Solve Relativistic Doppler Effect for Wavelengths of Lights

[Niles]

Homework Statement

I am told that to a stationary observer, the lights infront of a spaceship has the wavelength = 480 nm when it arrives, and the lights in the back have wavelength 640 nm when the ship is moving away. Both lights (front and back) are the same.

I have to find the wavelength of the two lights.

The Attempt at a Solution

I have tried to use T = T_0 * (1+w/c)/sqrt(1-w^2/c^2) and from there use T-T_1 ... so the w's would go out, but they don't.

When the spaceship is arriving, the speed is v - when it is moving away, it is -v. From there, I tried the above.

How do I find the wavelengths? (I have to calculate relativistic).

 

[Niles]

Is it possible to find it without using the velocity? (Because I am not told what it is)

 

[Niles]

Another thing, related to my first post, is: When the spaceship is arriving and the light is being sent forward - is there any change from that between when the spaceship is moving forward and the light is being emitted backwards?

 

[Doc Al]

I have tried to use T = T_0 * (1+w/c)/sqrt(1-w^2/c^2) and from there use T-T_1 ... so the w's would go out, but they don't.

This is an equation for period, which is 1/f. Express this in terms of wavelength by using $\lambda = c/f = cT$.

Now write both equations: one for an approaching source and one for a receding source.

Is it possible to find it without using the velocity? (Because I am not told what it is)

By combining the two equations, you'll be able to eliminate the velocity and solve for the source wavelength.

Another thing, related to my first post, is: When the spaceship is arriving and the light is being sent forward - is there any change from that between when the spaceship is moving forward and the light is being emitted backwards?

Nothing that's not already contained in the formula.

 

[Niles]

Hmm, it seems easier to solve for v first, and then for lambda?

 

[Doc Al]

If you combine the two equations correctly, you won't need to solve for v.

 

[Niles]

I have:

1) lamdba/480 = (1 + v/c)/sqrt(1-v^2/c^2)

2) lamdba/640 = (1 - v/c)/sqrt(1-v^2/c^2)

I isolated lamdbda, and inserted and found v. But I think I am meant to go the other way? Although it seems much harder

 

[Doc Al]

I have:

1) lamdba/480 = (1 + v/c)/sqrt(1-v^2/c^2)

2) lamdba/640 = (1 - v/c)/sqrt(1-v^2/c^2)

Good.

I isolated lamdbda, and inserted and found v. But I think I am meant to go the other way? Although it seems much harder

Not sure what you mean. If you mean to solve for v, then plug it into the other equation to get lamdba... WAY too hard.

Hint: Think of clever ways of combining these two equations. (Think of the basic math operations.)

 

[Niles]

Good.


Not sure what you mean. If you mean to solve for v, then plug it into the other equation to get lamdba... WAY too hard.

I can add them, so the v/c-part goes out, but the squareroot in the nominator can't? I tried subtracting, but that gave 2v/c, which doesn't seem much better.

 

[Doc Al]

Good! So addition and subtraction don't seem to help. What's next?

 

[Niles]

Ahh.. thanks Doc!