Prove Covariant Four Vector: \frac{\partial\phi}{\partial x^{\mu}}

In summary: However, if we're talking about a general transformation between frames in space, then Lambda will depend on x and x' (and possibly other variables). In that case, you would differentiate with respect to x'^\alpha.
  • #1
ercagpince
30
0

Homework Statement


Show that [tex]\frac{\partial\phi}{\partial x^{\mu}}[/tex] is a covariant four vector .


Homework Equations


All covariant four vector transformations .


The Attempt at a Solution


I really didn't understand what question implies . How can this vector be showed as being a covariant vector ?
If you could give me a hint at least about what is really asked .

Thanks .
 
Last edited:
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  • #2
ercagpince said:

Homework Statement


Show that [tex]\frac{\partial\phi}{\partial x^{\mu}}[/tex] is a covariant four vector .


Homework Equations


All covariant four vector transformations .


The Attempt at a Solution


I really didn't understand what question implies . How can this vector be showed as being a covariant vector ?
If you could give me a hint at least about what is really asked .

Thanks .

You must write the corresponding expression expressed in a different frame, viz

[tex]\frac{\partial\phi'(x')}{\partial x'^{\mu}}[/tex]
and then relate all the quantities to the unprimed coordinate system.
 
  • #3
How can i write it in a different frame since I don't know how to transform the derivate from one frame to another ?
 
  • #4
I would call the

[tex]\frac{\partial \phi}{\partial x^\mu}[/tex]

components of a covector field with respect to a particular basis, but physicists often think terms of components (or abstract index notation).

In any case, switch to a primed coordinate system and use the multivariable chain rule to express

[tex]\frac{\partial \phi}{\partial x'^\nu}[/tex]

in terms of

[tex]\frac{\partial \phi}{\partial x^\mu}.[/tex]

[edit]Was typing while nrqed's post came in. His notation is better than mine. Why?[/edit]
 
Last edited:
  • #5
I don't know how to switch from one inertial frame to another on differential form .
I have Lorentz transformation of (x0,x1,x2,x3) linear terms but I really have no clue how to obtain derivate form of that .
 
  • #6
ercagpince said:
I don't know how to switch from one inertial frame to another on differential form .
I have Lorentz transformation of (x0,x1,x2,x3) linear terms but I really have no clue how to obtain derivate form of that .

Are yuo working only with inertial coordinate systems in special relativity, and do you have something like (I might have the staggering wrong)

[tex]x^\nu = \Lambda^\nu{}_\mu x'^\mu?[/tex]

If so, differentiate this with respect to [itex]x'^\alpha[/itex] to get something that you can use in conjunction with the chain rule.
 
  • #7
Should superscript alpha be nu ? Or is alpha for another indice ?
 
  • #8
ercagpince said:
Should superscript alpha be nu ? Or is alpha for another indice ?

No to the first; yes to the second.

If you differentiate with respect to a nu index, nu will appear twice in each side of the expression, and there will be possible problems with the summation convention. If you differentiate with respect to mu, the mu will appear three times.

What does

[tex]\frac{\partial x'^\mu}{\partial x'^\alpha}[/tex]

equal?
 
  • #9
What about the lamdba term ? It contains some x0 to x3 term that should be derivated .

Actually , this is the key problem that confuses my mind from the beginning .
 
  • #10
ercagpince said:
What about the lamdba term ? It contains some x0 to x3 term that should be derivated .

Actually , this is the key problem that confuses my mind from the beginning .

Again, I'm not sure of the context in which we're working. If talking about Lorentz transformations between inertial coordinate systems, Lambda is a matrix of constants (i.e., the constants that multiply the x coordinates or x' coordinates). It has no x dependence
 

What is a covariant four vector?

A covariant four vector is a mathematical object used in the field of physics to represent a four-dimensional vector quantity. It is typically denoted as xμ and has components that are dependent on the choice of coordinate system.

What does it mean to prove a covariant four vector?

To prove a covariant four vector means to show that it follows the rules of transformation in different coordinate systems and maintains its vector nature. This can be done through mathematical equations and calculations.

What is the significance of the partial derivative symbol in "∂ϕ/∂xμ"?

The partial derivative symbol (∂) indicates that we are taking the derivative of a multivariable function, in this case, the function ϕ with respect to the variable xμ. It is used to express how a function changes in response to changes in one of its variables while holding the others constant.

What is the relationship between a covariant four vector and the partial derivative of a function?

A covariant four vector is often used to represent the gradient of a scalar function, which is calculated using partial derivatives. The components of the covariant four vector correspond to the partial derivatives of the function with respect to each coordinate variable, providing a way to express the gradient in a four-dimensional space.

How is the covariant four vector used in physics?

In physics, the covariant four vector is used in various fields, including relativity, electromagnetism, and quantum mechanics. It is a fundamental concept in these areas and is often used to describe and analyze physical phenomena, such as the motion of particles, electromagnetic fields, and energy-momentum transfer.

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