Misleading Frame of Reference in Special Relativity

[Sensei]

Misleading Frame of Reference in Special Relativity.

 

Suppose so we're on board of rocket R, traveling in cosmos from object B to object A.

To simplify let's assume that A and B are stationary each other.

 

 

Object A is sending red photons with wavelength 700 nm.

Object B is sending violet photons with wavelength 400 nm.

(you can choose any other pair)

 

Rocket after acceleration can reach such velocity, that red photons from source A are blueshifted to the same energy as violet photons from source B are redshifted.

 

Blue shift equation in Special Relativity:

[math]f=f_0 \sqrt{\frac{1+v}{1-v}} = f_0 (1+v) \gamma[/math]

 

Red shift equation in Special Relativity:

[math]f'=f_0 \sqrt{\frac{1-v}{1+v}} = f_0 (1-v) \gamma[/math]

 

We need situation when f=f' to solve for v.

 

[math]f_R \sqrt{\frac{1+v}{1-v}} = f_V \sqrt{\frac{1-v}{1+v}}[/math]

 

[math]f_R (1+v) \gamma = f_V (1-v) \gamma[/math]

 

[math]f_R (1+v) = f_V (1-v)[/math]

 

[math]f_R+f_R v = f_V - f_V v[/math]

 

[math]f_R v + f_V v = f_V - f_R[/math]

 

[math]v(f_R + f_V) = f_V-f_R[/math]

 

[math]v=\frac{f_V-f_R}{f_R + f_V}[/math]

 

Convert [math]\lambda_R \lambda_V[/math] to [math]f_R f_V[/math].

 

And solve equation.

 

It'll give v=0.27272727c

 

[math]f=f_0 \sqrt{\frac{1+0.27272727}{1-0.27272727}}= 529.15 nm[/math]

 

If somebody would wake up, have amnesia, born etc. in the middle of travel,

and would see green photon from A, and green photon from B,

 

or even entire blackbody spectrum from A, and from B,

but at completely different temperatures (thus different peak),

 

he/she could get (misleading) conclusion that there is no relative motion,

and either A, B and R are stationary each other.

[swansont]

It's only misleading if you assume you know the wavelength of the source is the same.

 

I also don't think it works for a blackbody. The blackbody spectrum is given by a width as well as a peak, so you could notice that the spectrum doesn't look right.

[J.C.MacSwell]

It's only misleading if you assume you know the wavelength of the source is the same.

 

I also don't think it works for a blackbody. The blackbody spectrum is given by a width as well as a peak, so you could notice that the spectrum doesn't look right.

Isn't that one of the things about the blue/red shifting of the blackbody spectrum...the peak shifts but all the points still look right?

[swansont]

Isn't that one of the things about the blue/red shifting of the blackbody spectrum...the peak shifts but all the points still look right?

 

 

Whether you see it depends on how big the shift is. Sensei is proposing a very large shift. It also depends on the temperature of the blackbody.

 

http://quantumfreak.com/wp-content/uploads/2008/09/black-body-radiation-curves.png

 

If you blue- or red-shifted these curves, they would not look identical.

[michel123456]

Misleading Frame of Reference in Special Relativity.

 

Suppose so we're on board of rocket R, traveling in cosmos from object B to object A.

To simplify let's assume that A and B are stationary each other.

 

Rocket Misleading FoR.png

 

Object A is sending red photons with wavelength 700 nm.

Object B is sending violet photons with wavelength 400 nm.

(you can choose any other pair)

 

Rocket after acceleration can reach such velocity, that red photons from source A are blueshifted to the same energy as violet photons from source B are redshifted.

 

Blue shift equation in Special Relativity:

[math]f=f_0 \sqrt{\frac{1+v}{1-v}} = f_0 (1+v) \gamma[/math]

 

Red shift equation in Special Relativity:

[math]f'=f_0 \sqrt{\frac{1-v}{1+v}} = f_0 (1-v) \gamma[/math]

 

We need situation when f=f' to solve for v.

 

[math]f_R \sqrt{\frac{1+v}{1-v}} = f_V \sqrt{\frac{1-v}{1+v}}[/math]

 

[math]f_R (1+v) \gamma = f_V (1-v) \gamma[/math]

 

[math]f_R (1+v) = f_V (1-v)[/math]

 

[math]f_R+f_R v = f_V - f_V v[/math]

 

[math]f_R v + f_V v = f_V - f_R[/math]

 

[math]v(f_R + f_V) = f_V-f_R[/math]

 

[math]v=\frac{f_V-f_R}{f_R + f_V}[/math]

 

Convert [math]\lambda_R \lambda_V[/math] to [math]f_R f_V[/math].

 

And solve equation.

 

It'll give v=0.27272727c

 

[math]f=f_0 \sqrt{\frac{1+0.27272727}{1-0.27272727}}= 529.15 nm[/math]

 

If somebody would wake up, have amnesia, born etc. in the middle of travel,

and would see green photon from A, and green photon from B,

 

or even entire blackbody spectrum from A, and from B,

but at completely different temperatures (thus different peak),

 

he/she could get (misleading) conclusion that there is no relative motion,

and either A, B and R are stationary each other.

Yes I think you are correct. If the rocket travels at constant velocity.

one will observe the red/blueshift because of the characteristic spectral lines of hydrogen.

[Strange]

Yes I think you are correct. If the rocket travels at constant velocity.

one will observe the red/blueshift because of the characteristic spectral lines of hydrogen.

 

The photons of 700 and 400 nm will not have the spectrum of hydrogen. They are single frequencies.

[michel123456]

The photons of 700 and 400 nm will not have the spectrum of hydrogen. They are single frequencies.

Dear Strange, you are so intelligent but sometimes I don't know what to say....

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Sorry, that must be me.

 

I mean, you need extra information to solve the problem. Only with frequencies you cannot determine your velocity.

Edited July 26, 2016 by michel123456

[swansont]

Dear Strange, you are so intelligent but sometimes I don't know what to say....

----------------------------------------------------

Sorry, that must be me.

 

I mean, you need extra information to solve the problem. Only with frequencies you cannot determine your velocity.

 

 

If you identified the hydrogen spectrum pattern you could, because then you know the frequency of the source.

[michel123456]

 

 

If you identified the hydrogen spectrum pattern you could, because then you know the frequency of the source.

Exactly.

 

Otherwise, without that information you will observe light (red, green blue wathever) without getting any information about the velocity of the source.

Where is Sensei?