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

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In summary, the conversation is about relating four formulas from Special Relativity, including the Lorentz Transform and the formula for the addition of velocities. The discussion centers around understanding and deriving the formula for W, which represents the velocity of an object in one reference frame as measured from another reference frame. The final result is W = { w+v \over 1 + { wv \over c^2 } }.
  • #1
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.

[tex]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 } }[/tex]

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:

[tex]w = { x^\prime \over t^\prime }[/tex]

[tex]\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 } } }[/tex]

[tex]\Rightarrow w = { x-vt \over t - { v \over c^2 } x }[/tex]

...so then...

[tex]W=v+{ x-vt \over t - { v \over c^2 } x }[/tex]

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

Can anyone help me here?
 
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  • #2
OneEye said:
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.

[tex]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 } }[/tex]

Dr. Einstein says:



(Hope that made sense.)

I have tried this, and got here:

[tex]w = { x^\prime \over t^\prime }[/tex]

[tex]\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 } } }[/tex]

[tex]\Rightarrow w = { x-vt \over t - { v \over c^2 } x }[/tex]

...so then...

[tex]W=v+{ x-vt \over t - { v \over c^2 } x }[/tex]

...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
[tex]w = { x-vt \over t - { v \over c^2 } x }[/tex]
divide the top and bottom by t
[tex]w = { \frac{x}{t}-v \over 1 - { v \over c^2 } \frac{x}{t} }[/tex]
[tex]w = { W-v \over 1 - { Wv \over c^2 } }[/tex]
Solve for W and you will get
[tex]W = { w+v \over 1 + { wv \over c^2 } }[/tex]
 
  • #3
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!
 
  • #4
OneEye said:
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.
 

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

What is a derivation?

A derivation is a process of obtaining a result or formula from other known results or formulas through logical steps and mathematical operations.

Why do I need help with derivation?

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How do I solve a derivation problem?

To solve a derivation problem, you need to understand the basic principles and rules of differentiation, which is the process of finding the rate of change of a function. You also need to have a strong foundation in algebra and calculus.

Can I use software to assist with derivation?

Yes, there are many software programs available that can assist with derivation. However, it is still important to have a good understanding of the concepts and steps involved in solving a derivation problem.

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Some common mistakes to avoid in derivation include not following the correct order of operations, forgetting to apply the chain rule, and making arithmetic errors. It is also important to double-check your work and use proper notation.

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