What Time Did Clock A Measure During Clock B's 4-Minute Journey?

[latentcorpse]

One tau to much here.
But partial fractions is a good way to go later.

Baah! I'm an idiot!

Ok so I get

t(\tau)=-4 \sqrt{2} \int \frac{1}{4 \sqrt{2}} \frac{d \tau}{\tau -2 - 2 \sqrt{2}} - 4 \sqrt{2} \int (- \frac{1}{4 \sqrt{2}}) \frac{d \tau}{\tau -2 + 2 \sqrt{2}}
=- \int \frac{d \tau}{\tau -2 -2 \sqrt{2}} + \int \frac{d \tau}{\tau -2 + 2 \sqrt{2}}
= \ln{( \tau -2 + 2 \sqrt{2})} - \ln{( \tau -2 -2 \sqrt{2})}

and so

t(4)=\ln{( 2+ 2 \sqrt{2})} - \ln{( 2 - 2 \sqrt{2})} = \ln{ \frac{ 2 + \sqrt{2}}{2 - \sqrt{2}}}

?

 

[betel]

Baah! I'm an idiot!

Ok so I get

t(\tau)=-4 \sqrt{2} \int \frac{1}{4 \sqrt{2}} \frac{d \tau}{\tau -2 - 2 \sqrt{2}} - 4 \sqrt{2} \int (- \frac{1}{4 \sqrt{2}}) \frac{d \tau}{\tau -2 + 2 \sqrt{2}}

How did you get the 1/4Sqrt(2) prefactor. Without that i agree

=- \int \frac{d \tau}{\tau -2 -2 \sqrt{2}} + \int \frac{d \tau}{\tau -2 + 2 \sqrt{2}}
= \ln{( \tau -2 + 2 \sqrt{2})} - \ln{( \tau -2 -2 \sqrt{2})}

and so

t(4)=\ln{( 2+ 2 \sqrt{2})} - \ln{( 2 - 2 \sqrt{2})} = \ln{ \frac{ 2 + \sqrt{2}}{2 - \sqrt{2}}}

?

I think you missed the lower limit.

 

[betel]

Sorry, i was a bit too quick. I get only a prefactor of 1/2Sqrt(2) and taking the lower limit gives me 4 times your result.

 

[betel]

We covered that point already.

 

[betel]

pfff, not my day. I missed the 2, so your prefactor is all fine. But you still have to include the lower limit.

 

[latentcorpse]

How did you get the 1/4Sqrt(2) prefactor. Without that i agree

I think you missed the lower limit.

\frac{dt}{d \tau} = \frac{\sqrt{2}}{-\frac{1}{4} \tau^2 + \tau + 1} = - 4 \sqrt{2} \frac{1}{ \tau^2 - 4 \tau - 4} = - 4 \sqrt{2} \frac{1}{( \tau -2 - 2 \sqrt{2})( \tau -2 + 2\sqrt{2})}

Now \frac{1}{( \tau -2 - 2 \sqrt{2})( \tau -2 + 2\sqrt{2})}=\frac{A}{\tau -2 - 2 \sqrt{2}} + \frac{B}{\tau -2 + 2 \sqrt{2}}

So to get B

\frac{1}{\tau - 2 - 2 \sqrt{2}} = \frac{A ( \tau - 2 + 2 \sqrt{2})}{ \tau -2 - 2 \sqrt{2}} + B
set \tau = 2 - 2 \sqrt{2} \Rightarrow B= -\frac{1}{4 \sqrt{2}}

and similarly we find A=\frac{1}{4 \sqrt{2}}

this means then that

t (\tau) = -4 \sqrt{2} \frac{1}{4 \sqrt{2}} \int_0^\tau \frac{d \tau}{\tau -2 - 2 \sqrt{2}} - 4 \sqrt{2} (-\frac{1}{4 \sqrt{2}}) \int_0^\tau \frac{d \tau}{\tau -2 + 2 \sqrt{2}}
=-\int_0^tau \frac{d \tau}{\tau -2 - 2 \sqrt{2}} + \int_0^\tau \frac{d \tau}{\tau -2 + 2 \sqrt{2}}
=-\ln{(\tau -2 + 2 \sqrt{2})} + \ln{(2 \sqrt{2} - 2)} + \ln{( \tau -2 - 2 \sqrt{2})} - \ln{(-2-2 \sqrt{2})}

So t(4)=\ln{2+ 2 \sqrt{2}} - \ln{2-2 \sqrt{2}} - \ln{ 2 \sqrt{2} - 2} + \ln{-2 - 2 \sqrt{2}}
=\ln{ \frac{(2 + \sqrt{2})(-2 - 2 \sqrt{2})}{(2-2 \sqrt{2})( 2 \sqrt{2} - 2 )}
=\ln{ \frac{-12 - 8 \sqrt{2}}{-12 + 8 \sqrt{2}}

How's that?

Thanks!

 

[betel]

\frac{dt}{d \tau} = \frac{\sqrt{2}}{-\frac{1}{4} \tau^2 + \tau + 1} = - 4 \sqrt{2} \frac{1}{ \tau^2 - 4 \tau - 4} = - 4 \sqrt{2} \frac{1}{( \tau -2 - 2 \sqrt{2})( \tau -2 + 2\sqrt{2})}

Now \frac{1}{( \tau -2 - 2 \sqrt{2})( \tau -2 + 2\sqrt{2})}=\frac{A}{\tau -2 - 2 \sqrt{2}} + \frac{B}{\tau -2 + 2 \sqrt{2}}

So to get B

\frac{1}{\tau - 2 - 2 \sqrt{2}} = \frac{A ( \tau - 2 + 2 \sqrt{2})}{ \tau -2 - 2 \sqrt{2}} + B
set \tau = 2 - 2 \sqrt{2} \Rightarrow B= -\frac{1}{4 \sqrt{2}}

and similarly we find A=\frac{1}{4 \sqrt{2}}

this means then that

t (\tau) = -4 \sqrt{2} \frac{1}{4 \sqrt{2}} \int_0^\tau \frac{d \tau}{\tau -2 - 2 \sqrt{2}} - 4 \sqrt{2} (-\frac{1}{4 \sqrt{2}}) \int_0^\tau \frac{d \tau}{\tau -2 + 2 \sqrt{2}}
=-\int_0^tau \frac{d \tau}{\tau -2 - 2 \sqrt{2}} + \int_0^\tau \frac{d \tau}{\tau -2 + 2 \sqrt{2}}
=-\ln{(\tau -2 + 2 \sqrt{2})} + \ln{(2 \sqrt{2} - 2)} + \ln{( \tau -2 - 2 \sqrt{2})} - \ln{(-2-2 \sqrt{2})}

So t(4)=\ln{2+ 2 \sqrt{2}} - \ln{2-2 \sqrt{2}} - \ln{ 2 \sqrt{2} - 2} + \ln{-2 - 2 \sqrt{2}}
=\ln{ \frac{(2 + \sqrt{2})(-2 - 2 \sqrt{2})}{(2-2 \sqrt{2})( 2 \sqrt{2} - 2 )}
=\ln{ \frac{-12 - 8 \sqrt{2}}{-12 + 8 \sqrt{2}}

How's that?

Thanks!

I think that is what I got. You should swap the denominator, otherwise you take log of something negative.

 

[latentcorpse]

I think that is what I got. You should swap the denominator, otherwise you take log of something negative.

well I can simplify it to

\ln{\frac{-3-2 \sqrt{2}}{-3+2 \sqrt{2}}}

Clearly that argument is negative which causes problems but what do you mean swap the denominator?

Also, why was it convenient (or necessary?) to rewrite the geodesic in that covariant way? Would it not have worked if we had just used the standard geodesic equation and if not, why not?
How did you know you had to rewrite it that way?

Thanks very much for your help on this question!

 

[betel]

well I can simplify it to

\ln{\frac{-3-2 \sqrt{2}}{-3+2 \sqrt{2}}}

Clearly that argument is negative which causes problems but what do you mean swap the denominator?

When getting log as the result of an integral you always have to take absolute values. So if you ignore it at the start, you have manually correct later.

[/QUOTE]
Also, why was it convenient (or necessary?) to rewrite the geodesic in that covariant way? Would it not have worked if we had just used the standard geodesic equation and if not, why not?
How did you know you had to rewrite it that way?

Thanks very much for your help on this question![/QUOTE]
Not neccessary, but this form of the geodesic equation is of more practical use in some cases. Otherwise you have to first calculate all the Christoffel symbols and then sum. Here if your metric does not depend on a couple of coordinates, because of the partial derivative the geodesic equation for such coordinates will be very easy.

And I knew it was a practical way to rewrite it because I gave the question to my students rececently.