Lorentz Transformation - Proof that t'2 - t'1 >0

[zacl79]

A 'cause' occurs at point 1 (x1, t1) and its 'effect' occurs at point 2 (x2, t2) as measured by observer O. Use Lorentz transformation to find t'2 - t'1 as measured by O' and show that t'2 - t'1 >= 0. that is Observer O' can never see the effect before the cause.


I know that is possible to prove this, but just having some difficulty in doing so.
I use:

t'1 = (gamma)(t1 - ux1 /c^2) => goes to zero as X1 and t1 are 0??
t'2 = (gamma)(t2 - ux2 / c^2)

Working through this i get:

t'2 - t'1 = gamma(t2 - ux2 /c^2)

Now i don't think that this is the correct proof that i require.

Any help to where i have gone wrong, or if i am overlooking something would be greatly appreciated.

Thanks

 

[Mr.A.Gibson]

since x2<ct2 as lower than the speed of light, you can substitute this into your equation to get.

ux2/c^2 < uct2/c^2 < t2

hence t2 - ux2 / c^2 > 0

so t'2 - t'1 > 0 also

 

[tiny-tim]

hi zacl79!

(try using the X₂ icon just above the Reply box )

you haven't used the causation condition yet …

A 'cause' occurs at point 1 (x1, t1) and its 'effect' occurs at point 2 (x2, t2) as measured by observer O

t'1 = (gamma)(t1 - ux1 /c^2) => goes to zero as X1 and t1 are 0??

you can put x₁ = t₁ = 0, but it's probably better not to mess about with the original question, in case you make a mistake "translating" it back

 

[Mr.A.Gibson]

you can put x₁ = t₁ = 0, but it's probably better not to mess about with the original question, in case you make a mistake "translating" it back

I think you should always mess with the question and use logic, if it makes the problem easier to solve. Since the physics is the same in all reference frames it doesn't matter where we assign the origin.

 

[tiny-tim]

I think you should always mess with the question and use logic, if it makes the problem easier to solve. Since the physics is the same in all reference frames it doesn't matter where we assign the origin.

Mr Gibson, please don't advise students to take steps which make it easier to make mistakes, and to lose both time and marks in exams.

 

[Mr.A.Gibson]

Mr Gibson, please don't advise students to take steps which make it easier to make mistakes, and to lose both time and marks in exams.

Sorry Tim, but I disagree with this statement. I thought the reasoning for the method was justified. If the student clearly reassigns the origin in their working their result are still valid. In fact I believe this method would save time and lead to less mistakes otherwise I would not have advised it.

You particularly stated that this could lead to error translating it back, and I agree with that statement, but it is not applicable to this problem. You are defining x₁, t₁ as the origin, no other origin has been defined so there is no need translate back.

Hence I would only understand your objection if the origin had already been defined.

 

[zacl79]

Thanks guys,

Ill give it a go!