Relativistic Doppler Shift and a Star breaking up

[TFM]

Homework Statement

A distant star at rest with respect to an observer on Earth emits light of frequency 6.690 x 10^14 Hz. The star breaks up into two remnants of equal mass, which are observed to emit light of frequency 7.135 x 10^14 Hz and 4.282 x 10^14 Hz

Find the velocities of the remnant and the angle between line of sight and their direction of motion.

Homework Equations

v^{prime} = v \gamma [1 - \beta cos \theta]

v^{prime} = v \sqrt{\frac{1 + \beta}{1 - \beta}}

The Attempt at a Solution

I have solved the first part, finding the speed, using:

v^{prime} = v \sqrt{\frac{1 + \beta}{1 - \beta}}

assuming that since the star was initally at rest, teh centre of momentum will remain in the same polace, hence the two parts will fly away from each other.

I got \beta to be 0.47, and the speed of the star remnants to be 0.235c

I am having a small problem with the second part, finding the angle. I am sure I need to use:

v^{prime} = v \gamma [1 - \beta cos \theta]

with the v primke being the orginal stars frequency. but I am nto sure which of the othe frequencies to use as v, and I am not sure how to get a value of beta to use in this part, since I am sure it will be different to the value in part A.

Any ideas will be greatly appreciated,

TFM

 

[Doc Al]

I have solved the first part, finding the speed, using:

v^{prime} = v \sqrt{\frac{1 + \beta}{1 - \beta}}

That's the formula for the longitudinal Doppler shift when the source is approaching. It assumes an angle of 180 degrees, so it's not relevant here.

You can't solve the two parts separately. Hint: How are the speed and angle of the two remnants related?

 

[TFM]

Would you need to use the equation c = \lambda * frequency

so:

frequecy = c/ \lamda

so:

\frac{c}{\lambda}^{prime} = \frac{c}{\lambda} \gamma [1 - \beta cos \theta]

Or I am going in the worng direction?

TFM

 

[Doc Al]

The equation to use is the first one you listed (I'll use f for frequency):

f' = f \gamma [1 - \beta cos \theta]

 

[TFM]

So:

f' = f \frac{[1 - \beta cos \theta]}{1 - \beta^2}

but we don't know beta or theta?

TFM

 

[Doc Al]

but we don't know beta or theta?

Of course you don't--that's what you're trying to find! Set up two equations (using the two observed frequencies) and you can solve for the two unknowns.

Don't neglect to answer my question: How are the speed and angle of the two remnants related?

 

[TFM]

Would the relationship between speed and angle be:

v_{obseerved} = v*sin \theta

?

TFM

 

[Doc Al]

How are the speed and angle of the two remnants related?

I meant this as two questions:
(1) How are the speeds of each remnant related?
(2) How are their angles related?

These should be easy to answer. Drawing a diagram will help.

 

[TFM]

Considereing coinservation of momentum, then the momentum before in the Earth Observer frame should be zero, since the star has zero speed. this means that the two stars remnants must have equal and opposite speeds, to keep the center of momentum frame the same. the angle between them must be 180 degrees (Pi)?

TFM

 

[Doc Al]

Good! So how will the angles that you use in the Doppler formula relate to each other? How will their cosines relate to each other?

 

[TFM]

Would one of the remnants hav an angle 0f 0, thus the cos would be 1, whilst the other would have 180, with a cos of -1?

TFM

 

[Doc Al]

Would one of the remnants hav an angle 0f 0, thus the cos would be 1, whilst the other would have 180, with a cos of -1?

No reason to think so. (But if one did have an angle of 0, then the other must certainly have an angle of 180.) Again, this is a more general case than the simple longitudinal Doppler shift.

Call the angle that one remnant makes \theta. (Draw that diagram!)

 

[TFM]

Is the daigram required?

TFM

 

[Doc Al]

That's a good start. Now mark the angle that each remnant makes with the line of sight. (Assume for all practical purposes that both remnants are located just where the original star was.)

 

[TFM]

Would that be like this?

TFM

 

[Astronuc]

Would that be like this?

Yes - so now find the velocity components of both stars with respect (parallel) to the line of sight, and relate them back to the velocities of the stars. Note that the star moving away is at the same angle \theta as the star moving toward the observer.

 

[TFM]

This would be breaking down into components, so, for star Remnant A:

Horizontal speed would be:

v*cos(\pi - \theta)

Vertical speed would be:

v*sin(\pi - \theta)

and for Remnant B:

Horizontal Speed would be:

v*cos(\theta)

Vertical speed would be:

v*sin(\theta)

?

TFM

 

[Doc Al]

Would that be like this?

Almost. The angle should be measured from the position vector that extends from Earth to the star.

In the general formula for Doppler shift you should recognize that \theta = 0 corresponds to the ordinary longitudinal formula when the source is receding.

There's no need to find parallel or perpendicular components of the velocity. The \beta in the Doppler formula represents the full velocity, not a component. (Any needed component is taken care of by the \cos\theta factor.)

Fix your diagram (marking the correct angle), then set up two equations, one for each remnant.

 

[TFM]

Should it be like this then, since I believe you measure angles clockwise

?

TFM

 

[TFM]

opps, wrong diagram:

Sorry,

TFM

 

[Doc Al]

Should it be like this then, since I believe you measure angles clockwise

The issue is not clockwise versus counterclockwise, but that you should be measuring the angle from the forward direction (a vector pointing towards the star). An angle of zero means the source is moving directly away from earth.

 

[TFM]

That makes sense, so:

?

TFM

 

[Doc Al]

Good. Now set up your two equations.

 

[TFM]

f' = f \gamma [1 - \beta cos \theta]


So, Equation for remnant A:

(6.690*10^{14}) = (4.282*10^{14}) \gamma [1 - \beta cos \theta]

For Remnant B:

(6.690*10^{14}) = (7.135*10^{14}) \gamma [1 - \beta cos (\pi - \theta)]

Are these the right equations?

TFM

 

[Doc Al]

For some reason you have f and f' reversed. Other than that, those are the right equations. Simplify this expression: cos (\pi - \theta).

 

[TFM]

Would that simplify down to sin(\theta)

TFM

 

[Doc Al]

No. Review your trig identities. Even better, recognize vertical angles in your diagram.

 

[TFM]

Would it be sin{\pi + \theta}

TFM

 

[Doc Al]

No. It's much simpler than that. Look up trig identities related to supplements.

 

[TFM]

Would it be:

-cos(\theta)

TFM

 

[Doc Al]

Yes. That's the one you want.

 

[TFM]

So for Remnant A, the equation is:

(4.282*10^{14}) = (6.690*10^{14}) \gamma [1 - \beta cos \theta]

And for remnant B:

(7.135*10^{14}) = (6.690*10^{14}) \gamma [1 + \beta cos (\theta)]

TFM

 

[Doc Al]

Good. Now combine and solve for \cos\theta and \beta.

 

[TFM]

I've canceled them down to:

Remnant A:

\beta cos(\theta) = 1 - 0.648\gamma

Remnant B:

\beta cos (\theta) = 1.0665 \gamma - 1

Does this look right?

TFM

 

[Doc Al]

There are many ways to combine these equations. (I didn't check your arithmetic.)

However, there seems to be a problem with the data. There doesn't seem to be a solution for the given numbers. Unless I'm making an error, I'd say there's a mistake in the supplied data.

 

[TFM]

to find beta would you do:

\beta = \frac{1 - 0.648\gamma}{cos\theta}

and

\beta = \frac{1.0665 \gamma - 1}{cos \theta}

and equate to get:

\frac{1 - 0.648\gamma}{cos\theta} = \beta = \frac{1.0665 \gamma - 1}{cos \theta}

Also, what seems to be the problem with the data?

TFM

 

[Doc Al]

Don't forget to express \gamma in terms of \beta.

Also, what seems to be the problem with the data?

Solve for \beta and see.

 

[TFM]

How am I going so far:

\frac{1 - 0.648\gamma}{cos\theta} = \frac{1.0665 \gamma - 1}{cos \theta}

Times both sides by cos theta

1 - 0.648\gamma = 1.0665 \gamma - 1

Put Gamma in:

1 - \frac{0.0648}{\sqrt{1 - \beta^2}} = \frac{1.0665}{\sqrt{1 - \beta^2}} - 1

2 - \frac{0.0648}{\sqrt{1 - \beta^2}} = \frac{1.0665}{\sqrt{1 - \beta^2}}

rearrange:

2 = \frac{1.0665}{\sqrt{1 - \beta^2}} + \frac{0.0648}{\sqrt{1 - \beta^2}}

How does this look?

TFM

 

[TFM]

2 = \frac{1.0665}{\sqrt{1 - \beta^2}} + \frac{0.0648}{\sqrt{1 - \beta^2}}

Which goes to

2 = \frac{1.1313}{\sqrt{1 - \beta^2}}

Does this look okay so far?

TFM

 

[Doc Al]

I've canceled them down to:

Remnant A:

\beta cos(\theta) = 1 - 0.648\gamma

Remnant B:

\beta cos (\theta) = 1.0665 \gamma - 1

Does this look right?

TFM

Redo these. I don't see how you got these from the equations in post #32.

 

[TFM]

Remnant A:

4.282*10^{14} = (6.690*10^{14})\gamma(1-\beta cos\theta)

\frac{ 4.282*10^{14}}{(6.690*10^{14})\gamma} = 1 - \beta cos\theta

\frac{ 4.282*10^{14}}{(6.690*10^{14})\gamma} - 1 = -\beta cos\theta

-(\frac{ 4.282*10^{14}}{(6.690*10^{14})\gamma} - 1)/ \beta = cos\theta

Remnant B:

7.135*10^{14} = (6.690*10^{14})\gamma(1+\beta cos\theta)

\frac{7.135*10^{14}}{(6.690*10^{14})\gamma} = 1+\beta cos\theta

\frac{7.135*10^{14}}{(6.690*10^{14})\gamma} - 1 = \beta cos\theta

(\frac{7.135*10^{14}}{(6.690*10^{14})\gamma} - 1)/ \beta = cos\theta

So:

(\frac{7.135*10^{14}}{(6.690*10^{14})\gamma} - 1)/ \beta = -(\frac{ 4.282*10^{14}}{(6.690*10^{14})\gamma} - 1)/ \beta

How does this look so far?

TFM

 

[Doc Al]

It looks OK to me, so long as you realize that \gamma is a function of \beta.

 

[TFM]

So

\frac{7.135*10^{14}}{(6.690*10^{14})\gamma} - 1)/ \beta = -(\frac{ 4.282*10^{14}}{(6.690*10^{14})\gamma} - 1)/ \beta

This is the same as:

\beta( \frac{7.135*10^{14}}{(6.690*10^{14})\gamma} - 1)) = -\beta(\frac{ 4.282*10^{14}}{(6.690*10^{14})\gamma} - 1)

And

\gamma = \frac{1}{1-\beta^2}

So

\beta( \frac{7.135*10^{14}}{(6.690*10^{14})(\frac{1}{1-\beta^2})} - 1)) = -\beta(\frac{ 4.282*10^{14}}{(6.690*10^{14})(\frac{1}{1-\beta^2})} - 1)

TFM

 

[TFM]

Thus goes to:

\beta( \frac{7.135*10^{14}}{((\frac{6.690*10^{14}}{1-\beta^2})} - 1)) = -\beta(\frac{ 4.282*10^{14}}{)(\frac{6.690*10^{14}}{1-\beta^2})} - 1)

Edit sorry, brackets slightly weong:

\beta( \frac{7.135*10^{14}}{(\frac{6.690*10^{14}}{1-\beta^2})} - 1)) = -\beta(\frac{ 4.282*10^{14}}{(\frac{6.690*10^{14}}{1-\beta^2})} - 1)


Okay so far?

TFM

 

[Doc Al]

Looks OK, but please simplify. To start, you can:
(1) Cancel the \betas on the outside.
(2) Cancel the exponents.

 

[TFM]

So, cancels:

\frac{7.135}{(\frac{6.690}{1-\beta^2})} - 1 = -\beta(\frac{ 4.282}{(\frac{6.690}{1-\beta^2})} - 1)

Look Okay?

Edit: missed a beta, sorry

\frac{7.135}{(\frac{6.690}{1-\beta^2})} - 1 = -\frac{ 4.282}{(\frac{6.690}{1-\beta^2})} - 1

TFM

 

[Doc Al]

\frac{7.135}{(\frac{6.690}{1-\beta^2})} - 1 = -\frac{ 4.282}{(\frac{6.690}{1-\beta^2})} - 1

Check signs on the right hand side.

 

[TFM]

Where is the wrongsign, because it seems to still kepp with:

\frac{7.135*10^{14}}{(6.690*10^{14})\gamma} - 1)/ \beta = -(\frac{ 4.282*10^{14}}{(6.690*10^{14})\gamma} - 1)/ \beta

?

TFM

 

[Doc Al]

-(a - 1) \ne -a - 1

 

[TFM]

\frac{7.135}{(\frac{6.690}{1-\beta^2})} - 1 = -\frac{ 4.282}{(\frac{6.690}{1-\beta^2})} + 1

Is this correct?

TFM