Special Relativity (1-dimensional Lorentz Transformation)

  • #1
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Homework Statement



During class today, I was told that:

x' = γ(x-vt)
y' = y
z' = z
t' = γ[t-(v/c2)x],
where γ=(1-v2/c2)-1/2

(This is just the standard Lorentz transformation.)

Then I was told that we could find dx/dt, the inertial velocity, by finding dx' and dt' and dividing them.

Homework Equations



dx' = dγ(x-vt)+γ(dx-v.dt-t.dv)
dt' = dγ[t-(v/c2)x]+γ[dt-(v/c2)x-(1/c2)x.dv)]
dγ = (-1/2)(1-v2/c2)-3/2(-2v.dv) = vγ3.dv

Hence,

dx' = [vγ3(x-vt)-γt]dv + γ(dx-v.dt)
dt' = [vγ3(t-v/c2x)-γt]dv + γ(dt-v/c2.dx)

We set dv=0, and we end up with

dx'/dt' = (dx-v.dt)/(dt-v/c2.dx)
= (dx/dt)[(1-v(dt/dx))/(1-v/c2(dx/dt))]
= [(dx/dt)-v]/[1-v/c2(dx/dt)]

and consequently:

dx/dt = [dx'/dt'+v]/(1+v/c2(dx'/dt')], which is what one would expect.

The Attempt at a Solution



But then I inquired about what happens when dv is non-zero, since the entire equation looks like:

dx' = [vγ3(x-vt)-γt]dv + γ(dx-v.dt)
dt' = [vγ3(t-v/c2x)-γt]dv + γ(dt-v/c2.dx)

My TA told me that the equations don't hold, and that people have actually proven that the equation doesn't hold unless dv=0. My question is: why? I feel that my TA was just trying to escape answering the question.
 

Answers and Replies

  • #2
BruceW
Homework Helper
3,611
119
I thought I would say that your derivation is not quite right. (This often happens in long derivations, it is almost certain to go wrong at some part).
Harrisonized said:
dt' = dγ[t-(v/c2)x]+γ[dt-(v/c2)x-(1/c2)x.dv)]
This isn't right, the term -γ(v/c2)x should be -γ(v/c2)dx But luckily in your next equation for dt', you do have the dx, so maybe it was just a mis-type that you forgot the d?

Also, dγ is not right. (You can tell because it is not dimensionally correct). But because dγ cancels out in the final expression, it does not matter, and you still get the right answer at the end.

About your question... v is the relative velocity between the two frames of reference. What do you know about the allowed frames of reference in special relativity? This will give the answer to why dv must be zero.
 
  • #3
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Good catch. It is a typo, but PF won't let me edit the first post.

Anyways, I don't see why terms can just be ignored if they are dimensionally incorrect. For example, I could argue that dx'/dt' is dimensionally correct when dx/dt=0. I don't agree with just erasing terms just because they're dimensionally incorrect. For example, I can write the motion of a particle on a line as

x=(t2-1).sin(t)

That definitely does not have the correct units, and if I wanted to find dx/dt, it still won't have the correct units.

Surely, there must be a better explanation for throwing it out? Maybe part of the construction?

About your question... v is the relative velocity between the two frames of reference. What do you know about the allowed frames of reference in special relativity? This will give the answer to why dv must be zero.
The only allowed reference frame is the inertial reference frame. But how do you know it's inertial, and what do you do for cases in which v is changing? I mean, in real life, v can change, right?
 
  • #4
BruceW
Homework Helper
3,611
119
Anyways, I don't see why terms can just be ignored if they are dimensionally incorrect. For example, I could argue that dx'/dt' is dimensionally correct when dx/dt=0. I don't agree with just erasing terms just because they're dimensionally incorrect.
Yes, I agree, I'm not saying that any terms should be erased because they are dimensionally incorrect. I just mean your equation for dγ is not right. So you should go back to γ, and find dγ again, and do the derivation again using the correct dγ.

Harrisonized said:
The only allowed reference frame is the inertial reference frame. But how do you know it's inertial, and what do you do for cases in which v is changing? I mean, in real life, v can change, right?
This webpage gives a good explanation about inertial frames: http://en.wikipedia.org/wiki/Inertial_frame_of_reference
And I'm not sure what you mean by v can change in real life... The frames of reference are what we define them to be. Do you mean "what if we want to use an inertial frame of reference which is rigidly attached to an accelerating observer"? Well, in special relativity, we can't. But this doesn't mean we can't make calculations about the motion of that accelerating person, it's just that we must use an inertial frame of reference to make the calculations with respect to.
 
  • #5
209
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Let me try that again.

γ=(1-v2/c2)-1/2
dγ=(-1/2)(1-v2/c2)-3/2(-2v/c2).dv)
dγ=γ3(v/c2).dv)

I see. It looks dimensionally correct now. γ is unit-less and v/c2).dv cancels into a unit-less quantity.
 
  • #6
BruceW
Homework Helper
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119
yep, that's right :)
 

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