Question about rearranging formula in Brian Cox's Why Does E=MC2

[DPR]

This has been driving me insane. I don't get how he went from s/c to t/y. If someone could explain step by step how you do it I would greatly appreciate it.

Recall that we arrived at an expression for the length of the momentum vector in three-dimensional space, mΔx/Δt. We have just argued that Δx should be replaced by Δs and Δt should be replaced by Δs/c to form the four-dimensional momentum vector, which has a seemingly rather uninteresting length of mc. Indulge us for one more paragraph, and let us write the replacement for Δt, i.e., Δs/c, in full. Δs/c is equal to [sqrt (cΔt)^2)-(xΔ)^2]/c. This is a bit of a mouthful, but a little mathematical manipulation allows us to write it in a simpler form, i.e., it can also be written as Δt/γ where y=1/[sqrt 1-v^2/c^2)]. To obtain that, we have used the fact that υ = Δx/Δt is the speed of the object. Now γ is none other than the quantity we met in Chapter 3 that quantifies the amount by which time slows down from the point of view of someone observing a clock fly past at speed.
pg 127

 

[DPR]

could someone help me out with this or give me some tips on what to study to be able to figure this out?

 

[Matterwave]

s is defined as s=\sqrt{-(\Delta x)^2+c^2(\Delta t)^2}

Now divide both sides by c\Delta t and you get:

\frac{s}{c\Delta t}=\sqrt{\frac{-(\Delta x)^2}{c^2(\Delta t)^2}+1}

Now notice that \frac{\Delta x}{\Delta t}=v to get:

\frac{s}{c\Delta t}=\sqrt{-\frac{v^2}{c^2}+1}

Move the t to the right hand side and use the definition of gamma to get:
\frac{s}{c}=\frac{\Delta t}{\gamma}

 

[DPR]

hey thanks! I really need to brush up on my math!

 

[Matterwave]

One additional point. We often call \frac{s}{c} the proper time \tau=\frac{s}{c}. You may see this proper time pop up more often depending on which sources you're reading.

 

[DPR]

One additional point. We often call \frac{s}{c} the proper time \tau=\frac{s}{c}. You may see this proper time pop up more often depending on which sources you're reading.

Thanks. How exactly did you go do this step? Now divide both sides by c\Delta t and you get:

\frac{s}{c\Delta t}=\sqrt{\frac{-(\Delta x)^2}{c^2(\Delta t)^2}+1}

I get everything after that.

 

[elfmotat]

Thanks. How exactly did you go do this step? Now divide both sides by c\Delta t and you get:

\frac{s}{c\Delta t}=\sqrt{\frac{-(\Delta x)^2}{c^2(\Delta t)^2}+1}

I get everything after that.

\frac{s}{c\Delta t}=\frac{\sqrt{-(\Delta x)^2+c^2(\Delta t)^2}}{\sqrt{(c\Delta t)^2}}=\sqrt{\frac{-(\Delta x)^2+c^2(\Delta t)^2}{c^2(\Delta t)^2}}=\sqrt{\frac{-(\Delta x)^2}{c^2(\Delta t)^2}+\frac{c^2(\Delta t)^2}{c^2(\Delta t)^2}}=\sqrt{\frac{-(\Delta x)^2}{c^2(\Delta t)^2}+1}

 

[Matterwave]

Just divide both sides like I said. You have to move the c\Delta t inside the squareroot and divide both sides by it.

EDIT: What elfmotat said.

 

[DPR]

thanks guys!