Length contraction - problems with transformation and time

[kapitan90]

Homework Statement

Hello, I have a question about length contraction transformation.
In my textbook it looks like this:
$$x_1=γ(x_1'+ut'), x_2=γ(x_2'+ut')$$ If the coordinates of the two events are $$(x_1,t_1), (x_2, t_2)$$, why is t used instead of t_1and t_2?

The second problem I have is with the length contraction transformation. If we use the formula in my textbook: $$x_1=γ(x_1'+ut'), x_2=γ(x_2+ut')$$ then indeed $$Δx=γΔx'$$ and $$Δx'=Δx/γ$$ so we do get length contraction.

But what about starting with: $$x_1'=γ(x_1-ut) x_2'=γ(x_2-ut)$$ Then $$Δx'=γΔx$$ and it looks like we obtain length expansion rather than contraction. What's wrong with my reasoning?

Homework Equations

1.Why is the time of two events equal to t not $$t_1$$and $$t_2$$
2. What's wrong with the second transformation for length contraction?

 

[Doc Al]

Homework Statement

Hello, I have a question about length contraction transformation.
In my textbook it looks like this:
$$x_1=γ(x_1'+ut'), x_2=γ(x_2'+ut')$$ If the coordinates of the two events are $$(x_1,t_1), (x_2, t_2)$$, why is t used instead of t_1and t_2?

For the general Lorentz transformation, you would have t_1' and t_2'. But here they are applying it to length contraction, so t_1' = t_2' = t'. To derive the length contraction formula, the positions must be measured at the same time in the 'moving' frame.

The second problem I have is with the length contraction transformation. If we use the formula in my textbook: $$x_1=γ(x_1'+ut'), x_2=γ(x_2+ut')$$ then indeed $$Δx=γΔx'$$ and $$Δx'=Δx/γ$$ so we do get length contraction.

But what about starting with: $$x_1'=γ(x_1-ut) x_2'=γ(x_2-ut)$$ Then $$Δx'=γΔx$$ and it looks like we obtain length expansion rather than contraction. What's wrong with my reasoning?

In the first application, you have essentially something at rest in the unprimed frame being measured by the primed frame. (Those measurements must take place at the same time in the primed frame.)

In the second application, things are reversed. It's the unprimed frame measuring the length of something at rest in the primed frame. (Each frame measures the length of something in the other frame to be contracted.)

 

[kapitan90]

Thanks for your reply. I understood the first problem, but I still have problems with the second one.
These are the coordinates of an object in a frame S', moving with velocity u with respect to S, in which the object is in rest: $$x_1'=γ(x_1-ut) x_2'=γ(x_2-ut)$$ Then $$Δx'=γΔx$$ so the object appears longer in the S' frame. I guess I didn't understand your explanation.

 

[Doc Al]

These are the coordinates of an object in a frame S', moving with velocity u with respect to S, in which the object is in rest: $$x_1'=γ(x_1-ut) x_2'=γ(x_2-ut)$$ Then $$Δx'=γΔx$$ so the object appears longer in the S' frame.

The unprimed coordinates represent the measurement of the ends of an object that is at rest in the primed frame.

Note the symmetry. If the object is at rest in the unprimed frame, its rest length would be Δx. Its measured length according to the primed frame would be Δx' = Δx/γ. (The measurement of the ends of a moving object must be done at the same time, so Δt' = 0.)

Now if the situation is reversed and the object is at rest in the primed frame, everything is swapped around: If the object is at rest in the primed frame, its rest length would be Δx'. Its measured length according to the unprimed frame would be Δx = Δx'/γ. (The measurement of the ends of a moving object must be done at the same time, so Δt = 0.)