emob2p
- 56
- 1
All derivations of the Lorentz Transformations I've seen assume a linear transformation between coordinates. Why must this be the case? Thanks.
Why must the Lorentz Transformations Be Linear?
- Context: Graduate
- Thread starter emob2p
- Start date
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
11 replies · 9K views
- 14,922
- 28
A transformation is linear if and only if it keeps the origin fixed, and it maps lines to lines.
- 7,465
- 2,986
In particular, a Lorentz transformation must preserve the nature of timelike lines, which represent inertial particles.
emob2p
- 56
- 1
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
- 988
- 1
linearity of the LET
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.emob2p said:
- 14,922
- 28
LET? Oh, you mean Lorentz-Einstein transforms, not Lorentz Ether Theory.
That happens with any 1-1 transformation...
bernhard.rothenstein
- 988
- 1
linearity
sine ira et studio
:
of couse! in any consistent theory.Hurkyl said:
sine ira et studio
:- 3,742
- 249
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 said:
emob2p
- 56
- 1
why no F=ma in SR?
emob2p
- 56
- 1
I guess it's more appropriate to say F=dp/dt.
- 3,742
- 249
a and dp/dt are related by:
[itex]\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)}].[/itex]
[itex]\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)}].[/itex]
leright
- 1,317
- 19
emob2p said:
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.
Last edited: