Understanding the Doppler Effect: Formula and Examples for Light Waves

[Stephanus]

The worldline (W) can be anywhere. I did not mark it but the bottom left corner is (0,0) as you deduced.

The calculation does not depend on the values of $x_1$ nor $x_2$ as long as they are not equal.

The calculation is correct so you have misunderstood something.

I didn't say the calculation is incorrect. It's the picture that I asked.
Shouldn't world line originate from (0,0). From your picture it originates not from (0,0)

 

[Mentz114]

I didn't say the calculation is incorrect. It's the picture that I asked.
Shouldn't world line originate from (0,0). From your picture it originates not from (0,0)

OK...

A worldline can originate at any event and continue to infinity if you like. W can be extended into the future or past but I have only shown the epoch where the signals are sent.

You must remember that the things are moving only in the +x or -x direction. The worldline describes motion in the x-axis - it is a path through spacetime, not space. If you want to see things moving try the 'Animate' feature'

 

[Stephanus]

Thanks.
So this formula is not for blue shifted/red shifted. It's for (proper time for rest)/(proper time for w)

 

[Mentz114]

Thanks.
So this formula is not for blue shifted/red shifted. It's for (proper time for rest)/(proper time for w)

$\tau_R/\tau_E$ is the Doppler shift. In this case it is a red-shift because the period is increasing and the frequency is decreasing.

Suppose the sender sends 100 pulses with 1 second ( by his clock) between them. Then they would be received with gaps of $\sqrt{(1+v)/(1-v)}$secs on the rest clock so the frequency is lower by the inverse of that.

 

[Stephanus]

Ahh, so when we emit the signal from rest with proper time $\Delta t$ to W. The ratio is $\sqrt{\frac{1+v}{1-v}}$
So $\Delta tau = \sqrt{\frac{1+v}{1=v}} \Delta t$
And if bounce back the signal from W to rest with proper tme $\Delta t'$. The ratio is the same.
$\Delta t' = \sqrt{\frac{1+v}{1-v}} \Delta tau = \sqrt{\frac{1+v}{1-v}} * \sqrt{\frac{1+v}{1-v}} \Delta t$
$\Delta t' = \frac{1+v}{1-v} \Delta t$
And for the frequency, ...

The Doppler shift component can be expressed as
$$F = \frac{1}{1+v} F_o$$ [EDIT by me. I omit c, we'll consider v is a factor of c to shorten the formula]

.Using the same convention, the time dilation component follows the expression

$$\sqrt{1-v^2}$$

Combining the two gives you

$$F = \sqrt{\frac{1-v}{1+v}} F_o$$

Where v is positive when the source is receding and negative when approaching.

So basically the formula for "period" and "frequency" factor is the same.
It's very clear for me now! Thanks everybody.
[Add]Conculsion:
One way trip:
For Period: $\sqrt{\frac{1+v}{1-v}}$, as in Mentz114 pdf
For Frequency: $\sqrt{\frac{1+v}{1-v}}$

Two way trip
For Period: $\frac{1+v}{1-v}$
For Frequency: $\frac{1+v}{1-v}$, the light is bounced back by a mirror.

 

[Stephanus]

$\tau_R/\tau_E$ is the Doppler shift. In this case it is a red-shift because the period is increasing and the frequency is decreasing.

Suppose the sender sends 100 pulses with 1 second ( by his clock) between them. Then they would be received with gaps of $\sqrt{(1+v)/(1-v)}$secs on the rest clock so the frequency is lower by the inverse of that.

YES. I understand. Thanks!

 

[PAllen]

No, the formula for period and frequency are reciprocals. Frequency = 1/period. Thus for red shift,

period: √( (c+v)/(c-v)) (bigger)
frequency: √( (c-v)/(c+v)) (smaller)

Both what I wrote and Mentz114 are correct and consistent. He derived period, I derived frequency. Note what Mentz114 said: "n this case it is a red-shift because the period is increasing and the frequency is decreasing."

Similarly, for the mirror case, period and frequency are reciprocals.

 

[Stephanus]

No, the formula for period and frequency are reciprocals. Frequency = 1/period. Thus for red shift,

period: √( (c+v)/(c-v)) (bigger)
frequency: √( (c-v)/(c+v)) (smaller)

Both what I wrote and Mentz114 are correct and consistent. He derived period, I derived frequency. Note what Mentz114 said: "n this case it is a red-shift because the period is increasing and the frequency is decreasing."

Similarly, for the mirror case, period and frequency are reciprocals.

Perhaps I stated myself incorrectly.
I mean with or withou square root.
Such as $\sqrt{\frac{1+v}{1-v}}$ vs $\frac{1+v}{1-v}$
Both are the same. Yes, I understand. Thanks.
And if I had been a little careful, I would realize for period it is $\frac{A}{B}$ and frequency is $\frac{B}{A}$

 

[Stephanus]

No, the formula for period and frequency are reciprocals. Frequency = 1/period. Thus for red shift,

period: √( (c+v)/(c-v)) (bigger)
frequency: √( (c-v)/(c+v)) (smaller)

Both what I wrote and Mentz114 are correct and consistent. He derived period, I derived frequency. Note what Mentz114 said: "n this case it is a red-shift because the period is increasing and the frequency is decreasing."

Similarly, for the mirror case, period and frequency are reciprocals.

Ah, you must have read my deleted post. Yes, I was too hasty to answer, before I realize that it's

[..]Then they would be received with gaps of $\sqrt{(1+v)/(1-v)}$secs on the rest clock so the frequency is lower by the inverse of that.

I didn't see "secs", I tought it was Hertz. So I deleted the post.
Again, I hastily answer you. Should have deleted my previous post post #43

 

[Mentz114]

Ah, you must have read my deleted post. Yes, I was too hasty to answer, before I realize that it's

I think you might be misunderstanding what a worldline represents. Here's another pic from my notes.

Imagine something traveling in a circle in the x-y plane. The worldline of that object would be a spiral coming out of the plane. We would have a 3D spacetime diagram with 2 spatial dimensions and one time.

In our ST plots we have only 1 spatial dimension and one time ( there's alway only one time dimwnsion, naturally).

 

[Stephanus]

@Mentz114
Actually my deleted post is a protest about your post:

$\tau_R/\tau_E$ is the Doppler shift. In this case it is a red-shift because the period is increasing and the frequency is decreasing.

Suppose the sender sends 100 pulses with 1 second ( by his clock) between them. Then they would be received with gaps of $\sqrt{(1+v)/(1-v)}$secs on the rest clock so the frequency is lower by the inverse of that.

I tought it should be $\sqrt{(1-v)/(1+v)}$. But I see that you wrote secs not hertz. So your formula is correct $\sqrt{(1+v)/(1-v)}$. That's why I deleted my post. But before

No, the formula for period and frequency are reciprocals. Frequency = 1/period. Thus for red shift,

period: √( (c+v)/(c-v)) (bigger)
frequency: √( (c-v)/(c+v)) (smaller)

Both what I wrote and Mentz114 are correct and consistent. He derived period, I derived frequency. Note what Mentz114 said: "n this case it is a red-shift because the period is increasing and the frequency is decreasing."

Similarly, for the mirror case, period and frequency are reciprocals.

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

I think you might be misunderstanding what a worldline represents. Here's another pic from my notes.

Imagine something traveling in a circle in the x-y plane. The worldline of that object would be a spiral coming out of the plane. We would have a 3D spacetime diagram with 2 spatial dimensions and one time.

In our ST plots we have only 1 spatial dimension and one time ( there's alway only one time dimwnsion, naturally).

Yes, you taught me ST diagram. It's 2D. But not 2D space, it's 1D space + 1D time. No height, just x distance.
I helps me much learning SR. Now, I think I have all the requirement to study Twins Paradox. But still can't make the logic out of it.

It's very "easy" to see that Green proper time is less then Blue's. The slanted green world line, the $sqrt{t^2-x^2}$, as seen in the double perpendicular Green triangles.
But the only thing that Blue and Green see is their bouncing signal, and their receiving time and emitting time.
Perhaps I should study it again really slow.

 

[Mentz114]

@Mentz114
But the only thing that Blue and Green see is their bouncing signal, and their receiving time and emitting time.
.

That is not necessarily true. What if green and blue are people walking down X street in daylight ? They can certainly see each other without using radar.

You have to describe the physical scenario that the WLs represent and ask questions appropriate to that.

Your diagram shows all you need to know about green and blue clock times. They start together and part and then meet and the green traveller is younger when they meet. That is the 'twin paradox'.

What don't you understand about it ?

 

[Stephanus]

That is not necessarily true. What if green and blue are people walking down X street in daylight ? They can certainly see each other without using radar.

Not quite. Green will see Blue as 3.33333E-08 seconds ago, assuming their distance is 10 metres. And the clock? My calculator gives 1 for gamma if green walks at 5 km/hour. Should use more precision variable type.

You have to describe the physical scenario that the WLs represent and ask questions appropriate to that.

Your diagram shows all you need to know about green and blue clock times. They start together and part and then meet and the green traveller is younger when they meet. That is the 'twin paradox'.

What don't you understand about it ?

That's the problem. I don't know what I don't know. As I said, by looking at the ST diagram, we'll spot at once the time dilation for green. But what happens in nature?
Still trying to make sense out of it. PeterDonis says my concept is wrong. https://www.physicsforums.com/threads/nature-cheats-twins-paradox.825093/#post-5180754
I have to think another one. Perhaps I can find the answer tomorrow.

 

[Mentz114]

OK. I don't mean to harass you. Please don't feel obliged to answer.

 

[PeterDonis]

by looking at the ST diagram, we'll spot at once the time dilation for green. But what happens in nature?

In nature, we never have complete information the way we do in artifically constructed scenarios like the ones you are drawing spacetime diagrams for. So in nature, we can make predictions that turn out to be wrong, because we were missing information. That's just the way it is, and as I said in the other thread, I don't see why it's a problem.

 

[Stephanus]

OK. I don't mean to harass you. Please don't feel obliged to answer.

Of course NO . The answer is for me. Why in the world that I keep asking about SR, while the thing that interest me is Cosmology. It's the Twins Paradox that I tought would be easy that got me stuck in this forum for 3 month. But thank for your invaluable help all this time.