for its propagation, but it needs space. So, in a sense, empty space is the medium for its propagation.

What speed do you have relative to this "medium"? None. If you move relative to a planet, your observation about the planet chages over time. If there is no change, you conclude that you are at rest. Empty space around you is always the same, empty space, no matter how you try to move, so you are always at rest relative to it. As the same applies to all observers, maybe it's not so surprising after all that light speed is constant. If air had the same property and was always at rest with every observer (which it's not) then speed of sound would also be constant and not affected by wind or motion.

[Disclaimer: This is just an aid for understanding and may have its own shortcomings. I find it somewhat useful in trying to understand why light speed is constant, but if anyone doesn't, don't use it. Usually light speed constancy is taken as a postulate.]

19. Jan 3, 2015

### ChrisVer

Well afterall, it's not Physics job to do that ... Physics would change "is" to "works"...the most successful Physical theory sees particles as representations of some groups...of course particles are not "that thing". But (amazingly) they work perfectly (up to now) as such.

Prove what theoretically? hmmm.... My answer would be that the Maxwell equations are invariant under the Lorentz transformations of Special Relativity.
So the speed parameter (c) that appears in the Maxwell equations for the EM wave:

$\frac{1}{c^2} \frac{\partial ^2 \vec{X} }{\partial t^2} = \nabla^2 \vec{X} ~~,~~ X= E ~\text{or}~ B$

will remain the same after you do a lorentz transformation.

I don't know that's the first/fast thing that came in my mind.

20. Jan 4, 2015

### alva

"No. That method would add other factors, like the propagation times of the signals in the cables" You can use two cables of the same length.
"The two way measurement requires just one clock". The scope has just one clock inside.

And I dont want to use mirrors.

21. Jan 5, 2015

### Staff: Mentor

Then you just have a more complicated way of doing a two-way measurement - you're measuring the time for a light signal to travel outbound to the photodetectors plus the time for the electrical signal from the detector to return through the wires of equal length.

You also have to know what the return time is, because (unlike the version with a mirror) you cannot assume that the inbound and outbound speeds are the same. In principle this calibration can be done; in practice the errors introduced here will render the experiment impractical.

More generally, any time that we have one clock the experiment must be either invalid or a two-way measurement because the light signal leaves the clock at time zero and we have to some way of getting some signal back to our clock or we don't have any time interval to calculate with.

22. Jan 5, 2015

### ghwellsjr

If the scope has just one clock inside, then it can only make a two way measurement.

Let's assume that your cables propagate their signals at the speed of light and that they are both exactly one-half meter in length. Let's put the scope midway between the two photodetectors which are one meter apart. Now a burst of light arrives at the first detector and generates a signal which propagates down the length of the cable right next to the burst of light which continues to travel along the outside of the cable at the same speed. They both arrive at the midpoint at the same time where the cable signal triggers the scope but the burst of light continues another half meter to the second photodetector which generates a second signal which travels back to the scope at the speed of light to generate the second signal.

Now you could have moved the first photo detector and connect it right at the first input of the scope and not used any cable length at all, it would have been the same thing since the light outside the cable travels at the same speed as the signal inside the cable.

So what have you actually measured? You have measured the length of time it takes for light to travel one half meter from the scope to the second photo detector plus the length of time it takes for the signal to travel one-half meter back to the scope. You have measured the roundtrip time it takes for light to travel one half meter in one direction plus the time it takes for light (or a signal) to travel one-half meter back and you're calling that the one way speed of light to traverse in one direction a full meter.

I've tried to make this simple bu if you want to able to change the cable lengths and move the scope to another position it will just complicate things but the analysis will end up being a roundtrip measurement that you are calling a one way measurement.

Last edited: Jan 5, 2015
23. Jan 5, 2015

### Khashishi

Maxwell's equations don't, by themselves, suggest that the speed of light is observer independent. What they do suggest is that there is a universal speed of light, c, but it wasn't clear right away that this applied to all observers (and not just one special observer in the absolute rest frame of the universe). Some historical context is useful. At the time, most people were working in the assumptions of Gallilean relativity, so this was a bit of a surprise, since the speed of light does not depend on the speed of the source according to Maxwell. This seemed to support a luminiferous aether theory and eventually led to Michelson and Morley's experiment to measure our motion through the aether. This experiment's null result is what strongly suggested that the speed of light was the same for all observers.

24. Jan 5, 2015

### alva

"Then you just have a more complicated way of doing a two-way measurement" OK, it is a two-way measurement. But part of the way happens inside my cables, that are at rest in my laboratory. And if I connect my two cables to the same detector I will see that both signals in the scope arrive at the same moment.
" you cannot assume that the inbound and outbound speeds are the same" I do not understand what you mean by inbound and outbound speeds.
" in practice the errors introduced here will render the experiment impractical" Have you ever used a scope ? The propagation time of the cables (probes) can be much bigger than the time delay you are measuring.

Last edited: Jan 5, 2015
25. Jan 5, 2015

### alva

Yes, it is a roundtrip measurement. Anyway ¿could you measure the speed of light from any source with this set-up?