An intriguing question came to mind:

Scenario:
A ship that has it's inner walls painted blue say something like the base blue used in computer graphics [hyperlinks], zips past you at 0.8c, what colour do the walls appear to be from our RF?

why inner walls?

why inner walls?

mainly because then the colour is a product of the light eminating from a ship that has a v =0.8c

Maybe we can modify the scenario by painting the outside the same blue as the inside and compare the difference [if any]
If the inside of the ship was rectagular with the length towards it's vector would the forward inner wall be a different colour to the other inner walls to the 1] ships observer and 2] to the RF observer?

Just curious.......

Maybe we can modify the scenario by painting the outside the same blue as the inside and compare the difference [if any]

it does not make any difference.

If the inside of the ship was rectagular with the length towards it's vector would the forward inner wall be a different colour to the other inner walls to the 1] ships observer and 2] to the RF observer?

1. ships observer would not notice any difference as he is at rest (or co-moving) with ship's walls.

2. The orientation of walls does not make any significant difference in velocity between observer and the ship. To the observer the ship's color after blue-shift would be UV.

Ok...no surprises there, thanks.....

it does not make any difference.



1. ships observer would not notice any difference as he is at rest (or co-moving) with ship's walls.

2. The orientation of walls does not make any significant difference in velocity between observer and the ship. To the observer the ship's color after blue-shift would be UV.


Try a different slant on this...Just a thought….

If the leading inside wall of the box were to be painted blue and there was a mirror on the inside trailing wall, then the beam of light leaving the mirror and heading forward would remain blue.

To the observer the ship's color after blue-shift would be UV.
Why blue-shift instead of red-shift due to "transverse doppler"?

I would have thought blue-shift during the "approach" phase, redshift during the "transverse" instant, and then more extreme red-shift during the "depart" phase.

If the resulting UV frequency during approach is outside of the visible spectrum, wouldn't the ship appear black?

Same question if the walls were originally red: If the red-shift during departure were to result in an IR frequesncy outside of the visible spectrum.

Try a different slant on this...Just a thought….

If the leading inside wall of the box were to be painted blue and there was a mirror on the inside trailing wall, then the beam of light leaving the mirror and heading forward would remain blue.
I'm not sure I agree with this, which is a shame, because it would be neat if it was true.

The observer looks at a mirror on the trailing wall, and sees an image of the leading wall. How is the light from the image of the leading wall different from the light from the real trailing wall? It appears just like the trailing wall, only farther back (like a wall trailing along behind the ship).

I would think the light from the image wall and the real wall would both be blue-shifted as the ship approaches.

You guys need to specify which frame you're in when you make these statements. Are you in the box looking at the wall/mirror? Or are you observing it from the outside?

From inside the box, the reflected light is unchanged. It's the same frequency of blue.

From outside the box, you can only observe the front wall as it receeds from you (you are behind the box) or the reflected light as it comes toward you (you are in front of the box).

Yes?

You guys need to specify which frame you're in when you make these statements. Are you in the box looking at the wall/mirror? Or are you observing it from the outside?

From inside the box, the reflected light is unchanged. It's the same frequency of blue.

From outside the box, you can only observe the front wall as it receeds from you (you are behind the box) or the reflected light as it comes toward you (you are in front of the box).

Yes?
I'd agree with that.

My comments about blue-shift and red-shift were from the reference frame of the relatively 'stationary' observer (who is watching the ship move past).

Try a different slant on this...Just a thought….

If the leading inside wall of the box were to be painted blue and there was a mirror on the inside trailing wall, then the beam of light leaving the mirror and heading forward would remain blue.

But the mirror is co-moving with the leading wall so the incident blue light is not red-shifted but remain blue. the reflected blue light will be blue-shifted to UV.

Why blue-shift instead of red-shift due to "transverse doppler"?

I would have thought blue-shift during the "approach" phase, redshift during the "transverse" instant, and then more extreme red-shift during the "depart" phase.

yes i took only approach phase.

This is where it gets weird.

I think we all agree that we (the observers) are standing in front of the box. Imagine the box has a hole in the front leading edge such that you can view a mirror placed on its trailing internal wall. The box is painted inside and out in the colour blue (except for the mirror.

If the box is stationary relative to us, we must agree that all colours look blue.

But as the box moves quickly towards us, the outside leading edge changes colour because it is blue-shifted, hence, we see UV. Agreed?

But this is the trick, the light leaving the front internal leading wall is emitted backwards (and hence red-shifted relative to the observer) towards the mirror. Normally, we would NEVER have the privilege of observing such light, but in this case we have a mirror.

The mirror is also travelling relative to the observer and has the effect of “correcting” the already shifted light back to blue.

Result….the internal front leading edge looks blue regardless of relative velocity.

What do you think guys?

But as the box moves quickly towards us, the outside leading edge changes colour because it is blue-shifted, hence, we see UV. Agreed?
the interesting thing is we can not see UV so does the wall disappear or is it just black? Obviously the wall isn't going to disappear.....xcing xcling xcling [x- files] ......or it it? :m: :p

actually I think it is a very valid question:
If the walls colour is shifted to UV what do we actually see?
The wall still exists yes? But it's not reflecting visable light and it's not just absorbing light ....so.....hmmmmmmm

But this is the trick, the light leaving the front internal leading wall is emitted backwards (and hence red-shifted relative to the observer) towards the mirror.

Yes, the light leaving the front internal leading wall is red-shifted relative to the stationary observer, not to the mirror which is co-moving with the leading wall. as such the mirror simply reflects back the blue light. If a mirror is stationary behind the moving box then you are right. That is not the case here.

Yes, the light leaving the front internal leading wall is red-shifted relative to the stationary observer, not to the mirror which is co-moving with the leading wall. as such the mirror simply reflects back the blue light. If a mirror is stationary behind the moving box then you are right. That is not the case here.

Hmmm, I'm sure you're wrong here, but let me think about it.

First you say "Yes, the light leaving the front internal leading wall is red-shifted relative to the stationary observer"

So, the mirror (relative to the observer) will correct this red-shift back to blue, won't it? :confused:

The blue light emitted by leading wall looks red-shifted to the stationary observer.
For the observer, the moving mirror should receive the blue-shifted red-shifted light.
For the mirror, no-shift, its blue light all the way from the wall.
Don't try to confuse me with your confusion of frames. :p

The blue light emitted by leading wall looks red-shifted to the stationary observer.

No it doesn't because it's travelling backwards and is thus impossible to EVER see.


For the observer, the moving mirror should receive the blue-shifted red-shifted light.

"blue-shifted red-shifted light." hahahaha

Think about it. You agree that the light leaving the front internal wall is red-shifted relative to the observer. Agree?

So a reflection of this red-shifted light (relative to the observer) will be blue-shifted by an object (the mirror) travelling toward the observer. Hence, back to blue. :p :p :p

No it doesn't because it's travelling backwards and is thus impossible to EVER see.
OMG, this is nothing but similar when the ship is moving away from the observer, you don't imagine properly??

"blue-shifted red-shifted light." hahahaha

I was laughing ROF while typing this exclusively needed for you.

Think about it. You agree that the light leaving the front internal wall is red-shifted relative to the observer. Agree?

Yes, but you only 'discovered' that the observer cannot see. :p

So a reflection of this red-shifted light (relative to the observer) will be blue-shifted by an object (the mirror) travelling toward the observer. Hence, back to blue.

It is red-shifted for the observer because the emitting wall is moving towards the observer. For the observer, the moving mirror recieves the 'blue light'. The observer understands what the mirror would recieve. what? blue light. its BLUE light. back to blue, ofcourse.

OMG, this is nothing but similar when the ship is moving away from the observer, you don't imagine properly??



I was laughing ROF while typing this exclusively needed for you.



Yes, but you only 'discovered' that the observer cannot see. :p



It is red-shifted for the observer because the emitting wall is moving towards the observer. For the observer, the moving mirror recieves the 'blue light'. The observer understands what the mirror would recieve. what? blue light. its BLUE light. back to blue, ofcourse.

Everneo, you can be soooo dense sometimes.

For one thing, I haven't the foggiest what your last paragraph is talking about. Please reconstruct it so we understand it.

You keep swapping reference frames, you knob-head.

1. Observer "knows" box internal trailing wall is blue, because he painted it.

2. Observer "knows" box is travelling towards him.

3. Observer "knows" that light travelling backwards will be red-shifted due to relative motion.

4. Observer "knows" there is now a beam of light travelling away from him with a longer wavelength than the original blue box.

5. Observer "knows" there is a moving mirror (moving towards him) which will "correct" the light beam back to blue i.e. blue-shift it back to blue once again RELATIVE TO THE OBSERVER.

You need more sleep :bugeye:

You keep swapping reference frames, you knob-head.
You are applying relative reciprocity where it is not applicable. Lets see who is the knob-head.

1. Observer "knows" box internal trailing wall is blue, because he painted it.

He painted it? it is a news for me. Anyway i will go with your assumption.

2. Observer "knows" box is travelling towards him.

Yes.

3. Observer "knows" that light travelling backwards will be red-shifted due to relative motion.

Yes.

4. Observer "knows" there is now a beam of light travelling away from him with a longer wavelength than the original blue box.

Yes.

Edit : forgot to type 'Yes'.

5. Observer "knows" there is a moving mirror (moving towards him) which will "correct" the light beam back to blue

No.

Observer "knows" there is a moving mirror (moving towards him) which will "correct" the light beam that is red-shifted with respect to the observer only. In the emitting Wall & Mirror frame the emitted light would not undergo any doppler-shift because the emitting leading wall and the mirror are at rest with each other. The leading wall emitts blue light as for as the mirror is concered.

As for the observer, the wall emitts 'red-shifted' light , the moving mirror (with respect to the observer) would recieve blue-shifted red-shifted (*ROFL*) that is now blue for the mirror as understood by the observer.


You need more sleep :bugeye:
You loose the sleep now.

You are applying relative reciprocity where it is not applicable. Lets see who is the knob-head.



He painted it? it is a news for me. Anyway i will go with your assumption.



Yes.



Yes.





No.

Observer "knows" there is a moving mirror (moving towards him) which will "correct" the light beam that is red-shifted with respect to the observer only. In the emitting Wall & Mirror frame the emitted light would not undergo any doppler-shift because the emitting leading wall and the mirror are at rest with each other. The leading wall emitts blue light as for as the mirror is concered.

As for the observer, the wall emitts 'red-shifted' light , the moving mirror (with respect to the observer) would recieve blue-shifted red-shifted (*ROFL*) that is now blue for the mirror as understood by the observer.



You loose the sleep now.



Ahrrghg, my head is going to explode.

Do you know what, I simply don't believe you at the moment.

I need to think about it some more.

What do other people think about this problem?

Does the observer see a blue reflection or a UV one?

I've edited my previous post. my answer to (4) is Yes. forgot to type the answer. Hope it might prevent your knob-head from exploding. :p

Everneo,

Please answer the following. You are only allowed to answer with the following answers: yes, no, blue, red-shifted, uv. Hope you are ok with this.

Firstly, let me describe the setup.

We have an observer on a planet A.
We have an observer on a Planet B.
Planet A and B are stationary with respect to each other.
There is a box travelling at a high velocity from A towards B.
The box is painted in BLUE on all the outside faces.
The box is also painted in BLUE on the inside.
The leading outside face (the one facing and travelling towards B) has a hole in it such that an observer at Planet B can see through the hole and onto the back inside trailing face, which has a mirror.
The rear face (the one facing but travelling away from A) also has a hole (in the mirror) such that Planet A observer can see through and onto the inside leading face.

Phew, get it?

Ok on to the questions….

1. When observer B looks at the outside leading face, what does he see?
2. When observer A looks at the outside trailing face, what does he see?
3. When an observer A looks through the back of the mirror and onto the front leading inside BLUE painted face, what does he see?
4. What colour is the light reaching A (the light beam from the front leading inside face)?
5. If A holds up a flag to tell B what colour he has seen, what colour is it?
6. A wants to prove to B what colour the light is so he places a mirror at A which reflects the light back to B. Is that ok?
7. The mirror at A is stationary relative to A and B and so there is no Doppler shift. Do you agree?
8. So what colour does B see in the mirror?
9. Suddenly the mirror at A starts moving very fast towards B. Does B see a Doppler shift?
10. What colour is the beam now?

If your answer to 10 is anything other than BLUE, please explain why.
If your answer for 10 is BLUE, please explain the difference between the original question and this one.

Thank You.

The way to "correct" the blue-shifted UV light back to its original frequency is to have a mirror image in which the moving box does not appear to be moving. This can be accomplished by having a mirror moving at half of the box's speed following along behind it.

Observer____________________Box___Mirror
Observer______________Box______Mirror
Observer_______Box__________Mirror

As long as the image in the mirror appears stationary to the observer, then there should be no doppler shift.:cool:

So even though the front of the moving box appears black, (assuming all of the UV light is outside of the visible spectrum) the image in the mirror trailing at v/2 would appear to be the normal shade of blue.

Firstly, let me describe the setup.

We have an observer on a planet A.
We have an observer on a Planet B.
Planet A and B are stationary with respect to each other.
There is a box travelling at a high velocity from A towards B.
The box is painted in BLUE on all the outside faces.
The box is also painted in BLUE on the inside.
The leading outside face (the one facing and travelling towards B) has a hole in it such that an observer at Planet B can see through the hole and onto the back inside trailing face, which has a mirror.
The rear face (the one facing but travelling away from A) also has a hole (in the mirror) such that Planet A observer can see through and onto the inside leading face.


1. When observer B looks at the outside leading face, what does he see?
UV
2. When observer A looks at the outside trailing face, what does he see?
3. When an observer A looks through the back of the mirror and onto the front leading inside BLUE painted face, what does he see?
4. What colour is the light reaching A (the light beam from the front leading inside face)?
5. If A holds up a flag to tell B what colour he has seen, what colour is it?

Red (at .9c relative velocity it should be IR but for this setup let us say Red)

6. A wants to prove to B what colour the light is so he places a mirror at A which reflects the light back to B. Is that ok?
7. The mirror at A is stationary relative to A and B and so there is no Doppler shift. Do you agree?

Yes for both.
8. So what colour does B see in the mirror?

Mirror at A reflects red light. B sees red.

9. Suddenly the mirror at A starts moving very fast towards B. Does B see a Doppler shift?

Yes.

10. What colour is the beam now?

It depends on the velocity between Mirror and B. lets say its .9c (same as that of the box).

Mirror does not recieves red light any more. Becasue of its movement towards the source of light it receives blue-shifted RED light, that is BLUE.
It reflects blue light but B receives it as UV because the mirror moves towards B.

Alternatively, if the mirror moves at v/2 (half the velocity of the box) B would receive BLUE, as Neddy Bate said.

So why is there a double doppler shift for the mirror but not for the front surface of the box?

I mean, the box is BLUE and because of its speed it shifts the frequency up by, say, 1 step to UV.

The mirror (if stationary) would reflect red light, so why when it is travelling at the same rate as the box doesn't it only step up one frequency back to BLUE?

What's the diff between a mirror and a painted surface?

Tell me, what does the moving mirror receive ? IR or Blue light :

1. from planet A' reference frame
2. from moving mirror's reference frame

Tell me, what does the moving mirror receive ? IR or Blue light :

1. from planet A' reference frame
2. from moving mirror's reference frame

This i where I get confused because I'm only concerned with what the mirror actually receives i.e answer to 2 is IR.

Then I'm only concerned with what the mirror actually bounces off ie Blue.

Therefore if mirror actually bounces off BLUE (already due to its motion relative to B) then B will see BLUE.

Got to go, be back later.

Look, Everneo,

In my 10 question scenario, you have stated that the mirror at Planet A reflects IR or red light. Agreed?

And you said that the observer at B sees the box as UV (travelling at 0.9c). Agreed?

You agreed that the observer at B sees the mirror as red. Yes?

So why when the mirror accelerates to 0.9c (the same as the box), why is the doppler shift twice that of expected? Surely a mirror only refelects the frequency change just as a painted surface would.

Is this puzzling or what?

You also said in your answer

"Mirror does not recieves red light any more. Becasue of its movement towards the source of light..."

But this is wrong because when the mirror is travelling with the box it is stationary relative to the painted light source.

What you're saying doesn't stack up!

I know you want the light from the mirror to always be the same as the wall to ALL observers, because if it didn't it would violate the laws of relativity. But you have to convince me why a mirror at Planet A, which accelerates to 0.9c, behaves differently to a painted surface doing the same thing.

Why are you doubling the doppler shift for a mirror?

You also said in your answer

"Mirror does not recieves red light any more. Becasue of its movement towards the source of light..."

But this is wrong because when the mirror is travelling with the box it is stationary relative to the painted light source.

What you're saying doesn't stack up!


When the mirror moves towards the source and keeps same velocity as the box then for the mirror there would not be any red-shifted light.

Is that clear now?

Now go the B's frame. The light going away from B from the box is red-shifted.
Though the observer at B sees this red-shifted light seems to hit the moving mirror as Red (we will leave the IR part, lets assume it is red) the observer also knows the mirror is moving. What is appearing to him as red will not be red for the moving approaching mirror. Agree?

When the mirror moves towards the source and keeps same velocity as the box then for the mirror there would not be any red-shifted light.

Is that clear now?

yES, but you keep jumping frames! But I'll let you off for now :rolleyes:


Now go the B's frame. The light going away from B from the box is red-shifted.
Though the observer at B sees this red-shifted light seems to hit the moving mirror as Red (we will leave the IR part, lets assume it is red) the observer also knows the mirror is moving. What is appearing to him as red will not be red for the moving approaching mirror. Agree?

Agreed.....and so it looks BLUE!!!

Why are you saying it looks UV? Why two steps up?

When the mirror moves towards the source and keeps same velocity as the box then for the mirror there would not be any red-shifted light.

Is that clear now?

yES, but you keep jumping frames! But I'll let you off for now :rolleyes:



Thanks for letting me off from the hell, devil57. This is a non-issue.


Now go the B's frame. The light going away from B from the box is red-shifted.
Though the observer at B sees this red-shifted light seems to hit the moving mirror as Red (we will leave the IR part, lets assume it is red) the observer also knows the mirror is moving. What is appearing to him as red will not be red for the moving approaching mirror. Agree?
Agreed.....and so it looks BLUE!!!

Why are you saying it looks UV? Why two steps up?

You are jumping too early. For the mirror the INCIDENT light is BLUE. So, it reflects BLUE. B receives UV after blue-shift because of mirror's movement towards B. Over.

Thanks for letting me off from the hell, devil57. This is a non-issue.



You are jumping too early. For the mirror the INCIDENT light is BLUE. So, it reflects BLUE. B receives UV after blue-shift because of mirror's movement towards B. Over.

No it's not over!


You clearly stated in answer to question 8 that "Mirror at A reflects red light. B sees red."

I repeat:

Why when the mirror moves towards A at 0.9c does the colour that B sees suddenly steps up twice to UV (LIKE YOU CLAIM)?

No it's not over!


You clearly stated in answer to question 8 that "Mirror at A reflects red light. B sees red."

go back & read properly. you clearly stated mirror is stationary at A till question (9).

go back & read properly. you clearly stated mirror is stationary at A till question (9).

Yes, I know that.

And so I reiterate (for the nth time):

When the mirror (which is emitting red light towards B when stationary at A) suddenly accelerates to 0.9c, why does it increase the doppler up to UV and not BLUE?

Then go and sit on the moving mirror and tell me
what is the color of light approaching the mirror (you on top of it) from the box.
RED or BLUE?

Then go and sit on the moving mirror and tell me
what is the color of light approaching the mirror (you on top of it) from the box.
RED or BLUE?

You're dodging answering the question by asking me questions aren't you!!!!!

Listen, I KNOW what the colour of the light is in the stationary mirror as seen fro B. We've gone through it 10 times and you and I AGREE that it is RED light leaving the mirror as seen from B.

Stop answering my questions with more questions. Instead, tell me why the mirror, when accelerated to 0.9c suddenly steps up to UV from the same point of view of observer B.

Then go and sit on the moving mirror and tell me
what is the color of light approaching the mirror (you on top of it) from the box.
RED or BLUE?

We're looking at this from B you double-trebble knob head!

Answer my previous post.

Try to answer my question. By luck you might get the facts right in the process. *yawn* take your own time i will be back after some time.

Try to answer my question. By luck you might get the facts right in the process. *yawn* take your own time i will be back after some time.

Something has dawned on me.....you don't know the answer do you! :eek:

You keep swapping frames when it suits you, so that you can get B to observe uv for both box and mirror. :bugeye:

You realise this is a paradox and you don't know the answer, because if you did, you've had answered by now.

We both agree that the mirror situated (motionless) at A is reflecting Red light and that B is viewing it as red light.

You KNOW that when the mirror accelerates to 0.9c you have got to account for it appearing UV, and you know THAT is impossible. If it doesn't, or rather you can't account for this happening, you know this violates the laws.

See if you can answer this question (BUT i WON'T BE FORGETTING THE ORIGINAL QUESTION):

Forget the box, but lets stick with Planet A and B.

Suppose B beams a red light ray at a mirror heading towards it at 0.9c. What colour does B see bouncing back? Blue or UV?

QQ: Sorry for hijacking your thread :rolleyes:

Not a problem, as I have already got my answer.....to the question.
What do we predict we will see of the colours in inside a ship with v=0.8c?

Confusion...is all I find is the answer....this of course is a common result when ever a relativity question is asked.
Allegations of frame swapping.
Scientific fraud and conspiracy.
all the result of the confusion created by short speak internet dialogues.

Dav, If you look really closely at the question and spend 3 years working through all the possibilities you will eventually conclude that the Game called debunk SRT is impossible to win at. It took me 3 years maybe it will take you less time.......

It's a bt like attempting to find Pepetual energy devices etc .......it is intoxicating and obsessive but with present technology and knowledge impossible.

If you really wish to show a problem with SRT using a scenario I suggest you concentrate on maintaining a Reference frame table in all discussions.

Add to all you claims a RF code. to maintain clarity, and hopefully you or the reader will not get lost in this labyrinth called SRT reasoning.

SRT offers the debunker a great opportiunity to learn how to argue consitently and in a way that doesn't get lost along the way. So drop those bread crumbs as you travel and maybe it will save you and others a lot of wasted time. If you are receptive you will make great strides in learning how to argue consistently a very complex arguement. Even though you will ultimately fail to debunk SRT you will have gained a valuable insight in the art of reasoning.

The SRT puzzle has been around for 100 years and no one has been able to break the riddle in all that time. Mainly because it isn't a riddle to begin with.

QQ,

Very, very nice post.

BTW, the colors as viewed inside the ship don't change. As viewed from outside the ship (moving at 0.8c wrt the ship that is) the colors are all blue shifted if it's approaching, red shifted if it's receding.

QQ,

Very, very nice post.

BTW, the colors as viewed inside the ship don't change. As viewed from outside the ship (moving at 0.8c wrt the ship that is) the colors are all blue shifted if it's approaching, red shifted if it's receding.
and if it is traveling orthagonal?

all three visable walls would have a different colour from outside the ship from a rest frame? [Assume one transperant wall so we can see inside the ship.]

BTW dav,
Maybe this exercise will help you if you take the time to do it.

Set yourself the following challenge:

Assume that lights velocity is invariant to all observers regardless of their velocity.

Now, what is the only possible theory you can come up with that fully supports such a notion?
btw no cheating allowed, you have to work it all out for yourself.....hmmmmmm......the transforms may be a bit tough but if you can work out their genesis you will achieved something most people would not achieve in a life time.

and if it is traveling orthagonal?

Sure. The red/blue shift is dependent on the angle the ship makes across your field of vision. From zero shift if it's completely perpendicular to your line of sight (LOS) (at the point where it just crosses your LOS) to 100% shift when it's completely parallel to your LOS.

SL,would you be prepared to analyse the video taken aboard the space shuttle in SRT terms?

"This was the inspiration behind the original question"

When we see video footage from the space shuttle [SS] do we correct that image here on Earth or do we just take it as it comes?

Say for example the shuttles camera is taking frames at 52 persecond. Do we see it at that rate or do we see it at 53 frames per second type of question?

Are the colours shown by those images subject to the relative velocity effects and if so how so?

The speed of the shuttle in orbit is so microscopically tiny compared to 'c' that there is zero detectable difference. They are gloriously "True to Life™ "

The speed of the shuttle in orbit is so microscopically tiny compared to 'c' that there is zero detectable difference. They are gloriously "True to Life™ "

The thought occurred after my last post:

If the video camer is sequenced by an Earth RF timer would we technically see any thing differently?

I understand your point though the v's are so small....however in a lot of ways this is similar to the GPS question I think.

Say we have two cameras one is timed by the Space shuttles on board clock and the other video is timed by a timing pulse from the ground.
Or colour vibration rates are measured by both on board controlled equipment and gound based equipment....is there any discernable way of showing SRT effects?

The shuttle's Ku band link works in the 15GHz to 17Ghz range. Gamma for the shuttle wrt the earth is 1/sqrt(1-5²/186282²) = 1.00000000036022 which results in a 5Hz (!) difference between the earth and shuttle transcievers. That's 5 billionths of the carrier frequency. Not detectable. No effect.

The timing devices on the shuttle are not accurate enough to show this kind of thing. In 2007 the PARCS (Primary Atomic Reference Clock in Space) will fly on the ISS. Fun, fun, fun.

The timing devices on the shuttle are not accurate enough to show this kind of thing. In 2007 the PARCS (Primary Atomic Reference Clock in Space) will fly on the ISS. Fun, fun, fun.

Super -b SL.
Thanks for that..... we look forward to 2007 with a stifled yawn of anticipation.....ha.....

Ha! Yeah. I've mentioned the PARCS thing a bunch of times. No one seems to care... Oh well.

"This was the inspiration behind the original question"

When we see video footage from the space shuttle [SS] do we correct that image here on Earth or do we just take it as it comes?

Say for example the shuttles camera is taking frames at 52 persecond. Do we see it at that rate or do we see it at 53 frames per second type of question?

Are the colours shown by those images subject to the relative velocity effects and if so how so?
Time dilation is supposed to apply to everything; all laws of physics; Time itself.

In terms of frame-rate, a video camera should function identically in all reference frames. Therefore, viewing the recording is analogous to being in the original frame of reference in which it was recorded. There should be no "slow motion" or "fast motion", nor any disagreement as to how much time elapsed during the recording*. The colors should not be shifted by any doppler effect; neither classical or relativistic.


* (as measured by the clock local to the reference frame of the ship)

I think he was talking about real-time transmissions. I could be mistaken...

If the shuttle were moving relativistically wrt the earth, the approaching horizon would appear blue shifted to them...

QQ,

Very, very nice post.

I agree. A very insightful post by QQ.

BTW, the colors as viewed inside the ship don't change. As viewed from outside the ship (moving at 0.8c wrt the ship that is) the colors are all blue shifted if it's approaching, red shifted if it's receding.
SuperLum-- what about the relativistic "translational" doppler affect, supposedly always red-shifted -- should that be summed in with the classical doppler shifts that you mention above?

If so, then you might wish to consider the quote below where you would have the effect changing from blueshift to redshift at the perpendicular:

Sure. The red/blue shift is dependent on the angle the ship makes across your field of vision. From zero shift if it's completely perpendicular to your line of sight (LOS) (at the point where it just crosses your LOS) to 100% shift when it's completely parallel to your LOS.

Neddy, I believe you are correct. The relativistic doppler must be considered which will uniformly lower all of the frequencies of light. But the transition of red to blue as the object passes your line of sight should remain. Yes? Does this seem correct?

Ha! Yeah. I've mentioned the PARCS thing a bunch of times. No one seems to care... Oh well.
I care! :D

Ok, QQ, to tell you the truth, I'm still puzzled by the discussion which took place between me and Everneo and I don't feel satisfied with the answer.

I don't know if you followed the discourse but I'm unhappy with the fact that my mirror at Planet A (when stationary) is receiving and emitting RED light, but when accelerated to the speed of the box Everneo says it will doppler shift up past BLUE and onto UV. Why does the mirror do that when the painted BLUE surface steps up only one step to UV? This doesn't make sense to me.

You need to read the 10 question scenario that I put to Everneo before you comment.

QQ,

Very, very nice post.

BTW, the colors as viewed inside the ship don't change. As viewed from outside the ship (moving at 0.8c wrt the ship that is) the colors are all blue shifted if it's approaching, red shifted if it's receding.

Yeah, but the mirror at Planet A in my 10 question scenario!!!! Everneo said that relative to B it looks red. How then does it become UV (up two steps) when accelerated to the same speed as the box?

If we just had a mirror reflecting red from B and accelerated it up to 0.9c(or whatever speed) you would say it would then look BLUE NOT UV. What's going on.

If I don't get an answer in this thread I'll have to start a new thread :(

Something all of you seem to be forgetting, or don't know. The precieved color of an
object does not necessarily change due to relativistic velocity. Yes, the light from an
object you are approaching at relativistic velocities does 'blue shift', but the whole spectra of the recieved light is frequency shortened. For instance, invisible infrared is
shifted to the visible red frequency, replacing the 'lost' red color that was also shifted.
ALL frequencies are shifted, so there is no visible difference in color unless the velocities involved are ultra fast. Velocities can become great enough, according to STR, that all EM radiation is shifted to gamma and radio wave frequencies only, detected as a 'point' in the direction of travel due to relativistic aberration, as mentioned by James R in another thread.

If I have a green tree I'm watching, and I boost it up to 0.Xc, the green will shift toward the red if receeding and the blue if approaching. Any color mix you see will shift accordingly and the percieved color will change, hence the ability to judge astronomical distances by redshift (I know it's cosmological in origin), and the rotation rates of galaxies by red/blue shift.

dav57, I don't even know what you're original problem was. Too much of a mess. If an object is receeding from you, no matter what it is (mirror, whatever) is appears red shifted.

by superluminal:

"Any color mix you see will shift accordingly and the percieved color will change, hence the ability to judge astronomical distances by redshift..."
================================================== =============

Sorry, superluminal, although that is the way Doppler shift is portrayed in the popular press and often on forums, that is not the way it works. Ask James R, I feel sure he can explain it better than myself. I am speaking of astronomical objects that we can see that emmit EM radiation in many frequencies. I don't know about a mythical blue
wall inside a spaceship or whatever, but the discussion seemed to be speaking of color
shifts due to velocity later in the thread.

Umm... whatever.

Look, unless the body is radiating uniformly in a very broad chunk of spectrum, the percieved color will shift.

ALL frequencies are shifted, so there is no visible difference in color unless the velocities involved are ultra fast. Velocities can become great enough, according to STR, that all EM radiation is shifted to gamma and radio wave frequencies only, detected as a 'point' in the direction of travel due to relativistic aberration, as mentioned by James R in another thread.
2Inq, now thats an interesting conindrum you are showing. Correct me if I am mistaken but what you are saying is that the whole EM spectrum is shifted.....sheesh, !! And that means that there would be no change in colour. Facinating ....hmmmmmmm.....so blue would shift down and uV would shift down to Blue.....red would shift down to IR.....and somethng would shift down to red......so no net effect is observed assuming a uniform spectrum.

Is this correct to what you are saying?

QQ,

Only if the object is emitting uniformly across enough of the spectrum to cover the shift. It would necessecarily appear white to us.

..............|...vis...|
_______------------------_____ white (unshifted)

__------------------_____ still white (red shifted)

________|___||________ purplish (unshifted)

______|___||________ green/yellow (red shifted)

Make any sense? (need sketch function...)

I think so, but it is certainly something I hadn't considered before.....

The color you see will be whatever spectral content falls in the "visible" window of the spectrum as the frequencies shift.

would you agree that every spacetime co-ordinant has a unique EM signature?
Maybe too difficult to answer but I thought I'd ask.

Well, since there are an infinite number of spacetime coordinates (x,y,z,t) I would say it's impossible to say.

Logically one could assume this I guess, like moving a TV aerial around in a room trying to get the best picture. Each position in the room recieving a unique to that location EM signal......hmmmmmmm

Ok.