# Deriving length contraction using spacetime

• Kaguro
Kaguro
Homework Statement
Derive time dilation and length contraction using invariance of spacetime interval.
Relevant Equations
##ds^2 = dx^2+dy^2+dz^2 - c^2 dt^2##
Deriving time dilation was easy:

Imagine two events in frame O' at the same location.
##ds^2 = -c^2 dt'^2##

The same viewed in O frame is:
##ds^2 = dx^2+dy^2 + dz^2 - c^2 dt^2##
##\Rightarrow dx^2+dy^2 + dz^2 - c^2 dt^2 = -c^2 dt'^2##
##\Rightarrow (\frac{dx}{dt})^2+(\frac{dy}{dt})^2+ (\frac{dz}{dt})^2 - c^2 = -c^2(\frac{dt'}{dt})^2##

But since these events are at the same location in O', the dx, dy, dz is hence the amount by which O' moves in O in time dt.
Therefore,
##v^2 -c^2 = -c^2 (\frac{dt'}{dt})^2##
##\Rightarrow (\frac{dt'}{dt})^2 = 1-\frac{v^2}{c^2}##
##\Rightarrow (\frac{dt'}{dt}) = \sqrt{1-\frac{v^2}{c^2}}##
##\Rightarrow dt = \gamma dt'##

But for length contraction

I choose two events in O which are simultaneous. So dt=0
##ds^2 = dx^2+dy^2 + dz^2 = dx'^2 + dy'^2 + dz'^2 - c^2 dt'^2##
Here I can not equate any of these with the amount by which O' moves in O.

P.S.: Why has the website become so weird ? The preview button is shifted and works differently, and I am asked to "submit homework statement" and "relevant equations" twice...

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I did it as attached. Last edited:
• Kaguro
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Homework Statement:: Derive time dilation and length contraction using invariance of spacetime interval.
Relevant Equations:: ##ds^2 = dx^2+dy^2+dz^2 - c^2 dt^2##

Deriving time dilation was easy:

Imagine two events in frame O' at the same location.
##ds^2 = -c^2 dt'^2##

The same viewed in O frame is:
##ds^2 = dx^2+dy^2 + dz^2 - c^2 dt^2##
##\Rightarrow dx^2+dy^2 + dz^2 - c^2 dt^2 = -c^2 dt'^2##
##\Rightarrow (\frac{dx}{dt})^2+(\frac{dy}{dt})^2+ (\frac{dz}{dt})^2 - c^2 = -c^2(\frac{dt'}{dt})^2##

But since these events are at the same location in O', the dx, dy, dz is hence the amount by which O' moves in O in time dt.
Therefore,
##v^2 -c^2 = -c^2 (\frac{dt'}{dt})^2##
##\Rightarrow (\frac{dt'}{dt})^2 = 1-\frac{v^2}{c^2}##
##\Rightarrow (\frac{dt'}{dt}) = \sqrt{1-\frac{v^2}{c^2}}##
##\Rightarrow dt = \gamma dt'##

But for length contraction

I choose two events in O which are simultaneous. So dt=0
##ds^2 = dx^2+dy^2 + dz^2 = dx'^2 + dy'^2 + dz'^2 - c^2 dt'^2##
Here I can not equate any of these with the amount by which O' moves in O.

P.S.: Why has the website become so weird ? The preview button is shifted and works differently, and I am asked to "submit homework statement" and "relevant equations" twice...
I suggest you take an object of proper length ##L## and describe the spacetime coordinates of each end in a second frame where the object is moving with constant velocity ##\vec v##.

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PS you can also use this idea to give a much simpler derivation of time dilation. You start with a clock in its rest frame having the coordinates: ##(t', 0, 0, 0)## and in other frame coordinates ##(t, v_xt, v_yt, v_zt)##.

Note: if the object does not pass through the spacetime origin, then we have coordinates ##(t', x'_0, y'_0, z'_0)## and ##(t, x_0 + v_xt, y_0 + v_yt, z_0 + v_zt)## and it all works out just as well.

Kaguro
PS you can also use this idea to give a much simpler derivation of time dilation. You start with a clock in its rest frame having the coordinates: ##(t', 0, 0, 0)## and in other frame coordinates ##(t, v_xt, v_yt, v_zt)##.

Note: if the object does not pass through the spacetime origin, then we have coordinates ##(t', x'_0, y'_0, z'_0)## and ##(t, x_0 + v_xt, y_0 + v_yt, z_0 + v_zt)## and it all works out just as well.
Wouldn't that be just plain derivation using Lorentz Transformations ?

Kaguro
I have a related question, if someone asks me to prove length contraction using four vectors, would that be similar to what I originally asked in this thread?

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Wouldn't that be just plain derivation using Lorentz Transformations ?
Not if you use the invariance rather than the explicit LT.

Kaguro
Okay..
Can you recommend me a good place/book which illustrates the use of 4-vectors to derive everything we have derived using LTs? Focusing heavily on applications of four vectors.

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Wouldn't that be just plain derivation using Lorentz Transformations ?
Let me show you what I mean.

We have two events described in two different reference frames by the coordinates:
$$(t'_0, 0, 0, 0), (t'_1, 0, 0, 0)$$ and $$(t_0, x_0, y_0, z_0), (t_1, x_0 + v_xt_1, y_0 + v_yt_1, z_0 + v_zt_1)$$
Instead of using the LT and the explicit relationship between these coordinates, we can use only the invariance of the spacetime interval, to give: $$(\Delta s')^2 = (\Delta s)^2$$ $$\Rightarrow \ -c^2(\Delta t')^2 = -c^2(\Delta t)^2 + (v_x^2 + v_y^2 + v_z^2)(\Delta t)^2 = -c^2(\Delta t)^2 + v^2(\Delta t)^2 = -c^2(1 - \frac{v^2}{c^2})(\Delta t)^2$$ And we can get the time dilation formula.

• Kaguro
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Okay..
Can you recommend me a good place/book which illustrates the use of 4-vectors to derive everything we have derived using LTs? Focusing heavily on applications of four vectors.
Most textbooks on SR should present four-vectors in all their glory. Although they may wait until you are studying the energy-momentum of particle collisions.

A critical point is that all four-vectors transform according to the Lorentz Transformation. E.g. for boosts in the x-direction: $$A'^t = \gamma(A^t - \frac v c A^x), \ A'^x = \gamma(A^x - \frac v c A^t), \ A'^y = A^y, A'^z = A^z$$ And you can check that ##(ct, x, y, z)## is a four-vector, for example.

Kaguro
Let me show you what I mean.

We have two events described in two different reference frames by the coordinates:
$$(t'_0, 0, 0, 0), (t'_1, 0, 0, 0)$$ and $$(t_0, x_0, y_0, z_0), (t_1, x_0 + v_xt_1, y_0 + v_yt_1, z_0 + v_zt_1)$$
Instead of using the LT and the explicit relationship between these coordinates, we can use only the invariance of the spacetime interval, to give: $$(\Delta s')^2 = (\Delta s)^2$$ $$\Rightarrow \ -c^2(\Delta t')^2 = -c^2(\Delta t)^2 + (v_x^2 + v_y^2 + v_z^2)(\Delta t)^2 = -c^2(\Delta t)^2 + v^2(\Delta t)^2 = -c^2(1 - \frac{v^2}{c^2})(\Delta t)^2$$ And we can get the time dilation formula.
Okay! So I could also say this in this way: Since (ct,x,y,z) is a four vector so I can say that the coordinates of the clock is also a four vector and its norm should remain invariant under different frames. And this is exactly the invariance of the spacetime interval.

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Okay! So I could also say this in this way: Since (ct,x,y,z) is a four vector so I can say that the coordinates of the clock is also a four vector and its norm should remain invariant under different frames. And this is exactly the invariance of the spacetime interval.
To be precise, I should have said that ##(ct, x, y, z)## are the components of a four-vector.

An important point is that there is an invariant quantity associated with every four-vector: $$A^2 = -c^2(A^t)^2 + (A^x)^2 + (A^y)^2 + (A^z)^2$$ The proof of invariance follows in the same way as the proof that the spacetime interval is invariant - by using the LT directly.

The other point is that the proper time shown on a clock is a measure of this invariant spacetime interval along the worldline of the clock.

And, of course, the invariant quantity associated with the energy-momentum four-vector of a particle is its mass. This is where SR pulls everything together - mass, energy & momentum - in a way that is not even hinted at in classical mechanics.

Kaguro
My book defines four-vectors as quantities with four components that transform via LTs under a change of frame.

The other point is that the proper time shown on a clock is a measure of this invariant spacetime interval along the worldline of the clock.
Yes! So the invariant quantity associated with position four vector is spacetime interval,
The invariant quantity associated with velocity four vector is speed of light,
The invariant quantity associated with energy-momentum four vector is rest mass energy!(hence mass)

Cool!