Relativistic Doppler Shift and a Star breaking up

[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 $\beta$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!