1
$\begingroup$

The squared line element in any spacetime is given as $$ds^{2}=g_{ab}dx^{a}dx^{b}.$$ The use of tensors helps us to infer that the line element in some other frame would be $$ds'^{2}=g'_{ab}dx'^{a}dx'^{b}$$ where simply $dx'^{a}=\frac{\partial x'^{a}}{\partial x^{b}} dx'^{b}$.

My question is, in special relativity, there is further a condition on the line element that it should $$c^{2}(s-t)^{2}-(x_{1}-y_{1})^{2}-(x_{2}-y_{2})^{2}-(x_{3}-y_{3})^{2}=c^{2}(s'-t')^{2}-(x'_{1}-y'_{1})^{2}-(x'_{2}-y'_{2})^{2}-(x'_{3}-y'_{3})^{2}$$ which gives us the Lorentz transformations. How can we prove this condition using the postulates of special relativity? Also where and how do we employ the condition that the frames we are transforming to are inertial?

$\endgroup$
  • $\begingroup$ Hi Naman. If I understand your question correctly this is just the invariance of the line element i.e. using your notation $ds^2 = ds'^2$. This is true for both special and general relativity. The only special thing about special relativity is that $g_{\alpha\beta}$ has the simple form $\text{diag}(-1, 1, 1, 1)$ $\endgroup$ – John Rennie Jan 9 '18 at 9:42
  • $\begingroup$ Then why are only linear transformations in special relativity? Because by the general formula of $ds^{2}$ all transformations are possible. $\endgroup$ – Naman Agarwal Jan 9 '18 at 13:08
  • $\begingroup$ Now I have framed my question in the right way: Why is the following relation true in special relativity $\eta _{ab}=\eta _{mn}\Lambda_{a}^{m}\Lambda_{b}^{n}$ $\endgroup$ – Naman Agarwal Jan 9 '18 at 14:30
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/12664/2451 and links therein. $\endgroup$ – Qmechanic Jun 8 '18 at 12:36
0
$\begingroup$

In fact THIS IS a postulate, the second postulate of Einsteins's special relativity, velocity of light must be invariant in any inertial frame of reference, so mathematically, for a light wave we have : $cdt^2-dx^2-dy^2-dz^2=0$ and $cdt'^2-dx'^2-dy'^2-dz'^2=0$ in another inertial frame, and so $ds^2=ds'^2$ ; this is the origin of your 'further condition'. General relativity agree with this just in local frames.

$\endgroup$
  • $\begingroup$ So is lorentz transformation an explicit condition which is our limit of exploring special relativity or does it emerge from somewhere in special relativity...if yes then how? $\endgroup$ – Naman Agarwal Jan 9 '18 at 13:15
  • $\begingroup$ As a consequence of the second postulate, Lorentz transformation (LT) tell us how to perform a change of (inertial) observer from an initial one if both must measure the same velocity of light first ; it's a mathematical calculation, as perform by Lorentz initially and Poincaré, and later by Einstein, which comes from the need of invariance of light rays. $\endgroup$ – nyx Jan 9 '18 at 14:19
  • $\begingroup$ Now I have framed my question in the right way: Why is the following relation true in special relativity $\eta _{ab}=\eta _{mn}\Lambda_{a}^{m}\Lambda_{b}^{n}$ $\endgroup$ – Naman Agarwal Jan 9 '18 at 14:31
  • $\begingroup$ As I have already said, the relation $\eta_{\mu\nu}=\eta_{mn}\Lambda^{m}_\mu\Lambda^{n}_\nu$is consequence of the second axiom of relativity of invariance of $c$ or as a point of view an axiom full. 1. Principle of relativity : the physical laws are covariates over a change of inertial frame ; 2. Principle of invariance of speed of light : we must have the relation $\eta_{\mu\nu}=\eta_{mn}\Lambda^{m}_\mu\Lambda^{n}_\nu$ so universe is a 4D space-time $\endgroup$ – nyx Jan 9 '18 at 14:43
0
$\begingroup$

Here why $\eta_{ab} = \eta_{mn} \Lambda^m_a \Lambda^n_b$ in SR (special relativity).
$ds^2 = \eta_{ab} dx^a dx^b$ (1)
$ds^2 = \eta_{mn} dx^m dx^n$ (2)
but:
$dx^m = \Lambda^m_a dx^a$ Lorentz transformation
so:
$\eta_{mn} dx^m dx^n = \eta_{mn} \Lambda^m_a \Lambda^n_b dx^a dx^b$ (3)
by equating (1) and (2), taking count of (3):
$\eta_{ab}=\eta_{mn}\Lambda_a^m\Lambda_b^n$
Note:
In SR you have linear transformations as Lorentz transformation is linear in the coordinates as a consequence of the principles of special relativity.

$\endgroup$
0
$\begingroup$

The reason you've been unable to find a derivation of the Lorentz transformation (relating, say, Bob's frame to Alice's) from the usual two postulates of special relativity is that the Lorentz transformation does not follow from those postulates. You're going to need some additional assumption.

You can, for example, add some assumption about homogenenity/isotropy (though it's not enough, of course, to just mumble the words "homogeneity" and "isotropy"; you need some careful statement of exactly what you're assuming), or you can start by assuming that Bob's measurement of Alice's velocity is constant over time. Or you can just assume that the transformation must be linear.

There might then be room for someone to argue that your new assumption is really just a consequence of the old assumptions (in particular that the laws of nature "look the same" to both Bob and Alice). Since the old assumption is pretty vague to begin with (never specifying exactly what counts as a law of nature), it's possible to defend this argument, but it's at least a stretch.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.