# Same equation, different units

1. Mar 12, 2005

### tony873004

Starting fromthe equation $$\alpha=a/d$$ with $$\alpha$$ in radians and a and d in meters, show that the equation is also valid if $$\alpha$$ is expressed in arcseconds, a is in AU and d is in parsecs.

Would this be the proper way to show this?

$$\alpha=a/d$$
$$radians=meters/meters$$
$$4.8481*10^{-6} radians / arcsecond = \frac{1.49598*10^{11}m/AU}{3.0857*10^{16}m/pc}$$

Divide the numbers and cancel the m's

$$4.8481*10^{-6} radians / arcsecond = 4.8481*10^{-6}AU/pc$$

Cancel the numbers
$$radians / arcsecond = AU/pc$$
But radians is still there in the left part of the formula! What did I do wrong?

Last edited: Mar 12, 2005
2. Mar 12, 2005

### dextercioby

U don't need any #-s.Just the definition of a parallaxis arcsecond:

$$1\mbox{parsec}=:\frac{1\mbox{AU}}{1\mbox{arcsecond}}$$

Daniel.

3. Mar 12, 2005

### tony873004

Thanks, Dex. The problem says we have to start with alpha in radians, a and d in meters and justify it that way. That's why I did it the way I did. I just don't know why the radians wont drop off, like the intuitive answer says they should.

4. Mar 12, 2005

### dextercioby

They do.Both radians & arcsecond are plane angle units...There's a connection between them

$$2\pi \ \mbox{radians}<--------------------->(180\cdot 3600) \ \mbox{arcseconds}$$...

Daniel.

5. Mar 12, 2005

### Data

Just write down the identities:

$$4.8481 \cdot 10^{-6} \mbox{rad} = 1 \mbox{arcsecond}, \; 1.49598 \cdot 10^{11} \mbox{m} = 1 \mbox{AU}, \; 3.0857 \cdot 10^{16} \mbox{m} = 1 \mbox{parsec}, \; \Longrightarrow \frac{\mbox{arcsecond}}{4.8481 \cdot 10^{-6}} = 1 \mbox{rad} = \frac{1 \mbox{m}}{1 \mbox{m}} = \frac{\left(\frac{\mbox{AU}}{1.49598 \cdot 10^{11}}\right)}{\left(\frac{\mbox{parsec}}{3.0857 \cdot 10^{16}}\right)}$$

$$\Longrightarrow 2.0627 \cdot 10^{5} \mbox{arcsecond} = 2.0627 \cdot 10^{5} \frac{\mbox{AU}}{\mbox{parsec}}$$

$$\Longrightarrow 1\mbox{arcsecond} = \frac{1\mbox{AU}}{1\mbox{parsec}}$$

Last edited: Mar 12, 2005