Confused by Lorentz transformation equation

[PeroK]

As an appropriate starting point, can we state the coordinates of the beacons A and B, relative to A, for example, such that A is at the origin, and B is 500 m away. Then we use the Lorentz transformation to change both coordinates of A and B. Am I on the right track?

Yes, the Lorentz Transformation is given in your OP.

$$\Delta x' = \gamma(\Delta x - v\Delta t) = \frac{5}{4}\cdot 500 = 625 \textnormal{ m}$$

In fact, they didn't even specify an origin. The origin is irrelevant to the distance between two events. To expand on that equation:

We have two simultaneous events with locations $x$ and $x + \Delta x$ in the platform frame ($\Delta x = 500m$). Simultaneous means $\Delta t = 0$.

The above is a legitimate form of the Lorentz Transformation, as:
$$x'_1 = \gamma(x_1 - vt_1), \ x'_2 = \gamma(x_2 - vt_2)$$$$\Rightarrow \Delta x' = x'_2 - x'_1 = \gamma(x_2 - vt_2) - \gamma(x_1 - vt_1) = \gamma([x_2-x_1] - v[t_2-t_1])$$$$\Rightarrow \Delta x'= \gamma(\Delta x - v\Delta t)$$And, you can start to see my point that the location of the train, the origin, any observers and any light signals are irrelevant. It's all about the coordinates!

 

[Sagittarius A-Star]

The answer is not backed up by any explanation. It literally reads as I've written in the first post. Just the delta equation, then replaced with numbers, with the final numerical answer. It also says underneath "One may use $L\cdot \gamma = L_0$".

That is literally...it.

From that cited partly answer, you can continue to answer question d) with the inverse LT:
$\Delta x = \gamma (\Delta x' + v \Delta t')$
$500m = \gamma (625m + v \Delta t')$
... and then solve for $\Delta t'$.

 

[Orodruin]

1) SR is a theory of spacetime transformations between reference frames. It does not depend on the observations by any particular observer. The location of the train is of no consequence. The only relevant point is that the train and the platform have a relative speed of 0.6c in the direction aligned with the platform. In other words, the train could be nowhere near that particular station and the answer would be the same - as long at the train is moving in the given direction.

(My emphasis)
I’d rather say SR is a theory of spacetime geometry. The transformations between inertial frames are of course part of that in the same manner that rotations form part of Euclidean geometry, but I would say that the fundamental part is the geometry.

 

[PeroK]

(My emphasis)
I’d rather say SR is a theory of spacetime geometry. The transformations between inertial frames are of course part of that in the same manner that rotations form part of Euclidean geometry, but I would say that the fundamental part is the geometry.

That's the next insight, of course. The initial insight is that the events don't have to be beacons emitting light signals. The LT doesn't care whether the events are luminous beacons or geiger-counter clicks in a sealed box.

 

[Ibix]

I hope I'm not asking for too much if you could perhaps show me how this coordinate transformation would take place (as per the homework guidelines, the solution is already apparent and I'm asking for an alternative explanation).

The fundamental physics hiding under the examples is as follows.

Spacetime is the background on which all physics happens.

A reference frame is just a way of "drawing" a coordinate system on that background. It makes it easier to do physics because you can assign numbers to different locations and express physical quantities as a function of those numbers.

A different reference frame is just a different way of drawing coordinates. Just like you can draw different grids on a piece of paper, you can define different grids on spacetime.

On a piece of paper, if we agree the origin and units for our grid, the coordinates of a point using one syatem are related to the coordinates in another by a rotation. On spacetime, if we agree the origin and units for our grids, the coordinates in two grids are related by the Lorentz transforms.

Any object you see around you is actually 4d - the 3d object you see now and an extent in time. A sphere, for example, is actually a 4d analog of a cylinder - a spherical cross section and a long length in the time direction.

Imagine drawing a very long narrow rectangle on a piece of paper. Draw a grid on it. What is the x-extent of the rectangle according to the grid? Different grids will measure different widths because their x axis makes a different angle to the rectangle. Now remember that frames in relativity are just different grids on spacetime. The extent of an object in the x direction depends on the choice of grid - that's length contraction.

Imagine two points drawn on a piece of paper at the same y coordinate. Using a different grid, will they be at the same y' coordinate? Similarly, two events in spacetime that have the same t coordinate won't have the same t' coordinate. This is the relativity of simultaneity.

Two points with different y coordinates will have some y-separation. Using a different grid they will have a different y'-separation. Similarly, two events in spacetime with some t-separation will have a different t'-separation. This is time dilation.

Note that all effects like time dilation and length contraction are just consequences of different choices of the imaginary grid you chose to use.

Go and look up Minkowski diagrams, which are just maps of spacetime drawn on a piece of paper. You can draw multiple grids on them and begin to build intuition for how measurements made with the grids relate.

 

[Herman Trivilino]

I just didn't expect there to be such an issue with the wording, but I guess such is the world of SR.

All areas of knowledge. I can't tell you how much confusion arises over the misunderstanding of a question. It leaves both the student and the helper unnecessarily confused.

 

[Sagittarius A-Star]

how on Earth can the perceived distance between these two lights be greater than the length of the platform?

Consider the two events "light beacons turning on". An event happens at a certain coordinate-location (x, y, z) and at a certain instance of coordinate-time (t). In SR, the squared spacetime-interval between two events is the same with reference to both inertial coordinate-systems.
For simplicity, consider the events with $y=z=0$. Assume, that the primed frame is the frame of the train.
$$(\Delta s)^2 = (c\Delta t)^2-(\Delta x)^2=(c\Delta t')^2-(\Delta x')^2$$Both events happen simultaneously with respect to the platform frame; $\Delta t=0$:

$0^2-(\Delta x)^2=(c\Delta t')^2-(\Delta x')^2$

$(\Delta x)^2 = (\Delta x')^2 [1-({ c\Delta t' \over\Delta x' })^2]= (\Delta x')^2 (1-v^2/c^2)$

$$\Delta x' = \gamma \Delta x \ \ \ \ \ \text{(condition:}\Delta t=0\text{)}$$.

Also, shouldn't the platform itself (as has been established) be contracted, ... ?

Yes. The "light beacons" are no events. They are objects.

Imagine, the $x'$ axis is represented by a long ruler in the train.
Now consider the following different two measurement events at the same coordinate-time $t'$:

• One person in the train reads the ruler scale at the nearby passing "light beacon 1".
• Another person in the train reads the ruler scale at the nearby passing "light beacon 2".

The distance between the two ruler scale positions is $\Delta x'$.
The coordinate-time interval between the two ruler scale reading events is $\Delta t'=0$.

$(c\Delta t)^2-(\Delta x)^2=0^2-(\Delta x')^2$

$(\Delta x')^2 = (\Delta x)^2 [1-({ c\Delta t \over\Delta x })^2]= (\Delta x)^2 (1-v^2/c^2)$

$$\Delta x' = {1 \over \gamma} \Delta x\ \ \ \ \ \text{(condition:}\Delta t'=0\text{)}$$.