I don't understnad how to get the 4-volume element for a Minkowksi space

In summary, the expression you gave is the square root of the determinant of the metric in Minkowski space-time. The determinant is -1, so the volume element in Minkowski space is just multiplied by one.
  • #1
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I see that it is [itex]\sqrt{|-g|}[/itex][itex]d^{4}[/itex]x but I'm not sure why it needs to be multiplied by the square root of the determinant of -g. It must be the Jacobian of [itex]g_{μ\nu}[/itex] or something right? So I guess I'm asking how do you calculate the Jacobian of the metric.
 
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  • #2
In Minkowski space the coordiates are orthonormal. The diagonal of the metric is (-1,1,1,1) with all other entries zero.

The expression you gave should be [itex]\sqrt{-g}[/itex], where [itex]g[/itex] is the determinate of the metric. It means you take the square root of the negative of the determinate of the metric. The determinate is -1, so the volume element in Minkowski space is just multipled by one.

[tex]d^4 x' = \sqrt{-g}d^4 x[/tex]

What we are doing with [itex]\sqrt{-g}[/itex] is translating between a volume element in x' coordinates to orthonormal minkowski volume elements in the unprimed frame.
 
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  • #3
Yes but why do we use the square root of the determinate of -g to transform between volume elements in different frames? I'd like to see the math of it.
 
  • #4
The volume element in the Minkowski metric is invariant under Lorentz transformations. You can show this by explicitly computing the Jacobian for a Lorentz Transformation.

That factor is redundant in Minkowski space-time.
 
  • #5
Consider the transformation x' = ax

Consider how the metric transforms under this transformation (from t,x,y,z to t,x',y,z), and how the volume element tranforms under this transformation.

Hint:

A unit volume 4- cube has dt, dy, and dz the same, but dx' = a dx.

We can use the tensor transformation law to transform the metric, or just algebra. To use just algebra, we write the line element in terms of dx,dy,dz and we use again use the formula dx' = a dx.

So a line element of -dt^2 + dx^2 + dy^2 + dz^2 transforms into -dt^2 + dx'^2/a^2 + dy^2 + dz^2

This isn't a completely general derivation , but should (hopefully) give you a good intiutive feeling for questions like "where did that square root come from"
 

1. How is the 4-volume element defined in Minkowski space?

The 4-volume element in Minkowski space is defined as the product of the 4 differential lengths in the space, dx1dx2dx3dx4. This represents the infinitesimal volume in the space and is used in integral calculations.

2. What is the significance of the 4-volume element in Minkowski space?

The 4-volume element is important in Minkowski space because it allows us to calculate the volume of a region in the space. This is crucial in many physics applications, such as calculating the energy-momentum tensor or finding the invariant interval between two events.

3. How is the 4-volume element related to the metric tensor in Minkowski space?

The 4-volume element can be written in terms of the metric tensor as √|g|d4x, where g is the determinant of the metric tensor. This shows that the metric tensor plays a crucial role in determining the 4-volume element in Minkowski space.

4. Is the 4-volume element the same in all reference frames in Minkowski space?

Yes, the 4-volume element is a Lorentz scalar and therefore remains the same in all reference frames in Minkowski space. This is because the metric tensor, which determines the 4-volume element, is also a Lorentz scalar.

5. How can the 4-volume element be used in calculations in Minkowski space?

The 4-volume element is used in integrals and calculations involving volume in Minkowski space. It can also be used to define the volume element for higher dimensional spaces, such as 5-dimensional Minkowski space. Additionally, it is used in the formulation of special relativity and other theories in physics.

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