Gradient ∇4: Generalizing for Spacetime and Proving its Four-Vector Properties

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In summary: That would clear up any ambiguity.Ah, maybe you're right. I don't know any relativity. I just looked up 4-vector in wikipedia.
  • #1
Unicorn.
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Hi,

Homework Statement


The gradient ∇3 can be generalized for spacetime as:
4 =(∇3 ,d/dct)=(d/dx,d/dy,d/dz,d/dct)
Show that ∇4 is a four-vector.

Homework Equations





The Attempt at a Solution


I just have to write that :
d/dx'=γ(d/dx-βd/dct)
d/dy'=d/dy
d/dz'=d/dz
d/dct'=γ(d/dct-βd/dx)
And
4 =(d/dx,d/dy,d/dz,d/dct)=(∇3 ,d/dct) ..?

Thanks
 
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  • #2
I don't understand the question. Of course it's a 4-vector. (a, b, c, d) is a 4-vector. Is there some wording missing?
 
  • #3
Hi,
No, this is the complete wording.
 
  • #4
What? The 4-gradient ##\nabla_{a}## is a derivative operator. It isn't a 4-vector of any kind. I don't think you are phrasing it correctly.
 
  • #5
WannabeNewton said:
What? The 4-gradient ##\nabla_{a}## is a derivative operator. It isn't a 4-vector of any kind. I don't think you are phrasing it correctly.
A vector operator is still a vector. You can add them, and multiply them by scalar values.
 
  • #6
haruspex said:
A vector operator is still a vector. You can add them, and multiply them by scalar values.
An operator acting on a vector isn't an element of the vector space that the vector belongs to. ##\nabla_{a}## is certainly not a 4-vector just because it is linear (it isn't a linear functional - it's just linear). How are you claiming that ##\nabla_{a}\in \mathbb{R}^{4}## (I used ##\mathbb{R}^{4}## because the OP is working in minkowski space-time). That doesn't make any sense.

In fact what is true is that ##\frac{\partial }{\partial x}|_{p},\frac{\partial }{\partial y}|_{p},\frac{\partial }{\partial z}|_{p},\frac{\partial }{\partial t}|_{p}## are 4-vectors in ##\mathbb{R}^{4}##.
 
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  • #7
haruspex said:
A vector operator is still a vector. You can add them, and multiply them by scalar values.

The OP is asking about 4-vectors. So saying that ∇ is a 4-vector is the same as saying that ∇ is an element of ##\mathbb{R}^4##. This is obviously nonsense.

Of course, we can add ∇ and multiply it by scalars. But that means that it is a vector in a vector space. But since the OP asks about 4-vectors, he's specifically asking about the vector space ##\mathbb{R}^4##. At least, that's how I interpret the question.
 
  • #8
The OP is supposed to show that ##\nabla## transforms like a four-vector. In other words, he or she needs to show that it makes sense to say that ##\partial'_\mu = \Lambda_\mu{}^\nu \partial_\nu##.
 
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  • #9
Actually, strictly speaking, shouldn't it transform like a 1-form?
 
  • #10
Chestermiller said:
Actually, strictly speaking, shouldn't it transform like a 1-form?
The coordinate vector fields ##\partial_{\mu}## do indeed transform like one-forms (hence why they are usually written with a lower index) but they are still 4-vectors as they form a basis for the tangent space when evaluated at each point. Also it doesn't make any sense to say ##\nabla## transforms like a 4-vector because it is not a vector in minkowski space-time, it is map on the ring of smooth functions on minkowski space-time. This bad notation is the kind you would see in Srednicki's QFT text.
 
  • #11
micromass said:
The OP is asking about 4-vectors. So saying that ∇ is a 4-vector is the same as saying that ∇ is an element of ##\mathbb{R}^4##. This is obviously nonsense.

Of course, we can add ∇ and multiply it by scalars. But that means that it is a vector in a vector space. But since the OP asks about 4-vectors, he's specifically asking about the vector space ##\mathbb{R}^4##. At least, that's how I interpret the question.
I gather from various posts that, within relativity theory at least, "4-vector" has a more specific meaning than I was aware of. It's not clear to me whether this is a matter of definition or just common usage.
 
  • #12
haruspex said:
I gather from various posts that, within relativity theory at least, "4-vector" has a more specific meaning than I was aware of. It's not clear to me whether this is a matter of definition or just common usage.

Ah, maybe you're right. I don't know any relativity. I just looked up 4-vector in wikipedia.

To avoid confusion, perhaps we should let the OP tell us how he defined "4-vector" in his course.
 

FAQ: Gradient ∇4: Generalizing for Spacetime and Proving its Four-Vector Properties

What is a gradient four-vector?

A gradient four-vector is a mathematical construct used in the field of physics to describe the rate of change of a scalar quantity in four-dimensional space-time. It is a four-dimensional vector with components representing the partial derivatives of a scalar field with respect to the four dimensions of space-time.

What is the significance of a gradient four-vector?

A gradient four-vector is significant because it allows us to mathematically describe how a scalar quantity changes in space and time. This is useful in many areas of physics, including electromagnetism, relativity, and fluid dynamics.

How is a gradient four-vector calculated?

To calculate a gradient four-vector, we take the partial derivatives of a scalar field with respect to each of the four dimensions of space-time, and then combine them into a four-dimensional vector. This can be written mathematically as ∇𝑓 = (∂𝑓/∂𝑡, ∂𝑓/∂𝑥, ∂𝑓/∂𝑦, ∂𝑓/∂𝑧), where ∇ is the gradient operator and 𝑓 is the scalar field.

What are some real-life applications of gradient four-vectors?

Gradient four-vectors have many applications in physics and engineering. They are commonly used in modeling fluid flow, heat transfer, and electromagnetic fields. They are also important in the study of general relativity and the motion of particles in space-time.

Can a gradient four-vector be used to describe changes in more than four dimensions?

No, a gradient four-vector is specifically used to describe changes in four dimensions of space-time. It cannot be extended to higher dimensions as it is a mathematical construct that is unique to four-dimensional space-time.

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