Relativity: the Special and General theories

[nadja]

"I stand at the window of a railway carriage which is travelling uniformly, and drop a stone on the embankment, without throwing it. Then, disregarding the influence of the air resistance, I see the stone descend in a straight line. A pedestrian who observes the misdeed from the footpath notices that the stone falls to earth in a parabolic curve. I now ask: Do the "positions" traversed by the stone lie "in reality" on a straight line or on a parabola? Moreover, what is meant here by motion "in space" ? From the considerations of the previous section the answer is self-evident."

This is a quote from
Albert Einstein: Relativity
Part I: The Special Theory of Relativity, Space and Time in Classical Mechanics.

Could someone please explain to me why, to the pedestrian, the stone falls in a parabolic curve and what type of parabolic curve it is? thank you

 

[phinds]

Could someone please explain to me why, to the pedestrian, the stone falls in a parabolic curve

Well, the vertical motion is proportional to y**2 (for both observers) and the horizontal motion in the reference frame of the pedestrian is proportional to the speed of the train , which is linear.

SO ... if it is indeed a parabola (and I haven't checked that but if it's what you read, it's likely correct) then I can only assume that if you combine the two motions and calculate the loci, you get the equation for a parabola.

and what type of parabolic curve it is? thank you

What do you mean, what kind of parabolic curve?

 

[phinds]

Do the "positions" traversed by the stone lie "in reality" on a straight line or on a parabola?

You are assuming that one of the two frames of reference gives some kind of absolute truth, which is a false assumption. The answer to "how is the motion really described" is "it depends on your frame of reference." Neither one is absolute. All motion is relative.

 

[Drakkith]

To the pedestrian, the stone is initially traveling with some horizontal velocity, let's call it V_x, and zero vertical velocity, let's call it V_y. Gravity begins to accelerate the stone downwards at a rate of about 9.8 m/s², while the horizontal velocity remains the same.

If we graph the velocities vs time, we see that since V_x remains constant and V_y accelerates at a rate proportional to time squared, the two combine to make a parabola. In other words, the equation that defines the trajectory is that of a parabola.

Well, the vertical motion is proportional to y**2 (for both observers)

I think that should be t², not y².

 

[nadja]

To the pedestrian, the stone is initially traveling with some horizontal velocity, let's call it V_x, and zero vertical velocity, let's call it V_y. Gravity begins to accelerate the stone downwards at a rate of about 9.8 m/s², while the horizontal velocity remains the same.

If we graph the velocities vs time, we see that since V_x remains constant and V_y accelerates at a rate proportional to time squared, the two combine to make a parabola. In other words, the equation that defines the trajectory is that of a parabola.

ohh ok i got it!! thank you very much for the explanation! I'm not too good at physics.

what would t^2 and y^2 mean here?

 

[Drakkith]

what would t^2 and y^2 mean here?

Those are variables, where t is time and y is the vertical position coordinate.

 

[phinds]

I think that should be t², not y².

 

[PeroK]

I'm not too good at physics.

It might be a good idea to study some classical kinematics before trying Einstein's book. You could look at something like the Khan Academy:

https://www.khanacademy.org/science/physics

 

[nadja]

It might be a good idea to study some classical kinematics before trying Einstein's book. You could look at something like the Khan Academy:

https://www.khanacademy.org/science/physics

hmm ok, will do! thank you!!

 

[vanhees71]

This is about Newtonian dynamics/kinematics. We can explain this in two different ways:

(a) The point of view of Einstein in the train:

He uses Cartesian coordinates $(x',y',z')$ with the $z'$ axis pointing upwards. The equation of motion is
$$m \ddot{\vec{x}}'=-m \vec{g}=-m (0,0,g).$$
Einstein drops the stone initial at rest from $\vec{x}_0'=(0,0,h)$. By integrating twice he finds
$$\vec{x}'(t)=-\vec{g}{2} t^2 + \vec{x}_0'=(0,0,h-g t^2/2).$$
The trajectory from Einstein's point of view is a straight line down.

How does the pedestrian on the embankment describe this? He uses coordinates
$$\vec{x}=\vec{x}'+\vec{v} t,$$
where $\vec{v}=\text{const}=(v,0,0)$ is the velocity of the train relative to him. So he describes the motion of the stone by
$$\vec{x}=(v t,0,h-g t^2/2),$$
which is a parabola. This becomes more obvious by describing the trajectory in the form $z=f(x)$. To that end you simply eliminate time by $x=v t$ or $t=x/v$:
$$z=h-g t^2/2 = h-g x^2/(2 v^2),$$
which obviously is a parabola in the $(x,z)$ plane.

(b) The point of view of the pedestrian

He describes the motion of the stone by the same equation as einstein,
$$m \ddot{\vec{x}}=-m \vec{g}=-m(0,0,g).$$
But now from his point of view the initial conditions are $\dot{\vec{x}}(0)=\vec{v}=(v,0,0)$, $\vec{x}(0)=\vec{x}_0=(0,0,h)$. Which gives
$$\vec{x}(t)=-\vec{g} t^2/2 + \vec{v} t +\vec{x}_0=(v t,0,h-g t^2/2),$$
which is the same result we got using Einstein's point of view.

The pedestrian will also conclude rightly what Einstein observes. He just uses the inverse transformation,
$$\vec{x}'=\vec{x}-\vec{v} t=(0,0,h-g t^2/2),$$
which is also the same as Einstein gets by solving the initial-value problem using his coordinates.

 

[Nugatory]

The parabola or straight line behavior that Einstein describes is an example of Galilean relativity (worth googling). Galilean relativity was, until Einstein developed special relativity, presumed to be the right mathematical relationship between positions described by observers moving relative to one another; as we might guess from the name, it was formalized by Galileo so has been around for a very long time.

Basically special relativity and the Lorentz transformation replaced Galilean relativity and the Galilean transform.