# B Can I get clarification on the constant speed of light

1. Jul 13, 2016

### Quandry

I do not have a problem with the concept of the constant speed of light as it has no mass and therefore no inertia and therefore no relationship to any IFR. However it seems to be expressed as constant in all IFR's which I do not understand. This seems to say that if I am traveling at 1/2c and I shine a torch forward the light moved away from me at c and if I shine the light backwards it also travels away from me at c.
This seems to say that, in the first case, the light is travelling away from the point in space that it is created (independent of my IFR), at 1.5c.
To clarify this for me, my question is:
If I am travelling towards a light source at .5c and I have two light detectors with a set space between them to measure the time that the light takes to transition from one to the other, will the time taken correspond to a speed of c? And, if I decrease my speed to .25c and do the same measurement will the measurement also indicate a speed of c?

2. Jul 13, 2016

### A.T.

No, it's traveling at c in all frames.

3. Jul 13, 2016

### Orodruin

Staff Emeritus
This is an experimental fact and so is not something you should set out to understand using your regular Galilei transformation. In fact, as you have discovered, it directly violates Galilei addition of velocities. The conclusion you should draw from this is that Galilei addition of velocities does not work in this extreme. Other conclusions such as the relativity of simultaneity follow in a relatively straight forward fashion.

4. Jul 13, 2016

### Quandry

So you are saying that regardless of my speed, the measurement between to two detectors is always the same?

5. Jul 13, 2016

### Ibix

The point in space is a frame-dependent concept. Say your flashlights are on a table on a train. To someone on the train, the point in space where the light was emitted is a point just above the table. To someone standing beside the track, the point in space where the light was emitted is some point above the track where the flashlights happened to be when they emitted.

The emission event is frame-independent, but you can't measure speed with respect to that.

6. Jul 13, 2016

### Quandry

I do not set out to understand using Galilei transformation. In fact if you refer to my post I am clear that this does not happen (although it seems a logical, but incorrect, conclusion to the speed of light being a constant in all frames of reference, which by logic says that it is different between two frames of reference). But if I am approaching a source of light which is travelling at c and I am travelling at .5c the relationship between my detector and the source of light is changing at 1.5c, even although the light is travelling at c. There are many incidences where relationships change faster than the speed of light - e.g. phase relationships between travelling waves can move down a waveguide faster than the speed of light.
There is no implication of adding velocities in this process.

7. Jul 13, 2016

### Quandry

It is not the point in space that is frame dependent. It is the observers definition of that point in space relative to other factors.

Which I guess is what I am saying - but if you are in the same frame of reference as the photons you can measure speed - but since you are not.....

8. Jul 13, 2016

### A.T.

How did you calculate that 1.5c then?

9. Jul 13, 2016

### Ibix

How do you define a point without reference to other factors?

An event is a point in spacetime. It doesn't have a velocity for you to compare an object's velocity to.

Light doesn't have a reference frame for you to be in, so measuring speed if you were in it doesn't make sense.

10. Jul 13, 2016

### Quandry

I am saying that that is the implication of the light travelling away from me at c, when it is travelling away from the point in space that it was created at c. These two things are incompatible and would require Galilei transformation to make it work.

11. Jul 13, 2016

### Quandry

The event happens at a point whether or not you can define it

That is correct, but if the event is defined as the emission of photons you can assume a velocity= c

I guess that's still what I am saying.

12. Jul 13, 2016

### Quandry

Just going back to my question, it seems that y'all are saying the answer is yes?

13. Jul 13, 2016

### Ibix

An event is a point in spacetime. There will come a time when you say that that event is in the past - it is not in the slice of spacetime you call "now". So you cannot measure a spatial distance to the event "now". You can only decide that some point in "now" is the same as the (spatial part of) the event. That choice is frame dependent - in fact, it's one definition of a choice of frame (up to spatial rotation).

This is pure nonsense. An event is a point in spacetime. Velocity is the slope of a line in spacetime. A point does not have a slope and there is no way to define one for it. Simply "assuming" an undefined quantity has a particular value is meaningless.

Then you are talking nonsense. There is no frame of reference for light - it's a contradiction in terms.

Last edited: Jul 13, 2016
14. Jul 13, 2016

### Ibix

Everyone will always measure a time between the reception events consistent with the motion of the detectors (as measured in their frame) and the constant speed of light. Between length contraction, time dilation the relativity of simultaneity, and the different motion of the detectors in any other frame, everyone will always be able to come up with a consistent explanation for why everyone else also comes up with the same invariant speed of light.

15. Jul 13, 2016

### A.T.

It's a numerical value. How did you calculate it exactly?

16. Jul 13, 2016

### Quandry

Your response is not relevant. If the event is the result of turning on a torch (as defined) it is reasonable to assume that the event results in photon emission.

I m sorry that you think this is nonsense. I was agreeing with you.

'nuff said!

17. Jul 13, 2016

### Quandry

Exactly? I added 1 and .5 and I got 1.5.

Again, 'nuff said.

18. Jul 13, 2016

### Mister T

That's Galilean relativity! It works only as an approximation in the limit of low speeds.

The correct way is $\frac{1+0.5}{1+(1)(0.5)}=\frac{1.5}{1.5}=1$.

19. Jul 14, 2016

### Orodruin

Staff Emeritus
In addition to what has been said already, one thing that confuses many people is the difference between separation speed and relative speed.

If you are moving to the left with 0.5c relative to an observer A and a light signal travels to the right, then A will see the distance between you and the light growing with 1.5c. This is separation speed.

The above does not mean that you will see the light moving at 1.5c! To draw that conclusion you must assume absolute space and time, which directly contradict SR. In fact, they are basic assumptions behind the Galilei transformation! The basic assumption in SR is that the light will have speed c in your inertial frame too. This is the relative speed.

Also note that there is no way you can objectively say "I am moving at 0.5c" as velocities are relative and change between inertial frames. You need to specify relative to what you move at 0.5c.

20. Jul 14, 2016

### Ibix

In that case your writing is imprecise, because I'm having trouble reading you as agreeing with me even when you say you are. I think I shall duck out of this conversation for now.

21. Jul 14, 2016

### A.T.

Hmm...

22. Jul 14, 2016

### Quandry

Thanks for the term separation speed. It clarifies the confusion caused by my use of relative speed to mean the same thing.
Can I therefore conclude that, if I am the observer, I see observer A moving right at .5 and, since light belongs to no IFR, I see the same light signal moving to the right and a separation speed between them of .5c

23. Jul 14, 2016

### Staff: Mentor

Yes, and there's nothing special about light in this specific case. If you see a spaceship approaching from the left at speed $v$# with $v<c$ and another spaceship approaching from the right with speed $u$ (also less than $c$), you will find that the distance between them is shrinking by $(u+v)<2c$ - that's a separation velocity and it's not the speed of anything in any frame. The right-moving spaceship on the left will consider himself to be at rest while you are moving at speed $v$ and the left-mover is moving at speed $(u+v)/(1+uv/c^2)$ which will always be less than $c$. Likewise, the left-moving ship on the right will consider himself to be at rest while you are moving at speed $u$ and the right-mover on the left is moving at the same speed $w$.

24. Jul 15, 2016

### Stephanus

@Quandry welcome to PF Forum,
It's hard to quote your questions and answer them at the same time.
So, I'll answer in my own format.
You: This seems to say that if I am traveling at 1/2c and I shine a torch forward the light moved away from me at c and if I shine the light backwards it also travels away from me at c.
You: that if I am traveling at 1/2c.
Answer: With respect to whom? You always have to say in SR that, "I'm traveling at some velocity with respect to something"
So let's just say that you're traveling 1/2c wrt Alice.

You: and I shine a torch forward the light moved away from me at c and if I shine the light backwards it also travels away from me at c.
Answer: Yes!, From your ifr you are at rest. From Alice ifr it's you who is traveling at 1/2c. You are always at rest from your ifr!

You: This seems to say that, in the first case, the light is travelling away from the point in space that it is created (independent of my IFR), at 1.5c.
Answer: No! The light travels at 1c.
You: the light is travelling away from the point in space that it is created (independent of my IFR)
This might be a little confusing. This is where the velocity addition formula comes to mind.
$w = \frac{u+v}{1+uv}$
Let's say you (u) travels at 0.5c and fires a missile 0.4c (v), so from Alice (w) ifr the speed of the missile is
$w = \frac{0.5+0.4}{1+0.5*0.4} = \frac{0.9}{1.2} = 0.75c$
So the speed of the missile in Alice ifr is 0.75c
From yours is (of course) 0.4c
Let's say you (u) travels at 0.5c and shines a light at, of course c or a.
$w = \frac{0.5+1}{1+0.5*1} = \frac{1.5}{1.5} = 1c$
From Alice ifr your light travels at c
from yours, it travels at c.

You: To clarify this for me, my question is:
If I am travelling towards a light source at .5c and I have two light detectors with a set space between them to measure the time that the light takes to transition from one to the other, will the time taken correspond to a speed of c?
Answer: a speed of c? Sure, why not. But it's blue-shifted.
You: And, if I decrease my speed to .25c and do the same measurement will the measurement also indicate a speed of c?
Answer: at a speed of c? Yes, But be careful here, how you do the measurement!

Let's have a detailed thought experiment here.
You (YY) are moving 0.5c to the east, and Alice(AA) from your east is at rest. Y2 is your friend far away from you at east and is comoving with you, that is he travels 0.5c to the east also.
Or Alice will say that it's you who are at rest, and she travels to the west at 0.5c
A2 is Alice friend's who is comoving with her. That is, A2 is traveling to the west at 0.5c
Let's say the distance between YY and Y2 is 1000 light seconds or 300 millions km wrt you!
The distance between A2 and AA is 1000 light seconds wrt you!
Wrt Alice A2 and AA is $1000/\sqrt{0.75} = 1154.7 \text{ seconds or } 346.4 \text{ millions km}$
But YY and Y2 wrt Alice is $1000*\sqrt{0.75} = 866.03 \text{ seconds or } 259.8 \text{ millions km}$

Be careful, there will be length contraction and relative simultaneity of event that will come later.
BB is Bob which we will use in later case.
Okay, so here's the picture.

.................BB.....................................B2
.................YY.....................................Y2
..........<------AA.....................................A2

Surely, you will see Alice who is moving toward you at 0.5c to the west and when A2 meets Y2 then A2 shines a light. This light will reach YY 1000 seconds later wrt you, blue shifted. If you don't move 0.25c.
But what if you move 0.25c to the west? There will be two of you here. One who was at rest, and the one who is traveling 0.25c to the west.
So, let's add Bob (BB) who is at rest in your first frame. And now you moves 0.25c to the west wrt Bob. Let's see when YY receives the signal
And let's put B1 at some distance where when YY receives the signal YY will meet B1.
...............<---YY...................................Y2 .......
...................B1....BB.............................B3......B2
....<------AA.....................................A2..............

This can be solved by simple algebra.
$0.25T + 1000 = T, T = 1333$
So when YY receives the signal the clock show will 1333. B1 clock not YY clock!
What does YY clock show?
Time dilation.
$\gamma = \frac{1}{\sqrt{1-0.25^2}} = 1.033$
$1333/1.033 = 1291$
YY clock will show 1291 not 1000 not 1333. And the light still travels at c wrt you (YY).
How come that be?
So the only explanation for this situation is this.
Facts that you and Bob both agree
1. When BB meets YY and B2 meets Y2, all of you synchronize your clock.
2. Then A2 meets Y2 and (of course B2) and A2 sends a signal to the west and at that time YY and Y2 moves to the west at 0.25c
3. Now this is tricky. You change frame.
In your new frame, it seems that the light comes further back from Y2.
I think this is rather advanced, although for some people this is very, very bacic. Perhaps you ask about read time dilation, length contraction and relative simultaneity of events.
And the answer about your two detector should come after that. There's a relative simultaneity of events there.

I hope I give the correct answer and the right answer, the answer that you're looking for.
Sincerely

25. Jul 18, 2016

### Quandry

Stephanus, thank you for taking the time to provide such a detailed answer. I was going to say complete, but clearly there is more. However, you have answered my question very well.
Now my challenge is to understand why an observer is required, and why the observer appears to require intelligence.
Thank you for getting me past the barrier,
Bill