Proving Approximation for Relativity Math Problem | T^2 << (c^2/alpha^2)

[faklif]

Homework Statement

This comes from a book on relativity but it basically comes down to a math problem. The problem is to prove that if
$$T^2 \ll\frac{c^2}{\alpha^2}$$
then
$${t}\approx{T}(1-\frac{\alpha^2{T^2}}{6c^2})$$
given
$$\frac{\alpha{T}}{c}=sinh(\frac{\alpha{t}}{c})$$

Homework Equations

See above.

The Attempt at a Solution

I've tried solving for t from the equation
$$\frac{\alpha{T}}{c}=sinh(\frac{\alpha{t}}{c})$$
which gives
$$t=\frac{c}{\alpha}\log(\frac{T\alpha}{c}+\sqrt{1+\frac{T^2\alpha^2}{c^2}})$$
I thought I'd be able to use maclaurin expansion at this point becuase of how the approximation looks but I keep making mistakes and I'm not getting anywhere at the moment so I'd really appreciate some help.

 

[cepheid]

What if you tried writing the hyperbolic sine function in terms of exponential functions first? Their Taylor expansions are easy...

 

[faklif]

What if you tried writing the hyperbolic sine function in terms of exponential functions first? Their Taylor expansions are easy...

Thanks for the reply!

Using the first four terms of each of the expansions I end up with what looks like the some kind of opposite to what I want
$${T}\approx{t}(1+\frac{\alpha^2{t^2}}{6c^2})$$
I can't see that I can get where I want from this and I haven't used all information in the problem to get here, maybe I'm missing something.

 

[Dick]

Thanks for the reply!

Using the first four terms of each of the expansions I end up with what looks like the some kind of opposite to what I want
$${T}\approx{t}(1+\frac{\alpha^2{t^2}}{6c^2})$$
I can't see that I can get where I want from this and I haven't used all information in the problem to get here, maybe I'm missing something.

You need to reverse the roles of t and T. You could also write at/c=arcsinh(aT/c) and use arcsinh(x)=x-x^3/3!+...

 

[HallsofIvy]

No reason in the world to be always writing hyperbolic functions as exponentials!

sihh(0)= 0.
(sinh(x))'= cosh(x) and cosh(0)= 1.
(sinh(x))"= (cosh(x))'= sinh(x) and sinh(0)= 0.
(sinh(x))'''= (sinh(x))'= cosh(x) and cosh(0)= 1, etc.

 

[Hurkyl]

I can't see that I can get where I want from this

Well, have you at least substituted that approximation into $T(1-\frac{\alpha^2{T^2}}{6c^2})$?

 

[faklif]

You need to reverse the roles of t and T. You could also write at/c=arcsinh(aT/c) and use arcsinh(x)=x-x^3/3!+...

Thanks a lot that does it! Also gives me another question and that is how to find that arcsinh(x)=x-x^3/3!+...? My thought of simply solving for t and then looking at what I got didn't really get me there.

Well, have you at least substituted that approximation into $T(1-\frac{\alpha^2{T^2}}{6c^2})$?

How can I do that substitution? I'm feeling very rusty here, I'm sorry.

 

[Billy Bob]

Also gives me another question and that is how to find that arcsinh(x)=x-x^3/3!+...?

Here's one way. The derivative of arcsinh x is $$(1+x^2)^{-1/2}$$ so use the binomial series to find the series for that, then integrate term by term. You only need a few terms anyway, not the whole series.

 

[Dick]

A physicist would just take
$${T}\approx{t}(1+\frac{\alpha^2{t^2}}{6c^2})$$
and say, hey, that means t=T with correction of higher order. So I can just replace t^2 with T^2 without making any errors I would care about. So I can change that into
$${T}\approx{t}(1+\frac{\alpha^2{T^2}}{6c^2})$$
Now move all the T's to one side and use 1/(1+x) is approximately equal to 1-x for x<<1.

 

[faklif]

Thank you everyone!