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The primed reference frame is moving relative to the unprimed frame. So if we were to take the lorentz transformation of point P from the unprimed to primed, would it be the point A or B that it returns ? Assuming that the Lorentz transformation is passive i.e. we are talking about the same event in two frames, my first guess was B, because only at B will the primed reference frame see P to occur. A on the other hand is what the unprimed reference frame sees when P occurs. Is this true ?

PA is parallel to the x axis and PB is parallel to the x' axis.

Qmechanic
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Ajaykrishnan R
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  • I do not understand the question. $P$ lays on the $t$-axis, so it's $x$ coordinate is 0. In the other coordinate frame, it will be at the same point (unless you re-draw the diagram with the new coordinates being graphically orthogonal), just that the coordinates will be different, with the $x'$ coordinate being non-zero. – Koschi Feb 17 '23 at 09:31

2 Answers2

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See below for an update to address the "passive transformation".

The [active] Lorentz-boost-transformation of event P is neither your events A nor B,
but an event between your A and B
that is on the hyperbola centered at the origin that passes through P.

The boosted event is joined to the original event by a line that is not parallel to either axis of these two frames.

From my spacetime diagrammer https://www.desmos.com/calculator/emqe6uyzha ,

robphy-spacetimeDiagrammer

the boosted events lie on a common hyperbola (which are events that "equidistant in future time from the meeting event" according to the Minkowski metric).

The spatial-axis of a frame is parallel to the tangent to that hyperbola (shown as a dotted line).

It may be instructive to consider the Euclidean version of the construction (move the slider to $E=-1$):

robphy-spacetimeDiagrammer-Euc

The rotated point is joined to the original point by a line that is not parallel to either axis of these two coordinate systems.

UPDATE:

Concerning the passive transformation... the events are left unchanged in spacetime. But now the passive Lorentz boost provides the coordinates of an event in another inertial frame.

In this case, your event B is the event simultaneous with P, according to the moving inertial observer.

The diagram below is for $v=(3/5)c$. So, the transformation would assign the event $P$ with red-coordinates $(t=1,x=0)$ with blue coordinates $(t=1.25, x=-0.75)$.

The passive transformation identifies event B as the event the moving inertial observer says is simultaneous with P and calculates the blue time-coordinate of P to be equal to the blue time-coordinate of B as $t=1.25$.

(The transformation will also identify an event (not shown in the image) on the moving x-axis (at blue time t=0) that is "at the same place as P" according to blue. The spatial-coordinate of that event is $x=-0.75$.)

robphy-spacetimeDiagrammer-passive

robphy
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    The Lotentz transform of P is still P. It just has different coordinates in the primed frame (sometimes it is called P’ for that reason). Can you clarify the sense in which you are connecting P with this other event? – Dale Feb 17 '23 at 05:59
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    @Dale In this diagram, these are active transformations. Given the event "1 tick after the meeting" along the "at-rest-in-this-frame" inertial observer's worldline, what are the other "1-tick after the meeting" events along other inertial-observer worldlines? – robphy Feb 17 '23 at 06:02
  • @robphy As said in this answer : https://physics.stackexchange.com/a/135133, the active lorentz transformation talks about 2 different physical realities, there arent 2 different coordinate frames when you talk about an active transformation just the single lab frame in which both the original and the different reality are described. The coordinates the primed frame gives to P is the coordinates that are simultaneous to the event P in the primed frame which is B. So is this conclusion valid if we interpret the transform in a passive form ? – Ajaykrishnan R Feb 17 '23 at 07:08
  • @AjaykrishnanR I have made an update to address the passive transformation. – robphy Feb 17 '23 at 10:07
  • @robphy Thanks! I see it now, just one more thing, isnt the active form of LT the one which has x' = Ɣ(x - vt) ? [The primed frame moves along the +ve x axis of the unprimed frame.]. Since Ɣ > 1 this means that the basis vector in the moving frame (written in the unprimed frame) is longer than the basis vector in the rest frame, right ? – Ajaykrishnan R Feb 17 '23 at 12:10
  • @AjaykrishnanR I find the prime-unprime notation ambiguous, partly because it's not used consistently among various authors. If I am interpreting your question correctly, I think you are describing the fact that the Minkowski-magnitude of the adjacent side (the base) of a Minkowski-right-triangle is as long or longer than the hypotenuse (since $\gamma=\cosh\theta\geq 1$, where $\theta$ is the rapidity such that $(v/c)=\tanh\theta$). – robphy Feb 17 '23 at 18:55
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enter image description here

Look at this Minkowski diagram point P move parallel to the x‘ axis , the event point is then $~E(ct‘,x‘)$

Eli
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