# Basic concept in special relativity

1. Dec 1, 2005

### Theoretician

I was asked (on another non physics site) how an observer can measure the speed of light to be the same as someone who is moving away/ towards the light source. I know that it is a fairly simple idea but I couldn't find the words to answer comprehensively. Here is what I wrote:

In physics, there is a wave equation. This describes how a disturbance propogates through space in time. The Maxwell equations give a form that looks like the wave equation (with the disturbance as the electric field and the magnetic field) that predicts that light always travels as 1 divided by the square root of the product of two fundamental physical constants (this is the speed of light).

There is no reason that the equations should be different for a man travelling at some speed. Hence, the equations suggest that light will always be measured to travel at about 300 000 000 m/s (or c) regardless of the speed of the observer.

As you have obviously considered [hence, the question], this is strange. A man moving away from (or toward) a source of light still sees the light moving towards him at c.

The strangeness of the idea can be taken away (and replaced with other strange ideas) when you realize that basic concepts such as absolute space, absolute time and simultaneity are not valid in the Universe in which we live. For example, if you watch a spaceship move away from a light source at close to the speed of light, from your perspective, the spaceship will shorten in the direction of travel and any processes taking place inside will appear to run in slow motion. In this example, you would 'see' the light approach the spaceship at the speed c-v and (naively) infer that the observers in the ship should see the light coming towards them at this speed also. The reason that they don't and that they actually see the light come towards them at c has to do with the fact that they are in a different timeframe and on a different lengthscale.

I haven't actually studied special relativity in awhile and you can tell that I am not sure of myself by the length of the text. Is my explanation basically correct (once again, this is not a physics site, just the general chat forum of a football site)? Have I missed a more succinct way of putting it? Any comments would be greatly appreciated.

Last edited: Dec 1, 2005
2. Dec 1, 2005

### El Hombre Invisible

I wouldn't say simultaneity is invalid, it is just relative - it depends on your reference frame. Two events may still be described as being simultaneous, there is just now the caveat that they occur simultaneous in certain frames only.

I know what you're saying, but your wording is misleading. You would 'see' the light approach the spaceship at the speed c, not c - v. You would see the rate of change of the distance between the front of the ray of light and the ship as c - v, and so (not knowing SR) would infer the ship would see the light approaching at this speed.

Again, I can see what you mean, but it is referred to as an 'inertial reference frame' in which a clock running at normal speed in that frame would appear to run slow in your inertial frame.

Perhaps a shorter way of putting it is this:

For the laws of physics to hold true in all inertial frames, two observers travelling at different velocities along an axis must observe the same ray of light propogating along that axis cover the same distance in equal time intervals. Since the two observers will disagree on the length of this distance in their frame because one is moving relative to the other, they must also disagree on how much time has passed.

3. Dec 1, 2005

### JesseM

This answer doesn't mention the different definitions of simultaneity that the different observers have, which is crucial in understanding how they can both measure the light to have the same velocity. Time dilation and length contraction alone will not explain it.

Here's something I wrote on another thread about the relativity of simultaneity, and how it relates to the fact that each observer assumes light travels at the same speed when synchronizing his own clocks:
Now consider what happens if observers in motion relative to each other use this assumption to synchronize their own clocks. If I'm on a rocket which is moving forward in your frame, I can synchronize my own clocks at either end of the rocket by setting off a flash at the midpoint of the rocket, and setting both clocks to read the same time at the moment the light from the flash reaches them. But in your frame, since the rocket is moving forward, the clock at the front of the rocket will be travelling away from the point where the flash happened, while the clock at the back will be travelling towards that point--so if you also make the assumption that each light beam must travel at the same speed in your own frame, then in your frame the light must catch up with the back clock before it catches up with the front clock, and therefore you will conclude my clocks are out-of-sync. Without this difference in our definitions of synchronization, there's no way we could both conclude that the light was travelling at the same speed in our own frames.

4. Dec 8, 2005

5. Dec 9, 2005

### Theoretician

Thank you for all replies.

I will have to work on my communication skills regarding science.

I often feel, when I return to special relativity having not considered it for awhile, that I have to partially learn the subject all over again (that, in a sense, being part of the reason that I was having trouble explaining some of the ideas clearly).

Last edited: Dec 9, 2005
6. Dec 12, 2005

### rbj

there was another thread about why the speed of light (or any EM radiation) would be the same for all observers. here was my response:

i think we can have a feel for why the speed of E&M propagation 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. so 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 (relative to you) 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 $1/ \sqrt{ \epsilon_0 \mu_0 }$? how is it different for an observer traveling with the flashlight or the one that the flashlight is moving past?

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. that is qualitatively different than waves that require a substantive medium to travel in, e.g. sound. if the wind is moving past you from left to right at 10 m/s, a measurement of the speed of sound coming from a source on your left will be 20 m/s faster than a sound emanating from a source on your right. but there is no (ether) wind for electromagneting radiation. so once the E&M wave disturbance has left its source, does it have any contact with it anymore? can it "remember" that it came from a high speed moving flashlight (relative to some observer) or a stationary flashlight? it can't. and, if there is no "wind" to carry E&M, what would be left for it to know it should move any faster, just because its source it came from was moving?

7. Dec 12, 2005

### Theoretician

Nicely put rbj. At first, I thought that it doesn't cover the situation where the light source is stationary and the observer is moving but, from the observers point of view, the source is moving not him and, so, to him, your argument should still apply. If it didn't, the laws of physics would have changed and we don't live in a Universe where the laws of physics change with local speed apparently.