Special relativity issue - doppler

In summary, two spaceships, one moving towards the observer at a speed of 0.6c and the other moving away at 0.8c, are observed. The first ship emits a signal with a frequency of 10^4 Hz in its own reference frame, and the observer measures a frequency of 2x10^4 Hz due to the signal being blue-shifted. This is calculated using the Lorentz transformations. The observer is always considered to be at rest and the source to be moving. The same formula can be used to calculate the frequency observed by the second ship if the signal is passed on by the observer. The relative motion of the source and observer is the only factor that affects the observed frequency.
  • #1
dingo_d
211
0

Homework Statement


The observer sees first spaceship 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?

Homework Equations



Lorentz transformations.

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 :)
 
Physics news on Phys.org
  • #2
dingo_d said:
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.
 
  • #3
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?
 
  • #4
dingo_d said:
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.
 
  • #5
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?
 
  • #6
dingo_d said:
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.
 
  • #7
Thank you very much ^^
 

Related to Special relativity issue - doppler

1. What is the Doppler effect in special relativity?

The Doppler effect in special relativity refers to the change in frequency of light or sound waves as the source of the waves and the observer move relative to each other. This effect is a result of the relative motion between the source and observer and is a fundamental principle in special relativity.

2. How does the Doppler effect affect the perception of time in special relativity?

The Doppler effect plays a crucial role in the perception of time in special relativity. As an object moves closer to an observer, the frequency of the light waves emitted from the object increases, causing time to appear to slow down for the observer. Conversely, as an object moves away from an observer, the frequency decreases, causing time to appear to speed up for the observer.

3. Can the Doppler effect be observed in all types of waves in special relativity?

Yes, the Doppler effect is observed in all types of waves, including light and sound waves, in special relativity. This effect is not limited to a specific type of wave and is a result of the relative motion between the source and observer.

4. How is the Doppler effect explained in the context of special relativity?

In special relativity, the Doppler effect is explained by the concept of time dilation. As an object moves closer to an observer, the time for the observer appears to slow down due to the increased frequency of the light waves. This is because the observer is experiencing time at a different rate than the moving object. Similarly, as an object moves away from an observer, the time appears to speed up due to the decreased frequency of the light waves.

5. Can the Doppler effect in special relativity be used to explain redshift and blueshift?

Yes, the Doppler effect in special relativity can be used to explain redshift and blueshift. As an object moves away from an observer, the frequency of the light waves decreases, causing the light to shift towards the red end of the spectrum. Similarly, as an object moves closer to an observer, the frequency increases, causing the light to shift towards the blue end of the spectrum. This is known as the redshift and blueshift phenomenon and is a result of the Doppler effect in special relativity.

Similar threads

  • Introductory Physics Homework Help
Replies
14
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
968
  • Introductory Physics Homework Help
Replies
6
Views
696
  • Advanced Physics Homework Help
Replies
9
Views
5K
  • Advanced Physics Homework Help
Replies
3
Views
2K
  • Advanced Physics Homework Help
Replies
4
Views
972
  • Special and General Relativity
Replies
20
Views
998
  • General Discussion
Replies
5
Views
1K
  • Special and General Relativity
Replies
7
Views
972
  • Classical Physics
Replies
16
Views
1K
Back
Top