The scalar product of 4-vectors in special relativity

[Kutuzov]

Homework Statement

I'm confused about the difference between the following two statements:
$$\mathbf{V_1}\mathbf{V_2}=V_1V_2\cosh (\phi)$$
and
$$\mathbf{V_1}\mathbf{V_2}=\gamma c^2$$

Where $\gamma$ is the Lorentz factor of the relative speed between the two vectors. Both vectors are time-like.

The second expression is derived by writing one of the vectors in standard form, (0, 0, 0, c) and the other vector is arbitrary. However, we can always choose a frame so that one vector is written in standard form. In my mind, the two expression should be equivalent. The only way I can see that they would not be equivalent, is if the hyperbolic angle transforms when going from one inertial fram to another. If it does, could someone explain how?

EDIT: Someone could also tell me how to use Latex on this website. It appears not to be working.

 

[Chestermiller]

First of all, I assume you are talking about 4-velocities, and not just arbitrary 4-vectors. The hyperbolic angle is a function only of the relative velocity between the two frames of reference, and is given by

cosh($\phi$) = $\gamma$

The magnitudes of the two 4-velocity vectors are both c. So the two equations are consistent with one another.

Another way of deriving the second result is to express each of the two 4-velocity vectors in terms of the exact same set of coordinate basis vectors. If you consider one frame of reference to be at rest (say with respect to yourself), then its 4-velocity components with respect to the coordinate basis vectors for your frame are (0, 0, 0, c). The components of the 4-velocity vector for the other frame of reference expressed in terms of the coordinate basis vectors for your frame of reference are (v_xγ ,v_yγ , v_zγ, cγ). If you take the dot product of these two 4-velocity vectors, you get γc².

 

[George Jones]

Homework Statement

I'm confused about the difference between the following two statements:
$$\mathbf{V_1}\mathbf{V_2}=V_1V_2\cosh (\phi)$$
and
$$\mathbf{V_1}\mathbf{V_2}=\gamma c^2$$

Where $\gamma$ is the Lorentz factor of the relative speed between the two vectors. Both vectors are time-like.

The second expression is derived by writing one of the vectors in standard form, (0, 0, 0, c) and the other vector is arbitrary. However, we can always choose a frame so that one vector is written in standard form. In my mind, the two expression should be equivalent. The only way I can see that they would not be equivalent, is if the hyperbolic angle transforms when going from one inertial fram to another. If it does, could someone explain how?

EDIT: Someone could also tell me how to use Latex on this website. It appears not to be working.

Since Chestermiller has answered your questions, just a couple of comments.

1) "hyperbolic angle" is often called "rapidity" or "velocity parameter".

2) to make LateX work, enclose mathematics by the tags

[itex] and [/itex]

(for in-line mathematics,), or

[tex] and [/tex]

(for stand-alone mathematics), instead of $ signs.

 

[Kutuzov]

The magnitudes of the two 4-velocity vectors are both c. So the two equations are consistent with one another.

But both velocities cannot be in the rest frame? I understand that one can always be written in this way, (0, 0, 0, $\gamma$c). But not both at the same time? And only a timelike velocity written in such a way will have the magnitude c, right?

 

[George Jones]

But both velocities cannot be in the rest frame? I understand that one can always be written in this way, (0, 0, 0, $\gamma$c). But not both at the same time? And only a timelike velocity written in such a way will have the magnitude c, right?

Do you mean, as in you first post, that in its rest frame, the 4-velocity of a particle is $\left( 0, 0, 0, c \right)$?

The 4-velocity of a particle not in the rest frame is $\left( 0, 0, \gamma v, \gamma c \right)$. What is the invariant length of this 4-velocity?

 

[Kutuzov]

Do you mean, as in you first post, that in its rest frame, the 4-velocity of a particle is $\left( 0, 0, 0, c \right)$?

The 4-velocity of a particle not in the rest frame is $\left( 0, 0, \gamma v, \gamma c \right)$. What is the invariant length of this 4-velocity?

Yes, in the rest frame we can write a 4-velocity as $\left( 0, 0, 0, c \right)$ by choosing the inertial frame appropriately.

My understanding is that the magnitude of $\left( 0, 0, \gamma v, \gamma c \right)$ would be $(\gamma^2 c^2-\gamma v^2)^{1/2}$. I also don't know why we can assume that the first two components are zero.

But my main concern is $(\gamma^2 c^2-\gamma^2 v^2)^{1/2} \neq c$.

The invariant length... I suppose a velocity squared should be a scalar which should be an invariant or something like that? Is this $c^2$? Is this what you mean?

 

[George Jones]

Yes, in the rest frame we can write a 4-velocity as $\left( 0, 0, 0, c \right)$ by choosing the inertial frame appropriately.

My understanding is that the magnitude of $\left( 0, 0, \gamma v, \gamma c \right)$ would be $(\gamma^2 c^2-\gamma v^2)^{1/2}$.

Yes. To what does this simplify?

I also don't know why we can assume that the first two components are zero.

A reference frame is a human-chosen convention. We can always choose the directions of our coordinate axes such that one moving particle moves in the direction of one of the coordinate axes.

But my main concern is $(\gamma^2 c^2-\gamma^2 v^2)^{1/2} \neq c$.

Yes, this is true, but, as I asked above, to what does the left side simplify?

 

[Kutuzov]

Yes, this is true, but, as I asked above, to what does the left side simplify?

We can break out $\gamma$ for one. That gives us

$$\gamma (c^2-v^2)^{1/2}=\gamma c(1-\frac{v^2}{c^2})^{1/2}=\gamma^2 c$$

Um wait... is this what we expect? Returning to my original quesiton, if this is the magnitude of $\mathbf{V_1}$ and $\mathbf{V_2}=c$, we get
$$V_1V_2cosh(\phi)=\gamma^3 c^2\neq \gamma c^2=\mathbf{V_1}\mathbf{V_2}$$

Where one $\gamma$ comes from cosh, two comes from the expression above.

The only thing I can think of that could save this is that we could choose the frame so that $\gamma=1$. But I don't know that we can.

 

[Kutuzov]

Posting again in case you want to keep helping, read this topic by email and got sent my first more positive response before I edited it.