# Speed of light

1. Aug 19, 2005

### LedZep_Kamal

I am sooo intrigued. The speed of light is independent to the speed of the observer. Why?????? I dont understand and is it the speed of light only that is independent to the speed of the observer? What makes it so special?

2. Aug 19, 2005

Its because when you begin to travel close to the speed of light, things begin to change: length contracts, mass increases and time dilates. These all contibute to you (the observer in the inertial reference frame) seeing the speed of light being constant no matter how fast you go. You gotta love special relativity.

Regards,

3. Aug 19, 2005

### Tide

The constancy of the speed of light (in vacuum) is a deduction based on observation. Because no one had ever been able to observe (and still cannot observe) any variation of the vacuum speed of light in any situation with the source and observer moving relative to each other, Einstein adopted it as a postulate in the formulation of his Theory of Relativity. He also adopted the principle that, written in the right way, the laws of physics are independent of the relative motion of observers.

4. Aug 19, 2005

### rbj

i think there are theoretical reasons for why the speed of E&M propagation (of which light is) should be the same for all inertial reference frames. it really just comes from Maxwell's Eqs. and the knowledge that there is no ether medium that E&M is propagated in.

for sound, this is wave propagation within air by minute compressions and rarefractions of the gas and we can tell if the medium, air, is moving past us or not. and if it is moving past us, we can observe a different speed of sound in the direction of the wind compared to the opposite direction.

but how do we tell the difference between a moving vacuum and a stationary vacuum? if we can't, if there really is no difference between a moving vacuum and a stationary vacuum, that such a concept is really meaningless, then whether the light that you are measuring originated from a flashlight mounted on a rocket moving past you at $c/2$ or from a stationary flashlight, how does that change the fact that a changing E field is causing a changing B field which is causing a changing E field, etc.? that propagation of an E field and B field disturbance, which has velocity $$c = 1 / \sqrt{ \epsilon_0 \mu_0 }$$?

how is it different? whether you are holding the flashlight or moving past it at high velocity, Maxwell's Eqs. say the same thing regarding the nature of E&M in the vacuum.

so then, there is no reason to expect a different observed speed of light for the different observers (the one on the rocket and the other who is "stationary").

does that get to your question or did i misunderstand it?

5. Aug 19, 2005

### rbj

Nenad, i think, at least pedagogically, that you have cause and effect mixed up. Because the speed of light (the very same beam of light) is the same for both observers that are moving relative to each other, this has the effect that both observers observe the other's clock to be ticking more slowly, both observe length contraction of the other, etc.

6. Aug 19, 2005

### quasar987

Also, one of the good physics teachers I had once said to me that the constancy of the speed of light can be derived from the postulate alone that the physical laws take the same form in every inertial frame. Is this true? If so, does the proof of that invokes Noether's theorem?

7. Aug 19, 2005

True, if you want to get technical.

Regards,

8. Aug 20, 2005

### HungryChemist

I am very interested to know if one can derive the speed of light being constant from the statement that the physical laws take the same form in every inertial frame. So far, I have learned when formulating special theory of relativity, we need the both postulates. By the way, I was wondering if someone can elaborate what we mean by 'physical laws take the same form'??

9. Aug 20, 2005

### robphy

It can be shown that the wave equation (say, obtained from the Maxwell Equations) is not invariant under the Galilean Transformations, although it is invariant under the Lorentz Transformations.

10. Aug 20, 2005

### quasar987

Nonsense was here.

Last edited: Aug 20, 2005
11. Aug 21, 2005

### learningphysics

'Physical laws take the same form' mean that if we perform an experiment in one frame... and perform exactly the same experiment in another frame (with the same equipment... everything is identical except which frame we are in) ... the results observed will be identical in both cases.

I'd just say the physical laws are the same instead of saying same form as that implies the form is the same yet something else is different.

I like the way Rindler derives the constancy of the speed of light in "Introduction to Special Relativity"... he takes the relativity principle:
"The laws of physics are identical in all inertial frames, or equivalently, the outcome of any physical experiment is the same when performed with identical initial conditions relative to any inertial frame."

and this axiom:
"There exists an inertial frame in which light signals in vacuum always travel rectilinearly at constant speed c, in all directions independently of the motion of the source."

And from these two axioms, he deduces:
"Light signals in vacuum are propagated rectilinearly, with the same speed c, at all times, in all directions, in all inertial frames."

(since we have one frame where the speed of light is constant at c... the principle of relativity dictates that in every frame the speed of light is constant at c... since all frames require equivalent outcomes of the same experiment)

12. Aug 21, 2005

### Perspicacious

The simple, primary, natural law responsible for this characteristic of spacetime is the homogeneity of time.

http://www.everythingimportant.org/relativity/
http://scitation.aip.org/getabs/servlet/GetabsServlet?prog=normal&id=AJPIAS000043000005000434000001 [Broken]
http://arxiv.org/PS_cache/physics/pdf/0302/0302045.pdf [Broken]

Last edited by a moderator: May 2, 2017
13. Aug 21, 2005

### robphy

These approaches essentially discuss a "maximum signal speed" of a theory of relativity, which is infinite for the Galilean case but finite for the Einsteinian case. So, along these lines, the natural question to ask is why is the maximum signal speed finite, and why is that speed that of light? Certainly, one can appeal to experimental observation. However, I'm not sure if the original poster would be satisfied with that.

Last edited by a moderator: May 2, 2017
14. Aug 21, 2005

### Perspicacious

None of those papers presuppose a "maximum signal speed." Relativity is a consequence of the homogeneity of time.

It's an irrelevant assumption. There are theories where light is thought of as not moving at the maximum possible speed.

If the original poster is dissatisfied with the limited knowledge of physicists, then he might want to entertain a short theological response: God selected the spacetime structure constant to be what it is for no unavoidable special reason. The assigned present value of c is just as workable as an infinite number of other possibilities.

15. Aug 21, 2005

### Perspicacious

Noether's theorem is not required. A maximum possible speed is an easy consequence of the relativity postulate.

There is no need for both postulates. The first postulate is sufficient and we can get by with an even weaker axiom. The answers to your questions are all contained in the first, second or third link that I offered.

16. Aug 21, 2005

### robphy

I never said that such a thing was presupposed.
As you say in your response to quasar987, "A maximum possible speed is an easy consequence of the relativity postulate."

As you are probably aware, as one attempts to boost an observer's 4-velocity, one finds an upper bound, which is infinite for the Galilean case and finite for the Einsteinian case. These corresponding speeds coincide with the eigenvectors of the Galilean and Lorentz boosts, respectively.

17. Aug 23, 2005

### dalamar96

Along these lines for both the question posted before and the fact that this is the first place I saw the speed of light discussed. I would like to post the following. Could anyone tell me if I am off on my thoughts here? Bear in mind that I did not keep the significant digits throughout (obviously)

If we are moving away from a star at .5c and are at a distance of 1lyr when it goes nova, what will the distance between us and the nova when the light reaches us?

.5c = 539,626,424.4kph
time to reach us = t = 1lyr
1 year = 8,765.81277075hr

.5c * 8,765.81277075 = 4,730,264,202,440km = distance added since light left source

4,730,264,202,440km + 9,460,528,404,880km = 14,190,792,607,300km = 1.5lyr
1.5lyr = .000000459892177752mpc

1.5lyr * hubble’s constant = .0000326523446204km/s

1yr = 31556925.9747s

.0000326523446204km/s * 21556925.9747s = 1,030.40762209km

1.5lyr + 1,030.40762209km = 14,190,792,608,330km

14,190,792,608,330km = 1.50000000011lyr

The nova would be 14,190,792,608,330km from us when the light reached us.

18. Aug 24, 2005

### pervect

Staff Emeritus
I don't think that Hubble's constant belongs in the answer. Before the question can be answered, we need to know - is the ship 1 light year away from the star in the star's frame, or is the star 1 light year away from the ship in the ship's frame?

A similar remark applies to the distance when the nova light reaches the ship, it wil depend on the frame used. Here, though, there is a reasonably good reason to prefer the star frame - because the explosion will appear to be symmetrical only in that frame. Thus if we want to calculate how much radiation the ship will receive, that's the logical frame to use.

I will now proceed to make the assumption that all distances are being measured in the star frame, because that's the frame in which the explosion will appear to be symmetrical. Then we can write the following equations

x_ship = 1 + .5*c*t

This gives the position of the ship at time 't' where time 't' is defined in the star's frame of reference.

x_light = c*t

This gives the position of the nova light at time 't'.

Setting them equal, we solve for the time t when x_ship = x_light and we get

1 + .5*c*t = c*t

Solving this, we find that t=2, and that x_ship = x_light = 2. Therfore the ship will be 2 light years away from the star when the nova light reaches it

19. Aug 24, 2005

### dalamar96

I was going more on the frame of reference of earth moving away from another galaxy at .5c and being 1lyr away from the galaxy when the nova occurred.