# Observed speed of light shone back from 0.9c forward motion

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1. Jun 1, 2015

### abrogard

been answered numerous times i guess but i couldn't find it.

i am an ignoramus.

if spaceship doing 0.9c shines light forward we - from another inertial frame - see that light as proceeding forwards at 0.1c, do we not? But 'they' see it as proceeding forwards at c.

fine.

But if they shine a light backwards. Then what do we see?

'They' see it going back at c, I suppose. Fine.

But we see - what?

2. Jun 1, 2015

### A.T.

No, every inertial frame observes the light at c.

See above.

3. Jun 1, 2015

### harrylin

We measure light as propagating at c, because we first adjusted our clocks in such a way that we measure light propagating at c in both directions wrt our chosen reference system, the one in which the spaceship is doing 0.9c.

Perhaps you mean that the difference in velocity (also called "relative velocity" in a number of books) is 0.1c according to us, but c according them? That's correct:
We measure and calculate: c - 0.9c = 0.1c
They measure and calculate: c - 0 = c
Simple:
- the velocity of the light according to us still c.
- consequently, their relative velocity (velocity difference) according to our chosen reference system is now c - (-0.9c) = 1.9c.

There is nothing odd about that: nothing can go faster than c implies that the velocity difference can be maximally 2c.
That's the case for light rays in opposite directions:
<----
-c
----->
+c

4. Jun 1, 2015

### Staff: Mentor

If I am moving relative to you with speed $u$, and set something in motion at speed $v$ relative to me (either or both speeds can be negative/backwards), you will find that its speed is NOT the $(u+v)$ that you'd expect, it is $(u+v)/(1+uv)$ (measuring distance in light-seconds and time in seconds so the speed of light is equal to 1.0). It's a good exercise to try this formula for the case in which $v=1$ (I'm shining a light beam) and also for cases in which $u$ and $v$ are both very small compared to the speed of light (missiles launched from a jet fighter).

5. Jun 1, 2015

### Janus

Staff Emeritus
Let's make it simple. The ship emits a short pulse of light in all directions. They see the light expanding outward at c away from them in all directions as a sphere with themselves at the center at all times.
What we see from a frame that the ship is traveling at 0.9c relative to is the light expanding outward at c away from the point where the light was emitted (which remains stationary with respect to us.) while the ship continues on at 0.9c relative to and away from this point.

6. Jun 1, 2015

### abrogard

Yep, I meant we see their forward going beam moving relative to them at 0.1c, of course. Sorry not to make that clear.

So - 1.9c. That's what I thought it should be. But I was thinking I must be wrong.

So we can't go faster than c but we can leave a beam of light, move away from a beam of light with an effective separation speed up to a max of 2c.

Simple enough, thank you.

I'm not sure about Nugatory's remark, though. Why would that be? (u+v)/(1+ uv) ? That looks very similar to the Lorentz transform I learned about from Ramamurti Shankar's Youtube videos. I worked through that - it's down at my level - and I can see how it comes out, see how it works out. Can't understand it but can see it and believe it because I can work it for myself.

But I don't see this one. How does this one come about?

It means we won't see that light beam as separating from the ship at 1.9c doesn't it?

7. Jun 1, 2015

### Staff: Mentor

As Nugatory said, google for "relativistic velocity addition". Or look in any textbook on SR. This is a very basic concept in relativity.

8. Jun 1, 2015

### Staff: Mentor

There are a bunch of good derivations out there on the web if you search for "relativistic velocity addition derivation", but you can work it out for yourself pretty quickly if you already understand the Lorentz transformations.

The general idea is that if I am rest in the primed frame and I throw an object at speed $v$ relative to me, then its position relative to me at time $t'$ will be $x'(t')=vt'$. However, its position relative to you will be the Lorentz transform of the point $(vt',t')$ and its speed relative to you will be the first derivative of that position with respect to $t$ (not $t'$!). The chain rule and a fair amount of algebra will get you from there to the velocity addition formula.

9. Jun 2, 2015

### harrylin

That is correct, it's derived from the Lorentz transforms and tells you directly how velocities are measured differently in the different reference systems. It tells you thus that the light propagates at c relative to each reference system according to measurements in both systems.
No, it only tells you how the speeds of moving entities are measured in each frame:

- we will measure the ship to move at +0.9c
- we will measure the light to propagate at -c
Consequently we measure the separation speed as 1.9c

If the ship sets up a reference system in which the ship is assumed to be in rest:

- they will measure the ship to move at 0c
- they will measure the light to propagate at -c
Consequently they measure the separation speed as c.

10. Jun 2, 2015

### abrogard

Thanks for all that.

I just got a message from Nugatory saying that this forum has a policy not to give complete answers.

It is here to 'help people work things out for themselves'.

I wasn't aware of that. Didn't read the rule I suppose. Must be hard sometimes to figure out how much you can tell and how much you can't.

Must border on the absurd sometimes - how about you refute or contest some statement but forbear to explain your refutation on the grounds that they've got to 'work it out' ?

I told you I was a bit thick. And I am. Doesn't depress me too much because, of course, there's much I can comprehend and enjoy in life - and there even appear to be a fair few even thicker than myself, so I can indulge my baser instincts sometimes.....

But, yes, thick. And so I don't really understand all that's been said to me.

My take on it is that its all just reiteration of the Lorentz transform principle which I've stated I've got some understanding of (well, follow the basic maths) and which is implicit anyway in my question - where I state that we'll see different speeds to that which the people in the spaceship see.

And I'm thinking now that Nugatory's formula is simply the Lorentz transform minus the 'divided by c²' bit - which I've seen other people neglect on the grounds at 'it is just a constant'. i.e., I suppose they're saying that the relationships indicated in the formula remain true without the constant which merely changes quantities - moves lines on graph? Something like that?

I've always been interested in that, though, ever since I first met a guy saying stuff like that. I don't see it as 'just a constant' simply because the number is the speed of light. Why that particular speed, see?

And because it is a speed. A speed has time intimately involved in it. In that term there I see time as being involved in the e=mc² and I like that, think it's right.

Anyway, I'm just blathering, which I'm prone to do.

Then it looked for a while like Nugatory was saying it wouldn't be 1.9c because of his formula.

Now, because of Harrylin, who repeats everything I first thought and said but who will be, I assume, as mathematically learned and capable as most on these forums, it looks again like 1.9c.

So I came with a question, was illuminated with the light of knowledge, then thrown into the darkness of uncertainty and finally relased into the light again.

Happy days.

:)

p.s.

I will try not ask any questions to which I need the answer.

11. Jun 2, 2015

### A.T.

You are confusing two different things:
- separation speed, which is the velocity difference 1.9c in this case
- relative speed which is always c for light

12. Jun 2, 2015

### harrylin

You're welcome

BTW, these kind of things were rather well explained already in Einstein's famous 1905 paper, which gives an elaborate overview of what he later named special relativity.

In the derivation of the Lorentz transformation, in §3, he clarifies for a point that is moving at velocity v in the same direction as a light ray that is supposed to propagate at velocity c:

" the ray moves relatively to the initial point of k, when measured in the stationary system, with the velocity c-v "
- http://fourmilab.ch/etexts/einstein/specrel/www/

And in §5 he derived the equation that Nugatory presented here; it's the system transformation equation for velocities. That equation is particularly useful if you want to quickly calculate how the velocity of the spaceship will be measured in different frames.

13. Jun 2, 2015

### Staff: Mentor

It's even simpler than that. If I measure distances in light-seconds and time in seconds the speed of light will come out to be 1.0 - and multiplying and dividing by one doesn't change anything so I can simplify $v^2/c^2=v^2/(1)^2=v^2$ and reduce the clutter in my equations. We can't always make a constant go away by choosing different units, but it's worth doing whenever we can.

You were doing the same thing when you started this thread, perhaps without even noticing, when you said ".9c forward motion" instead of "$2.698\times{10}^8$ meters/second forward motion".

14. Jun 2, 2015

### abrogard

Nugatory: Got it. Thankyou. Didn't realise it at the time. It is an indication of just how much of a non-mathematician I am. Unfortunately.

A.T.: I don't think I'm really confusing them. Like I'm not confused about them in my head at all. I may well be confusing them in posing my question.
It is the apparent speed of separation between the backward moving light wave front or light packet or whatever and the space ship that I was concerned with.

Harrylin: Thanks for the link to the 1905 paper. I'd not seen it before. Everything I know I got from Ramamurti Shankar's Youtube lectures and a couple of similar. I'm surprised Einstein is so easy for the uneducated such as myself to read. At the beginning. It's not long before I start getting bogged down in the maths. It'll do me good if I can find a way to work through the paper though. For sure. If I can.

:)