B Cox & Forshaw: "Why E=mc^2" - A Newbie's Question

[Ste7e]

Hi All

My first post here and I'm sure it's going to be one of the easiest for you to answer.

I've just read Einstein's book on the theory of relativity and now I'm currently reading 'Why does e=mc² (and why should we care)' by Brian Cox and Jeff Forshaw.

My Math isn't up to much but I have a inquisitive mind... I'm stumped by something rather basic and early on in this book.

On page 51/52(ish) - not 100% sure as I'm reading the Kindle version - there's the first bit of maths which seems to be getting to the Lorentz transformations starting from good old Pythagoras.

The authors get us to T² = 1/ (c²—υ²) which is all good.

But then we get this:

Taking the square root of our equation above for T², and multiplying by 2, we find that 2T= 2/√c²υv²

I'm baffled by where the 'v' has come from. I thought it was a problem with the Kindle fonts first of all - the 'u' is printed in italics and the 'v' not, wasn't sure if they were the same thing - but I found a PDF of the book and it's the same in there.

Can someone walk me through this?

Thanks

Steve

 

[jtbell]

On page 51/52(ish)

Of what?
[added] Oh, never mind... it's the book named in the title of your post. Carry on... Unfortunately I've never seen that book. However, I don't recognize the equation you ended up with.

 

[Ste7e]

Sorry, should have been clearer. I put the title of the book as the thread name and then referred to it in the body as 'this book'.

The book is: Cox and Forshaw: Why E=mc^2 (and why we should care)

Thanks for pointing this out - just edited the original post to be clearer.

Steve

 

[Battlemage!]

Hi All

My first post here and I'm sure it's going to be one of the easiest for you to answer.

I've just read Einstein's book on the theory of relativity and now I'm currently reading 'Why does e=mc² (and why should we care)' by Brian Cox and Jeff Forshaw.

My Math isn't up to much but I have a inquisitive mind... I'm stumped by something rather basic and early on in this book.

On page 51/52(ish) - not 100% sure as I'm reading the Kindle version - there's the first bit of maths which seems to be getting to the Lorentz transformations starting from good old Pythagoras.

The authors get us to T² = 1/ (c²—υ²) which is all good.

But then we get this:

Taking the square root of our equation above for T², and multiplying by 2, we find that 2T= 2/√c²υv²

I'm baffled by where the 'v' has come from. I thought it was a problem with the Kindle fonts first of all - the 'u' is printed in italics and the 'v' not, wasn't sure if they were the same thing - but I found a PDF of the book and it's the same in there.

Can someone walk me through this?

Thanks

Steve

Could you post verbatim what they said? I'd like to be able to derive it myself but I don't know the context.

 

[jtbell]

Tip: you can write equations more legibly using LaTeX:

https://www.physicsforums.com/help/latexhelp/

 

[Ste7e]

I'll try copying and pasting from the PDF. It's a discussion about the light-clock thought experiment...

If the train is moving at a speed, υ, then the clock moves a distance υT each half-tick.
Again we did nothing except use “distance = speed x time.” This distance
is the length of the base of a right-angled triangle and because we know the length of the longest side,
we can go ahead and figure out the distance between the two mirrors using Pythagoras’ theorem. But
we know what that distance actually is already—it is 1 meter. So Pythagoras’ theorem tells us that
(cT)²= 1²+ (υT)².

We are nearly done now. We know c, the speed of light, and let’s presume to know the speed of the
train, υ. Then we can use this equation to figure out T. The crudest way to do it would be to guess a
value of Tand see if it solves the equation. More often than not the guess will be wrong and we’ll
need to try another guess. After a while we might hone in on the right answer. Fortunately, we can
avoid that tedious process because the equation can be “solved.” The answer is T²= 1/ (c²—υ²),
which means, “first work out c²—υ² and then divide 1 by that number.”

..., we can now write the time taken for one tick of the clock as
determined by someone on the platform: It is the time for light to travel up to the top mirror and back
down again—that is 2T. Taking the square root of our equation above for T², and multiplying by 2, we
find that 2T= 2/√c²υv² This equation allows us to work out the time taken for one tick as measured
by the person on the platform, knowing the speed of the train, the speed of light, and the distance
between the two mirrors (1 meter). But the time for one tick according to someone sitting on the train
next to the clock is simply equal to 2/c, because for them the light simply travels 2 meters at a speed
c (distance = speed x time, so time = distance / speed). Taking the ratio of these two time intervals tells
us by how much the clock on the train is running slow, as measured by someone on the platform; it is
running slow by a factor of c/√c²υv², which can also be written, with a little more mathematical
rearranging, as 1/ √1 - v²υc². This is a very important quantity in relativity theory, and it is usually
represented by the Greek letter γ, pronounced “gamma.” Notice that γ is always larger than 1 as long
as the clock is flying along at less than the speed of light, because υ/c will be smaller than 1. When υ
is very small compared to the speed of light (i.e., for most ordinary speeds, since in units more
familiar to motorists the speed of light is 671 million miles per hour), γ is very close to 1 indeed. Only when
υ becomes a significant fraction of the speed of light does γ start to deviate appreciably
from 1.

 

[stevendaryl]

I can't make any sense of that. But what the answer is supposed to be is:

\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} where v is the speed of the clock, and c is the speed of light.

 

[PeroK]

Looks like a simple misprint.

 

[DrClaude]

It is a misprint. It should look like

 

[Ste7e]

That's great. Thanks everyone. I got to the same result with my remembered algebra from eons ago but didn't have enough confidence in that to assume it was a misprint. It's been bothering me too much to go further into the book until I'd got to the bottom of it. I can now carry on :)

Steve