Relativistic Doppler Shift and a Star breaking up

Click For Summary

Homework Help Overview

The discussion revolves around a problem in relativistic physics, specifically focusing on the Doppler shift of light emitted from a star that breaks apart into two remnants. The original poster presents the frequencies of light emitted by the star and its remnants, seeking to determine their velocities and the angle of motion relative to the observer on Earth.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply the relativistic Doppler shift equations to find the velocities and angles of the remnants. Some participants question the appropriateness of the equations used, particularly regarding the angle and the relationship between the two remnants. Others suggest considering the conservation of momentum and the implications of the angles involved.

Discussion Status

The discussion is ongoing, with participants exploring various approaches to relate the speeds and angles of the remnants. Guidance has been offered regarding the setup of equations based on the observed frequencies, and there is an acknowledgment of potential issues with the provided data. Multiple interpretations of the problem are being examined.

Contextual Notes

Participants note that the problem may contain inconsistencies in the data, which could affect the ability to find a solution. There is also a focus on ensuring that angles and velocities are correctly represented in the context of the Doppler effect.

  • #31
Yes. That's the one you want.
 
Physics news on Phys.org
  • #32
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
 
  • #33
Good. Now combine and solve for \cos\theta and \beta.
 
  • #34
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
 
  • #35
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.
 
  • #36
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
 
  • #37
Don't forget to express \gamma in terms of \beta.
TFM said:
Also, what seems to be the problem with the data?
Solve for \beta and see.
 
  • #38
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
 
  • #39
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
 
  • #40
TFM said:
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.
 
  • #41
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
 
  • #42
It looks OK to me, so long as you realize that \gamma is a function of \beta.
 
  • #43
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
 
  • #44
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
 
  • #45
Looks OK, but please simplify. To start, you can:
(1) Cancel the \betas on the outside.
(2) Cancel the exponents.
 
  • #46
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
 
  • #47
TFM said:
\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.
 
  • #48
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
 
  • #49
-(a - 1) \ne -a - 1
 
  • #50
\frac{7.135}{(\frac{6.690}{1-\beta^2})} - 1 = -\frac{ 4.282}{(\frac{6.690}{1-\beta^2})} + 1

Is this correct?

TFM
 
  • #51
Looks OK.
 
  • #52
So now:

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

And:

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

Does this look okay?

TFM
 
  • #53
Sure. But let's not go through each step. Take it home. Isolate (1 - \beta^2), then solve for \beta. (Show each step, but don't wait for confirmation before continuing.)
 
  • #54
Rght, so

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

Putting the 1 + beta squared on top:

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

Add together:

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

Times by 6.690

7.135(1-\beta^2) + (1-\beta^2)4.282 = 2*6.69

11.417(1-\beta^2) = 13.38

So

1-\beta^2 = 13.38/11.417

1-\beta^2 = 1.172

1 = 1.172 + \beta^2

\beta^2 = 1-1.172

\beta^2 = -0.17

I have I done something wrong? you cannot square root a negative number?

TFM
 
  • #55
TFM said:
I have I done something wrong? you cannot square root a negative number?
Remember what I said (in post #35) about there being something wrong with the data? That's what I'm talking about.
 
  • #56
Would the best thing to do in this case be remove the minus sign, and then indicate on my work I have done so and the reason why?

TFM
 
  • #57
TFM said:
Would the best thing to do in this case be remove the minus sign, and then indicate on my work I have done so and the reason why?
What would be the reason why?

I would present your work clearly (and concisely) and show that it leads to impossible results, which indicates that the problem is flawed. (If your instructor thinks the problem is OK, then I'd like to see his solution.)
 
  • #58
Well, the question is worth ten marks, so I thought just make it a magnitude (by squaring then square rooting the negative number) , so that you can still get a answer, but if you think it would be better to leave it as it is, I will do so.

Thanks,

TFM
 
  • #59
Hey,

I was reading this thread and spotted this,

TFM said:
So


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


TFM

I've probably missed something but shouldn't this be,

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

or

\gamma = \sqrt{\frac{1}{1-\beta^2}}
 
  • #60
Vuldoraq said:
Hey,

I was reading this thread and spotted this,



I've probably missed something but shouldn't this be,

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

or

\gamma = \sqrt{\frac{1}{1-\beta^2}}
Absolutely. Good catch!
 

Similar threads

Doppler effect: why do I find this exercise so difficult?
  • · Replies 5 ·
Replies
5
Views
1K
How fast is the star moving? (Doppler effect)
  • · Replies 5 ·
Replies
5
Views
2K
Speed of a star in Special Relativity
  • · Replies 6 ·
Replies
6
Views
1K
Can Light Wavelength Remain Unchanged at High Velocities?
  • · Replies 8 ·
Replies
8
Views
2K
Relativistic doppler effect for light
  • · Replies 2 ·
Replies
2
Views
1K
Relativistic Doppler Shift for Transverse Movement
  • · Replies 1 ·
Replies
1
Views
2K
Blood flow velocity via Doppler effect
  • · Replies 1 ·
Replies
1
Views
2K
Transverse Doppler Effect for a Moving Star
  • · Replies 1 ·
Replies
1
Views
1K
Moving vehicles and Doppler Effect
  • · Replies 9 ·
Replies
9
Views
7K
How can the Doppler effect be used as a motion detector?
  • · Replies 16 ·
Replies
16
Views
2K