How can I relate four formulas from special relativity to find a solution?

[OneEye]

[SOLVED] Need help with derivation

I have four formulas from SR, and need to relate them. Two are the Lorentz Transform, one is a simple formula for velocity, and the fourth is the formula for the addition of velocity in SR.

$$1... x^\prime = { x-vt \over \sqrt { 1-{v^2 \over c^2} } } \quad 2... t^\prime = { { t-{v \over c^2 } x } \over \sqrt { 1-{v^2 \over c^2 } } } \quad 3... x^\prime = wt^\prime \quad 4... W={ v+w \over 1+{vw \over c^2 } }$$

Dr. Einstein says:

Relativity, page 39:

In the equation x'=wt' we must then express x' and t' in terms of x and t, making use of the first and fourth equations of the Lorentz transformation [equations (1) and (2), above)]. Instead of W=v+w, we then obtain the equation [(4) above].

(Hope that made sense.)

I have tried this, and got here:

$$w = { x^\prime \over t^\prime }$$

$$\Rightarrow w = { { x-vt \over \sqrt { 1- { v^2 \over c^2 } } } \over { { t - { v \over c^2 } x } \over \sqrt { 1- { v^2 \over c^2 } } }$$

$$\Rightarrow w = { x-vt \over t - { v \over c^2 } x }$$

...so then...

$$W=v+{ x-vt \over t - { v \over c^2 } x }$$

...and that's as far as I got. I am quite a ways away from equation (4), above.

Can anyone help me here?

 

[DW]

I have four formulas from SR, and need to relate them. Two are the Lorentz Transform, one is a simple formula for velocity, and the fourth is the formula for the addition of velocity in SR.

$$1... x^\prime = { x-vt \over \sqrt { 1-{v^2 \over c^2} } } \quad 2... t^\prime = { { t-{v \over c^2 } x } \over \sqrt { 1-{v^2 \over c^2 } } } \quad 3... x^\prime = wt^\prime \quad 4... W={ v+w \over 1+{vw \over c^2 } }$$

Dr. Einstein says:



(Hope that made sense.)

I have tried this, and got here:

$$w = { x^\prime \over t^\prime }$$

$$\Rightarrow w = { { x-vt \over \sqrt { 1- { v^2 \over c^2 } } } \over { { t - { v \over c^2 } x } \over \sqrt { 1- { v^2 \over c^2 } } }$$

$$\Rightarrow w = { x-vt \over t - { v \over c^2 } x }$$

...so then...

$$W=v+{ x-vt \over t - { v \over c^2 } x }$$

...and that's as far as I got. I am quite a ways away from equation (4), above.

Can anyone help me here?

Yeah, as he said W is NOT v+w. Instead W is x/t. So continuing from
$$w = { x-vt \over t - { v \over c^2 } x }$$
divide the top and bottom by t
$$w = { \frac{x}{t}-v \over 1 - { v \over c^2 } \frac{x}{t} }$$
$$w = { W-v \over 1 - { Wv \over c^2 } }$$
Solve for W and you will get
$$W = { w+v \over 1 + { wv \over c^2 } }$$

 

[OneEye]

DW,

Thanks for a quick and thorough response. I was kind of afraid to go there, because it seems to me that there is a pun here between v=x/t and W=x/t.

Clearly both are true in SOME sense, and it is also clear that v<>W (though in the abstract, W is a kind of v).

Do you think that you can spare me another moment and clear that up for me?

Thanks!

 

[DW]

DW,

Thanks for a quick and thorough response. I was kind of afraid to go there, because it seems to me that there is a pun here between v=x/t and W=x/t.

Clearly both are true in SOME sense, and it is also clear that v<>W (though in the abstract, W is a kind of v).

Do you think that you can spare me another moment and clear that up for me?

Thanks!

In this context, v is not x/t. W is. There are three velocities being related. There is w which is the velocity of some "thing" according to measurements made from the primed coodinate system. There is W which is the velocity of that same "thing" according to measurements made from the unprimed coordinate system. And then there is v which is the speed of one of the coordinate systems according to measurements made from the other.