Tensor Calculation & Lorentz Transformation: Understanding Relations

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The discussion focuses on the relationship between the Minkowski metric and the line segment in spacetime, specifically how the equation ds² = dx² + dy² + dz² relates to ds² = ηαβ dxα dxβ. It clarifies that the correct definition of the interval is ds² = c²dt² - dx² - dy² - dz², emphasizing that dx² + dy² + dz² represents spatial separation and is not invariant under Lorentz transformations. The participants also discuss the conditions under which the equivalence ds² = ds'² holds true, linking it to the invariance of proper time. Lastly, there is a clarification on the manipulation of terms involving the Lorentz transformation matrix and vectors, confirming that they can be interchanged due to the properties of matrix multiplication. Understanding these concepts is essential for grasping the principles of special relativity.
laguna
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Hi,
I have trouble understanding why the following relations hold true. Given the Minkowski metric \eta_{\alpha\beta}=diag(1,-1,-1,-1) and the line segment ds^2 = dx^2+dy^2+dz^2, then how can i see that this line segment is equal to ds^2 = \eta_{\alpha\beta}dx^\alpha dx^\beta. Further, we want the line segment to be unchanged under this metric. And i don't understand why the following equivalences hold true: ds^2 = ds'^2 if and only if c^2d\tau^2 = c^2d\tau'^2
and \Lambda^{\alpha}{}_{\gamma} \Lambda^{\beta}{}_{\delta} \eta_{\alpha}{\beta} = \eta_{\gamma}{\delta} \iff \Lambda^T \eta \Lambda = \eta<br /> <br />
Thank you.
 
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laguna said:
Hi,
I have trouble understanding why the following relations hold true. Given the Minkowski metric \eta_{\alpha\beta}=diag(1,-1,-1,-1) and the line segment ds^2 = dx^2+dy^2+dz^2, then how can i see that this line segment is equal to ds^2 = \eta_{\alpha\beta}dx^\alpha dx^\beta. Further, we want the line segment to be unchanged under this metric. And i don't understand why the following equivalences hold true: ds^2 = ds&#039;^2 if and only if c^2d\tau^2 = c^2d\tau&#039;^2
and \Lambda^{\alpha}{}_{\gamma} \Lambda^{\beta}{}_{\delta} \eta_{\alpha}{\beta} = \eta_{\gamma}{\delta} \iff \Lambda^T \eta \Lambda = \eta<br /> <br />
Thank you.

Why would you think that ##dx^2+dy^2+dz^2 = c^2 dt^2 - dx^2-dy^2-dz^2##?
 
if the first relation is true (which I of course believe but do not understand) then ds^2 = dx^2 + dy^2 + dz^2 = \eta_{\alpha\beta}dx^\alpha dx^\beta = c^2dt^2 -dr^2, since dx^{\alpha} = (cdt,dx,dy,dz) is the 4-vector and we treat dx^{\alpha}dx^{\beta} like the scalar product to get (c^2 dt^2,dx^2,dy^2,dz^2).
After multiplying this with the metric eta we would get your equation, is that correct?
 
laguna said:
Given... the line segment ds^2 = dx^2+dy^2+dz^2
That's not the right definition of the interval; it's supposed to be (using your sign convention for the metric) ##ds^2 = dt^2-dx^2-dy^2-dz^2##.

The quantity dx^2+dy^2+dz^2 is something different. It's the square of the spatial separation in a particular coordinate system between two events that happen to have the same time coordinate in that coordinate system. It is not unchanged under Lorentz transformations, and it has no physical significance except when the coordinate system is such that ##dt=0## so that we can interpret it as the square of a spatial distance.
 
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vanhees71 said:
Ah, i see. We just define it this way. Thank you so much. But in equation (1.2.8), in the last equality, how do I formally see that we can swap \bar{e_\mu} with \Lambda^{\nu}{}_{\sigma}? I mean is it not a matrix multiplication?
 
The ##{\Lambda^{\nu}}_{\sigma}## is a number and by definition thus commutes with vectors. This of course holds also under the sum, which is implied through the Einstein summation equation.
 
Thank you all for taking your time helping me. I think i understand this part now. Thanks.
 

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