Why must the Lorentz Transformations Be Linear?

[emob2p]

All derivations of the Lorentz Transformations I've seen assume a linear transformation between coordinates. Why must this be the case? Thanks.

 

[Hurkyl]

A transformation is linear if and only if it keeps the origin fixed, and it maps lines to lines.

 

[robphy]

In particular, a Lorentz transformation must preserve the nature of timelike lines, which represent inertial particles.

 

[emob2p]

So if the transformation were not linear, then an object would appear to be accelerating in one inertial frame but moving at a constant velocity in another. This would mean F=ma doesn't hold in both frames, a violation of a relativity assumption. Does that argument sound good to you guys?

 

[bernhard.rothenstein]

linearity of the LET

All derivations of the Lorentz Transformations I've seen assume a linear transformation between coordinates. Why must this be the case? Thanks.

I used to say to my students that the LET should be linear because to a pair of space-time coordinates in one inertial reference frame should correspond a single pair of space-time coordinates in an other one.

 

[Hurkyl]

LET? Oh, you mean Lorentz-Einstein transforms, not Lorentz Ether Theory.

a pair of space-time coordinates in one inertial reference frame should correspond a single pair of space-time coordinates in an other one.

That happens with any 1-1 transformation...

 

[bernhard.rothenstein]

linearity

LET? Oh, you mean Lorentz-Einstein transforms, not Lorentz Ether Theory.


That happens with any 1-1 transformation...

of couse! in any consistent theory.
sine ira et studio :

 

[Meir Achuz]

So if the transformation were not linear, then an object would appear to be accelerating in one inertial frame but moving at a constant velocity in another. This would mean F=ma doesn't hold in both frames, a violation of a relativity assumption. Does that argument sound good to you guys?

Yes, the LT is linear so that if there is no accelartion in one LF, there will no acceleration in any other LF. But, don't talk about F=ma in SR.

 

[emob2p]

why no F=ma in SR?

 

[emob2p]

I guess it's more appropriate to say F=dp/dt.

 

[Meir Achuz]

a and dp/dt are related by:
\frac{d\bf p}{dt}=\frac{d}{dt}(m{\bf v}\gamma)<br /> = m\frac{d}{dt}\left[\frac{\bf v}<br /> {\sqrt{1-{\bf v}^2}}\right]<br /> =m\gamma^3[{\bf a}+{\bf v\times(v\times a)}].

 

[leright]

So if the transformation were not linear, then an object would appear to be accelerating in one inertial frame but moving at a constant velocity in another. This would mean F=ma doesn't hold in both frames, a violation of a relativity assumption. Does that argument sound good to you guys?

What? How do you figure that a frame can be accelerated to one inertial frame but not accelerated to another inertial frame. This makes no sense. F=ma holds in all inertial frames. However, as mentioned above, mass can vary. If mass did not vary then, due to the speed limit of c, a given force in an inertial frame @ .9999c would obviously yield a much smaller acceleration than that same force at .00001c, and if only acceleration changed due to a given force then the postulate of special relativity would be violated. But, at very high speeds the lack of acceleration is made up for in a large increase in mass, so regardless of speed, a given force yields the same momentum.