Tương tác mạnh

Tương tác mạnh

Bài viết chưa xemgửi bởi dangvanda » Thứ 7 Tháng 1 02, 2010 12:26 am

Trong tương tác mạnh. Hai hạt tương tác với nhau bằng việc trao đổi gluon. Mà có tới 8 hạt gluon. Ai có thể kể tên 8 loại hạt đó được không?. hãy giúp tôi với.
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Ngày tham gia: Thứ 7 Tháng 1 02, 2010 12:19 am

Re: Tương tác mạnh

Bài viết chưa xemgửi bởi Cry » Thứ 7 Tháng 1 02, 2010 4:56 am

Trích từ Wikipedia

Eight gluon colors

There are eight remaining independent color states, which correspond to the "eight types" or "eight colors" of gluons. Because states can be mixed together as discussed above, there are many ways of presenting these states, which are known as the "color octet." One commonly used list is:
1.(r\bar{b}+b\bar{r})/\sqrt{2}

2.-i(r\bar{b}-b\bar{r})/\sqrt{2}

3.(r\bar{g}+g\bar{r})/\sqrt{2}

4.-i(r\bar{g}-g\bar{r})/\sqrt{2}

5.(b\bar{g}+g\bar{b})/\sqrt{2}

6.-i(b\bar{g}-g\bar{b})/\sqrt{2}

7.(r\bar{r}-b\bar{b})/\sqrt{2}

8.(r\bar{r}+b\bar{b}-2g\bar{g})/\sqrt{6}

These are equivalent to the Gell-Mann matrices; the translation between the two is that red-antired is the upper-left matrix entry, red-antiblue is the left middle entry, blue-antigreen is the bottom middle entry, and so on. The critical feature of these particular eight states is that they are linearly independent, and also independent of the singlet state; there is no way to add any combination of states to produce any other. (It is also impossible to add them to make r\bar{r}, g\bar{g}, or b\bar{b}; otherwise the forbidden singlet state could also be made.) There are many other possible choices, but all are mathematically equivalent, at least equally complex, and give the same physical results.
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Re: Tương tác mạnh

Bài viết chưa xemgửi bởi hoanganhtuan_baby » Thứ 7 Tháng 1 02, 2010 12:41 pm

có lẽ cái này dễ đọc hơn. Bạn tham khảo bằng tiếng việt nhé xeko
No promises !
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Re: Tương tác mạnh

Bài viết chưa xemgửi bởi Nhoc_PT » Thứ 7 Tháng 1 02, 2010 8:18 pm

hoanganhtuan_baby đã viết:có lẽ cái này dễ đọc hơn. Bạn tham khảo bằng tiếng việt nhé xeko

Tuấn đăng trang wiki bằng Tiếng Việt đúng là dễ đọc hơn thật nhưng lại chưa chỉ ra được cụ thể cuối cùng 8 loại gluon ấy tên là gì. Bên wiki tiếng Anh như trong câu trả lời của Linh thì đã liệt kê ra 8 loại thật nhưng mà vẫn mơ hồ quá! :(
Tớ hiểu đơn giản thế này nhé, mỗi gluon mang 1 "màu" và "đối màu" ("màu điện tích" nhá). Ta có 3 "màu" là đỏ (red), xanh lơ (green), xanh lam (blue) và 3 "đối màu" là đối đỏ (antired), đối xanh lơ (antigreen) và đối xanh lam (antiblue). Như vậy sẽ có các tổ hợp: r\bar{r} (red-antired hay đỏ-đối đỏ nhé), r\bar{g}, r\bar{b}, g\bar{r}, g\bar{g}, g\bar{b}, b\bar{r}, b\bar{g}, b\bar{b}, tức là có 9 tổ hợp. Nhưng vì 1 số sự kết hợp cho ta màu trắng nên rốt cuộc chỉ còn lại 8 tổ hợp, nhưng mà tớ chẳng hiểu cuối cùng làm sao ra 8 cái tổ hợp ấy phunlua
Tớ tìm được bài giải thích tại sao lại là 8 loại bằng tiếng Anh, nhưng đọc chưa hiểu hết, nếu có ai đó biết thì chỉ giáo với nha :D
As far as I know, gluons have a color charge, it is the combination
of a color ith an anticolor.

I have read in many places that due to SU(3) simmetry properties,
there are only 8 gluons, but I can't figure out wich one are them.

Using a simple combination, gluons could be red-antiblue, red-
antigreen, red-antired, blue-antired, blue-antigreen, blue-antiblue,
green-antiblue, green-antired and green-antigreen. (9 kinds of gluons)

Since three of them have a white color (red-antired, blue-antiblue
and green-antigreen) I can eliminate them, but then I only have 6
gluons.

Where does the number 8 come from?

To be truly white, the gluon must be red-antired, green-antigreen, AND
blue-antiblue. Based on the mathematics of wave function probability, the
white gluon is (red-antired + blue-antiblue + green-antigreen)/sqrt(3). The
square root of three is to keep things the correct size. There are tow wave
functions that are not actually white: (red-antired -
green-antigreen)/sqrt(6) and (red-antired + green-antigreen -
2*blue-antiblue)/sqrt(6). These two gluons can interact without changing
the color of a quark, but they are not truly white. A white particle can
have no preference for red, green, or blue.

Cái này thì còn lôi cả ma trận vào cơ:
Why are there eight gluons and not nine?

According to QCD and the standard model of particle physics, quarks carry an SU(3) "color" charge which can be "red", "blue" or "green". The strong nuclear force which binds these together inside the nucleons is mediated by gluons which must carry a color-anticolor charge. This seems to give 9 types of gluon:

green-antigreen, green-antired, green-antiblue,
red-antired, red-antiblue, red-antigreen,
blue-antiblue, blue-antired, blue-antigreen.

Why then are there only eight gluons?

Rather than start with the SU(3) theory, consider first what our knowledge of nature is--upon which we will base the theory. We know that the matter we believe to be composed of quarks (the hadrons) comes in two types: mesons (short lived particles, pions etc.) and baryons (protons, neutrons etc.), both of which have to be strong force neutral.

Consider the evidence: scattering experiments strongly suggest a meson to be composed of a quark anti-quark pair and a baryon to be composed of three quarks. The famous 3R experiment also suggests that whatever force binds the quarks together has 3 types of charge (called the 3 colors).

Now, into the realm of theory: we are looking for an internal symmetry having a 3 dimensional representation which can give rise to a neutral combination of 3 particles (otherwise no color neutral baryons). The simplest such statement is that a linear combination of each type of charge (red + green + blue) must be neutral, and following William of Occam we believe that the simplest theory describing all the facts must be the correct one. We now postulate that the photons (called gluons) must occur in color anti-color units (i.e. nine of them). BUT, red + blue + green is neutral, which means that the linear combination red anti-red + blue anti-blue + green anti-green must be non-interacting, since otherwise the colorless baryons would be able to emit these gluons and interact with each other via the strong force--contrary to the evidence. So, there can only be EIGHT gluons. This is just Occam's razor again: a hypothetical particle that can't interact with anything, and therefore can't be detected, doesn't exist.

The simplest theory describing the above is the SU(3) one with the gluons as the basis states of the Lie algebra. That is, gluons transform in the adjoint representation of SU(3), which is 8-dimensional.

The physics of color is not understandable if all one knows is that there are 3 colors. One must really understand something about SU(3). SU(3) is the group of 3 x 3 unitary matrices with determinant 1. This is the symmetry group of the strong force. What this means is that, as far as the strong force is concerned, the state of a particle is given by a vector in some vector space on which elements of SU(3) act as linear (in fact unitary) operators. We say the particle "transforms under some representation of SU(3)". For example, since elements of SU(3) are 3 x 3 matrices, they can act on column vectors by matrix multiplication. This gives a 3-dimensional representation of SU(3). Quarks transform under this representation of SU(3), and because it's 3-dimensional we say quarks come in 3 colors: red, green and blue. This is just an amusing way of talking about the 3 column vectors

1
0
0

0
1
0

and

0
0
1.

Alternatively, we could let elements of SU(3) act on row vectors by multiplication on the right. Antiquarks transform under that representation, and since it is also 3-dimensional we say they come in three colors as well: antired, antiblue, and antigreen. This is just an amusing way of talking about the 3 row vectors

100

010

and

001.

Alternatively, we could let elements of SU(3) act on 3 x 3 hermitian matrices with trace equal to 0, by letting the element g of SU(3) act on the matrix T to give the new matrix gTg-1. The space of 3 x 3 hermitian matrices with trace equal to zero is 8-dimensional. Gluons transform under this representation, so there are 8 gluons.

Some examples of a 3 x 3 hermitian matrix with trace zero are

0 1 0 0 i 0
1 0 0 -i 0 0
0 0 0 0 0 0

If we allow ourselves to take complex linear combinations of these we can get things like

0 1 0
0 0 0
0 0 0

which we could call "red-antiblue" since it has a 1 in the red row (the first row) and the blue column (the second column). Or we could get

0 0 0
1 0 0
0 0 0

which we could call "blue-antired".

But, no matter how we take complex linear combinations of trace-zero hermitian matrices, we cannot get

1 0 0
0 0 0
0 0 0

since this has trace 1. (I.e., the sum of the diagonal entries is 1: that's what the trace means, the sum of the diagonal entries.) So we cannot really get red-antired. The closest we can get are things like

1 0 0
0 -1 0
0 0 0

or

1 0 0
0 0 0
0 0 -1

or

0 0 0
0 1 0
0 0 -1

But note, these three are not linearly independent: any one of them is a linear combination of the other two. So we can get stuff like

red-antired - blue-antiblue

and so on, but not 3 linearly independent things of this sort, only two. One less than you might expect.

If you are wondering what the hell I am doing subtracting particles from each other, well, that's quantum mechanics.

This may have made things seem more, rather than less, mysterious, but in the long run I'm afraid this is what one needs to think about.

Nói tóm lại ta cứ ghi nhớ luôn là có 8 loại cho đỡ phức tạp :D
1.(r\bar{b}+b\bar{r})/\sqrt{2}<br /><br />2.-i(r\bar{b}-b\bar{r})/\sqrt{2}<br /><br />3.(r\bar{g}+g\bar{r})/\sqrt{2}<br /><br />4.-i(r\bar{g}-g\bar{r})/\sqrt{2}<br /><br />5.(b\bar{g}+g\bar{b})/\sqrt{2}<br /><br />6.-i(b\bar{g}-g\bar{b})/\sqrt{2}<br /><br />7.(r\bar{r}-b\bar{b})/\sqrt{2}<br /><br />8.(r\bar{r}+b\bar{b}-2g\bar{g})/\sqrt{6}
Sửa lần cuối bởi Nhoc_PT vào ngày Thứ 7 Tháng 1 02, 2010 8:31 pm với 1 lần sửa trong tổng số.
Nguyên nhân: sửa tex
Đường đi khó không phải vì ngăn sông, cách núi
mà bởi vì lòng người ngại núi, e sông
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