# Meaning and derivation of 4-vector

1. Oct 8, 2012

### simoncks

Meaning of ct in Lorentz transformation -
In Lorentz transformation matrix, the first column is defined as - ct, not t itself. Is it because ct satisfies the units of x, y, z? Or, simpler Lorentz transformation matrix will be derived? The idea of 'ct', instead of t, is quite abstract for me. Not sure whether it is conceptually correct to consider ct as -
a meter of x*(0) corresponds to the time it takes light to travel 1 meter in vacuum.

Derivation and use of 4-vector scalar product -
Why is the scalar product defined in such way, where square of 'ct' has a minus sign? How to prove its invariant property for any inertial frame? Already in a big mystery....
How about the momentum 4-vector? What does p0 in the first row mean?

Thanks a lot.

Last edited: Oct 8, 2012
2. Oct 8, 2012

### Muphrid

A meter along the $ct$ axis corresponds to the time it takes for light to travel one meter, yes.

There are only a few geometries that have an invariant speed and obey a relativity principle. The geometry where $(ct)^2$ picks up a minus sign happens to be the best candidate for modeling our world.

The 0th component of the four-momentum is the energy (divided by $c$).

3. Oct 8, 2012

### bcrowell

Staff Emeritus
Relativists typically work in units where c=1, so the distinction between t and ct is irrelevant.

I assume you mean the first component of the vector, rather than the first vector?

It's not true generically that the 0th component of the position vector has that minus sign. Are you working in a context where covariant and contravariant vectors like xk and xk are defined? Or are you talking about old-fashioned treatments where the timelike component is ict (not -ct)?

4. Oct 8, 2012

### simoncks

Isn't the velocity of an object varying in different inertial frames? Can you explain further about why this geometry is the best candidate? It is quite abstract...

Is there indeed a reason why 0th component is E/c? Or, it is simply defined in such way?

Yes. Sorry, to be confusing.

I am working in covariant and contravariant vectors, but what confuse me is the scalar product. For 3-vector, the scalar product is x2+y2+z2. The 4-vector one is rather different -
-(ct)2+x2+y2+z2
It is a problem why the scalar product is defined such an 'abnormal' way, with a minus sign.

Last edited: Oct 8, 2012
5. Oct 8, 2012

### bcrowell

Staff Emeritus
Does this help?

For a 3-vector, it's x2+y2+z2 that is the same for all observers, i.e., regardless of how you rotate the x, y, and z axes.

For a 4-vector, it's easiest to think in just one spacelike dimension and one timelike one, so we have only t and x. Then the Lorentz transformation looks like figure k here: http://www.lightandmatter.com/html_books/lm/ch23/ch23.html [Broken] . What stays the same for all observers is -t2+x2 (in units with c=1). This is because the Lorentz transformation doesn't change the diagonal line x=t, which represents motion at the speed of light.

Last edited by a moderator: May 6, 2017
6. Oct 10, 2012