# Homework Help: Invariance of c puzzle

1. May 6, 2010

### stever

My friend is moving in my inertial frame and receives light in the two directions parallel to her movement: from behind and from ahead. The photons that reach her travel at c in my frame, so I presume they will approach my friend at different speeds (as I view it): faster from ahead and slower from behind. But we know from SR that my friend would measure all photons to be moving at speed c. Explain (away) the seeming contradiction so that the photons coming from behind and in front both measure at c on board her frame.

2. May 7, 2010

### D H

Staff Emeritus
Thread moved to the Homework Section

Why do you think there's a contradiction? What do you think the answer is?

3. May 7, 2010

### stever

Are you saying you cannot see a contradiction?

To this puzzling question I have no answer. I have been looking for the answer for 25 years and have yet to see it?

I would like to know where the people are who can answer this question.

4. May 7, 2010

### D H

Staff Emeritus
It only seems contradictory because you have certain expectations based on living in a slow-moving world regarding how velocities add. You are assuming that the same rules pertain at very high velocities. They don't.

For example, suppose a police officer driving 80 mph detects someone via radar going 30 mph faster than the police car. We assume that this can only mean that the speedster was going 110 mph. That isn't exactly correct. The speeding car is going 109.99999999999941 mph relative to an observer who is standing still on the side of the road. No radar can measure that tiny, tiny difference of 5.9×10-13 mph. As far as we are concerned, the speeding car is going 110. In the example above, if you replace the cars with spaceships and change the speeds from 80 mph and 30 mph to 0.8c and 0.3c, the combined speed 0.887c rather than 1.1c.

The rules that we thought were the rules of the universe are just an approximation of the true rules, and an approximation that is valid at very small speeds. What is a rule of the universe (as far as we know) is that the speed of light is the same to all inertial observers. This is a very well-observed fact.

5. May 7, 2010

### vela

Staff Emeritus
At everyday, low speeds, time and space seem to be independent. Time for me appears to be the same as time for you; similarly, we would agree on the spatial distance between two points. It turns out, though, that time and space aren't actually independent. Moving clocks run more slowly, and moving objects shrink in size. Those effects, however, only become apparent as speeds approach that of light.

Special relativity tells us that one observer's space and time dimensions are a mixture of another's space and time dimensions and that how they mix depends on their relative motion. This mixing occurs in exactly the right way so that both observers would see that a light ray moves at speed c.

6. Jun 30, 2010

### stever

These responses don't really answer the question do they?

7. Jun 30, 2010

### Hurkyl

Staff Emeritus
Here is the problem:

8. Jun 30, 2010

### DaveC426913

Well, these answers don't teach Special Relativity. They answer the question if you know SR already. You can;t learn that from a forum thread.

The simplest answer is that, at the very essence of relativity is the relativity of simultaneity. Simply put: two observers, travelling in different FoRs will have different ideas of simultaneous. One will see two events occurring simultaneously, the other will see them occurring separately.

Classic thought experiments about this involve a train moving at relativistic velocities with two lights at either end of a passenger car. Both lights are switched on "simultaneously". Depending on who you ask (guy on train versus guy on platform), the lights may or may not have turned on simultaneously.

There are literally countless examples of this. Google "relativity simultaneity train" or any other combo and you'll get more hits than you know what to do with.

Last edited: Jun 30, 2010
9. Jun 30, 2010

### stever

Dave, are you saying the phenomenon is just a measurement problem because of the time it takes light to reach different observers??

Last edited: Jun 30, 2010
10. Jun 30, 2010

### stever

Hurkyl, try to explain why light arrives at a moving object at speed c.

11. Jun 30, 2010

### DaveC426913

Prove what? A postulate of relativity??

Do you need an explanation more comprehensive than any book on relativity basics would deliver? Not sure if an answer needs to be recut from whole cloth when it's readily available.

Last edited: Jun 30, 2010
12. Jun 30, 2010

### DaveC426913

OK, before this gets much farther:

stever: do you think that relativity is wrong, or do you think perhaps you just don't understand it?

i.e. are you looking for enlightenment, or are you looking for an argument?

13. Jun 30, 2010

### stever

Dave, are you saying the speed of light is constant in different frames because of the different times it takes light to reach different observers?

14. Jun 30, 2010

### Hurkyl

Staff Emeritus
Why not first try to explain why light wouldn't "arrive at a moving object at speed c".

Actually, you should first clarify what you mean by that phrase. The most important aspect is what you mean by "arrives at speed c".

15. Jun 30, 2010

### DaveC426913

No. The two spaceships could pass millimetres from each other (negating time lag) and it would not change the results.

I'm saying simultaneity is not absolute. There is no preferred frame in the universe in which one person is right and the other wrong.

16. Jun 30, 2010

### stever

What I don't understand is the answer to the question I originally posed. I have not seen an answer to it anywhere, including SR101. If you have an idea about the answer I would like to hear it.

17. Jun 30, 2010

### stever

"No. The two spaceships could pass millimetres from each other (negating time lag) and it would not change the results."

Yes. I agree. So what then is happening?

18. Jun 30, 2010

### Hurkyl

Staff Emeritus
We cannot explain away the "seeming contradiction" without first understanding why it seems contradictory to you. That requires you to explain why you think it's contradictory.

There's also may be a Socratic thing going on here, depending on your responses. If you explain your own reasoning in greater detail, you might spot the gaps/errors yourself.

19. Jun 30, 2010

### Hurkyl

Staff Emeritus
To illustrate by example, the thing following your words "I presume" could mean a couple different things. The thing I originally thought you meant was in error. However, I've thought of another thing you might have meant which is correct and reasonable, but makes a later statement in error (and when explained it makes the later error easier to spot) (assuming I've correctly interpreted everything else you've said and have left unsaid).

20. Jun 30, 2010

### stever

I don't see how the light arrives at c everywhere in space regardless of movement of the emitter or receiver. How does the light adjust its speed or how does the measurer make a measurement such that the speed is always c?

21. Jun 30, 2010

### Hurkyl

Staff Emeritus
It doesn't. Its (coordinate) speed is always c in an inertial reference frame. No adjustments necessary.

They don't do anything special; they just measure (coordinate) displacement over (coordinate) time, like you would expect them to.

22. Jun 30, 2010

### DaveC426913

How much reading have you done?

I think we're getting the impression that you expect us to do all the work - essentially teach you SR.

23. Jun 30, 2010

### zzzoak

Copy from http://en.wikipedia.org/wiki/Special_relativity" [Broken]
If the observer in S sees an object moving along the x axis at velocity w, then the observer in the S' system, a frame of reference moving at velocity v in the x direction with respect to S, will see the object moving with velocity w' where
$$w'=\frac{w-v}{1-\frac{wv}{c^2}}$$

So you see w=+c. The light is going to +x direction and reaches your friend from behind.
Your friend sees this light with speed
$$w'=\frac{c-v}{1-\frac{cv}{c^2}}=\frac{c-v}{\frac{c-v}{c}}=c$$

You see another light w=-c is going to -x direction and reaches your friend from front.
Your friend sees this light with speed
$$w'=\frac{-c-v}{1+\frac{cv}{c^2}}=\frac{-c-v}{\frac{c+v}{c}}=-c$$

So you see the light going to your friend with speed +c and -c.
Your friend sees this light with the same speed +c and -c.
This doesnt depend on the speed v your friend is moving with.

Last edited by a moderator: May 4, 2017
24. Jul 1, 2010

### stever

Yes I know. It is a given from the experimental data. What we are trying to explain, or at least I am, is why it is at c.

If two observers are moving relative to each other, or in relation to a light source, something is happening such that they both can measure light at c. Yes it happens. But it does not just happen. Even death and taxes can be explained to a certain degree, but light...?

Last edited: Jul 1, 2010
25. Jul 1, 2010

### stever

I am impressed by how neatly the math works out. Is there something hidden in the math that explains to you how the light can go at c, in space, for all observers?

Last edited by a moderator: May 4, 2017
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