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How can I prove whether or not d4x is a 0th rank tensor? It seems strange that I should be so when it is the product of a 0th, 1st, 2nd and 3rd component, dx0dx1dx2dx3. I heard that the proof involves the Jacobian. I don't get it.
Originally posted by matt grime
I'm notm sure my interpretation is correct, but in differential forms, d^2 is identically zero. so d^4(x) is zero.
Originally posted by matt grime
thank you
You don't think that would be confused with the 4th component of a tensor (superscript instead of exponent)? I think the notation itself is confusing (the one physicists use). That's why I'm trying to make the transition to using the mathematician's notation, and then instigate a revolution to make all physicists use it under penalty of death by strenuous physical activity.Originally posted by lethe
in fact, a better notation for the measure would probably be dx^4. but we are stuck with this notation.
Originally posted by turin
You don't think that would be confused with the 4th component of a tensor (superscript instead of exponent)? I think the notation itself is confusing (the one physicists use). That's why I'm trying to make the transition to using the mathematician's notation, and then instigate a revolution to make all physicists use it under penalty of death by strenuous physical activity.
OK, so d4x' = |∂(xμ')/∂(xν)|d4x
?
It says the latex source is invalid. I see a \vol here. I'm assuming that is supposed to be some sort of a "v" or a fancy "vol."Originally posted by lethe
anyway, to make a fair comparison, i will tell you the mathematicians notation for the volume form: [itex]\vol[/itex]
Originally posted by turin
OK, so d4x' = |∂(xμ')/∂(xν)|d4x
?
I'm understanding
|∂(xμ')/∂(xν)|
to be the Jacobian.
Is this the determinant of the transformation matrix? I took multivariable calculus years ago, and I don't remember this stuff.
Originally posted by turin
Do you have any idea what is making me so confused? Am I just being incredibly stubborn about this? Is the distinction trivial or something?
A 4-volume tensor is a mathematical object used in the field of physics to describe the behavior of physical quantities in four-dimensional space-time. It is a rank (0,0) tensor, meaning it has no indices and is a scalar quantity.
A 4-volume tensor is different from other tensors in that it has no indices, while other tensors have one or more indices. This means that a 4-volume tensor does not transform under coordinate transformations and is invariant in all frames of reference.
A 4-volume tensor is significant in physics because it allows us to describe physical quantities, such as energy and momentum, in four-dimensional space-time. This is essential for understanding and solving problems in relativity and other areas of physics.
In general relativity, a 4-volume tensor is used to describe the curvature of space-time due to the presence of mass and energy. It is a key component in Einstein's field equations, which relate the curvature of space-time to the distribution of matter and energy.
No, a 4-volume tensor cannot be visualized or represented graphically because it is a scalar quantity with no indices. It is a mathematical concept used to represent physical quantities in four-dimensional space-time, and cannot be depicted in a two or three-dimensional format.