Difference between revisions of "4-vectors"
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− | <center><math>\ | + | <center><math>\underline{\textbf{Navigation}}</math> |
− | [[ | + | [[Relativistic_Units|<math>\vartriangleleft </math>]] |
− | [[VanWasshenova_Thesis# | + | [[VanWasshenova_Thesis#Moller_Scattering|<math>\triangle </math>]] |
− | [[ | + | [[4-momenta|<math>\vartriangleright </math>]] |
</center> | </center> | ||
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− | + | Since <math>ds^2 </math> is nothing more than a dot product of a vector with itself, we should expect the components of the indices to follow a similar relationship. Following the rules of matrix multiplication, the dot product of two 4-vectors should follow the form: | |
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<center><math>(ds)^2\equiv d \mathbf{x_{\mu}} d \mathbf{x^{\mu}}</math></center> | <center><math>(ds)^2\equiv d \mathbf{x_{\mu}} d \mathbf{x^{\mu}}</math></center> | ||
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\end{bmatrix}</math></center> | \end{bmatrix}</math></center> | ||
+ | This gives the desired results as expected. | ||
<center><math>(ds)^2 \equiv (dx^0)^{2}-(dx^1)^2-(dx^2)^2-(dx^3)^2=(dx^0)^{'2}-(dx^1)^{'2}-(dx^2)^{'2}-(dx^3)^{'2}</math></center> | <center><math>(ds)^2 \equiv (dx^0)^{2}-(dx^1)^2-(dx^2)^2-(dx^3)^2=(dx^0)^{'2}-(dx^1)^{'2}-(dx^2)^{'2}-(dx^3)^{'2}</math></center> | ||
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− | <center><math>ds^2 \equiv \eta_{\mu | + | <center><math>ds^2 \equiv \eta_{\nu}^{ \mu} \mathbf{dx^{\mu}} \mathbf{dx^{\mu}}=d\mathbf{R} \cdot d\mathbf{R}</math></center> |
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+ | <center><math>\mathbf R \cdot \mathbf R = s = \eta_{\nu}^{ \mu} \mathbf{x^{\mu}} \mathbf{x^{\mu}}</math></center> | ||
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+ | <center><math>\mathbf R_1 \cdot \mathbf R^1 = x_0^2-(x_1^2+x_2^2+x_3^2)</math></center> | ||
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+ | Similarly, for two different 4-vectors, | ||
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− | <center><math> | + | <center><math>\mathbf R_1 \cdot \mathbf R^2 = x_{0_1}x^{0_2}-(x_{1_1}x^{1_2}+x_{2_1}x^{2_2}+x_{3_1}x^{3_2})</math></center> |
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+ | This is useful in that it shows that the scalar product of two 4-vectors is an invariant since the time-space interval is an invariant. | ||
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---- | ---- | ||
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− | <center><math>\ | + | <center><math>\underline{\textbf{Navigation}}</math> |
− | [[ | + | [[Relativistic_Units|<math>\vartriangleleft </math>]] |
− | [[VanWasshenova_Thesis# | + | [[VanWasshenova_Thesis#Moller_Scattering|<math>\triangle </math>]] |
− | [[ | + | [[4-momenta|<math>\vartriangleright </math>]] |
</center> | </center> |
Latest revision as of 18:47, 15 May 2018
4-vectors
Using index notation, the time and space coordinates can be combined into a single "4-vector"
, that has units of length(i.e. ct is a distance).
We can express the space time interval using the index notation
Since
is nothing more than a dot product of a vector with itself, we should expect the components of the indices to follow a similar relationship. Following the rules of matrix multiplication, the dot product of two 4-vectors should follow the form:
This gives the desired results as expected.
The change in signs in the covariant term,
from the contravarient term
Comes from the Minkowski metric
Similarly, for two different 4-vectors,
This is useful in that it shows that the scalar product of two 4-vectors is an invariant since the time-space interval is an invariant.