# 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

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#### 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

Science Advisor
Gold Member
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

Science Advisor
Gold Member
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

Science Advisor
Gold Member
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.

thanks guys!

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