Length Contraction: Need Help Understanding Lorentz Transformation

[Piyu]

Hello! need some help with length contraction.

So according to lorentz transformation we got

I don't know how to put symbols so ill use Y as gamma since they look alike :)

dx' = Y dx - u Y dt

So proper length refers to the frame where dt = 0 since u are measuring the ends at the same time. Therefore, U get

dx' = Y dx

x' = Y x

L = Y L_0

Which is wrong.

Would appreciate some help please :)

 

[Meir Achuz]

The way you have written it, dt=0. This is in the unprimed frame.
You should write dx=dx'/Y.

 

[Piyu]

But in any case, L_0 will still be dx while L is dx' isn't it? since the dx frame is the one that measures the proper length

 

[Dale]

So proper length refers to the frame where dt = 0 since u are measuring the ends at the same time.

Careful here, dt=0 and dt'=0. They are measuring DIFFERENT pairs of events.

I can be more explicit, but hopefully that is enough.

 

[Piyu]

Mind explaining it in abit more detail? I'll appreciate it. This part has been hurting my brains for the past few weeks and i still haven't been able to come up with a reasonable argument or mindset in assigning the t_0 and L_0 of time dilation and length contraction.

All I've concluded so far is as long as for time dilation, u use t_0 as the person who is moving with the clock. (eg, a guy on the rocket and you on earth, he's measures proper time of him traveling from Earth to point X)

For length contraction, L_0 is the person who is can measure the distance at the same time meaning he is not moving with respect with the object.

Is that correct?

 

[GrayGhost]

Hello! need some help with length contraction.

So according to lorentz transformation we got

I don't know how to put symbols so ill use Y as gamma since they look alike :)

dx' = Y dx - u Y dt

So proper length refers to the frame where dt = 0 since u are measuring the ends at the same time. Therefore, U get

dx' = Y dx

x' = Y x

L = Y L_0

Which is wrong.

Would appreciate some help please :)

Well, you are considering "the proper length", which means the measurer is at rest with that being measured. So it's not that dt=0, but rather that (the velocity) u=0. So the term uYdt vanishes because u=0. The variable dt would be the time it takes light to travel from one end of the rod to the other end. Here's the thing though ... when u=0, then (gamma) Y = 1. And your final equation then becomes this ...

dx' = Y dx - u Y dt
dx' = (1) dx - (0) Y dt
L = L_0​

Does that help?

GrayGhost

 

[Piyu]

Well, you are considering "the proper length", which means the measurer is at rest with that being measured. So it's not that dt=0, but rather that (the velocity) u=0. So the term uYdt vanishes because u=0. The variable dt would be the time it takes light to travel from one end of the rod to the other end. Here's the thing though ... when u=0, then (gamma) Y = 1. And your final equation then becomes this ...

dx' = Y dx - u Y dt
dx' = (1) dx - (0) Y dt
L = L_0​

Does that help?

GrayGhost

Wow, i guess i was totally wrong in my understanding of the equation.

Btw, how could i use this then to derive length contraction?

In this case, the u will not be zero and hence it'll look like

dx'=Y(dx - u dt)

 

[GrayGhost]

Wow, i guess i was totally wrong in my understanding of the equation.

Btw, how could i use this then to derive length contraction?

In this case, the u will not be zero and hence it'll look like

dx' = Y(dx - u dt)

Piyu,

That is the equation to use, ie ...

dx'=Y(dx - u dt)​

dx' is the contracted length, and the greater the u, the greater dx' is contracted wrt dx as the reference. Multiplying it out ...

dx' = Y(dx - u dt)

dx' = Ydx - Yudt​

Now, you'd think that dx' > dx since Y>1 when u>0 ... assuming you ignore the right term on the right side of the eqn. And you'd be right! However, the right term (-Yudt) is a also function of time, and as it turns out, the magnitude of Yudt (which is subtracted from Ydx) will always be enough to ensure dx' < dx. Pretty cool, huh?

GrayGhost

 

[Piyu]

Ah icic, but how do i turn that equation to form the length contraction expression of L = L_0 / Y

 

[GrayGhost]

Ah icic, but how do i turn that equation to form the length contraction expression of L = L_0 / Y

The transformation is ...

dx' = Y(dx - u dt)​

Per the observer who moves relatively, the moving contracted length is L = x - u*t, which means dL = dx - u*dt, and so ...

dx' = Y(dx - u dt)​

dx' = YdL​

or ...

dL = dx'/Y​

since dx' = dL₀ ...

dL = dL₀/Y​

So over the entire interval ...

L = L₀/Y​

GrayGhost

 

[Dale]

For length contraction, L_0 is the person who is can measure the distance at the same time meaning he is not moving with respect with the object.

Is that correct?

No, this is not correct. Let me explain conceptually what we are doing in the length contraction and time dilation formulas and then walk you through the derivation.

In time dilation we have one worldline, the clock, and we pick two events on that worldline and calculate the time between those events in the frame where the clock is at rest and a frame where the clock is moving. That is relatively simple to understand, and the frame where it is at rest always records the minimum amount of time.

Length contraction is not so simple. In length contraction we have two parallel worldlines, the two ends of the rod, and each frame picks two events which are simultaneous (dt=0 and dt'=0) in that frame and calculates the distance between those different pairs of events. It is critically important in deriving length contraction that you realize that the measurements are simultaneous in both frames and that they are therefore different pairs of events in each frame.

So for the derivation let A be one end of the rod and B be the other end of the rod. Then, in the rest frame we can write:
x_A(t)=0
x_B(t)=L_0
x_B(0)-x_A(0)=L_0

Now the Lorentz transform gives us
x&#039;_A(t)=\gamma t v
t&#039;_A(t)=\gamma t
x&#039;_B(t)=\gamma (L_0 + t v)
t&#039;_B(t)=\gamma (t + L_0 v/c^2)

Here is where you are making your mistake. You are calculating
x&#039;_B(0)-x&#039;_A(0)=\gamma L_0
but note that
t&#039;_B(0)-t&#039;_A(0) \ne 0
therefore the calculated distance is not the length of the rod in the primed frame since the positions of the ends are not measured at the same time and the rod has moved between the two measurements. To calculate the length of the rod we must use simultaneous events in the primed frame. We do this by using each expression for t' to eliminating the t from the corresponding expressions for x'.
x&#039;_A(t&#039;) = t&#039; v
x&#039;_B(t&#039;) = t&#039; v + L_0/\gamma
x&#039;_B(0&#039;)-x&#039;_A(0&#039;)=L_0/\gamma

Hope that clears it up.