Square of the difference of four-vectors

In summary: The issue is that for everyone who has responded on this thread so far, your problem is trivial, yet you seem to have misunderstandings. We feel it is much more useful to help you resolve these than simply give you the... solution.In summary, what is the correct way to expand (p3-p4)2 where p3 and p4 are 4-vectors, with metric gmu nu=diag[1,-1,-1,-1], p = [wp, p], where p is 3-vector, and wp= (p2+m2)(1/2)?You don't need to do that. The 4-vector product distributes over addition and subtraction
  • #1
Arny_Toynbee
8
0
What is the correct way to expand (p3-p4)2 where p3 and p4 are 4-vectors, with metric gmu nu=diag[1,-1,-1,-1], p = [wp, p], where p is 3-vector, and wp= (p2+m2)(1/2)
 
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  • #2
Do you know how to take the dot product of two vectors when the metric isn't trivial? (If I'm allowed to phrase it that way.)
 
  • #3
p3.p4 = (p30, p3) . (p40, p4)

and p30= wp3 etc.
 
  • #4
And do you know how to expand that? In Euclidean 3-space ##\vec a.\vec b=a_xb_x+a_yb_y+a_zb_z##. Do you know the equivalent Minkowski 4-space expression?
 
  • #5
Ibix said:
And do you know how to expand that? In Euclidean 3-space ##\vec a.\vec b=a_xb_x+a_yb_y+a_zb_z##. Do you know the equivalent Minkowski 4-space expression?

p3.p4 = = p30 p40 - p3.p4
 
  • #6
Right. So when you replace the vector with a difference of two vectors what do you get?
 
  • #7
Ibix said:
Right. So when you replace the vector with a difference of two vectors what do you get?
The question is
(p3 - p4)^2 - whether it is (wp3-wp4)2 + (p3 - p4)2

or is it

(wp3+wp4)2 - (p3 + p4)2

Hmmmmmm...what do I get?
 
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  • #8
Arny_Toynbee said:
.what do I get?

Instead of guessing, write it out explicitly: ##(p_3 - p_4)^2 = (p_3 - p_4) \cdot (p_3 - p_4)##, and then just expand out the product and do the algebra.
 
  • #9
PeterDonis said:
Instead of guessing, write it out explicitly: ##(p_3 - p_4)^2 = (p_3 - p_4) \cdot (p_3 - p_4)##, and then just expand out the product and do the algebra.

How would you write out (p3 - p4)?

There's a bit of subtlety here, and not guessing, see for e.g. Equation 46.29 here
 
  • #10
Arny_Toynbee said:
How would you write out (p3 - p4)?

You don't need to do that. The 4-vector product distributes over addition and subtraction just like the ordinary scalar product does.
 
  • #11
Arny_Toynbee said:
How would you write out (p3 - p4)?

Arny_Toynbee said:
see for e.g. Equation 46.29

You do realize that that equation answers the question you just asked?
 
  • #12
Arny_Toynbee said:
How would you write out (p3 - p4)?

There's a bit of subtlety here, and not guessing, see for e.g. Equation 46.29 here
First, vector subtraction against an orthonormal basis is always just what you would expect.

Second, you don't need to do this anyway because dot product distributes over vector addition/subtraction.
 
  • #13
Arny_Toynbee said:
The question is
(p3 - p4)^2 - whether it is (wp3-wp4)2 + (p3 - p4)2

or is it

(wp3+wp4)2 - (p3 + p4)2

Hmmmmmm...what do I get?
You wrote down an expression for the dot product of two vectors in terms of the components. So if the vectors, instead of ##p_3## and ##p_4##, are both ##p##, and ##p=p_3-p_4## what do you get?
 
  • #14
PeterDonis said:
You do realize that that equation answers the question you just asked?

Actually, you may not realize, there might be a surprise on the way!
 
  • #15
Arny_Toynbee said:
Actually, you may not realize, there might be a surprise on the way!
Perhaps it would be helpful if you either completed the algebra we've been discussing or said what you think is wrong with 4.29 in the pdf you linked.
 
  • #16
Ibix said:
Perhaps it would be helpful if you either completed the algebra we've been discussing or said what you think is wrong with 4.29 in the pdf you linked.
The link provided may be using a different convention. Until now, I have provided all the algebra and the question. It was suggested that "...realize that the link you provided...has the solution...?" The original question has two alternatives, and it is useful to see what the forum members come up with. This is not a homework that is to be submitted to the forum in all its completeness, for it to be graded.
 
  • #17
Arny_Toynbee said:
The link provided may be using a different convention. Until now, I have provided all the algebra and the question. It was suggested that "...realize that the link you provided...has the solution...?" The original question has two alternatives
You mean your expressions in #7? Neither appears to be correct, as PeterDonis said in #8. How did you get them? I haven't tried to replicate 4.29 from your link, but it looks plausible at a quick glance.
 
  • #18
Arny_Toynbee said:
The link provided may be using a different convention. Until now, I have provided all the algebra and the question. It was suggested that "...realize that the link you provided...has the solution...?" The original question has two alternatives, and it is useful to see what the forum members come up with. This is not a homework that is to be submitted to the forum in all its completeness, for it to be graded.
The issue is that for everyone who has responded on this thread so far, your problem is trivial, yet you seem to have misunderstandings. We feel it is much more useful to help you resolve these than simply give you the answer.
 
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  • #19
Ibix said:
You mean your expressions in #7? Neither appears to be correct, as PeterDonis said in #8. How did you get them? I haven't tried to replicate 4.29 from your link, but it looks plausible at a quick glance.
I did replicate that equation, just doing the algebra in my head.
 
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  • #20
PAllen said:
I did replicate that equation, just doing the algebra in my head.
Thank you. I was confident up to sign errors on the cross terms - it's been a long day...
 
  • #21
Arny_Toynbee said:
Until now, I have provided all the algebra

Have you done the algebra I asked for in post #8? If so, post it. If not, do it and post it.

Arny_Toynbee said:
The original question has two alternatives

Do you mean the question you asked in post #7? Neither of those alternatives is correct.

Now go do the algebra I asked you to do in post #8, so you can compute what the right answer is instead of guessing.

Arny_Toynbee said:
This is not a homework that is to be submitted to the forum in all its completeness, for it to be graded

No, but that doesn't mean we're just going to tell you an answer that you could derive for yourself by doing basic algebra. PF is not an oracle that you can feed guesses to and get yes/no answers back.
 
  • #22
Ibix said:
Thank you. I was confident up to sign errors on the cross terms - it's been a long day...
Sorry, sign errors in which equation? By the way, the way it is derived in the link is not necessarily consistent.

What is the correct answer (of the alternatives posted)?
 
  • #23
Arny_Toynbee said:
What is the correct answer (of the alternatives posted)?

Do the algebra yourself and find out what the correct answer is. As I said, PF is not an oracle.

Thread closed.
 
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What is a four-vector?

A four-vector is a mathematical representation of a physical quantity in four-dimensional space-time. It is typically denoted by a symbol with four components, such as (ct, x, y, z), where c is the speed of light and t, x, y, and z represent the coordinates in space and time.

What is the square of a four-vector?

The square of a four-vector is a mathematical operation that involves multiplying the vector by itself. This results in a scalar quantity. In other words, the square of a four-vector is the sum of the squares of its components.

What is the difference of two four-vectors?

The difference of two four-vectors is a mathematical operation that involves subtracting the components of one vector from the components of another vector. This results in a new four-vector that represents the displacement between the two original vectors.

What is the square of the difference of two four-vectors?

The square of the difference of two four-vectors is a mathematical operation that involves first finding the difference between the two vectors and then squaring the resulting vector. This results in a scalar quantity that represents the square of the distance or displacement between the two original vectors.

What is the significance of the square of the difference of four-vectors in physics?

The square of the difference of four-vectors is commonly used in physics to calculate quantities such as the spacetime interval between two events or the energy-momentum of a particle. It is also used in special relativity to determine the speed of an object in different reference frames.

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