Modern Physics/Math:Four Vectors

Special Relativity
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Math Tutorial - Four-VectorsEdit

We have seen that a wave is described by four numbers, the components of the spatial vector k, and the frequency ω

In special relativity these four numbers form a four-vector

It is called a four-vector because it has 3 spacelike components, forming a vector, and one timelike component when there are 3 space dimensions. It is called a four-vector because of the way it behaves when we change reference frames.

The spacelike component of the wave four-vector is just ${\displaystyle \mathbf {k} }$  when there are 3 space dimensions, while the timelike component is ${\displaystyle \omega /c}$  where the c is in the denominator to give the timelike component the same dimensions as the spacelike component.

Let us define some terminology. We indicate a four-vector by underlining and write the components in the following way: ${\displaystyle {\underline {k}}=(k,\omega /c)}$ , where ${\displaystyle {\underline {k}}}$  is the wave four-vector, ${\displaystyle k}$  is its spacelike component, and ${\displaystyle \omega /c}$  is its timelike component. For three space dimensions, where we have a wave vector rather than just a wavenumber, we write ${\displaystyle {\underline {k}}=(k,\omega /c)}$ .

Another example of a four-vector is simply the position vector in spacetime, ${\displaystyle {\underline {x}}=(x,ct)}$ , or ${\displaystyle {\underline {x}}=(\mathbf {x} ,ct)}$  in three space dimensions. The ${\displaystyle c}$  multiplies the timelike component in this case, because that is what is needed to give it the same dimensions as the spacelike component.

In three dimensions we define a vector as a quantity with magnitude and direction. Extending this to spacetime, a four-vector is a quantity with magnitude and direction in spacetime. Implicit in this definition is the notion that the vector's magnitude is a quantity independent of coordinate system or reference frame. We have seen that the invariant interval in spacetime is

${\displaystyle I={\sqrt {x^{2}-c^{2}t^{2}}}}$ ,

so it makes sense to identify this as the magnitude of the position vector. This leads to a way of defining a dot product of four-vectors.

Given two four-vectors

${\displaystyle {\underline {A}}=(\mathbf {A} ,A_{t})\quad {\underline {B}}=(\mathbf {B} ,B_{t})}$

then the dot product is

${\displaystyle {\underline {A}}\cdot {\underline {B}}=\mathbf {A} \cdot \mathbf {B} -A_{t}B_{t}\quad {\mbox{(dot product in spacetime)}}}$

This is consistent with the definition of invariant interval if we set

${\displaystyle {\underline {A}}={\underline {B}}={\underline {x}}}$ ,

since then

${\displaystyle {\underline {x}}\cdot {\underline {x}}=x^{2}-c^{2}t^{2}=I^{2}}$ .

Now, the key point about dot products for 3-vectors is that they are scalars, independent of the observer. They do not change if the axes are rotated, as was proven earlier.

For our definition of the dot product of four-vectors to be useful, it should also be independent of the observer. In particular, it should not depend on the observers velocity, else it would violate the principle of relativity.

We can easily check that our definition does satisfy this criterion.

It's clear that its independent of rotation, since it is the difference between a dot product and the product of two scalars, both of which terms aren't affected by rotating the coordinates.

Is it also frame velocity independent?

To check, first we need to be able to write down our four-vectors in the new reference frame. We know how to do this for the position vector — use the Lorentz transform. It can be shown that the same transform must hold for all vectors , so the components of a four vector in the new reference frame, moving at velocity v along the x-axis with respect to the previous one, are

${\displaystyle {\begin{matrix}A'_{x}&=&\gamma \left(A_{x}-{\frac {v}{c}}A_{t}\right)\\A'_{y}&=&A_{y}\\A'_{z}&=&A_{z}\\A'_{t}&=&\gamma \left(-{\frac {v}{c}}A_{x}+A_{t}\right)\end{matrix}}}$

The dot product in this frame is

${\displaystyle {\underline {A'}}\cdot {\underline {B'}}=A'_{x}B'_{x}+A'_{y}B'_{y}+A'_{z}B'_{z}-A'_{t}B'_{t}}$

Simplifying, we get

${\displaystyle {\begin{matrix}{\underline {A'}}\cdot {\underline {B'}}&=&\gamma ^{2}\left(A_{x}B_{x}-{\frac {v}{c}}(A_{x}B_{t}+A_{t}B_{x})+{\frac {v^{2}}{c^{2}}}A_{t}B_{t}\right)\\&&+A_{y}B_{y}+A_{z}B_{z}&\\&&-\gamma ^{2}\left({\frac {v^{2}}{c^{2}}}A_{x}B_{x}-{\frac {v}{c}}(A_{x}B_{t}+A_{t}B_{x})+A_{t}B_{t}\right)\\&=&\left(1-{\frac {v^{2}}{c^{2}}}\right)^{-1}\left(A_{x}B_{x}+{\frac {v^{2}}{c^{2}}}A_{t}B_{t}\right)\\&&+A_{y}B_{y}+A_{z}B_{z}\\&&-\left(1-{\frac {v^{2}}{c^{2}}}\right)^{-1}\left({\frac {v^{2}}{c^{2}}}A_{x}B_{x}-A_{t}B_{t}\right)\\&=&A_{x}B_{x}+A_{y}B_{y}+A_{z}B_{z}-A_{t}B_{t}\end{matrix}}}$

which is just the dot product in the original frame, exactly as we wanted.

We now know that the dot product of two four-vectors is a scalar result, i. e., its value is independent of coordinate system. This can be used to advantage on occasion.

In the odd geometry of spacetime it is not obvious what perpendicular means. We therefore define two four-vectors ${\displaystyle {\underline {A}}}$  and ${\displaystyle {\underline {B}}}$  to be perpendicular if their dot product is zero, in the same way as with three-vectors.

${\displaystyle {\underline {A}}\cdot {\underline {B}}=0}$

Because the dot product is a scalar, if vectors are perpendicular in one frame, they will be perpendicular in all frames.

We can also consider the dot product of a four-vector ${\displaystyle {\underline {A}}}$  which resolves into ${\displaystyle (A_{x},A_{t})}$  in the unprimed frame. Let us further suppose that the spacelike component is zero in some primed frame, so that the components in this frame are (0, At' ) The fact that the dot product is independent of coordinate system means that

${\displaystyle {\underline {A}}\cdot {\underline {A}}=A_{x}^{2}-A_{t}^{2}=-A_{t}'^{2}}$

This constitutes an extension of the spacetime Pythagorean theorem to four-vectors other then the position four-vector. Thus, for instance, the wavenumber for some wave may be zero in the primed frame, which means that the wavenumber and frequency in the unprimed frame are related to the frequency in the primed frame by ${\displaystyle k^{2}-\omega ^{2}/c^{2}=-\omega '^{2}/c^{2}}$ .

Proper timeEdit

Classically, the temporal derivative, d/dt acts like a scalar so we can multiply a vector by it, and get another vector.

In relativity t is part of a four-vector, which means d/dt also is, so we can't simply differentiate vectors with respect to t and expect to expect to get vectors.

For example, the position of a stationary particle is (0, ct).

Viewed from a frame moving at v to the right, its position becomes (-vτ, cτ), where τ=γt is the time as measured in the moving frame.

If we differentiate with respect to τ the velocity would be (-v, c)

If we differentiate with respect to t, we get (0, c) in the stationary frame, which would be (using the Lorentz transform) (-γv, -γc) in the moving frame, if this were a four vector.

These two expressions differ by a factor of γ, when measured in the same frame, so this can not be a four vector.

However, if the moving observer divides by γ, which is the time dilation, they will get the same vector as the stationary observer.

Doing this is equivalent to differentiating by the time in the particles own rest frame. Since this works for the position vector, we can expect it to work for all vectors.

The time measured in a particle's rest frame is called its proper time.

Differentiating a vector with respect to proper time gives another vector, which is the relativistic equivalent of the temporal derivative.