Transformation of acceleration between two reference frames

[Paul Black]

hello

i want to derive the Transformation of acceleration between two reference frames
i searched in internet and found a book but i don't understand just one step

i attach a picture so you can see what i found in the internet

my problem is eq. (1.10) \begin{align}du=\frac{dU}{\gamma ^{2}(1-U\frac{v}{c^{2}})^{2}}\end{align}
how did he get this factor
can you please give me a tip


thank you very much

 

[Bill_K]

There's a misprint. Calculus rule is d(x/y) = (y dx - x dy)/y² where x = U - v and y = 1 - Uv/c², so the first denominator in Eq.(1.10) should be (1 - U v/c²)². The numerator is dU(1 - v²/c²), which reduces to dU(1/γ²).

 

[Paul Black]

thank you for your answer
where is this misprint? is it the first denominator? the square is missing right?
my problem is this gamma in eq 1.10 how did he get it? i don't understand it
also the next step with gamma
can you show me how to get it?
thank you

 

[xox]

thank you for your answer
where is this misprint? is it the first denominator? the square is missing right?
my problem is this gamma in eq 1.10 how did he get it? i don't understand it
also the next step with gamma
can you show me how to get it?
thank you

There are TWO errors, not one in the book:

$$dU=\frac{dU(1-Uv/c^2)-(U-v)(-dUv/c^2)}{(1-Uv/c^2)^2}$$

The final result is correct, the derivation is riddled with errors.

 

[Paul Black]

xox thank you for your answer but I am sorry you are wrong
the positive sign is wrong. it must be negative
therefore the only one error is the square in the denominator

but this is not my problem. i just want to know how to get the next step with gamma square.
hope you can help me

 

[xox]

xox thank you for your answer but I am sorry you are wrong
the positive sign is wrong. it must be negative
therefore the only one error is the square in the denominator

but this is not my problem. i just want to know how to get the next step with gamma square.
hope you can help me

There are TWO minus signs, they indeed cancel each other, so I was a little pedantic. $$du=\frac{dU(1-Uv/c^2)-(U-v)(-dUv/c^2)}{(1-Uv/c^2)^2}=\frac{dU(1-v^2/c^2)}{(1-Uv/c^2)^2}$$

But $$1-v^2/c^2=\frac{1}{\gamma^2}$$

 

[Paul Black]

yes you are right
sorryyyy

this gamma confused me xD

 

[xox]

yes you are right
sorryyyy

this gamma confused me xD

You are welcome!

 

[Paul Black]

xox
thank you very much. i get it
so now i have to divide both sides by dt to get du/dt=a
but again with this gamma. eq 1.11 i don't understand it

 

[xox]

xox
thank you very much. i get it
so now i have to divide both sides by dt to get du/dt=a
but again with this gamma. eq 1.11 i don't understand it

OK, let's help u a little more:

We agree that:
$$du=\frac{dU}{\gamma^2(1-Uv/c^2)^2}$$

Now,

$$dt=\gamma(dT-dXv/c^2)$$

Therefore:

$$a=\frac{du}{dt}=\frac{dU}{\gamma^3(1-Uv/c^2)^2(dT-dXv/c^2)}=\frac{dU/dT}{\gamma^3(1-Uv/c^2)^2(1-dX/dTv/c^2)}=\frac{dU/dT}{\gamma^3(1-Uv/c^2)^3}=\frac{A}{\gamma^3(1-Uv/c^2)^3}$$

If $u=V$ then $a=\gamma^3A$

 

[Paul Black]

thank you i got it finally
thank you very much

 

[xox]

thank you i got it finally
thank you very much

Glad I could help.