Ok, I know just enough in my hobby-esque interest of the universe, et al, to grasp some general concepts and to confusedly believe that I know the answer to something I really don't. So, I thought I'd ask those who do!

If I am travelling at 0.99c past you, who are a stationary frame of reference, and I shine a light directly in front of me, what happens?

Now, I think that the answer is that from your stand point as a stationary observer, you would see the light inching away from me at just 0.01c faster than I am travelling. But from my standpoint I would see the light moving away from me at 1.0c. (While I think this is right, I'm certainly not betting any big money on it.)

But, if that's right, I'm definitely shady on the explanation. Is there a Lorentz contraction working from my frame of reference? Something else? Does the ether actually exist?

Thanks!

You are correct. You would see the light traveling away from you at c. I would see the light traveling at c relative to me, and thus 0.01c relative to you.

And yes, there is Lorentz contraction and time dilation in effect here, as well as a simultaneity difference.

How detailed of an explanation would you like?

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SeanF

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Well, in response, I'm pretty familiar with Lorentz contraction and time dilation, but in my (limited) readings, I don't think I've ever heard of a simultaneity difference. So that might clue you into my knowlege base.


(That, and while I've read a book or 2 on it, if the answer involves quantum physics, it'll be mostly greek to me)



And thanks!

The simultaneity difference is due to the limit of c as a information transfer speed and the result of people seeing the same thing at different times. If the sun blows up, an observer on Alpha Centauri (4 LY away) will see it happen before an observer on Sirius (8 LY away).

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I'm a bit rusty on my special relativity, but doesn't the simultaneity difference have to do with people having different reference frames rather than different distance from the object under observation?

Well, the different distances are equivalent to different frames. I just put it that way because, due to his comments, I felt he may understand it better.

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Some try to tell me, thoughts they cannot defend,... - Moody Blues.

But as I pointed out in the other thread - this only happens as each 'reference' frame has a slower/faster clock speed (time) to allow it to happen.

By light appearing to move at one speed or the other is really only due to relative time slowing down when the reference frame is at speed.

Nick

Yep, I'm with madamwitty and Nick on this--I don't think different distances are equivalent to different frames. In what sense?

Okay, here's the simultaneity thing.

Picture two clocks located some distance apart. Clock A sends out a light pulse towards clock B, which bounces off of B and returns to A. Let's say that the signal leaves A at time A=12:00, hits B at time B=12:05, and arrives back at A at time A=12:10

Now, to an observer who is stationary relative to the two clocks, the light would take the same amount of time going from A to B as it does going from B to A. Thus this observer concludes that that light hit B at time A=12:05. Thus time A=12:05 was simultaneous with time B=12:05, and the two clocks are synchronized.

Now, let's consider an observer who is in motion relative to the two clocks. This observer, of course, considers himself to be at rest, with the two clocks themselves in motion. Thus, this observer sees the light moving away from clock A at less than c and towards clock A at greater than c (assuming the clocks are moving in that direction). So, to this observer, the light hitting B must occur at, for example, time A=12:07 (that is, because of the motion of the clocks, it takes seven minutes for the light to get from A to B and only three minutes to get from B back to A). He then concludes that time A=12:07 was simultaneous with time B=12:05 and the clocks are not synchronized.

Does that help?

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SeanF

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An easy way to put it.

You are at 0 speed. Light travels at c. For arguments sake, lets say it travels 10 units a second.

You're at speed 99% c. Your time frame is slooower (clocks tick less). It still travels 10 units a second, but as your second is longer than the observer, it moves the same speed to both observers.

Light isn't the question here - time is.

Nick

It's not that simple. If you send out a pulse in both directions - in front of you and behind you - you will see both pulses as moving away from you at the same velocity. An observer in motion relative to you will see the backwards pulse moving away from you faster than the forward pulse is. Simple time dilation can't account for that.

But a simultaneity difference can.

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SeanF

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Yep clears it up. Thanks.

I read, I think a Brian Greene book where they use an example of a train with a lightbulb in the middle, passing by a platform. If you turn on the bulb, senors on the front and back of the train show the light striking both simultaneously. But from the platform's frame of reference, the light struck the back sensor on the train first. Never caught the name of the concept, I guess.



And, I think I sort of have the concept of the original question down: But someone please correct if necessary!


(For the sake of my ease of math, I'm going to assume that at 0.75c, the time dilation & Lorentz contraction is 1/2. I know this is way too high, but it's trouble enough getting my head uncoupled from classical physics; I'm trying to keep the math as easy as possible)


Assuming I am looking to go to Alpha Centari. From Earth, we all agree it's 4 LY away and that in my 0.75c ship, it will take me 6 years.

Once I get in the ship and immediately go to 0.75c (it's a magic, liquid-schwartz-fueled ship), the trip only winds up taking me 3 years. Of course, from my frame of reference, it turns out that Alpha Centauri was just 2 LY away, due to Lorentz.

If as soon as I instantaneously hit 0.75c (but now at constant velocity), I shine a beam of light at Alpha Centauri, you, on stationary Earth, will see the light travelling at c, as it always does, but only moving away from my ship at 0.25c faster. But, since it's travelling at c, you will see it hit Alpha Centauri in 4 years, or in 2/3 of the time that it takes me in my ship to reach the star.

From my vantage on the ship, the light left me travelling a full c faster than my ship. And travelling at c, for a distance of 2LY (which according to me is the distance), it will hit Alpha Centauri in 2 years, or, still in 2/3 of the time that it takes me to get there.

If I'm close, thanks all, for helping push me in the direction!

The logic's right, but the math isn't (even ignoring the .75c=.5 contraction issue, 4 light-years at .75c would only take 5.333... years, not 6).

BTW, I do all my Special Relativity thought experiments with a velocity of .6c, because that gives a .8 contraction. That's easy enough to work with.

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SeanF

"Ask to understand, but don't challenge unless you have the knowledge."--NEOWatcher

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How can you see the light moving away from you and how can you measure its velocity?

I suppose one could have a string of mirros in both directions and measure how long it takes the reflection to come back from each mirror.

I'm probably in a minority here, but these physical descriptions of special relativity just give me a headache. By contrast, I can understand the Minkowski-style linear algebra formulation instantly.

That works, though it requires you to assume that the light moves at the same speed in both directions (since it gets reflected back). You could also set up a detector ahead of time in the right position, attached to a clock to measure the time the light reaches it. That requires the assumption that a clock which is at rest relative to you, but separated by some physical distance, will keep the same time as an identical one right next to you. Both of these methods only calculate an average speed of light, of course, but one could imagine send out a large number of parallel beams that are all intercepted at varying distances, and use that to give you the instantaneous velocity, with the assumption that identical setups would behave identically.

Yeah, I'm with them too. I spaced this one (I should know better than to try and get a quick reply in at work when I haven't work with something in a while) Thanks to madamwitty, Nick and hhEb09'1

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Some try to tell me, thoughts they cannot defend,... - Moody Blues.

Son of a...I guess I wasn't kidding about needing to keep the math simple!

I figured out how to explain simultaneity differences on a basic level, yet I forgot how basic multiplication/division works. Oh well, I guess there's always www.badmathematics.com for that!

Thanks, all!

I just want to make the comment that, within Special Relativity, the constant speed of light (in all directions, regardless of the motion of the observer) is a postulate. In other words, it's assumed to be true.

If one is looking for proof that the speed of light is constant, one must look outside the framework of SR. SR merely explains how the universe works (or would work, if you insist) with a constant speed of light.

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SeanF

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