Special Relativity, transversal movement and the first postulate

[md65536]

From whose perspective? think about it.

theta_s is the angle in the source frame.

 

Your error is here:

Small correction:

 

[math]\frac{f_o}{f_s}= \gamma (1-\beta cos \theta_s )[/math]

 

Based on the above, redshift occurs for all:

 

[math]\frac{f_o}{f_s}<1[/math]

 

meaning for all

 

[math]\theta>\theta_{critical}=arccos \frac{1-\sqrt{1-\beta^2}}{\beta}[/math]

How can redshift occur for smaller values of f_o, but greater values of theta_s?

Try increasing theta_s. f_o increases. Do you understand what that means?

The formula you solved is for theta_s, not theta_o.

[xyzt]

theta_s is the angle in the source frame.

 

Your error is here:

How can redshift occur for smaller values of f_o, but greater values of theta_s?

Try increasing theta_s. f_o increases. Do you understand what that means?

 

I asked you a question, you didn't answer: from whose perspective is the redshift being measured? Let me make it easier for you to answer:

-for [math]\theta=\frac{\pi}{2}[/math] [math]f_0=f \gamma[/math] i.e. [math]f=\frac{f_0}{\gamma}[/math]

 

 

 

The formula you solved is for theta_s, not theta_o.

 

Yes, the formula clearly says [math]f_o=f_s \gamma (1-\beta \theta_s)[/math] as pointed out several times. [math]f_0[/math] represents the proper frequency (the frequency in the frame of the emitter). See Moller, "Theory of Relativity", top of page 62, formula (90). The confusion may be caused by the fact that this notation is opposite to the one used by wiki.You seem not to understand that blueshift from the perspective of one frame means redshift from the perspective of the other frame, as in [math]f'=f \gamma <=> f=\frac{f'}{\gamma}[/math].

Edited January 27, 2014 by xyzt

[md65536]

I asked you a question, you didn't answer: from whose perspective is the redshift being measured? Let me make it easier for you to answer:

-for [math]\theta=\frac{\pi}{2}[/math] [math]f_0=f \gamma[/math] i.e. [math]f=\frac{f_0}{\gamma}[/math]

Fix your errors before trying to educate people. You've switched from [math]f_o[/math] to [math]f_0[/math], I no longer know what you're talking about. Post #17 doesn't make sense. Please correct it before quizzing me, thanks.

 

 

----

 

Trying to get this thread back on track...

 

The angle between the direction to the source and the direction of movement of the source is different in the source frame and the observer frame, due to abberation of light.

 

When the angle according to the source, [math]\theta_s[/math], is 90 degrees, that light will be observed as blueshifted.

 

When the angle according to the observer, [math]\theta_o[/math], is 90 degrees, that light will be observed as redshifted.

 

See http://en.wikipedia.org/wiki/Relativistic_Doppler_effect#Motion_in_an_arbitrary_direction

 

The equations of post #17 are according to the source frame, I'm guessing with a mistaken attempt to apply them to the observer's frame, messing up the math.

Edited January 27, 2014 by md65536

[DimaMazin]

That should be all

[math]\theta_s < \theta_{s\mathrm{\ critical}}=\arccos \frac{1-\sqrt{1-\beta^2}}{\beta}[/math].

 

Your theta_s is the angle in the source frame. At 90 degrees light will be blueshifted. The angle in the observer's frame (theta_o, using http://en.wikipedia.org/wiki/Relativistic_aberration_formula) will be greater, due to aberration... it seems the critical theta_o will always be greater than 90 degrees (light coming from the side is red-shifted).

For 0.8c theta_{s critical} = 60 degrees

90 degrees > 60 degrees

30 degrees < 60 degrees

If theta_s=90 degrees then light has blue shift, I am agreed.

Does light have blue shift if theta_s=30 degrees ? How does your formula work here?

[md65536]

For 0.8c theta_{s critical} = 60 degrees

90 degrees > 60 degrees

30 degrees < 60 degrees

If theta_s=90 degrees then light has blue shift, I am agreed.

Does light have blue shift if theta_s=30 degrees ? How does your formula work here?

If the critical angle is 60 degrees, that's the angle where the light is neither blue nor red shifted. At smaller angles, it's red-shifted.

 

The angle is between the direction from observer to light source (when the light was emitted), and the relative direction of the movement of the light source. Or, equivalently I think, the angle between the direction from source to observer, and the relative velocity of the observer, which might be easier to think about since theta_s is measured in the source's frame. (Also these are measured at the time that the light is emitted. ???) So what that means is that with smaller angles, the light source and observer are moving more away from each other. The smaller the angle, the greater the redshift.

 

 

This might make more sense in the reference frame of the observer? In that case,

[math]\frac{f_o}{f_s} = \frac{1}{\gamma ( 1 + \beta \cos \theta_o )}[/math].

(http://en.wikipedia.org/wiki/Relativistic_Doppler_effect#Motion_in_an_arbitrary_direction)

 

Then I think that

[math]\theta_{o\mathrm{\ critical}}=\arccos \frac{1/\gamma - 1}{\beta} = \arccos \frac{\sqrt{1-\beta^2}-1}{\beta}[/math].

 

For beta = 0.8, that would be 120 degrees. Again, smaller angles are red-shifted, and light coming in at 90 degrees to the observer is red-shifted.

I am trying to get grip on the special relativity theory and therefore I have created a thoughtexperiment.

Your experiment's over my head, but I was trying to think of one to wrap my head around these equations. Perhaps this is similar?

 

Consider 2 observers, where one is inertial and the other is rapidly circling the other at a constant distance. In the frame of the inertial observer as a source, any light sent to the other will arrive at an angle of theta_s = 90 degrees relative to its motion around the circle. That light will then be blue-shifted, F = gamma. Meanwhile, any light from the other is emitted at theta_o = 90 degrees relative to the direction of its motion (still in the inertial observer's frame here, now as the observer of light). That light will be red-shifted, F = 1/gamma.

 

This agrees with other calculations, where we expect to find that the traveling observer's clock is slowed by a factor of gamma relative to the inertial observer's. And it can be seen; the inertial observer will appear blue-shifted to the traveling observer, and the traveling observer will appear red-shifted (slowed) to the inertial observer.

 

 

To do the same calculations in the observer's frame we'd have to account for aberration and acceleration?, I wouldn't know how...

Edited January 28, 2014 by md65536

[Rettich]

Thanks for the answer md65536. I already had concluded the same.

[DimaMazin]

The point was that it is smaller than [math]\frac{\pi}{2}[/math].

Seems I start to understand why that is important. When theta > pi/2 then positive speed turns into negative speed.

For 0.8c theta_{s critical} = arccos((1 - 0.6)/(-0.8))=arccos(-0.5)=120 degrees

The observer sees blue shift when theta_s < 120 degrees

The observer sees red shift when theta_s > 120 degrees

If the critical angle is 60 degrees, that's the angle where the light is neither blue nor red shifted. At smaller angles, it's red-shifted.

 

The angle is between the direction from observer to light source (when the light was emitted), and the relative direction of the movement of the light source. Or, equivalently I think, the angle between the direction from source to observer, and the relative velocity of the observer, which might be easier to think about since theta_s is measured in the source's frame. (Also these are measured at the time that the light is emitted. ???) So what that means is that with smaller angles, the light source and observer are moving more away from each other. The smaller the angle, the greater the redshift.

 

 

 

 

Then I think that

[math]\theta_{o\mathrm{\ critical}}=\arccos \frac{1/\gamma - 1}{\beta} = \arccos \frac{\sqrt{1-\beta^2}-1}{\beta}[/math].

 

For beta = 0.8, that would be 120 degrees. Again, smaller angles are red-shifted, and light coming in at 90 degrees to the observer is red-shifted.

Your experiment's over my head, but I was trying to think of one to wrap my head around these equations. Perhaps this is similar?

 

 

For 0.8c theta_{o critical} ~81.34 degrees

Edited January 29, 2014 by DimaMazin

[md65536]

Seems I start to understand why that is important. When theta > pi/2 then positive speed turns into negative speed.

For 0.8c theta_{s critical} = arccos((1 - 0.6)/(-0.8))=arccos(-0.5)=120 degrees

The observer sees blue shift when theta_s < 120 degrees

The observer sees red shift when theta_s > 120 degrees

Those values work out for -0.8c...

 

But I don't think you would change the sign of velocity unless you also rotated your reference frame by 180 degrees (or flipped it)? Anyway by changing the sign of v you've changed the meaning of the angle.

 

I take it that you mean that "moving away at a negative speed" is the same as "approaching". But also I think you could say that "moving away at a positive speed at an angle greater than pi/2" is the same as "approaching". Also then, "moving away at a negative speed at an angle greater than pi/2" becomes "receding" again.

 

If you start with a fixed velocity v and slowly increase the angle theta_s from 0 up, you'll hit the critical angle at pi/3 (or 60 deg). If you change the sign of v at pi/2, then to make the light source increasingly oriented towards the observer, the angle must now decrease from pi/2. You won't hit 2 critical angles, one at 60 deg and one at 120 deg.

 

At -0.8c the critical angle in the source frame is 120 degrees. Having a velocity of -0.8c at an angle of 120 degrees is the same as having a velocity of 0.8c at an angle of 60 degrees, the Doppler factor will be the same (1, in this case).

 

 

If this doesn't make sense or isn't a conventional explanation, hopefully someone who knows this better will correct me.

[DimaMazin]

 

At -0.8c the critical angle in the source frame is 120 degrees. Having a velocity of -0.8c at an angle of 120 degrees is the same as having a velocity of 0.8c at an angle of 60 degrees, the Doppler factor will be the same (1, in this case).

 

http://en.wikipedia.org/wiki/Relativistic_aberration_formula

Yes, but theta_{s critical}=120 degrees

theta_{o critical}=60 degrees

blue shift : theta<theta_critical

 

red shift : theta>theta_critical

[md65536]

http://en.wikipedia.org/wiki/Relativistic_aberration_formula

Yes, but theta_{s critical}=120 degrees

theta_{o critical}=60 degrees

blue shift : theta<theta_critical

 

red shift : theta>theta_critical

That looks okay to me.

 

If you keep the angle the same but change the sign of velocity, the source is now moving in the opposite direction and a previous red-shift will become a blue-shift. Post #17 becomes correct for negative velocity.

 

Whether you use a positive or negative velocity, there is still a blue shift when theta_s = 90 deg, and a red shift when theta_o = 90 deg.

 

 

It's interesting/puzzling that it works out so nicely, with some symmetries.

[DimaMazin]

That looks okay to me.

 

If you keep the angle the same but change the sign of velocity, the source is now moving in the opposite direction and a previous red-shift will become a blue-shift. Post #17 becomes correct for negative velocity.

 

Whether you use a positive or negative velocity, there is still a blue shift when theta_s = 90 deg, and a red shift when theta_o = 90 deg.

 

 

It's interesting/puzzling that it works out so nicely, with some symmetries.

Only critical angles have symmetries relative to 90 degrees

Because theta_{s critical} +theta_{o critical} = 180 degrees