Relativistic Doppler Shift and a Star breaking up

[TFM]

Lets see where I went wrong:

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

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

So:

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

Also, can't Gamma be expressed as:

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

??

TFM

 

[Vuldoraq]

Yeah gamma can have that form too, I would have written like that but I couldn't get the latex to do it for me.

 

[TFM]

So, first cancel out the exp. and the beta's:

$$\frac{7.135}{(6.690)\gamma} - 1 = -\frac{ 4.282}{(6.690)\gamma} + 1$$

Reinsert gamma:

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

Multiiply out:

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

Move over the 1:

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

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

Does this look okay?

TFM

 

[TFM]

This can go to:

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

get rid of denominator:

$$7.135(\sqrt{1 - \beta^2}) + 4.282( \sqrt{1 - \beta^2}) = 2*6.690$$

This can go to:

$$11.417 (\sqrt{1 - \beta^2}) = 13.38$$

and:

$$\sqrt{1 - \beta^2} = 13.38/11.417$$

$$\sqrt{1 - \beta^2} = 1.172$$

so square out the root:

$$1 - \beta^2 = 1.373$$

and:

$$1 - 1.373 = \beta^2$$

Is the data still wrong, or have I done something wrong, because I now get beta squared as
-0.373

??

TFM

 

[Vuldoraq]

Your math looks fine, your answer is the same as my own. It's as Doc Al said, the data is wrong. Slightly annoying because I think it makes it impossible to find the angle, as requested in the second part, though I may be wrong.

 

[Doc Al]

Yes, the data is wrong. When I checked it earlier, I used my own calculations (with correct expression for $\gamma$).

 

[Vuldoraq]

Hey all,

Is there a possibility that the answer is imaginary because one velocity is the negative of the other?

Vuldoraq

 

[TFM]

I finally have the answer sheet for the question:

Quoted from Sheet:

Since the fragments have equal masses, they must have equal velocities. Using the general Doppler shift relation, we find that for one of the fragments

$$\frac{v prime}{v} = \frac{6.690*10^{-14}}{7.135*10{-14}} = 0.9376 = \gamma - \beta \gamma cos \theta$$

whereas for the other

$$\frac{v prime}{v} = \frac{6.690*10^{-14}}{4.282*10{-14}} = 1.562 = \gamma + \beta \gamma cos \theta$$

Adding these, we obtain 2γ =2.5 so γ =1.25=5/4 and so

$$\beta^2 = 1 - \frac{1}{\gamma^2} = 1 - \frac{16}{25} = \frac{9}{25}$$

and therefore

$$\beta = \frac{3}{5} = 0.6$$

which gives a velocity of 0.6 × 3×108 = 1.83 × 108 m/s.
Substituting back into the first equation of this part, we find

$$0.936 = 1.25(1 - 0.6cos \theta)$$

$$0.75 = 1 - 0.6cos \theta$$

$$cos \theta = \frac{0.25}{0.6}0.4167$$

and so the angle is θ = 65.4°.

End Sheet Quote

TFM

 

[Doc Al]

Yeah, I really messed this one up big time. I had reversed the primed and unprimed frequencies (mixing up source and observer). D'oh!

My apologies.