Deriving length contraction using spacetime

In summary: A^t - \beta A^x)$$ $$A'^x... = \gamma(A^x - \beta A^t)$$ $$A'^y... = A^y$$ $$A'^z... = A^z$$In summary, using the invariance of the spacetime interval, we can derive the time dilation formula without explicitly using the Lorentz transformations. This can also be applied to derive the length contraction formula, by considering two simultaneous events in one frame and their coordinates in another frame. Four-vectors can also be used to derive these formulas, transforming according to the Lorentz transformations. Most textbooks on special relativity cover the use of four-vectors in detail.
  • #1
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.

Please help.

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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  • #2
I did it as attached.

2021-04-17 13.40.43.jpg
 
Last edited:
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  • #3
Kaguro said:
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.

Please help.

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##.
 
  • #4
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.
 
  • #5
  • #6
PeroK said:
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 ?
 
  • #7
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?
 
  • #8
Kaguro said:
Wouldn't that be just plain derivation using Lorentz Transformations ?
Not if you use the invariance rather than the explicit LT.
 
  • #9
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.
 
  • #10
Kaguro said:
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.
 
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  • #11
Kaguro said:
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.
 
  • #12
PeroK said:
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.
 
  • #13
Kaguro said:
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.
 
  • #14
My book defines four-vectors as quantities with four components that transform via LTs under a change of frame.

PeroK said:
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! :biggrin:

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!
 

1. What is length contraction in the context of spacetime?

Length contraction is a phenomenon in which an object appears shorter in the direction of motion when observed from a different frame of reference. This is a consequence of the theory of special relativity, which states that the laws of physics are the same for all inertial observers.

2. How is length contraction derived using spacetime?

Length contraction can be derived using the Lorentz transformation equations, which describe how measurements of space and time change between two frames of reference moving at constant velocities relative to each other. By applying these equations to the concept of simultaneity, it can be shown that the length of an object will appear shorter in the direction of motion to an observer in a different frame of reference.

3. What factors affect the amount of length contraction observed?

The amount of length contraction observed depends on the relative velocity between the two frames of reference, as well as the direction of motion of the object. The closer the object's velocity is to the speed of light, the greater the amount of length contraction will be. Additionally, length contraction only occurs in the direction of motion, so an object moving perpendicular to the observer's line of sight will not appear shorter.

4. Does length contraction have any practical applications?

Yes, length contraction has been observed and measured in various experiments, and it is a crucial factor in the operation of particle accelerators and other high-speed technologies. It also helps explain phenomena such as the muon decay paradox, in which high-energy particles called muons are able to travel much further than expected before decaying.

5. Is length contraction the only effect of special relativity on measurements of space and time?

No, there are several other effects of special relativity, including time dilation, which is the slowing of time for objects in motion, and the relativity of simultaneity, which states that two events that appear simultaneous to one observer may not appear simultaneous to another. These effects are all interconnected and can be derived using the same principles of spacetime.

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