Tensor Calculation & Lorentz Transformation: Understanding Relations

In summary, the line segment ds^2 = dx^2+dy^2+dz^2 is equal to the eta-vector ds^2 = \eta_{\alpha\beta}dx^\alpha dx^\beta . These equivalences hold true because the metric eta is a symmetric matrix and c^2d\tau^2 = c^2d\tau'^2.
  • #1
laguna
9
0
Hi,
I have trouble understanding why the following relations hold true. Given the Minkowski metric [itex] \eta_{\alpha\beta}=diag(1,-1,-1,-1) [/itex] and the line segment [itex] ds^2 = dx^2+dy^2+dz^2[/itex], then how can i see that this line segment is equal to [itex] ds^2 = \eta_{\alpha\beta}dx^\alpha dx^\beta [/itex]. Further, we want the line segment to be unchanged under this metric. And i don't understand why the following equivalences hold true: [itex] ds^2 = ds'^2 [/itex] if and only if [itex] c^2d\tau^2 = c^2d\tau'^2[/itex]
and [itex] \Lambda^{\alpha}{}_{\gamma} \Lambda^{\beta}{}_{\delta} \eta_{\alpha}{\beta} = \eta_{\gamma}{\delta} \iff \Lambda^T \eta \Lambda = \eta

[/itex]
Thank you.
 
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  • #2
laguna said:
Hi,
I have trouble understanding why the following relations hold true. Given the Minkowski metric [itex] \eta_{\alpha\beta}=diag(1,-1,-1,-1) [/itex] and the line segment [itex] ds^2 = dx^2+dy^2+dz^2[/itex], then how can i see that this line segment is equal to [itex] ds^2 = \eta_{\alpha\beta}dx^\alpha dx^\beta [/itex]. Further, we want the line segment to be unchanged under this metric. And i don't understand why the following equivalences hold true: [itex] ds^2 = ds'^2 [/itex] if and only if [itex] c^2d\tau^2 = c^2d\tau'^2[/itex]
and [itex] \Lambda^{\alpha}{}_{\gamma} \Lambda^{\beta}{}_{\delta} \eta_{\alpha}{\beta} = \eta_{\gamma}{\delta} \iff \Lambda^T \eta \Lambda = \eta

[/itex]
Thank you.

Why would you think that ##dx^2+dy^2+dz^2 = c^2 dt^2 - dx^2-dy^2-dz^2##?
 
  • #3
if the first relation is true (which I of course believe but do not understand) then [itex] ds^2 = dx^2 + dy^2 + dz^2 = \eta_{\alpha\beta}dx^\alpha dx^\beta = c^2dt^2 -dr^2[/itex], since [itex] dx^{\alpha} = (cdt,dx,dy,dz) [/itex] is the 4-vector and we treat [itex] dx^{\alpha}dx^{\beta} [/itex] like the scalar product to get [itex] (c^2 dt^2,dx^2,dy^2,dz^2) [/itex].
After multiplying this with the metric eta we would get your equation, is that correct?
 
  • #4
laguna said:
Given... the line segment [itex] ds^2 = dx^2+dy^2+dz^2[/itex]
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 [itex] dx^2+dy^2+dz^2[/itex] 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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  • #6
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 [itex] \bar{e_\mu} [/itex] with [itex] \Lambda^{\nu}{}_{\sigma} [/itex]? I mean is it not a matrix multiplication?
 
  • #7
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.
 
  • #8
Thank you all for taking your time helping me. I think i understand this part now. Thanks.
 

FAQ: Tensor Calculation & Lorentz Transformation: Understanding Relations

What are tensors and why are they important in physics?

Tensors are mathematical objects that represent physical quantities and their relationships in a coordinate-independent way. They are important in physics because they allow us to describe and understand the behavior of physical systems in a way that is consistent across different coordinate systems.

How are tensors used in relativity and the Lorentz transformation?

In relativity, tensors play a crucial role in describing the spacetime fabric and the relationship between space and time. The Lorentz transformation, which is a fundamental equation in relativity, uses tensors to describe how physical quantities such as distance, time, and momentum change when observed from different inertial frames of reference.

What is the difference between covariant and contravariant tensors?

Covariant tensors represent quantities that transform in the same way as the coordinates, while contravariant tensors represent quantities that transform in the opposite direction. In other words, covariant tensors change when we change our perspective, while contravariant tensors stay the same.

How do tensors help us understand the relationships between physical quantities?

Tensors allow us to express physical quantities in terms of their components and the transformation rules between different coordinate systems. This helps us to understand how different physical quantities are related and how they behave in different situations.

What are some real-life applications of tensor calculation and the Lorentz transformation?

Tensor calculation and the Lorentz transformation have many practical applications in modern physics, including the study of general relativity, electromagnetism, and quantum mechanics. They are also used in engineering and technology, such as in the design of GPS systems and particle accelerators.

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