Relativistic Doppler Shift and a Star breaking up

[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 $\beta$s 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

 

[Doc Al]

Looks OK.

 

[TFM]

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

 

[Doc Al]

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.)

 

[TFM]

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

 

[Doc Al]

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.

 

[TFM]

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

 

[Doc Al]

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.)

 

[TFM]

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

 

[Vuldoraq]

Hey,

I was reading this thread and spotted this,

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

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

 

[Doc Al]

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!

 

[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). :yuck: D'oh!

My apologies.