11 Jan '18 18:49

Can someone explain the Relativistic Doppler Effect to me in the most simple way possible? Particularly the time dilation aspect of it.

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podunk, PA12 Jan '18 03:14

What specific question about time dilation do you have? How it is derived maybe? Perhaps we can learn together.*Originally posted by @metal-brain***Can someone explain the Relativistic Doppler Effect to me in the most simple way possible? Particularly the time dilation aspect of it.**- Joined
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slatington, pa, usa12 Jan '18 14:495 edits

T1=T/(1-v^2/C^2)^0.5 is the dilation part, If you do the math you see at 0.9C the time dilates about 2.4 to one, that is where we on Earth see 2.4 years go by, in the spacecraft doing 0.9C only 1 year goes by on board and that is real time to them, if 24 years go by on Earth only 10 years goes by in the ship and if they come back home the crew will be 10 years younger than say their twin on Earth. It's a real time shift. T is Earth time and T1 is shipboard time.*Originally posted by @metal-brain***Can someone explain the Relativistic Doppler Effect to me in the most simple way possible? Particularly the time dilation aspect of it.**

At 0.99C the shift is about 7 to one so if you go to a star 7 light years away, Earthy clocks say 7 years went by but on the ship, it only took 1 year to get there so they effectively (as far as their clocks are concerned) have gone 7 times the speed of light.- Joined
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13 Jan '18 16:01

Since c is constant (more or less) how is there a Doppler Effect? How does the wavelength change?*Originally posted by @joe-shmo***What specific question about time dilation do you have? How it is derived maybe? Perhaps we can learn together.**- Joined
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13 Jan '18 16:02

That doesn't explain why wavelength changes.*Originally posted by @sonhouse***T1=T/(1-v^2/C^2)^0.5 is the dilation part, If you do the math you see at 0.9C the time dilates about 2.4 to one, that is where we on Earth see 2.4 years go by, in the spacecraft doing 0.9C only 1 year goes by on board and that is real time to them, if 24 years go by on Earth only 10 years goes by in the ship and if they come back home the crew will be 10 y ...[text shortened]... so they effectively (as far as their clocks are concerned) have gone 7 times the speed of light.**- Joined
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podunk, PA13 Jan '18 19:053 edits

It falls out of some surprisingly simple trigonometry. Pretend you are on a train moving with constant velocity "v" and your back is against a wall. There is a mirror on the wall opposite to you. You turn on a flashlight aimed at the mirror. The time the beam takes to reach back to you from your perspective inside the moving train is given by:*Originally posted by @metal-brain***Since c is constant (more or less) how is there a Doppler Effect? How does the wavelength change?**

t = 2*L/c

L is the distance between you and the mirror

c is the speed of light

t is the time interval you measure between you turning on the light and the light arriving back to you.

Now, at the exact same set of events is witnessed by me who is standing outside the train on the ground as it passes by.

I see you turn on the light, and I too see it return. However, I see it return and say it has traveled a further distance. Namely the distance 2*D. With a constant speed of light it means the time interval I measure is given by t' = 2*D/c, where D = √( ( ½*v*t )² + L² ) from Pythagoreans Theorem.

So that gives you three equations to solve simultaneously to get t' in terms of t ( or the time interval I measure in terms of the time interval you measure) which is given by what sonhouse already wrote:

t' = t / √( 1- (v/c)² )

So if your riding on the train at a constant speed "v" approaching "c" and we perform the experiment again: You measure the time interval 2*L/c and I never measure the interval when the beam returns to you...I think.- Joined
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Germany15 Jan '18 07:58

Here's an explanation:*Originally posted by @metal-brain***That doesn't explain why wavelength changes.**

https://en.wikipedia.org/wiki/Relativistic_Doppler_effect- Joined
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16 Jan '18 17:33

That was the first website I looked at. It didn't help me understand at all. It says it involves time dilation but doesn't explain why.*Originally posted by @kazetnagorra***Here's an explanation:**

https://en.wikipedia.org/wiki/Relativistic_Doppler_effect

Comparing it to a baseball player or train is not helpful to me. The speed of sound is not constant like light is. The train is irrelevant. Why does a star orbiting around a black hole or another star shift from red to blue? Don't say because it moves away and toward us as it orbits. I know that.

If c is constant (more or less according to Feynman in QED) why does the wavelength change? What does time dilation and light aberration have to do with it? Wikipedia does a horrible job of explaining it. Pathetic really.- Joined
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podunk, PA16 Jan '18 19:051 edit

The wavelength change arises from λ=c / f and the fact that c is equal in every frame of reference which gives the following relationship.*Originally posted by @metal-brain***That was the first website I looked at. It didn't help me understand at all. It says it involves time dilation but doesn't explain why.**

Comparing it to a baseball player or train is not helpful to me. The speed of sound is not constant like light is. The train is irrelevant. Why does a star orbiting around a black hole or another star shift from re ...[text shortened]... aberration have to do with it? Wikipedia does a horrible job of explaining it. Pathetic really.

λ * f = λ' * f '

λ is the wavelength , and f is the frequency of the light with respect to the person moving with the train in my example above. λ' and f ' are the same quantities in the reference frame of the person on the ground.

Also f = 1/t , and f ' = 1/t'

substitute and solve for λ'

λ' = t' / t * λ = 1/ √ ( 1 - ( v / c ) ² ) * λ

so as v → c , λ' → ∞

That is the wavelength increases.- Joined
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16 Jan '18 19:56

All you did was point out the inverse relationship between wavelength and frequency. I am already aware of that. You have not helped, but thanks for trying.*Originally posted by @joe-shmo***The wavelength change arises from λ=c / f and the fact that c is equal in every frame of reference which gives the following relationship.**

λ * f = λ' * f '

λ is the wavelength , and f is the frequency of the light with respect to the person moving with the train in my example above. λ' and f ' are the same quantities in the reference frame of th ...[text shortened]... λ = 1/ √ ( 1 - ( v / c ) ² ) * λ

so as v → c , λ' → ∞

That is the wavelength increases.- Joined
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slatington, pa, usa16 Jan '18 21:51

You know how doppler works in audio frequencies, right? You move into a sound and you will hear the frequency go up, wavelength go down and if you move away from the sound the frequency goes down and wavelength goes up? APPARENT frequency, not actual frequency, but it would be able to be picked up and recorded on instruments like microphones, you hear the sound change when a train goes by tooting its horn?*Originally posted by @metal-brain***All you did was point out the inverse relationship between wavelength and frequency. I am already aware of that. You have not helped, but thanks for trying.**

Same thing happens when you are in space moving at a good clip, even at velocities of human space travel, say 5 miles per second or so there will be a definite change in the perceived frequency of RF, say a light beam, say a laser at 542 Nm (green) and you travel towards the laser beam it will shift ever so slightly to the blue but if you have a detector looking at the laser beam and you are going away from it the frequency will get smaller, the color shifting now towards red (700 odd Nm). It happens because you are plowing into some RF field and your detector is now going from peak to peak of the RF signal, whether it is 5 hertz, or 5 terahertz or higher. It follows a linear rule at low velociities like humans use getting to the moon or Mars. It gets tricky when a spacecraft gets close to the speed of light because now you are messing with time flow itself. That follows the formula two of us have shown, so at 90% of c, effective time slows down so you THINK you are going 2 or 3 times the speed of light so doppler shift will follow the same way. Relativistic doppler is a result of time flow shifts when you approach the speed of light. At .99c the effect is you think you are going 7 times c and doppler shift will follow this new rule like time flow change at velocities close to c.- Joined
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podunk, PA18 Jan '18 13:173 edits

Well, I'm sorry to hear that. However, I do not believe you are sincere in your attempts to understand. How do you expect to understand the warping of space-time without any road map? ( i.e. equations of special relativity ) These are not phenomenon in which we have any standard sense of in our non-relativistic perspectives. For instance while I was trying to reconcile "Blue Shift" (the star coming at you) with the equation I provided above I realized the equation above could not be correct. There needed to be a term to account for velocity ( speed and direction ). The relation I provided was not capable of providing that ( it failed when trying to explain "Blue Shift" ) so I was forced to investigate further. I found that actual relationship for t' is not what I or sonhouse wrote.*Originally posted by @metal-brain***All you did was point out the inverse relationship between wavelength and frequency. I am already aware of that. You have not helped, but thanks for trying.**

t' = t/ √( 1 - (v/c)² )

but was in fact,

t' = ( 1 - v/c)*t/ √( 1 - (v/c)² )

which simplifies to

t' = √ ( [ 1 - (v/c)] / [ 1 + (v/c)] ) * t

The relationship for wavelength shift is actually:

λ' = √ ( [ 1 + (v/c)] / [ 1 - (v/c)] ) * λ

The factor "( [ 1 + (v/c)] / [ 1 - (v/c)] )" is greater than one ( red shifted ) for observers moving apart ( relative velocity is positive ) , and less than one for observers moving toward each other ( relative velocity is negative ).

I don't believe there is anyway to understand special relativity outside of the framework and formal language of special relativity. So if this didn't help, no offense... you'll probably never know.

Oh, and by the way; The link Kazetnagora provided describes it well, much more complete than myself. So if your serious about understanding it I'd get to work on understanding algebra, because that is all the more sophisticated the math ( the language through which it is understood) required is to describe it.- Joined
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slatington, pa, usa18 Jan '18 13:46

How did the V/C^2 function disappear? So in relativistic doppler, v is plus for approaching an emitter and minus for leaving an emitter? Does the equation work if v is zero?*Originally posted by @joe-shmo***Well, I'm sorry to hear that. However, I do not believe you are sincere in your attempts to understand. How do you expect to understand the warping of space-time without any road map? ( i.e. equations of special relativity ) These are not phenomenon in which we have any standard sense of in our non-relativistic perspectives. For instance while I was tr ...[text shortened]... ophisticated the math ( the language through which it is understood) required is to describe it.**- Joined
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18 Jan '18 14:157 edits

exactly what doesn't explain why 'wavelength changes'?*Originally posted by @metal-brain***That doesn't explain why wavelength changes.**

The 'wavelength changes' can be deduced from nothing more than the principle of relativity combined with just a few pretty basic facts about the properties of light such as photons have wavelengths etc. Einstein made this deduction. If you want to understand how, just look it up yourself. It actually isn't difficult for most people to understand but I guess given your appalling past record of absence of understanding and/or gross misunderstanding of even the very simplest scientific concepts (such as how the evidence for dark matter, assuming it exists, proves it cannot be mainly black holes. I have already explained to you how so), that don't require you to know the first thing about maths to understand it, indicates you won't.- Joined
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podunk, PA18 Jan '18 16:09

"How did the (V/C)^2 function disappear?"*Originally posted by @sonhouse***How did the V/C^2 function disappear? So in relativistic doppler, v is plus for approaching an emitter and minus for leaving an emitter? Does the equation work if v is zero?**

It didn't disappear, its just some algebraic manipulation

We have a factor ( 1 - v/c) in the numerator, and √( 1-(v/c)^2 ) in the denominator. we bring the numerator under the root by squaring it.

( 1 - v/c)*1/√( 1-(v/c)^2 ) = √( (1-v/c)²/( 1-(v/c)² ) ) = √( (1-v/c)*(1-v/c)/( 1-(v/c)² ) )

Then apply the difference of squares to break up the denominator:

1-(v/c)^2 = (1+v/c)*(1-v/c)

The factor of (1 - v/c) cancels and your left with √( (1 - v/c) / (1+v/c) ).

"So in relativistic doppler, v is plus for approaching an emitter and minus for leaving an emitter?"

The opposite; Minus for approaching, plus for receding.

"Does the equation work if v is zero?"

Plug in v = 0 and tell me what you get?