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Special relativity issue - doppler

by dingo_d
Tags: doppler, issue, relativity, special
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dingo_d
#1
Mar4-12, 09:26 AM
P: 211
1. The problem statement, all variables and given/known data
The observer sees first space ship that is moving from the right towards him with the speed [itex]v_r=0.6c[/itex], and the second ship that is coming from the left with the speed [itex]v_l=0.8c[/itex].

If the first ship emits a signal with the frequency of [itex]10^4[/itex] Hz (measured in the reference frame of the ship - where the ship is standing still), what is the frequency of the signal that the observer measures?

2. Relevant equations

Lorentz transformations.

3. The attempt at a solution

I am wondering if I am thinking right on this one. Since everything is relative, I can look at this as if the ship is standing still, and the observer is moving towards him with the speed of 0.6 c. And then, since it's moving towards the signal it will see it being blue shifted - coming towards him, right?

So using the formula:

[itex]\nu'=\nu\sqrt{\frac{1+\beta}{1-\beta}}[/itex]

I get that the frequency measured by observer is [itex]\nu'=2\times 10^4[/itex] Hz.
(higher frequency means blue shifted, right?).

Is this correct, or did I messed sth up?

Thanks :)
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Doc Al
#2
Mar4-12, 09:34 AM
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Quote Quote by dingo_d View Post
I am wondering if I am thinking right on this one. Since everything is relative, I can look at this as if the ship is standing still, and the observer is moving towards him with the speed of 0.6 c.
The observer always views himself as being at rest and the source as moving.
And then, since it's moving towards the signal it will see it being blue shifted - coming towards him, right?
The source is moving towards the observer, so the signal will be blue shifted.

So using the formula:

[itex]\nu'=\nu\sqrt{\frac{1+\beta}{1-\beta}}[/itex]

I get that the frequency measured by observer is [itex]\nu'=2\times 10^4[/itex] Hz.
(higher frequency means blue shifted, right?).
Good.
dingo_d
#3
Mar4-12, 09:47 AM
P: 211
But can't I say:

Before there is the observer O, which is at rest, and ship A which is moving with a certain speed.
Now I put myself in a system where ship is at rest, and then the point O is moving towards him.

That kind of thinking seems legitimate, no?

Doc Al
#4
Mar4-12, 09:54 AM
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Special relativity issue - doppler

Quote Quote by dingo_d View Post
But can't I say:

Before there is the observer O, which is at rest, and ship A which is moving with a certain speed.
Now I put myself in a system where ship is at rest, and then the point O is moving towards him.

That kind of thinking seems legitimate, no?
Sure, an observer in ship A can view himself as at rest with O moving towards him. But here you want what the observer O sees, not what ship A would see.
dingo_d
#5
Mar4-12, 10:00 AM
P: 211
Oh! I see :D

Now, if the observer passes that signal with the frequency of [itex]2\times 10^4[/itex]Hz, to the ship on the left, since the ship is moving towards the stationary observer, I can use the same formula to calculate what frequency will the ppl on the ship measure, right?

What kinda bothers me is, since I am working with relative values, I have no need for coordinate systems, and therefore I can use the same formula. The only thing I need to look out is whether the ships are moving away or towards the observer, right?
Doc Al
#6
Mar4-12, 10:10 AM
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Quote Quote by dingo_d View Post
Now, if the observer passes that signal with the frequency of [itex]2\times 10^4[/itex]Hz, to the ship on the left, since the ship is moving towards the stationary observer, I can use the same formula to calculate what frequency will the ppl on the ship measure, right?
Not sure what you mean by "passes that signal". But if O sends out a signal that matches what it observes from A, and B receives that signal, then sure you can use the same formula to figure out what B will observe. (O becomes the source and B the observer.)
What kinda bothers me is, since I am working with relative values, I have no need for coordinate systems, and therefore I can use the same formula. The only thing I need to look out is whether the ships are moving away or towards the observer, right?
Right. All that matters is the relative motion of source and observer.
dingo_d
#7
Mar4-12, 10:13 AM
P: 211
Thank you very much ^^


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