# Speed of Light Always The Same?

1. May 12, 2009

### antd

I know this has been beaten to death in this forum and other places -> The problem is that I have not come across an explanation which I can understand!!

I understand the most basic things of time dilation. And it makes sense to me.

However, I cannot grasp the concept of how light always travels at roughly the same speed in a vacuum.

If a spaceship travelling at 99% of the speed of light is racing after a photon of light, how is it that the photon is moving away from the spaceship at 100% of the speed of light...?
Shouldn't the spaceship be only a small distance from the photon which it is chasing?

It seems though that this is wrong. And that the photon should be moving away from the spaceship at c... meaning the photon would be extremely far away from the spaceship?!

I have read many explanations, but none do it for me >.<
Can someone explain this in a very simple manner?

excuse my stupidity on the matter..

2. May 12, 2009

### HallsofIvy

Staff Emeritus
Not "roughly" the same speed- exactly the same speed!

No, the photon moves away from the space ship at 100% the speed of light. That would be true in clasical physics. In classical physics, a stationary person would say that the photon is moving, relative to him, at 199% the speed of light because he would use the "speed addition formula" u+ v. But according to relativity (and verified by experiment) for very high speeds the correct formula is
$$\frac{u+ v}{1+ \frac{uv}{c^2}}$$
In your example u= c. That formula gives
$$\frac{v+ c}{1+ \frac{vc}{c^2}}= \frac{v+c} {\frac{c+v}{c}}= c$$

That works out that way for any speed, not just .99c. That's why the speed of light, relative to anyone, no matter how fast, the spaceship that originates the light is moving relative to them, is c.

Last edited: May 14, 2009
3. May 12, 2009

### antd

Hmm sorry, I don't understand the mathematics.

I accept what you told me is true :) But it doesn't help me explaining how it is true... >.<

For example, I don't know why the spaceship would see the photon far away from them

But someone not moving and watching would see the photon close to the spaceship?

Or the other way around? hmm... I don't understand. Surely the photon is either close or far away from the spaceship... but not both?

What if we froze time and the spectator walks up to the photon... where would it be then?! close to the spaceship or far away?

(if you have a problem with freezing time, let's say we have a simulation of the universe in a computer. And we froze that instead.) :p

4. May 12, 2009

### JesseM

Different observers measure the speed of anything using rulers and clocks which are at rest relative to themselves, and synchronized in their own frame using the Einstein clock synchronization convention. But the rulers of different observers don't measure distance the same way (each observer measures rulers that are moving relative to themselves to be shrunk by a factor of sqrt(1 - v^2/c^2) relative to their own ruler/clock system, the effect known as length contraction), the clocks of different observers don't measure time the same way (each observer measures clocks that are moving relative to themselves to have the time between ticks expanded by a factor of 1/sqrt(1 - v^2/c^2), the effect known as time dilation), and clocks that are synchronized in their own frame will be measured to be out-of-sync in other frames (clocks that are synchronized and have a distance x between them in their own rest frame will be out-of-sync by vx/c^2 in another frame where they are moving at speed v, due to what's called the relativity of simultaneity). So, it's because of all these differences that two observers can each think light is moving at c relative to themselves.

For a numerical example of how all these effects come together to ensure both observers measure a single light ray to move at c, see my post #6 on this thread.

5. May 13, 2009

### granpa

you are confused for the same reason that all newbies get confused. 'relativity of simultaneity'. once you get the idea its not as hard as it sounds.

Last edited: May 13, 2009
6. May 13, 2009

### neopolitan

This wording is going to lead to cos-like discussion about what is "real" and "true". It's not just a case of each of the observers thinking that light is moving at c relative to themselves, light is moving at c relative to themselves.

The scenario raised here is almost precisely the one which got me interested in this area and I am not sure that someone who has approached from the OP's starting point will immediately appreciate the "relativity of simultaneity" approach.

The visualisation which I came up with to explain the phenomenon worked quite well for me, but months of effort have shown it is difficult to explain to others.

cheers,

neopolitan

7. May 13, 2009

### granpa

the best advice I can give you is to learn to draw a spacetime diagram.

8. May 13, 2009

### granpa

9. May 13, 2009

### JesseM

Well, it's only moving at c relative to themselves in a particular choice of coordinate system. Coordinate systems are human inventions, are they not? That's why I said "thinks", although I could equally well have said something like "measures using their own ruler/clock system constructed in the particular way defined by Einstein".
The relativity of simultaneity was only part of my answer, the more intuitive difference is that they measure length differently because of length contraction, and time differently because of time dilation (as well as because of the relativity of simultaneity if they are measuring the one-way speed of light, although you can dispense with simultaneity issues altogether if one of them is measuring the two-way speed of light and therefore needs only one clock), and each one defines "speed" in terms of their own distance/time measurements.

10. May 14, 2009

### antd

Hmm, but it seems as if the same photon is in two or more places at once... depending on how many observers there are?!

11. May 14, 2009

### sylas

Not at two places at once... rather that for any fixed event (place and instant) the distances and times to other events will depend on the observer.

It's not that the photon is in two places at once, but that the single place-and-instant is at many different distances at once, depending on how you look at it.

12. May 14, 2009

### Fredrik

Staff Emeritus
I agree with these comments, but I would like to add that even though it's not as hard as it sounds (mathematically), it's even more counterintuitive than it sounds. (When you hear about it the first time, you aren't even aware of some of the more bizarre consequences of the invariance of the speed of light). That's why spacetime diagrams are so useful. The only way to develop some sort of intuition for these things is to starting thinking in terms of spacetime diagrams.

13. May 14, 2009

### Fredrik

Staff Emeritus
A photon is the worst example you can pick, because photons aren't really localized in space. (That's a very difficult quantum mechanical result that I don't fully understand myself). The detection of a photon is a better example, because that's an event. It's something that happens at a specific time at a specific place. The set of all events is called "spacetime". All observers agree about what events are a part of spacetime. They just disagree about what coordinates to assign to the events. Each observer uses his/her own coordinate system, which is just a function that assigns four numbers to each event.

14. May 14, 2009

### neopolitan

Can you demonstrate a coordinate system in which c is not moving at c relative to an observer? (Stress is on the relative and note I write c, not the figure which is a consequence of the standard metre and the standard second, so c in terms of fathoms per fortnight or furlongs per full moon is still c.)

cheers,

neopolitan

15. May 14, 2009

### neopolitan

The OP specifically had a photon based scenario.

cheers,

neopolitan

16. May 14, 2009

### DaveC426913

I don't know - to me it's far simpler than everyone else is trying to say.

The observer on Earth watches a photon leave the spaceship and travel 186,000 miles in one second. The spacehip in that time has moved 99% of that distance, and so has moved 184.2 thousand miles. (The photon is only 1.86 miles ahead of the spaceship)

The observer in the spaceship is time dilated by, say a factor of 10*. i.e. compared to his Earth counterpart, he's moving ten times as slow.

So, the photon is now 1.8 miles in front of his ship - but only 0.01 seconds has passed in his spaceship. He measures the photon's speed as 1.8miles per 0.01 seconds or 186,000mph.

(Alternately, he waits for his appropriate one second, then measures the photon to be 186,000 miles beyond his ship. However, on Earth, ten seconds has passed, so they see the photon 1,860,000 miles in front of the ship.)

*numbers in this example are picked for their clarity of explanation, not for their accuracy

17. May 14, 2009

### daniel350

I need help with this as well; I'm trying to get my head around it and its so much... fun and frustration at the same time, fun because its so interesting, frustrating because I don't understand it as much as I wish too.

Here is my similar problem:

A spacecraft is moving at 80% the speed of light (0.8c) from planet X, a radio station sends a light beam parallel to the spacecraft.
What would the observer on planet X and the spacecraft pilot see in regards to 'watching' this light beam move towards the spacecraft.

In both situations light would obviously move at 1c, but how does this work for the spacecraft and observer... damn Counter-Intuitive addition of velocities... where does Time dilation and the Length Contraction come into effect, and how.

regards

18. May 14, 2009

### granpa

due to relativity of simultaneity he will measure distances differently too. not just time (the length of an object is the distance between the front and back at one simultaneous moment)

this is why learning to draw spacetime diagrams is so important

19. May 14, 2009

### Fredrik

Staff Emeritus
I guess I'm just nitpicking the word "photon". The OP's question is fine if we interpret "photon" as "massless classical particle", which is probably what he had in mind anyway. It's a convenient way to think about light in SR. (The word "photon" is actually defined by a quantum field theory which says that photons aren't localized).

20. May 14, 2009

### JesseM

Do you mean a coordinate system in which light is not moving at c (or even at a constant speed) relative to the observer? If so you could pick something like Rindler coordinates which are commonly used for an observer experiencing constant proper acceleration (also discussed here). But really you have total freedom in defining non-inertial coordinate systems any way you like, for example if x,y,z,t represent the coordinates of an inertial coordinate system I could define a non-inertial coordinate system in terms of any transformation I can think of, a simple one would be:

x' = x - at^2
y' = y
z' = z
t' = t

But you could also do a crazy one like:

x' = x * sin(t/t0)
y' = y - t*c
z' = pi*z - 1008
t' = t + 3x/c + 7y/c + 0.5z/c