I’m hoping that someone will forgive my ignorance and solve this little conundrum for me. I have lifted this problem from another thread due to the muddle the thread got in to.

Basically, I’ve got two separate experiments / scenarios which I’d like to explain and both involve two stationary planets with respect to each other, a mirror and a bit of Doppler shifting.

For the sake of argument, I’d like to stick to just three frequencies: UV, BLUE and INFRARED. It doesn’t really matter whether this is strictly correct for the sake of my experiments. What matters is how many steps the frequency changes in a particular scenario.

Ok.

Experiment 1.

We have a planet A and a Planet B stationary with respect to each other.
There is a mirror sitting stationary at planet A.
Planet B beams an IR light beam at the mirror.
The mirror reflects the IR beam back to B, who view the beam as IR.

Ok so far?

We now accelerate the mirror towards B at 0.9c (or some arbitrary high velocity – doesn’t matter if the figures don’t work properly, it’s the principle that counts)
The IR beam that B has emitted is reflected back from the mirror with a Doppler shift. Therefore B sees BLUE. THIS IS AN IMORTANT FACT THAT I NEED CLARIFYING PLEASE. Basically, due to the mirror’s velocity of 0.9c, lets say that the IR beam is turned into BLUE as far as B is concerned.

Right.

Experiment 2.

We now have a box painted BLUE inside and out heading at 0.9c from A to B. The box has a mirror on its inside trailing face (so the mirror is facing B).
There is a hole in the front and back of the box such that the mirror can be viewed from B and that a light beam can be sent straight out the back if so desired.
The light leaving the front internal leading face is Doppler shifted to IR as it heads backwards towards the mirror. We KNOW it is shifted to IR because planet A receives it as IR and can reflect it back to B from a STATIONARY mirror to prove it is IR.
Now, this light beam (which is IR) is reflected in the moving mirror (exactly as in experiment 1) and thus is Doppler shifted by the same amount (up to BLUE) which is then received by B. I can see no difference between this experiment and experiment 1.
Planet B views the front of the box as having Doppler shifted from BLUE up to UV due to its relative motion.
Therefore, Planet B sees the box as being UV but the mirror as being BLUE.


Is this right. If not, what in Heavens name is the difference between Exp 1 and Exp 2 ? :confused:

The light leaving the front internal leading face is Doppler shifted to IR as it heads backwards towards the mirror. We KNOW it is shifted to IR because planet A receives it as IR and can reflect it back to B from a STATIONARY mirror to prove it is IR.

Prove it according to who?

A sees the entire box (mirror and all) red shifted to the IR. Yes indeed. And B sees the entire box (mirror and all) blue shifted to the UV.

But to passengers in the box, A is seen red shifted to IR and B is seen blue shifted to UV and all is normal in the box. Peaceful blue. Ever hear a fire engine go by? Notice the strong doppler shift? Do the firefighters hear a dopple shift on the engine? Nope. Even sound reflected from the fire engine is "red" and "blue" shifted to bystanders.

The wavelength of light is a measure of its energy. Shorter wavelength, higher energy. Now energy is relative. Don't think so? How much energy does a rock have in your hand, wrt you? Basically zero right? How much does it have wrt your buddy at the bottom of a cliff? Drop it and ask him. Right. So, the light reflected from the box loses/gains energy wrt A and B, but how much energy does the light lose/gain wrt passengers moving with the box? Zero.

Hope this was a bit helpful.

Prove it according to who?

A sees the entire box (mirror and all) red shifted to the IR. Yes indeed. And B sees the entire box (mirror and all) blue shifted to the UV.

But to passengers in the box, A is seen red shifted to IR and B is seen blue shifted to UV and all is normal in the box. Peaceful blue. Ever hear a fire engine go by? Notice the strong doppler shift? Do the firefighters hear a dopple shift on the engine? Nope. Even sound reflected from the fire engine is "red" and "blue" shifted to bystanders.

The wavelength of light is a measure of its energy. Shorter wavelength, higher energy. Now energy is relative. Don't think so? How much energy does a rock have in your hand, wrt you? Basically zero right? How much does it have wrt your buddy at the bottom of a cliff? Drop it and ask him. Right. So, the light reflected from the box loses/gains energy wrt A and B, but how much energy does the light lose/gain wrt passengers moving with the box? Zero.

Hope this was a bit helpful.

Superluminal,

Thanks, but I'm not sure you've quite seen my point.

Firstly, I'm only interested in what A and B sees. I'm not really interested in what is viewed at the box.

Everneo says that an observer standing on Planet B sees the front of the box blue-shifted from BLUE to UV (which I agree with). He also maintains that the mirror at the back of the box reflects the front internal BLUE wall and observer at B sees this shifted also to UV (which I don't agree with because of what happens in experiment 1).

Now that is my problem you see, because the ray of light leaving the front internal wall of the box is RED-SHIFTED from BLUE to INFRARED. We know this is the case because an observer at A views this beam as INFRARED and could reflect it back to B as INFRARED to "prove" it was INFRARED. So it MUST be INFRARED according to planet A and B.

Therefore, as in experiment 1, the moving mirror will doppler shift the INFRARED light back to BLUE. So B sees BLUE in the mirror, NOT UV.

Do you agree that experiment 1 turns the INFRARED light into BLUE and NOT UV using a mirror travelling at 0.9c?

Do you agree that the moving box emits an INFRARED beam back to A and that this is viewed by A as being IR?

So if we simply introduce a 0.9c moving mirror (which happens to be on the back of the box) this should turn the IR beam back to BLUE (just as it did in experiment 1).

But Everneo says it doesn't and I can't see the difference here and I'm fed up :(

Ok.

Firstly, I'm only interested in what A and B sees. I'm not really interested in what is viewed at the box.
Cool.

Everneo says that an observer standing on Planet B sees the front of the box blue-shifted from BLUE to UV (which I agree with). He also maintains that the mirror at the back of the box reflects the front internal BLUE wall and observer at B sees this shifted also to UV (which I don't agree with because of what happens in experiment 1).
But he is right nevertheless. In your experiment 1, the light was emitted from B as IR and gained energy from reflecting off of the moving mirror becoming blue. I don't see the problem.

Now that is my problem you see, because the ray of light leaving the front internal wall of the box is RED-SHIFTED from BLUE to INFRARED.

Only according to A.

We know this is the case because an observer at A views this beam as INFRARED and could reflect it back to B as INFRARED to "prove" it was INFRARED. So it MUST be INFRARED according to planet A and B.
Fine.

Therefore, as in experiment 1, the moving mirror will doppler shift the INFRARED light back to BLUE. So B sees BLUE in the mirror, NOT UV.
It's not infrared in the box. It's blue. It only appears infrared to A because it's doppler shifted wrt A.

Do you agree that experiment 1 turns the INFRARED light into BLUE and NOT UV using a mirror travelling at 0.9c?
Yes.

Do you agree that the moving box emits an INFRARED beam back to A and that this is viewed by A as being IR?
Yes.

So if we simply introduce a 0.9c moving mirror (which happens to be on the back of the box) this should turn the IR beam back to BLUE (just as it did in experiment 1).
No. The mirror is moving with the box and so reflects blue light from the box.

You are forgetting that energy (wavelength) is relative. Remember the fire truck?

A and B can only see what comes out of the box. And what happens in the box is all blue. To A, every wavelength is red shifted, and to B they are blue shifted.

You are thinking that because A says the light coming from the box is IR that it's IR in the box. Not true. And that it should be shifted to blue for B observers. Nope. It's blue in the box. When it comes out in B's direction it's UV. When it comes out in A's direction, it's IR.

But Everneo says it doesn't and I can't see the difference here and I'm fed up

Don't get fed up. We'll putz around with this for weeks till we get it right! :)

Hi dav,
I think we need to do the maths to be sure, but here's my suspicion in a nutshell:

If an IR beam from B is reflected by the box's mirror and doppler shifted to BLUE according to B...

...then BLUE light emitted from the box to A will not be shifted all the way down to IR. It will only get to RED.


In other words, I think the problem with your scenario is with this premise:
The light leaving the front internal leading face is Doppler shifted to IR as it heads backwards towards the mirror. We KNOW it is shifted to IR because planet A receives it as IR and can reflect it back to B from a STATIONARY mirror to prove it is IR.
Try this instead:
"The light leaving the front internal leading face is Doppler shifted to RED as it heads backwards towards the mirror. We KNOW it is shifted to RED because planet A receives it as RED and can reflect it back to B from a STATIONARY mirror to prove it is RED."

Formulas we could use (I don't pretend to understand these properly - I've only just now looked them up):

Doppler shift formula for light reflected from a uniformly moving mirror:

fo = fs (1 - 2v/c + v²/c²) / (1 - v²/c²)

fo = observed frequency
fs = source frequency
v = velocity of the mirror relative to the observer
c = light speed


Doppler shift formula for light emitted from a uniformly moving source:

fo = fs (1 + v/c) / (1 - v/c)

fo = observed frequency
fs = source frequency
v = velocity of the light source relative to the observer
c = light speed

Pete,

Confirm me on this. In the box all is blue. Any blue radiation from the box will be seen as IR by A and UV by B. No matter how it is reflected or otherwise escapes th box. Right? (again, the actual IR/UV values are just for argument's sake).

Yes, the radiation from the box will be redder to A and bluer to B.

Ok, thanks for the help guys. I hope you understand my problem, but just to be sure (and for the reader) I’ll reiterate the main crux of what I want to get to the bottom of…..

You see, in both experiments, one way or another there is an IR ray of light beamed backwards to observer A. i.e. Observer A views the inside BLUE wall of the moving box as IR, or indeed in Exp 1, a simple IR ray beamed from Observer B. Regardless, the beams according to A and B are IR.

So at this point, do you understand that Observer A always sees IR light, regardless of which experiment is undertaken? Everneo has agreed with this.

So if Observer A then reflects this IR light back to B (regardless of which experiment created the IR light), B sees IR light. Yes? Everneo agreed with this.

Now for the tricky bit….

If in Experiment 1 (without the box) we simply introduce a 0.9c mirror travelling away from A and facing B, the reflected IR light would be Doppler shifted to BLUE and B would view BLUE. Ok? Everneo agreed with this.

Now for the REALLY tricky bit….

So in experiment 2 (with the box), we also KNOW we have an IR beam because A and B have seen this IR beam and this IR beam is indistinguishable from the IR beam in Exp 1, according to A or B. And again Everneo agreed with this.

The bit I have the massive problem with is that the mirror in Exp 2 is travelling at EXACTLY the same speed (0.9c) as in Exp 1 and is also receiving an IR beam (all relative to A and B), but due to Doppler it should treat the IR beam exactly the same as in Exp 1, thus B should see the reflection as BLUE, not UV.

So,
Same IR beam according to A and B,
Same mirror,
Same speeds involved
Same observers

So why does Everneo say that for Exp 1 the IR beam is reflected back to B as BLUE (by the moving mirror), and yet in Exp 2 the IR beam (which A and B see as IR) is reflected back as UV?

Something weird is going on here.

I'm not sure everneo is on the money... or perhaps I'm mistaken.
We don't always agree, you know - we're learning relativity together.

You see, in both experiments, one way or another there is an IR ray of light beamed backwards to observer A. i.e. Observer A views the inside BLUE wall of the moving box as IR, or indeed in Exp 1, a simple IR ray beamed from Observer B. Regardless, the beams according to A and B are IR.

With those conditions, and the condition that v (the velocity of the mirror/box) is the same for experiment 1 and experiment 2, then perhaps the reflected beam in experiment 1 will be shifted to UV, not to BLUE?

Shall we do the maths and find out?

Or, perhaps you want to dictate conditions like this:

Experiment one is the calibration for the box colour.
B emits IR, and receives BLUE from the mirror moving at v. BLUE is reported to A as the colour to paint the box.

Experiment two:
The mirror is sent off in the BLUE box at v once more.


Now what?
Now, we don't know what A sees. Perhaps the BLUE light is red-shifted to RED, not to IR?

If so, then for B the BLUE direct light is shifted to UV, and the RED light hiting the moving mirror is also shifted to UV.



Do you see? You can't dictate all the conditions you're trying to... some of them must be determined in some way, by experiment or theory.

Perhaps the key is that light reflected from an approaching mirror is blueshifted more than light emitted from an approaching source?
That's kind of loose language, so I hope you know what I mean.

So why does Everneo say that for Exp 1 the IR beam is reflected back to B as BLUE (by the moving mirror),

I think, you are missing something here. The moving mirror just reflects what it recieves, that is BLUE. It is B who sees the reflected BLUE light as UV after blueshift.

For B the incident light on the moving mirror is IR but not for the moving mirror. Thats why i asked you to sit on top of the moving mirror (it is thought experiment anyway) and note what is the color of light approaching you.

The incident light on the moving mirror is BLUE for you, IR for B.
But the mirror does not lie. It just reflects what it recieves.
So, after reflection it is again BLUE for you but UV for B after blue-shift.

That is why you should never see a really fast moving mirror that reflects something else than what you expect!

Hope this removes the ghost.

I'm not sure everneo is on the money.



With those conditions, and the condition that v (the velocity of the mirror/box) is the same for experiment 1 and experiment 2, then perhaps the reflected beam in experiment 1 will be shifted to UV, not to BLUE?

Shall we do the maths and find out?

Pete, I will be forever in your debt if you can help me sort this out :cool:


I'm perhaps not so concerned with the figures but rather the fact that the two scenarios should have the same outcome as far as observer A and B are concerned. For the sake of argument, I'm happy to stick with BLUE, IR and UV as the doppler shifts as a result of, say 0.9c, or whatever.

Also, I'm not the foggiest bit interested in what the box sees. All we need to know is that it is painted BLUE.

Perhaps the key is that light reflected from an approaching mirror is blueshifted more than light emitted from an approaching source?
That's kind of loose language, so I hope you know what I mean.

Now, you see my problem. I'm wondering if this is the case too. But WHY?

Or, perhaps you want to dictate conditions like this:

Experiment one is the calibration for the box colour.
B emits IR, and receives BLUE from the mirror moving at v. BLUE is reported to A as the colour to paint the box.

Experiment two:
The mirror is sent off in the BLUE box at v once more.


Now what?
Now, we don't know what A sees. Perhaps the BLUE light is red-shifted to RED, not to IR?

If so, then for B the BLUE direct light is shifted to UV, and the RED light hiting the moving mirror is also shifted to UV.



Do you see? You can't dictate all the conditions you're trying to... some of them must be determined in some way, by experiment or theory.


Pete, you're getting all complicated!

We have two stationary observers, a box painted BLUE (cuz I painted it), and a couple of mirrors. One mirror is in the possession of A and one mirror is fixed to the back of the box and facing B.

Both mirrors recieve doppler-shifted light which A receives as IR.

A reflects this light back to B as IR radiation.

When mirrors move at 0.9c, they are both reflecting IR light so both experiments should render the same outcome i.e. the IR beam is turned back to BLUE for B to view as BLUE.

I think, you are missing something here. The moving mirror just reflects what it recieves, that is BLUE. It is B who sees the reflected BLUE light as UV after blueshift.

For B the incident light on the moving mirror is IR but not for the moving mirror. Thats why i asked you to sit on top of the moving mirror (it is thought experiment anyway) and note what is the color of light approaching you.

The incident light on the moving mirror is BLUE for you, IR for B.
But the mirror does not lie. It just reflects what it recieves.
So, after reflection it is again BLUE for you but UV for B after blue-shift.

That is why you should never see a really fast moving mirror that reflects something else than what you expect!

Hope this removes the ghost.

Thanks for joining us Ev,

I understand exactly what you are saying but ALL my reference points and frames of reference are ALWAYS taken from A or B's perspective.

The fact is that the beam travelling backwards from the inside leading face of the box is undisputedly IR, as seen from A or B.

This beam of light would look exactly the same as a normal IR beam directly beamed from B to A.

Thus, a mirror travelling at 0.9c would doppler shift the normal beam to BLUE.

So why not the other beam?

And I promise not to call you a knob-head from now on :rolleyes:

Thus, a mirror travelling at 0.9c would doppler shift the normal beam to BLUE.

To avoid merry-go-around, please answer, what is the color of the incident & reflected light from the frame of moving mirror? Forget the planets A and B for the time being.

Pete, you're getting all complicated!

We have two stationary observers, a box painted BLUE (cuz I painted it), and a couple of mirrors. One mirror is in the possession of A and one mirror is fixed to the back of the box and facing B.

Both mirrors recieve doppler-shifted light which A receives as IR.

A reflects this light back to B as IR radiation.

When mirrors move at 0.9c, they are both reflecting IR light so both experiments should render the same outcome i.e. the IR beam is turned back to BLUE for B to view as BLUE.

OK, now you have a different set of conditions:

1) The box is BLUE (it emits BLUE light)
2) The box is moving at a speed such that the BLUE light emitted to A is red-shifted to IR. Note that this isn't necessarily the same as the speed of the mirror in experiment one! (I think it must be faster, if the doppler theory is consistent.)

So now you don't know in advance what B sees... the blue-shift of the reflected IR beam must be determined by the first two conditions.
Wouldn't it make sense that the reflected IR beam is blue-shifted all the way to UV?

To avoid merry-go-around, please answer, what is the color of the incident & reflected light from the frame of moving mirror? Forget the planets A and B for the time being.

Everneo,

I know that from within the box the beam hitting the mirror is BLUE. And I wholeheartedly agree 100%.

But that’s not the crux of my problem because I have never wanted to take any measurements from the box’s frame of reference and I don’t intend to.

You see, all my observations are made from either A or B.

We KNOW that the beam leaving the front internal box and heading backwards towards A is red-shifted to IR and leaves the box as IR with respect to A.

We KNOW this is true because both A and B would see this beam as IR whether it be viewed directly or via a mirror sitting stationary at A.

Basically, from the frame of reference of either A or B, the beam leaving the back of the box is undisputedly IR.

This IR beam is identical in appearance to that of a regular IR beam emitted directly from B to A. A would simply reflect it back to B using a stationary mirror.

So why then, when we introduce a 0.9c mirror are you saying that one mirror Doppler-shifts the IR to BLUE (as in experiment 1) but to UV when considering the box?

Remember, we have two indistinguishable IR beams arriving at A and you’re saying that a moving mirror will affect them differently and I need to go out and get hammered!

So why then, when we introduce a 0.9c mirror are you saying that one mirror Doppler-shifts the IR to BLUE (as in experiment 1) but to UV when considering the box?

I remember saying, the mirror moving at 0.9c would no more recieve IR but BLUE. It recieves BLUE. It just reflects back BLUE that would appear to be UV for B. It is simple.

No matter whether the mirror is inside the box or co-moving at same velocity with the box at some distance. effects are same.

I understand your dilemma. The IR moving towards the moving mirror gets a double blue-shift and appears UV to B, right? You can imagine, each blue-shift happens before and after reflection because of the movement of mirror.

Thats why i warned you, never trust a moving mirror, it might not reflect what you expect. ;)

I remember saying, the mirror moving at 0.9c would no more recieve IR but BLUE. It recieves BLUE. It just reflects back BLUE that would appear to be UV for B. It is simple.

No matter whether the mirror is inside the box or co-moving at same velocity with the box at some distance. effects are same.

I understand your dilemma. The IR moving towards the moving mirror gets a double blue-shift and appears UV to B, right? You can imagine, each blue-shift happens before and after reflection because of the movement of mirror.

Thats why i warned you, never trust a moving mirror, it might not reflect what you expect. ;)

Ok, so the mirror introduces a double blue-shift and turns the IR into UV. Ok, I can live with that.

So, in experiment 1, Does the same apply? Does the IR beam (which is beamed straight from B) hit the 0.9c mirror as IR and get double shifted to UV?

the IR beam (which is beamed straight from B) hit the 0.9c mirror as IR as viewed from B, but the moving mirror gets the blue-shifted* IR, say BLUE and reflects the same, and B gets the reflection as UV after another blue-shift**

*first blue-shift is for the moving mirror.
**second blue-shift is for B due to the movement of mirror.

there was no need of a blue box here, white would have done better