shape of a relativistic wheel

[Tach]

No it's not. It is not always necessary to calculate two quantities to show that one is larger than the other. In this case it simply follows from the fact that energy increases strictly monotonically with speed.

I have already proven your claim false.

This calculation that's so simple you never do it?

So you cannot read the post just above? You have had the tools to do this calculation all by yourself for a couple of days now. Had you done it instead of waving your arms, you would have known that your claims are empty.

 

[Tach]

BTW, Tach, this "zero Doppler shift" is only valid for mirrors. An object such as a colored, spoked wheel would indeed appear Doppler shifted. Your comment is therefore irrelevant and in the context of this thread, false.

did your armchair teach you that as well? you got to stop taking lessons from it , it is giving you bad information

 

[RJBeery]

did your armchair teach you that as well?

Yes, sometimes the things I find under this cushion are quite valuable. I've said it before, you have an unusual blindness if you MUST see the math to see the truth.

 

[Tach]

Yes, sometimes the things I find under this cushion are quite valuable.

well, you should fire your armchair as your tutor, it's been teaching you wrong stuff

 

[RJBeery]

You're killing me with the post edits. I don't know if I should reply by editing my responses or making a new post...

Anyway, what I say is true.

 

[Tach]

You're killing me with the post edits. I don't know if I should reply by editing my responses or making a new post...

Anyway, what I say is true.

The fact that it is wrong should not preclude you from posting more of the same, it has never stopped you in the past. seriously, you should stop taking cues from your armchair.

 

[RJBeery]

Behold the power of thought: colored objects are colored because they only reflect certain frequencies. Subtle shifts in the the spectrum of white light used to illuminate it will not change the emitted color of the object under consideration. Only a mirror would perfectly reflect the incoming frequencies.

 

[przyk]

I have already proven your claim false.

What claim? That energy increases monotonically with velocity? You haven't proven anything false - just dismissed everything I said out of hand.

So you cannot read the post just above?

Yes. What about it? You haven't done any calculations. Just asserted results. And from your results it's clear that either you've gotten things horribly messed up or you aren't even considering the same problem everyone else is. The energies of the top and bottom halves should not be time-dependant for instance.

In any case I'm at work now, so I'll take a closer look at it later.

 

[Tach]

Yes. What about it? You haven't done any calculations. Just asserted results.

Err , all the steps are in the previous posts, you only need to do a simple integral. You have been provided the formulas and their derivations. You had about two days to complete the LAST step, instead of waving your arms you could have done so.

And from your results it's clear that either you've gotten things horribly messed up or you aren't even considering the same problem everyone else is. The energies of the top and bottom halves should not be time-dependant for instance.

Because you say so? The tangential speed is time dependent, therefore, so is the total energy. In the frame of the axle, the energies are equal and they are not time dependent. In any other frame, the energies ARE time dependent. They oscillate in time , as I pointed out to you quite a few posts ago. But you wouldn't listen.

In any case I'm at work now, so I'll take a closer look at it later.

Please do so. You had all the time in the world and you had the derivation up to the last step. I am sure that you will come back with a last gasp effort to prove yourself right and me wrong. I am going skiing, so I will be out until Monday, so you have at least two days to figure how you could possibly miss. I won't be responding until Monday.

 

[Tach]

Behold the power of thought: colored objects are colored because they only reflect certain frequencies.

And those frequencies still behave like good citizens and they do get Doppler shifted, no?

Subtle shifts in the the spectrum of white light used to illuminate it will not change the emitted color of the object under consideration. Only a mirror would perfectly reflect the incoming frequencies.

You are confusing "not reflecting all frequencies" with "non having the reflected frequencies Doppler shifted". Why do you keep pushing your foot deeper and deeper in your mouth? I am going skiing, so I won't be answering. In the meanwhile, reflect on the errors in your posts. Maybe it is time to take a break?

 

[RJBeery]

And those frequencies still behave like good citizens and they do get Doppler shifted, no?



You are confusing "not reflecting all frequencies" with "non having the reflected frequencies Doppler shifted". Why do you keep pushing your foot deeper and deeper in your mouth? I am going skiing, so I won't be answering. In the meanwhile, reflect on the errors in your posts. Maybe it is time to take a break?

Jesus Tach if we can't agree on this then you're truly lost. A green wheel will emit green light, period. Move the wheel relative to a light source and you're only shifting the average frequency of the illuminating white light. As long as the wheel continues to be illuminated by the proper (green) frequency of light it will continue to emit that green light, and absorb the rest.

The apparent color of the wheel is indeed dependent on the the movement of the observer (i.e. camera) relative to the wheel. Conclusion: the objects under consideration in this thread do appear Doppler shifted. Enjoy the snow!

 

[Tach]

Jesus Tach if we can't agree on this then you're truly lost. A green wheel will emit green light, period.

Correction, when illuminated with white light it will REFLECT ONLY frequencies in the green spectrum.

Move the wheel relative to a light source and you're only shifting the average frequency of the illuminating white light.

White light is made of ALL frequencies, INCLUDING green.

Conclusion: the objects under consideration in this thread do appear Doppler shifted.

Nope, because the frequency (green , in this case) is equally blueshifted when reaching the object and redshifted (by the same amount) when arriving at the camera, as I explained in the file that pete had so much trouble understanding.

Enjoy the snow!

Thank you, bye. Reflect on what I've been showing you.

 

[RJBeery]

Nope, because the frequency (green , in this case) is equally blueshifted when reaching the object and redshifted (by the same amount) when arriving at the camera, as I explained in the file that pete had so much trouble understanding.

I hesitate to respond until I know you're done fiddling with what you've written...

Anyway, the incident blueshifted green light would not get reflected! The color that got reflected would be the incident yellow light (for example) that happened to get blueshifted into the proper frequency so as to appear green to the wheel. The wheel would emit green, period. The camera would determine the wheel's color based upon the Doppler shifting caused by the relative motion between the objects.

 

[przyk]

Err , all the steps are in the previous posts, you only need to do a simple integral.

No Tach, you don't just post "steps" and tell people what calculations to do. Giving people orders, which you are in absolutely no position to do, does not constitute a proof or justification of any kind. If anything I'm inclined not to do them just because it's incredibly arrogant (you show no openness to the possibility you might be doing the wrong calculations in the first place) and plain rude.

Because you say so? The tangential speed is time dependent, therefore, so is the total energy.

Funny how you leave out any qualifications in that. The tangential speed of some given element of the wheel, as you follow it around while the wheel rolls along, has a constantly varying speed. But the velocity profile of the wheel is static. If the wheel is rolling at velocity $$v$$, the point at the top of the wheel always has a speed $$\frac{2v}{1+\frac{v^{2}}{c^{2}}}$$ while the point at the bottom in contact with the ground has zero speed for instance.

 

[RJBeery]

I find it difficult to believe that TACH believes that anyone was actually discussing different energies for each ROTATING half. Intentional obfuscation.

 

[Neddy Bate]

Trying to intellectually engage Tach is an exercise in futility. Those who have dared to try have my utmost sympathy.

He has made it clear in this thread that he does not understand the difference between visual effects, and Relativity of Simultaneity (ROS) effects. The reason there are more spokes in the top half of the wheel, (according to the reference frame of the road), is because of ROS, not because of visual (ray tracing) effects.

I'm not even sure if he believes in the Doppler effect at all, based on his inane argument about mirrors.

He also seems to think that, according to reference frame of the road, there is just as much energy in the top half of the wheel as there is in the bottom.

It's rather amazing, because he certainly is excellent at maths. How he can do the maths without understanding the concepts behind them is a mystery to me.

 

[Masterov]

Can you show a fixed wheel which rotates in the plane of the monitor screen?

The animations given earlier are the best I can provide, but these not only account length contraction but other effects such as terrell rotation and ray tracing. This wheel looks 100% normal.

How much change diameter and length of rim of the static wheel for variable speed of rotation (in the plane of the monitor screen) of its.

 

[Pete]

Good point, Mr Beery.

Proposition for you, Tach.

As you have pointed out with all the humility and politeness we've come to expect from you, my criticism in this thread of your doppler shift analysis has been woefully inadequate. Clearly, I have something to learn.

So, after my exams are done, how about you and I explore this moving mirror doppler shift problem properly in the formal debates forum?
I'll take the position that light reflected from a moving mirror will generally be doppler shifted.
You take the position that light reflected from a moving mirror is never doppler shifted.
Loser (as judged by a mutually agreed adjudicator - please suggest someone) has to be perfectly polite to all posters at all times for 3 months, or be banned for that time.
Or perhaps you can think of a different penalty involving avatar images and associated text?

What say you?

 

[James R]

Does Tach really think that light reflected from a moving mirror (say one that is heading directly towards the source) will not be Doppler shifted when it arrives back at the source?

Really?

I'll take Tach on right now in a formal debate on that topic. It will only take one post to demolish him and a few lines of maths.

 

[Pete]

His proof, such as it is, is here:
https://docs.google.com/viewer?a=v&...N2E4MC00NDc3LWE1ZjctNDA5YjVjZmM5NWIy&hl=en_US

After a couple of half-assed dismissals, I had a better look and said...

In your diagram, you have the object moving at velocity V.
Since you said you're working in the object rest frame, I'll call that a typo and pretend that V is attached to the camera and the light source.
Also, you have the primary ray pointing at the camera instead of the source.
And you don't need separate scenarios for approaching and receding... it's OK to let $$\phi$$ vary from 0 to 2pi.

Now, $$\phi$$ is the angle between V and the primary ray, which I think you've correctly used in your formula for $$f_{mirror}$$.

But you have used $$\phi$$ again in your formula for $$f_{s'}$$.
This is not right.
In the $$f_{s'}$$ equation, you should be using $$\phi + 2\theta$$, the angle between V and the secondary ray.

But don't take my word for it... ask your professor. I'm sure they will be happy to help you with the doppler shift from a moving mirror.

Tach's response is in [post=2850347]post 214[/post]. Excerpt:

The angle between the speed V and the secondary ray and the angle between the speed V and the primary ray are the SAME ($$\phi$$) due to geometric symmetry. Definitely not the $$\phi + 2\theta$$ you are claiming. Claiming such a thing shows that you don't even understand the basics of the relativistic Doppler effect.

He also gave the impressions that's it's impossible for the mirror to be approaching both the source and the camera ([post=2850195]post 194[/post], but I didn't get that clarified.
He might also try to hedge by talking about raytracers vs reality.