# Active vs Passive Lorentz transformation

• I
Hi. First, excuse my English.

In my lecture notes on classical electrodynamics, we are introduced to the Lorentz transformations: a system S' moves relative to a system S with positive veloticy v in the x axis (meassured in S), spacial axis are parallel, origin of times t and t' coincide, etc., which gives the well-known transformation of coordinates of one relative to the other: http://www.relativitycalculator.com...cities/Lorentz_Equations_in_S_and_S-prime.png (sorry, I don't know how to write equations here).

As I understand, this would be a passive transformation. This transforms the same event from S coordinates to S' coordinates.

However, some pages later it says: "Active Lorentz boosts transform the (x, y, z) spacial coordinates and instant t in which it happens into (x', y', z') and t' given in the same reference system S with the expresions:" and proceeds to write the same expressions as before EXCEPT the sign of v is changed. So like if the primed and non-primed were exchanged in the picture. It further states that the change of sign is due to the active nature of these transformations as opposed to the passive one of those.

But I'm not getting why. As I understand it, an active transformation can be interpreted as bringing what another system S' sees to our system S; either way, the components are the same. Obviously, this doesn't match the change in sign in the velocity. Any help?

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I read that, @Orodruin, but my questions remain.

Orodruin
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One changes the coordinates in space-time, the other physically changes what is in the space-time but keep the coordinates fixed. There really is nothing more to it. Whether you get a sign or not depends on your conventions for which active transformation you compare with which passive.

Going to a coordinate system whose origin moves at v gives different coordinates than physically boosting all velocities by v. (It gives the same coordinates as physically boosting all velocities by -v.)

• vanhees71 and voila
I know that going to a coordinate system whose origin moves at v gives the same coordinates as physically boosting everything by -v and different to v. But is this ("physically boosting everything by v") what we mean when we say "active boost"? Seems to me rather strange, for active and passive transformations should give the same numerical coordinates, and we would be making a difference here.

Orodruin
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Seems to me rather strange, for active and passive transformations should give the same numerical coordinates, and we would be making a difference here.
They only give the same coordinates if you do the transformations in opposite directions. Compare with the rotation case. You get the same coordinates if you rotate the axes 10 degrees counter clockwise as you do if you rotate the object 10 degrees clockwise.

• voila
Yes, @Orodruin, I know passive means changing the basis and active means changing the vector in exactly the inverse way, such that numerical coordinates match, but that's not what's happening here as far as I can see. The explanation I understand from the notes gives different numerical coordinates.

As I see from your lecture cite, the word 'active' for the lorentz transformation is not adequate. The lorentz transformation is always a passive transformation because it only gives you the coordinates in a different reference frame, but the event remains the same. I wouldn't trust what it says in your lecture notes: Maybe it was mistake from your teacher.

It's definitely not a mistake, @Seba Gonzalez M; he is a terrific and scrupulous theoretical physicist, and he goes on later on the notes using it.

Orodruin
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Yes, @Orodruin, I know passive means changing the basis and active means changing the vector in exactly the inverse way, such that numerical coordinates match, but that's not what's happening here as far as I can see. The explanation I understand from the notes gives different numerical coordinates.
I already told you the point. You only get the same coordinates if the active transformation boosts in the opposite direction wrt the direction in which the passive moves the axes. Your teacher is talking about the transformation that boosts in the same direction. If you do two transformations, one active and one passive, boosting and then changing the coordinate axes in the same direction, you should end up with the same coordinates you started with. There is nothing more to see here.

Ibix
A passive Lorentz transformation I understand - I choose a different set of coordinates. But what does an active Lorentz transformation mean, physically? An active rotation I understand - you get hold of an object in space and rotate it through x degrees. The analogous operation in spacetime would seem to be a hyperbolic rotation. But how does one rotate a worldline?

Orodruin
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A passive Lorentz transformation I understand - I choose a different set of coordinates. But what does an active Lorentz transformation mean, physically? An active rotation I understand - you get hold of an object in space and rotate it through x degrees. The analogous operation in spacetime would seem to be a hyperbolic rotation. But how does one rotate a worldline?
There is nothing conceptually different with an active Lorentz transformation wrt an active rotation. In fact, an active rotation is a special case of an active Lorentz transformation.

• vanhees71
@Orodruin help me get back on track, please, because I think I'm making a non-needed mess out of this, which seems rather simple. Let's state the facts clear:

The typical Lorentz transformations which appear in the image on my first comment transform the coordinates of an event in system S to the coordinates of an event in system S' which moves with velocity v (in the base S) with respect to S. So, it's a passive transformation of one event from one system to the other.

Then, what my question addresses, is that it later says "an active transformation behaves this way:" an proceeds to write the same equations with a change in the sign of v (so, the inverse transformation). But I really don't know what we are doing there. What the primed and non-primed coordinates mean, etc.

Orodruin
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You are building a problem where there is none. The active and passive transformations take exactly the same form if you do them in opposite directions. The actual change in the coordinates under the transformations is a different thing. Both of the transformations listed in your picture are Lorentz transformations and can be active or passive, depending on which direction the active boost or passive transformation is taken. If you take the left one as a passive transformation and the right one as an active, they are in the same direction (boosting and changing axes in the same direction and so obviously one is the inverse of the other - if you boost something at rest and then change coordinates to the new rest frame coordinates, you should end up with the thing being at rest again).

Ibix
There is nothing conceptually different with an active Lorentz transformation wrt an active rotation.
So - a passive LT gives me the answer to a question like "the coordinates of one end of a rod are ##(x, t)## in frame S; what are they in S'?". An active LT gives me the answer to a question like "the coordinates of one end of a rod at rest in frame S are ##(x,t)## in that frame; what would they be in S if I boosted the rod to be at rest in S'?"

Do I have that right? If so, what was worrying me was that I was interpreting the active transformation as "you set up your axes and then rotate the block universe under them", which struck me as unphysical. But that's not what we're doing, as noted in the previous paragraph.

In fact, an active rotation is a special case of an active Lorentz transformation.
Really? Are you using "Lorentz transform" to include both purely spatial rotations and boosts here, are you talking Thomas precession, or am I missing something again?

• voila
Orodruin
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An active LT gives me the answer to a question like "the coordinates of one end of a rod at rest in frame S are (x,t)(x,t)(x,t) in that frame; what would they be in S if I boosted the rod to be at rest in S'?"

Do I have that right?
Right.

Really? Are you using "Lorentz transform" to include both purely spatial rotations and boosts here, are you talking Thomas precession, or am I missing something again?
The set of Lorentz transformations is the space-time transformations that preserve the Minkowski metric, this includes spatial rotations by definition. If you just consider boosts your group is not closed (or rather, you do not have a group as groups are closed by definition).

• voila and vanhees71
Ibix